StudyGuideRealAnalysis (2)

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TEMPLE UNIVERSITYMATHEMATICS DEPARTMENTA Study Guide For Real Analysis

By: Ziad Adwan

REAL ANALYSIS(451 Completely Solved problems)

Copyright c 2005 Ziad Adwan. All rights reserved

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TEMPLE UNIVERSITYMATHEMATICS DEPARTMENTStudy Guide for REAL ANALYSIS

The ProblemsBy: Ziad Adwan

———————————————————————————————————————————————–1. If A and B are sets, show that A B if and only if A \ B = A :———————————————————————————————————————————————–2. Prove the Second De Morgan Law : If A, B , and C are sets, then A n(B \ C ) = ( A nB ) [ (A nC ) :———————————————————————————————————————————————–3. For each n 2N; let A n = f(n + 1) k : k 2Ng:(a) What is A 1 \ A 2?(b) Determine the sets [fA n : n 2Ng and \ fA n : n 2Ng:———————————————————————————————————————————————–4. Draw diagrams in the plane of the Cartesian products A B for the given sets A and B .(a) A = fx 2R : 1 x 2 or 3 x 4g; B = fx 2R : x = 1 or x = 2g:(b) A = f1; 2; 3g; B = fx 2R : 1 x 3g———————————————————————————————————————————————–5. Let f (x) = 1 =x2; x 6= 0 ; x 2R:(a) Determine the direct image f (E ) where E = fx 2R : 1 x 2g:(b) Determine the inverse image f 1(G ) where G = fx 2R : 1 x 4g:———————————————————————————————————————————————–6. Let g(x) = x2 and f (x) = x + 2 for x 2R; and let h be the composite function h = g f:(a) Find the direct image h(E ) of E = fx 2R : 0 x 1g:(b) Determine the inverse image h 1(G ) where G = fx 2R : 0 x 4g:———————————————————————————————————————————————–7. Let f (x) = x2 for x 2R; and let E = fx 2R : 1 x 0g and F = fx 2R : 0 x 1g:Show that E \ F = f0g and f (E \ F ) = f0g; while f (E ) = f (F ) = fy 2R : 0 y 1g:Hence, f (E \ F ) is a proper subset of f (E ) \ f (F ): What happens if 0 is deleted from the sets E and F ?———————————————————————————————————————————————–

8. Show that if f :A

!B

and E , F are subsets of A , then f (E

[F

) = f (E

)[ f (F

) and f (E

\F

) f (E

)\ f (F

):———————————————————————————————————————————————–9. Show that if f : A ! B and G , H are subsets of B, then f 1(G [ H ) = f 1(G ) [ f 1(H ) andf 1(G \ H ) = f 1(G ) \ f 1(H ):———————————————————————————————————————————————–10 . Show that the function f de…ned by f (x) = x=p x2 + 1 ; x 2R; is a bijection of R onto fy : 1 < y < 1g:———————————————————————————————————————————————–11 . (a) Show that if f : A ! B is injective and E A ; then f 1(f (E )) = E : Give an example to showthat equality need not hold if f is not injective.(b) Show that if f : A ! B is surjective and H B ; then f (f 1(H )) = H : Give an example to show thatequality need not hold if f is not surjective.———————————————————————————————————————————————–12 . (a) Suppose that f is an injection. Show that f 1 f (x) = x for all x 2D (f ) and that

f f 1(y) = y for all y 2R(f ):(b) If f is a bijection of A onto B , show that f 1 is a bijection of B onto A .

———————————————————————————————————————————————–13 . Let f : A ! B and g : B ! C be functions.(a) Show that if g f is injective, then f is injective.(b) Show that if g f is surjective, then g is surjective.———————————————————————————————————————————————–14. Prove that 1=1 2 + 1 =2 3 + + 1 =n(n + 1) = n= (n + 1) for all n 2N:———————————————————————————————————————————————–15. Prove that 3 + 11 + + (8 n 5) = 4 n2 n for all n 2N:

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———————————————————————————————————————————————–16. Prove that 12 22 + 3 2 42 + + ( 1)n +1 n2 = ( 1)n +1 n(n + 1) =2 for all n 2N:———————————————————————————————————————————————–17. Prove that 52n 1 is divisible by 8 for all n 2N:———————————————————————————————————————————————–18. Conjecture a formula for the sum of the …rst n odd natural numbers 1 + 3 + 5 + + (2 n 1); and proveyour formula using mathematical induction.———————————————————————————————————————————————–19. Prove the following Second Version of the Principle of Mathematical Induction : Let n0 2N andlet P (n) be a statement for each natural number n n 0: Suppose that(1) The statement P (n0) is true.(2) For all k n0 ; the truth of P (k) implies the truth of P (k + 1) :Then P (n) is true for all n n 0:———————————————————————————————————————————————–20. Prove that 2n < n ! for all n 4; n 2N:———————————————————————————————————————————————–21 . Prove that a nonempty set T 1 is …nite if and only if there is a bijection from T 1 onto a …nite set T 2 :———————————————————————————————————————————————–22 . (a) Prove that if A is a set with m

2N elements and C A is a set with 1 element, then A

nC is

a set with m 1 elements.(b) Prove that if C is an in…nite set and B is a …nite set, then C nB is an in…nite set.———————————————————————————————————————————————–23 . Exhibit a bijection between N and a proper subset of itself.———————————————————————————————————————————————–24 . Prove that a set T 1 is denumerable if and only if there is a bijection from T 1 onto a denumerable set T 2 .———————————————————————————————————————————————–25 . Give an example of a countable collection of …nite sets whose union is not …nite.———————————————————————————————————————————————–26 . Prove in detail that if S and T are denumerable, then S [ T is denumerable.———————————————————————————————————————————————–27. Use mathematical induction to prove that if a set S has n elements, then P (S ) has 2n elements.

———————————————————————————————————————————————–28 . Prove that if a; b 2R; then(a) (a + b) = ( a) + ( b) ; (b) ( a) ( b) = a b;(c) 1=( a) = (1=a); (d) (a=b) = ( a)=b if b 6= 0 :———————————————————————————————————————————————–29 . Solve the following equations, justifying each step by referring to an appropriate property or theorem.(a) 2x + 5 = 8 ; (b) x2 = 2 x;(c) x2 1 = 3 ; (d) (x 1)(x + 2) = 0 :———————————————————————————————————————————————–30 . If a 2R satis…es a a = a; prove that either a = 0 or a = 1 :———————————————————————————————————————————————–31 . Show that there does not exist a rational number t such that t2 = 3 :———————————————————————————————————————————————–

32 . (a) Show that if x; y are rational numbers, then so are x + y and xy:(b) Prove that if x is a rational number and y is an irrational number, then x + y is an irrational number.If, in addition, x 6= 0 ; then show that xy is an irrational number.———————————————————————————————————————————————–33 . Let K = fs + tp 2 : s; t 2Qg: Show that K satis…es the following:(a) If x1 ; x2 2K ; then x1 + x2 2K and x1x2 2K :(b) If x 6= 0 and x 2K ; then 1=x 2K :(Thus, the set K is a sub…eld of R. With the order inherited from R, the set K is an ordered …eld that liesbetween Q and R).———————————————————————————————————————————————–

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34. (a) If a < b and c d; prove that a + c < b + d:(b) If 0 < a < b and 0 c d; prove that 0 ac bd:———————————————————————————————————————————————–35. If a; b 2R; show that a2 + b2 = 0 if and only if a = b = 0 :———————————————————————————————————————————————–36. If 0 a < b; show that a2 ab < b2 : Show by example that it does not follow that a2 < ab 3x + 4 ; (b) 1 < x 2 < 4;(c) 1=x < x; (d) 1=x < x 2:———————————————————————————————————————————————–38. Let a; b 2R; and suppose that for every " > 0 we have a b + ": Show that a b:———————————————————————————————————————————————–39. (a) If 0 < c < 1; show that 0 < c 2 < c < 1:(b) If 1 < c; show that 1 < c < c 2 :———————————————————————————————————————————————–40. Use mathematical induction to show that if a 2R and m; n 2N; then am + n = am an and (am )n = amn :———————————————————————————————————————————————–41 . If a; b

2R; show that

ja + b

j=

ja

j+

jb

j if and only if ab 0:

———————————————————————————————————————————————–42 . If x;y;z 2R and x z; show that x y z if and only if jx yj+ jy zj= jx zj:Interpret this geometrically.———————————————————————————————————————————————–43 . Show that jx aj< if and only if a < x < a + :———————————————————————————————————————————————–44 . If a < x < b and a < y < b; show that jx yj< b a: Interpret this geometrically.———————————————————————————————————————————————–45 . Find all x 2R that satisfy the following inequalities:(a) j4x 5j 13; (b) x2 1 3:———————————————————————————————————————————————–46 . Find all x 2R that satisfy both j2x 3j< 5 and jx + 1 j> 2 simultaneously.

———————————————————————————————————————————————–47. Determine and sketch the set of pairs (x; y) 2R R that satisfy:(a) jxj= jyj; (b) jxj+ jyj= 1 ;(c) jxyj= 2 ; (d) jxj jyj= 2 :———————————————————————————————————————————————–48 . Let > 0 and > 0; and a 2R: Show that V (a) \ V (a) and V (a) [ V (a) are neighborhoodsof a for appropriate values of :———————————————————————————————————————————————–49 . Show that if a; b 2R; and a 6= b; then there exists neighborhoods U of a and V of b such thatU \ V = ? :———————————————————————————————————————————————–50 . Let S 1 = fx 2R : x 0g: Show that the set S 1 has lower bounds, but no upper bounds.Show that inf S 1 = 0 :

———————————————————————————————————————————————–51 . Let S 2 = fx 2R : x > 0g: Does S 2 have lower bounds? Does S 2 have upper bounds?Does inf S 2 exist? Does sup S 2 exist? Prove your statements.———————————————————————————————————————————————–52 . Let S 3 = f1=n : n 2Ng: Show that sup S 3 = 1 and inf S 3 0:———————————————————————————————————————————————–53 . Let S be a nonempty subset of R that is bounded below. Prove that inf S = supf s : s 2S g:———————————————————————————————————————————————–54. If a set S R contains one of its upper bounds, show that this upper bound is the supremum of S:———————————————————————————————————————————————–

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55. Let S R be nonempty. Show that u 2R is an upper bound of S if and only if the conditions t 2Rand t > u imply that t =2S:———————————————————————————————————————————————–56. Let S R be nonempty. Show that if u = sup S; then for every number n 2N the number u 1=nis not an upper bound of S; but the number u + 1 =n is an upper bound of S:———————————————————————————————————————————————–57. Let S be a bounded set in R and let S 0 be a nonempty subset of S:Show that inf S inf S 0 sup S 0 sup S:———————————————————————————————————————————————–58. Let S R and suppose that s = sup S 2S: If u =2S; show that sup(S [ fug) = sup fs ; ug:———————————————————————————————————————————————–59. Use the previous exercise and mathematical induction to show that a nonempty …nite set S Rcontains its supremum.———————————————————————————————————————————————–60. Show that a number u is the supremum of a nonempty set S R if and only if u satis…es the conditions:(1) s u for all s 2S;(2) if v < u; then there exists s0 2S such that v < s 0:———————————————————————————————————————————————–61 . If S =

f1=n 1=m : n; m

2Ng; …nd inf S and sup S:

———————————————————————————————————————————————–62 . Let S R be nonempty. Suppose that a number u in R has the properties:(i) For every n 2N the number u 1=n is not an upper bound of S; and(ii) For every n 2N the number u + 1 =n is an upper bound of S:Prove that u = sup S:———————————————————————————————————————————————–63 . Let X be a nonempty set and let f : X ! R have bounded range in R: Let a 2R: Assuming the fact thatif S is a nonempty subset of R that is bounded above, then sup(a + S ) = a + sup S; Show that(i) supfa + f (x) : x 2X g= a + sup ff (x) : x 2X g; and(ii) inf fa + f (x) : x 2X g= a + inf ff (x) : x 2X g:———————————————————————————————————————————————–64 . Let A and B be bounded nonempty subsets of R; and let A + B = fa + b : a 2A; b 2Bg: Prove that:

(i) sup(A + B ) = sup( A) + sup( B ); and(ii) inf(A + B ) = inf( A) + inf( B ):———————————————————————————————————————————————–65 . Let X be a nonempty set, and let f and g be de…ned on X and have bounded ranges in R: Prove that:(i) supff (x) + g(x) : x 2X g supff (x) : x 2X g+ sup fg(x) : x 2X g; and(ii) inf ff (x) : x 2X g+ inf fg(x) : x 2X g inf ff (x) + g(x) : x 2X g:Give examples to show that each of these inequalities can be either equalities or strict inequalities.———————————————————————————————————————————————–66 . Let X = Y = fx 2R : 0 < x < 1g: De…ne h1 ; h2 : X Y ! R by

h1(x; y) = 2 x + y; and h2(x; y) = 0 if x < y1 if x y :

(a) For each x

2X; …nd f 1(x) = sup

fh1(x; y) : y

2Y

g and f 2(x) = sup

fh2(x; y) : y

2Y

g; then …nd

inf ff 1(x) : x 2X g and inf ff 2(x) : x 2X g:(b) For each y 2Y; …nd g1(y) = inf fh1(x; y) : x 2X g and g2(y) = inf fh2(x; y) : x 2X g; then …ndsupfg1(y) : y 2Y g and supfg2(y) : y 2Y g:———————————————————————————————————————————————–67 . Let X and Y be nonempty sets and let h : X Y ! R have bounded range in R: Let f : X ! R andg : Y ! R be de…ned by f (x) = sup fh(x; y) : y 2Y g; and g(y) = inf fh(x; y) : x 2X g: Prove that

supfg(y) : y 2Y g inf ff (x) : x 2X g:———————————————————————————————————————————————–

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68 . Show that if a > 0; then there exists a positive real number z such that z2 = a:———————————————————————————————————————————————–69 . If u > 0 is any real number and x < y; show that there exists a rational number r such that x < ru < y:———————————————————————————————————————————————–70 . If S R is nonempty, show that S is bounded if and only if there exists a bounded closed interval I such that S

I:

———————————————————————————————————————————————–71 . If S R is a nonempty bounded set, and I S = [inf S; sup S ]; show that S I S : Moreover, if J is anyclosed bounded interval containing S; show that I S J:———————————————————————————————————————————————–

72 . Let I n = [0; 1=n] for n 2N: Prove that1

\n =1I n = f0g:

———————————————————————————————————————————————–

73 . Let K n = ( n; 1 ) for n 2N: Prove that1

\n =1

K n = ? :

———————————————————————————————————————————————–74 . Give the two binary representations of 3

8 and 716 :

———————————————————————————————————————————————–75 . Show that if ak ; bk

2 f0; 1;:::; 9

g and if

a 110 + a 2

10 2 + + a n10 n = b1

10 + b210 2 + + bm

10 m 6= 0 ;

then n = m; and ak = bk for k = 1 ; 2;:::;n:———————————————————————————————————————————————–76 . For any b 2R, prove that lim

n !1bn = 0 :

———————————————————————————————————————————————–77 . Use the de…nition of a limit of a sequence to establish the following limits:(a) lim

n !1 n

n 2 +1 = 0 ; (b) limn !1

2nn +1 = 2 ; (c) lim

n !13n +12n +5 = 3

2 ; (d) limn !1

n 2 12n 2 +3 = 1

2 :———————————————————————————————————————————————–78 . Show that:(a) lim

n

!1

1p n +7 = 0 ; (b) limn

!1

2nn +2 = 2 ; (c) lim

n

!1

p nn +1 = 0 ; (d) lim

n

!1

( 1) n nn 2 +1 = 0 :

———————————————————————————————————————————————–79 . Let xn = 1ln( n +1) for n 2N:

(a) Use the de…nition of limit to show that limn !1

(xn ) = 0 :(b) Find a speci…c value of K ( ) as required in the de…nition of limit for each of (i) = 1=2; and (ii) = 1 =10:———————————————————————————————————————————————–80 . Prove that lim

n !1(xn ) = 0 if and only if lim

n !1(jxn j) = 0 : Give an example to show that the convergence

of (jxn j) need not imply the convergence of (xn ) :———————————————————————————————————————————————–81 . Prove that if lim

n !1(xn ) = x and if x > 0; then there exists M 2N such that xn > 0 for all n M:

———————————————————————————————————————————————–82 . Let b 2R satisfy 0 < b < 1: Show that lim

n !1(nbn ) = 0 : (Hint : Use the binomial theorem).

———————————————————————————————————————————————–

83 . For xn given by the following formulas, establish either the convergence or the divergence of the sequence X = ( xn ):(a) xn = n

n +1 ; (b) xn = ( 1) n nn +1 ; (c) xn = n 2

n +1 ; (d) xn = 2n 2 +3n 2 +1 :

———————————————————————————————————————————————–84 . Give an example of two divergent sequences X and Y such that:(a) Their sum converges,(b) Their product converges.———————————————————————————————————————————————–85 . Show that the following sequences are not convergent:(a) (2n ) ; (b) ( 1)n n2 :

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———————————————————————————————————————————————–86 . Find the limits of the following sequences:(a) lim

n !1(2 + 1 =n)2 ; (b) lim

n !1( 1) n

n +2 ; (c) limn !1

p n 1p n +1 ; (d) limn !1

n +1n p n :

———————————————————————————————————————————————–87 . If (bn ) is a bounded sequence and lim

n !1(an ) = 0 ; show that lim

n !1(an bn ) = 0 :

———————————————————————————————————————————————–88 . If 0 < a < b; determine lim

n !1a n +1 + bn +1

a n + bn :———————————————————————————————————————————————–89 . Use the squeeze Theorem to determine the limits of the following:(a) n1=n 2

; (b) (n!)1=n 2

:———————————————————————————————————————————————–90 . Recall the following Theorem: "Let (xn ) be a sequence of positive real numbers such thatL = lim

n !1(xn +1 =xn ) exists. If L < 1; then (xn ) converges and lim

n !1(xn ) = 0" :

Apply this theorem to the following sequences, where a; b satisfy: 0 < a < 1; b > 1:(a) (an ) ; (b) (bn =2n ) ; (c) (n=bn ) ; (d) 23n =32n :———————————————————————————————————————————————–91 . Let X = ( xn ) be a sequence of positive real numbers such that lim

n !1(xn +1 =xn ) = L > 1:

Show that X is not a bounded sequence and hence is not convergent.———————————————————————————————————————————————–92 . Let (xn ) be a sequence of positive real numbers such that lim

n !1x1=n

n = L < 1:Show that there exists a number r with 0 < r < 1 such that 0 < x n < r n for all su¢ciently large n 2N:Use this to show that lim

n !1(xn ) = 0 :

———————————————————————————————————————————————–93 . Suppose that (xn ) is a convergent sequence and (yn ) is such that for any > 0 there exists M suchthat jxn yn j< for all n M: Does it follow that (yn ) is convergent?———————————————————————————————————————————————–94 . Let x1 = 8 and xn +1 = 1

2 xn + 2 for n 2N. Show that (xn ) is bounded and monotone. Find limn !1

(xn ) :———————————————————————————————————————————————–95 . Let x1 > 1 and xn +1 = 2 1=xn for n 2N. Show that (xn ) is bounded and monotone. Find lim

n !1(xn ) :

———————————————————————————————————————————————–

96 . Let x1 2 and xn +1 = 1 + p xn 1 for n 2N. Show that (xn ) is decreasing and bounded below by 2.Find limn !1

(xn ) :———————————————————————————————————————————————–97 . Let y1 = p p; where p > 0; and yn +1 = p p + yn for n 2N. Show that (yn ) converges and …nd lim

n !1(yn ) :

(Hint : One upper bound is 1 + 2p p):———————————————————————————————————————————————–98 . Let (an ) be an increasing sequence and (bn ) be a decreasing sequence, and assume that an bn 8n 2N:Show that lim

n !1(an ) lim

n !1(bn ) ; and thereby deduce the Nested Intervals Property from the Monotone

Convergence Theorem.———————————————————————————————————————————————–99 . Let A be an in…nite subset of R that is bounded above and let u = sup A : Show that there existsan increasing sequence (xn ) with xn 2A for all n 2N such that u = lim

n !1(xn ) :

———————————————————————————————————————————————–100 . Let (xn ) be a bounded sequence, and for each n 2N let sn = sup fxk : k ng and tn = inf fxk : k ng:Prove that (sn ) and (tn ) are monotone and convergent. Also, prove that if lim

n !1(sn ) = lim

n !1(tn ) ; then

(xn ) is convergent.———————————————————————————————————————————————–101 . Establish the convergence or divergence of the sequence (yn ) ; where

yn = 1n +1 + 1

n +2 + + 12n for n 2N:

———————————————————————————————————————————————–102 . Let xn = 1 =12 + 1 =22 + + 1 =n2 for each n 2N: Prove that (xn ) is increasing and bounded, and hence

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converges. [Hint : Note that if k 2; then 1=k2 1=k(k 1) = 1 =(k 1) 1=k:]:———————————————————————————————————————————————–103 . Establish the convergence and …nd the limits of the following sequences:(a) (1 + 1 =n)n +1 ; (b) (1 + 1 =n)2n ; (c) (1 + 1

n +1 )n ; (d) ((1 1=n)n ) :———————————————————————————————————————————————–104 . Calculate p 2; correct to within 4 decimals.———————————————————————————————————————————————–105 . Give an example of an unbounded sequence that has a convergent subsequence.———————————————————————————————————————————————–106 . Show that the following sequences are divergent.(a) (1 ( 1)n + 1 =n) ; (b) (sin( n =4)) :———————————————————————————————————————————————–107 . Let X = ( xn ) and Y = ( yn ) be given sequences, and let the "shu ed" sequence Z = ( zn ) be de…nedby: z1 = x1 ; z2 = y1 ; ; z2n 1 = xn ; z2n = yn ; : Show that Z is convergent if and only if both X and Y are convergent and lim X = lim Y:———————————————————————————————————————————————–108 . Establish the convergence and …nd the limits of the following sequences:(a) (1 + 1 =n2)n 2

; (b) ((1 + 1 =2n)n ) ; (c) (1 + 1 =n2)2n 2; (d) ((1 + 2 =n)n ) :

———————————————————————————————————————————————–109 . Determine the limits of the following:(a) (3n)1=2n ; (b) (1 + 1 =2n)3n :———————————————————————————————————————————————–110 . Suppose that every subsequence of X = ( xn ) has a subsequence that converges to 0. Show thatlim X = 0 :———————————————————————————————————————————————–111 . Show that if (xn ) is unbounded, then there exists a subsequence (xn k ) such that lim

k!1(1=xn k ) = 0 :

———————————————————————————————————————————————–112 . Let (xn ) be a bounded sequence and let s = sup fxn : n 2Ng: Show that if s =2 fxn : n 2Ng; thenthere is a subsequence of (xn ) that converges to s:———————————————————————————————————————————————–113 . Let (I n ) be a nested sequence of closed bounded intervals. For each n

2N; let xn

2I n : Use the Bolzano-

Weierstrass Theorem to show that there exists a number 2R such that 2I n 8n 2N:[That is, Use the Bolzano-Weierstrass Theorem to prove the Nested Intervals Property].———————————————————————————————————————————————–114 . Show directly from the de…nition that the following are Cauchy sequences.(a) n +1

n ; (b) 1 + 12! + + 1

n ! :———————————————————————————————————————————————–115 . Show directly from the de…nition that the following are not Cauchy sequences.(a) (( 1)n ) ; (b) n + ( 1) n

n ; (c) (ln n)———————————————————————————————————————————————–116 . Show directly from the de…nition that if (xn ) and (yn ) are Cauchy sequences, then (xn + yn ) and (xn yn )are Cauchy sequences.———————————————————————————————————————————————–

117 . If xn = p n; show that (xn ) satis…es limn !1 jxn +1 xn j= 0 ; but that it is not a Cauchy sequence.———————————————————————————————————————————————–118 . Let (xn ) be a Cauchy sequence such that xn is an integer for every n 2N: Show that (xn ) is ultimatelyconstant.———————————————————————————————————————————————–119 . Show directly that a bounded, monotone increasing sequence is a Cauchy sequence.———————————————————————————————————————————————–120 . If 0 < r < 1 and jxn +1 xn j< r n for all n 2N; show that (xn ) is a Cauchy sequence.———————————————————————————————————————————————–121 . Show that if (xn ) is an unbounded sequence, then there exists a properly divergent subsequence.

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———————————————————————————————————————————————–122 . Give examples of properly divergent sequences (xn ) and (yn ) with yn 6= 0 for all n 2N such that(a) (xn =yn ) is convergent ; (b) (xn =yn ) is properly divergent :———————————————————————————————————————————————–123 . Show that if xn > 0 for all n 2N; then lim


n !1(1=xn ) = + 1 :

———————————————————————————————————————————————–124 . Establish the proper divergence of the following sequences:(a) (p n) ; (b) p n + 1 ; (c) p n 1 ; (d) n=p n + 1 :———————————————————————————————————————————————–125 . Is the sequence (n sin n) properly divergent?———————————————————————————————————————————————–126 . Let (xn ) be properly divergent and let (yn ) be such that lim

n !1(xn yn ) belongs to R: Show that (yn )

converges to 0.———————————————————————————————————————————————–127 . Let (xn ) and (yn ) be sequences of positive numbers such that lim

n !1(xn =yn ) = 0 :

(a) Show that if limn !1

(xn ) = + 1 ; then limn !1

(yn ) = + 1 :(b) Show that if (yn ) is bounded, then lim

n !1(xn ) = 0 :

———————————————————————————————————————————————–128 . Investigate the convergence or divergence of the following sequences:(a) p n2 + 2 ; (b) p n= n2 + 1 ; (c) p n2 + 1 =p n ; (d) (sin p n) :———————————————————————————————————————————————–129 . Let (xn ) and (yn ) be sequences of positive numbers such that lim

n !1(xn =yn ) = + 1 :


(yn ) = + 1 ; then limn !1

(xn ) = + 1 :(b) Show that if (xn ) is bounded, then lim

n !1(yn ) = 0 :

———————————————————————————————————————————————–130 . Show that if lim

n !1(an =n) = L; where L > 0; then lim

n !1(an ) = + 1 :

———————————————————————————————————————————————–131 . Show that the convergence of a series is not a¤ected by changing a …nite number of its terms.(of course, the value of the sum may be changed).———————————————————————————————————————————————–

132 . If 1Xn =1

xn and 1Xn =1

yn are convergent, show that 1Xn =1

(xn + yn ) is convergent.

———————————————————————————————————————————————–

133 . (a) Show that the series1

Xn =1cos n is divergent.

(b) Show that the series1

Xn =1

cos nn 2 is convergent.

———————————————————————————————————————————————–

134 . Show that the series1

Xn =1

( 1) n

p n is convergent.

———————————————————————————————————————————————–

135 . If 1

Xn =1

an with an > 0 is convergent, then is1

Xn =1

a2n always convergent?

Either prove or give a counterexample.———————————————————————————————————————————————–

136 . If 1

Xn =1an with an > 0 is convergent, then is

1

Xn =1

p an always convergent?


137 . If 1

Xn =1an with an > 0 is convergent, and if bn = (a 1 + + a n )

n for n 2N; then show that1

Xn =1bn is always

9



divergent.———————————————————————————————————————————————–138 . Determine a condition on jx 1j that will assure that:(a) x2 1 < 1

2 ; (b) x2 1 < 1=103 ;(c) x2 1 < 1

n for a given n 2N; (d) x3 1 < 1n for a given n 2N:

———————————————————————————————————————————————–139 . Let c be a cluster point of A R and let f : A ! R: Prove that limx! cf (x) = L if and only if limx! c jf (x) Lj= 0 :———————————————————————————————————————————————–140 . Let f : R! R and let c 2R: Show that lim

x! cf (x) = L if and only if lim

x! 0f (x + c) = L:

———————————————————————————————————————————————–141 . Let I = (0 ; a) where a > 0; and let g(x) = x2 for x 2I: For any points x; c 2I; show thatg(x) c2 2a jx cj: Use this inequality to prove that lim

x! cx2 = c2 for any c 2I:

———————————————————————————————————————————————–142 . Let I be an interval in R; let f : I ! R, and let c 2I: Suppose there exists constants K and L such that

jf (x) Lj< K jx cj for x 2I: Show that limx! c

f (x) = L:———————————————————————————————————————————————–143 . Show that lim

x

!cx3 = c3 for any c 2R:

———————————————————————————————————————————————–144 . Show that limx! c

p x = p c for any c > 0:———————————————————————————————————————————————–145 . Use the de…nition of limit to show that(a) lim

x! 2x2 + 4 x = 12 ; (b) lim

x! 1x +5

2x +3 = 4 :———————————————————————————————————————————————–146 . Show that the following limits do not exist:(a) lim

x! 01

x 2 (x > 0); (b) limx! 0

1p x (x > 0);

(c) limx! 0

(x+ sgn(x)) ; (d) limx! 0

sin 1=x2

———————————————————————————————————————————————–147 . Suppose the function f : R ! R has limit L at 0; and let a > 0: If g : R! R is de…ned byg(x) = f (ax ) for x 2R; show that lim

x! 0g(x) = L:

———————————————————————————————————————————————–148 . Let f : R! R be de…ned by: f (x) = x if x 2Q

0 if x =2Q :

(a) Show that f has a limit at x = 0 :(b) Use a sequential argument to show that if c 6= 0 ; then f does not have a limit at c:———————————————————————————————————————————————–149 . Determine the following limits:

(a) limx! 1

(x + 1)(2 x + 3) ( x 2R); (b) limx! 1

x 2 +2x 2 2 (x > 0);

(c) limx! 2

1x +1 1

2x (x > 0); (d) limx! 0

x +1x 2 +2 (x 2R):

———————————————————————————————————————————————–150 . Determine the following limits:

(a) limx! 2q 2x +1

x +3 (x > 0); (b) limx! 2

x 2 4x 2 (x > 0);

(c) limx! 0

(x +1) 2 1x (x > 0); (d) lim

x! 1

p x 1x 1 (x > 0):

———————————————————————————————————————————————–151 . Find lim

x! 0

p 1+2 x p 1+3 xx +2 x 2 where x > 0:

———————————————————————————————————————————————–152 . Prove that lim

x! 0cos(1=x) does not exist but that lim

x! 0x cos(1=x) = 0 :

———————————————————————————————————————————————–

10



153 . Let f ; g be de…ned on A R to R; and let c be a cluster point of A: Suppose that f is boundedon a neighborhood of c and that lim

x! cg(x) = 0 : Prove that lim

x! cf (x) g (x) = 0 :

———————————————————————————————————————————————–154 . Use the de…nition of the limit to prove that if f ; g are de…ned on A R to R; c is a cluster point of A;limx! c

f (x) = L; and limx! c

g(x) = M; then limx! c

[f (x) + g(x)] = L + M:———————————————————————————————————————————————–155 . Let n 2N be such that n 3: Derive the inequality x2 xn x2 for 1 < x < 1: Then use the factthat lim

x! 0x2 = 0 to show that lim

x! 0xn = 0 :

———————————————————————————————————————————————–156 . Give examples of functions f and g such that f and g do not have limits at a point c; but such that bothf + g and f g have limits at c:———————————————————————————————————————————————–157 . Determine whether the following limits exist in R :

(a) limx! 0

sin 1=x2 (x 6= 0) ; (b) limx! 0

x sin 1=x2 (x 6= 0) ;

(c) limx! 0

sgn(sin (1 =x)) (x 6= 0) ; (d) limx! 0

p x sin 1=x2 (x > 0):

———————————————————————————————————————————————–

158 . Let A R; let f : A ! R; and let c 2R be a cluster point of A: In addition, suppose that f (x) 0for all x 2A; and let p f be the function de…ned for x 2A by p f (x) = p f (x):If lim

x! cf (x) exists, prove that lim

x! cp f (x) = q limx! c

f (x):———————————————————————————————————————————————–159 . Let f (x) = jxj

1=2 for x 6= 0 : Show that limx! 0+

f (x) = limx! 0

f (x) = + 1 :———————————————————————————————————————————————–160 . Let c 2R and let f be de…ned for x 2(c;1 ) and f (x) > 0 for all x 2(c;1 ):Show that lim

x! cf (x) = 1 if and only if lim

x! c1=f (x) = 0 :

———————————————————————————————————————————————–161 . Evaluate the following limits, or show that they do not exist.

(a) limx! 1+

xx 1 (x 6= 1) ; (b) lim

x! 1x

x 1 (x 6= 1) ;

(c) limx! 0+ x +2p x (x > 0); (d) limx!1 x +2p x (x > 0);(e) lim

x! 0

p x +1x (x > 1); (f) lim

x!1p x +1

x (x > 0);

(g) limx!1

p x 5p x +3 (x > 0); (h) limx!1

p x xp x + x (x > 0):———————————————————————————————————————————————–162 . Suppose that f and g have limits in R as x ! 1 and that f (x) g(x) for all x 2(a; 1 ):Prove that lim

x!1f (x) lim

x!1g(x):

———————————————————————————————————————————————–163 . Let f be de…ned on (0;1 ) to R: Prove that lim

x!1f (x) = L if and only if lim

x! 0+f (1=x) = L:

———————————————————————————————————————————————–164 . Show that if f : (a; 1 ) ! R is such that lim

x!1xf (x) = L; where L 2R; then lim

x!1f (x) = 0 :

———————————————————————————————————————————————–165 . suppose that lim

x! cf (x) = L; where L > 0; and that lim

x! cg(x) =

1: Show that lim

x! cf (x)g(x) =

1:

If L = 0 ; show by example that this conclusion may fail.———————————————————————————————————————————————–166 . Find functions f and g de…ned on (0;1 ) such that lim

x!1f = 1 and lim

x!1g = 1 ; and lim

x!1(f g) = 0 :

Can you …nd such functions, with g(x) > 0 for all x 2(0;1 ); such that limx!1

f=g = 0?———————————————————————————————————————————————–167 . Prove the following Sequential Criterion for Continuity : A function f : A ! R is continuous ata point c 2A if and only if for every sequence (xn ) in A that converges to c; the sequence (f (xn ))converges to f (c):———————————————————————————————————————————————–

11



168 . Let f be de…ned for all x 2R; x 6= 2 ; by f (x) = ( x2 + x 6)=(x 2): Can f be de…ned at x = 2in such a way that f is continuous at this point?———————————————————————————————————————————————–169 . Let f : R ! R be continuous at c and let f (c) > 0: Show that there exists a neighborhood V (c)of c such that if x 2V (c); then f (x) > 0:———————————————————————————————————————————————–170 . Let f : R! R be continuous on R and let S = fx 2R : f (x) = 0 g be the zero set of f : If (xn )is in S and x = lim

n !1(xn ) ; show that x 2S:

———————————————————————————————————————————————–171 . Let K > 0 and let f : R ! R satisfy the condition jf (x) f (y)j K jx yj for all x; y 2R: Showthat f is continuous at every point c 2R:———————————————————————————————————————————————–172 . Suppose that f : R ! R is continuous on R and that f (r ) = 0 for every rational number r: Provethat f (x) = 0 for all x 2R:———————————————————————————————————————————————–173 . Determine the points of discontinuity of the following functions and state which theorems are usedin each case.(a) f (x) = x2 +2 x +1

x 2 +1 (x 2R) (b) g(x) =

p x + p x (x 0)

(c) h(x) =p 1+

jsin x

jx (x 6= 0) (d) k(x) = cos p 1 + x2 (x 2R)———————————————————————————————————————————————–174 . Show that if f : A ! R is continuous on A R and if n 2N; then the function f n de…ned by:f n (x) = ( f (x))n for x 2A; is continuous on A:———————————————————————————————————————————————–175 . Let g be de…ned on R by g(1) = 0 ; and g(x) = 2 if x 6= 1 ; and let f (x) = x + 1 for all x 2R:Show that lim

x! 0g f 6= g f (0) : Why doesn’t this contradict Theorem 5.2.6?

———————————————————————————————————————————————–176 . Give an example of a function f : [0; 1] ! R that is discontinuous at every point of [0; 1] butsuch that jf j is continuous on [0; 1]:———————————————————————————————————————————————–177 . Let f ; g be continuous from R to R; and suppose that f (r ) = g(r ) for all rational numbers r: Is ittrue that f (x) = g(x) for all x

2R?

———————————————————————————————————————————————–178 . Let f : R! R is continuous on R; and let P = fx 2R : f (x) > 0g: If c 2P; show that thereexists a neighborhood V (c) P:———————————————————————————————————————————————–179 . Let I = [a; b] and let f : I ! R be a continuous function such that f (x) > 0 for each x 2I:Prove that there exists a number > 0 such that f (x) for each x 2I:———————————————————————————————————————————————–180 . Let f be continuous on the interval [0; 1] to R and such that f (0) = f (1) : Prove that there existsa point c 2[0; 1

2 ] such that f (c) = f (c + 12 ): Conclude that there are, at any time, antipodal points

on the earth’s equator that have the same temperature. ( Hint : consider g(x) = f (x) f (x + 12 )) :

———————————————————————————————————————————————–181 . Show that the equation x = cos x has a solution on the interval [0; =2]: Use the Bisection methodand a calculator to …nd an approximate solution of this equation, with error less than 10 3 :———————————————————————————————————————————————–182 . Let I = [a; b], let f : I ! R be continuous on I ; and assume that f (a) < 0; f (b) > 0:Let W = fx 2I : f (x) < 0g; and let w = sup W: Prove that f (w) = 0 :———————————————————————————————————————————————–183 . Suppose that f : R ! R is continuous on R and that lim

x! 1f = 0 and lim

x!1f = 0 :

Prove that f is bounded on R and attains either a maximum or a minimum on R:Give an example to show that both a maximum and a minimum need not be attained.———————————————————————————————————————————————–184 . Show that the function f (x) = 1

x 2 is uniformly continuous on A = [1;1 ); but that it is not uniformly

12



continuous on B = (0 ;1 ):———————————————————————————————————————————————–185 . Show that if f and g are uniformly continuous on A R and if they are both bounded on A; thentheir product f g is uniformly continuous on A:———————————————————————————————————————————————–186 . Prove that if f and g are each uniformly continuous on R; then their composite function f

g is

uniformly continuous on R:———————————————————————————————————————————————–187 . If g(x) = p x for x 2[0; 1]; show that there does not exist a constant K such that jg(x)j K jxj for allx 2[0; 1]: Conclude that the uniformly continuous function g is not a Lipschitz function on [0; 1]:———————————————————————————————————————————————–188 . Show that if a function f : R! R is Lipschitz on R, then f is uniformly continuous on R:———————————————————————————————————————————————–189 . Show that the function f (x) = 1 =x is uniformly continuous on the set A = [a; 1 ); where a is a positiveconstant.———————————————————————————————————————————————–190 . Show that the function f (x) = x2 is not uniformly continuous on the set A = [0;1 ):———————————————————————————————————————————————–191 . Show that the function f (x) = sin(1 =x) is not uniformly continuous on the set B = (0 ;

1):

———————————————————————————————————————————————–192 . Show that the function f (x) = 1

1+ x 2 for x 2R is uniformly continuous on R:———————————————————————————————————————————————–193 . Show that if f and g are uniformly continuous on A R; then f + g is uniformly continuous on A:———————————————————————————————————————————————–194 . If f is uniformly continuous on A R; and jf (x)j k > 0 for all x 2A; show that 1=f is uniformlycontinuous on A:———————————————————————————————————————————————–195 . Prove that if f is uniformly continuous on a bounded set A R; then f is bounded on A:———————————————————————————————————————————————–196 . Show that if f is continuous on [0;1 ) and uniformly continuous on [a; 1 ) for some positive constant a;then f is uniformly continuous on [0;1 ):

———————————————————————————————————————————————–197 . Let A R and suppose that f : A ! R has the following property:

8 > 0 9g : A ! R such that g is uniformly continuous on A and jf (x) g (x)j< for all x 2A:Show that f is uniformly continuous on A:———————————————————————————————————————————————–198 . (a) Show that if f is uniformly continuous on (a; b] and on [b; c); then f is also uniformlycontinuous on (a; c):(b) Prove that any function f : R! R that is continuous and periodic must be uniformly continuous.———————————————————————————————————————————————–199 . Show that if f : R ! R is continuous and such that lim

x! 1f (x) and lim

x!1f (x) are …nite, then f is

uniformly continuous on R.———————————————————————————————————————————————–200 . A function f : R

!R is continuous at 0 and satis…es the following conditions:

f (0) = 0 and f (x + y) f (x) + f (y) for any x; y 2R:Prove that f is uniformly continuous on R:———————————————————————————————————————————————–201 . A point x 2R is said to be an interior point of A R in case there is a neighborhood V of x suchthat V A: Show that a set A R is open if and only if every point of A is an interior point of A:———————————————————————————————————————————————–202 . A point x 2R is said to be a boundary point of A R in case every neighborhood V of x containspoints in A and points in Ac : Show that a set A and its complement Ac have exactly the same boundarypoints.

13



———————————————————————————————————————————————–203 . Show that a set G R is open if and only if it does not contain any of its boundary points.———————————————————————————————————————————————–204 . Show that a set F R is closed if and only if it contains all of its boundary points.———————————————————————————————————————————————–205 . If A R; let A be the union of all open sets that are contained in A; the set A is called the interiorof A: Show that:(a) A is an open set,(b) A is the largest open set contained in A; and(c) A point z belongs to A if and only if z is an interior point of A:———————————————————————————————————————————————–206 . If A R; let A be the intersection of all closed sets containing A; the set A is called the closure of A:Show that:(a) A is a closed set,(b) A is the smallest closed set containing A; and(c) A point w belongs to A if and only if w is either an interior point or a boundary point of A:———————————————————————————————————————————————–207. Exhibit an open cover of N that has no …nite subcover.———————————————————————————————————————————————–208 . Exhibit an open cover of the set A = f1=n : n 2Ng that has no …nite subcover.———————————————————————————————————————————————–209 . Prove that if K 1 and K 2 are compact sets in R; then K = K 1 [ K 2 is compact.———————————————————————————————————————————————–210 . Prove that the intersection of an arbitrary collection of compact sets in R is compact.———————————————————————————————————————————————–211 . Let K 6= ? be compact in R and let c 2R:(a) Prove that there exists a point a 2K such that jc aj= inf fjc xj: x 2K g:(b) Prove that there exists a point b 2K such that jc bj= sup fjc xj: x 2K g:———————————————————————————————————————————————–212 . Let h : R ! R be de…ned by h(x) = 1 if 0 x 1

0 otherwise : Find an open set G such that h 1(G) is

not open, and a closed set F such that h 1(F ) is not closed.———————————————————————————————————————————————–213 . Prove that f : R ! R is continuous if and only if for each closed set F in R; the inverse image f 1(F )is closed in R:———————————————————————————————————————————————–214 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) < g is open in R for each 2R:———————————————————————————————————————————————–215 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) g is closed in R for each 2R:———————————————————————————————————————————————–216 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) = kg is closed in R for each k 2R:———————————————————————————————————————————————–217 . Give an example of a function f : R! R such that the set fx 2R : f (x) = 1 g is neither opennor closed in R:

———————————————————————————————————————————————–218. Let I = [a; b] and let f : I ! R and g : I ! R be continuous functions on I : Show that the set

fx 2R : f (x) = g(x)g is closed in R:———————————————————————————————————————————————–219 . Show that each of the following functions is a metric on the given space:(a) S = R2 ; d1(P 1 ; P 2) = jx1 x2j+ jy1 y2j; where P 1 = ( x1 ; y1) and P 2 = ( x2 ; y2):(b) S = R2 ; d1 (P 1 ; P 2) = sup fjx1 x2j; jy1 y2jg; where P 1 = ( x1 ; y1) and P 2 = ( x2 ; y2):

(c) S = C [0; 1]; d1(f; g ) =1

Z 0

jf gj:

14



(d) S = C [0; 1]; d1 (f; g ) = sup fjf (x) g(x)j: x 2[0; 1]g:

(e) S = Any nonempty set ; d(s; t ) = 0 if s = t1 if s 6= t : This is called the discrete metric on S .

———————————————————————————————————————————————–220 . If P n = ( xn ; yn ) 2R2 and d1 is the metric in the previous problem part (b), show that (P n ) convergesto P = ( x; y) with respect to this metric if and only if (xn ) and (yn ) converge to x and y, respectively.

———————————————————————————————————————————————–221 . Let S be a nonempty set and let d be the discrete metric de…ned above. Show that in the metric space(S; d); a sequence (xn ) in S converges to x if and only if there is a K 2N such that xn = x for all n K:———————————————————————————————————————————————–222 . Show that if d is the discrete metric on a set S; then every subset of S is both open and closed in (S; d):———————————————————————————————————————————————–223 . Prove that in any metric space (S; d), an neighborhood of a point is an open set.———————————————————————————————————————————————–224 . Let f : R! R be de…ned by f (x) = x2 if x 2Q

0 if x =2Q : Show that f is di¤erentiable at x = 0 ; and

…nd f 0(0) :———————————————————————————————————————————————–225. Let g : R

!R be de…ned by g(x) = x2 sin(1=x2) if x 6= 0

0 if x = 0: Show that g is di¤erentiable for

all x 2R; Also, show that the derivative g0 is not bounded on the interval [ 1; 1]:———————————————————————————————————————————————–226. Assume that there exists a function L : (0;1 ) ! R such that L0(x) = 1 =x for x > 0: Calculate thederivatives of the following functions:(a) f (x) = L(2x + 3) for x > 0;(b) g(x) = L(x2) 3 for x > 0;(c) h(x) = L(ax ) for a > 0; x > 0;(d) k(x) = L(L(x)) when L(x) > 0; x > 0:———————————————————————————————————————————————–227. Given that the function h(x) = x3 +2 x +1 for x 2R has an inverse h 1 on R; …nd the value of h 1 0 (y)at the points corresponding to x = 0 ; 1; 1:———————————————————————————————————————————————–228. Given that the restriction of the cosine function cos to I = [0 ; ] is strictly decreasing and that cos 0 = 1;cos = 1; let J = [ 1; 1]; and let arccos : J ! R be the function inverse to the restriction of cos to I :Show that arccos is di¤erentiable on ( 1; 1) and D (arccos( y)) = 1p 1 y 2

for y 2( 1; 1):

Show that arccos is not di¤erentiable at 1 and 1:———————————————————————————————————————————————–229. Let f : I ! R be di¤erentiable at c 2I: Establish the Straddle Lemma:Given > 0; there exists > 0 such that if u; v 2I satisfy c < u < c < v < c + ;then we have that jf (v) f (u) (v u)f 0(c)j (v u) :(Hint: add and subtract the term f (c) cf 0(c)) :———————————————————————————————————————————————–230. Prove that if f : R! R is an even function [that is, f ( x) = f (x) for all x 2R] and has a derivativeat every point, then the derivative f 0 is an odd function [that is, f 0( x) = f 0(x) for all x 2R]: Also,

prove that if g : R! R is a di¤erentiable odd function, then g0 is an even function.———————————————————————————————————————————————–231. If f : R! R is di¤erentiable at c 2R; show that f 0(c) = lim

n !1(n ff (c + 1 =n) f (c)g) : However, show

by an example that the existence of the limit of this sequence does not imply the existence of f 0(c):———————————————————————————————————————————————–232. Use the Mean Value Theorem to prove that jsin x sin yj jx yj for all x; y 2R:———————————————————————————————————————————————–233. If h(x) = 0 if x < 0

1 if x 0 ; prove there does not exist a function f : R! R such that f 0(x) = h(x) for

all x 2R: Give examples of two functions, not di¤ering by a constant, whose derivatives equal h(x) for

15



all x 6= 0 :———————————————————————————————————————————————–234. Let I be an interval. Prove that if f is di¤erentiable on I and if the derivative f 0 is bounded on I ; thenf satis…es a Lipschitz condition on I :———————————————————————————————————————————————–235. Let f ; g be di¤erentiable functions on R and suppose that f (0) = g(0) and f 0(x) g0(x) for all x 0:Show that f (x) g(x) for all x 0:———————————————————————————————————————————————–236. Suppose that f : [0; 2] ! R is continuous on [0; 2] and di¤erentiable on (0; 2); and that f (0) = 0 ;f (1) = 1 ; f (2) = 1 :(a) Show that there exists c1 2(0; 1) : f 0(c1) = 1 :(b) Show that there exists c2 2(1; 2) : f 0(c2) = 0 :(c) Show that there exists c 2(0; 2) : f 0(c) = 1 =3:———————————————————————————————————————————————–237. Use the Mean Value Theorem to prove that x 1

x < ln x < x 1 for x > 1:[Hint: Use the fact that D ln x = 1=x for x > 0]:———————————————————————————————————————————————–238. Let f : [a; b] ! R be continuous on [a; b]; and di¤erentiable in (a; b): Show that if lim

x! af 0(x) = A;

then f 0(a) exists and is equal to A:———————————————————————————————————————————————–239. Let f : R! R be de…ned by f (x) = 2x4 + x4 sin(1=x) if x 6= 0

0 if x = 0 : Show that f has an absolute

minimum at x = 0 ; but that its derivative has both positive and negative values in every neighborhood of 0.———————————————————————————————————————————————–240. Give an example of a uniformly continuous function on [0; 1] that is di¤erentiable on (0; 1) but whosederivative is not bounded on (0; 1):———————————————————————————————————————————————–241 . Suppose that f and g are continuous on [a; b], di¤erentiable on (a; b), that c 2[a; b]; and that g(x) 6= 0for x 2[a; b]; x 6= c: Let A = lim

x! cf (x) and B = lim

x! cg(x): If B = 0 ; and if lim

x! cf (x )g(x ) exists in R; show that we

must have A = 0 : [Hint : f (x) = f (x )g(x ) g(x)]:

———————————————————————————————————————————————–242. In addition to the suppositions of the preceding exercise, let g(x) > 0 for x 2[a; b]; x 6= c: If A > 0 andB = 0 ; prove that we must have lim

x! cf (x )g(x ) = 1 : If A < 0 and B = 0 ; prove that we must have lim

x! cf (x )g(x ) = 1:

———————————————————————————————————————————————–243. Let f (x) = x2 for x 2Q

0 for x =2Q ; and let g(x) = sin x for x 2R: Use Theorem 6.3.1 to show that

limx! 0

f (x )g(x ) = 0 : Explain why L’Hospital’s Rule I cannot be used.

———————————————————————————————————————————————–244. Evaluate the following limits:(a) lim

x!1ln xx 2 (0;1 ); (b) lim

x!1ln xp x (0;1 );

(c) limx! 0

x lnsin x (0; ); (d) limx!1

x +ln xx ln x (0;1 ):

———————————————————————————————————————————————–245. Evaluate the following limits:(a) limx! 0+ x2x (0;1 ); (b) limx! 0(1 + 3x )x (0;1 );(c) lim

x!1(1 + 3

x )x (0;1 ); (d) limx! 0+

1x 1

arctan x (0;1 ):———————————————————————————————————————————————–246. Let g(x) = x3 for x 2R: Find g0(x) and g00(x) for x 2R; and g000(x) for x 6= 0 : Show that g000(0) doesnot exist.———————————————————————————————————————————————–247. Use induction to prove Leibniz’s rule for the n-th derivative of a product:

(fg )(n ) (x) =n

Xk=0

nk f (n k) (x)g(k ) (x):

16



———————————————————————————————————————————————–248. If x > 0; show that (1 + x)1=3 1 + 1

3 x 19 x2 5

81 x3 : Use this inequality to approximate 3p 1:2and 3p 2:———————————————————————————————————————————————–249. If f (x) = ex ; show that the remainder term in Taylor’s Theorem converges to 0 as n ! 1 ; for each…xed x0 and x:———————————————————————————————————————————————–250. Let h(x) = e 1=x 2

for x 6= 00 for x = 0

: Show that h (n ) (0) = 0 for all n 2N: Conclude that the remainder

term in Taylor’s Theorem for x0 = 0 does not converge to 0 as n ! 1 ; for x 6= 0 : [Hint: By L’Hospital’sRule, lim

x! 0h (x )x k = 0 for any k 2N: Use Leibniz’s Rule to calculate h (n ) (x) for x 6= 0] :

———————————————————————————————————————————————–251. We wish to approximate sin by a polynomial on [ 1; 1] so that the error is less than 0:001. Show thatwe have sin x x x3

6 + x5

120 < 15040 for jxj 1:

———————————————————————————————————————————————–252 . If f 2 R[a; b] and jf (x)j M for all x 2[a; b]; show that R

ba f M (b a):

———————————————————————————————————————————————–253. Suppose f is bounded on [a; b] and that there exists two sequences of tagged partitions of [a; b] such that

:

P n ! 0 and:

Qn ! 0; but such that limn !1 S (f ;:

P n ) 6= limn !1 S (f ;:

Qn ): Show that f =2 R[a; b]:———————————————————————————————————————————————–254. Let f be the Dirichlet function de…ned by f (x) = 1 if x 2Q\ [0; 1]

0 if x 2[0; 1] Q : Show that f =2 R[0; 1]:———————————————————————————————————————————————–255. Suppose that f : [a; b] ! R and that f (x) = 0 except for a …nite number of points c1 ;:::;cn in [a; b]:Prove that f 2 R[a; b] and that R b

a f = 0 :———————————————————————————————————————————————–256. If g 2 R[a; b] and if f (x) = g(x) except for a …nite number of points in [a; b]; prove that f 2 R[a; b]and that R

ba f = R

ba g:

———————————————————————————————————————————————–257. Let 0 a < b; let f (x) = x2 for x 2[a; b]; and let P = f[x i 1; x i ]gn

i=1 be a partition of [a; b]: For each i;let q

i =

q 1

3x2

i + x

i 1x

i + x2

i 1:

(a) Show that q i satis…es 0 x i 1 q i x i :(b) Show that f (q i )(x i x i 1) = 1

3 x3i x3

i 1 :(c) If

:

Q is the tagged partition with the same subintervals as P and the tags q i ; show thatS (f ;

:

Q) = 13 b3 a3 :

(d) Show that f 2 R[a; b] and R b

a f = R b

a x2dx = 13 b3 a3 :

———————————————————————————————————————————————–258. Consider the function h(x) = x + 1 if x 2Q\ [0; 1]

0 if x 2[0; 1] Q : Show that h =2 R[0; 1]:———————————————————————————————————————————————–259. Let H (x) = k if x = 1

k ; k 2N0 if x 2[0; 1] f1

k : k 2Ng: Show that H =2 R[0; 1]:

———————————————————————————————————————————————–

260. If S (f ;

:

P ) is any Riemann sum of f : [a; b] ! R, show that there exists a step function ' : [a; b] ! Rsuch that R b

a ' = S (f ;:

P ):———————————————————————————————————————————————–261. Suppose that f is continuous on [a; b]; that f (x) 0 for all x 2[a; b] and that R

ba f = 0 : Prove that

f (x) = 0 for all x 2[a; b]:———————————————————————————————————————————————–262. Show that the continuity hypothesis in the preceeding exercise cannot be dropped.———————————————————————————————————————————————–263. If f and g are continuous on [a; b] and if R

ba f = R

ba g; prove that there exists c 2[a; b] such that

f (c) = g(c):

17



———————————————————————————————————————————————–264. Give an example of a function f : [a; b] ! R such that f 2 R[c; b] for each c 2(a; b) but f =2 R[a; b]:———————————————————————————————————————————————–265. (Mean Value Theorem for Integrals ) If f is continuous on [a; b]; a < b; show that there exists c 2[a; b]such that

R b

a f = f (c)(b a):———————————————————————————————————————————————–266 . If : [a; b] ! R takes on only a …nite number of distinct values. is a step function?———————————————————————————————————————————————–267. If (x) = x and ! (x) = x and if (x) f (x) ! (x) for all x 2[0; 1]; does it follow from the squeezetheorem that f 2 R[0; 1]?———————————————————————————————————————————————–268. If f is bounded by M on [a; b] and if the restriction of f to every interval [c; b] where c 2(a; b) isRiemann integrable, show that f 2 R[a; b] and that lim

c! a + R b

c f = R b

a f :

[Hint: Let c(x) = M for x 2[a; c)f (x) for x 2[c; b] and ! c(x) = M for x 2[a; c)

f (x) for x 2[c; b] :

Apply the squeeze theorem for c su¢ciently near a].———————————————————————————————————————————————–

269. Show that g(x) = sin(1=x) for x

2(0; 1]

0 for x = 0 belongs to R[0; 1]:———————————————————————————————————————————————–270. If f is bounded and there is a …nite set E such that f is continuous at every point of [a; b]nE; show thatf 2 R[a; b]:———————————————————————————————————————————————–271. Let f be continuous on [a; b]; let f (x) 0 for x 2[a; b]; and let

M n = R b

a f n1=n

:

Show that limn !1

M n = supx2[a;b ] ff (x)g:

———————————————————————————————————————————————–272. If n 2N and H n (x) = xn +1

n +1 for x 2[a; b]; show that the …rst form of the fundamental theorem of

calculus implies that R ba xn dx = b

n +1a

n +1

n +1 : What is the set E for which H 0n (x) 6= xn ?———————————————————————————————————————————————–273. If g(x) = x for jxj 1

x for jxj< 1 and if G(x) = 12 x2 1 ; show that R

32 g(x)dx = G(3) G( 2) = 5 =2:

———————————————————————————————————————————————–274. Let B (x) = 1

2 x2 for x < 012 x2 for x 0 : Show that R

ba jxjdx = B(b) B (a):

———————————————————————————————————————————————–275. If f 2 R[a; b] and if c 2[a; b]; the function de…ned by F c(z) = R

zc f for z 2[a; b] is called the inde…nite

integral of f with basepoint c: Find a relation between F a and F c :———————————————————————————————————————————————–

276. Let f : [0; 3] ! R be de…ned by f (x) = 8<:

x for 0 x < 11 for 1 x < 2x for 2 x 3

9=;

: Obtain formulas for F (x) =

R x

0 f and

sketch the graphs of f and F: Where is F di¤erentiable? Evaluate F 0(x) at all such points.———————————————————————————————————————————————–277. If f : R! R is continuous and c > 0; de…ne g : R! R by g(x) = R x + c

x c f (t)dt. Show that g isdi¤erentiable on R and …nd g0(x):———————————————————————————————————————————————–278. If f : [0; 1] ! R is continuous and R

x0 f = R

1x f for all x 2[0; 1]; show that f (x) = 0 for all x 2[0; 1]:

———————————————————————————————————————————————–279. (a) If Z 1 and Z 2 are null sets, show that Z 1 [ Z 2 is a null set.

(b) More generally, if Z n is a null set for each n 2N; show that 1

Sn =1Z n is a null set.

18



———————————————————————————————————————————————–280. Let f ; g 2 R[a; b]:(a) If t 2R; show that R

ba (tf g)2 0:

(b) Use (a) to show that 2 R b

a f g tR b

a f 2 + 1t R

ba g2 for t > 0:

(c) If

R b

a f 2 = 0 ; show that

R b

a f g = 0 ; and if

R b

a g2 = 0 ; show that

R b

a f g = 0 :(d) Prove the following Schwartz Inequality :

R ba f g

2

R ba jfg j

2

R ba f 2 R b

a g2

———————————————————————————————————————————————–281 . Show that lim

n !1x

x + n = 0 for all x 2R; x 0:———————————————————————————————————————————————–282 . Show that lim

n !1nx

1+ n 2 x 2 = 0 for all x 2R:———————————————————————————————————————————————–283 . Evaluate lim

n !1nx

1+ nx for x 2R; x 0:———————————————————————————————————————————————–284 . Evaluate lim

n !1x n

1+ x n for x 2R; x 0:———————————————————————————————————————————————–285 . Evaluate lim

n !1sin( nx )1+ nx for x 2R; x 0:

———————————————————————————————————————————————–286 . Evaluate lim

n !1e nx for x 2R; x 0:

———————————————————————————————————————————————–287 . Show that lim

n !1xe nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–288 . Show that lim

n !1x2e nx = 0 and lim

n !1n2x2e nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–289 . Show that lim

n !1(cos x)2n exists for all x 2R. What is its limit?

———————————————————————————————————————————————–290 . Show that if a > 0; then the convergence of the sequence n x

x + n o1n =1

is uniform on [0; a]; but notuniform on [0;

1):

———————————————————————————————————————————————–291 . Show that if a > 0; then the convergence of the sequence n nx

1+ n 2 x 2 o1n =1

is uniform on [a; 1 ); but notuniform on [0;1 ):———————————————————————————————————————————————–292 . Show that if 0 0; then the convergence of the sequence nsin( nx )

1+ nx o1n =1

is uniform on [a; 1 ); but notuniform on [0;1 ):———————————————————————————————————————————————–294 . Show that the sequence x2e nx converges uniformly on [0;1 ):———————————————————————————————————————————————–295 . Show that if a > 0; then the sequence n2x2e nx 1n =1 converges uniformly on [a; 1 ); but it does notconverge uniformly on [0;1 ):———————————————————————————————————————————————–296 . Show that if (f n ) and (gn ) converge uniformly on the set A to f and g; respectively, then (f n + gn )converges uniformly on A to f + g:———————————————————————————————————————————————–297 . Show that if f n (x) = x + 1

n and f (x) = x for x 2R; then (f n ) converges uniformly on R to f ; but thesequence f 2n does not converge uniformly on R. (Thus, the product of uniformly convergent sequencesof functions may not converge uniformly).

19



———————————————————————————————————————————————–298 . Let (f n ) and (gn ) be sequences of bounded functions on A that converge uniformly on A to f and g;respectively. Show that (f n gn ) converges uniformly on A to f g:———————————————————————————————————————————————–299 . Let (f n ) be a sequence of functions that converges uniformly to f on A and that satis…es jf n (x)j M for all n

2N and all x

2A: If g is continuous on the interval [ M; M ]; show that the sequence (g f n )

converges uniformly to g f on A:———————————————————————————————————————————————–300 . Show that the sequence xn

1+ x n does not converge uniformly on [0; 2] by showing that the limitfunction is not continuous on [0; 2]:———————————————————————————————————————————————–301 . Construct a sequence of functions on [0; 1] each of which is discontinuous at each point of [0; 1]and which converges uniformly to a function that is continuous at every point.———————————————————————————————————————————————–302 . Suppose (f n ) is a sequence of continuous functions on an interval I that converges uniformly on I toa function f : If (xn ) I converges to x0 2I; show that lim

n !1(f n (xn )) = f (x0) :

———————————————————————————————————————————————–303 . Let f : R! R be uniformly continuous on R and let f n (x) = f (x + 1

n ) for x 2R: Show that (f n )

converges uniformly on R to f :———————————————————————————————————————————————–304 . Let f n (x) = 1

(1+ x ) n for x 2[0; 1]: Find the pointwise limit f of the sequence (f n ) on [0; 1]:Does (f n ) converge uniformly to f on [0; 1]?———————————————————————————————————————————————–305 . Suppose that the sequence (f n ) converges uniformly to f on the set A; and suppose that each f nis bounded on A: (That is, for each n there is a constant M n such that jf n (x)j M n for all x 2A):Show that the function f is bounded on A:———————————————————————————————————————————————–306 . Let f n (x) = nx

1+ nx 2 for x 2A = [0;1 ): Show that each f n is bounded on A; but the pointwise limit f of the sequence is not bounded on A: Does (f n ) converge uniformly to f on A?———————————————————————————————————————————————–307 . Let f n (x) = xn

n for x

2[0; 1]: Show that the sequence (f n ) of di¤erentiable functions converges uniformly

to a di¤erentiable function f on [0; 1]; and that the sequence (f 0n ) converges on [0; 1] to a function g ; butthat g(1) 6= f 0(1) :———————————————————————————————————————————————–308 . Let gn (x) = e nx =n for x 0; n 2N: Examine the relation between lim

n !1(gn ) and lim

n !1(g0n ) :

———————————————————————————————————————————————–309 . Show that lim

n !1 R 2

1 e nx 2dx = 0 :

———————————————————————————————————————————————–310 . Let f n (x) = nx

1+ nx for x 2[0; 1]: Show that (f n ) converges nonuniformly to an integrable functionf and that R

10 f (x)dx = lim

n !1 R 1

0 f n (x)dx:———————————————————————————————————————————————–311 . Let gn (x) = nx (1 x)n for x 2[0; 1]; n 2N: Discuss the convergence of (gn ) and R

10 gn (x)dx :

———————————————————————————————————————————————–312 . Let

fr

1; r

2; ; r

n;

g be an enumeration of the rational numbers in [0; 1]; and let

f n (x) = 1 if x 2 fr 1 ; r 2 ; ; rn g0 otherwise :

Show that f n 2 R[0; 1] for each n 2N; that f 1(x) f 2(x) f n (x) ; and that f (x) = limn !1

(f n (x))is the Dirichlet function which is not Riemann integrable on [0; 1]:———————————————————————————————————————————————–313 . Show that if a convergent series contains only a …nite number of negative terms, then it is absolutelyconvergent.

20



———————————————————————————————————————————————–314 . Show that if a series is conditionally convergent, then the series obtained from its positive terms isdivergent and the series obtained from its negative terms is divergent.———————————————————————————————————————————————–315 . If 1

Pn =1

xn is conditionally convergent, give an argument to show that there exists a rearrangement

whose partial sums diverge to 1:———————————————————————————————————————————————–316 . Find an explicit expression for the n -th partial sum of 1

Pn =2ln 1 1

n 2 and show that this series

converges to ln 2: Is this convergence absolute?———————————————————————————————————————————————–317 . (a) If 1

Pn =1an is absolutely convergent and (bn ) is a bounded sequence, show that 1

Pn =1an bn is absolutely

convergent.

(b) Give an example to show that if the convergence of 1

Pn =1an is conditional and (bn ) is a bounded

sequence, then 1

Pn =1an bn may diverge.

———————————————————————————————————————————————–

318 . Give an example of a convergent series 1Pn =1an such that 1Pn =1

a2n is not convergent.———————————————————————————————————————————————–319 . Give an example of a divergent series 1

Pn =1an with (an ) decreasing and such that lim

n !1(na n ) = 0 :

———————————————————————————————————————————————–320 . If (an ) is a sequence and if lim

n !1n 2an exists in R; show that 1

Pn =1an is absolutely convergent.

———————————————————————————————————————————————–321 . If (an k ) is a subsequence of (an ), then the series 1

Pn =1an k is called a subseries of 1

Pn =1an : Show that

1

Pn =1an is absolutely convergent if and only if every subseries of it is convergent.

———————————————————————————————————————————————–322 . Let a > 0: Show that the series 1

Pn =1

11+ a n is divergent if 0 < a 1 and is convergent if a > 1:

———————————————————————————————————————————————–323 . Establish the convergence or divergence of the series whose n th term is:(a) (n (n + 1)) 1=2 ;(b) n2 (n + 1) 1=2 ;(c) n!=nn ;(d) ( 1)n n= (n + 1) :———————————————————————————————————————————————–324 . If a;b > 0; show that 1

Pn =1(an + b) p converges if p > 1 and diverges if p 1:

———————————————————————————————————————————————–325 . Discuss the series whose nth term is:(a) n !

3 5 7 (2 n +1) ;

(b) (n !)2

(2 n )! ;(c) 2 4 (2 n )

3 5 (2 n +1) ;

(d) 2 4 (2 n )5 7 (2 n +3) :

———————————————————————————————————————————————–326 . Show that the series 1 + 1

2 13 + 1

4 + 15 1

6 + + is divergent.———————————————————————————————————————————————–327 . For n 2N; let cn = 1

1 + 12 + + 1

n ln n: Show that (cn ) is a decreasing sequence of positivenumbers. The limit C of this sequence is called Euler’s Constant and is approximately equal to 0:577.Show that if we put

21



bn = 11 1

2 + 13 1

2n

then the sequence (bn ) converges to ln 2: [Hint: bn = c2n cn + ln 2 :]———————————————————————————————————————————————–328 . Test the following series for convergence and for absolute convergence:

(a) 1

Pn =1

( 1) n +1

n2

+1 ; (b) 1

Pn =1

( 1) n +1

n +1 ; (c) 1

Pn =1

( 1) n +1 n

n +2 ; (d) 1

Pn =1

( 1)n +1 ln n

n :

———————————————————————————————————————————————–329 . Give an example to show that the Alternating Series Test may fail if (an ) is not a decreasing sequence.———————————————————————————————————————————————–330 . Consider the series

1 12 1

3 + 14 + 1

5 16 1

7 + + ;

where the signs come in pair. Does it converge?———————————————————————————————————————————————–331 . Let an 2R for n 2N and let p < q: If the series 1

Pn =1

a nn p is convergent, show that the series 1

Pn =1

a nn q is

also convergent.———————————————————————————————————————————————–332 . If p and q are positive numbers, show that 1Pn =2

( 1)n (ln n ) pn q is a convergent series.

———————————————————————————————————————————————–333 . Discuss the series whose nth term is:(a) ( 1)n n n

(n +1) n +1 ; (b) n n

(n +1) n +1 ; (c) ( 1)n (n +1) n

n n ; (d) (n +1) n

n n +1 :———————————————————————————————————————————————–334 . If the partial sums of 1

Pn =1an are bounded, show that the series 1

Pn =1an e nt converges for t > 0:

———————————————————————————————————————————————–335 . If the partial sums sn of 1


Pn =1

a nn converges to 1

Pn =1

s nn (n +1) :

———————————————————————————————————————————————–336. Show that the hypothesis that the sequence (xn ) is decreasing in the Dirichlet’s Test can be replaced

by the hypothesis that 1

Pn =1 j

xn xn +1

j is convergent.

———————————————————————————————————————————————–337. Show that if the partial sums sn of the series 1

Pk =1ak satisfy jsn j Mn r for some r < 1; then the

series 1

Pk =1

a kk converges.

———————————————————————————————————————————————–338. Suppose that 1

Pn =1an is a convergent series of real numbers. Either prove that 1

Pn =1bn converges or

give a counter-example, where we de…ne bn by:(a) an

n ; (b) p a nn (an 0) (c) p a n

n (an 0); (d) n1=n an :———————————————————————————————————————————————–339. Discuss the convergence and uniform convergence of the series 1

Pn =1

f n where f n is given by:

(a) x2 + n21

(b) (nx )2

(x 6= 0) (c) sin(x=n2) (d) (xn + 1)1

(x 0)———————————————————————————————————————————————–340. If 1

Pn =1an is an absolutely convergent series, show that the series 1

Pn =1an sin nx is absolutely and

uniformly convergent.———————————————————————————————————————————————–341. Determine the radius of convergence of the power series 1

Pn =1an xn , where an is given by:

(a) 1=nn ; (b) n =n!; (c) n n =n!; (d) (ln n) 1 (n 2); (e) (n!)2 =(2n)!;(f) n p n :

22



———————————————————————————————————————————————–342. If an = 1 if n = k2

0 otherwise and bn = 1 if n = k!0 otherwise for k 2N, …nd the radius od convergence

of the power series 1

Pn =1an xn and of 1

Pn =1bn xn :

———————————————————————————————————————————————–

343. If 0 < p jan j q for all n 2N; …nd the radius of convergence of 1

Pn =1 an xn :———————————————————————————————————————————————–344. Let f (x) = 1

Pn =1an xn for jxj< R: If f (x) = f ( x) for all jxj< R; show that an = 0 for all n odd.

———————————————————————————————————————————————–345. Prove that if f is de…ned for jxj< r and if there exists a constant B such that f (n ) (x) B for all

jxj< r and n 2N; then the Taylor series expansion

1

Pn =1

f ( n ) (0)n ! xn

converges to f (x) for jxj< r:———————————————————————————————————————————————–346. Find a series expansion for

R x

0 e t 2dt for x

2R:

———————————————————————————————————————————————–347. Let f : R! R be a Lebesgue integrable function, i.e., R R fdm < 1 .(a) Prove that m(fx 2R : f (x) = 1g) = 0 :(b) Prove that 8 > 0; m(fx 2R : jf (x)j g) < 1 :(c) Prove that the Lebesgue integral is absolutely continuous with respect to the Lebesgue measure, i.e.,if f is Lebesgue integrable on A , then 8 > 0 9 > 0 :R B jf jdm < whenever B A with m(B ) < :(d) Prove that 8 > 0; 9 a compact set K R : R RnK jf jdm < :(e) Prove that 8 > 0; 9 M > 0 and a measurable set A R : jf (x)j M on A ; and R RnA jf jdm < :———————————————————————————————————————————————–348. Show that if f is a Lebesgue integrable function on A and A n = fx 2A : jf (x)j ng; thenlim

n !1n m (A n ) = 0 :

———————————————————————————————————————————————–

349. Show that if f 0 on a set A , m(A

) > 0; and R A fdm = 0 ; then f = 0 a.e. on A .———————————————————————————————————————————————–

350. Show that if R B fdm = 0 for every measurable subset B of A ; m(A ) > 0; then f = 0 a.e. on A .———————————————————————————————————————————————–351. Show that if f is Lebesgue integrable on A and R A fdm = R A jf jdm; then either f 0 a.e. on A ,or f 0 a.e. on A .———————————————————————————————————————————————–352. Let ff n g be a sequence of nonnegative and measurable functions on A such that lim

n !1 R A f n dm = 0 :

Show that f nm

! 0: ( f nm

! 0 means f n converges to 0 in measure).———————————————————————————————————————————————–353. Show that in problem 6, convergence in measure cannot be replaced by a.e. convergence.———————————————————————————————————————————————–354. Let

ff n

gbe a sequence of measurable functions on a set A , m(A ) <

1: Show that lim

n !1 R A jf n j

1+ jf n jdm = 0

if and only if f nm

! 0:———————————————————————————————————————————————–355. Show that the assumption m(A ) < 1 is essential in problem 8.———————————————————————————————————————————————–356. Suppose that f is nonnegative and measurable on A , m(A ) < 1 : Let A k = fx 2A : k f (x) < k +1 g:

Prove that f is Lebesgue integrable on A if and only if 1

Xk=0

km (A k ) < 1 :

———————————————————————————————————————————————–357. Suppose that f is nonnegative and measurable on A , m(A ) < 1 : Let B k = fx 2A : f (x) kg:

23




Xk=0

m(B k ) < 1 :

———————————————————————————————————————————————–

358. Suppose that f is nonnegative and integrable on A , m(A ) < 1 : For > 0; de…ne S ( ) =1

Xk =0

k m(A k );

where A

k = fx 2A

: k f (x) < (k + 1) g: Prove that lim! 0S ( ) = R A fdm:———————————————————————————————————————————————–

359. Let ff n g be a sequence of nonnegative functions converging to a function f on R; and suppose thatlim

n !1 R R f n dm = R R fdm < 1 : Show that for each measurable set A R; limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–360. Suppose that ff n g is a sequence of measurable functions on A such that jf n j g; where g is integrableon A . Show that R A limf n dm limR A f n dm limR A f n dm R A limf n dm:———————————————————————————————————————————————–361. Suppose that ff n g is a sequence of real-valued functions integrable on A , m(A ) < 1 : Show thatif f n

Af , then lim

n !1 R A f n dm = R A fdm: ( f n A

f means f n converges uniformly to f on A ).———————————————————————————————————————————————–362. Let f n (x) = nx n 1 (n + 1) xn ; x 2(0; 1):

(a) Show that R (0 ;1)

1

Xn =1 f n dm 6=1

Xn =1 R (0 ;1) f n dm:

(b) Show that1

Xn =1 R (0 ;1) jf n jdm = 1 :

———————————————————————————————————————————————–

363. Let ff n g be a sequence of measurable functions on A such that1

Xn =1 R A jf n jdm < 1 :

(a) Show that the function f =1

Xn =1

f n is Lebesgue integrable on A .

(b) Show that R A

1

Xn =1f n dm =

1

Xn =1 R A f n dm:

———————————————————————————————————————————————–

364. Compute limn !1 R 10sin( x

n )(1+ x

n ) n dx and justify your answer.———————————————————————————————————————————————–365. Compute lim

n !1 R n

0 (1 xn )n eix dx and justify your answer.

———————————————————————————————————————————————–366. Let f n (x) = n for 0 < x < 1

n0 for 1

n x < 1 :

(a) Find limn !1

f n (x):

(b) Find limn !1 R 1

0 f n (x)dx:

(c) Is there a function g(x) 2L1(0; 1) : g(x) f n (x) for all n?———————————————————————————————————————————————–367. (a) Let (a; b) (0; 2 ) be an open interval. Show that lim

n !1 R b

a cos(nx )dx = 0 :(b) Let A (0; 2 ) be a measurable set. Show that lim

n!1 R

A cos(nx )dx = 0 :———————————————————————————————————————————————–368. Let f n (x) = n

1+ n 2 x 2 :(a) Does f n ! 0 a.e. on [0; 1]?(b) Does f n ! 0 in L1[0; 1]?(c) Find lim

n !1 R 1

0sin x

x f n (x)dx and justify your answer.———————————————————————————————————————————————–369. Compute lim

n !1 R 11

sin x1+ nx 2 dx and justify your answer.

———————————————————————————————————————————————–370. Let f be a nonnegative Lebesgue integrable function on (0;1 ): Prove that lim

n !11n R

n0 xf (x)dx = 0 :

24



———————————————————————————————————————————————–371. Let f be nonnegative and measurable on [0; 1]: Let A = fx 2[0; 1] : f (x) > 1g:Show that lim

n !1 R 1

0 (f (x))n dx exists if and only if m(A ) = 0 :———————————————————————————————————————————————–372. Show that if f is Riemann integrable on [a; b] and f (x) = 0 for x 2[a; b] \ Q; then

R b

a f (x)dx = 0 :———————————————————————————————————————————————–373. Prove the following Schwartz inequality :If f and g are Lebesgue integrable on (a; b); then R

ba f (x)g(x)dx

2

R b

a (f (x))2 dx R b

a (g(x))2 dx :———————————————————————————————————————————————–374. Let f : [a; b] ! R be nonnegative and Lebesgue integrable.

Prove that R ba f (x)cos xdx

2+ R b

a f (x)sin xdx2

R ba f (x)dx

2:

———————————————————————————————————————————————–375. Find R [0;1] fdm; where f (x) = x2 for x 2[0; 1]nQ

1 for x 2[0; 1] \ Q : Is f Riemann integrable on [0; 1]?———————————————————————————————————————————————–376. Let C be the Cantor set .

(a) Find

R [0;1] fdm; where f (x) = 8

<:sin( x) for x 2[0; 1=2]nC

cos( x) for x 2[1=2; 1]nC

x2

for x 2C

9=;

:

(b) Let f be de…nd on [0; 1] as follows: f (x) = 8><>:0 for x 2C

n for x 22n 1

[k=1

C n;k9>=>;

;where C n;k is a removed

interval of length 13n : Find R [0;1] fdm:

———————————————————————————————————————————————–377. De…ne the Cantor-Lebesgue function ' : [0; 1] ! [0; 1] as follows:

If x is an element of the Cantor set C and x =1

Xn =1

a n3n with an = 0 or 2; then we put

' (x) = ' (1

Xn =1

a n3n ) =

1

Xn =1

a n2

12n ;

that is, if an is the n -th ternary digit for x; then the n-th binary digit for ' (x) is a n2 :

We extend ' to [0; 1] by setting:

' (x) = sup f' (y) : y 2C ; y < xg(a) Show that the Cantor function maps the Cantor set C onto [0; 1].(b) Show that ' is increasing and continuous on [0; 1]:(c) Show that ' 0(x) = 0 a.e. on [0; 1]:(d) Compute R [0;1] 'dm and justify your answer.———————————————————————————————————————————————–378. Let A [0; 1] be a perfect nowhere dense set with m(A ) > 0: Show that A (x) = 1 for x 2A

0 for x =2A

is not Riemann integrable on [0; 1] but it is Lebesgue integrable on [0; 1].———————————————————————————————————————————————–379. Let fr n g be the sequence of rational numbers and de…ne f (x) = Xfn :r n <x g

12n :

(a) Show that f is continuous on the irrationals.(b) Show that f is discontinuous on the rationals.(c) Compute R [0;1] fdm and justify your answer.———————————————————————————————————————————————–

380. Let fr n g be the sequence of rational numbers in [a; b] and de…ne f (x) =1

Xn =1

12n

1p jx r n j:

25



(a) Show that R b

a f (x)dx < 1 :(b) Show that R

ba (f (x))2 dx = 1 :

———————————————————————————————————————————————–381. Let fxn g be a sequence in [0; 1] and fan g be a sequence of nonnegative real numbers such that1

Xn =1

an <

1: Show that

1

Xn =1

a n

p jx x n jconverges for almost all x

2[0; 1]:

———————————————————————————————————————————————–382. Suppose f : [0; 1] ! [0;1 ) is measurable and satis…es m(fx 2[0; 1] : f (x) yg) 1

y(log y) 2 8 y 2:Show that f is Lebesgue integrable on [0; 1]:———————————————————————————————————————————————–383. For p > 0; let B= ff : R! R such that f is measurable and R R jf j

p dm 1g:Suppose that f n 2 B; and f n

m

! f on R: Prove that f 2 B:———————————————————————————————————————————————–384. True or False: If f is Lebesgue integrable on R; then there exists a continuous function g : R ! Rsuch that R R jf gjdm = 0 :———————————————————————————————————————————————–385. We say that functions f n ; n = 1 ; 2; 3; :::; integrable on A are equi-integrable if 8 > 0 9 > 0 such thatfor every measurable subset B of A,

R B

jf n

jdm < for n = 1 ; 2; 3; :::; whenever m(B ) < :

Show that if ff n g is a convergent sequence, say f n ! f; of equi-integrable functions on a set A , m(A ) < 1 ;then lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–386. (Fatou in measure) Show that if f n 0 and f n

m

! f on A , then R A fdm limR A f n dm:———————————————————————————————————————————————–387. (Monotone Convergence in measure) Show that if f n 0; f n f n +1 ; and f n

m

! f on A ,then lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–388. (Lebesgue DCT in measure) Show that if f n

m

! f on A , and there exists a function g integrableon A such that jf n (x)j g(x) for all n 2N and for all x 2A ; then lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–389. (Equi-integrability in measure) Show that if ff n g is a sequence of equi-integrable functions ona set A , m(A ) <

1; and f

n

m

!f on A , then lim

n !1 R A

f n

dm =

R A

fdm:———————————————————————————————————————————————–390. Let f ; f n : R ! R be integrable functions such that f n

a.e.

! f on R:Show that lim

n !1 R R jf n f jdm = 0 if and only if limn !1 R R jf n jdm = R R jf jdm:

———————————————————————————————————————————————–391. Use Fubini’s theorem to prove that R R n e jxj2 dx = n= 2 :———————————————————————————————————————————————–392. Let a > 1: Prove that R

10

x sin x1+( nx ) a dx = o( 1

n ):———————————————————————————————————————————————–393. Compute lim

n !1 R 10

dx(1+ x

n ) n x 1 =n and justify your answer.———————————————————————————————————————————————–394. Compute lim

n !1 R 10

log( x + n )n e x cos x dx and justify your answer.

———————————————————————————————————————————————–

395. (a) Compute limn !1 R 1

0nx

1+ n 2 x 2 dx and justify your answer.(b) Compute lim

n !1 R 1

0n 3 = 2 x

1+ n 2 x 2 dx and justify your answer.———————————————————————————————————————————————–

396. Compute1

Xn =0 R =2

0 1 p sin xn

cos xdx and justify your answer.

———————————————————————————————————————————————–397. Compute lim

n !1 R 1

0 (1 + xn ) n sin( x

n )dx and justify your answer.———————————————————————————————————————————————–398. Compute lim

n !1 R 1

01+ nx

(1+ x ) n dx and justify your answer.

26



———————————————————————————————————————————————–399. Show that f (x) = 1 for x = 1

n0 otherwise is Riemann integrable on [0; 1] and compute R

10 f (x)dx:

———————————————————————————————————————————————–400. (Variant of Fatou’s Theorem) Let fgn g be a sequence of integrable functions on A such thatgn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such that f n

a.e.

! f on A ,

and f n gn a.e. for each n 2N: Prove that lim R A f n dm R

A limf n dm = R A fdm:———————————————————————————————————————————————–

401. (Variant of Fatou’s Theorem in measure) Let fgn g be a sequence of integrable functions on Asuch that gn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such thatf n

m

! f on A , and f n gn a.e. for each n 2N: Prove that lim R A f n dm R A limf n dm:———————————————————————————————————————————————–402. (Variant of Lebesgue’s DCT) Let fgn g be a sequence of integrable functions on A such thatgn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such that f n

a.e.

! f on A ,and jf n j gn a.e. for each n 2N: Prove that lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–403. (Variant of Lebesgue’s DCT in measure) Let fgn g be a sequence of integrable functions on Asuch that gn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such thatf n

m

!f on A , and

jf n

jgn a.e. for each n

2N: Prove that lim

n !1 R A f n dm =

R A fdm:

———————————————————————————————————————————————–404. Let 1 p < 1 ; f n 2L p(R) such that f n

a.e.

! f on R: Suppose that(i) 9n1 > 0 and A R with m(A ) < 1 : R RnA jf n j

p dm 1 8n n1 ; and(ii) 9n0 > 0 and 0 < < 1 : whenever m(F ) < ; R F jf n j

p dm 1 8n n0 :Show that f 2L p(R):———————————————————————————————————————————————–405. Prove the following Tchebyshev’s Inequality : Let f be nonnegative and measurable on A :If > 0; then m(fx 2A : f (x) > g) 1 R A fdm:———————————————————————————————————————————————–406. Let f n (x) = p ne nx on [0; 1]: Prove that:(a) f n (x) ! 0 pointwise in (0; 1]:(b)

R 1

0 (f n (x))2dx C for all n 2N:

(c) f n does not converge in L2(0; 1):

(d) R 1

0 f n (x)g(x)dx ! 0 for each g 2L2(0; 1):———————————————————————————————————————————————–407. Let f n (x) = 1

jx 1n j

1 = 2 on (0; 1): Prove that:

(a) f n (x) converges pointwise on (0; 1):(b) f n (x) converges in measure on (0; 1):(c) f n converges in L1(0; 1):(d) @ g 2L1(0; 1) : f n (x) g(x) for a.e. x 2(0; 1) and for all n 2N:———————————————————————————————————————————————–408. Let p 1: Suppose f n 2L p(R), sup n jf n j 2L p(R); and f n

a.e.

! f in R. Prove that f 2L p(R) andthat f n ! f in L p(R):———————————————————————————————————————————————–

409. Let A

R be measurable with m(A

) < 1 : Let ff n g be a sequence of functions in L1

(A

) such thatf nm

! 0 on A , and kf n kL 1 ( A ) M for all n 2N. Prove that limn !1 R A jf n gj

1=2 dm = 0 for all g 2L1(A ):———————————————————————————————————————————————–410. Let f be a measurable function on [0; 1] and let A = fx 2[0; 1] : f (x) 2Zg:Prove that A is measurable and that lim

n !1 R 10 [cos( f (x))]2n dx = m(A ):

———————————————————————————————————————————————–411. Show that if m(A ) < 1 ; and 0 < p 1 < p 2 < 1 ; then L p2 (A ) L p1 (A ); and the inequality

kf k p1 kf k p2(m(A ))

1p 1

1p 2

27



holds for f 2L p2 (A ):———————————————————————————————————————————————–412. Given f 2L2([0; 1]) de…ne K (x) = 1

x 4 = 3 R x

0 f (t)dt:Show that kK k1 C kf k2 where C is a constant independent of f :———————————————————————————————————————————————–413. Recall that the convolution of two integrable functions f and g is de…ned by:

(f g) (x) = R R f (x y)g(y)dy:

Let f 0 such that R R f (x)dx = A < 1: De…ne the sequence f n = f f f

| {z } n times

:

Prove that f n 2L1(R) for each n 2N, and f n ! 0 in L1(R):———————————————————————————————————————————————–414. Let ff n g be a sequence of measurable functions on A , m(A ) < 1 ; such that f n

m

! f on A ,and jf n (x)j C for x 2A , and n = 1 ; 2; 3; :::: Show that if g is continuous on [ C; C ]; thenlim

n !1 R A g(f n )dm = R A g(f )dm:———————————————————————————————————————————————–415. Let ff n g be a sequence of measurable functions on A , m(A ) < 1 ; such that f n

m

! f on A .

Then limn !1 R A sin( f n )dm = R

A sin(f )dm:———————————————————————————————————————————————–416. Suppose that g satis…es a Lipschitz condition on R; i.e., 9C > 0 : 8x; y 2R; jg(x) g(y)j C jx yj:Show that if f is Lebesgue integrable on [a; b]; then g(f ) is Lebesgue integrable on [a; b]:———————————————————————————————————————————————–417. Suppose that g satis…es a Lipschitz condition on R; i.e., 9C > 0 : 8x; y 2R; jg(x) g(y)j C jx yj:Show that if ff n g is a sequence of measurable functions on [a; b] such that f n

m

! f and there is a Lebesgueintegrable function G such that jf n (x)j G(x); then lim

n !1 R [a;b ] g(f n )dm = R [a;b ] g(f )dm:———————————————————————————————————————————————–418. Let ff n gbe a sequence of functions converging to f on A such that R A jf n j

p dm < 1 and R A jf j p dm < 1 ;

1 p < 1 : Show that limn !1 R A jf n f j

p dm = 0 if and only if limn !1 R A jf n j

p dm = R A jf j p dm:

———————————————————————————————————————————————–419. Let

ff n

g be a sequence of functions in L p[a; b]; 1 0 : kf n k p C for n = 1 ; 2; 3;:::: Prove that 8g 2Lq[a; b]; 1 p + 1q = 1 ;

limn !1 R [a;b ] f n gdm = R [a;b ] fgdm:———————————————————————————————————————————————–420. Let ff n g be a sequence of functions in L2(R): Suppose that kf n kL 2 (R ) M for all n 2N; andf n

a.e.

! f on R: Prove that limn !1 R R f n gdm = R R fgdm for all g 2L2(R):

———————————————————————————————————————————————–421. Let ff n g be a sequence of functions in L p[a; b]; 1 p < 1 ; which converges in norm to f 2L p[a; b];and let fgn g be a sequence of measurable functions such that jgn j C for n = 1 ; 2; 3;:::; and gn

a.e.

! g:Show that f n gn ! fg in L p[a; b]:———————————————————————————————————————————————–422. Let f ; f n : R ! R be measurable functions such that f n

a.e.

! f on R and there exists a function gintegrable on R such that

jf n (x)

jg(x) a.e. for all n

2N. Show that f n

!f almost uniformly on R:

(That is, 8 > 0; 9 a measurable set A R : m(RnA ) < and f n A

f ):———————————————————————————————————————————————–423. Suppose that f is nonnegative and measurable on a set A , where 0 < m (A ) < 1 : Suppose also thatthere are positive and such that 1

m ( A ) R A fdm and 1m ( A ) R A f 2dm :

Show that if 0 < < 1 and A = fx 2A : f (x) g; then m(A ) m(A )(1 )2 2 :

———————————————————————————————————————————————–424. We say that a measurable function f on a set A is essentially bounded if m(fx 2A : jf (x)j> r g) = 0for some real number r: In this case, we de…ne the essential supremum of f by:

kf k1 = inf fr : m(fx 2A : jf (x)j> r g) = 0 g:

28



Show that if f is essentially bounded on [a; b]; then lim p!1 R [a;b ] jf j

p dm1=p

= kf k1 :———————————————————————————————————————————————–425. Prove the following Hölder’s Inequality : If 1 p 1 and 1

p + 1 p0 = 1 ; then kfgk1 kf k p kgk p0 ;

that is,

R A jfg jdm

R A jf j

p dm 1=p

R A jgj

p0

dm1=p 0

; and

R A jfg jdm kf k1

R A jgjdm:

———————————————————————————————————————————————–426. (a) Prove the following Minkowski’s Inequality : If 1 p 1; then kf + gk p kf k p + kgk p ;

that is, R A jf + gj p dm 1=p

R A jf j p dm 1=p + R A jgj

p dm 1=p ; and kf + gk1 kf k1 + kgk1 :(b) Show that Minkowski’s Inequality fails for 0 < p < 1:———————————————————————————————————————————————–427. Let f : [0; 1] ! R be positive and continuous. For p 6= 0 ; let N ( p) = R

10 f p(x)dx

1=p:

(a) Compute lim p!1

N ( p):

(b) Compute lim p! 1

N ( p):

(c) Compute lim p! 0

N ( p):———————————————————————————————————————————————–428. Suppose f 2C 1([a; b]); f (a) = f (b) = 0 ; and

R b

a f 2(x)dx = 1 :

(a) Show that R b

a xf (x)f 0(x)dx =12 :

(b) Show that 14 R

ba (f 0(x))2 dx R

ba x2f 2(x)dx :

———————————————————————————————————————————————–429. (a) Find min

f 2AR 10 (1 + x2)f 2(x)dx; where A= ff 2C ([0; 1]) :R 1

0 f (x)dx = 1g; and …nd a function for

which the minimum is attained.(b) Let A= ff 2C ([0; 1]) :R

10 f (x)dx = 3 ; and R

10 xf (x)dx = 2g: Find min

f 2AR 1

0 f 2(x)dx; and a function for

which the minimum is attained.———————————————————————————————————————————————–430. Show that if f is continuous on [0; 1]; di¤erentiable on (0; 1); f (0) = 0 ; and 0 < f 0(x) 1 on (0; 1);

then R 1

0 f (x)dx2

R 1

0 (f (x))3 dx: Show also that equality occurs if and only if f (x) = x:———————————————————————————————————————————————–431. Show that if f is a convex function on [a; b]; then f a + b

2(b a)

R b

a f (x)dx f (a )+ f (b)

2 (b a):

———————————————————————————————————————————————–432. Show that if f 2C ([a; b]) is positive and strictly concave on [a; b]; then R b

a f (x)dx > 12 (b a) max

x2[a;b ]f (x):

———————————————————————————————————————————————–433. Prove the following Opial’s inequality : If f is continuously di¤erentiable on [0; a] and f (0) = 0 ; then

R a

0 jf (x)f 0(x)jdx a2 R

a0 (f 0(x))2 dx: Show also that the constant a

2 is the best possible.———————————————————————————————————————————————–434. Let f 0 be measurable on [0; 1] and f n gbe a sequence of positive numbers such that n #0 as n ! 1 :Prove that R

10 f (x)dx < 1 if and only if lim

n !1 R 1n

f (x)dx < 1 :———————————————————————————————————————————————–435. Assume that f 2C 1([0;1 )) ; f 0 is monotonic on [0;1 ); and lim

x!1f (x) = l < 1 :

Show that for 0 < a < b;

R 1

0f (bx ) f (ax )

x dx = ( l f (0)) ln ba :

———————————————————————————————————————————————–436. (a) Find R 10

e bx e ax

x dx; 0 < a < b:(b) Find R 1

0cos( bx ) cos( ax )

x 2 dx; 0 < a < b:———————————————————————————————————————————————–

437. (a) Show that if f 2C [0; 1] and jf (x)j 1 8x 2[0; 1]; then R 1

0 p 1 f 2(x)dx r 1 R 1

0 f (x)dx2:

(b) Show that if f 2C 1[0;1 ) and limx!1

f (x) = 0 ; then R 10 f 2(x)dx 2 R 1

0 x2f 2(x)dx1=2

R 10 (f 0(x)) 2 dx

1=2:

———————————————————————————————————————————————–438. Prove the following Hölder inequality : If f 1 ; f 2 ;:::;f n are nonegative and Riemann integrable on [a; b];

29



and if 1 ; 2 ;:::; n > 0 are such thatn

Xi =1i = 1 ; then R

ba

n

Yi =1

f ii (x)!dx

n

Yi=1 R b

a f i (x)dxi

:

———————————————————————————————————————————————–439. Suppose that f and g are nonnegaive and Riemann integrable on [a; b]: For p 6= 0 ; Let q be its conjugate,i.e., 1

p + 1q = 1 : Prove that:

(a) If p > 1; then R b

a f (x)g(x)dx R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:

(b) If 0 < p < 1; and f and g are positive, then R b

a f (x)g(x)dx R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:———————————————————————————————————————————————–440. Suppose that f is continuous on [0; 1] and there is a > 0 such that 0 f (x) a2=3 8x 2[0; 1]:Show that if R

10 f (x)dx = a; then R

10 p f (x)dx a2=3 :

———————————————————————————————————————————————–441. Prove the following Minkowski inequality for integrals : If f 1 ; f 2;:::;f n are nonnegative and Riemannintegrable on [a; b]; then


a (f 1(x) + + f n (x)) p dx1=p

R b

a f p1 (x)dx1=p

+ + R b

a f pn (x)dx1=p

:

(b) If 0 < p < 1; and f i 0, then

R b

a (f 1(x) + + f n (x)) p dx1=p

R b

a f p1 (x)dx1=p

+ +

R b

a f pn (x)dx1=p

:———————————————————————————————————————————————–442. Assume that f 1 ; f 2 ;:::;f n are nonnegative and Riemann integrable on [a; b]:(a) If p > 1; then R

ba (f 1(x) + + f n (x)) p dx R

ba f p1 (x)dx + + R

ba f pn (x)dx:

(b) If 0 0; is strictly increasing on [0; c] and f (0) = 0 : Let f 1(x) denotethe inverse of f : Show that for x 2[0; c]; R

x0 f (t)dt + R

f (x )0 f 1(t)dt = xf (x):

———————————————————————————————————————————————–444. Prove the following Young inequality : Suppose that f 2C ([0; c]); c > 0; is strictly increasing on [0; c]and f (0) = 0 : Let f 1(x) denote the inverse of f :(a) Show that for any a 2[0; c] and for any b 2[0; f (c)]; we have R

a0 f (t)dt + R

b0 f 1(t)dt ab:

(b) Show that equality holds in part (a) if and only if b = f (a):———————————————————————————————————————————————–445. Show that for a; b 0; (1 + a)ln(1 + a) (1 + a) + ( eb b) ab:———————————————————————————————————————————————–446. Suppose that f is positive and Riemann integrable on [a; b]:(a) Show that (b a)2 R b

a f (x)dx R ba

1f (x ) dx :

(b) Show that if 0 < m f (x) M; then R b

a f (x)dx R b

a1

f (x ) dx (m + M ) 2

4mM (b a)2 :———————————————————————————————————————————————–

447. Given an in…nite sequence of complex numbers (cn ) ; let f (x) =1

Xn =1

cn einx : Show:

(a) If 1

Xn =1jcn j< 1 ; then f is continuous on R:

(b) If 1

Xn =1n jcn j< 1 ; then f is continuously di¤erentiable on R:

———————————————————————————————————————————————–

448. Show that the function f (t) =1

Z 1

e tx 2dx is …nite and continuous for t 1; di¤erentiable for t > 1;

and satis…es the di¤erential equation 2tf 0(t) + f (t) = 0 :———————————————————————————————————————————————–449. Let f be continuous on [0; 1]; and suppose that g is a nonnegative continuous function on R with

30



g(x + 1) = g(x): Show that limn !1

1

Z 0

f (x)g(nx )dx = 0@1

Z 0

f (x)dx1A0@1

Z 0

g(x)dx1A:

———————————————————————————————————————————————–450. Let f 2L1(R). Show that there exists a sequence xn ! 1 such that xn f (xn ) ! 0:———————————————————————————————————————————————–

451. Let g : Rd

! R be nonnegative with R R d g(y)dy = 1 ; let g (x) =

dg(x= ) for > 0; and de…ne

f (x) = ( g f ) (x) = R R d g (y)f (x y)dy: If f : Rd ! R is continuous and bounded, then show thatfor any compact subset K Rd , f

K f as ! 0:

————————————————————————————————————————————————

31



TEMPLE UNIVERSITYMATHEMATICS DEPARTMENTStudy Guide for REAL ANALYSIS

The SolutionsBy: Ziad Adwan

———————————————————————————————————————————————–1. If A and B are sets, show that A B if and only if A \ B = A :———————————————————————————————————————————————–Proof. (=) ) Suppose that A B . We want to show that A \ B = A :Let x 2A \ B : Then x 2A and x 2B : Hence, A \ B A (1) :On the other hand, if we let x 2A ; then since we are assuming that A B ; we deduce that x 2B :Hence, x 2A \ B and thus, A A \ B (2) :Now, (1) and (2) imply that A \ B = A :(( =) Suppose that A \ B = A : We want to show that A B :Let x 2A : Since we are assuming that A \ B = A ; we get x 2A \ B and consequently, x 2B :

———————————————————————————————————————————————–2. Prove the Second De Morgan Law : If A, B , and C are sets, then A n(B \ C ) = ( A nB ) [ (A nC ) :

———————————————————————————————————————————————–Proof. ( ) Let x 2A n(B \ C ) and suppose that x =2A nB : Since A = ( A nB ) [ (A \ B ) ; we get thatx 2A \ B : Now, x 2A n(B \ C ) and x 2A \ B =) x =2A \ C : Since A = ( A nC ) [ (A \ C ) ; weget that x 2A nC :( ) Note that A nB A n(B \ C ) and A nC A n(B \ C ): Thus, (A nB ) [ (A nC ) A n(B \ C ):Hence, A n(B \ C ) = ( A nB ) [ (A nC ) :

———————————————————————————————————————————————–3. For each n 2N; let A n = f(n + 1) k : k 2Ng:(a) What is A 1 \ A 2?(b) Determine the sets [fA n : n 2Ng and \ fA n : n 2Ng:———————————————————————————————————————————————–Solution. (a) A 1 = f(1 + 1) k : k 2Ng= f2k : k 2Ng= f2; 4; 6; 8; 10;:::g; andA 2 =

f(2 + 1) k : k

2N

g=

f3k : k

2N

g=

f3; 6; 9; 12; 15;:::

g:

Thus, A 1 \ A 2 = f6; 12; 18; 24;:::g= f6k : k 2Ng= A 5 :(b) [fA n : n 2Ng= Nnf1g and \ fA n : n 2Ng= ? :———————————————————————————————————————————————–4. Draw diagrams in the plane of the Cartesian products A B for the given sets A and B .(a) A = fx 2R : 1 x 2 or 3 x 4g; B = fx 2R : x = 1 or x = 2g:(b) A = f1; 2; 3g; B = fx 2R : 1 x 3g———————————————————————————————————————————————–Solution. (a)

32



(b)

———————————————————————————————————————————————–5. Let f (x) = 1 =x2; x 6= 0 ; x 2R:(a) Determine the direct image f (E ) where E = fx 2R : 1 x 2g:

(b) Determine the inverse image f 1(G ) where G = fx 2R : 1 x 4g:———————————————————————————————————————————————–

Solution. The graph of f (x) = 1 =x2 is:

Note that f is even, strictly increasing on ( 1; 0) and strictly decreasing on (0; + 1 ):

(a) Since f is strictly decreasing on [1; 2]; f ([1; 2]) = [f (2) ; f (1)] = [1 =4; 1]:(b) f 1(G ) = [ 1; 1=2][ [1=2; 1]:———————————————————————————————————————————————–6. Let g(x) = x2 and f (x) = x + 2 for x 2R; and let h be the composite function h = g f:(a) Find the direct image h(E ) of E =

fx

2R : 0 x 1

g:

(b) Determine the inverse image h 1(G ) where G = fx 2R : 0 x 4g:———————————————————————————————————————————————–Solution. h(x) = g f (x) = g(f (x)) = ( x + 2) 2 : The graph of h(x) is:

33



Note that h is strictly decreasing on ( 1; 2) and strictly increasing on ( 2; + 1 ):

(a) Since h is increasing on [0; 1]; h([0; 1]) = [h(0); h(1)] = [4 ; 9]:

(b) h 1([0; 4]) = [ 4; 0]:———————————————————————————————————————————————–7. Let f (x) = x2 for x 2R; and let E = fx 2R : 1 x 0g and F = fx 2R : 0 x 1g:Show that E \ F = f0g and f (E \ F ) = f0g; while f (E ) = f (F ) = fy 2R : 0 y 1g:Hence, f (E \ F ) is a proper subset of f (E ) \ f (F ): What happens if 0 is deleted from the sets E and F ?———————————————————————————————————————————————–Proof. E \ F = fx 2R : 1 x 0 and 0 x 1g= f0g; f (E \ F ) = f (0) = 0 2 = 0 :f (E ) = fy = f (x) : 1 x 0g= fy = x2 : 1 x 0g

f is decreasing on E= fy 2R : 0 y 1g:

Also, f (F ) = fy = f (x) : 0 x 1g= fy = x2 : 0 x 1gf is increasing on F

= fy 2R : 0 y 1g:If 0 is deleted from the sets E and F ; we get that E \ F = ? ; f (E \ F ) = ? ; andf (E ) = f (F ) = fy 2R : 0 < y 1g:

———————————————————————————————————————————————–

8. Show that if f : A ! B and E , F are subsets of A , then f (E [ F ) = f (E )[ f (F ) and f (E \ F ) f (E )\ f (F ):———————————————————————————————————————————————–Proof. Let us show that f (E [ F ) = f (E ) [ f (F ):( ) Let y 2f (E [ F ): Then there is an x 2E [ F such that y = f (x): Since x 2E [ F ; x 2E or x 2F :Hence, f (x) 2f (E ) or f (x) 2f (F ): Thus, y 2f (E ) or y 2f (F ) and so y 2f (E ) [ f (F ):( ) Since E E [ F and F E [ F ; we get that f (E ) f (E [ F ) and f (F ) f (E [ F ):Hence, f (E ) [ f (F ) f (E [ F ):

Next, let us show that f (E \ F ) f (E ) \ f (F ):Let y 2f (E \ F ): Then there is an x 2E \ F such that y = f (x): Since x 2E \ F ; x 2E and x 2F :Hence, f (x) 2f (E ) and f (x) 2f (F ): Thus, y 2f (E ) and y 2f (F ) and so y 2f (E ) \ f (F ):

———————————————————————————————————————————————–9. Show that if f : A

!B and G , H are subsets of B, then f 1(G

[H ) = f 1(G )

[f 1(H ) and

f 1(G \ H ) = f 1(G ) \ f 1(H ):———————————————————————————————————————————————–Proof. Let us show that f 1(G [ H ) = f 1(G ) [ f 1(H ):( ) Let x 2f 1(G [ H ): Then there is a y 2G [ H such that x 2f 1(y): Since y 2G [ H ; y 2G or y 2H :Hence, x 2f 1(G ) or x 2f 1(H ) and so x 2f 1(G ) [ f 1(H ):( ) Since G G [ H and H G [ H ; we get that f 1(G ) f 1(G [ H ) and f 1(H ) f 1(G [ H ):Hence, f 1(G ) [ f 1(H ) f 1(G [ H ):

Next, let us show that f 1(G \ H ) = f 1(G ) \ f 1(H ):( ) Since G \ H G and G \ H H ; we get that f 1(G \ H ) f 1(G ) and f 1(G \ H ) f 1(H ):

34



Hence, f 1(G \ H ) f 1(G ) \ f 1(H ):( ) Let x 2f 1(G ) \ f 1(H ): Then x 2f 1(G ) and x 2f 1(H ): Since f is a function, f assigns a uniqueelement to each x in its domain. Consequently, there is a y 2G \ H such that x 2f 1(y):But f 1(y) f 1(G \ H ).

———————————————————————————————————————————————–10 . Show that the function f de…ned by f (x) = x=p x2 + 1 ; x

2R; is a bijection of R onto

fy : 1 < y < 1

g:

———————————————————————————————————————————————–Proof. Here is a graph of f (x):

Note that for all x 2R; jf (x)j= x=p x2 + 1 = p x2=p x2 + 1 < 1:Hence, 1 < f (x) < 1 for all x 2R:To show that f is one to one, pick 0 < x 1 < x 2 : Then x2

1 < x 22 =) 1

x 21

> 1x 2

2=) 1 + 1

x 21

> 1 + 1x 2

2

=) x2

1 +1x 2

1> x2

2 +1x 2

2=)

x21

x 21 +1 < x2

2x 2

2 +1 =) f (x1) < f (x2): Thus, f is strictly increasing on (0; + 1 ):Since f is odd, f is also strictly increasing on ( 1; 0) and so f is strictly increasing on R:Hence, f is one to one on R:To show that f is onto, pick y 2( 1; 1): Then x = yp 1 y 2

is a well-de…ned element in R:

Moreover, f (x) = f ( y

p 1 y 2 ) = y:Therefore, f is a bijection of R onto fy : 1 < y < 1g:

———————————————————————————————————————————————–11 . (a) Show that if f : A ! B is injective and E A ; then f 1(f (E )) = E : Give an example to showthat equality need not hold if f is not injective.(b) Show that if f : A ! B is surjective and H B ; then f (f 1(H )) = H : Give an example to show thatequality need not hold if f is not surjective.———————————————————————————————————————————————–Proof. (a) ( ) Let x 2f 1(f (E )) : Then f (x) 2f (E ) and since f is injective, x 2E :( ) Let x 2E : Then f (x) 2f (E ) and so x 2f 1(f (x)) 2f 1(f (E )) :Thus, f 1(f (E )) = E :

Example for (a). Let f (x) = x2

on R and let E

= [0 ; 1]: Then f (E

) = [0 ; 1] but f 1

(f (E

)) = [ 1; 1] 6=E

:(b) ( ) This is clear.( ) Let x 2H : Since f is surjective, there exists y 2A such that f (y) = x:Now, f (y) 2H =) y 2f 1(H ) =) x 2f (f 1(H )) :Thus, f (f 1(H )) = H :

Example for (b). Let A = B = f0; 1gand de…ne f : A ! B by f (0) = f (1) = 0 : Note that f is not surjective.Then if H = B ; we get that f (f 1(H )) = f (A ) = f0g$ H :———————————————————————————————————————————————–

35



12 . (a) Suppose that f is an injection. Show that f 1 f (x) = x for all x 2D (f ) and thatf f 1(y) = y for all y 2R(f ):(b) If f is a bijection of A onto B , show that f 1 is a bijection of B onto A .———————————————————————————————————————————————–Proof. (a) This follows immediately from the previous problem with E = fxg and F = fyg:

(b) We have to prove that f 1 is both injective and surjective. f 1 is injective: Let x; y 2B : Then f 1(x) = f 1(y) =) f (f 1(x)) = f (f 1(y))

(a )=) x = y:

f 1 is surjective: Let x 2A : Then y = f (x) 2B and so, by (a), f 1(y) = f 1(f (x)) = x:Hence, f 1 is a bijection of B onto A .

———————————————————————————————————————————————–13 . Let f : A ! B and g : B ! C be functions.(a) Show that if g f is injective, then f is injective.(b) Show that if g f is surjective, then g is surjective.———————————————————————————————————————————————–Proof. (a) Let x; y 2A : Then f (x) = f (y) =) g(f (x)) = g(f (y))

g f is injective=) x = y:

Thus, f is injective.

(b) Let z 2C : Since g f is surjective, we can …nd x 2A such that g(f (x)) = z:Let y = f (x): Then g(y) = z and so g is surjective.

———————————————————————————————————————————————–14. Prove that 1=1 2 + 1 =2 3 + + 1 =n(n + 1) = n= (n + 1) for all n 2N:———————————————————————————————————————————————–Proof. We use induction. The statement is true for n = 1 since 1=1 2 = 1 =(1 + 1) = 1 =2:Suppose the statement is true for n 1; i.e., 1=1 2 + 1 =2 3 + + 1 =(n 1)n = ( n 1) =n:Then 1=1 2 + 1 =2 3 + + 1 =n(n + 1) = [1 =1 2 + 1 =2 3 + + 1 =(n 1)n] + 1=n(n + 1)= [( n 1) =n] + 1=n(n + 1) = n (n 2 1)+ n

n 2 (n +1) = n 3

n 2 (n +1) = n= (n + 1) :Thus, by the induction theorem, 1=1 2 + 1 =2 3 + + 1 =n(n + 1) = n= (n + 1) for all n 2N:

———————————————————————————————————————————————–15. Prove that 3 + 11 + + (8 n 5) = 4 n2 n for all n 2N:

———————————————————————————————————————————————–Proof. We use induction. The statement is true for n = 1 since 8(1) 5 = 4(1) 2 1 = 3 :Suppose the statement is true for n 1; i.e., 3 + 11 + + (8( n 1) 5) = 4( n 1)2 (n 1):Then 3 + 11 + + (8 n 5) = [3 + 11 + + (8( n 1) 5)] + (8 n 5)= 4(n 1)2 (n 1) + (8 n 5) = 4 n2 n:Thus, by the induction theorem, 3 + 11 + + (8 n 5) = 4 n2 n for all n 2N:

———————————————————————————————————————————————–16. Prove that 12 22 + 3 2 42 + + ( 1)n +1 n2 = ( 1)n +1 n(n + 1) =2 for all n 2N:———————————————————————————————————————————————–Proof. We use induction. The statement is true for n = 1 since ( 1)1+1 (1)2 = ( 1)1+1 1(1 + 1) =2 = 1 :Suppose the statement is true for n 1; i.e., 12 22 + 3 2 42 + + ( 1)n (n 1)2 = ( 1)n (n 1)n=2:Then 12 22 + 3 2 42 + + ( 1)n +1 n2 = 12 22 + 3 2 42 + + ( 1)n (n 1)2 + ( 1)n +1 n2

= [( 1)n (n 1)n=2]+( 1)n +1 n2 = ( 1)n +1 n2 (n 1)n=2 = ( 1)n +1 (n(n + 1) =2) = ( 1)n +1 n(n +1) =2:

Thus, by the induction theorem, 12 22 + 3 2 42 + + ( 1)n +1 n2 = ( 1)n +1 n(n + 1) =2 for all n 2N: ———————————————————————————————————————————————–17. Prove that 52n 1 is divisible by 8 for all n 2N:———————————————————————————————————————————————–Proof. We use induction. The statement is true for n = 1 since 52 1 = 24 = 8 3 is divisible by 8.Suppose the statement is true for n 1; i.e., 52( n 1) 1 is divisible by 8, say 52( n 1) 1 = 8k for some k 2N:Then 52n 1 = 5 2n 2+2 1 = 5 2( n 1)+2 1 = 25(5 2( n 1) ) 1 = 25(5 2( n 1) 1 + 1) 1 = 25(8 k + 1) 1= 25(8 k) + 24 = 8(25 k + 3) is divisible by 8.Thus, by the induction theorem, 52n 1 is divisible by 8 for all n 2N:

———————————————————————————————————————————————–

36



18. Conjecture a formula for the sum of the …rst n odd natural numbers 1 + 3 + 5 + + (2 n 1); and proveyour formula using mathematical induction.———————————————————————————————————————————————–Proof. Since 1+2+3+ +2 n = (2 n)(2n +1) =2 = n(2n +1) , and 2+4+6+ +2 n = 2(1+ 2+ 3+ + n)= 2 [n(n + 1) =2] = n(n + 1) ; we conjecture that 1 + 3 + 5 + + (2 n 1) = n(2n + 1) n(n + 1) = n2 for alln

2N: To prove this, we use induction. The statement is true for n = 1 since 2(1) 1 = 1 2 = 1 :

Suppose the statement is true for n 1; i.e., 1 + 3 + 5 + + (2( n 1) 1) = ( n 1)2

Then 1 + 3 + 5 + + (2 n 1) = [1 + 3 + 5 + + (2( n 1) 1)] + (2 n 1) = ( n 1)2 + (2 n 1)= n2 2n + 1 + (2 n 1) = n2 :Thus, by the induction theorem, 1 + 3 + 5 + + (2 n 1) = n2 for all n 2N:

———————————————————————————————————————————————–19. Prove the following Second Version of the Principle of Mathematical Induction : Let n0 2N andlet P (n) be a statement for each natural number n n 0: Suppose that(1) The statement P (n0) is true.(2) For all k n0 ; the truth of P (k) implies the truth of P (k + 1) :Then P (n) is true for all n n 0:———————————————————————————————————————————————–Proof. Let S = fn 2N : P (n0 1 + n) is trueg: Then S is a subset of N with the following properties:(1) The number 1

2S: This is because P (n0 1 + 1) = P (n0) is true.

(2) For all k 2N; if k 2S; then P (n0 1 + k) is true and so P ((n0 1 + k)+ 1) = P (n0 1 + ( k +1)) is true.Hence, k + 1 2N:Therefore, the principle of mathematical induction implies that S = N:That is, P (n0 1 + n) is true for all n 2N and so, P (n) is true for all n n 0:

———————————————————————————————————————————————–20. Prove that 2n < n ! for all n 4; n 2N:———————————————————————————————————————————————–Proof. We use induction. The statement is true for n = 4 since 24 = 16 < 24 = 4!:Suppose the statement is true for n 1; i.e., 2n 1 < (n 1)!: (Here, n 5):Then 2n = 2(2 n 1) < 2 ((n 1)!) < n ((n 1)!) = n!:Thus, by the induction theorem, 2n < n ! for all n 4; n 2N:

———————————————————————————————————————————————–

21 . Prove that a nonempty set T 1 is …nite if and only if there is a bijection from T 1 onto a …nite set T 2 :———————————————————————————————————————————————–Proof. (=) ) Suppose that T 1 is …nite. We need to show that there is a …nite set T 2 and a bijection' : T 1 ! T 2 : Take T 2 = T 1 and ' = identity. Then T 2 is …nite and ' is a bijection from T 1 onto T 2 :(( =) Suppose that there is a bijection ' from T 1 onto a …nite set T 2 : Write T 2 = fy1 ; y2 ;:::;yn g:Consider the set A = f' 1(y1); ' 1(y2);:::;' 1(yn )g T 1 : We will show that A = T 1 :For this, let x 2T 1: Then ' (x) 2T 2 and so ' (x) = yj for some 1 j n:Since ' is bijective, we get that x = ' 1(' (x)) = ' 1(yj ): Thus, T 1 A :Hence, T 1 = A and so T 1 is …nite.

———————————————————————————————————————————————–22 . (a) Prove that if A is a set with m 2N elements and C A is a set with 1 element, then A nC isa set with m 1 elements.(b) Prove that if C is an in…nite set and B is a …nite set, then C nB is an in…nite set.

———————————————————————————————————————————————–Proof. (a) Write A = fa1 ; a2 ;:::;a m g and let C A be a set with 1 element.Then C = fa j g for some 1 j m:Now, A nC = fx 2A : x =2C g= fx 2A : x 6= a j g= fa1 ; a2 ;:::;a j 1 ; a j +1 ;:::;a m g is a set with m 1 elements.

(b) Let C be an in…nite set and B be a …nite set. Write B = fb1 ; b2 ;:::;bm g:We proceed by contradiction. Suppose that C nB is a …nite set, say C nB = fc1 ; c2 ;:::;cn g:Then C = C nB [ B = fc1 ; c2 ;:::;cn ; b1; b2 ;:::;bm g is a set with n + m elements

37



This means that C is a …nite set, a contradiction!Therefore, C is an in…nite set.

———————————————————————————————————————————————–23 . Exhibit a bijection between N and a proper subset of itself.———————————————————————————————————————————————–Solution. Let A =

f2; 4; 6; 8;:::

g=

f2n : n

2Ng

N be the set of all even natural numbers.Since 1 =2A ; A is a proper subset of N.Now, de…ne ' : A ! N as follows:

' (2n) = n; for n 2N:

We claim that ' is bijective, and here is the proof:

(i) ' is injective. Let x 6= y 2A ; say x = 2 n1 and y = 2 n2 where n 1 ; n2 2N:Then 2n1 6= 2 n2 =) ' (x) = n1 6= n2 = ' (y): Thus, ' is injective.

(ii) ' is surjective. Let n 2N, and let x = 2 n: Then ' (x) = ' (2n) = n: Thus, ' is surjective.

Therefore, ' is bijective and we are done.

———————————————————————————————————————————————–24 . Prove that a set T 1 is denumerable if and only if there is a bijection from T 1 onto a denumerable set T 2 .———————————————————————————————————————————————–Proof. (=) ) Suppose that T 1 is denumerable. We need to show that there is a denumerable set T 2 anda bijection ' : T 1 ! T 2 : Take T 2 = T 1 and ' = identity. Then T 2 is denumerable and ' is a bijectionfrom T 1 onto T 2:(( =) Suppose that there is a bijection ' from T 1 onto a denumerable set T 2 : Write T 2 = fy1 ; y2 ;:::;yn ;:::g:Consider the set A = f' 1(y1); ' 1(y2);:::;' 1(yn );:::g T 1 : Then A is a denumerable subset of T 1 .We will show that A = T 1 : For this, let x 2T 1 : Then ' (x) 2T 2 and so ' (x) = yj for some j 2N.Since ' is bijective, we get that x = ' 1(' (x)) = ' 1(yj ): Thus, T 1 A :Hence, T 1 = A and so T 1 is denumerable.

————————————————————————–

Alternatively for (( =) , since T 2 is denumerable, there exists a bijection : N! T 2 and since we havea bijection ' : T 2 ! T 1 ; we get that the composition ' : N! T 1 is a bijection.Thus, T 1 is denumerable.

———————————————————————————————————————————————–25 . Give an example of a countable collection of …nite sets whose union is not …nite.———————————————————————————————————————————————–Solution . Let A 1 = f1g; A 2 = f2g; A 3 = f3g; :::; A n = fng; ::::

Then each A n is …nite. Yet,1

[n =1

A n =1

[n =1fng= f1; 2; 3;:::g= N is in…nite.

———————————————————————————————————————————————–26 . Prove in detail that if S and T are denumerable, then S [ T is denumerable.———————————————————————————————————————————————–Proof. Write S =

fs1 ; s2 ;:::;s n ;:::

g and T =

fr 1 ; r 2 ;:::;r n ;:::

g:

Let A = S \ T: We have three cases.

Case 1 T nA is in…nite. Write T nA = ft1 ; t2 ;:::;t n ;:::g; and note that S \ (T nA) = ? :Then S [ T = S [ (T nA) = fs1 ; t1 ; s2 ; t2 ;:::;s n ; tn ;:::g:

De…ne ' : N! S [ T as follows:

' (n) = s(n +1) =2 if n is oddtn= 2 if n is even :

38



Since S \ (T nA) = ? ; ' is injective, and it is pretty clear that ' is surjective. Hence, ' is bijective.

Case 2 T nA is …nite. Write T nA = ft1 ; t2 ;:::;t m g; and note that S \ (T nA) = ? :Then S [ T = S [ (T nA) = ft1 ; t2 ;:::;t m ; s1 ; s2 ;:::;s n ;:::g:

De…ne ' : N!

S

[T as follows:

' (n) = tn if n = 1 ; 2;:::;msn m if n m + 1 :

Since S \ (T nA) = ? ; ' is injective, and it is pretty clear that ' is surjective. Hence, ' is bijective.

Case 3 T nA = ? . In this case, T = S and so S [ T = S is denumerable.

———————————————————————————————————————————————–27. Use mathematical induction to prove that if a set S has n elements, then P (S ) has 2n elements.———————————————————————————————————————————————–Proof. If S has 1 element, say S = fsg; then P (S ) = f? ;fsg= S g has 21 elements.Suppose that the result is true for any set with n 1 elements, that is, if a set has n 1 elements, thenits power set has 2n 1 elements.

Let S be a set with n elements, say S = fs1 ; s2 ;:::;s n g; and let A = fs1 ; s2 ;:::;s n 1g: Then S = A [ fsn g:By the induction hypothesis, P (A) has 2n 1 elements, say P (A) = fA1 = ? ; A2 ; A3 ;:::;A2n 1 1 ; A2n 1 = Ag:Now, P (S ) = P (A) [ fA1 [ fsn g; A2 [ fsn g;:::;A2n 1 1 [ fsn g; A2n 1 [ fsn gg= P (A) [ ffsn g; A2 [ fsn g;:::;A2n 1 1 [ fsn g; S g has 2n 1 + 2 n 1 = 2 n elements.Thus, by the induction theorem, we conclude that if a set S has n elements, then P (S ) has 2n elements.

Another elementary proof . Let S = fs1 ; s2 ;:::;s n g: Then an element in P (S ) is a subset of S:We adopt the following convention: For each A 2 P (S ); we assign the indicator ( 1 ; 2 ;:::; n );

where j = 0 if s j =2A1 if s j 2A : For example, (0; 0;:::; 0) is assigned to ? ; (1; 1; :::; 1) is assigned to S;

(1; 0; 1; 0; 0; :::; 0) is assigned to fs1 ; s3g; and so on. Let I be the set of all possible indicators.It is easy to check that the function ' : P (S ) ! I is a bijection, and so P (S ) is …nite and has the same numberof elements as the set I : But I has 2 2 2 2

| {z } n times

= 2 n elements.

———————————————————————————————————————————————–28 . Prove that if a; b 2R; then(a) (a + b) = ( a) + ( b) ; (b) ( a) ( b) = a b;(c) 1=( a) = (1=a); (d) (a=b) = ( a)=b if b 6= 0 :———————————————————————————————————————————————–Proof. (a) We have (a + b) + (( a) + ( b)) = a + b + ( a) + ( b) = a + ( a) + b + ( b)= ( a + ( a)) + ( b + ( b)) = 0 + 0 = 0 : Thus, (a + b) = ( a) + ( b) :

(b) We have ( a) ( b) = ( 1 a) ( 1 b) = ( 1) a ( 1) b = (( 1) ( 1)) (a b)= 1 (a b) = a b: (Note that 1 = 1 ( 1) = ( 1) ( 1) and so ( 1) ( 1) = ( 1) = 1) :

(c) Since ( 1) ( 1) = 1 ; we get that 1 = 1( 1) : Thus, 1

( a ) = 1( 1) a = 1

( 1) 1a = 1 1

a

= (1 1a ) = ( 1a ):

(d) ab = 1 a

b = 1 a 1b = ( 1 a) 1

b = ( a) 1b = ( a )

b :

———————————————————————————————————————————————–29 . Solve the following equations, justifying each step by referring to an appropriate property or theorem.(a) 2x + 5 = 8 ; (b) x2 = 2 x;(c) x2 1 = 3 ; (d) (x 1)(x + 2) = 0 :———————————————————————————————————————————————–Solution. (a) 2x + 5 = 8 = ) 2x + 5 + ( 5) = 8 + ( 5) =) 2x + 0 = 8 5 =) 2x = 3 = ) ( 1

2 ) 2x = ( 12 ) 3

39



=) 1 x = 32 =) x = 3

2 :

(b) x2 = 2 x , x2 + ( 2x) = 2 x + ( 2x) =) x2 2x = 0 = ) x (x + ( 2)) = 0 = ) x = 0 or x = ( 2) = 2 :

(c) x2 1 = 3 =) x2 1 + ( 3) = 3 + ( 3) =) x2 4 = 0 =) (x 2) (x + 2) = 0=

)x 2 = 0 or x + 2 = 0 =

)x = 2 or x = 2:

(d) (x 1)(x + 2) = 0 = ) x 1 = 0 or x + 2 = 0 = ) x = 1 or x = 2:

———————————————————————————————————————————————–30 . If a 2R satis…es a a = a; prove that either a = 0 or a = 1 :———————————————————————————————————————————————–Proof. a a = a =) (a a) + ( a) = a + ( a) =) (a a) + ( a ( 1)) = 0 = ) a (a + ( 1)) = 0=) a (a 1) = 0 = ) a = 0 or a 1 = 0 =) a = 0 or a = 1 :

———————————————————————————————————————————————–31 . Show that there does not exist a rational number t such that t2 = 3 :———————————————————————————————————————————————–Proof. Suppose, to the contrary, that there is a rational number t such that t2 = 3 :Write t = p

q ; where p and q 6= 0 are relatively prime.

Then t2

= 3 = ) ( pq )

2= 3 = ) p

2= 3 q

2=) p

2is divisible by 3

see remark below

=) p is divisible by 3 ( )

=) p = 3 k for some k 2Z p2 =3 q2

=) 9k2 = 3 q 2 =) q 2 = 3 k2 =) q 2 is divisible by 3 =) q is divisible by 3 ( ) .Thus, ( ) and ( ) =) p and q are not relatively prime, a contradiction. Hence, t cannot be rational.

Remark . Let us show that if p is not divisible by 3, then so is p2 .Proof. p is not divisible by 3 =) p = 3k + 1 or p = 3k + 2 for some k 2Z=) p2 = 9 k2 + 6 k + 1 or p2 = 9 k2 + 12 k + 4 for some k 2Z=) p2 = 3(3 k2 + 2 k) + 1 or p2 = 3( k2 + 4 k + 1) + 1 for some k 2Z=) p2 is not divisible by 3.Thus, if p2 is divisible by 3, then so is p.

———————————————————————————————————————————————–32 . (a) Show that if x; y are rational numbers, then so are x + y and xy:

(b) Prove that if x is a rational number and y is an irrational number, then x + y is an irrational number.If, in addition, x 6= 0 ; then show that xy is an irrational number.———————————————————————————————————————————————–Proof. (a) Suppose that x = p

q and y = rs are rational numbers. Of course, q; s 2Znf0g:

Then qs 2Znf0g; pr 2Z; and ps + qr 2Z:Thus, x + y = p

q + rs = ps + qr

qs 2Q; and xy = pq

rs = pr

qs 2Q:

(b) Suppose to the contrary that x + y is a rational number. Then, since x is rational, so is x.Thus, by (a), y = ( x + y) + ( x) is a rational number, a contradiction. Thus, x + y is an irrational number.

For the second statement, suppose to the contrary that xy is a rational number.Then, since x is a nonzero rational number, so is 1

x .Thus, by (a), y = ( 1

x )(xy) is a rational number, a contradiction. Thus, xy is an irrational number.

———————————————————————————————————————————————–33 . Let K = fs + tp 2 : s; t 2Qg: Show that K satis…es the following:(a) If x1 ; x2 2K ; then x1 + x2 2K and x1x2 2K :(b) If x 6= 0 and x 2K ; then 1=x 2K :(Thus, the set K is a sub…eld of R. With the order inherited from R, the set K is an ordered …eld that liesbetween Q and R).———————————————————————————————————————————————–Proof. Before we start, keep the results of the previous problem in mind.

40



(a) Let x1 ; x2 2K ; say x1 = s1 + t1p 2; and x2 = s2 + t2p 2 for s1 ; t1 ; s2 ; t2 2Q:Then x1 + x2 = s1 + t1p 2 + s2 + t2p 2 = ( s1 + s2) + ( t1 + t2)p 2 2K since s1 + s2 ; t1 + t2 2Q:Also, x1x2 = s1 + t1p 2 s2 + t2p 2 = s1s2 + s1 t2p 2 + s2 t1p 2 + t1 t2 p 2 2

= ( s1s2 + 2 t1 t2) + ( s1 t2 + s2t1) p 2 2K since s1s2 + 2 t1 t2 ; s1 t2 + s2 t1 2Q:

(b) Let 0 6= x 2K

: Then x = s + tp

2 for s; t 2Q and s and t are not simultaneously 0. If s = 0 and t 6= 0 ; then 1x = 1

tp 2 = 12t p 2 2K since 1

2t 2Q: If s 6= 0 and t = 0 ; then 1

x = 1s 2K since 1

s 2Q: If s 6= 0 and t 6= 0 ; then 1

x = 1s + tp 2 = 1

s + tp 2s tp 2s tp 2 = s tp 2

s 2 2t 2 = ss 2 2t 2 + t

s 2 2t 2p 2:

Note that s2 2t2 6= 0 since p 2 =2Q: Also, s2 2t2 2Q and so ss 2 2t 2 and t

s 2 2t 2 2Q:

Thus, 1x = s

s 2 2t 2 + ts 2 2t 2

p 2 2K :

———————————————————————————————————————————————–34. (a) If a < b and c d; prove that a + c < b + d:(b) If 0 < a < b and 0 c d; prove that 0 ac bd:———————————————————————————————————————————————–Proof. (a) Since a 1b :Since ab > 0, we get that 1

ab > 0 (because ab ( 1ab ) = 1 > 0 and ab > 0)

Thus, 0 < a 0; we have that ac 0:Thus, 0 ac bd:

———————————————————————————————————————————————–35. If a; b 2R; show that a2 + b2 = 0 if and only if a = b = 0 :———————————————————————————————————————————————–Proof. (=) ) Suppose that a 6= 0 : Then a2 > 0 and since b2 0; we get that a2 + b2 > 0:Similarly, if b 6= 0 ; then a2 + b2 > 0: Thus, if a2 + b2 = 0 ; then a = b = 0 :

(( =) If a = b = 0 ; then a2 + b2 = 0 2 + 0 2 = 0 + 0 = 0 :

Thus, a2 + b2 = 0 if and only if a = b = 0 :

———————————————————————————————————————————————–36. If 0 a < b; show that a2 ab < b2 : Show by example that it does not follow that a2 < ab 0> a 0; and so ab a2 :

ab b2 . This follows because b2 ab = b(b a) b a> 0

> b > 0; and so ab < b2:

Example. Let a = 0 and b = 1 : Then a2 = 0 2 = 0 ; ab = 0 1 = 0 ; and b2 = 1 2 = 1 :Thus, in this case, a2 = ab:———————————————————————————————————————————————–37. Find all real numbers x that satisfy the following inequalities.(a) x2 > 3x + 4 ; (b) 1 < x 2 < 4;(c) 1=x < x; (d) 1=x < x 2:———————————————————————————————————————————————–Proof. (a) x2 > 3x + 4 () x2 3x 4 > 0 () (x 4)(x + 1) > 0

() Either (x 4) > 0 and (x + 1) > 0;or (x 4) < 0 and (x + 1) < 0

() Either x > 4 and x > 1; or x < 4 and x < 1

() Either x > 4, or x < 1

() x 2Rn[ 1; 4]:

41



(b) 1 < x 2 < 4 () x2 > 1 and x2 < 4 () x2 1 > 0 and 4 x2 > 0

() (x 1) (x + 1) > 0 and (2 x) (2 + x) > 0:

Now, (x 1) (x + 1) > 0 () Either (x 1) > 0 and (x + 1) > 0;or (x 1) < 0 and (x + 1) < 0

() Either x > 1 and x > 1; or x < 1 and x < 1

() Either x > 1, or x < 1() x 2Rn[ 1; 1]:

Also, (2 x) (2 + x) > 0 () Either (2 x) > 0 and (2 + x) > 0;or (2 x) < 0 and (2 + x) < 0

() Either x < 2 and x > 2;or x > 2 and x < 2

() x 2( 2; 2) [ ? = ( 2; 2) :

Note that (Rn[ 1; 1]) \ ( 2; 2) = ( 2; 1) [ (1; 2): Thus, 1 < x 2 < 4 () x 2( 2; 1) [ (1; 2):

(c) If x > 0; then 1x < x () x2 > 1 () x2 1 > 0

(b )

() x 2(1:1 ):

If x < 0; then 1x < x () 1 > x 2 () 1 x2 > 0

as in (b)

() x 2( 1; 0):Thus, 1

x < x

()x

2(1:

1)

[( 1; 0):

(d) Note that this is true for all x < 0 since x2 > 0:

If x > 0; then 1x < x 2 () x3 1 > 0 () (x 1)(x2 + x + 1) > 0

x2 + x +1 > 0, for x> 0

() x 1 > 0

() x > 1 () x 2(1;1 ):Thus, 1

x < x 2 () x < 0 or x > 1 () x 2Rn[0; 1]:

———————————————————————————————————————————————–38. Let a; b 2R; and suppose that for every " > 0 we have a b + ": Show that a b:———————————————————————————————————————————————–Proof. Suppose to the contrary that a > b: Then if we take "0 = 1

2 (a b) > 0; we have b 0 such that a > b + "0; a contradiction. Hence, a > b is false and so a b:

———————————————————————————————————————————————–39. (a) If 0 < c < 1; show that 0 < c 2 < c < 1:

(b) If 1 < c; show that 1 < c < c2

:———————————————————————————————————————————————–Proof. (a) Since c > 0 and c < 1; c c < 1 c and so c2 < c: Also, since c > 0; c2 > 0 c = 0 :Thus, 0 < c2 < c < 1:

(b) Since c > 1; c c > 1 c and so c2 > c: Thus, 1 < c < c 2 :

———————————————————————————————————————————————–40. Use mathematical induction to show that if a 2R and m; n 2N; then am + n = am an and (am )n = amn :———————————————————————————————————————————————–Proof. Fix m 2N:We will prove that if a 2R and n 2N; then am + n = am an and (am )n = amn :We use mathematical induction for both arguments simultaneously.For n = 1 ; am +1 = am a1 and (am )1 = am 1 :Suppose the argument is true for n 1; that is, am +( n 1) = am an 1 and (am )n 1 = am (n 1) :

Then am + n = am +( n 1) a = am an 1 a = am an ; and (am )n

= ( am )n 1

(am ) = am (n 1) am = amn :Thus, by the induction theorem am + n = am an and (am )n = amn for all n 2N:Similarly, if we …x n 2N, then we get that am + n = am an and (am )n = amn for all m 2N:Thus, for any m; n 2N; am + n = am an and (am )n = amn :

———————————————————————————————————————————————–41 . If a; b 2R; show that ja + bj= jaj+ jbj if and only if ab 0:———————————————————————————————————————————————–Proof. (=) ) If ab < 0; then either a < 0 and b > 0; or a > 0 and b < 0:If a < 0 and b > 0; we get that jaj= a and jbj= b: Thus, jaj+ jbj= a + b > b > ja + bj:If a > 0 and b < 0; we get that jaj= a and jbj= b: Thus, jaj+ jbj= a b > a > ja + bj:

42



(( =) ab 0 =) a 0 and b 0; or a 0 and b 0:If a 0 and b 0; then a + b 0 and so jaj= a; jbj= b; and ja + bj= a + b:Thus, ja + bj= a + b = jaj+ jbj:If a 0 and b 0; then a + b 0 and so jaj= a; jbj= b; and ja + bj= (a + b) = a b:Thus, ja + bj= a b = jaj+ jbj:

———————————————————————————————————————————————–42 . If x;y;z 2R and x z; show that x y z if and only if jx yj+ jy zj= jx zj:Interpret this geometrically.———————————————————————————————————————————————–Proof. (=) ) x y z =) y x 0; z y 0; z x 0 =) j x yj= y x; jy zj= z y; jz xj= z x=) j x yj+ jy zj= y x + z y = z x = jx zj:(( =) Suppose jx yj+ jy zj= jx zj: Since x z; we have that jx zj= z x:Let y < x: Then y < z; and so jx yj+ jy zj= x y + z y 6= z x = jx zj (or else, y = x):Let y > z: Then y > x; and so jx yj+ jy zj= y x + y z 6= z x = jx zj (or else, y = z):Thus, x y z:

———————————————————————————————————————————————–43 . Show that jx aj< if and only if a < x < a + :———————————————————————————————————————————————–Proof.

jx a

j< if and only if

< x a < if and only if a < x < a + :

———————————————————————————————————————————————–44 . If a < x < b and a < y < b; show that jx yj< b a: Interpret this geometrically.———————————————————————————————————————————————–Proof. Suppose without loss of generality that a < x < y < b:

Since a < x; we get that x < a: Thus, jx yj= y xy<b< b x < b a:

Geometrically, this means that if two points lie between two …xed points, then the distance betweenthe two points in the middle is less than the distance between the two …xed points.———————————————————————————————————————————————–45 . Find all x 2R that satisfy the following inequalities:(a) j4x 5j 13; (b) x2 1 3:———————————————————————————————————————————————–Solution. (a)

j4x 5

j13 =

)13 4x 5 13 =

)8 4x 18 =

)2 x 4:5

=) x 2[ 2; 4:5]:

(b) x2 1 3 =) 3 x2 1 3 =) 2 x2 4 x2 0=) x2 4 =) j xj 2 =) 2 x 2=) x 2[ 2; 2]:

———————————————————————————————————————————————–46 . Find all x 2R that satisfy both j2x 3j< 5 and jx + 1 j> 2 simultaneously.———————————————————————————————————————————————–Proof. j2x 3j< 5 and jx + 1 j> 2 =) 5 < 2x 3 < 5 and fx + 1 > 2 or x + 1 < 2g=) 2 < 2x < 8 and fx > 1 or x < 3g=) 1 < x < 4 and fx > 1 or x < 3g=) x 2( 1; 4) \ f (1;1 ) [ ( 1; 3)g=) x 2 f( 1; 4) \ (1;1 )g [ f( 1; 4) \ ( 1; 3)g=) x 2 f(1; 4)g [ f? g= (1 ; 4):

———————————————————————————————————————————————–47. Determine and sketch the set of pairs (x; y) 2R R that satisfy:

(a) jxj= jyj; (b) jxj+ jyj= 1 ;(c) jxyj= 2 ; (d) jxj jyj= 2 :———————————————————————————————————————————————–Solution. (a) jxj= jyj=) x = y or x = y =) f (x; y) 2R R : jxj= jyjg= f(x; x ) 2R Rg [ f(x; x) 2R Rg: This is sketched below:

(b) jxj+ jyj= 1 : This is sketched below:

43



(c) jxyj= 2 = ) xy = 2 or xy = 2 =) y = 2x or y = 2

x=) f (x; y) 2R R : jxyj= 2 g= f(x; 2

x ) 2R Rg [ f(x; 2x ) 2R Rg:

This is sketched below:

(d) jxj jyj= 2 : jxj= x x 0x x < 0 and jyj= y y 0

y y < 0 : Thus, jyj= y y 0y y < 0 :

Hence, jxj jyj= 8>><>>:

x y if x 0 and y 0x + y if x 0 and y < 0x y if x < 0 and y 0x + y if x 0 and y < 0

9>>=>>;:

Thus, f(x; y) 2R R : jxj jyj= 2 g= f(x; y) 2R+ R+ : x y = 2g [ f(x; y) 2R+ R : x + y = 2g[f(x; y) 2R R+ : x y = 2g [ f(x; y) 2R R : x + y = 2g:This is sketched below:

———————————————————————————————————————————————–

48 . Let > 0 and > 0; and a 2R: Show that V (a) \ V (a) and V (a) [ V (a) are neighborhoodsof a for appropriate values of :———————————————————————————————————————————————–Proof. V (a) \ V (a) = fx 2R : a < x < a + g \ fx 2R : a < x < a + g= ( a ; a+ ) \ (a ; a + ) = ( a ; a + ) where = min f ; g:Thus, V (a) \ V (a) is a neighborhood of a:V (a) [ V (a) = fx 2R : a < x < a + g [ fx 2R : a < x < a + g= ( a ; a+ ) [ (a ; a + ) = ( a ; a + ) where = max f ; g:Thus, V (a) [ V (a) is a neighborhood of a:

———————————————————————————————————————————————–49 . Show that if a; b 2R; and a 6= b; then there exists neighborhoods U of a and V of b such thatU \ V = ? :———————————————————————————————————————————————–Proof. Since a 6= b; either a 0 and so we can …nd an > 0such that 0 < < b a: Let =

2 ; and consider the neighborhoods U = V (a) of a and V = V (b)of b: Now, 0 < < b a =) 0 < + < b a =) a + < b :Since V (a) = ( a ; a + ); V (b) = ( b ; b+ ); and a + < b ; we get that(a ; a + ) \ (b ; b+ ) = ? :

———————————————————————————————————————————————–50 . Let S 1 = fx 2R : x 0g: Show that the set S 1 has lower bounds, but no upper bounds.Show that inf S 1 = 0 :———————————————————————————————————————————————–Proof. Any number y 0 is a lower bound for S 1 : An upper bound for S 1 , if it exists, must be positive.But if we add 1 to this supposedly upper bound, then we are still in S 1 : Thus, S 1 has no upper bounds.Now, 0 is a lower bound for S 1 since s 0 for all s 2S 1 : Moreover, if y is a lower bound for S 1 ; theny

0: (Note that y cannot be positive because then y0 = y y

2 is positive, y0

2S 1 ; and y0 < y ):

Thus, inf S 1 = 0 :

———————————————————————————————————————————————–51 . Let S 2 = fx 2R : x > 0g: Does S 2 have lower bounds? Does S 2 have upper bounds?Does inf S 2 exist? Does sup S 2 exist? Prove your statements.———————————————————————————————————————————————–Solution. Any number y 0 is a lower bound for S 2 : Also, S 2 cannot have upper bounds by the samereasoning as in the previous problem. As in the previous problem, inf S 2 = 0 : Since S 2 does not have anyupper bounds, sup S 2 does not exist.

———————————————————————————————————————————————–

44



52 . Let S 3 = f1=n : n 2Ng: Show that sup S 3 = 1 and inf S 3 0:———————————————————————————————————————————————–Proof. 1 2S and if we choose any v < 1; we get that 1 is an element of S for which v < 1:Thus, sup S 3 = 1 :Now, 0 < 1=n for all n 2N: hence, 0 is a lower bound for S 3 : Moreover, if t is any lower bound for S 3 ;then t 0 ( If t > 0; then we can …nd n large enough so that 1=n < t and so t is not a lower bound).Hence, inf S 3 = 0 : ———————————————————————————————————————————————–53 . Let S be a nonempty subset of R that is bounded below. Prove that inf S = supf s : s 2S g:———————————————————————————————————————————————–Proof. Since S is bounded below, u = inf S exists and v = sup( S ) exists.Thus, s v for all s 2S and so s v for all s 2S: Since u = inf S; we get that v u ( ) :Also, s u for all s 2S and so s u for all s 2S . Since v = sup( S ); we get that v u andso v u ( ) :Now, ( ) and ( ) =) u = v:

———————————————————————————————————————————————–54. If a set S R contains one of its upper bounds, show that this upper bound is the supremum of S:———————————————————————————————————————————————–

Proof. Let u 2S be an upper bound for S: Then if v is any upper bound for S; we get that s v for alls 2S: In particular, u v: Hence, u is an upper bound of S and u v for any upper bound v of S .Therefore, sup S = u:

———————————————————————————————————————————————–55. Let S R be nonempty. Show that u 2R is an upper bound of S if and only if the conditions t 2Rand t > u imply that t =2S:———————————————————————————————————————————————–Proof. (=) ) Suppose that u 2R is an upper bound of S: Then s u for all s 2S:Consequently, for any t 2R such that t > u; we have s u < t for all s 2S: Hence, t =2S:(( =) Suppose that the conditions t 2R and t > u imply that t =2S: Let s 2S: Then s u or else s =2S:That is, if s 2S; then s u: Thus, u is an upper bound of S .

———————————————————————————————————————————————–56. Let S

R be nonempty. Show that if u = sup S; then for every number n

2N the number u 1=n

is not an upper bound of S; but the number u + 1 =n is an upper bound of S:———————————————————————————————————————————————–Proof. u 1=n u 1=n: Hence, for every number n 2Nthe number u 1=n is not an upper bound of S:Now, since u = sup S; u is an upper bound for S and so s u for all s 2S: Consequently, since u < u + 1 =nfor all n 2N, we get that s u < u + 1 =n for all s 2S and for all n 2N:

———————————————————————————————————————————————–57. Let S be a bounded set in R and let S 0 be a nonempty subset of S:Show that inf S inf S 0 sup S 0 sup S:———————————————————————————————————————————————–Proof. Since S is bounded and S 0 is a nonempty subset of S; we get that S 0 is also bounded.Thus, u = inf S; v = sup S; u0 = inf S 0 ; and v0 = sup S 0 all exist.Now, s u for all s

2S and so, in particular, s u for all s

2S 0:

Hence, by de…nition of in…mum, we get that u0 u:Since u0 s v0 for all s 2S 0 ; we get that u0 v0 :Now, s v for all s 2S and so, in particular, s v for all s 2S 0 :Hence, by de…nition of supremum, we get that v0 v:Thus, u u0 v0 v:

———————————————————————————————————————————————–58. Let S R and suppose that s = sup S 2S: If u =2S; show that sup(S [ fug) = sup fs ; ug:———————————————————————————————————————————————–Proof. Let S # = S [ fug: We consider two cases:

45



Case 1 . s u: In this case, s u for all s 2S # and since u 2S # ; we get that u = sup S # :

Case 2 . s > u: In this case, s s for all s 2S # and since s 2S # ; we get that s = sup S # :

Thus, cases (1) and (2) =

)sup( S # ) = sup

fs ; u

g:

———————————————————————————————————————————————–59. Use the previous exercise and mathematical induction to show that a nonempty …nite set S Rcontains its supremum.———————————————————————————————————————————————–Proof. If S = fsg contains only one element, then sup S = s 2S:Suppose the result is true for n 1; and let S be a set which contains n elements, say S = fs1 ;:::;s n 1 ; sn g:Then S = S 0[ fsn g where S 0 = fs1 ;:::;s n 1g: By the induction hypothesis, sup S 0 2S 0 S:Since sn =2S 0; the previous problem implies that sup(S 0[ fsn g) = sup fsup S 0; sn g 2S:

———————————————————————————————————————————————–60. Show that a number u is the supremum of a nonempty set S R if and only if u satis…es the conditions:(1) s u for all s 2S;(2) if v < u; then there exists s0 2S such that v < s 0:———————————————————————————————————————————————–Proof. (=) ) Suppose that u is the supremum of a nonempty set S R: By de…nition, this means that(1’) u is an upper bound of S , and(2’) If v is any upper bound of S; then u v:Note that condition (1’) implies that s u for all s 2S which is exactly condition (1).Now, Let v < u and suppose that s v for all s 2S: Then v is another upper bound for S:Hence, by condition (2’), we get that u v; a contradiction. Hence, there must be an s0 2S such that v < s 0:Thus, condition (2) holds, too.(( =) Condition (1) implies that u is an upper bound of S:Now, let v be any upper bound of S and suppose that v < u: Then condition (2) implies that there existss0 2S such that v < s 0; contradicting the assumption that v is an upper bound of S:Hence, if v is any upper bound of S; then u v: This shows that condition (2) is satis…ed.

———————————————————————————————————————————————–

61 . If S = f1=n 1=m : n; m 2Ng; …nd inf S and sup S:———————————————————————————————————————————————–Solution. n 2N=) 0 < 1

n 1; and m 2N=) 0 < 1m 1 =) 1 1

m < 0:Thus, 0 1 < 1=n 1=m < 1 + 0 =) 1 < 1=n 1=m < 1 for any n; m 2N:Hence, 1 is a lower bound for S and 1 is an upper bound for S:We claim that inf S = 1 and sup S = 1 ; and here is the proof:Let x > 1: Then x + 1 > 0 and so, by the Archimedean property, 9 n 2N : 0 < 1

n < x + 1 :Hence, 1

n 1 < x: But 1n 1 2S (for m = 1) : Therefore, 1 = inf S:

Let x < 1: Then 1 x > 0 and so, by the Archimedean property, 9 m 2N : 0 < 1m < 1 x:

Hence, x < 1 1m : But 1 1

m 2S (for n = 1) : Therefore, 1 = sup S:

———————————————————————————————————————————————–62 . Let S R be nonempty. Suppose that a number u in R has the properties:(i) For every n 2N the number u 1=n is not an upper bound of S; and

(ii) For every n 2N the number u + 1 =n is an upper bound of S:Prove that u = sup S:———————————————————————————————————————————————–Proof. Let s 2S: Then s u + 1 =n for all n 2N; and so s u: Thus, u is an upper bound for S:Now, let v < u: We have to show that there exists s 2S : v < s:For this, note that u v > 0; and so by the Archimedean property, 9 n 2N : 0 < 1

n < u v:Hence, v < u 1

n : But u 1=n is not an upper bound of S for every n 2N=) v is not an upper bound of S:Thus, there exists s0 2S : v < s 0; and consequently, u = sup S:

———————————————————————————————————————————————–63 . Let X be a nonempty set and let f : X ! R have bounded range in R: Let a 2R: Assuming the fact that

46



if S is a nonempty subset of R that is bounded above, then sup(a + S ) = a + sup S; Show that(i) supfa + f (x) : x 2X g= a + sup ff (x) : x 2X g; and(ii) inf fa + f (x) : x 2X g= a + inf ff (x) : x 2X g:———————————————————————————————————————————————–Proof. Let S = f (X ) = ff (x) : x 2X g: Then S is a nonempty subset of R that is bounded above. Hence,(i) sup(a + f (X )) = a + sup f (X ) =

)sup

fa + f (x) : x

2X

g= a + sup

ff (x) : x

2X

g:

(ii) inf fa + f (x) : x 2X g= supf (a + f (x)) : x 2X g= [supf a + ( f (x)) : x 2X g]by (i)

= [ a + sup f f (x) : x 2X g] = a supf f (x) : x 2X g= a + inf ff (x) : x 2X g:

———————————————————————————————————————————————–64 . Let A and B be bounded nonempty subsets of R; and let A + B = fa + b : a 2A; b 2Bg: Prove that:(i) sup(A + B ) = sup( A) + sup( B ); and(ii) inf(A + B ) = inf( A) + inf( B ):———————————————————————————————————————————————–Proof. We use the fact that if S is a nonempty subset of R that is bounded, and a 2R; thensup( a + S ) = a + sup S and inf(a + S ) = a + inf S: This was proved in the previous problem.

(i) Let u = sup A and v = sup B and …x a 2A: Then sup(a + B ) = a + sup B = a + v:Now, sup(A + B ) = sup fa + v : a 2Ag= sup fA + v : a 2Ag= (sup A) + v = u + v:

(ii) Let u = inf A and v = inf B and …x a 2A: Then inf(a + B ) = a + inf B = a + v:Now, inf(A + B ) = inf fa + v : a 2Ag= inf fA + v : a 2Ag= (inf A) + v = u + v:

———————————————————————————————————————————————–65 . Let X be a nonempty set, and let f and g be de…ned on X and have bounded ranges in R: Prove that:(i) supff (x) + g(x) : x 2X g supff (x) : x 2X g+ sup fg(x) : x 2X g; and(ii) inf ff (x) : x 2X g+ inf fg(x) : x 2X g inf ff (x) + g(x) : x 2X g:Give examples to show that each of these inequalities can be either equalities or strict inequalities.———————————————————————————————————————————————–Proof. Since f and g have bounded ranges in R; u = sup f (X ); v = sup g(X ); u0 = inf f (X ); andv0 = inf g(X ) all exist in R.

(i) For all x

2X; f (x) u and g(x) v =

)f (x) + g(x) u + v and so, since this happens

8x

2X;

we deduce that sup( f (X ) + g(X )) u + v = sup f (X ) + sup g(X ):

(ii) For all x 2X; f (x) u0 and g(x) v0 =) f (x) + g(x) u0 + v0 and so, since this happens 8x 2X;we deduce that inf(f (X ) + g(X )) u0 + v0 = inf f (X ) + inf g(X ):

Examples. (a) Let X = R; and let f 1(x) = g1(x) = 0 8x 2X: Then f 1(x) + g1(x) = 0 8x 2X:Hence, supff 1(x) + g1(x) : x 2X g= 0 and supff 1(x) : x 2X g+ sup fg1(x) : x 2X g= 0 + 0 = 0 :Also, inf ff 1(x) + g1(x) : x 2X g= 0 and inf ff 1(x) : x 2X g+ inf fg1(x) : x 2X g= 0 + 0 = 0 :

(b) Let X = R; and let f 2(x) = 1 x 01 x > 0 ; and g2(x) = 1 x 0

1 x > 0 :

Then f 2(x) + g2(x) = 0 8x 2X:Hence, sup

ff 1(x) + g1(x) : x

2X

g= 0 and sup

ff 1(x) : x

2X

g+ sup

fg1(x) : x

2X

g= 1 + 1 = 2 :

Also, inf ff 1(x) + g1(x) : x 2X g= 0 and inf ff 1(x) : x 2X g+ inf fg1(x) : x 2X g= 1 + 1 = 2: ———————————————————————————————————————————————–66 . Let X = Y = fx 2R : 0 < x < 1g: De…ne h1 ; h2 : X Y ! R by

h1(x; y) = 2 x + y; and h2(x; y) = 0 if x < y1 if x y :

(a) For each x 2X; …nd f 1(x) = sup fh1(x; y) : y 2Y g and f 2(x) = sup fh2(x; y) : y 2Y g; then …ndinf ff 1(x) : x 2X g and inf ff 2(x) : x 2X g:(b) For each y 2Y; …nd g1(y) = inf fh1(x; y) : x 2X g and g2(y) = inf fh2(x; y) : x 2X g; then …nd

47



supfg1(y) : y 2Y g and supfg2(y) : y 2Y g:———————————————————————————————————————————————–Solution. (a) Let x 2X: Then f 1(x) = sup fh1(x; y) : y 2Y g= sup f2x + y : y 2Y g= sup f2x + Y g= 2 x + sup Y = 2 x + 1 : f 2(x) = sup fh2(x; y) : y 2Y g= sup f0; 1g= 1 :

Hence, inf

ff 1(x) : x

2X

g= inf

f2x + 1 : x

2(0; 1)

g= 1 ; and inf

ff 2(x) : x

2X

g= inf

f1

g= 1 :

(b) Let y 2Y: Then g1(y) = inf fh1(x; y) : x 2X g= inf f2x + y : x 2X g= inf f2X + yg= y + inf(2 X ) = y + 0 = y: g2(y) = inf fh2(x; y) : x 2X g= inf f0; 1g= 0 :

Hence, supfg1(y) : y 2Y g= sup fy : y 2(0; 1)g= 1 ; and supfg2(y) : y 2Y g= sup f0g= 0 :

———————————————————————————————————————————————–67 . Let X and Y be nonempty sets and let h : X Y ! R have bounded range in R: Let f : X ! R andg : Y ! R be de…ned by f (x) = sup fh(x; y) : y 2Y g; and g(y) = inf fh(x; y) : x 2X g: Prove that

supfg(y) : y 2Y g inf ff (x) : x 2X g:———————————————————————————————————————————————–Proof. Let x0 2X and y0 2Y: Then g(y0) = inf fh(x; y0) : x 2X g h(x; y0) for all x 2X:So, in particular, g(y0) h(x0 ; y0) sup

fh(x0 ; y) : y

2Y

g= f (x0):

Thus, g(y) f (x) for any x 2X and any y 2Y:Hence, supfg(y) : y 2Y g f (x) 8x 2X:Therefore, supfg(y) : y 2Y g inf ff (x) : x 2X g:

———————————————————————————————————————————————–68 . Show that if a > 0; then there exists a positive real number z such that z2 = a:———————————————————————————————————————————————–Proof. Let S = fs 2R : 0 s; s 2 < a g: Since 0 2S; S 6= ? : Also, S is bounded above by a + 1 ;because if t > a + 1 ; then t2 > (a + 1) 2 = a2 + 2 a + 1 > a so that t =2S:Therefore, z = sup S exists in R: Note that z > 0:We will prove that z2 = a by ruling out the other two possibilities: z2 < a and z2 > a:

Suppose that z2 < a: We will show that this assumption contradicts the fact that z = sup S by …nding

an n 2N such that z + 1 =n 2S; thus implying that z is not an upper bound for S:Note that (z + 1 =n)2 = z2 + 2zn + 1

n 2 z2 + 1n (2z + 1) :

Hence, if we can choose n so that 1n (2z + 1) < a z2 ; then we get (z + 1 =n)2 < a:

By assumption, we have a z2 > 0 so that a z 2

2z +1 > 0 and by the Archimedean property 9n 2N : 1n < a z 2

2z +1 :Thus, (z + 1 =n)2 < a and so z + 1 =n 2S:Hence, z is not an upper bound for S; a contradiction. Therefore, we cannot have z2 < a:

Suppose that z2 > a: We will show that it is impossible to …nd m 2N : z 1=m is also an upper boundfor S; Contradicting the fact that z = sup S:Note that (z 1=m)2 = z2 2z

m + 1m 2 > z 2 2z

m :Hence, if we can choose m so that 2z

m < z 2 a; then we get (z 1=m)2 > a:By assumption, we have z2 a > 0 so that z2 a

2z > 0 and by the Archimedean property 9m 2N : 1m < z2 a

2z :Thus, (z 1=m)2 > a:Now, if s 2S; then s2 < a < (z 1=m)2 and so s < z 1=m =) z 1=m is an upper bound for S:This contradicts the fact that z = sup S . Therefore, we cannot have z2 > a:

Thus, we ruled out the two possibilities z2 < a and z2 > a; and so z2 = a:

———————————————————————————————————————————————–69 . If u > 0 is any real number and x < y; show that there exists a rational number r such that x < ru < y:———————————————————————————————————————————————–Proof. x < y and u > 0 =) x

u < yu =) 9 a rational number r : x

u < r < yu :

Hence, x < ru < y:

48



———————————————————————————————————————————————–70 . If S R is nonempty, show that S is bounded if and only if there exists a bounded closed interval I such that S I:———————————————————————————————————————————————–Proof. (=) ) Suppose that S is bounded. Then u = inf S and v = sup S both exist in R:Hence, u s v for all s

2S; and so S

[u; v] = I ; which is a closed interval.

(( =) Suppose there exists a bounded closed interval I = [a; b] such that S I:Then a s b for all s 2S and so S is bounded.

———————————————————————————————————————————————–71 . If S R is a nonempty bounded set, and I S = [inf S; sup S ]; show that S I S : Moreover, if J is anyclosed bounded interval containing S; show that I S J:———————————————————————————————————————————————–Proof. The …rst part of this problem was done in the previous exercise. Now, let J = [ a; b] be any closedbounded interval such that S J: Let x 2I S : Then inf S x sup S:Since S J; we get that a s b for all s 2S: Thus, by de…nition of inf and sup, we get thata inf S and sup S b: That is, a inf S x sup S b and so x 2[a; b] = J:Hence, I S J:

———————————————————————————————————————————————–

72 . Let I n = [0; 1=n] for n 2N: Prove that 1\n =1I n = f0g:

———————————————————————————————————————————————–

Proof. Clearly, 0 21

\n =1

I n : Let 0 < x 1. By the Archimedean property, 9m 2N : 1m < x:

Thus, x =2I n for all n m: Hence, x =21

\n =1I n ; and so

1

\n =1I n = f0g:

———————————————————————————————————————————————–

73 . Let K n = ( n; 1 ) for n 2N: Prove that1

\n =1K n = ? :

———————————————————————————————————————————————–Proof. Let x > 0: By the Archimedean property,

9m

2N : x < m: Thus, x =

2K n for all n m:

Therefore, x =2 1\n =1K n : It is also clear that K n only contains positive numbers.

Hence,1

\n =1

K n = ? :

———————————————————————————————————————————————–74 . Give the two binary representations of 3

8 and 716 :

———————————————————————————————————————————————–

Solution. (i) Since 38 1

4 = 18 =

1

Xn =4

12n ; we get that

38 = 1

4 + 18 = 1

22 + 123 ; and 3

8 = 14 +

1

Xn =4

12n = 1

22 +1

Xn =4

12n :

Hence, 38 = (0 :011000 0 )2 = (0 :010111 1 )2 :

(ii) Since 716 1

4 = 316 and 3

16 18 = 1

16 =1

Xn =5

12n ; we get that

716 = 1

4 + 18 + 1

16 = 122 + 1

23 + 124 ; and 7

16 = 14 + 1

8 +1

Xn =5

12n = 1

22 + 123 +

1

Xn =5

12n :

Hence, 716 = (0 :0111000 0 )2 = (0 :0110111 1 )2 :

———————————————————————————————————————————————–75 . Show that if ak ; bk 2 f0; 1;:::; 9g and if

49



a 110 + a 2

10 2 + + a n10 n = b1

10 + b210 2 + + bm

10 m 6= 0 ;

then n = m; and ak = bk for k = 1 ; 2;:::;n:———————————————————————————————————————————————–Proof. Suppose that m < n:Then we can write b1

10 + b210 2 + + bm

10 m = b110 + b2

10 2 + + bm10 m + bm +1

10 m +1 + + bn10 n ;

where bm + j = 0 for all j = 1 ;:::;n m:Now, we are given that a 1

10 + a210 2 + + a n

10 n = b110 + b2

10 2 + + bn10 n :

Assume that there is a j 6= bj : Let j be the …rst index in f1;:::;ng so that a j 6= bj .Then a1 = b1 ; a2 = b2 ; :::; aj 1 = bj 1 ; but a j 6= bj . Say, a j < b j :Then a j

10 j + a j +110 j +1 + + a n

10 n a j10 j + 9

10 j +1 + + 910 n < bj

10 j + bj +110 j +1 + + bn

10 n for any bj + k 2 f0; :::; 9g:This contradicts our assumption that a 1

10 + a 210 2 + + a n

10 n = b110 + b2

10 2 + + bn10 n :

Hence, a j = bj for all j = 1 ; 2;:::;n:Thus, am + j = bm + j = 0 for all j = 1 ;:::;n m; and so m = n:

———————————————————————————————————————————————–76 . For any b 2R, prove that lim

n !1bn = 0 :

———————————————————————————————————————————————–

Proof. If b = 0 ; then bn = 0 for all n 2N and so limn !1

bn = 0 : Hence, assume that b 6= 0 :

Let > 0 be given. Then

jbj > 0 and so, by the Archimedean property, we can …nd N 2N large enough

so that 0 < 1N <

jbj: Thus, 1n <

jbj for all n N; and so jbjn = bn < for all n N:

Since our choice of > 0 was arbitrary, we deduce that limn !1

bn = 0 :

———————————————————————————————————————————————–77 . Use the de…nition of a limit of a sequence to establish the following limits:(a) lim

n !1 n

n 2 +1 = 0 ; (b) limn !1

2nn +1 = 2 ; (c) lim

n !13n +12n +5 = 3

2 ; (d) limn !1

n 2 12n 2 +3 = 1

2 :———————————————————————————————————————————————–Proof. (a) Let > 0 be given. Then, by the Archimedean property, we can …nd N 2N large enoughso that 0 < 1

N < : Thus, 1n < for all n N:

Note that since n 2 < n 2 + 1 we have nn 2 +1 < 1

n for all n 2N:But then n

n 2 +1 < for all n N:


nn 2 +1 = 0 :

(b) Let > 0 be given. Then 2 > 0 and so, by the Archimedean property, we can …nd N 2N large enough

so that 0 < 1N <

2 : Thus, 1

n +1 < 1n <

2 for all n N and so 2n +1 < for all n N:

Thus, 2nn +1 2 = 2

n +1 = 2n +1 < for all n N:


2nn +1 = 2 :

(c) Let > 0 be given. Then 13 > 0 and so, by the Archimedean property, we can …nd N 2N large enough

so that 0 < 1N <

13 : Thus, 1

4n +10 < 1n <


Thus, 3n +12n +5 3

2 = 134n +10 = 13

4n +10 < for all n N:


3n +1

2n +5= 3

2:

(d) Let > 0 be given. Then 5 > 0 and so, by the Archimedean property, we can …nd N 2N large enough

so that 0 < 1N <

5 : Thus, 1

4n 2 +6 < 1n <

5 for all n N and so 54n 2 +6 < for all n N:

Thus, n 2 12n 2 +3 1

2 = 54n 2 +6 = 5

4n 2 +6 < for all n N:


n 2 12n 2 +3 = 1

2 :

———————————————————————————————————————————————–78 . Show that:(a) lim

n !1 1p n +7 = 0 ; (b) lim

n !1 2nn +2 = 2 ; (c) lim

n !1p nn +1 = 0 ; (d) lim

n !1( 1) n n

n 2 +1 = 0 :

50



———————————————————————————————————————————————–Proof. (a) Let > 0 be given. Then 2 > 0 and so, by the Archimedean property, we can …nd N 2Nlarge enough so that 0 < 1

N < 2 : Thus, 1n +7 < 1

n < 2 for all n N and so 1p n +7 < for all n N:


1p n +7 = 0 :

(b) Let > 0 be given. Then 4 > 0 and so, by the Archimedean property, we can …nd N 2N large enoughso that 0 < 1

N < 4 : Thus, 1

n +2 < 1n <


Thus, 2nn +2 2 = 4

n +2 = 4n +2 < for all n N:


2nn +2 = 2 :

(c) Let > 0 be given. Then 2 > 0 and so, by the Archimedean property, we can …nd N 2N large enoughso that 0 < 1

N < 2 : Thus, n(n +1) 2 < 1

n < 2 for all n N and so p nn +1 < for all n N:


p nn +1 = 0 :

(d) Let > 0 be given. Then, by the Archimedean property, we can …nd N 2N large enough so that0 < 1

N < : Thus, ( 1) n nn 2 +1 = n

(n +1) 2 < 1n < for all n N .


( 1) n nn 2 +1 = 0 :

———————————————————————————————————————————————–79 . Let xn = 1

ln( n +1) for n 2N:(a) Use the de…nition of limit to show that lim

n !1(xn ) = 0 :

(b) Find a speci…c value of K ( ) as required in the de…nition of limit for each of (i) = 1=2; and (ii) = 1 =10:———————————————————————————————————————————————–Proof. (a) Let > 0 be given. Then, by the Archimedean property, we can …nd N 2N large enoughso that e1= 1 < N: This implies that e1= 1 < n for all n N:Hence, for all n N; we get n > e 1= 1 =) n + 1 > e 1= =) ln(n + 1) > 1= =) 1

ln( n +1) < :


1ln( n +1) = 0 :

(b) From what we did in (a), K (1=2) > e 2 1 = 6 : 39 =)

K (1=2) = 7 :Also, K (1=10) > e 10 1 = 22025 : 47 =) K (1=10) = 22026 :

———————————————————————————————————————————————–80 . Prove that lim


n !1(jxn j) = 0 : Give an example to show that the convergence

of (jxn j) need not imply the convergence of (xn ) :———————————————————————————————————————————————–Proof. (=) ) Suppose lim

n !1(xn ) = 0 and let > 0 be given. Then there exists N 2N large enough so that

jxn 0j= jxn j< for all n N:Thus, since jjxn j 0j= jxn j; and since our choice of > 0 was arbitrary, we deduce that lim

n !1(jxn j) = 0 :

(( =) Suppose limn !1

(jxn j) = 0 and let > 0 be given. Then there exists N 2N large enough so that

jjxn j 0j= jxn j< for all n N:Since our choice of > 0 was arbitrary, we deduce that lim

n

!1(xn ) = 0 :

Let xn = ( 1)n : Then (jxn j) = f1; 1; 1;:::g and so limn !1

(jxn j) = 1 ; yet (xn ) does not converge.

———————————————————————————————————————————————–81 . Prove that if lim

n !1(xn ) = x and if x > 0; then there exists M 2N such that xn > 0 for all n M:

———————————————————————————————————————————————–Proof. Since x > 0; we can …nd > 0 such that 0 < < x: Since lim

n !1(xn ) = x; we can …nd N 2N

large enough so that jxn xj< for all n N:Hence, < xn x < =) x < xn < x + for all n N:But x > 0:

51



———————————————————————————————————————————————–82 . Let b 2R satisfy 0 1 and so 1=b = 1 + k for some k > 0:Thus, nbn = n

(1 =b) n = n(1+ k )n :

Now, (1 + k)n =

n

Xi=0ni ki = 1 + nk + n2 k2 + + kn = 1 + nk + n (n 1)2 k2 + + kn :

Note that for n 2; we have n(1+ k ) n < n

(n2 )k 2 = n

n ( n 1)2 k2 = 2

k 2 (n 1) :

Let > 0 be given. Then k2

2 > 0 and so, by the Archimedean property, we can …nd N 2N large enoughso that 1

N < k2

2 : Hence, for all n N + 1 ; we have 1n 1 < k2

2 ; and so 2k 2 (n 1) < :

Thus, jnbn j= nbn = n(1+ k) n < 2

k 2 (n 1) < for all n N + 1 :Since our choice of > 0 was arbitrary, we deduce that lim

n !1(nbn ) = 0 :

———————————————————————————————————————————————–83 . For xn given by the following formulas, establish either the convergence or the divergence of the sequence X = ( xn ):(a) xn = n

n +1 ; (b) xn = ( 1) n nn +1 ; (c) xn = n 2

n +1 ; (d) xn = 2n 2 +3n 2 +1 :

———————————————————————————————————————————————–Proof. (a) We claim that lim

n !1 nn +1 = 1 : Let > 0 be given. Then by the Archimedean property,

we can …nd N 2N large enough so that 1N +1 < : Hence, 1

n +1 < for all n N:

Thus, nn +1 1 = 1

n +1 = 1n +1 < for all n N:


nn +1 = 1 :

(b) First note that jxn j= ( 1) n nn +1 = n

n +1 12 for all n 2N:

Thus, at least we know that the limit is not 0.Since (xn ) has an in…nite number of positive terms and an in…nite number of negative terms and the limitof (xn ) is not 0, we conclude that the limit does not exist and so the sequence diverges.

(c) We will show that the sequence (xn ) = n 2

n +1 is not bounded and so is not convergent.Let R > 0 be any positive real number. Then, by the Archimedean property, we can …nd n 2N largeenough so that R < n 1: Then R(n + 1) < (n 1)(n + 1) = n2 1 < n 2 :Thus, R < n 2

n +1 ; and since our choice of R > 0 was arbitrary, we conclude that the sequence (xn ) = n 2

n +1is not bounded and so is not convergent.

(d) We will show that the sequence (xn ) = 2n 2 +3n 2 +1 converges to 2.

Note that 2n 2 +3n 2 +1 = n 2 +1+ n 2 +1+1

n 2 +1 = (n 2 +1 )+ (n 2 +1 )+1n 2 +1 = 1 + 1 + 1

n 2 +1 = 2 + 1n 2 +1 :

Let > 0 be given. Then by the Archimedean property, we can …nd N 2N large enough so that 1N < :

Hence, 1n < for all n N:

Thus, 2n 2 +3n 2 +1 2 = 2 + 1

n 2 +1 2 = 1n 2 +1 < 1

n < for all n N:

Since our choice of > 0 was arbitrary, we deduce that limn !12n 2 +3n 2 +1 = 2 :

———————————————————————————————————————————————–84 . Give an example of two divergent sequences X and Y such that:(a) Their sum converges,(b) Their product converges.———————————————————————————————————————————————–Solution. (a) Let X = (( 1)n ) and Y = ( 1)n +1 : Then X and Y are both divergent sequences.

Yet, X + Y = ( 1)n + ( 1)n +1 = (0) converges to 0.

52



(b) Let X = (( 1)n ) and Y = ( 1)n +1 : Then X and Y are both divergent sequences.

Yet, X Y = ( 1)n ( 1)n +1 = ( 1) converges to 1.

———————————————————————————————————————————————–85 . Show that the following sequences are not convergent:(a) (2n ) ; (b) ( 1)n n2 :

———————————————————————————————————————————————–Proof. (a) Let R > 1 be any positive real number greater than 1. We will show that we can …nd n 2Nso that 2n > R: Choose, by the Archimedean property, n 2N large enough so that n > ln R

ln 2 :Then n ln 2 > ln R =) ln 2n > ln R =) 2n > R:Since our choice of R > 1 was arbitrary and we were able to …nd a member of the sequence that is largerthan R; we conclude that (2n ) is not bounded and so is not convergent.

(b) Let R > 1 be any positive real number greater than 1. We will show that we can …nd n 2Nso that ( 1)n n2 > R: Choose, by the Archimedean property, an even integer n 2N large enough so thatn2 > R: Then ( 1)n n2 = n2 > R: (Here, since n is even, ( 1)n = 1) :Since our choice of R > 1 was arbitrary and we were able to …nd a member of the sequence that is largerthan R; we conclude that ( 1)n n 2 is not bounded and so is not convergent.

———————————————————————————————————————————————–86 . Find the limits of the following sequences:(a) lim

n !1(2 + 1 =n)2 ; (b) lim

n !1( 1) n

n +2 ; (c) limn !1

p n 1p n +1 ; (d) limn !1

n +1n p n :

———————————————————————————————————————————————–Solution. (a) lim

n !1(2 + 1 =n)2 = lim

n !14 + 2 =n + 1 =n2 because they converge= lim

n !1(4)+ lim

n !1(2=n)+ lim

n !11=n2

= 4 + 0 + 0 = 4 :

(b) Since limn !1

1n +2 = 0 ; and 1

n +2 = ( 1) n

n +2 ; we conclude, by problem 80, that limn !1

( 1) n

n +2 = 0 :

(c) Note that p n 1p n +1 = p n +1 2p n +1 = 1 2p n +1 :

If we show that limn !1

2p n +1 = 0 ; then we will have limn !1

p n 1p n +1 = 1 :

Let > 0 be given. Then 2

4 > 0 and so, by the Archimedean property, we can …nd N 2N large enough

so that 1N +1 < 2

4 ; and so 1p n +1 < 2 for all n N:Since 1p n +1 1p n +1 for all n 2N; we get that 1p n +1 <

2 for all n N:Hence, 2p n +1 < for all n N .Since our choice of > 0 was arbitrary, we deduce that lim

n !12p n +1 = 0 :

Thus, limn !1

p n 1p n +1 = 1 :

(d) limn !1

n +1n p n = lim

n !1 1p n + 1

n p nbecause they converge

= limn !1

1p n + limn !1

1n p n = 0 + 0 = 0 :

———————————————————————————————————————————————–87 . If (bn ) is a bounded sequence and lim

n !1(an ) = 0 ; show that lim

n !1(an bn ) = 0 :

———————————————————————————————————————————————–Proof. Since (bn ) is bounded, we can …nd M > 0 : jbn j< M for all n 2N:

Let > 0 be given and proceed as follows:Since limn !1

(an ) = 0 ; we can …nd N 2N large enough so that jan j< M for all n N:

Hence, for all n N; we have jan bn j= jan j jbn j< M M = :


(an bn ) = 0 :

———————————————————————————————————————————————–88 . If 0 < a < b; determine lim

n !1a n +1 + bn +1

a n + bn :———————————————————————————————————————————————–Solution. One claims that lim

n !1a n +1 + bn +1

a n + bn = b: To show this, one proceeds as follows:

Let r = ab < 1: Then a n +1 + bn +1

a n + bn b = bn +1

bnr n +1 +1

r n +1 b = b r n +1 +1r n +1 b = b r n +1 +1

r n +1 1

53



= b r n +1 r n

r n +1 = brn r 1r n +1 0 be given. Since limn !1

r n = 0 ; we can …nd N 2N large enough so that r n 0 was arbitrary, we deduce that limn

!1

a n +1 + bn +1

a n + bn = b:

———————————————————————————————————————————————–89 . Use the squeeze Theorem to determine the limits of the following:(a) n1=n 2

; (b) (n!)1=n 2

:———————————————————————————————————————————————–Proof. (a) We have that 8n 2N; 1 n nn =) 1 n1=n 2

(nn )1=n 2

= ( n)1=n! 1 as n ! 1 :

Thus, by the squeeze Theorem, limn !1

n1=n 2= 1 :

(b) Note that n! = n(n 1) 3 2 1 n n n n n

| {z } n times

= nn for all n 2N:

Thus, 8n 2N; 1 n! nn =) 1 (n!)1=n 2

(nn )1=n 2

= ( n)1=n! 1 as n ! 1 :

Thus, by the squeeze Theorem, limn !1

(n!)1=n 2

= 1 :

———————————————————————————————————————————————–

90 . Recall the following Theorem: "Let (xn ) be a sequence of positive real numbers such thatL = limn !1

(xn +1 =xn ) exists. If L < 1; then (xn ) converges and limn !1

(xn ) = 0" :Apply this theorem to the following sequences, where a; b satisfy: 0 < a < 1; b > 1:(a) (an ) ; (b) (bn =2n ) ; (c) (n=bn ) ; (d) 23n =32n :———————————————————————————————————————————————–Proof. (a) Here xn = an ; L = lim

n !1(xn +1 =xn ) = lim

n !1an +1 =an = lim

n !1(a) = a < 1:

Thus, by the above theorem, (an ) converges and limn !1

(an ) = 0 :

(b) Here xn = bn =2n ; L = limn !1

(xn +1 =xn ) = limn !1

bn +1 =2n +1

bn =2n = limn !1

b2 = b

2 :Hence, L < 1 if and only if b < 2:Thus, by the above theorem, (bn =2n ) converges for b < 2 and lim

n !1(bn =2n ) = 0 :

(c) Here xn = n=bn

; L = limn !1 (xn +1 =xn ) = limn !1n +1 =bn +1

n=b n = limn !1n +1bn =

1b < 1:

Thus, by the above theorem, (n=bn ) converges and limn !1

(n=bn ) = 0 :

(d) Here xn = 2 3n =32n = (8 =9)n ; L = limn !1

(xn +1 =xn ) = limn !1

(8 =9) n +1

(8 =9) n = limn !1

(8=9) = 8 =9 < 1:

Thus, by the above theorem, 23n =32n converges and limn !1

23n =32n = 0 :

———————————————————————————————————————————————–91 . Let X = ( xn ) be a sequence of positive real numbers such that lim

n !1(xn +1 =xn ) = L > 1:

Show that X is not a bounded sequence and hence is not convergent.———————————————————————————————————————————————–Proof. Since L > 1; we can …nd a real number R with 1 < R < L .Since lim

n !1(xn +1 =xn ) = L; we can …nd N 2N large enough so that R < xn +1

x n< L for all n N:

Thus, in particular, xN +1x N

> R; and so xN +1 > Rx N :Also, xN +2

x N +1 > R; and so, xN +2 > Rx N +1 > R 2xN :Also, xN +3

x N +2> R; and so, xN +3 > Rx N +2 > R 3xN :

...Continuing in this fashion, we get that xN + n > R n xN for all n 2N; and so, xn > R n N xN for all n N + 1 :But R > 1 =) Rn ! 1 as n ! 1 :Hence, X = ( xn ) is not bounded and hence is not convergent.

———————————————————————————————————————————————–92 . Let (xn ) be a sequence of positive real numbers such that lim

n !1x1=n

n = L < 1:

54



Show that there exists a number r with 0 < r < 1 such that 0 < x n < r n for all su¢ciently large n 2N:Use this to show that lim

n !1(xn ) = 0 :

———————————————————————————————————————————————–Proof. Since L < 1; we can …nd a real number r such that L < r < 1:Since lim

n !1x1=n

n = L < 1; we can …nd N 2N large enough so that x1=nn < r 8n N: (Here, you can

take = r L and apply the de…nition of a limit ):Hence, 8n N; 0 < x 1=nn < r =) 0 < x n < r n :

Since r < 1; limn !1

r n = 0 ; and hence, by the squeeze theorem, limn !1

(xn ) = 0 :

———————————————————————————————————————————————–93 . Suppose that (xn ) is a convergent sequence and (yn ) is such that for any > 0 there exists M suchthat jxn yn j< for all n M: Does it follow that (yn ) is convergent?———————————————————————————————————————————————–Proof. Yes, and we will actually show that if lim

n !1(xn ) = x; then lim

n !1(yn ) = x:

Let > 0 be given. Then: There exists M such that jxn yn j<

2 for all n M (by hypothesis) There exists N such that jxn xj<

2 for all n N (by de…nition of limit).Let K = max fM; N g:Then,

jyn x

j=

jyn xn + xn x

j jyn xn

j+

jxn x

j<

2 + 2 = for all n K:


(yn ) = x:

———————————————————————————————————————————————–94 . Let x1 = 8 and xn +1 = 1

2 xn + 2 for n 2N. Show that (xn ) is bounded and monotone. Find limn !1

(xn ) :———————————————————————————————————————————————–Proof. Direct calculation shows that x2 = 6 : Hence, we have 4 < x 2 < x 1:We show, by induction, that 4 < x n for all n 2N: It is true for n = 1 ; 2:If xk > 4 holds for some k 2N; then xk+1 = 1

2 xk + 2 > 12 (4) + 2 = 4 :

Therefore, by the induction theorem, xn > 4 for all n 2N:We now show, by induction, that xn +1 < x n for all n 2N:The truth of this assertion has been veri…ed for n = 1 :Now suppose that xk+1 < x k for some k; then xk+2 = 1

2 xk+1 + 2 < 12 xk + 2 = xk+1 :

Thus, xk+1 < x k implies xk+2 < x k +1 : Therefore, xn +1 < x n for all n 2N:We have thus shown that the sequence (x

n) is decreasing and bounded below by 4.

It follows from the Monotone Convergence Theorem that (xn ) converges to a limit that is at least 4:Let L = lim

n !1xn : Then L = lim

n !1xn +1 :

Now, xn +1 = 12 xn + 2 =) lim

n !1xn +1 = 1

2 limn !1

xn + 2 = ) L = 12 L + 2 = ) L = 4 :

Therefore, limn !1

xn = 4 :

———————————————————————————————————————————————–95 . Let x1 > 1 and xn +1 = 2 1=xn for n 2N. Show that (xn ) is bounded and monotone. Find lim

n !1(xn ) :

———————————————————————————————————————————————–Proof. Since x1 > 1; 1=x1 < 1 and so x2 = 2 1=x1 > 1:We show, by induction, that xn > 1 for all n 2N: It is true for n = 1 ; 2:If xk > 1 holds for some k 2N; then xk+1 = 2 1=xk > 2 1 = 1 :Therefore, by the induction theorem, xn > 1 for all n 2N:

We now show, by induction, that xn +1 < x n for all n 2N:For n = 1 ; we have 0 < (x1 1)2 = x21 2x1 + 1 =) 2x1 < x 2

1 + 1 =) 2 < x 1 + 1 =x1=) 2 1=x1 < x 1 : But x2 = 2 1=x1 : Hence, x2 < x 1 :Now suppose that xk+1 < x k for some k: Then 1=xk +1 > 1=xk =) 1=xk+1 < 1=xk=) 2 1=xk +1 < 2 1=xk : Therefore, xk +2 = 2 1=xk+1 < 2 1=xk = xk+1 :Thus, xk+1 < x k implies xk+2 < x k +1 : Therefore, xn +1 < x n for all n 2N:We have thus shown that the sequence (xn ) is decreasing and bounded below by 1.It follows from the Monotone Convergence Theorem that (xn ) converges to a limit that is at least 1:Let L = lim

n !1xn : Then L = lim

n !1xn +1 :

55



Now, xn +1 = 2 1=xn =) limn !1

xn +1 = 2 1

limn !1

x n

=) L = 2 1L =) (L 1)2 = 0 = ) L = 1 :

Therefore, limn !1

xn = 1 :

———————————————————————————————————————————————–96 . Let x1 2 and xn +1 = 1 + p xn 1 for n 2N. Show that (xn ) is decreasing and bounded below by 2.Find lim

n!1

(xn ) :———————————————————————————————————————————————–Proof. Since x1 2; x1 1 1 and so x2 = 1 + p x1 1 2:We show, by induction, that xn 2 for all n 2N: It is true for n = 1 ; 2:If xk 2 holds for some k 2N; then xk+1 = 1 + p xk 1 > 1 + 1 = 2 :Therefore, by the induction theorem, xn 2 for all n 2N:We now show, by induction, that xn +1 xn for all n 2N:For n = 1 ; we have x1 1 p x1 1 =) x1 1 + p xn 1 = x2 :Now suppose that xk+1 xk for some k: Then xk +1 1 xk 1 =) p xk+1 + 1 p xk + 1=) 1 + p xk+1 + 1 1 + p xk + 1 : But 1 + p xk +1 + 1 = xk+2 and 1 + p xk + 1 = xk +1 :Thus, xk+1 xk implies xk+2 xk +1 : Therefore, xn +1 xn for all n 2N:We have thus shown that the sequence (xn ) is decreasing and bounded below by 2.It follows from the Monotone Convergence Theorem that (xn ) converges to a limit that is at least 2:

Let L = limn !1 xn : Then L = limn !1 xn +1 :Now, xn +1 = 1 + p xn 1 =) lim

n !1xn +1 = 1 + q lim

n !1xn 1 =) L = 1 + p L 1 =) (L 1)2 = L 1

=) L = 2 : Therefore, limn !1

xn = 2 :

———————————————————————————————————————————————–97 . Let y1 = p p; where p > 0; and yn +1 = p p + yn for n 2N. Show that (yn ) converges and …nd lim

n !1(yn ) :

(Hint : One upper bound is 1 + 2p p):———————————————————————————————————————————————–Proof. Let us show that 1 + 2p p is an upper bound for (yn ) : We use induction.For n = 1 ; y1 = p p < 1 + 2p p:Suppose that the result is true for n; that is, yn 1 + 2p p:

Then yn +1 = p p + yn

p p + 1 + 2 p p =

q p p + 1 2 = p p + 1 < 2p p + 1 :

Thus, by the induction theorem, we deduce that yn

2

p p + 1 for all n

2N:

Now, we prove by induction that the sequence (yn ) is increasing.Note that y2 = p p + y1 = p p + p p > p p = y1 :Also, yn > y n 1 =) yn +1 = p p + yn > p p + yn 1 = yn :Thus, by the induction theorem, yn < y n +1 for all n 2N:Hence, by the Monotone Convergence Theorem, the sequence (yn ) converges. Say, lim

n !1(yn ) = L:

Then L = p p + L =) L2 = p + L =) L2 L p = 0 = ) L = 1 p 1+4 p2

Since L> 0=) L = 1+ p 1+4 p2 :

———————————————————————————————————————————————–98 . Let (an ) be an increasing sequence and (bn ) be a decreasing sequence, and assume that an bn 8n 2N:Show that lim

n !1(an ) lim

n !1(bn ) ; and thereby deduce the Nested Intervals Property from the Monotone

Convergence Theorem.———————————————————————————————————————————————–

Proof. Since (an ) is increasing, we have a1 a2 a3 an :Since (bn ) is decreasing, we have b1 b2 b3 bn :Now, by above and since an bn 8n 2N; we have that a1 an bn b1 8n 2N:Thus, both sequences are bounded and since they are monotone, they both converge.Let L1 = lim

n !1(an ) ; and L2 = lim

n !1(bn ) : Suppose that L2 < L 1 : Then 9 > 0 : L2 + < L1 :

Since L1 = limn !1

(an ) ; 9N 1 2N large enough so that L2 + < an L1 for all n N 1 ( ) :

Since L2 = limn !1

(bn ) ; 9N 2 2N large enough so that bn L2 + for all n N 2 ( ) :Combining ( ) and ( ), one gets: bn < a n for all n maxfN 1 ; N 2g; a contradiction.Hence, L1 L2 and one is done.

56



———————————————————————————————————————————————–99 . Let A be an in…nite subset of R that is bounded above and let u = sup A : Show that there existsan increasing sequence (xn ) with xn 2A for all n 2N such that u = lim

n !1(xn ) :

———————————————————————————————————————————————–Proof. If u 2A ; then we can take the sequence (xn ) = fu;u;u;::: g: Thus, we may assume that u =2A :Since u = sup A ; we can …nd an element x1

2A : (u 1; u):

Since u = sup A ; we can …nd an element x2 2A : x2 > x 1 and x2 2(u 1=2; u):Since u = sup A ; we can …nd an element x3 2A : x3 > x 2 and x3 2(u 1=3; u):...Since u = sup A ; we can …nd an element xn 2A : xn > x n 1 and xn 2(u 1=n;u):...Note that xn is an increasing sequence of points in A by our construction, and that lim

n !1(xn ) = u:

———————————————————————————————————————————————–100 . Let (xn ) be a bounded sequence, and for each n 2N let sn = sup fxk : k ng and tn = inf fxk : k ng:Prove that (sn ) and (tn ) are monotone and convergent. Also, prove that if lim

n !1(sn ) = lim

n !1(tn ) ; then

(xn ) is convergent.———————————————————————————————————————————————–

Proof. First note that (sn ) is decreasing since the sup is taken over a "smaller" set as n increases.Similarly, (tn ) is increasing. Since (xn ) is bounded, we can …nd a; b 2R : a xn b for all n 2N:Thus, by de…nition of sup and inf ; we get that a tn xn sn b for all n 2N:Hence, the sequences (sn ) and (tn ) are monotone and bounded and so are convergent.Now, if lim

n !1(sn ) = lim

n !1(tn ) = L; then by the squeeze theorem, lim

n !1(xn ) = L:

———————————————————————————————————————————————–101 . Establish the convergence or divergence of the sequence (yn ) ; where

yn = 1n +1 + 1

n +2 + + 12n for n 2N:

———————————————————————————————————————————————–Proof. We …rst note that yn = 1

n +1 + 1n +2 + + 1

2n < 1n +1 + 1

n +1 + + 1n +1 = n

n +1 < 1 for all n 2N:Thus, the sequence (yn ) is bounded above. We will show that (yn ) is also increasing, thus convergent.To do so, we have to prove that yn

yn +1 for all n

2N:

y1 = 12 ; and y2 = 13 + 14 = 712 > 12 = y1 :Note that yn +1 = 1

n +2 + 1n +3 + + 1

2n +2 = h 1n +2 + 1

n +3 + + 12n i+ 1

2n +1 + 12n +2 = hyn

1n +1 i+ 1

2n +1 + 12n +2 :

But note that 12n +1 + 1

2n +2 1n +1 = 1

2n +1 12n +2 = 1

(2 n +1)(2 n +2) > 0:Hence yn +1 > y n for all n 2N, and one is done.

———————————————————————————————————————————————–102 . Let xn = 1 =12 + 1 =22 + + 1 =n2 for each n 2N: Prove that (xn ) is increasing and bounded, and henceconverges. [Hint : Note that if k 2; then 1=k2 1=k(k 1) = 1 =(k 1) 1=k:]:———————————————————————————————————————————————–Proof. Since xn +1 = xn + 1

(n +1) 2 ; It is clear that (xn ) is increasing.Now we show that (xn ) is bounded above.Indeed, xn = 1 =12 + 1 =22 + 1 =32 + + 1 =(n 1)2 + 1 =n2 = 1 + 1=22 + 1=32 + + 1=(n 1)2 + 1=n2

by the Hint

1 +11

12 +

12

13 + + h

1n 2

1n 1

i+ h 1n 1

1n

i= 1 + 1 1n = 2

1n 2 for all n 2N:

Thus, the sequence (xn ) is increasing and bounded, and hence converges.

———————————————————————————————————————————————–103 . Establish the convergence and …nd the limits of the following sequences:(a) (1 + 1 =n)n +1 ; (b) (1 + 1 =n)2n ; (c) (1 + 1

n +1 )n ; (d) ((1 1=n)n ) :———————————————————————————————————————————————–Proof. (a) (1 + 1 =n)n +1 = ((1 + 1 =n)n ) ((1 + 1 =n)) = X Y:Note that lim X = e and lim Y = 1 : Thus, lim X Y = (lim X ) (lim Y ) = e 1 = e:

57



(b) (1 + 1 =n)2n = ((1 + 1 =n)n ) ((1 + 1 =n)n ) = X X:Note that lim X = e. Thus, lim X X = (lim X ) (lim X ) = e e = e2 :

(c) (1 + 1n +1 )n = (1 + 1

n +1 )n +1 (1 + 1n +1 ) 1 = X Y:

Note that lim X = e and lim Y = 1 : Thus, lim X Y = (lim X ) (lim Y ) = e 1 = e:

(d) Note that (1 1=n)n = ( n 1n )n = (1 = n

n 1 )n = 1 = 1 + 1n 1

n :

Hence, ((1 1=n)n ) = 1= 1 + 1n 1

n= 1= 1 + 1

n 1

n 11= 1 + 1

n 1 = X Y:

Note that lim X = 1=e and lim Y = 1 : Thus, lim X Y = (lim X ) (lim Y ) = (1 =e) 1 = 1=e:

———————————————————————————————————————————————–104 . Calculate p 2; correct to within 4 decimals.———————————————————————————————————————————————–Solution. Recall that the sequence (sn ) that converges to p 2 was constructed as follows:Let s1 > 0 and set sn +1 = 1

2 sn + 2s n

for n 2N:

Recall the inequality 0 sn p 2 sn 2s n

= s2n 2s n

for all n 2:

Now, sn +1 = 12 sn + 2

sn

= s2n +22s

n

1 for all n 1: (since (sn 1)2 + 1 1)Hence, s2 1 for all n 2 and so 1

s n1 for all n 2:

Thus, 0 sn p 2 s2n 2s n

s2n 2 for all n 2:

To make sn p 2 < 0:00001; it then su¢ces, by the above inequality, to make s2n 2 < 0:00001:

That is, we have to …nd n 2N : s2n < 2:00001:

Let s1 = 1 : Then s2 = 12 1 + 2

1 = 32 ; s3 = 1

232 + 2

32

= 1712 ; s4 = 1

21712 + 2

1712

= 577408 :

Note that s24 = 577

4082 2: 000006007 < 2:00001:

Thus, p 2 s4 = 577408 = 1 : 414215686 is correct to within 4 decimals. (Actually, 5 decimals).

On your calculator, p 2 = 1 : 414213562 :

———————————————————————————————————————————————–105 . Give an example of an unbounded sequence that has a convergent subsequence.———————————————————————————————————————————————–Solution. Let xn = 0 if n is even

n if n is odd :

Then (xn ) is clearly unbounded, and the subsequence (x2n ) converges to 0.

———————————————————————————————————————————————–106 . Show that the following sequences are divergent.(a) (1 ( 1)n + 1 =n) ; (b) (sin( n =4)) :———————————————————————————————————————————————–Proof. (a) Let xn = 1 ( 1)n + 1 =n:Then for n k = 2k; we have xn k = 1 ( 1)2k + 1 =2k = 1 1 + 1 =2k = 1 =2k for all k 1:Thus, lim

k!1(xn k ) = lim

k!1(1=2k) = 0 :

Now, for n k = 2k + 1 ; we have xn k = 1 ( 1)2k+1 + 1 =2k = 1 + 1 + 1 =2k = 2 + 1 =2k for all k 1:Thus, lim

k

!1(xn k ) = lim

k

!1(2 + 1 =2k) = 2 :

Hence, we found two subsequences with di¤erent limits, and so (xn ) is divergent.

(b) Let yn = sin( n =4):Then for n k = 4k; we have yn k = sin(4 k =4) = sin( k ) = 0 for all k 1:Thus, lim

k!1(yn k ) = lim

k!1(0) = 0 :

Now, for n k = 8k + 6 ; we have yn k = sin((8 k + 6) =4) = sin(2 k + 32 ) = 1 for all k 1:

Thus, limk!1

(yn k ) = limk!1

( 1) = 1:Hence, we found two subsequences with di¤erent limits, and so (yn ) is divergent.

———————————————————————————————————————————————–

58



107 . Let X = ( xn ) and Y = ( yn ) be given sequences, and let the "shu ed" sequence Z = ( zn ) be de…nedby: z1 = x1 ; z2 = y1 ; ; z2n 1 = xn ; z2n = yn ; : Show that Z is convergent if and only if both X and Y are convergent and lim X = lim Y:———————————————————————————————————————————————–Proof. (=) ) Suppose that Z is convergent. Say lim Z = L; and let > 0 be given.Then

9N

2N large enough so that

jzn L

j< for all n N:

In particular, jxn Lj< for all n N and jyn Lj< for all n N:Thus, both X and Y are convergent and lim X = lim Y = L = lim Z:

(( =) Suppose that both X and Y are convergent and lim X = lim Y = L: Let > 0 be given.Then 9N 1 ; N 2 2N : jxn Lj< for all n N 1 and jyn Lj< for all n N 2 :Let N = max fN 1 ; N 2g: Then jzn Lj< for all n 2N + 2 :Thus, Z is convergent and lim Z = L:

———————————————————————————————————————————————–108 . Establish the convergence and …nd the limits of the following sequences:(a) (1 + 1 =n2)n 2

; (b) ((1 + 1 =2n)n ) ; (c) (1 + 1 =n2)2n 2; (d) ((1 + 2 =n)n ) :

———————————————————————————————————————————————–Proof. (a) (1 + 1 =n2)n 2

is a subsequence of the sequence ((1 + 1 =n)n ) :

Since limn !1 ((1 + 1 =n)n ) = e; we conclude that limn !1 (1 + 1 =n2)n2

= e:

(b) [(1 + 1=2n)n ]2 = (1 + 1 =2n)2n e for all n 2N:Thus, ((1 + 1 =2n)n ) is increasing and bounded above by p e: So it converges. Say, L = lim

n !1((1 + 1 =2n)n ) :

Now, as in (a), limn !1

(1 + 1 =2n)2n = e since (1 + 1 =2n)2n is a subsequence of the sequence ((1 + 1 =n)n ) :

Thus, (1 + 1 =2n)n = (1+1 =2n ) 2 n

(1+1 =2n ) n =) L = eL =) L2 = e =) L = p e:

But (1 + 1 =2n)n 1 > 0 for all n 2N=) L = p e:

(c) Let X = (1 + 1 =n2)2n 2: Then (1 + 1 =n2)2n 2

= (1 + 1 =n2)n 2(1 + 1 =n2)n 2

= X X:

By (a), limn !1

X = e; and so limn !1

(1 + 1 =n2)2n 2= e e = e2 :

(d) ((1 + 2 =n)n ) is increasing and bounded above, so it converges.Moreover, (1 + 2 =2n)2n = (1 + 1 =n)2n is a subsequence of ((1 + 2 =n)n ) that converges to e2 :Hence, lim

n !1((1 + 2 =n)n ) = e2 :

———————————————————————————————————————————————–109 . Determine the limits of the following:(a) (3n)1=2n ; (b) (1 + 1 =2n)3n :———————————————————————————————————————————————–Proof. (a) (3n)1=2n = p 31=n

n1=2n = X Y:

Since 1 p 31=nn1=n for all n 2; the squeeze theorem implies that lim X = 1 :

Since 1 n1=2n n 1=n for all n 1; the squeeze theorem implies that lim Y = 1 :Hence, lim

n !1(3n)1=2n = (lim X ) (lim Y ) = 1 1 = 1 :

(b) (1 + 1 =2n)3n = (1 + 1 =2n)2n ((1 + 1 =2n)n ) = X Y:As in the previous exercise, lim X = e; and lim Y = p e:Thus, lim

n !1(1 + 1 =2n)3n = (lim X ) (lim Y ) = e p e = e3=2 :

———————————————————————————————————————————————–110 . Suppose that every subsequence of X = ( xn ) has a subsequence that converges to 0. Show thatlim X = 0 :———————————————————————————————————————————————–Proof. We proceed by contradiction. Suppose that lim X 6= 0 : Then we can …nd a subsequence (xn k ) of X such that, for some r > 0 , jxn k j r > 0 for all k 1: Thus, any subsequence xn k m of (xn k ) also satis…es

59



xn k m r > 0 for all m 1: Hence, limm !1

xn k m 6= 0 :Therefore, we have constructed a subsequence of X with no subsequence that converges to 0.This contradicts our assumption that every subsequence of X = ( xn ) has a subsequence that converges to 0.Hence, lim X = 0 :

———————————————————————————————————————————————–111 . Show that if (xn ) is unbounded, then there exists a subsequence (xn

k) such that lim

k!1(1=xn

k) = 0 :

———————————————————————————————————————————————–Proof. Since (xn ) is unbounded, we can …nd xn 1 2(xn ) : jxn 1 j> 1; xn 2 2(xn ) : jxn 2 j> maxfjxn 1 j; 2g;xn 3 2(xn ) : jxn 3 j> maxfjxn 2 j; 3g; ; xn k 2(xn ) : jxn k j> maxf xn k 1 ; kg; :Thus, 0 < 1

x n k< 1

k for all k = 1 ; 2; 3;:::; and so by the squeeze theorem, we get that limk!1

(1=xn k ) = 0 :

———————————————————————————————————————————————–112 . Let (xn ) be a bounded sequence and let s = sup fxn : n 2Ng: Show that if s =2 fxn : n 2Ng; thenthere is a subsequence of (xn ) that converges to s:———————————————————————————————————————————————–Proof. Since s = sup fxn : n 2Ng; we can …nd an element xn 1 : xn 1 2(s 1; s):Since s = sup fxn : n 2Ng; we can …nd an element xn 2 : n2 > n 1 ; xn 2 > x n 1 and xn 2 2(s 1=2; s):Since s = sup fxn : n 2Ng; we can …nd an element xn 3 : n3 > n 2 ; xn 3 > x n 2 and xn 3 2(s 1=3; s):...Since s = sup fxn : n 2Ng; we can …nd an element xn k : nk > n k 1 ; xn k > x n k 1 and xn k 2(s 1=k;s):...Note that xn k is an increasing sequence of points by our construction, and that lim

k!1(xn k ) = s:

———————————————————————————————————————————————–113 . Let (I n ) be a nested sequence of closed bounded intervals. For each n 2N; let xn 2I n : Use the Bolzano-Weierstrass Theorem to show that there exists a number 2R such that 2I n 8n 2N:[That is, Use the Bolzano-Weierstrass Theorem to prove the Nested Intervals Property].———————————————————————————————————————————————–Proof. Since I 1 I 2 I n ; we get that xn 2I 1 for all n 2N:Hence, the sequence (xn ) is a bounded sequence, and so, by the Bolzano-Weierstrass Theorem, there isa subsequence (xn k ) of (xn ) such that (xn k ) converges. Say, lim

k!1(xn k ) = :

We claim that 2I n for all n 2N, and here is the proof:Suppose, by contradiction, that there is N 2N : =2I N :Then, since the sequence of intervals is nested, we get that =2I n for all n N:Say, I N = [ aN ; bN ]: Since =2I N ; we can …nd > 0 small enough so that ( ; + ) \ I N = ? :Again, since the sequence of intervals is nested, ( ; + ) \ I n = ? for all n N ( ) :Since lim

k!1(xn k ) = ; we can …nd K 2N large enough so that jxn k j< for all k K:

That is, xn k 2( ; + ) for all k K: But xn k 2I n k for all k 2N; a contradiction to ( ).Hence, our claim is proved, and so we are done.

———————————————————————————————————————————————–114 . Show directly from the de…nition that the following are Cauchy sequences.(a) n +1

n ; (b) 1 + 12! + + 1

n ! :———————————————————————————————————————————————–

Proof. (a) Let > 0 be given, and let N 2N be such that 1N <

2 : Then for all n > m N; we have:jxn xm j= n +1

n m +1m = 1

n 1m

1n + 1

m 1N +

1N =

2N < :

Thus, n +1n is a Cauchy sequence.

(b) Let > 0 be given, and let N 2N be such that 1N <

2 : Then for all n > m N; we have:

jxn xm j= 1 + 12! + + 1

n ! 1 + 12! + + 1

m ! = 1(m +1)! + 1

(m +2)! + + 1n !

1m (m +1) + 1

(m +1)( m +2) + + 1(n 1) n = 1

m 1m +1 + 1

m +1 1m +2 + + 1

n 1 1n = 1

m 1n

1m + 1

n < 1N +

1N = 2

N < :Thus, 1 + 1

2! + + 1n ! is a Cauchy sequence.

60



———————————————————————————————————————————————–115 . Show directly from the de…nition that the following are not Cauchy sequences.(a) (( 1)n ) ; (b) n + ( 1) n

n ; (c) (ln n)———————————————————————————————————————————————–Proof. (a) Let 0 = 1 ; and N 2N be any positive integer. We will show that we can always …nd n; m N such that

j( 1)n ( 1)m

j> 1: Indeed, let n N be any even number, and let m = n + 1 :

Then j( 1)n ( 1)m j= j1 ( 1)j= 2 > 1 = 0 :Thus, (( 1)n ) is not a Cauchy sequence.

(b) Let 0 = 1 =2; and N 2N be any positive integer. We will show that we can always …nd n; m N such that n + ( 1) n

n m + ( 1) m

m > 1=2: Indeed, let n N be any even number, with 1n < 1=4;

and let m = n + 1 :Then n + ( 1) n

n m + ( 1) m

m = n + ( 1) n

n n + 1 + ( 1) n +1

n +1 = n + 1n n + 1 1

n +1

= 1n 1 + 1

n +1 = 1 + 1n (n +1) 1 1

n (n +1) > 1 1n > 1 1=4 = 3=4 > 1=2 = 0 :

Thus, n + ( 1) n

n is not a Cauchy sequence.

(b) Let 0 = 1 ; and N

2N be any positive integer. We will show that we can always …nd n; m N

such that jln n ln mj> 1: Indeed, let m = N , and choose n = 4 N:Hence, jln n ln mj= jln 4N ln N j= ln 4N

N = ln 4 > 1 = 0 :Thus, (ln n) is not a Cauchy sequence.

———————————————————————————————————————————————–116 . Show directly from the de…nition that if (xn ) and (yn ) are Cauchy sequences, then (xn + yn ) and (xn yn )are Cauchy sequences.———————————————————————————————————————————————–Proof. First, we show that (xn + yn ) is a Cauchy sequence. Let > 0 be given.Then since (xn ) and (yn ) are Cauchy sequences, we can …nd N 2N large enough so that whenever n; m N;we have jxn xm j< =2 and jyn ym j< =2:Thus, for all n; m N; we have:

j(xn + yn ) (xm + ym )j= j(xn xm ) + ( yn ym )j jxn xm j+ jyn ym j< =2 + =2 = :

Thus, (xn + yn ) is a Cauchy sequence.

Next, we show that (xn yn ) is a Cauchy sequence.Before we do this, let us show that the sequences (xn ) and (yn ) are bounded.Let = 1 . Then since (xn ) and (yn ) are Cauchy sequences, we can …nd N 2N large enough so that whenevern; m N; we have jxn xm j< 1 and jyn ym j< 1:In particular, jxn xN j< 1 and jyn yN j< 1 for all n N:Thus, jxn j< 1 + jxN j and jyn j< 1 + jyN j for all n N:Let M 1 = max fjx1j; jx2j; ; jxN 1j; 1 + jxN jg and M 2 = max fjy1j; jy2j; ; jyN 1j; 1 + jyN jg:Then jxn j M 1 and jyn j M 2 for all n 2N:Thus, the sequences (xn ) and (yn ) are bounded.Now, let > 0 be given. Then since (xn ) and (yn ) are Cauchy sequences, we can …nd N 2N large enoughso that whenever n; m N; we have jxn xm j<

2M 2 and jyn ym j< 2M 1 :

Hence, for all n; m N; jxn yn xm ym j= jxn yn xn ym + xn ym xm ym j= jxn (yn ym ) + ym (xn xm )jjxn j jyn ym j+ jym j jxn xm j M 1 jyn ym j+ M 2 jxn xm j< M 1 2M 1 + M 2 2M 2 = 2 +

2 = :Thus, (xn yn ) is a Cauchy sequence.

———————————————————————————————————————————————–117 . If xn = p n; show that (xn ) satis…es lim

n !1 jxn +1 xn j= 0 ; but that it is not a Cauchy sequence.———————————————————————————————————————————————–Proof. lim

n !1 jxn +1 xn j= limn !1

p n + 1 p n = limn !1

p n + 1 p n p n +1+ p np n +1+ p n= lim

n !11p n +1+ p n = 0 :

Now, let 0 = 1 =2; and N 2N be any positive integer. We will show that we can always …nd n; m N

61



such that jxn xm j> 1=2: Indeed, for n = N 2 and m = ( N + 1) 2 ; we have:

jxn xm j= p N 2 q (N + 1) 2 = jN (N + 1) j= 1 > 1=2 = 0 :

Thus, (xn ) is not a Cauchy sequence.

———————————————————————————————————————————————–118 . Let (xn ) be a Cauchy sequence such that xn is an integer for every n

2N: Show that (xn ) is ultimately

constant.———————————————————————————————————————————————–Proof. We are asked to show that after …nitely many terms, the sequence (xn ) becomes constant.Suppose not. This means that for any N 2N; we can …nd n > N : xn 6= xN :But since xn and xN are integers, and xn 6= xN ; we get that jxn xN j 1:Thus, if we let 0 = 1 =2; we get that (xn ) is not a Cauchy sequence.(choose, e.g., m = N; and n > N as above).Hence, if (xn ) is a Cauchy sequence such that xn is an integer 8n 2N; then (xn ) is ultimately constant.

———————————————————————————————————————————————–119 . Show directly that a bounded, monotone increasing sequence is a Cauchy sequence.———————————————————————————————————————————————–Proof. A bounded monotone increasing sequence is convergent and hence is Cauchy.

———————————————————————————————————————————————–120 . If 0 < r < 1 and jxn +1 xn j< r n for all n 2N; show that (xn ) is a Cauchy sequence.———————————————————————————————————————————————–Proof. Recall from an earlier exercise that the sequence (nr n ) converges to 0.Let > 0 be given. Then 9N 2N : nr n < r for all n N; and so nr n 1 < for all n N:Now, for any n > m N; we have:

jxn xm j=n 1

Xk = m

(xk +1 xk )n 1

Xk= mjxk+1 xkj<

n 1

Xk = m

r k < (n m)r n 1 < nr n 1 < :

Hence, (xn ) is a Cauchy sequence.

———————————————————————————————————————————————–121 . Show that if (xn ) is an unbounded sequence, then there exists a properly divergent subsequence.———————————————————————————————————————————————–Proof. Since (xn ) is an unbounded sequence, we can …nd a positive integer n1 :

jxn 1

j> 1:

Since (xn ) is an unbounded sequence, we can …nd a positive integer n2 : n2 > n 1; and jxn 2 j> maxf2; jxn 1 jg:Since (xn ) is an unbounded sequence, we can …nd a positive integer n3 : n3 > n 2; and jxn 3 j> maxf3; jxn 2 jg:...Since (xn ) is an unbounded sequence, we can …nd a positive integer nk : nk > n k 1 ; and jxn k j> maxfk; xn k 1 g:...Continuing in this fashion, we get a subsequence (xn k ) of (xn ) such that lim

k!1 jxn k j= + 1 :

Now, limk!1 jxn k j= + 1 =) (xn k ) has a subsequence xn k m : Either lim

m !1xn k m = + 1 ; or lim

m !1xn k m = 1:

Say, limm !1

xn k m = + 1 : Then xn k m is a properly divergent subsequence of (xn ) :Thus, every unbounded sequence has a properly divergent subsequence.

———————————————————————————————————————————————–

122 . Give examples of properly divergent sequences (xn ) and (yn ) with yn 6= 0 for all n 2N such that(a) (xn =yn ) is convergent ; (b) (xn =yn ) is properly divergent :———————————————————————————————————————————————–Proof. (a) Let (xn ) = ( n) and (yn ) = ( n) : Then both (xn ) and (yn ) are properly divergent.Yet, (xn =yn ) = (1) which is convergent.

(a) Let (xn ) = n2 and (yn ) = ( n) : Then both (xn ) and (yn ) are properly divergent.Also, (xn =yn ) = ( n) which is properly divergent :

———————————————————————————————————————————————–123 . Show that if xn > 0 for all n 2N; then lim


n !1(1=xn ) = + 1 :

62



———————————————————————————————————————————————–Proof. Let R > 0 be a large real number. (Note that since xn > 0 for all n 2N; we have jxn j= xn ):Then, we have:lim

n !1(xn ) = 0 , 9 N 2N : xn < 1

R for all n N , 1x n

> R for all n N R arbitrary

, limn !1

(1=xn ) = + 1 :

———————————————————————————————————————————————–124 . Establish the proper divergence of the following sequences:(a) (p n) ; (b) p n + 1 ; (c) p n 1 ; (d) n=p n + 1 :———————————————————————————————————————————————–Proof. Recall that a sequence (xn ) of positive real numbers is properly divergent if and only if 8R > 0

9N = N (R) 2N : xn > R for all n N:This is the idea that we will use to show proper divergence in what follows.

(a) Let R > 0 be a large real number. Then there exists N 2N : N > R 2 :Hence, for all n N; we have n N > R 2 and so p n > p R2 = R:Thus, lim

n !1(p n) = + 1 :

(b) Let R > 0 be a large real number. Then there exists N 2N : N + 1 > R 2 :Hence, for all n N; we have n + 1 N + 1 > R 2 and so p n + 1 > p R2 = R:

Thus, limn !1 p n + 1 = + 1 :

(c) Let R > 0 be a large real number. Then there exists N 2N : N 1 > R 2 :Hence, for all n N; we have n 1 N 1 > R 2 and so p n 1 > p R2 = R:Thus, lim

n !1p n 1 = + 1 :

(d) Note …rst that for any positive integer n; n2 > n 2 1 = ( n 1)(n + 1) and so n > p n 1p n + 1 :Hence, np n +1 > p n 1 for all n 2N:Let R > 0 be a large real number. Then there exists N 2N : N 1 > R 2 :Hence, for all n N; we have n 1 N 1 > R 2 and so p n 1 > p R2 = R:Thus, np n +1 > p n 1 > R for all n N:Thus, lim

n !1p n 1 = + 1 :

———————————————————————————————————————————————–125 . Is the sequence (n sin n) properly divergent?———————————————————————————————————————————————–Proof. No. Recall that the function sin x is > 0 for x 2((2k 2) ; (2k 1) ) , k 2N

< 0 for x 2((2k 1) ; 2k ) , k 2N :

Here is a graph to refresh your memory:

Since the length of each of the above intervals is ; we can …nd an integer n within each interval.Hence, the sequence (n sin n) has in…nitely many positive terms and in…nitely many negative terms.This means that the sequence (n sin n) cannot be properly divergent.

———————————————————————————————————————————————–126 . Let (xn ) be properly divergent and let (yn ) be such that limn !1

(xn yn ) belongs to R: Show that (yn )converges to 0.———————————————————————————————————————————————–Proof. Say, lim

n !1(xn yn ) = a 2R: Suppose to the contrary that (yn ) does not converge to 0.

Then 9 a subsequence (yn k ) of (yn ) such that jyn k j r > 0 for some r and for all k 2N:Since (xn ) is properly divergent, lim

n !1(jxn j) = + 1 ; and so lim

k!1(jxn k j) = + 1 :

Thus, we can …nd K 2N large enough so that jxn k j> ja j+1r for all k K:

Now, for all k K; we have jxn k yn k j r jxn k j> jaj+ 1 :

63



Hence, jxn k yn k j jaj> 1 for all k K; and so jxn k yn k aj> 1 for all k K: (Recall: jx yj jxj jyj):Thus, we found a subsequence (xn k yn k ) of (xn yn ) that does not converge to a; a contradiction.Therefore, (yn ) converges to 0.

———————————————————————————————————————————————–127 . Let (xn ) and (yn ) be sequences of positive numbers such that lim

n !1(xn =yn ) = 0 :


(xn

) = +

1; then lim

n !1(y

n) = +

1:

(b) Show that if (yn ) is bounded, then limn !1

(xn ) = 0 :———————————————————————————————————————————————–Proof. (a) Let zn = 1

yn: Then we have that (xn ) is properly divergent, and lim

n !1(xn zn ) = 0 belongs to R:

Hence, by the previous exercise, limn !1

(zn ) = 0 : Thus, by another previous exercise, we know that if we havea sequence (zn ) of positive numbers, then lim

n !1(zn ) = 0 , lim

n !1(1=zn ) = + 1 :

But 1=zn = yn : Hence, limn !1

(yn ) = + 1 :

(b) Since (yn ) is bounded, there is a positive real number M : yn M for all n 2N:Let > 0 be given. Since lim

n !1(xn =yn ) = 0 ; 9N 2N large enough so that xn =yn <

M for all n N:Thus, xn <

M yn M M = for all n N:

This is exactly what it means for limn !1

(xn ) to be 0.

———————————————————————————————————————————————–128 . Investigate the convergence or divergence of the following sequences:(a) p n2 + 2 ; (b) p n= n2 + 1 ; (c) p n2 + 1 =p n ; (d) (sin p n) :———————————————————————————————————————————————–Proof. (a) We have n2 + 2 > n 2 for all n 2N and so p n2 + 2 > n for all n 2N:But lim

n !1n = + 1 : Hence, lim

n !1p n2 + 2 = + 1 :

(b) We have 0 p n:

But limn!1

p n = +

1: Hence, lim

n!1

p n 2 +1p n = +

1:

(d) This sequence is bounded and divergent.

———————————————————————————————————————————————–129 . Let (xn ) and (yn ) be sequences of positive numbers such that lim

n !1(xn =yn ) = + 1 :


(yn ) = + 1 ; then limn !1

(xn ) = + 1 :(b) Show that if (xn ) is bounded, then lim

n !1(yn ) = 0 :

———————————————————————————————————————————————–Proof. Note that since lim

n !1(xn =yn ) = + 1 ; and xn =yn > 0; we have that lim

n !1(yn =xn ) = 0 :

Thus, this problem is a reformulation of the previous to the previous problem.

———————————————————————————————————————————————–130 . Show that if lim

n !1(an =n) = L; where L > 0; then lim

n !1(an ) = + 1 :

———————————————————————————————————————————————–Proof. Since limn !1 (an =n) = L > 0; we can assume, without loss of generality, that an > 0 for all n 2N:Also, since lim

n !1(an =n) = L; we can …nd N 2N large enough so that a n

n L < 12 L for all n N:

Hence, L a nn < a n

n L < 12 L for all n N; and so, a n

n > 12 L for all n N:

Thus, an > 12 Ln for all n N:

But limn !1

12 Ln = + 1 : Hence, lim

n !1(an ) = + 1 :

———————————————————————————————————————————————–131 . Show that the convergence of a series is not a¤ected by changing a …nite number of its terms.(of course, the value of the sum may be changed).———————————————————————————————————————————————–

64



Proof. Let1

Xn =1xn be a convergent series, say

1

Xn =1xn = L:

Pick any arbitrary …nite terms, say xn 1 ; xn 2 ;:::;x n k ; where n 1 < n 2 < < n k : Change them as you wish.

Let1

Xn =1

yn be the resulting series. (of course, yn = xn for all n nk + 1) :

Then 1Xn =1yn =

n k

Xn =1yn + 1Xn = n k +1

yn =n k

Xn =1yn + 1Xn = n k +1

xn =n k

Xn =1yn + L

n k

Xn =1xn!=

n k

Xn =1(yn xn )!+ L:

Now,n k

Xn =1(yn xn ) is a …nite sum, (it might not be 0), say

n k

Xn =1(yn xn ) = a:

Thus,1

Xn =1

yn = a + L < 1 :

———————————————————————————————————————————————–

132 . If 1

Xn =1xn and

1

Xn =1yn are convergent, show that

1

Xn =1(xn + yn ) is convergent.

———————————————————————————————————————————————–Proof. Let sn = x1 + + xn and tn = y1 + + yn .

That is, (sn ) and (tn ) are the sequences of partial sums of 1

Xn =1xn and

1

Xn =1yn ; respectively.

Note that the sequence of partial sums, (un ) ; of 1

Xn =1(xn + yn ) is given by:

un = ( x1 + y1) + + ( xn + yn ) = ( x1 + + xn ) + ( y1 + + yn ) = sn + tn :

Thus, since1

Xn =1xn and

1

Xn =1yn are convergent, the sequences (sn ) and (tn ) are convergent.

Hence, (un ) = ( sn + tn ) = ( sn ) + ( tn ) is convergent, and so1

Xn =1(xn + yn ) is convergent.

Note that we have also proved that1

Xn =1

(xn + yn ) =1

Xn =1

xn +1

Xn =1

yn :

———————————————————————————————————————————————–133 . (a) Show that the series X1n =1

cos n is divergent.

(b) Show that the series X1n =1

cos nn 2 is convergent.

———————————————————————————————————————————————–Proof. (a) The sequence (cos n) is divergent as we saw in an earlier exercise (might have been (sin n)) :Hence, lim

n !1cos n 6= 0 ; and so the series cannot converge.

(b) We have1

Xn =1

cos nn 2

1

Xn =1

1n 2 < 1 : Thus, the series

1

Xn =1

cos nn 2 converges absolutely, and hence,

is convergent.

———————————————————————————————————————————————–

134 . Show that the series1

Xn =1

( 1) n

p n is convergent.

———————————————————————————————————————————————–Proof. Instead of following the book, let us be more general and prove a much more general result:—————————————————————————–

Theorem. Suppose that1

Xn =1

an and1

Xn =1

bn satisfy the following properties:

(i) The sequence of partial sums, sn =n

Xk=1

ak ; is a bounded sequence; and

(ii) The sequence (bn ) is decreasing and limn !1

(bn ) = 0 :

65



Then1

Xn =1an bn converges.

—————————————————————————–Proof. Since (sn ) is bounded, we can …nd M : jsn j M for all n 2N:Let > 0 be given. Then 9N 2N : bn <

2M for all n N:Now, for any k > m > N; we have:

k

Xn =1an bn

m

Xn =1an bn =

k

Xn = m +1an bn

see Remark (1) below=

k 1

Xn = m +1sn (bn bn +1 )!+ sk bk sm bm +1

M k 1

Xn = m +1

(bn bn +1 ) + bk + bm +1 = M j2bm +1 j= 2 Mbm +1 < 2M 2M = :

Thus, by Cauchy’s Criterion, the sequence of partial sums of 1

Xn =1an bn is convergent and so

1

Xn =1an bn is convergent.

—————————————————————————–Now, in our situation, (an ) = (( 1)n ) ; (bn ) = 1p n :

Note that sn = a1 + + an = 1 if n is odd0 if n is even : Thus, jsn j 1:

Also, (bn ) is decreasing and limn !1

(bn ) = 0 : Thus, the theorem applies to our case.

Remark (1).k

Xn = m +1an bn =

k

Xn = m +1(sn sn 1)bn =

k

Xn = m +1(sn sn 1)bn =

k

Xn = m +1sn bn

k

Xn = m +1sn 1bn

= k

Xn = m +1sn bn! k 1

Xn = msn bn +1 != sk bk +

k 1

Xn = m +1sn bn! sm bm +1 +

k 1

Xn = m +1sn bn +1 !

= k 1

Xn = m +1

sn bn

k 1

Xn = m +1

sn bn +1 !+ sk bk sm bm +1 = k 1

Xn = m +1

sn (bn bn +1 )!+ sk bk sm bm +1 :

(This result is known as the Abel partial summation formula ).

Remark (2). Note that the above theorem implies, in particular, that1

Xn =1

( 1) n

n a is convergent 8a > 0:

———————————————————————————————————————————————–

135 . If 1

Xn =1

an with an > 0 is convergent, then is1

Xn =1

a2n always convergent?


Proof. Yes. Since an > 0 and1

Xn =1an is convergent, we get that lim

n !1an = 0 :

Let N

2N be large enough so that 0 < a n < 1 for all n N:

Then 0 < a 2n < a n < 1 for all n N:

Thus,1

Xn = N

a2n <

1

Xn = N

an < 1 ; and so1

Xn =1a2

n =N 1

Xn =1a2

n +1

Xn = N

a2n < 1 :

This can also be done using the limit comparison test; since limn !1

a 2n

a n= lim

n !1an = 0 ; and

1

Xn =1an < 1 ;

we conclude that1

Xn =1

a2n < 1 :

———————————————————————————————————————————————–

66



136 . If 1

Xn =1an with an > 0 is convergent, then is

1

Xn =1

p an always convergent?


Solution. No. Take an = 1n 2 : Then

1

Xn =1

an =1

Xn =1

1n 2 is convergent, but

1

Xn =1

p an =1

Xn =1

1n is divergent.

———————————————————————————————————————————————–

137 . If 1

Xn =1an with an > 0 is convergent, and if bn = (a 1 + + a n )

n for n 2N; then show that1

Xn =1bn is always

divergent.———————————————————————————————————————————————–Proof. bn = (a 1 + + a n )

n =) nbn = a1 + + an and since an > 0 for all n; we get that a1 < nb n 8n 2N:Thus, bn > a 1

n 8n 2N; and so, b1 + b2 + + bn > a11 + a 1

2 + + a1n = a1(1 + 1

2 + + 1n ) 8n 2N:

Thus,1

Xn =1

bn > a 11

Xn =1

1n and so, by the limit comparison test,

1

Xn =1

bn = + 1 :

———————————————————————————————————————————————–138 . Determine a condition on jx 1j that will assure that:(a) x2 1 < 1

2 ; (b) x2 1 < 1=103 ;(c) x2 1 < 1n for a given n 2N; (d) x3 1 < 1n for a given n 2N:———————————————————————————————————————————————–Solution. Note that if jx 1j< 1; then x < 2 and so jx + 1 j< 3:Since x2 1 = jx 1j jx + 1 j; we get that x2 1 < 3 jx 1j for all x 2(0; 2):Also, if jx 1j< 1; then x < 2 and so x2 + x + 1 x2+ jxj+ 1 < 4 + 2 + 1 = 7 :Since x3 1 = jx 1j x2 + x + 1 ; we get that x3 1 < 7 jx 1j for all x 2(0; 2):

(a) From what we did above, let jx 1j< 16 :

(b) From what we did above, let jx 1j< 13 10 3 :

(c) From what we did above, let jx 1j< 13n :

(d) From what we did above, let jx 1j< 17n :

———————————————————————————————————————————————–139 . Let c be a cluster point of A

R and let f : A

! R: Prove that lim

x! cf (x) = L if and only if

limx! c jf (x) Lj= 0 :———————————————————————————————————————————————–Proof. (=) ) Suppose that lim

x! cf (x) = L and let > 0 be given.

Then there exists > 0 : 0 < jx cj< =) j f (x) Lj< :Since jf (x) Lj= jjf (x) Lj 0j; we get that 0 < jx cj< =) jjf (x) Lj 0j< :This is exactly what it means for lim

x! c jf (x) Lj to be 0.

(( =) Suppose that limx! c jf (x) Lj= 0 and let > 0 be given.

Then there exists > 0 : 0 < jx cj< =) jjf (x) Lj 0j< :Since jf (x) Lj= jjf (x) Lj 0j; we get that 0 < jx cj< =) j f (x) Lj< :This is exactly what it means for lim

x! cf (x) to be L.

———————————————————————————————————————————————–140 . Let f : R! R and let c 2R: Show that limx! c

f (x) = L if and only if limx! 0

f (x + c) = L:———————————————————————————————————————————————–Proof. (=) ) Suppose that lim

x! cf (x) = L, let > 0 be given, and set y = x c.

Then there exists > 0 : 0 < jx cj< =) j f (x) Lj< ( )

:In terms of y; ( ) can be rewritten as: 0 < jyj= jy 0j< =) j f (y + c) Lj< :Thus, lim

y! 0f (y + c) = L: Note that we can replace y by x since they are dummy variables.

(( =) Suppose that limx! 0

f (x + c) = L, let > 0 be given, and set y = x + c.

67



Then there exists > 0 : 0 < jx 0j< =) j f (x + c) Lj< ( )

:In terms of y; ( ) can be rewritten as: 0 < jy cj< =) j f (y) Lj< :Thus, lim

y! cf (y) = L: Note that we can replace y by x since they are dummy variables.

———————————————————————————————————————————————–141 . Let I = (0 ; a) where a > 0; and let g(x) = x2 for x 2I: For any points x; c 2I; show that

g(x) c2

2a jx cj: Use this inequality to prove that limx! cx2

= c2

for any c 2I:———————————————————————————————————————————————–Proof. For any x; c 2I = (0 ; a); we have that jx + cj jxj+ jcj= x + c < a + a = 2 a:Hence, for any x; c 2I = (0 ; a); g(x) c2 = x2 c2 = jx + cj jx cj< 2a jx cj:Now, …x c 2I; let > 0 be given and set =

2a > 0:Then 0 < jx cj< =

2a =) x2 c2 < 2a jx cj< 2a 2a = :

This is exactly what it means for limx! c

x2 to be c2 :

———————————————————————————————————————————————–142 . Let I be an interval in R; let f : I ! R, and let c 2I: Suppose there exists constants K and L such that

jf (x) Lj< K jx cj for x 2I: Show that limx! c

f (x) = L:———————————————————————————————————————————————–Proof. Let > 0 be given and set =

K .Than for all x

2I with 0 <

jx c

j< ; we have that

jf (x) L

j< K

jx c

j< K = :

This is exactly what it means for limx! cf (x) to be L: ———————————————————————————————————————————————–143 . Show that lim

x! cx3 = c3 for any c 2R:

———————————————————————————————————————————————–Proof. Let c 2R: Then c is a cluster point of the intervel I = ( c 1; c + 1) :Also, for all x 2I; we have that x2 + cx + c2 x2 + jcj jxj+ c2 < (c + 1) 2 + jcj jc + 1 j+ c2 < 3 c2 + jcj+ 1 :Note that x3 c3 = x2 + cx + c2 jx cj:Now, if we let L = c3 and K = 3 c2 + jcj+ 1 ; then we have: x3 L < K jx cj for all x 2I:Thus, by the previous exercise, we get that lim

x! cx3 = c3 :

———————————————————————————————————————————————–144 . Show that lim

x! cp x = p c for any c > 0:

———————————————————————————————————————————————–Proof. Let c > 0: Then c is a cluster point of the intervel I = (0 ; 2c):Also, for all x 2I; we have that p x + p c > 0 + p c = p c; and so 1p x + p c < 1p c .Note that jp x p cj= 1p x + p c jx cj:Now, if we let L = p c and K = 1p c ; then we have: jp x Lj< K jx cj for all x 2I:Thus, by the previous to the previous exercise, we get that lim

x! cp x = p c:

———————————————————————————————————————————————–145 . Use the de…nition of limit to show that(a) lim

x! 2x2 + 4 x = 12 ; (b) lim

x! 1x +5

2x +3 = 4 :———————————————————————————————————————————————–Proof. (a) Note that 2 is a cluster point of the intervel I = (1 ; 3); and that x2 + 4 x 12 = jx + 6 j jx 2j:For all x 2I; we have jx + 6 j< 9:Let > 0 be given and set =

9 : Then whenever x 2I with 0 < jx 2j< ; we have that

x2

+ 4 x 12 = jx + 6 j jx 2j< 9 = : Thus, limx! 2 x2

+ 4 x = 12 :

(b) Note that 1 is a cluster point of the intervel I = ( 2; 0); and that x +52x +3 4 = 7

j2x +3 j jx + 1 j:For all x 2I; we have j2x + 3 j> j2( 2) + 3 j= 1 ; and so 1

j2x +3 j < 1:Let > 0 be given and set =

7 : Then whenever x 2I with 0 < jx + 1 j< ; we have thatx +5

2x +3 4 = 7

j2x +3 j jx + 1 j< 7 = : Thus, limx! 1

x +52x +3 = 4 :

———————————————————————————————————————————————–146 . Show that the following limits do not exist:(a) lim

x! 01

x 2 (x > 0); (b) limx! 0

1p x (x > 0);

68



(c) limx! 0

(x+ sgn(x)) ; (d) limx! 0

sin 1=x2

———————————————————————————————————————————————–Proof. (a) Let f (x) = 1

x 2 for x > 0: Let xn = 1n ; n 2N: Then lim

n !1(xn ) = 0 :

However, the sequence (f (xn )) = 11=n 2 = n2 does not converge in R:

Hence, limx! 0

1x 2 does not exist.

(b) Let f (x) = 1p x for x > 0: Let xn = 1n 2 ; n 2N: Then lim

n !1(xn ) = 0 :

However, the sequence (f (xn )) = 1p 1=n 2= ( n) does not converge in R:

Hence, limx! 0

1p x does not exist.

(c) Let f (x) = x+ sgn(x): Let xn = ( 1) n

n ; n 2N: Then limn !1

(xn ) = 0 :

However, the sequence (f (xn )) = ( xn + sgn(xn )) = ( 1) n

n + ( 1)n has a subsequence that convergesto 1 and a subsequence that converges to 1, for n even and n odd, respectively.Hence, (f (xn )) does not converge in R:Thus, lim

x! 0(x+ sgn(x)) does not exist.

(d) Let f (x) = sin 1=x2 : Let xn = 1p 2 n ; and yn = 1p 2 n + 2; n 2N: Then lim

n !1(xn ) = lim

n !1(yn ) = 0 :

However, the sequence (f (xn )) = sin 1=x2n = (sin2 n) = (0) , and the sequence (f (yn )) = sin 1=y2

n= sin 2 n +

2 = (1) :Hence, lim

x! 0sin 1=x2 does not exist.

———————————————————————————————————————————————–147 . Suppose the function f : R ! R has limit L at 0; and let a > 0: If g : R! R is de…ned byg(x) = f (ax ) for x 2R; show that lim

x! 0g(x) = L:

———————————————————————————————————————————————–Proof. Let (xn ) be any sequence of points such that lim

n !1xn = 0 : Then (ax n ) also converges to 0.

Since limx! 0

f (x) = L; we get, by the sequential criterion for limits, that limx! 0

f (ax ) = L:but lim

x! 0

f (ax ) = limx! 0

g(x):

———————————————————————————————————————————————–148 . Let f : R! R be de…ned by: f (x) = x if x 2Q

0 if x =2Q :

(a) Show that f has a limit at x = 0 :(b) Use a sequential argument to show that if c 6= 0 ; then f does not have a limit at c:———————————————————————————————————————————————–Proof. (a) We claim that lim

x! 0f (x) = 0 : Let > 0 be given and set = :

Then for all x 2R with 0 < jxj= jx 0j< ; we have that jf (x) f (0)j= jf (x) 0j= jf (x)j jxj< = :Hence, lim

x! 0f (x) = 0 :

(b) Recall that both Q and RnQ are dense in R. Thus, if c 6= 0 ; then we may choose two sequences (xn ) and(yn ) of points converging to c so that xn 2Q and yn =2Q for all n 2N:

Then limn !1 f (xn ) Since xn

2Q

= limn !1 xn = c: Also, limn !1 f (yn ) Since yn =

2Q

= limn !1 0 = 0 :Since c 6= 0 ; we get that f does not have a limit at c:

———————————————————————————————————————————————–149 . Determine the following limits:(a) lim

x! 1(x + 1)(2 x + 3) ( x 2R); (b) lim

x! 1x 2 +2x 2 2 (x > 0);

(c) limx! 2

1x +1 1

2x (x > 0); (d) limx! 0

x +1x 2 +2 (x 2R):

———————————————————————————————————————————————–Solution. (a) Since lim

x! 1x = 1 ; lim

x! 11 = 1 ; lim

x! 12x = 2 lim

x! 1x = 2 ; and lim

x! 13 = 3 ; we get that

69



limx! 1

(x + 1) = limx! 1

x + limx! 1

1 = 1 + 1 = 2 ; and limx! 1

(2x + 3) = 2 limx! 1

x + limx! 1

3 = 2 + 3 = 5 :

Thus, limx! 1

(x + 1) (2 x + 3) = limx! 1

(x + 1) limx! 1

(2x + 3) = 2 5 = 10 :

(b) Note that 1 is a cluster point of the interval (0;p 2):

Also, limx! 1 x2 + 2 = limx! 1x2 + limx! 12 = 1 + 2 = 3 ; and limx! 1 x2 2 = limx! 1x2 limx! 12= 1 2 = 1 6= 0 :

Hence, limx! 1

x 2 +2x 2 2 =

limx ! 1(x 2 +2 )lim

x ! 1(x 2 2) = 3

1 = 3:

(c) Note that 2 is a cluster point of the interval (0;1 ):

Also, limx! 2

1x +1 =

limx ! 2

1

limx ! 2

(x +1) = 13 ; and lim

x! 21

2x =lim

x ! 21

limx ! 2

(2 x ) = 14 :

Hence, limx! 2

1x +1 1

2x = limx! 2

1x +1 lim

x! 21

2x = 13 1

4 = 112 :

(d) Note that x2 + 2 > 0 for all x

2R; lim

x! 0(x + 1) = 1 ; and lim

x! 0x2 + 2 = 2 :

Hence, limx! 0

x +1x 2 +2 =

limx ! 0

(x +1)

limx ! 0

(x 2 +2) = 12 :

———————————————————————————————————————————————–150 . Determine the following limits:(a) lim

x! 2q 2x +1x +3 (x > 0); (b) lim

x! 2x 2 4x 2 (x > 0);

(c) limx! 0

(x +1) 2 1x (x > 0); (d) lim

x! 1

p x 1x 1 (x > 0):

———————————————————————————————————————————————–

Solution. (a) Using an exercise below, limx! 2q 2x +1

x +3 = q limx! 22x +1x +3 = r limx ! 2

(2 x +1)

limx ! 2

(x +3) = q 55 = 1 :

(b) limx! 2

x 2 4x 2 = lim

x! 2(x 2)( x +2)

x 2 = limx! 2

(x + 2) = 4 :

(c) limx! 0(x +1) 2 1

x = limx! 0x 2 +2 x +1 1

x = limx! 0x (x +2)

x = limx! 0(x + 2) = 2 :

(d) limx! 1

p x 1x 1 = lim

x! 1

p x 1(p x 1)(p x +1 ) = lim

x! 11p x +1 =

limx ! 1

1

q limx ! 1x + lim

x ! 11

= 11+1 = 1

2 :

———————————————————————————————————————————————–151 . Find lim

x! 0

p 1+2 x p 1+3 xx +2 x 2 where x > 0:

———————————————————————————————————————————————–Solution. lim

x! 0

p 1+2 x p 1+3 xx +2 x 2 = lim

x! 0

p 1+2 x p 1+3 xx +2 x 2

p 1+2 x + p 1+3 xp 1+2 x + p 1+3 x = limx! 0

(1+2 x ) (1+3 x )(x +2 x 2 )(p 1+2 x + p 1+3 x)

= limx! 0

xx (1+2 x )(p 1+2 x + p 1+3 x) = lim

x! 0 1

(1+2 x )(p 1+2 x + p 1+3 x) = 0@lim

x ! 01

limx ! 0

(1+2 x ) q limx ! 0(1+2 x )+ q limx ! 0

(1+3 x )1A= 1

2:

———————————————————————————————————————————————–152 . Prove that lim

x! 0cos(1=x) does not exist but that lim

x! 0x cos(1=x) = 0 :

———————————————————————————————————————————————–Proof. Let f (x) = cos(1 =x) : Let xn = 1

2 n ; and yn = 12 n + 2

; n 2N: Then limn !1

(xn ) = limn !1

(yn ) = 0 :However, the sequence (f (xn )) = (cos (1 =xn )) = (cos 2 n) = (1) , and the sequence (f (yn )) = (cos (1 =yn ))= cos 2 n +

2 = (0) : Hence, limx! 0

cos(1=x) does not exist.

Now, note that for all x 2R; we have 1 cos(1=x) 1 and so x x cos(1=x) x:Since lim

x! 0x = 0 ; and lim

x! 0x = 0 ; the squeeze theorem implies that lim

x! 0x cos(1=x) = 0 :

70



———————————————————————————————————————————————–153 . Let f ; g be de…ned on A R to R; and let c be a cluster point of A: Suppose that f is boundedon a neighborhood of c and that lim

x! cg(x) = 0 : Prove that lim

x! cf (x) g (x) = 0 :

———————————————————————————————————————————————–Proof. Let V be a neighborhood of c on which f is bounded, i.e., jf (x)j M for all x 2V:Let > 0 be given. Since lim

x! c

g(x) = 0 ;

9 > 0 : whenever 0 <

jx c

j< ; we have that

jg(x)

j<

M :We may choose small enough so that (c ; c+ ) V:Hence, whenever 0 < jx cj< ; we have jf (x)g(x)j M jg(x)j< M

M = :Therefore, lim

x! cf (x) g (x) = 0 :

———————————————————————————————————————————————–154 . Use the de…nition of the limit to prove that if f ; g are de…ned on A R to R; c is a cluster point of A;limx! c

f (x) = L; and limx! c

g(x) = M; then limx! c

[f (x) + g(x)] = L + M:———————————————————————————————————————————————–Proof. Let > 0 be given.Since lim

x! cf (x) = L; 9 1 > 0 : whenever 0 < jx cj< 1 ; we have that jf (x) Lj<

2 :Since lim

x! cg(x) = M; 9 2 > 0 : whenever 0 < jx cj< 2 ; we have that jf (x) M j<

2 :Let = min f 1 ; 2g:Then whenever 0 <

jx c

j< ; we have that

j(f (x) + g(x)) (L + M )

j=

j(f (x) L) + ( g(x) M )

jjf (x) Lj+ jg(x) M j< 2 +

2 = :Therefore, lim

x! c[f (x) + g(x)] = L + M:

———————————————————————————————————————————————–155 . Let n 2N be such that n 3: Derive the inequality x2 xn x2 for 1 < x < 1: Then use the factthat lim

x! 0x2 = 0 to show that lim

x! 0xn = 0 :

———————————————————————————————————————————————–Proof. Note that if jxj 1; and k 2N; then xk = jxj

k 1 and so 1 xk 1:Thus, x2 xk+2 x2 for all k 2N and all x 2( 1; 1):Now, lim

x! 0x2 = 0 ; and lim

x! 0x2 = 0 :

Therefore, by the squeeze theorem, limx! 0

xn = 0 for all n 3:

———————————————————————————————————————————————–156 . Give examples of functions f and g such that f and g do not have limits at a point c; but such that both

f + g and f g have limits at c:———————————————————————————————————————————————–Solution. (i) Let f (x) = 1

x and g(x) = 1x :

Then limx! 0

f (x) and limx! 0

g(x) do not exist, but limx! 0

(f + g)(x) = limx! 0

1x 1

x = limx! 0

0 = 0 :

(ii) Let f (x) = g(x) = 1 for x < 01 for x 0 :

Then limx! 0

f (x) = limx! 0

g(x) = 1; and limx! 0+

f (x) = limx! 0+

g(x) = 1 : Thus, limx! 0

f (x) and limx! 0

g(x) do not exist.

Yet, f g (x) = ( 1) ( 1) = 1 for x < 0(1)(1) = 1 for x 0 1; and so lim

x! 0fg = lim

x! 01 = 1 exists.

———————————————————————————————————————————————–157 . Determine whether the following limits exist in R :(a) lim

x

!0

sin 1=x2 (x

6= 0) ; (b) lim

x

!0x sin 1=x2 (x

6= 0) ;

(c) limx! 0

sgn(sin (1 =x)) (x 6= 0) ; (d) limx! 0p x sin 1=x2 (x > 0):

———————————————————————————————————————————————–Solution. (a) Let f (x) = sin 1=x2 for x 6= 0 : Let xn = 1p 2 n ; and yn = 1p 2 n + 2

; n 2N:

Then limn !1

(xn ) = limn !1

(yn ) = 0 :However, the sequence (f (xn )) = sin 1=x2n = (sin 2 n) = (0) ,

and the sequence (f (yn )) = sin 1=y2n = sin 2 n +

2 = (1) :

Hence, limx! 0

sin 1=x2 does not exist .

Here is a graph of f (x) = sin 1=x2 :

71



(b) Let f (x) = x sin 1=x2 for x 6= 0 : Since 1 sin z 1 for all z 2R; we have the inequality

jxj f (x) = x sin 1=x2 jxj for all x 2R; x 6= 0 :

Since limx

!0 jxj= 0 ; the squeeze theorem implies that lim

x

!0x sin 1=x2 = 0 :

Here is a graph of f (x) = x sin 1=x2 :

(c) Let f (x) = sgn(sin (1 =x)) for x 6= 0 : Let xn = 12 n + 2

; and yn = 12 n + 3

2; n 2N:

Then limn !1

(xn ) = limn !1

(yn ) = 0 :

However, the sequence (f (xn )) = ( sgn (sin (1 =xn ))) = sgn sin 2 n + 2 = ( sgn (1)) = (1) ,

and the sequence (f (yn )) = ( sgn (sin (1 =yn ))) = sgn sin 2 n + 32 = ( sgn ( 1)) = ( 1) :

Hence, limx! 0

sgn(sin (1 =x)) does not exist .

(d) Let f (x) = p x sin 1=x2 for x > 0: Since

1 sin z

1 for all z

2R; and p x > 0 for x > 0;

we have the inequality p x f (x) = p x sin 1=x2 p x for all x 2R; x > 0:Since lim

x! 0p x = q limx! 0

x = 0 ; the squeeze theorem implies that limx! 0

p x sin 1=x2 = 0 :

Here is a graph of f (x) = p x sin 1=x2 :

72



———————————————————————————————————————————————–158 . Let A R; let f : A ! R; and let c 2R be a cluster point of A: In addition, suppose that f (x) 0for all x 2A; and let p f be the function de…ned for x 2A by p f (x) = p f (x):If lim

x! cf (x) exists, prove that lim

x! cp f (x) = q limx! c

f (x):———————————————————————————————————————————————–

Proof. Say, limx! cf (x) = L: Since f (x) 0 for all x 2A; L 0: Thus,p

L is well-de…ned.

Let > 0 be given.Since lim

x! cf (x) = L; we can …nd > 0 : whenever 0 < jx cj< ; we have jf (x) Lj< p L :

Now, jf (x) Lj= p f (x) p L p f (x) + p L :

Also, p f (x) + p L = p f (x) + p L p L since f (x) 0 for all x 2A:

Hence, p f (x) p L = jf (x ) L jp f (x )+ p L 1p L jf (x) Lj for all x 2A:

Therefore, whenever 0 < jx cj< ; we have p f (x) p L 1p L jf (x) Lj< 1p L p L = :

This is exactly what it means for limx! c

p f (x) to be p L:

———————————————————————————————————————————————–159 . Let f (x) = jxj 1=2 for x 6= 0 : Show that limx! 0+ f (x) = limx! 0 f (x) = + 1 :———————————————————————————————————————————————–Proof. Let > 0 be given and let = 1

2 > 0: Then for all x with 0 < jxj< = 12 ; we have that

0 : Thus, lim

x! 0+f (x) = lim

x! 0f (x) = + 1 :

Here is a graph of f (x) = jxj1=2

73



———————————————————————————————————————————————–160 . Let c 2R and let f be de…ned for x 2(c;1 ) and f (x) > 0 for all x 2(c;1 ):Show that lim

x! cf (x) = 1 if and only if lim

x! c1=f (x) = 0 :

———————————————————————————————————————————————–Proof. (=) ) Suppose that lim

x! cf (x) = 1 and let > 0 be given.

Then 9 > 0 : whenever 0 < jx cj< ; we have f (x) > 1

=) 1

f (x ) < :Thus, limx! c

1=f (x) = 0 :

(( =) Suppose that limx! c

1=f (x) = 0 and let > 0 be given.

Then 9 > 0 : whenever 0 < jx cj< ; we have 1f (x ) < 1 =) f (x) > :

Thus, limx! c

f (x) = 1 :

———————————————————————————————————————————————–161 . Evaluate the following limits, or show that they do not exist.(a) lim

x! 1+x

x 1 (x 6= 1) ; (b) limx! 1

xx 1 (x 6= 1) ;

(c) limx! 0+

x +2p x (x > 0); (d) limx!1

x +2p x (x > 0);

(e) limx! 0

p x +1x (x > 1); (f) lim

x!1p x +1

x (x > 0);

(g) limx!1 p x 5p x +3 (x > 0); (h) limx!1 p x xp x + x (x > 0):———————————————————————————————————————————————–Solution. (a) We claim that lim

x! 1+x

x 1 = 1 and so it does not exist in R:

To prove this, let > 1 be given. Set = 11 > 0:

Then if 0 < x 1 < = 11 ; we have 1

x 1 > 1 =) 1x 1 + 1 > =) x

x 1 > :Thus, lim

x! 1+x

x 1 = 1 :

(b) By what we did in part (a), we deduce that limx! 1

xx 1 does not exist.

(c) We claim that limx! 0+

x +2p x = 1 and so it does not exist in R:

To prove this, note that for x > 0; x+2p x > 2p x : Let > 0 be given. Set = 42 > 0:

Then if 0 < x < = 4

2 ; we have p x < 2

=) 2

p x > =) x+2

p x > :Thus, limx! 0+

x +2p x = 1 :

(d) We claim that limx!1

x +2p x = 1 and so it does not exist in R:

To prove this, note that for x > 0; x+2p x = p x + 2p x > p x: Let > 0 be given.Then for all x > 2 ; we have that p x > =) x+2p x > Thus, lim

x!1x +2p x = 1 :

(e) We claim that limx! 0

p x +1x does not exist in R:

To prove this, note that for x > 0; we have that p x +1x > 1p x : (This is because p x2 + 1 > p x2 = x):

By a similar argument to the one given in part (c), we conclude that limx

!0+

p x +1x = 1 :

(f) We claim that limx!1

p x +1x = 0 :

To prove this, note that for x > 0; we have that 0 < p x +1x < p x

x = 1p x :

Since limx!1

1p x = 0 ; the squeeze theorem implies that limx!1

p x +1x = 0 :

(g) We claim that limx!1

p x 5p x +3 = 1 :

Note that p x 5p x +3 1 = 8p x +3 = 8p x +3 :

To prove our claim, let 0 < < 1 be given and let = 8 3 2 :

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Then for x > = 8 3 2 ; we have that p x > 8 3 =)p x +3

8 > 1 =) 8p x +3 < :

Thus, for x > ; p x 5p x +3 1 = 8p x +3 < :

Therefore, limx!1

p x 5p x +3 = 1 :

(h) We claim that limx!1

p x xp x + x = 1:

Note that p x xp x + x ( 1) = p x xp x + x + 1 = 2p xp x + x = 2p xp x + x :

To prove our claim, let 0 < < 1 be given and let = 2 1 2 :Then for x > = 2 1 2 ; we have that p x > 2 1 =)

p x +12 > 1 =) 2p x +1 < =) 2p x +1

p xp x <

=) 2p xp x + x < : Thus, for x > ; p x xp x + x ( 1) = 2p xp x + x < :

Therefore, limx!1

p x xp x + x = 1:

———————————————————————————————————————————————–162 . Suppose that f and g have limits in R as x ! 1 and that f (x) g(x) for all x 2(a; 1 ):Prove that lim

x!1f (x) lim

x!1g(x):

———————————————————————————————————————————————–Proof. Let L = lim

x

!1f (x); and M = lim

x

!1g(x): Suppose to the contrary that L > M:

Then 9 > 0 : M + < L : Since L = limx!1f (x) and M = limx!1

g(x); we can …nd K 1 > a and K 2 > a :whenever x > K 1 we have jf (x) Lj< and whenever x > K 2 we have jg(x) M j< :Thus, for all x > K = max fK 1 ; K 2g; we have that jf (x) Lj< and jg(x) M j< :Hence, for all x > K; we have that f (x) > L > M + > g(x).This is a cotradiction to the fact that f (x) g(x) for all x 2(a; 1 ):Therefore, L M:

———————————————————————————————————————————————–163 . Let f be de…ned on (0;1 ) to R: Prove that lim

x!1f (x) = L if and only if lim

x! 0+f (1=x) = L:

———————————————————————————————————————————————–Proof. (=) ) Suppose that lim

x!1f (x) = L and let > 0 be given.

Then 9K > 0 : whenever x > K; we have that jf (x) Lj< : Let = 1K :

Thus, whenever 0 < x < ; we have 1x > K and so f ( 1

x ) L < :This is exactly what it means for lim

x! 0+f (1=x) to be L:

(( =) Suppose that limx! 0+

f (1=x) = L and let > 0 be given.

Then 9 > 0 : whenever 0 < x < ; we have that jf (1=x) Lj< : Let K = 1 :Thus, whenever x > K; we have 1

x < and so jf (x) Lj< :This is exactly what it means for lim

x!1f (x) to be L:

———————————————————————————————————————————————–164 . Show that if f : (a; 1 ) ! R is such that lim

x!1xf (x) = L; where L 2R; then lim

x!1f (x) = 0 :

———————————————————————————————————————————————–Proof. Since lim

x!1xf (x) = L; we can …nd > 0 : whenever x > ; jxf (x) Lj< 1:

Since jxf (x)j jLj< jxf (x) Lj; we deduce that whenever x > ; jxf (x)j jLj< 1 =) j xf (x)j< 1 + jLj=

) jf (x)

j< 1+ jL j

jxj :

Summing up, we have shown that whenever x > ; we have 0 < jf (x)j< 1+ jL jjxj :

Now, since limx!1

1+ jL jjx j = 0 ; the squeeze theorem implies that lim

x!1f (x) = 0 :

———————————————————————————————————————————————–165 . suppose that lim

x! cf (x) = L; where L > 0; and that lim

x! cg(x) = 1 : Show that lim

x! cf (x)g(x) = 1 :

If L = 0 ; show by example that this conclusion may fail.———————————————————————————————————————————————–Proof. Let > 0 be given.Since lim

x! cf (x) = L and L > 0; 9 1 > 0 : 0 < jx cj< 1 =) f (x) > 0:

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Since limx! c

g(x) = 1 ; 9 2 > 0 : 0 < jx cj< 2 =) g(x) > :

Let = min f 1 ; 2g: Then 0 < jx cj< =) f (x)g(x) > = :Thus, lim

x! cf (x)g(x) = 1 :

For an example, let f (x) = 0 and g(x) = 1

jxj for x 6= 0 : Then limx! 0

f (x) = 0 ; limx! 0

g(x) = 1 :

Yet, for all x 6= 0 ; f (x)g(x) = 0 ; and so limx! 0f (x)g(x) = 0 : ———————————————————————————————————————————————–166 . Find functions f and g de…ned on (0;1 ) such that lim

x!1f = 1 and lim

x!1g = 1 ; and lim

x!1(f g) = 0 :

Can you …nd such functions, with g(x) > 0 for all x 2(0;1 ); such that limx!1

f=g = 0?———————————————————————————————————————————————–Solution. Let f (x) = g(x) = x: Then lim

x!1f = 1 , lim

x!1g = 1 ; and lim

x!1(f g) = lim

x!10 = 0 :

The answer to the question is no . If g(x) > 0 for all x 2(0;1 ); limx!1

f = 1 , limx!1

g = 1 ; and limx!1

(f g) = 0 ;

then we can write f g = f g

g + 1 :Now, Since lim

x!1(f g) = 0 and lim

x!1g = 1 ; we get that lim

x!1f g

g = 0 :Hence, lim

x!1f=g = 0 + 1 = 1 6= 0 :

———————————————————————————————————————————————–

167 . Prove the following Sequential Criterion for Continuity : A function f : A ! R is continuous ata point c 2A if and only if for every sequence (xn ) in A that converges to c; the sequence (f (xn ))converges to f (c):———————————————————————————————————————————————–Proof. A function f : A ! R is continuous at a point c 2A if and only if lim

x! cf (x) = f (c):

Now, by the sequential criterion for limits, we get that limx! c

f (x) = f (c) if and only if for every sequence (xn )in A that converges to c; the sequence (f (xn )) converges to f (c):

———————————————————————————————————————————————–168 . Let f be de…ned for all x 2R; x 6= 2 ; by f (x) = ( x2 + x 6)=(x 2): Can f be de…ned at x = 2in such a way that f is continuous at this point?———————————————————————————————————————————————–Proof. Note that f (x) = ( x2 + x 6)=(x 2) = ( x 2)(x + 3) =(x 2) = x + 3 : This suggests that we de…nef (2) = 5 and f will be continuous at x = 2 :

———————————————————————————————————————————————–169 . Let f : R ! R be continuous at c and let f (c) > 0: Show that there exists a neighborhood V (c)of c such that if x 2V (c); then f (x) > 0:———————————————————————————————————————————————–Proof. Let = f (c)

2 > 0: Since f : R ! R be continuous at c; there exists > 0 : whenever jx cj< ; wehave that jf (x) f (c)j< = f (c)

2 : In other words, for all x 2V (c); we have f (c)

2 < f (x) f (c) < f (c)2 or

equivalently, f (c)2 < f (x) < 3 f (c)

2 and so f (x) > 0:

———————————————————————————————————————————————–170 . Let f : R! R be continuous on R and let S = fx 2R : f (x) = 0 g be the zero set of f : If (xn )is in S and x = lim

n !1(xn ) ; show that x 2S:

———————————————————————————————————————————————–Proof. Since x = lim

n !1(xn ) and f is continuous at x; the sequential criterion for continuity implies that

f (x) = limn !1

(f (xn )) : But xn 2S =) f (xn ) = 0 for all n 2N: Hence, f (x) = limn !1

(f (xn )) = limn !1

(0) = 0 :This is exactly what it means for x to be in S:

———————————————————————————————————————————————–171 . Let K > 0 and let f : R ! R satisfy the condition jf (x) f (y)j K jx yj for all x; y 2R: Showthat f is continuous at every point c 2R:———————————————————————————————————————————————–Proof. Let c 2R and let > 0 be given. Then for =

K ; the following is true:

8 x 2R with jx cj< ; we have that jf (x) f (c)j K jx cj< K K = : Thus, f is continuous at c:

But since our choice of c 2R was arbitrary, we conclude that f is continuous at every point c 2R:

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———————————————————————————————————————————————–172 . Suppose that f : R ! R is continuous on R and that f (r ) = 0 for every rational number r: Provethat f (x) = 0 for all x 2R:———————————————————————————————————————————————–Proof. Suppose, to the contrary, that there is a point x0 2R : f (x0) 6= 0 : Since f is continuous on R; weget that, in particular, f is continuous at x0 : Let (r n ) be a sequence of rational numbers that converges to x0 :Note that f (r n ) = 0 for all n 2N:Now, by the sequential criterion for continuity, f (x0) = lim

n !1(f (r n )) = lim

n !1(0) = 0 ; a contradiction.

Hence, f (x) = 0 for all x 2R:

———————————————————————————————————————————————–173 . Determine the points of discontinuity of the following functions and state which theorems are usedin each case.(a) f (x) = x2 +2 x +1

x 2 +1 (x 2R) (b) g(x) = p x + p x (x 0)

(c) h(x) =p 1+ jsin xj

x (x 6= 0) (d) k(x) = cos p 1 + x2 (x 2R)———————————————————————————————————————————————–Proof. (a) Since both f 1(x) = x2 + 2 x + 1 and f 2(x) = x2 + 1 are continuous everywhere on R; andx2 + 1 1; we get that f (x) is continuous everywhere on R:

(b) Since both g1(x) = x + p x and g2(x) = p x are continuous on [0;1 ) and x + p x 0 for all x 0;we get that g2 g1 = g is continuous on [0;1 ):

(c) Since h1(x) = 1 + jsin xj is continuous on R with h1(x) 1; and h2(x) = p x is continuous on [0;1 );we get that h2 h1 is continuous on R:

Now, since h3(x) = x is continuous on R; we get that h(x) = h 2 h 1 (x )h 3 (x ) =

p 1+ jsin x jx is continuous for x 6= 0 :

(d) Since k1(x) = 1 + x2 is continuous on R with k1(x) 1; and k2(x) = p x is continuous on [0;1 );we get that k2 k1 is continuous on R:Now, since k3(x) = cos x is continuous on R; we get that k(x) = k3 k2 k1(x) = cos p 1 + x2 iscontinuous for x 6= 0 :

———————————————————————————————————————————————–

174 . Show that if f : A ! R is continuous on A R and if n 2N; then the function f n

de…ned by:f n (x) = ( f (x))n for x 2A; is continuous on A:———————————————————————————————————————————————–Proof. Since the function g(x) = xn is continuous on R for all n 2N; and since f : A ! R is continuouson A R and f (A) R; we get that g f (x) = f n (x) = ( f (x)) n is continuous on A:

———————————————————————————————————————————————–175 . Let g be de…ned on R by g(1) = 0 ; and g(x) = 2 if x 6= 1 ; and let f (x) = x + 1 for all x 2R:Show that lim

x! 0g f 6= g f (0) : Why doesn’t this contradict Theorem 5.2.6?

———————————————————————————————————————————————–Proof. g f (x) = g(x + 1) = 0 if x = 0

2 if x 6= 0 : Thus, limx! 0

g f (x) = 2 :

Now, g f (0) = g(1) = 0 : Thus, limx! 0

g f 6= g f (0):This doesn’t contradict Theorem 5.2.6 because g is not continuous on f (R) = R:

———————————————————————————————————————————————–176 . Give an example of a function f : [0; 1] ! R that is discontinuous at every point of [0; 1] butsuch that jf j is continuous on [0; 1]:———————————————————————————————————————————————–Proof. Let f (x) = 1 if x 2[0; 1] \ Q

1 if x =2[0; 1]nQ : Then, by the sequential criterion for continuity, we see that

f is not continuous at any point x 2[0; 1]: (If x is rational, approach it by a sequence of irrational points andif x is irrational, approach it with a sequence of rational points).Note that jf (x)j= 1 for all x 2[0; 1] is a continuous function.

———————————————————————————————————————————————–

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177 . Let f ; g be continuous from R to R; and suppose that f (r ) = g(r ) for all rational numbers r: Is ittrue that f (x) = g(x) for all x 2R?———————————————————————————————————————————————–Proof. Yes, it is true and here is the proof. Let h(x) = ( f g) (x) = f (x) g(x) for x 2R:Then h is continuous on R since both f and g are. Moreover, h(x) = 0 for all x 2Q:Hence, by problem 172 above, we must have h(x) = 0 for all x

2R:

That is, f (x) = g(x) for all x 2R:———————————————————————————————————————————————–178 . Let f : R! R is continuous on R; and let P = fx 2R : f (x) > 0g: If c 2P; show that thereexists a neighborhood V (c) P:———————————————————————————————————————————————–Proof. Let c 2P: Then f (c) > 0: We saw, in problem 169, that there exists a neighborhood V (c) of csuch that if x 2V (c); then f (x) > 0: Thus, if x 2V (c); then x 2P: Hence, V (c) P:

———————————————————————————————————————————————–179 . Let I = [a; b] and let f : I ! R be a continuous function such that f (x) > 0 for each x 2I:Prove that there exists a number > 0 such that f (x) for each x 2I:———————————————————————————————————————————————–Proof. By the Maximum-Minimum Theorem, f attains its minimum on I = [a; b]; say at a point c 2[a; b]:(There might, of course, be more than one global minimum).Thus, f (x) f (c) > 0 for all x 2[a; b]: Let = f (c):

———————————————————————————————————————————————–180 . Let f be continuous on the interval [0; 1] to R and such that f (0) = f (1) : Prove that there existsa point c 2[0; 1

2 ] such that f (c) = f (c + 12 ): Conclude that there are, at any time, antipodal points

on the earth’s equator that have the same temperature. ( Hint : consider g(x) = f (x) f (x + 12 )) :

———————————————————————————————————————————————–Proof. Let g(x) = f (x) f (x + 1

2 ): Then g(0) = f (0) f ( 12 ) and g( 1

2 ) = f ( 12 ) f (1) :

If f ( 12 ) = f (1) ; then we are done and our point c = 0 or c = 1

2 : (both will work).

If f ( 12 ) 6= f (1) ; then, since f (0) = f (1) ; either g(0) < 0 < g ( 1

2 ) or g(0) > 0 > g ( 12 ):

Thus, by the Location of Roots Theorem, there exists a number c 2(0; 12 ) : g(c) = 0 :

In other words, there exists a number c 2(0; 12 ) : f (c) = f (c +

12 ):

Now, let R be the radius of earth. A standard parametrization of the circle x2 + y2 = R2 ; "the equator",is given by r (x) = ( R cos(2 x) ; R sin(2 x)) for x 2[0; 1]:The Temperature T : Equator ! R is a continuous function and hence, f = T r : [0; 1] ! R is continuous.Note that f (0) = f (1) and so, by above, there exists a point c 2[0; 1

2 ] such that f (c) = f (c + 12 ):

That is, T r (c) = T r (c + 12 ) =) T ((R cos(2 c) ; R sin(2 c))) = T ((R cos (2 c + ) ; R sin(2 c + ))) :

But cos(2 c + ) = cos(2 c) ; and sin(2 c + ) = sin(2 c) :Hence, T ((R cos(2 c) ; R sin(2 c))) = T (( R cos(2 c) ; R sin(2 c))) :That is, the Temperature at the point (R cos(2 c) ; R sin(2 c)) on the equator is the same as the Temperatureat the point ( R cos(2 c) ; R sin(2 c)) :Note that the two points above are antipodal.

———————————————————————————————————————————————–

181 . Show that the equation x = cos x has a solution on the interval [0; =2]: Use the Bisection methodand a calculator to …nd an approximate solution of this equation, with error less than 10 3 :———————————————————————————————————————————————–Proof. Let f (x) = x cos x: Then 1 = f (0) < 0 < f ( 2 ) =

2 :Thus, by the Location of Roots Theorem, there exists a number c 2(0; =2) : f (c) = 0 :In other words, there exists a number c 2(0; =2) : c = cos c:Using the Bisection method, we get that the solution has to be in the interval [0:7390; 0:7391]:

———————————————————————————————————————————————–182 . Let I = [a; b], let f : I ! R be continuous on I ; and assume that f (a) < 0; f (b) > 0:Let W = fx 2I : f (x) < 0g; and let w = sup W: Prove that f (w) = 0 :

78



———————————————————————————————————————————————–Proof. By continuity of f ; we get that f (w) 0: To show that f (w) = 0 ; assume not, i.e., f (w) < 0:Then, by continuity of f ; we can …nd a neighborhood V (w) such that f (x) < 0 for all x 2V (w) ( ) :Now, since w = sup W; we must have f (x) 0 for all x > w:This contradicts ( ): Hence, f (w) = 0 :

———————————————————————————————————————————————–183 . Suppose that f : R ! R is continuous on R and that limx! 1

f = 0 and limx!1

f = 0 :Prove that f is bounded on R and attains either a maximum or a minimum on R:Give an example to show that both a maximum and a minimum need not be attained.———————————————————————————————————————————————–Proof. Since lim

x! 1f = 0 and lim

x!1f = 0 ; there exists M > 0 large enough so that jf (x)j< 1 for all jxj> M:

Now, let I = [ M; M ]: Since f is continuous on I , f is bounded on I ; say jf (x)j L for all x 2I:Let M 0 = max f1; Lg: Then jf (x)j< M 0 for all x 2R: Hence, f is bounded on R:Suppose that f does not attain a minimum on R: Then f (x) > 0 for all x 2R: (why?)Now, let c 2I = [ M; M ] be such that f (c) = L: (This exists since I is closed, bounded and f is continuous).What we will do now is technical. Let K > M be such that jf (x)j< L

2 for all jxj> K:Then consider the closed, bounded interval J = [ K; K ]: Since f is continuous on J and we know for surethat c

2J (why?), we get that f attains its global maximum at a point (not necessarily c) in J:

By our choice of J; we are sure that this point is a global maximum for f on R:A similar argument may be applied in the case where f does not attain a maximum on R:

Example. Let f (x) = e x 2for x 2R: We know that f (x) > 0 for x 2R; that lim

x! 1f = 0 and lim

x!1f = 0 :

Also, as above, f can only have a global maximum but not a global minimum.

Here is a graph of f (x) = e x 2: (There are an uncountable number of examples, of course).


x 2 is uniformly continuous on A = [1;1 ); but that it is not uniformlycontinuous on B = (0 ;

1):

———————————————————————————————————————————————–Proof. For all 1 x < y < 1 ; jf (x) f (y)j= 1

x 2 1y 2 = y2 x 2

x 2 y2 = (y x )( y+ x )x 2 y 2 = jx yj

(y+ x )x 2 y 2

2 jx yj: Thus, f (x) is Lipschitz on [1;1 ) and hence is uniformly continuous on [1;1 ):

Now, let = 1 : We will show that for all 0 < < 1; we can always …nd x; y 2(0;1 ) : jf (x) f (y)j> 1:So, …x 0 < < 1 and let x 2(0; ): Let y = x

2 : Then y 2(0; ) and so jx yj< :Now, jf (x) f (y)j= 1

x 2 1y2 = 1

x 2 4x 2 = 3

x 2 > 3; since x < 1:

———————————————————————————————————————————————–185 . Show that if f and g are uniformly continuous on A R and if they are both bounded on A; thentheir product f g is uniformly continuous on A:———————————————————————————————————————————————–Proof. Say, jf j M 1 and jgj M 2 on A: Let M = max fM 1 ; M 2g: Then jf j M and jgj M on A:

Let > 0 be given. Since f and g are uniformly continuous on A, 9 > 0 : whenever x; y 2A withjx yj< ; we have that jf (x) f (y)j<

2M and jg(x) g(y)j< 2M :

Now, whenever x; y 2A with jx yj< ; we have that

jf (x)g(x) f (y)g(y)j= jf (x)g(x) f (x)g(y) + f (x)g(y) f (y)g(y)j jf (x)g(x) f (x)g(y)j+ jf (x)g(y) f (y)g(y= jf (x)j jg(x) g(y)j+ jg(y)j jf (x) f (y)j M

2M + M 2M = :

This is exactly what it means for the product function f (x)g(x) to be uniformly continuous on A:

———————————————————————————————————————————————–186 . Prove that if f and g are each uniformly continuous on R; then their composite function f g isuniformly continuous on R:———————————————————————————————————————————————–

79



Proof. Let > 0 be given and proceed as follows:Since f is uniformly continuous, 9 0 > 0 : whenever x; y 2R with jx yj< 0; we have that jf (x) f (y)j< :Since g is uniformly continuous, 9 > 0 : whenever x; y 2R with jx yj< ; we have that jg(x) g(y)j< 0:Thus, whenever x; y 2R with jx yj< ; we have that jg(x) g(y)j< 0 and so jf (g(x)) f (g(y))j< :This is exactly what it means for the composite function f g to be uniformly continuous on R:

———————————————————————————————————————————————–187 . If g(x) = p x for x 2[0; 1]; show that there does not exist a constant K such that jg(x)j K jxj for allx 2[0; 1]: Conclude that the uniformly continuous function g is not a Lipschitz function on [0; 1]:———————————————————————————————————————————————–Proof. For x 2(0; 1]; we have that jg(x )j

jxj = p xx = 1p x and so, lim

x! 0+g(x )

x = + 1 :

Hence, g(x )x is not bounded on (0; 1] and so that there does not exist a constant K such that jg(x)j K jxjfor all x 2[0; 1]: Hence, g is not a Lipschitz function on [0; 1]:

Now, since [0; 1] is a closed and bounded interval and g is continuous on [0; 1], we conclude that g isuniformly continuous on [0; 1]:

———————————————————————————————————————————————–188 . Show that if a function f : R! R is Lipschitz on R, then f is uniformly continuous on R:———————————————————————————————————————————————–

Proof. Since f is Lipschitz on R; there exists K > 0 : jf (x) f (y)j K jx yj for all x; y 2R:Let > 0 be given and set =

K : Then for all x; y 2R with jx yj< ; we have that

jf (x) f (y)j K jx yj< K = :This is exactly what it means for f to be uniformly continuous on R:

———————————————————————————————————————————————–189 . Show that the function f (x) = 1 =x is uniformly continuous on the set A = [a; 1 ); where a is a positiveconstant.———————————————————————————————————————————————–Proof. Note that for all x; y 2[a; 1 ); one has 1

x 1a ; 1

y 1a ; and therefore,

jf (x) f (y)j= 1x 1

y = 1xy jx yj 1

a 2 jx yj:Hence, f is Lipschitz on [a; 1 ) and thus uniformly continuous on [a; 1 ):

———————————————————————————————————————————————–

190 . Show that the function f (x) = x2 is not uniformly continuous on the set A = [0;1 ):———————————————————————————————————————————————–Proof. Note …rst that for any x; y 2[0;1 ); jf (x) f (y)j= x2 y2 = jx yj jx + yjLet = 1 : Then for all > 0; set x = + 1 and y = x +

2 = 32 + 1 :

Then jx yj= 2 < ; but jf (x) f (y)j=

252 + 2 = 5

4 2 + 1 > 1 = :This is exactly what it means for f (x) = x2 not to be uniformly continuous on [0;1 ):

———————————————————————————————————————————————–191 . Show that the function f (x) = sin(1 =x) is not uniformly continuous on the set B = (0 ;1 ):———————————————————————————————————————————————–Proof. Let = 1

2 ; (xn ) = 12 n and (yn ) = 1

2 n + 2: Then:

(i) (xn yn ) = =2[2 n ][2 n + 2 ] ! 0 as n ! 1 :

(ii) jf (xn ) f (yn )j= sin(2 n) sin(2 n + 2 ) = j0 1j= 1

12 = :This is exactly what it means for f (x) = sin(1 =x) not to be uniformly continuous on (0;1 ):


1+ x 2 for x 2R is uniformly continuous on R:———————————————————————————————————————————————–Proof. We will show that f (x) = 1

1+ x 2 is Lipschitz on R and so uniformly continuous on R:

For this, note that for any x; y 2R; one has jf (x) f (y)j= 11+ x 2 1

1+ y 2 = jx 2 y2 j(1+ x 2 )(1+ y 2 ) = jx + yjjx yj(1+ x 2 )(1+ y 2 ) :

Now, jx + yj(1+ x 2 )(1+ y 2 ) jxj(1+ x 2 )(1+ y2 ) + jyj(1+ x 2 )(1+ y2 ) 1 + 1 = 2 :That is, for any x; y 2R; one has jf (x) f (y)j 2 jx yj:

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Hence, f (x) = 11+ x 2 is Lipschitz on R and so uniformly continuous on R:

———————————————————————————————————————————————–193 . Show that if f and g are uniformly continuous on A R; then f + g is uniformly continuous on A:———————————————————————————————————————————————–Proof. Let > 0 be given. Then there exists > 0 such that for all x; y 2A with jx yj< ; we havethat

jf (x) f (y)

j<

2 and

jg(x) g(y)

j<

2 :Thus, for all x; y 2A with jx yj< ; we have that j(f + g) (x) (f + g) (y)j= jf (x) + g(x) f (y) g(y)j= jf (x) f (y) + g(x) g(y)j jf (x) f (y)j+ jg(x) g(y)j<

2 + 2 = :

This is exactly what it means for f + g to be uniformly continuous on A:

———————————————————————————————————————————————–194 . If f is uniformly continuous on A R; and jf (x)j k > 0 for all x 2A; show that 1=f is uniformlycontinuous on A:———————————————————————————————————————————————–Proof. Let > 0 be given. Since f is uniformly continuous on A; there exists > 0 : whenever x; y 2Awith jx yj< ; we have jf (x) f (y)j< k2 :Thus, whenever x; y 2A with jx yj< ; we have 1

f (x ) 1f (y ) = 1

jf (x )jjf (y)j jf (x) f (y)j 1k 2 k2 = :

This is exactly what it means for 1=f to be uniformly continuous on A:

———————————————————————————————————————————————–

195 . Prove that if f is uniformly continuous on a bounded set A R; then f is bounded on A:———————————————————————————————————————————————–Proof. Since f is uniformly continuous on A; there exists > 0 : whenever x; y 2A with jx yj< ;we have jf (x) f (y)j< 1: Since A is bounded, we can …nd a …nite number of points x1 < x 2 < < x n

in A such that An

[j =1

[x j ; xj + ] :

Let M = max f1 + jf (x1)j; 1 + jf (x2)j; :::; 1 + jf (xn )jg and let x 2A:Then x 2[x j ; xj + ] for some 1 j n; and so jx x j j< :Hence, jf (x) f (x j )j< 1 =) j f (x)j< 1 + jf (x j )j M:Since our choice of x 2A was arbitrary, we conclude that jf (x)j< M for all x 2A:Thus, f is bounded on A:

———————————————————————————————————————————————–

196 . Show that if f is continuous on [0;1 ) and uniformly continuous on [a; 1 ) for some positive constant a;then f is uniformly continuous on [0;1 ):———————————————————————————————————————————————–Proof. f is continuous on [0;1 ) =) f is continuous on [0; a + 1] :Since [0; a + 1] is closed and bounded, f is uniformly continuous on [0; a + 1] :Let > 0 be given. Then there exists 0 < 1 < 1 : whenever x; y 2[0; a + 1] with jx yj< 1 ;we have that jf (x) f (y)j< :Also, since f is uniformly continuous on [a; 1 ); there exists 0 < 2 < 1 : whenever x; y 2[a; 1 )with jx yj< 2 ; we have that jf (x) f (y)j< :Let = min f 1 ; 2g and note that for all x; y 2R with jx yj< ; we have that either x; y 2[0; a + 1] ;or x; y 2[a; 1 ):Therefore, for all x; y 2R with jx yj< ; we have that jf (x) f (y)j< :This is exactly what it means for f to be uniformly continuous on [0;1 ):

———————————————————————————————————————————————–197 Let A R and suppose that f : A ! R has the following property:

8 > 0 9g : A ! R such that g is uniformly continuous on A and jf (x) g (x)j< for all x 2A:Show that f is uniformly continuous on A:———————————————————————————————————————————————–Proof. Let > 0 be given. Then there exists a function g =3 : A ! R such that g =3 is uniformly continuouson A and f (x) g =3(x) < =3 for all x 2A:Since g =3 is uniformly continuous on A; there exists > 0 such that whenever x; y 2A with jx yj< ;we have that g =3(x) g =3(y) < =3:

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Thus, for all x; y 2A with jx yj< ; we have that

jf (x) f (y)j= f (x) g =3(x) + g =3(x) g =3(y) + g =3(y) f (y)f (x) g =3(x) + g =3(x) g =3(y) + g =3(y) f (y)

< =3 + =3 + =3 = :Therefore, f is uniformly continuous on A:

———————————————————————————————————————————————–198 . (a) Show that if f is uniformly continuous on (a; b] and on [b; c); then f is also uniformlycontinuous on (a; c):(b) Prove that any function f : R! R that is continuous and periodic must be uniformly continuous.———————————————————————————————————————————————–Proof. (a) Given > 0; there are 1 > 0 and 2 > 0 such that jf (x) f (b)j<

2 if 0 b x < 12

and jf (y) f (b)j< 2 if 0 y b < 2

2 :Setting = min f 1 ; 2g; we get that for any x 2(b

2 ; b] and any y 2[b; b+ 2 ); jf (x) f (y)j< ( ) :

For x; y 2(a; b] or x; y 2[b; c); ( ) is clearly satis…ed with some positive > 0:

(b) Let p be the period of f : Then f is continuous on each interval [kp; (k + 1) p] for all k 2Z:Since [kp; (k + 1) p] is closed and bounded, f is uniformly continuous on each of those intervals.So, by the solution of (a), one can show that f is uniformly continuous on R.

———————————————————————————————————————————————–199 . Show that if f : R ! R is continuous and such that lim

x! 1f (x) and lim

x!1f (x) are …nite, then f is

uniformly continuous on R.———————————————————————————————————————————————–Proof. Set lim

x!1f (x) = L and lim

x! 1f (x) = l:

Then, given > 0; there is A > 0 : jf (x) Lj< 2 for x A and jf (x) lj<

2 for x A:This implies that if x; y 2[A; 1 ) or x; y 2( 1; A]; then jf (x) f (y)j< :Of course, since [ A; A] is closed and bounded, f is uniformly continuous on [ A; A]:Finally, as in the solution of part (a) of the previous problem, one can show that f is uniformlycontinuous on R.

———————————————————————————————————————————————–200 . A function f : R

!R is continuous at 0 and satis…es the following conditions:

f (0) = 0 and f (x + y) f (x) + f (y) for any x; y 2R:Prove that f is uniformly continuous on R:———————————————————————————————————————————————–Proof. By the continuity of f at 0, given > 0 there is > 0 : jf (x)j< for jxj< :Hence, the subadditivity of f implies that ; for x 2R and jtj< ;

f (x + t) f (x) f (t) < and f (x) f (x + t) f ( t) < Thus, jf (x + t) f (x)j< ; which proves the uniform continuity of f on R.

———————————————————————————————————————————————–201 . A point x 2R is said to be an interior point of A R in case there is a neighborhood V of x suchthat V A: Show that a set A R is open if and only if every point of A is an interior point of A:———————————————————————————————————————————————–Proof. (=) ) Let A R be open and let x 2A: By de…nition of open-ness, there is a neighborhood V of x such that V

A: This means that x is an interior point of A:

(( =) Suppose that every point of A is an interior point of A and let x 2A: Being an interior point of A,there is a neighborhood V of x such that V A: Thus, for each x 2A there is a neighborhood V of xsuch that V A: Thus, A R is open.

———————————————————————————————————————————————–202 . A point x 2R is said to be a boundary point of A R in case every neighborhood V of x containspoints in A and points in Ac : Show that a set A and its complement Ac have exactly the same boundarypoints.———————————————————————————————————————————————–Proof. x is a boundary point of A if and only if every neighborhood V of x contains points in A and points

82



in Ac if and only if every neighborhood V of x contains points in A = ( Ac)c and points in Ac if and only if x is a boundary point of A c :

———————————————————————————————————————————————–203 . Show that a set G R is open if and only if it does not contain any of its boundary points.———————————————————————————————————————————————–Proof. (=

)) Suppose that G R is open and suppose that x is a boundary point of G: Then every

neighborhood V of x contains points in G and points in Gc : Thus, no neighborhood of x is containedentirely in G: Hence, x is not an interior point of G and thus x =2G:(( =) Suppose that G does not contain any of its boundary points and let x 2G: Not being a boundarypoint, x has a neighborhood V that does not intersect Gc : But V \ Gc = ? () V G: Hence, x is aninterior point of G: Since this is true for any x 2G; we conclude that G is open.

———————————————————————————————————————————————–204 . Show that a set F R is closed if and only if it contains all of its boundary points.———————————————————————————————————————————————–Proof. We can use the previous two exercises as follows:A set F R is closed

By De…nition

() F c is open By previous Exercise

() F c does not contain any of its boundarypoints

F an d F c have the same boundary points

() F contains all of its boundary points.

———————————————————————————————————————————————–

205 . If A R; let A be the union of all open sets that are contained in A; the set A is called the interiorof A: Show that:(a) A is an open set,(b) A is the largest open set contained in A; and(c) A point z belongs to A if and only if z is an interior point of A:———————————————————————————————————————————————–Proof. (a) A is a union of open sets and so is an open set. Note that, by de…nition, A A:(b) Let G be an open set contained in A: Since A is the union of all open sets that are contained in A; weconclude that G A . Thus, A is the largest open set contained in A:(c) z 2A () z 2V A for some open set V A () z is an interior point of A:

———————————————————————————————————————————————–206 . If A R; let A be the intersection of all closed sets containing A; the set A is called the closure of A:Show that:(a) A is a closed set,(b) A is the smallest closed set containing A; and(c) A point w belongs to A if and only if w is either an interior point or a boundary point of A:———————————————————————————————————————————————–Proof. (a) A is the intersection of closed sets and so is closed. Note that, by de…nition, A A:(b) Let F be a closed set containing A: Since A is the intersection of all closed sets containing A; weconclude that A F: Thus, A is the smallest closed set containing A:(c) A is closed () A

cis open () Every point x 2 A

cis an interior point of A

c

() Every pointx 2 A

chas a neighborhood V x A

cAc () Every point x 2 A

cis neither an interior point nor

a boundary point of A () Every point x 2A is either an interior point or a boundary point of A:

———————————————————————————————————————————————–207. Exhibit an open cover of N that has no …nite subcover.

———————————————————————————————————————————————–Solution. For n = 1 ; 2; 3;::: let Gn = ( n; n ): Note that G1 G2 Gn is an increasing

sequence of sets. We claim that G =1

[n =1Gn = R.

Obviously, G R: Now, let x 2R: By the Archimedian property, 9N 2N large enough so thatx < N: Thus, x 2( N; N ) G: Since x was arbitrary, we conclude that R G: Hence, G = R.In particular, N G and so G is an open cover of N:Suppose that N can be covered by a …nite number of the Gn ’s, say N Gn 1 [ Gn 2 [ [Gn k :Let m = max fn1 ; n 2 ;:::;n kg: Then, since the Gn ’s form an increasing sequence, we conclude thatGn 1 [ Gn 2 [ [Gn k = Gm = ( m; m ):

83



Thus, N ( m; m ); a contradiction since m 2N but m =2( m; m ): Thus, G has no …nite subcover.

Remark. Another solution is obtained by taking Gn = ( n 1; n + 1) for n 1:

Then note that the only Gn \ N= fng for every n and thus, G =1

[n =1Gn N is an open cover.

By our construction, it cannot reduce to a …nite subcover because each point n 2N is contained inexactly one Gn ; namely, Gn itself.———————————————————————————————————————————————–208 . Exhibit an open cover of the set A = f1=n : n 2Ng that has no …nite subcover.———————————————————————————————————————————————–Solution. We will follow the same method we used in the remark of the previous exercise. This methodis not applicable all the time. Actually, it is using the fact that the sets are discrete with no limit points.We want to …nd a sequence of sets Gn such that Gn \ A = f1=ng and nothing else from A:This is easy, let Gn = ( 1

n +1 ; 1n 1 ) for n 1: Now, proceed as in the remark of the previous exercise.

Remark. Try the same exercise but with A = f0g [ f1=n : n 2Ng: Can you …nd an open cover that hasno …nite subcover?———————————————————————————————————————————————–

209 . Prove that if K 1 and K 2 are compact sets in R; then K = K 1 [ K 2 is compact.———————————————————————————————————————————————–Proof. Let fG g2A be an open cover of K; i.e., K G = [2A

G and each G is open in R:

Then K 1 G and K 2 G: By compactness of K 1 and K 2 ; we conclude that we can …nd

two …nite subcovers G0 =k

[n =1Gn and G00 =

k +1+ m

[n = k+1

Gn for some k; m 2N such that K 1 G0

and K 2 G00: Thus, K k +1+ m

[n =1Gn is a …nite subcover of K: Hence, K = K 1 [ K 2 is compact.

———————————————————————————————————————————————–210 . Prove that the intersection of an arbitrary collection of compact sets in R is compact.———————————————————————————————————————————————–

Proof. Suppose that K is compact in R For in some index set A: Let K = \2AK :

We will show that K is closed and bounded and hence, by Heine-Borel Theorem, compact.Since each K is compact, Heine-Borel Theorem =) each K is closed and bounded.Since arbitrary intersection of closed sets is closed, K is closed .Since K K for each 2A; …x an in A; call it 0 :Now, K 0 is bounded and K K 0 =) K is bounded .Thus, K is closed and bounded and hence compact.

———————————————————————————————————————————————–211 . Let K 6= ? be compact in R and let c 2R:(a) Prove that there exists a point a 2K such that jc aj= inf fjc xj: x 2K g:(b) Prove that there exists a point b 2K such that jc bj= sup fjc xj: x 2K g:———————————————————————————————————————————————–Proof. (a) The in…mum exists because, by boundedness of K; fjc xj: x 2K g is bounded in R.Let = inf fjc xj: x 2K g:Let n 2N: By de…nition of inf ; there exists a point an 2K : + 1

n > jc an j:By the Bolzano-Weierstrass Theorem, there is a subsequence (an k ) of (an ) : a = lim

k!1an k 2K:

Since > jc an k j 1n k

for every k; limk!1 jc an k j 1

n k= jc aj:

Of course, being the inf and because a 2K; jc aj:Hence, = jc aj:

84



(b) The supremum exists because, by boundedness of K; fjc xj: x 2K g is bounded in R.Let = sup fjc xj: x 2K g:Let n 2N: By de…nition of sup ; there exists a point bn 2K : 1

n < jc bn j:By the Bolzano-Weierstrass Theorem, there is a subsequence (bn k ) of (bn ) : b = lim

k!1bn k 2K:

Since < jc bn k j 1n k

for every k; limk

!1jc bn k j 1

n k= jc bj:

Of course, being the sup and because b 2K; jc bj:Hence, = jc bj:

———————————————————————————————————————————————–212 . Let h : R ! R be de…ned by h(x) = 1 if 0 x 1

0 otherwise : Find an open set G such that h 1(G) is

not open, and a closed set F such that h 1(F ) is not closed.———————————————————————————————————————————————–Solution. Let G = (1 =2; 3=2): Then h 1(G) = fx 2R : 1=2 < f (x) < 3=2g= fx 2R : 0 x 1g= [0 ; 1]:Hence, G is open but h 1(G) is closed.Let F = [ 1=2; 1=2]: Then h 1(F ) = fx 2R : 1=2 f (x) 1=2g= fx 2R : x =2[0; 1]g= Rn[0; 1]:Hence, F is closed but h 1(F ) is open.

———————————————————————————————————————————————–213 . Prove that f : R

!R is continuous if and only if for each closed set F in R; the inverse image f 1(F )

is closed in R:———————————————————————————————————————————————–Proof. First note that for any subset G R; f 1(RnG) = Rnf 1(G): This is an elementary result.In what follows, we will use the fact that f : R! R is continuous if and only if for each open set G in R;the inverse image f 1(G) is open in R:

(=) ) suppose that f : R! R is continuous, and let F be closed in R: Then G = RnF is open and sof 1(G) = f 1(RnF ) = Rnf 1(F ) is open. Thus, f 1(F ) is closed in R:

(( =) Suppose that for each closed set F in R; the inverse image f 1(F ) is closed in R: Let G be open in R:Then RnG is closed and so f 1(RnG) is closed in R: But f 1(RnG) = Rnf 1(G):Thus, f 1(G) is open and so f is continuous.

———————————————————————————————————————————————–214 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) < g is open in R for each 2R:———————————————————————————————————————————————–Proof. Note that for any 2R; the set fx 2R : f (x) < g= f 1( 1; ) is open since f is continuous,and ( 1; ) is open.

———————————————————————————————————————————————–215 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) g is closed in R for each 2R:———————————————————————————————————————————————–Proof. Note that for any 2R; the set fx 2R : f (x) g= f 1( 1; ] is closed since f is continuous,and ( 1; ] is closed.

———————————————————————————————————————————————–216 . Show that if f : R ! R is continuous, then the set fx 2R : f (x) = kg is closed in R for each k 2R:———————————————————————————————————————————————–

Proof. Note that for any k 2R; the set fx 2R : f (x) = kg= f 1

fkg is closed since f is continuous,and fkg is closed.

———————————————————————————————————————————————–217 . Give an example of a function f : R! R such that the set fx 2R : f (x) = 1 g is neither opennor closed in R:———————————————————————————————————————————————–Proof. Let f (x) = 1 for x 2Q

0 for x =2Q : Then the set fx 2R : f (x) = 1 g= Q which is neither open

nor closed. Note that Q is not open because of the density of the irrationals, and Q is not closed becauseof its density in R.

85



———————————————————————————————————————————————–218. Let I = [a; b] and let f : I ! R and g : I ! R be continuous functions on I : Show that the set

fx 2R : f (x) = g(x)g is closed in R:———————————————————————————————————————————————–Proof. Let h(x) = f (x) g(x): Then h : I ! R is continuous.Moreover, the set

fx

2R : f (x) = g(x)

g=

fx

2R : h(x) = 0

g= h 1

f0

g which is closed in I since h

is continuous. ———————————————————————————————————————————————–219 . Show that each of the following functions is a metric on the given space:(a) S = R2 ; d1(P 1 ; P 2) = jx1 x2j+ jy1 y2j; where P 1 = ( x1 ; y1) and P 2 = ( x2 ; y2):(b) S = R2 ; d1 (P 1 ; P 2) = sup fjx1 x2j; jy1 y2jg; where P 1 = ( x1 ; y1) and P 2 = ( x2 ; y2):

(c) S = C [0; 1]; d1(f; g ) =1

Z 0

jf gj:

(d) S = C [0; 1]; d1 (f; g ) = sup fjf (x) g(x)j: x 2[0; 1]g:

(e) S = Any nonempty set ; d(s; t ) = 0 if s = t1 if s 6= t : This is called the discrete metric on S .

———————————————————————————————————————————————–Proof. (a) (i) (Positivity ) d1(P 1 ; P 2) =

jx1 x2

j+

jy1 y2

j0 for all P 1 ; P 2

2R2 :

(ii) ( De…niteness ) d1(P 1 ; P 2) = 0 = ) j x1 x2j+ jy1 y2j= 0 = ) jx1 x2j= 0 and jy1 y2j= 0=) x1 x2 = 0 and y1 y2 = 0 = ) x1 = x2 and y1 = y2 =) P 1 = P 2 :(iii) ( Symmetry ) d1(P 1 ; P 2) = jx1 x2j+ jy1 y2j= jx2 x1j+ jy2 y1j= d1(P 2 ; P 1) for all P 1; P 2 2R2 :(iv) ( Triangle Inequality ) Let P 1 = ( x1 ; y1), P 2 = ( x2 ; y2); and P 3 = ( x3 ; y3): Then d1(P 1 ; P 2) =

jx1 x2j+ jy1 y2j= jx1 x3 + x3 x2j+ jy1 y3 + y3 y2j jx1 x3j+ jx3 x2j+ jy1 y3j+ jy3 y2j= jx1 x3j+ jy1 y3j+ jx3 x2j+ + jy3 y2j= d1(P 1 ; P 3) + d1(P 3; P 2):

(b) (i) (Positivity ) d1 (P 1 ; P 2) = sup fjx1 x2j; jy1 y2jg 0 for all P 1 ; P 2 2R2 :(ii) ( De…niteness ) d1 (P 1 ; P 2) = 0 = ) sup fjx1 x2j; jy1 y2jg= 0 = ) jx1 x2j= 0 and jy1 y2j= 0=) x1 x2 = 0 and y1 y2 = 0 = ) x1 = x2 and y1 = y2 =) P 1 = P 2 :(iii) ( Symmetry ) d1 (P 1 ; P 2) = sup fjx1 x2j; jy1 y2jg= sup fjx2 x1j+ jy2 y1jg= d1 (P 2 ; P 1)for all P 1 ; P 2 2R2 :(iv) ( Triangle Inequality ) Let P 1 = ( x1 ; y1), P 2 = ( x2 ; y2); and P 3 = ( x3 ; y3):Then jx1 x2j jx1 x3j+ jx3 x2j sup fjx1 x3j; jy1 y3jg+ sup fjx3 x2j; jy3 y2jg; and

jy1 y2j jy1 y3j+ jy3 y2j sup fjx1 x3j; jy1 y3jg+ sup fjx3 x2j; jy3 y2jg:Hence, sup fjx1 x2j; jy1 y2jg sup fjx1 x3j; jy1 y3jg+ sup fjx3 x2j; jy3 y2jg:This is exactly d1 (P 1 ; P 2) d1 (P 1 ; P 3) + d1 (P 3 ; P 2):

(c) (i) (Positivity ) d1(f; g ) =1

Z 0

jf gj 0 for all f ; g 2C [0; 1]:

(ii) ( De…niteness ) d1(f; g ) = 0 = )1

Z 0

jf gj= 0 since jf gj is continuous and 0

=) j f gj 0 =) f g 0

=) f g:

(iii) ( Symmetry ) d1(f; g ) =1

Z 0jf gj=

1

Z 0jg f j= d1(g; f ) for all f ; g 2C [0; 1]:

(iv) ( Triangle Inequality ) Let f ;g ;h 2C [0; 1]: Then d1(f; g ) =1

Z 0

jf gj=1

Z 0

jf h + h gj1

Z 0

jf hj+1

Z 0

jh gj= d1(f; h ) + d1(h; g):

86



(d) (i) (Positivity ) d1 (f; g ) = sup fjf (x) g(x)j: x 2[0; 1]g 0 for all f ; g 2C [0; 1]:(ii) ( De…niteness ) d1 (f; g ) = 0 =) supfjf (x) g(x)j: x 2[0; 1]g= 0 = ) j f (x) g(x)j= 0 for all x 2[0; 1]=) f g 0 =) f g:(iii) ( Symmetry ) d1 (f; g ) = sup fjf (x) g(x)j: x 2[0; 1]g= sup fjg(x) f (x)j: x 2[0; 1] = d1 (g; f )for all f ; g

2C [0; 1]:

(iv) ( Triangle Inequality ) Let f ;g ;h 2C [0; 1]: Then for all x 2[0; 1]; jf (x) g(x)j jf (x) h(x)j+ jh(x) g(x)j supfjf (x) h(x)j: x 2[0; 1]g+ sup fjh(x) g(x)j: x 2[0; 1]g:Hence, supfjf (x) g(x)j: x 2[0; 1]g supfjf (x) h(x)j: x 2[0; 1]g+ sup fjh(x) g(x)j: x 2[0; 1]g:This can be written as d1 (f; g ) d1 (f; h ) + d1 (h; g):

(e) (i) (Positivity ) d(s; t ) = 0 if s = t1 if s 6= t 0 for all s; t 2S:

(ii) ( De…niteness ) d(s; t ) = 0 = ) s = t:(iii) ( Symmetry ) Clearly, d(s; t ) = d(t; s ):(iv) ( Triangle Inequality ) Let s;t; u 2S:If d(s; t ) = 0 = ) s = t =) d(s; u ) = d(t; u ) and each is either 0 or 1. Since 0 0 and 0 2; we are done.If d(s; t ) = 1 = ) s 6= t =) For any u 2S; either u = s; u = t; or u =2 fs; tg:In the …rst case, we get that d(s; u ) = 0 ; d(u; t ) = 1 and so d(s; t ) = 1 0 + 1 = d(s; u ) + d(u; t ):In the second case, we get that d(s; u ) = 1 ; d(u; t ) = 0 and so d(s; t ) = 1 1 + 0 = d(s; u ) + d(u; t ):In the third case, we get that d(s; u ) = 1 ; d(u; t ) = 1 and so d(s; t ) = 1 2 = 1 + 1 = d(s; u ) + d(u; t ):Hence, the triangle inequality holds.

———————————————————————————————————————————————–220 . If P n = ( xn ; yn ) 2R2 and d1 is the metric in the previous problem part (b), show that (P n ) convergesto P = ( x; y) with respect to this metric if and only if (xn ) and (yn ) converge to x and y, respectively.———————————————————————————————————————————————–Proof. (=) ) Suppose that (P n ) converges to P = ( x; y) with respect to d1 ; and let > 0 be given.Then there exists K 2N : P n 2V (P ) = fz 2R2 : d1 (z; P ) < g for all n K:Now, fz 2R2 : d1 (z; P ) < g= fz 2R2 : sup fjz1 xj; jz2 yjg< g:Since P n 2V (P ) for all n K; we conclude that sup fjxn xj; jyn yjg< for all n K:This means that jxn xj< and jyn yj< for all n K:Thus, (xn ) and (yn ) converge to x and y, respectively.

(( =) Suppose that (xn ) and (yn ) converge to x and y, respectively. Let > 0 be given.Then we can …nd K 2N : jxn xj< and jyn yj< for all n K:Thus, sup fjxn xj; jyn yjg< for all n K:Hence, P n 2V (P ) for all n K and so (P n ) converges to P = ( x; y) with respect to d1 :

———————————————————————————————————————————————–221 . Let S be a nonempty set and let d be the discrete metric de…ned above. Show that in the metric space(S; d); a sequence (xn ) in S converges to x if and only if there is a K 2N such that xn = x for all n K:———————————————————————————————————————————————–Proof. (=) ) Suppose that in the metric space (S; d); a sequence (xn ) in S converges to x:Then for = 1 =2; 9 K 2N : xn 2V 1=2(x) = fy 2S : d(y; x) < 1=2g

By de…nition of d= fxg for all n K:

Hence, xn = x for all n K:

(( =) Suppose that there is a K 2N such that xn = x for all n K and let > 0 be given.Then d(xn ; x) = 0 for all n K and so xn 2V (x) for all n K . Hence, (xn ) converges to x:

———————————————————————————————————————————————–222 . Show that if d is the discrete metric on a set S; then every subset of S is both open and closed in (S; d):———————————————————————————————————————————————–Proof. We will show that each singleton fx0g is an open set when we have the discrete metric.This is because for each x 2 fx0g; x = x0 and V 1=2(x) = fxg= fx0g fx0g:Thus, we have shown that 8x 2 fx0g there is a neighborhood U = V 1=2(x) of x such that U fx0g:This is exactly what it means for the set fx0g to be open in S:

87



Now, Let A S be any subset. Then we can write A = [x2Afxg and since fxg is open and an arbitrary union

of open sets is open, we conclude that A is open.

Now, let B

S be any set. Then S

nB

S and so by what we did above, S

nB is open.

Hence, B is closed, being the complement of an open set.

Thus, if d is the discrete metric on a set S; then every subset of S is both open and closed in (S; d):

———————————————————————————————————————————————–223 . Prove that in any metric space (S; d), an neighborhood of a point is an open set.———————————————————————————————————————————————–Proof. Let x0 2S and consider V (x0) = fx 2S : d(x; x 0) < g:Let x 2V (x0) and let = d(x; x 0) > 0:We claim that V (x) V (x0). To prove this, let y 2V (x):Then d(y; x) < =) d(y; x0) d(y; x) + d(x; x 0) < + d(x; x 0) = d(x; x 0) + d(x; x 0) = :Hence, y 2V (x0) and so V (x) V (x0).Thus, we have shown that 8x 2V (x0) there is a neighborhood U = V (x) of x such that U V (x0):Therefore, V (x0) is open.

———————————————————————————————————————————————–224 . Let f : R! R be de…ned by f (x) = x2 if x 2Q

0 if x =2Q : Show that f is di¤erentiable at x = 0 ; and

…nd f 0(0) :———————————————————————————————————————————————–Proof. We note that jf (x)j x2 for all x 2Rnf0g; and so f (x )

x jxj for all x 2Rnf0g:

We want to show that f 0(0) = 0 ; i.e., limx! 0

f (x ) f (0)x 0 = 0 or equivalently, since f (0) = 0 ; lim

x! 0f (x )

x = 0 :

Let > 0 be given. Then for = ; we have that 8jxj< ; f (x )x jxj< :

Hence, f 0(0) = limx! 0

f (x ) f (0)x 0 = 0 :

———————————————————————————————————————————————–225. Let g : R

!R be de…ned by g(x) = x2 sin(1=x2) if x 6= 0

0 if x = 0: Show that g is di¤erentiable for

all x 2R; Also, show that the derivative g0 is not bounded on the interval [ 1; 1]:———————————————————————————————————————————————–Proof. For x 6= 0 ; we get, by the product rule and chain rule, g0(x) = 2 x sin(1=x2) + x2( 2

x 3 cos(1=x2))= 2 x sin(1=x2) 2

x cos(1=x2) which exists for all x 6= 0 : Hence, g is di¤erentible for all x 6= 0 :For x = 0 ; we compute the limit, g0(0) = lim

x! 0g(x ) g(0)

x 0 = limx! 0

x 2 sin(1 =x 2 ) 0x = lim

x! 0x sin(1=x2) = 0 :

(Because x sin(1=x2) jxj for all x 2R):To show that g0 is not bounded on the interval [ 1; 1]; we consider the sequence (xn ) = 1p 2 n :

All the terms of this sequence are in [ 1; 1]; and g0(xn ) = 2p 2 n sin(2 n)

| {z } =0 for all n

2p 2 n cos(2 n)

| {z } =1 for all n

= 2p 2 n which is not bounded since it goes to 1 as n ! 1 :

———————————————————————————————————————————————–226. Assume that there exists a function L : (0;1 ) ! R such that L0(x) = 1 =x for x > 0: Calculate thederivatives of the following functions:(a) f (x) = L(2x + 3) for x > 0;(b) g(x) = L(x2) 3 for x > 0;(c) h(x) = L(ax ) for a > 0; x > 0;(d) k(x) = L(L(x)) when L(x) > 0; x > 0:———————————————————————————————————————————————–Solution. (a) Using the chain rule, f 0(x) = L0(2x + 3) 2 = 2

2x +3 :(b) Using the chain rule, g0(x) = 3( L(x2))2 L0(x2) 2x = 3( L(x2))2 1

x 2 2x = 6x (L(x2))2 :

88



(c) Using the chain rule, h0(x) = L0(ax ) a = 1ax a = 1

x :(d) Using the chain rule, k0(x) = L0(L(x)) L0(x) = 1

L (x ) 1x = 1

xL (x ) :

———————————————————————————————————————————————–227. Given that the function h(x) = x3 +2 x +1 for x 2R has an inverse h 1 on R; …nd the value of h 1 0 (y)at the points corresponding to x = 0 ; 1; 1:———————————————————————————————————————————————–Solution. h0(x) = 3 x2 + 2 ; h(0) = 1 ; h(1) = 4 ; and h( 1) = 2:Hence, h 1 0 (1) = h 1 0 (h(0)) = 1

h 0(h 1 (h (0))) = 1h 0(0) = 1

2 ;

h 1 0 (4) = h 1 0 (h(1)) = 1h 0(h 1 (h (1))) = 1

h 0(1) = 15 ;

h 1 0 ( 2) = h 1 0 (h( 1)) = 1h 0(h 1 (h ( 1))) = 1

h 0( 1) = 15 :

———————————————————————————————————————————————–228. Given that the restriction of the cosine function cos to I = [0 ; ] is strictly decreasing and that cos 0 = 1;cos = 1; let J = [ 1; 1]; and let arccos : J ! R be the function inverse to the restriction of cos to I :Show that arccos is di¤erentiable on ( 1; 1) and D (arccos( y)) = 1p 1 y 2

for y 2( 1; 1):

Show that arccos is not di¤erentiable at 1 and 1:———————————————————————————————————————————————–Proof. Let y 2( 1; 1): Then y = cos x for a unique x 2(0; ) and x = arccos y:

Thus, for y 2( 1; 1); D(arccos( y)) x =arccos y

= 1D (cos x ) = 1sin x(sin x )> 0 on (0 ; )

= 1p 1 cos 2 xy=cos x

= 1p 1 y 2 :Since 1p 1 y 2 ! 1 as y ! 1; we conclude that arccos is not di¤erentiable at 1 and 1:

Here is a graph of g(x) = arccos x:

———————————————————————————————————————————————–229. Let f : I ! R be di¤erentiable at c 2I: Establish the Straddle Lemma:Given > 0; there exists > 0 such that if u; v 2I satisfy c 0 be given. Since f : I

!R be di¤erentiable at c

2I; there exists > 0 such that

if x 2I satis…es 0 < jx cj< ; then f (x ) f (c)x c f 0(c) < :

Now, both u and v satisfy 0 < ju cj< ; 0 < jv cj< :Thus, jf (v) f (u) (v u)f 0(c)j

by the hint= jf (v) f (u) (v u)f 0(c) (f (c) cf 0(c)) + ( f (c) cf 0(c))j= jf (v) f (u) vf 0(c) + uf 0(c) (f (c) cf 0(c)) + ( f (c) cf 0(c))jjf (v) vf 0(c) (f (c) cf 0(c))j+ j f (u) + uf 0(c) + ( f (c) cf 0(c))j= jf (v) f (c) + ( c v) f 0(c)j+ jf (c) f (u) + ( u c) f 0(c)j= v cv c jf (v) f (c) + ( c v) f 0(c)j+ c u

c u jf (c) f (u) + ( u c) f 0(c)j= ( v c) f (v) f (c)

v c f 0(c) + ( c u) f (c) f (u )c u f 0(c)

89



= ( v c) f (v) f (c)v c f 0(c) + ( c u) f (u ) f (c)

u c f 0(c)< (v c) + ( c u) = ( v u) :

———————————————————————————————————————————————–230. Prove that if f : R! R is an even function [that is, f ( x) = f (x) for all x 2R] and has a derivativeat every point, then the derivative f 0 is an odd function [that is, f 0( x) = f 0(x) for all x 2R]: Also,

prove that if g : R! R is a di¤erentiable odd function, then g0 is an even function.———————————————————————————————————————————————–Proof. Suppose that f : R! R is a di¤erentiable even function. Let c 2R:Then f 0(c) = lim

h! 0f (c+ h ) f (c)

hf is even

= limh ! 0

f ( c h ) f ( c)h = lim

h! 0f ( c h ) f ( c)

h = limh! 0

f ( c+ h ) f ( c)h

= f 0( c):Since our choice of c 2R was arbitrary, we conclude that f 0(x) = f 0( x) for all x 2R:Hence, f 0 is an odd function.

Now, suppose that g : R ! R is a di¤erentiable odd function. Let c 2R:Then g0(c) = lim

h ! 0g(c+ h ) g(c)

hg is odd

= limh! 0

g( c h )+ g( c)h = lim

h! 0g( c h ) g( c)

h = limh! 0

g( c+ h ) g( c)h = g0( c):

Since our choice of c 2R was arbitrary, we conclude that g0(x) = g0( x) for all x 2R:Hence, g0 is an even function.

———————————————————————————————————————————————–231. If f : R! R is di¤erentiable at c 2R; show that f 0(c) = lim

n !1(n ff (c + 1 =n) f (c)g) : However, show

by an example that the existence of the limit of this sequence does not imply the existence of f 0(c):———————————————————————————————————————————————–Solution. Since f is di¤erentiable at c; f 0(c) = lim

h ! 0f (c+ h ) f (c)

h exists.

Hence, for any sequnce of points (xn ) such that xn ! 0; we must have that limn !1

f (c+ x n ) f (c)x n

= f 0(c):

In particular, let (xn ) = 1 =n: Then limn !1

f (c+1 =n ) f (c)1=n = f 0(c):

For an example, let f (x) = jxj: Then limn !1

f (0+1 =n ) f (0)1=n = lim

n !1j1=n j 0

1=n = limn !1

1=n1=n = 1 exists.

Yet, f 0(0) does not exist.

———————————————————————————————————————————————–

232. Use the Mean Value Theorem to prove that jsin x sin yj jx yj for all x; y 2R:———————————————————————————————————————————————–Proof. Let f (x) = sin x: Then for any x < y; the Mean Value Theorem implies that there exists c 2(x; y) :sin y sin x = (cos c) (y x) : Thus, jsin y sin xj= jcos cj jy xj jy xj: Since our choice of x; y 2Rwas arbitrary, we conclude that jsin x sin yj jx yj for all x; y 2R:

———————————————————————————————————————————————–233. If h(x) = 0 if x < 0

1 if x 0 ; prove there does not exist a function f : R! R such that f 0(x) = h(x) for

all x 2R: Give examples of two functions, not di¤ering by a constant, whose derivatives equal h(x) forall x 6= 0 :———————————————————————————————————————————————–Proof. If such an f exists, then it should be di¤erentiable everywhere with f 0(x) = h(x): Thus, f isdi¤erentiable on [ 1; 1]: Also, f 0( 1) = h( 1) = 0 ; and f 0(1) = h(1) = 1 : Hence, by Darboux’s

theorem, there must exist a c 2( 1; 1) : f 0(c) = h(c) = 1 =2: But this is absurd by de…nition of h:Let f (x) = 1 if x < 0

x if x 0 and g(x) = 3 if x < 0x + 1 if x 0 :

———————————————————————————————————————————————–234. Let I be an interval. Prove that if f is di¤erentiable on I and if the derivative f 0 is bounded on I ; thenf satis…es a Lipschitz condition on I :———————————————————————————————————————————————–Proof. Say jf 0(x)j M for all x 2I: Let x < y 2I: Then by the Mean Value Theorem, there existsc 2(x; y) : f 0(c) = f (y ) f (x )

y x =)f (y) f (x )

y x = jf 0(c)j M =) j f (y) f (x)j M jy xj:Since our choice of x and y was arbitrary, we conclude that f satis…es a Lipschitz condition on I with

90



Lipschitz constant M:

———————————————————————————————————————————————–235. Let f ; g be di¤erentiable functions on R and suppose that f (0) = g(0) and f 0(x) g0(x) for all x 0:Show that f (x) g(x) for all x 0:———————————————————————————————————————————————–Proof. Let h(x) = g(x) f (x): Then h(0) = 0 ; h0(x) = g0(x) f 0(x) 0 for all x 0:For x = 0 ; it is given that f (0) = g(0) ; so let x > 0:The by the Mean Value Theorem, 9c 2(0; x) : h0(c) = h (x ) h (0)

x 0 = h(x )x 0 =) h(x) 0 =) g(x) f (x):

Since our choice of x > 0 was arbitrary, we conclude that f (x) g(x) for all x 0:

———————————————————————————————————————————————–236. Suppose that f : [0; 2] ! R is continuous on [0; 2] and di¤erentiable on (0; 2); and that f (0) = 0 ;f (1) = 1 ; f (2) = 1 :(a) Show that there exists c1 2(0; 1) : f 0(c1) = 1 :(b) Show that there exists c2 2(1; 2) : f 0(c2) = 0 :(c) Show that there exists c 2(0; 2) : f 0(c) = 1 =3:———————————————————————————————————————————————–Proof. (a) By the mean value theorem, 9c1 2(0; 1) : f 0(c1) = f (1) f (0)

1 0 = 1 01 0 = 1 :

(b) By the mean value theorem, 9c2 2(1; 2) : f 0(c2) = f (2) f (1)2 1 = 1 1

2 1 = 0 :

(c) f is di¤erentiable on [c1 ; c2]; and f 0(c2) = 0 < 1=3 < 1 = f 0(c1): Therefore, by Darboux’s Theorem,there exists c 2(c1 ; c2) (0; 2) : f 0(c) = 1 =3:

———————————————————————————————————————————————–237. Use the Mean Value Theorem to prove that x 1

x < ln x < x 1 for x > 1:[Hint: Use the fact that D ln x = 1=x for x > 0]:———————————————————————————————————————————————–Proof. Let x > 1: Then f (x) = ln x is continuous on [1; x] and di¤erentiable on (1; x) and so by the MeanValue Theorem, 9 c 2(1; x) : 1

c = f 0(c) = f (x ) f (1)x 1 = ln x 0

x 1 = ln xx 1 :

Since 1 < c < x; we get that 1x < 1

c < 1:Thus, 1

x < ln xx 1 < 1 and so, x 1

x < ln x < x 1:Since our choice of x > 1 was arbitrary, we conclude that x 1

x < ln x < x 1 for all x > 1:

———————————————————————————————————————————————–

238. Let f : [a; b] ! R be continuous on [a; b]; and di¤erentiable in (a; b): Show that if limx! a f 0(x) = A;then f 0(a) exists and is equal to A:———————————————————————————————————————————————–Proof. Recall that f 0(a) = lim

x! af (x ) f (a )

x a : Let x 2(a; b): Then, by the Mean Value Theorem, 9cx 2(a; x ) :

f 0(cx ) = f (x ) f (a )x a : Taking limits, A = lim

x! af 0(cx ) = lim

x! af (x ) f (a )

x a = f 0(a):

———————————————————————————————————————————————–239. Let f : R! R be de…ned by f (x) = 2x4 + x4 sin(1=x) if x 6= 0

0 if x = 0 : Show that f has an absolute

minimum at x = 0 ; but that its derivative has both positive and negative values in every neighborhood of 0.———————————————————————————————————————————————–proof. For any x 6= 0 ; f (x) = 2 x4 + x4 sin(1=x) 2x4 x4 = x4 > 0: Since f (0) = 0 ; we conclude thatf (x) 0 for all x 2R: Hence, f has an absolute minimum at x = 0 :Now, f 0(x) = 8 x3 + 4 x3 sin(1=x) x2 cos(1=x):Let (xn ) = 1

2 n and yn = 12 n + 2

: Then both xn ! 0 and yn ! 0 as n ! 1 :Thus, every neighborhood of 0 contains an in…nite number of elements from both sequences.Note that f 0(xn ) = 8 1

2 n3 1

2 n2 < 0 for all n large enough.

Also, f 0(yn ) = 8 12 n + 2

3+ 4 1

2 n + 2

3> 0 for all n:

Thus, f 0 has both positive and negative values in every neighborhood of 0.

———————————————————————————————————————————————–240. Give an example of a uniformly continuous function on [0; 1] that is di¤erentiable on (0; 1) but whosederivative is not bounded on (0; 1):

91



———————————————————————————————————————————————–Solution. For x 0; let f (x) = p x: Then f is continuous on [0; 1] and since [0; 1] is compact, f is uniformlycontinuous on [0; 1]: Also, f 0(x) = 1

2p x which is not bounded on (0; 1):

———————————————————————————————————————————————–241 . Suppose that f and g are continuous on [a; b], di¤erentiable on (a; b), that c 2[a; b]; and that g(x) 6= 0for x

2[a; b]; x

6= c: Let A = lim

x! cf (x) and B = lim

x! cg(x): If B = 0 ; and if lim

x! c

f (x )

g(x ) exists in R; show that we

must have A = 0 : [Hint : f (x) = f (x )g(x ) g(x)]:

———————————————————————————————————————————————–Proof. Say, lim

x! cf (x )g(x ) = L 2R:

By the hint, A = limx! c

f (x) = limx! c

f (x )g(x ) g(x) = hlim

x! cf (x )g(x )ihlimx! c

g(x)i= LB = 0 :

———————————————————————————————————————————————–242. In addition to the suppositions of the preceding exercise, let g(x) > 0 for x 2[a; b]; x 6= c: If A > 0 andB = 0 ; prove that we must have lim

x! cf (x )g(x ) = 1 : If A < 0 and B = 0 ; prove that we must have lim

x! cf (x )g(x ) = 1:

———————————————————————————————————————————————–Proof. Let M > 0 be given. Since A = lim

x! cf (x) > 0; 9 1 > 0 : f (x) > A

2 > 0 for all x 2(c 1 ; c + 1) nfcg:

Since g(x) > 0 for x 2[a; b]nfcg; and B = limx! c

g(x) = 0 ; we conclude that 9 2 > 0 : 0 < g (x) < A2M for all

x 2(c 2 ; c + 2) nfcg: Thus, 1g(x ) >

2M A for all x 2(c 2 ; c + 2) nfcg:Let = min f 1 ; 2g:

Then for all x 2(c ; c+ ) fcg; we have f (x )g(x ) = ( f (x)) 1

g(x ) > A2

2M A = M:

Since our choice of M > 0 was arbitrary, we conclude that limx! c

f (x )g(x ) = 1 :

The other case is very similar.

———————————————————————————————————————————————–243. Let f (x) = x2 for x 2Q

0 for x =2Q ; and let g(x) = sin x for x 2R: Use Theorem 6.3.1 to show that

limx! 0

f (x )g(x ) = 0 : Explain why L’Hospital’s Rule I cannot be used.

———————————————————————————————————————————————–Proof. We have:(1) f and g are de…ned on [0;

2 ]:

(2) f (0) = g(0) = 0 :(3) g(x) 6= 0 for all x 2(0;

2 ):(4) f is di¤erentiable at a = 0 : This is because f (x ) f (0)

x 0 = f (x )x jxj for all x 6= 0 : Moreover, f 0(0) = 0 :

(5) g is di¤erentiable at a = 0 and g0(0) = cos(0) = 1 :

Hence, by Theorem 6.3.1 , we conclude that limx! 0+

f (x )g(x ) = f 0(0)

g0(0) = 01 = 0 :

Similarly, limx! 0

f (x )g(x ) = f 0(0)

g0(0) = 01 = 0 : Therefore, lim

x! 0f (x )g(x ) = 0 :

L’Hospital’s Rule I cannot be used because f is not continuous at any point di¤erent from 0, so it cannotbe di¤erentiable in any neighborhood of 0.


x!1ln xx 2 (0;1 ); (b) lim

x!1ln xp x (0;1 );

(c) limx! 0

x lnsin x (0; ); (d) limx!1

x +ln xx ln x (0;1 ):

———————————————————————————————————————————————–Solution. (a) lim

x!1ln xx 2

L’Hospital’s Rule= lim

x!11=x2x = lim

x!11

2x 2 = 1

1 = 0 :

(b) limx!1

ln xp xL’Hospital’s Rule= lim

x!11=x

1=2p x = limx!1

2p x = 2

1 = 0 :

(c) limx! 0

x lnsin x = limx! 0

lnsin x1=x

L’Hospital’s Rule= limx! 0

x 2 sin xcos x = 0

1 = 0 :

92



(d) limx!1

x +ln xx ln x


x!11+ 1

x1+ln x = 1

1 = 0 :


x! 0+x2x (0;1 ); (b) lim

x! 0(1 + 3

x )x (0;1 );

(c) limx

!1

(1 + 3x )x (0;1 ); (d) lim

x

!0+

1x 1

arctan x (0;1 ):

———————————————————————————————————————————————–Solution. (a) lim

x! 0+x2x = lim

x! 0+e2x ln x = e

limx ! 0+

(2 x ln x )= e

limx ! 0+

( 2 ln x1 =x ) L’Hospital’s Rule= e

limx ! 0+

2=x1 =x 2

= e0 = 1 :

(b) limx! 0

(1 + 3x )x = lim

x! 0ex ln(1+ 3

x ) = elim

x ! 0(x ln(1+ 3x )) = e

limx ! 0

ln(1+ 3x )

1 =x L’Hospital’s Rule= e

limx ! 0

3x 2

x 2 +3 x

= elim

x ! 0( 3xx +3 ) = e0 = 1 :

(c) limx!1

(1 + 3x )x = lim

x!1ex ln(1+ 3

x ) = e limx !1

ln(1+ 3x )

1 =x L’Hospital’s Rule= e limx !1

3x 2

x 2 +3 x = e3:

(d) limx

!0+

1x

1

arctan x = limx

!0+

(arctan x ) xx arctan x


x

!0+

11+ x 2 1

x1+ x 2 +arctan x

= limx! 0+

x 2

x +arctan x + x 2 arctan xL’Hospital’s Rule

= limx! 0+

2x1+ 1

1+ x 2 +2 x arctan x + x 21+ x 2

= 01+0+0+0 = 0 :

———————————————————————————————————————————————–246. Let g(x) = x3 for x 2R: Find g0(x) and g00(x) for x 2R; and g000(x) for x 6= 0 : Show that g000(0) doesnot exist.———————————————————————————————————————————————–Proof. g(x) = x3 for x 0

x3 for x < 0 : Let us …nd g0(0):

limx! 0+

g(x ) g(0)x 0 = lim

x! 0+x 3 0

x = limx! 0+

x2 = 0 :

limx! 0

g(x ) g(0)x 0 = lim

x! 0x 3 0

x = limx! 0

x2 = 0 :

Thus, g0(0) = limx! 0g(x ) g(0)x 0 exists and is equal to 0.

Hence, g0(x) = 3x2 for x 03x2 for x < 0 :

limx! 0+

g0(x ) g0(0)x 0 = lim

x! 0+3x 2 0

x = limx! 0+

3x = 0 :

limx! 0

g0(x ) g0(0)x 0 = lim

x! 03x 2 0

x = limx! 0

3x = 0 :

Thus, g00(0) = limx! 0

g0(x ) g0(0)x 0 exists and is equal to 0.

Hence, g00(x) = 6x for x 06x for x < 0 :

limx! 0+

g00(x ) g00(0)x 0 = lim

x! 0+6x 0

x = limx! 0+

6 = 6 :

limx! 0

g00(x ) g00(0)x 0 = lim

x! 06x 0

x = limx! 0

6 = 6:

Thus, g000(0) does not exist.

Hence, g000(x) = 6 for x > 06 for x < 0 and g000(0) does not exist.

———————————————————————————————————————————————–247. Use induction to prove Leibniz’s rule for the n-th derivative of a product:

93



(fg )(n ) (x) =n

Xk=0

nk f (n k) (x)g(k ) (x):

———————————————————————————————————————————————–

Proof. For n = 0 ; (fg )(0) (x) = f g (x) =0

Xk=0

0k f (0 k) (x)g(k ) (x):

For n = 1 ; (fg )0 (x) = f 0(x)g(x) + g0(x)f (x) =1

Xk =0

1k f (1 k ) (x)g(k ) (x):

Now, suppose the result is true for n: Then we have to prove it for n + 1 :

Well, (fg )(n +1) (x) = ddx h(fg )(n ) (x)iinduction hypothesis

= ddx " n

Xk=0

nk f (n k ) (x)g(k ) (x)#

=n

Xk =0

nk f (n k+1) (x)g(k ) (x) + f (n k ) (x)g(k +1) (x)

= f (n +1) (x)g(x) + f (n ) (x)g0(x) +n

Xk=1

nk f (n k +1) (x)g(k ) (x) + f (n k) (x)g(k+1) (x)

Now, we use the fact that for 1 k n; we have n +1k = n

k + nk 1

( ) : We continue:

= f (n +1) (x)g(x) + f (n ) (x)g0(x) +n

Xk=1 hn +1

k nk 1 i f (n k +1) (x)g(k ) (x) + f (n k ) (x)g(k+1) (x)

= f (n +1) (x)g(x) + f (n ) (x)g0(x) +n

Xk=1

n +1k f (n k +1) (x)g(k ) (x) + f (n k ) (x)g(k+1) (x)

n

Xk=1

nk 1 f (n k+1) (x)g(k ) (x) + f (n k) (x)g(k +1) (x)

= f (n +1) (x)g(x) + f (n ) (x)g0(x) +n

Xk=1

n +1k f (n k+1) (x)g(k ) (x) +

n

Xk=1

n +1k f (n k ) (x)g(k +1) (x)

n

Xk=1

nk 1 f (n k+1) (x)g(k ) (x)

n

Xk=1

nk 1 f (n k) (x)g(k +1) (x)

= f (n ) (x)g0(x) + "f (n +1) (x)g(x) +n

Xk =1

n +1k f (n k +1) (x)g(k ) (x)#+

n

Xk=1

n +1k f (n k) (x)g(k+1) (x)

n 1

Xk =0

nk f (n k ) (x)g(k+1) (x)

n

Xk =1

nk 1 f (n k ) (x)g(k+1) (x)

= f (n ) (x)g0(x) + " n

Xk=0

n +1k f (n k+1) (x)g(k ) (x)#+

n

Xk =1

n +1k f (n k ) (x)g(k +1) (x)

"f (n ) (x)g0(x) +n 1

Xk=1

nk f (n k) (x)g(k+1) (x)#"nf (x)g(n +1) (x) +

n 1

Xk =1

nk 1 f (n k) (x)g(k +1) (x)#

=

" n

Xk=0

n +1k f (n k+1) (x)g(k ) (x)

#+

n

Xk=1

n +1k f (n k) (x)g(k +1) (x)

"n 1

Xk =1 hnk + n

k 1 if (n k ) (x)g(k+1) (x)# nf (x)g(n +1) (x)

( )= " n

Xk=0

n +1k f (n k+1) (x)g(k ) (x)#+ "(n + 1) f (x)g(n +1) (x) +

n 1

Xk=1

n +1k f (n k ) (x)g(k+1) (x)#

"n 1

Xk =1

n +1k f (n k ) (x)g(k +1) (x)#nf (x)g(n +1) (x)

94



= " n

Xk=0

n +1k f (n k+1) (x)g(k ) (x)#+ ( n + 1) f (x)g(n +1) (x) nf (x)g(n +1) (x)

= " n

Xk=0

n +1k f (n k+1) (x)g(k ) (x)#+ f (x)g(n +1) (x)

=n +1

Xk =0

n +1k f (n k+1) (x)g(k ) (x):

———————————————————————————————————————————————–248. If x > 0; show that (1 + x)1=3 1 + 1

3 x 19 x2 5

81 x3 : Use this inequality to approximate 3p 1:2and 3p 2:———————————————————————————————————————————————–Proof. Let f (x) = (1 + x)1=3 =) f 0(x) = 1

3 (1 + x) 2=3 =) f 00(x) = 29 (1 + x) 5=3 :

Thus, f (0) = 1 ; f 0(0) = 13 ; and f 00(0) = 2

9 :Let x > 0: Then f ; f 0; and f 00 exist and are continuous on [0; x + 1] :Note that f 000(x) = 10

27 (1 + x) 8=3 exists in the interval (0; x + 1) :Hence, by Taylor’s theorem, f (x) = P 2(x) + R2(x) = 1 + 1

3 x 19 x2 + R2(x);

where R 2(x) = 13! f 000(c)x3 = 5

81 (1 + c) 8=3x3 for some point 0 < c < x:Note that if c > 0; then 1

1+ c < 1 and so (1 + c) 8=3 < 1:

Thus, (1 + x)1=3 1 + 13 x 1

9 x2 = R2(x) = 581 (1 + c) 8=3x3 < 5

81 x3 :Now, 3p 1:2 = f (0:2) 1 + 1

3 (0:2) 19 (0:2)2 = 1 : 062222 with an error 5

81 (0:2)3 = 0 :0005 :Also, 3p 2 = f (1) 1 + 1

3 19 = 1 : 2222 with an error 5

81 (1)3 = 0 :062:Compare these to 3p 1:2 = 1 : 062658 and 3p 2 = 1 : 259921 ; which we get if we use a calculator.

———————————————————————————————————————————————–249. If f (x) = ex ; show that the remainder term in Taylor’s Theorem converges to 0 as n ! 1 ; for each…xed x0 and x:———————————————————————————————————————————————–Proof. Fix x0 and x 2R: Then, by Taylor’s theorem, jRn (x)j= ec (x x 0 ) n +1

(n +1)! = ec jx x 0 jn +1

(n +1)! :

Now, let L = limn !1

jR n +1 (x )jjR n (x )j : Then L = lim

n !1ec jx x 0 jn +2 =(n +2)!ec jx x 0 jn +1 =(n +1)! = lim

n !1jx x 0 jn +2 = 0 < 1:

Hence, by Theorem 3.2.11, we get that limn

!1 jRn (x)

j= 0 and so, lim

n

!1Rn (x) = 0 :

———————————————————————————————————————————————–250. Let h(x) = e 1=x 2

for x 6= 00 for x = 0

: Show that h (n ) (0) = 0 for all n 2N: Conclude that the remainder

term in Taylor’s Theorem for x0 = 0 does not converge to 0 as n ! 1 ; for x 6= 0 : [Hint: By L’Hospital’sRule, lim

x! 0h (x )x k = 0 for any k 2N: Use Leibniz’s Rule to calculate h (n ) (x) for x 6= 0] :

———————————————————————————————————————————————–Proof. For x 6= 0 ; h(x) = e 1=x 2

=) h0(x) = 2x 3 e 1=x 2

= P 1( 1x )h(x); where P 1( 1

x ) is a polynomial in 1x :

Now, we use induction on n to prove that h(n ) (x) = P n ( 1x )h(x) ( ) ; where P n ( 1


Suppose that h (n 1) (x) = P n 1( 1x )h(x); where P n 1( 1


Note that the derivative P n 1( 1x ) 0 = Qn ( 1

x ) is another polynomial in 1x :

Then h (n ) (x) = P n 1( 1x ) 0h(x) + P n 1( 1

x )h0(x) = Qn ( 1x )h(x) + P n 1( 1

x )P 1( 1x )h(x)

= Qn (1

x ) + P n 1(1

x )P 1(1

x ) h(x) = P n (1

x )h(x); where P n (1

x ) = Qn (1

x ) + P n 1(1

x )P 1(1

x ) isa polynomial in 1x : Thus, ( ) follows.

Note that h0(0) = limx! 0

h (x ) h (0)x 0 = lim

x! 0e 1 =x 2

x = limy! 1

e y 2

1=y = limy! 1

yey 2

L’Hospital= lim

y! 11

2ye y 2 = 0 :

Actually, for all k 2N; limx! 0

h (x )x k = lim

x! 0e 1 =x 2

x k = limy! 1

e y 2

1=y k = limy! 1

yk

ey 2L’Hospital k times

= 0 :

Hence, limx! 0

P n ( 1x )h(x) = 0 for any polynomial P n ( 1

x ) ( ) :

Now, we use induction to prove that h (n ) (0) = 0 for all n 2N:

95



For n = 1 ; we showed it above. So, suppose h (n 1) (0) = 0Then, h (n ) (0) = lim

x! 0h ( n 1) (x ) h ( n 1) (0)

x 0

= limx! 0

h ( n 1) (x ) 0x 0 By induction hypothesis

= limx

!0

h ( n 1) (x )x

= limx! 0

P n 1 ( 1x )h (x )x For some polynomial P n 1( 1

x )

= limx! 0

Qn 1( 1x )h(x) Where Qn 1( 1

x ) = P n 1 ( 1x )

x

= 0 as in ( ) above.

———————————————————————————————————————————————–251. We wish to approximate sin by a polynomial on [ 1; 1] so that the error is less than 0:001. Show thatwe have sin x x x3

6 + x5

120 < 15040 for jxj 1:

———————————————————————————————————————————————–Proof. We have f (x) = sin x; f 0(x) = cos x; f 00(x) = sin x; f 000(x) = cos x; f (4) (x) = sin x;f (5) (x) = cos x; f (6) (x) = sin x; and f (7) (x) = cos x:Thus, f (0) = 0 ; f 0(0) = 1 ; f 00(0) = 0 ; f 000(0) = 1; f (4) (0) = 0 ; f (5) (0) = 1 ; and f (6) (0) = 0 :

Let 1 x 1:Then, by Taylor’s theorem, f (x) = P 6(x) + R6(x) =

6

Xk =0

f ( k ) (0)

k ! xk

+ 17! f

(7)

(c)x7

; forsome c between 0 and x:Now, jR6(x)j= jf (x) P 6(x)j= f (x) x x3

6 + x5

120 :

But, jR6(x)j= 17! f (7) (c)x7 = 1

7! j cos(c)j jxj7 1

7! = 15040 for all 1 x 1:

Thus, approximating sin by the Taylor’s 6th-polynomial on [ 1; 1] gives an error 17! = 0 :0001 0:001:

———————————————————————————————————————————————–252 . If f 2 R[a; b] and jf (x)j M for all x 2[a; b]; show that R b

a f M (b a):———————————————————————————————————————————————–Proof. We have M f (x) M for all x 2[a; b] and so R

ba M R

ba f (x) R

ba M:

(Recall that constant functions are Riemann integrable and R b

a M = M (b a)).Hence, M (b a)

R b

a f M (b a) and so

R b

a f M (b a):

———————————————————————————————————————————————–253. Suppose f is bounded on [a; b] and that there exists two sequences of tagged partitions of [a; b] such that:

P n ! 0 and:

Qn ! 0; but such that limn !1

S (f ;:

P n ) 6= limn !1

S (f ;:

Qn ): Show that f =2 R[a; b]:———————————————————————————————————————————————–Proof. Suppose that f 2 R[a; b] with R

ba f = L and let > 0 be given. Then, by de…nition, there exists > 0

such that for any tagged partition:

P with:

P < ; we have S (f ;:

P ) L < :

Since:

P n ! 0 and:

Qn ! 0; we can …nd N 2N ::

P n < and:

Qn < for all n N:

Hence, S (f ;:

P n ) L < and S (f ;:

Qn ) L < for all n N:

Since our choice of > 0 was arbitrary, we conclude that limn !1

S (f ;:

P n ) = L = limn !1

S (f ;:

Qn ):

———————————————————————————————————————————————–254. Let f be the Dirichlet function de…ned by f (x) = 1 if x 2Q\ [0; 1]

0 if x 2[0; 1] Q: Show that f =

2 R[0; 1]:

———————————————————————————————————————————————–Proof. Let P n = f[ j 1

n ; jn ]gn

j =1 be the partition of [0; 1] into n closed subintervals of equal length (= 1n ):

Then kP n k= 1n ! 0 as n ! 1 :

Now, let:

P n = P n with tags q j 2Q and let:

Qn = P n with tags r j =2Q:Then

:

P n ! 0,:

Qn ! 0:

Moreover, for all n 2N; we have S (f ;:

P n ) =n

Pj =1f (q j )( j

n j 1n ) =

n

Pj =1

1n = 1 ; and

96



S (f ;:

Qn ) =n

Pj =1f (r j )( j

n j 1n ) =

n

Pj =10 1

n = 0 :

Hence, limn !1

S (f ;:

P n ) = 1 6= 0 = limn !1

S (f ;:

Qn ):Thus, by the previous exercise, we conclude that f =2 R[0; 1]:

———————————————————————————————————————————————–

255. Suppose that f : [a; b] ! R and that f (x) = 0 except for a …nite number of points c1 ;:::;cn in [a; b]:Prove that f 2 R[a; b] and that R b

a f = 0 :———————————————————————————————————————————————–Proof. Let M = max fjf (c1)j; :::;jf (cn )jg and let > 0 be given. Here, there are n points where f is not 0;each of which can belong to 2 subintervals in a given tagged partition

:

P : Only these terms will make a nonzerocontribution to S (f ;

:

P ): Therefore, we choose = 2nM :

If :

P < ; let:

P 0be the subset of :

P with tags di¤erent from c1 ;:::;cn ; and let:

P 1be the subset of :

P with

tags at these points. Since S (f ;:

P 0) = 0 ; it is seen that S (f ;:

P ) = S (f ;:

P 1):Since there are at most 2n terms in S (f ;

:

P 1) and each term is between M and M ; we conclude that2n( M ) < S (f ;

:

P 1) < 2n(M ):Choosing =

2nM ; we get that < S (f ;:

P 1) = S (f ;:

P ) < and so S (f ;:

P ) 0 < :

Since our choice of the tagged partition :P with :P < was arbitrary, we conclude that S (f ; :P ) = 0

for all tagged partitions:

P of [a; b] with:

P < :

Hence, f 2 R[a; b] and that R b

a f = 0 :

———————————————————————————————————————————————–256. If g 2 R[a; b] and if f (x) = g(x) except for a …nite number of points in [a; b]; prove that f 2 R[a; b]and that R

ba f = R

ba g:

———————————————————————————————————————————————–Proof. Let h(x) = f (x) g(x): Then h : [a; b] ! R and h(x) = 0 except for a …nite number of points in [a; b]:Hence, by the previous exercise, h 2 R[a; b] and that R

ba h = 0 :

Now, f (x) = h(x) + g(x) and g ; h 2 R[a; b] =) f 2 R[a; b] and

R b

a f =

R b

a g +

R b

a h =

R b

a g:

———————————————————————————————————————————————–257. Let 0 a < b; let f (x) = x2 for x 2[a; b]; and let P = f[x i 1; x i ]gni=1 be a partition of [a; b]: For each i;let q i = q 1

3 x2i + x i 1x i + x2

i 1 :(a) Show that q i satis…es 0 x i 1 q i x i :(b) Show that f (q i )(x i x i 1) = 1

3 x3i x3

i 1 :(c) If

:

Q is the tagged partition with the same subintervals as P and the tags q i ; show thatS (f ;

:

Q) = 13 b3 a3 :

(d) Show that f 2 R[a; b] and R b

a f = R b

a x2dx = 13 b3 a3 :

———————————————————————————————————————————————–Proof. (a) We will show that 0 x2

i 1 q 2i x2i and since the function f (x) = x2 is increasing on [0;1 );

we will conclude that 0 x i 1 q i x i :Now, since 0 a x i 1 < x i b; we have that 2x2

i 1 x2i + x i x i 1 =) 2

3 x2i 1 1

3 x2i + x i x i 1

=) x2i 1

13 x

2i 1

13 x

2i + x i x i 1 =) x

2i 1

13 x

2i + x i x i 1 + x

2i 1 = q

2i :Also, x2

i + x i x i 1 2x2i =) 1

3 x2i + x i x i 1

23 x2

i =) 13 x2

i + x i x i 1 x2i 1

3 x2i

=) q 2i = 13 x2

i + x i x i 1 + x2i x2

i :Hence, 0 x2

i 1 q 2i x2i and so 0 x i 1 q i x i :

(b) Recall that a3 b3 = ( a b)(a2 + ab + b2):We have f (q i )(x i x i 1) = q 2i (x i x i 1) = 1

3 x2i + x i x i 1 + x2

i (x i x i 1) = 13 x3

i x3i 1 :

(c) We have S (f ;:

Q) =n

Pi=1f (q i )(x i x i 1)

( b)=

n

Pi=1

13 x3

i x3i 1 = 1

3

n

Pi =1x3

i x3i 1 = 1

3 (x3n x3

0)

97



= 13 b3 a3 :

(d) Let > 0 be given. Let:

P = f(I i ; t i )gni =1 be an arbitrary tagged partition of [a; b] with

:

P < ;(where > 0 is to be determined later) so that x i x i 1 < for i = 1 ;:::;n:Let

:

Q be the tagged partition with the same subintervals as:

P but with the tags q i ; i.e.,:

Q= f(I i ; q i )gni=1

with q i is in the statement of the exercise.Since both t i and q i 2I i and:

P =:

Q < ; we conclude that jt i q ij< for all 1 i n:Using the triangle inequality, one gets that

S (f ;:

P ) S (f ;:

Q) =n

Pi =1f (t i )(x i x i 1)

n

Pi =1f (q i )(x i x i 1) =

n

Pi =1[f (t i ) f (q i )] (x i x i 1)

=n

Pi=1t2i q 2i (x i x i 1)

n

Pi =1t2i q 2i (x i x i 1) =

n

Pi =1 jt i q ij(t i + q i ) (x i x i 1)n

Pi =1 (a + b) (x i x i 1) = (a + b)

n

Pi=1(x i x i 1) = (a + b) (b a) = (b2 a2):

Thus, choosing < b2 a 2 ; one gets that S (f ;

:

P ) S (f ;:

Q) < :

Since, by (c) , S (f ;:

Q) = 13 b3 a3 ; we conclude that S (f ;

:

P ) 13 b3 a3 < :

Hence, since our choice of the tagged partition:

P with:

P < was arbitrary, we conclude thatf 2 R[a; b] and R

ba f = R

ba x2dx = 1

3 b3 a3 :

———————————————————————————————————————————————–258. Consider the function h(x) = x + 1 if x 2Q\ [0; 1]

0 if x 2[0; 1] Q : Show that h =2 R[0; 1]:———————————————————————————————————————————————–Proof. Let 0 = 1

2 and let n 2N: Let P n = j 1n ; j

nnj =1 and consider the two tagged partitions

:

P n = j 1n ; j

n ; t jnj =1 and

:

Qn = j 1n ; j

n ; r jnj =1 ; where each t j 2Q and each r j is irrational.

Then,:

P n 1n and

:

Qn 1n :

Moreover, S (h;:

P n ) =n

Pj =1

h(t j )( jn j 1

n ) = 1n

n

Pj =1

(t j + 1) = 1n

n

Pj =1

t j!+ 1 1:

On the other hand, S (h; :Qn ) =n

Pj =1h(r j )( j

n j 1n ) = 0 :

Hence, S (h;:

P n ) S (h;:

Qn ) = " 1n

n

Pj =1t j!+ 1#0 1 > 0 :

This implies that h =2 R[0; 1]:

———————————————————————————————————————————————–259. Let H (x) = k if x = 1

k ; k 2N0 if x 2[0; 1] f1

k : k 2Ng: Show that H =2 R[0; 1]:

———————————————————————————————————————————————–Proof. Let 0 = 1



:

P n = j 1n ; j

n ; t jnj =1 and

:

Qn = j 1n ; j

n ; r jnj =1 ; where t1 = 1

n ; and all other t j are irrationals,and each r j is irrational.

Then,:

P n 1n and:

Qn 1n :

Moreover, S (H ;:

P n ) =n

Pj =1H (t j )( j

n j 1n ) = 1

n H (t1) = 1 :

On the other hand, S (H ;:

Qn ) =n

Pj =1H (r j )( j

n j 1n ) = 0 :

Hence, S (H ;:

P n ) S (H ;:

Qn ) = j1 0j= 1 > 0 :This implies that H =2 R[0; 1]:

———————————————————————————————————————————————–

98



260. If S (f ;:

P ) is any Riemann sum of f : [a; b] ! R, show that there exists a step function ' : [a; b] ! Rsuch that R

ba ' = S (f ;

:

P ):———————————————————————————————————————————————–Proof. Say, S (f ;

:

P ) =n

Pj =1f (t j )(x j x j 1): De…ne ' : [a; b] ! R by

' (x) = f (t j ) for x 2[x j 1 ; x j ] 1 j nThen ' is a step function and so ' 2 R[a; b]:

Thus, R b

a ' =n

Pj =1 R x j

x j 1' =

n

Pj =1 R x j

x j 1f (t j )dx =

n

Pj =1f (t j )R

x j

x j 1dx =

n

Pj =1f (t j )(x j x j 1) = S (f ;

:

P ):

———————————————————————————————————————————————–261. Suppose that f is continuous on [a; b]; that f (x) 0 for all x 2[a; b] and that R

ba f = 0 : Prove that

f (x) = 0 for all x 2[a; b]:———————————————————————————————————————————————–Proof. Suppose that there is an x0 2[a; b] such that f (x0) > 0: Then, by continuity of f ; we can …nda closed interval [c; d] [a; b] such that x0 2[c; d] and f (x) > 1

2 f (x0) for all x 2[c; d]:Now, 0 =

R

ba f =

R

ca f +

R

dc f +

R

bd f 0 +

R

dc f + 0 >

R

dc

12 f (x0)dx = 1

2 f (x0) (d c) > 0; a contradiction.Hence, f (x) = 0 for all x

2[a; b]:

———————————————————————————————————————————————–262. Show that the continuity hypothesis in the preceeding exercise cannot be dropped.———————————————————————————————————————————————–Proof. Let f (x) = 1 if x = a

0 if x 2(a; b] : Then, f (x) 0 for all x 2[a; b], and R b

a f = 0 :

Nevertheless, f is not identically 0 on [a; b]:

———————————————————————————————————————————————–263. If f and g are continuous on [a; b] and if R

ba f = R

ba g; prove that there exists c 2[a; b] such that

f (c) = g(c):———————————————————————————————————————————————–Proof. Let h(x) = f (x) g(x) for x 2[a; b]: Then h is continuous on [a; b] and R b

a h = 0 :

If h(x) 0 for all x

2[a; b]; then, by the above exercise, h(x) = 0 for all x

2[a; b].

Hence, f (x) = g(x) for all x 2[a; b]:

If h(x) 0 for all x 2[a; b]; then, similarly, h(x) = 0 for all x 2[a; b]:Hence, f (x) = g(x) for all x 2[a; b]:

If h(x1) < 0 and h(x2) > 0 for some x1 ; x2 2[a; b]; then, by Bolzano’s intermediate value property forcontinuous functions, 9x1 < c < x 2 such that h(c) = 0 : Hence, f (c) = g(c).

———————————————————————————————————————————————–264. Give an example of a function f : [a; b] ! R such that f 2 R[c; b] for each c 2(a; b) but f =2 R[a; b]:———————————————————————————————————————————————–Solution. Let f (x) = 0 if x = 0

1x if x 2(0; 1] : Then f is continuous on each closed interval [c; 1] where

c

2(0; 1) and so f

2 R[c; 1] for each c

2(0; 1):

Now, f is not bounded on [0; 1] and hence f =2 R[0; 1]: ———————————————————————————————————————————————–265. (Mean Value Theorem for Integrals ) If f is continuous on [a; b]; a < b; show that there exists c 2[a; b]such that R

ba f = f (c)(b a):

———————————————————————————————————————————————–Proof. Let m = inf

x2[a;b ]f (x) and M = sup

x2[a;b ]f (x): (They exist because f is continuous and [a; b] is compact).

Then R b

a f R b

a M = M (b a) and R b

a f R b

a m = m(b a):

Hence, m(b a) R b

a f M (b a) =) m 1b a R

ba f M

( )

:

99



But since f is continuos on [a; b] and m f (x) M , we get by Bolzano’s intermediate value theorem thatf assumes all the values between m and M: In particular, from ( ), we see that there exists c 2[a; b] suchthat f (c) = 1

b a R b

a f =) R b

a f = f (c)(b a):

———————————————————————————————————————————————–266 . If : [a; b] ! R takes on only a …nite number of distinct values. is a step function?———————————————————————————————————————————————–Proof. Not necessarily. Consider the Dirichlet function (x) = 1 for x 2[0; 1] \ Q

0 for x 2[0; 1]nQ :

Then takes on the values 0 and 1 on [0; 1], but cannot be a step function or else it has to be Riemannintegrable and we showed in a previous exercise that the Dirichlet function is not Riemann integrable.

———————————————————————————————————————————————–267. If (x) = x and ! (x) = x and if (x) f (x) ! (x) for all x 2[0; 1]; does it follow from the squeezetheorem that f 2 R[0; 1]?———————————————————————————————————————————————–Solution. No. Let f (x) = x for x 2[0; 1] \ Q

0 for x 2[0; 1]nQ: Then x f (x) x for all x 2[0; 1]:

Now, we show that f =2 R[0; 1]:Let 0 = 1



:

P n =j 1

n ; jn ; t j

n

j =1 and

:

Qn =j 1

n ; jn ; r j

n

j =1 ; where each t j = jn 2Q and each r j is irrational.

Then,:

P n 1n and

:

Qn 1n :

Moreover, S (f ;:

P n ) =n

Pj =1f (t j )( j

n j 1n ) = 1

n

n

Pj =1

jn = 1

n 2

n

Pj =1 j = n (n +1)

2n 2 = 12 + 1

2n 12 :

On the other hand, S (f ;:

Qn ) =n

Pj =1f (r j )( j

n j 1n ) = 0 :

Hence, S (f ;:

P n ) S (f ;:

Qn ) = 12 + 1

2n 0 12 = 0 :

This implies that f =2 R[0; 1]:

———————————————————————————————————————————————–268. If f is bounded by M on [a; b] and if the restriction of f to every interval [c; b] where c 2(a; b) isRiemann integrable, show that f 2 R[a; b] and that lim

c! a +

R b

c f =

R b

a f :

[Hint: Let c(x) = M for x 2[a; c)f (x) for x 2[c; b] and ! c(x) = M for x 2[a; c)

f (x) for x 2[c; b] :

Apply the squeeze theorem for c su¢ciently near a].———————————————————————————————————————————————–Proof. Let > 0 be given. Then there exists c 2(a; b) such that c a <

2M :

Let c(x) = M for x 2[a; c)f (x) for x 2[c; b] and ! c(x) = M for x 2[a; c)

f (x) for x 2[c; b] :

Then both c and ! c are Riemann integrable on [a; b] by the given.Also, since M f M on [a; b]; we have that c(x) f (x) ! c(x) for all x 2[a; b]:Moreover,

R b

a (! c c) =

R c

a (! c c) +

R b

c (! c c) =

R c

a (M ( M )) +

R b

c (f f ) =

R c

a 2M = 2 M (c a) < 2M

2M = :Hence, by the squeeze theorem, f 2 R[a; b]:

Now, R b

a f R b

c f = R c

a f R c

a M = M (c a) < M 2M =

2 < :

Thus, limc! a + R b

c f = R ba f :

———————————————————————————————————————————————–269. Show that g(x) = sin(1=x) for x 2(0; 1]

0 for x = 0 belongs to R[0; 1]:———————————————————————————————————————————————–Proof. Note that jg(x)j 1 for all x 2[0; 1] and that for any c 2(0; 1); g is continuous on [c; 1] and hence

100



g 2 R[c; 1]: Thus, by the previous exercise, we conclude that g 2 R[0; 1]:

———————————————————————————————————————————————–270. If f is bounded and there is a …nite set E such that f is continuous at every point of [a; b]nE; show thatf 2 R[a; b]:———————————————————————————————————————————————–Proof. Say, E =

fc1 ;:::;cn

gwhere a < c 1 < c 2 < < c n < b: (If E

\fa; b

g 6= ? ; then the proof is the same).

We will be done, by the additivity theorem, if we show that f 2 R[a; c1]; f 2 R[cn ; b]; and f 2 R[cj 1 ; cj ]for 2 j n. Since the idea of the following proof will be the same for proving all of the above statements,we will be done if we show that f 2 R[c1 ; c2]: So, let us do this.Fix 2(c1 ; c2): Then f is continuous on [ ; ] for every 2(c1 ; ) and so f 2 R[ ; ] for all 2(c1 ; ):Also, f is bounded on [c1; ]: Hence, by a previous exercise, f 2 R[c1 ; ] for every 2(c1 ; c2):Also, f is bounded on [c1; c2]: Therefore, by the same previous exercise, f 2 R[c1 ; c2]:

———————————————————————————————————————————————–271. Let f be continuous on [a; b]; let f (x) 0 for x 2[a; b]; and let

M n = R b

a f n1=n

:

Show that limn

!1

M n = supx

2[a;b ] ff (x)g:

———————————————————————————————————————————————–Proof. Since f is continuous on [a; b] and [a; b] is compact, M = sup

x2[a;b ] ff (x)g exists at some point c 2[a; b]:

If M = 0 ; then M n = 0 and so we are done. Thus, suppose that M > 0: Assume that c 2[a; b):(A similar argument to what follows takes care of the case c = b):

Let 0 < < M be given. Since f is continuous on [a; b]; we can …nd a > 0 such that 8x 2[c; c + ] [a; b];we have that f (x) M > 0 =) f n (x) (M )n for all x 2[c; c + ]:

Thus, R b

a f n R c+

c f n R c+

c (M )n = ( M )n =) M n (M ) 1=n for all n 2N (1) :

Now, f (x) M for all x 2[a; b] =) f n (x) M n for all x 2[a; b] =) R b

a f n R b

a M n = M n (b a)

=) M n M (b a)1=n for all n 2N (2) :

Therefore, (M ) 1=n M n M (b a)1=n for all n 2N:Hence, by the squeeze theorem for sequences, M lim

n !1M n M:

Since our choice of 0 < < M was arbitrary, we conclude that limn !1

M n = M:

———————————————————————————————————————————————–272. If n 2N and H n (x) = xn +1

n +1 for x 2[a; b]; show that the …rst form of the fundamental theorem of calculus implies that R b

a xn dx = bn +1 a n +1

n +1 : What is the set E for which H 0n (x) 6= xn ?———————————————————————————————————————————————–Proof. We have that:

(a) H n (x) = xn +1

n +1 is a polynomial and so is continuous on R: In particular, H n (x) is continuous on [a; b]:(b) H 0n (x) = xn for all x 2R: In particular, H 0n (x) = xn for all x 2[a; b]:

(c) h(x) = xn

is continuous on [a; b] and so h(x) 2 R[a; b]:Thus, by the …rst form of the fundamental theorem of calculus, we deduce that

R b

a xn dx = R b

a h(x)dx = H n (b) H n (a) = bn +1 a n +1

n +1 :

———————————————————————————————————————————————–273. If g(x) = x for jxj 1

x for jxj< 1 and if G(x) = 12 x2 1 ; show that R 3

2 g(x)dx = G(3) G( 2) = 5 =2:———————————————————————————————————————————————–Proof. We have that:

101



(a) G(x) = 12 x2 1 is continuous on [ 2; 3] since it is a composition of continuous functions on R.

(b) G0(x) = g(x) for x 2[ 2; 3]nf 1; 1g:(c) g(x) is continuous on [ 2; 3]nf 1; 1g and so g(x) 2 R[ 2; 3]:

Thus, by the …rst form of the fundamental theorem of calculus, we deduce that

R 3

2 g(x)dx = G(3) G( 2) = 4 (3=2) = 5 =2:

———————————————————————————————————————————————–274. Let B (x) = 1

2 x2 for x < 012 x2 for x 0 : Show that R

ba jxjdx = B(b) B (a):

———————————————————————————————————————————————–Proof. We have that:

(a) B (x) is continuous at x = 0 ; and so B (x) is continuous on [a; b]:

(b) B 0(x) = x for x < 0x for x > 0 = jxj for x 2[a; b]nf0g:

(c) jxj is continuous on [a; b] and so jxj 2 R[a; b]:

Thus, by the …rst form of the fundamental theorem of calculus, we deduce that

R b

a

jx

jdx = B(b) B (a):

———————————————————————————————————————————————–275. If f 2 R[a; b] and if c 2[a; b]; the function de…ned by F c(z) = R

zc f for z 2[a; b] is called the inde…nite

integral of f with basepoint c: Find a relation between F a and F c :———————————————————————————————————————————————–Proof. We have F a (z) = R

za f = R

ca f + R

zc f = R

ca f + F c(z): Hence, F c(z) = F a (z) R

ca f :

———————————————————————————————————————————————–

276. Let f : [0; 3] ! R be de…ned by f (x) = 8<:x for 0 x < 11 for 1 x < 2x for 2 x 3

9=;: Obtain formulas for F (x) = R

x0 f and

sketch the graphs of f and F: Where is F di¤erentiable? Evaluate F 0(x) at all such points.———————————————————————————————————————————————–Proof. If 0 x < 1; then F (x) =

R x

0 f =

R x

0 tdt = 12 x2 :

If 1 x < 2; then F (x) =

R x

0 f =

R 1

0 f +

R x

1 f =

R 1

0 tdt +

R x

1 dt = 1

2 + ( x 1) = x 1

2:

If 2 x 3; then F (x) = R x

0 f = R 1

0 f + R 2

1 f + R 3

2 f = R 1

0 tdt + R 2

1 dt + R x

2 tdt = 12 + 1 + x2

2 2 = 12 x2 1 :

Hence, F (x) = 8<:12 x2 for 0 x < 1

x 12 for 1 x < 2

12 x2 1 for 2 x 3

9=;and it is easy to see that it is continuous.

Here are the graphs of f and F :

f (x) F (x)

F is di¤erentiable everywhere on [0; 3] except when x = 2 :

102



This is because limh! 0+

F (2+ h ) F (2)h = lim

h ! 0+

12 ((2+ h ) 2 1) 3

2h = lim

h! 0+

2h + 12 h 2

h = limh! 0+

2 + 12 h = 2 ;

and limh ! 0

F (2+ h ) F (2)h = lim

h! 0+

(2+ h 12 ) 3

2h = lim

h ! 0+hh = lim

h! 0+1 = 1 :

Finally, F 0(x) =

8<:x for 0 x < 11 for 1 x < 2x for 2 < x 3 9=;

:

———————————————————————————————————————————————–277. If f : R! R is continuous and c > 0; de…ne g : R! R by g(x) = R

x + cx c f (t)dt. Show that g is

di¤erentiable on R and …nd g0(x):———————————————————————————————————————————————–Proof. we have g(x + h) g(x) = R

x + h + cx + h c f (t)dt R

x + cx c f (t)dt = R

x + c+ hx + c f (t)dt R

x c+ hx c f (t)dt:

Let > 0 be given. Since f is continuous at t = x c and t = x + c; we have that

For all t 2[x c; x c + h]; f (x c) 2 f (t) f (x c) +

2 ; and For all t 2[x + c; x + c + h]; f (x + c)

2 f (t) f (x + c) + 2 :

Thus, we ge that

R x c+ h

x c f (x c) 2 dt R x c+ h

x c f (t)dt R x c+ h

x c f (x c) + 2 dt; and

R x + c+ h

x + c f (x + c) 2 dt R

x + c+ hx + c f (t)dt R

x + c+ hx + c f (x + c) +

2 dt:

This can be rewritten as

f (x c) 2 h R

x c+ hx c f (t)dt f (x c) +

2 h; andf (x + c)

2 h R x + c+ h

x + c f (t)dt f (x + c) + 2 h:

Therefore,

f (x c) + 2 h R

x c+ hx c f (t)dt f (x c)

2 h; andf (x + c)

2 h

R x + c+ h

x + c f (t)dt f (x + c) + 2 h:

Adding the two inequalities above and dividing by h we get:

(f (x + c) f (x c) ) hR x + c+ h

x + c f (t)dt R x c+ h

x c f (t)dti=h (f (x + c) f (x c) + )

But hR x + c+ h

x + c f (t)dt R x c+ h

x c f (t)dti=h = g(x + h ) g(x )h :

Hence, g(x + h ) g(x )

h (f (x + c) f (x c)) and so g(x + h ) g(x )h (f (x + c) f (x c)) :

Since our choice of > 0 was arbitrary, we conclude that g0(x) = limh! 0

g(x + h ) g(x )h = f (x + c) f (x c):

———————————————————————————————————————————————–278. If f : [0; 1]

!R is continuous and

R x

0 f =

R 1

x f for all x

2[0; 1]; show that f (x) = 0 for all x

2[0; 1]:

———————————————————————————————————————————————–Proof. we have R

10 f = R

11 f = 0 : Also, R

10 f = R

x0 f + R

1x f = R

x0 f + R

x0 f = 2 R

x0 f:

Hence, R x

0 f = 0 for all x 2[0; 1]:Let F (x) = R

x0 f (t)dt: Then, since f is continuous on [0; 1]; F is an antiderivative for f , i.e.,

F 0(x) = f (x) for all x 2[0; 1]:Moreover, F (x) = 0 for all x 2[0; 1] =) F 0(x) = 0 for all x 2(0; 1):Thus, f (x) = 0 for all x 2(0; 1) and by continuity of f ; we get that f (x) = 0 for all x 2[0; 1]:

———————————————————————————————————————————————–279. (a) If Z 1 and Z 2 are null sets, show that Z 1 [ Z 2 is a null set.

103



(b) More generally, if Z n is a null set for each n 2N; show that 1


———————————————————————————————————————————————–Proof. Since (a) is a special case of (b) where Z n = ? for all n 3; it su¢ces to prove (b).Let > 0 be given. Since Z n is a null set, there exists a countable collection of open intervals fJ nk g1k=1

such that Z n

1

Sk =1

J nk and 1

Pk =1 j

J nk

j<

2n : (Here,

jJ nk

j denotes the length of the open interval J nk ):

Thus, 1

Sn =1Z n

1

Sn;k =1J nk and 1

Pn;k =1 jJ nk j= 1

Pn =1

1

Pk=1 jJ nk j < 1

Pn =12n = 1

Pn =1

12n = :

Hence, 1


———————————————————————————————————————————————–280. Let f ; g 2 R[a; b]:(a) If t 2R; show that R

ba (tf g)2 0:

(b) Use (a) to show that 2 R b

a f g tR b

a f 2 + 1t R

ba g2 for t > 0:

(c) If R ba f 2 = 0 ; show that R b

a f g = 0 ; and if R ba g2 = 0 ; show that R b

a f g = 0 :(d) Prove the following Schwartz Inequality :

R ba f g

2

R ba jfg j

2

R ba f 2 R

ba g2

———————————————————————————————————————————————–Proof. (a) Let t 2R: Since f 2 R[a; b]; tf 2 R[a; b]; and since g 2 R[a; b]; tf g 2 R[a; b]:By the product rule for integrals, we get that (tf g)2

2 R[a; b]:Since (tf g)2 0; we have that R

ba (tf g)2 0:

(b) We have 0 R b

a (tf + g)2 = R b

a t2f 2 + 2 tfg + g2 = t2 R b

a f 2 + 2 tR b

a f g + R b

a g2 :Hence, since t > 0; we can divide by t to get:

0 tR b

a f 2 + 2 R b

a f g + 1t R

ba g2 =) 2R

ba f g tR

ba f 2 + 1

t R b

a g2 (1)

:

Also, we have 0

R b

a (tf

g)2 =

R b

a t2f 2 2tfg + g2 = t2

R b

a f 2 2t

R b

a f g +

R b

a g2 :Hence, since t > 0; we can divide by t to get:

0 tR ba f 2 2R b

a f g + 1t R b

a g2 =) 2R ba f g tR b

a f 2 + 1t R b

a g2(2)

:

By (1) and (2) , we get that 2 R b

a f g tR b

a f 2 + 1t R

ba g2 :

(c) Suppose that R b

a f 2 = 0 : Then by (b), R b

a f g 12t R

ba g2 for all t > 0: Letting t ! 1 ; we get that

R ba f g = 0 and so R b

a f g = 0 :

Suppose that R b

a g2 = 0 : Then by (b), R b

a f g t2 R

ba f 2 for all t > 0: Letting t ! 0; we get that

R ba f g = 0 and so R b

a f g = 0 :

(d) If R b

a f 2

= 0 or R b

a g2

= 0 ; then, by (c), R b

a f g = 0 and the inequality follows trivially.So, suppose that R

ba f 2 > 0 and R

ba g2 > 0:

Let t = (R ba g 2 )1 = 2

(R ba t 2 )1 = 2 : Then t > 0; and so by (b), 2 R

ba f g tR

ba f 2 + 1

t R b

a g2 = 2 R b

a f 21=2

R b

a g21=2

:

Thus, R b

a f g R b

a f 21=2

R b

a g21=2

:

Squaring, we get R b

a f g2

R b

a f 2 R b

a g2 :

104



Now, R b

a f g R b

a jfg j=) R b

a f g2

R b

a jfg j2

( )

:

Also, applying the same argument to jf j and jgj; we get that

R b

a jfg j2

=

R b

a jfg j2

=

R b

a jf j jgj2

R b

a jf j2

R b

a jgj2 =

R b

a f 2

R b

a g2 :

Hence, R b

a jfg j 2

R b

a f 2 R b

a g2 ( ) :

Hence, by ( ) and ( ), we get that R b

a f g2

R b

a jfg j2

R b

a f 2 R b

a g2 :

———————————————————————————————————————————————–281 . Show that lim

n !1x

x + n = 0 for all x 2R; x 0:———————————————————————————————————————————————–Proof. If x = 0 ; then x

x + n = 0 for all n 2N and so limn !1

xx + n = 0 :

If x > 0; then 0 < xx + n x

n and since limn !1

xn = x lim

n !11n = 0 ; the squeeze theorem implies that

limn !1

xx + n = 0 : Thus, lim

n !1x

x + n = 0 for all x 2R; x 0:

———————————————————————————————————————————————–282 . Show that lim

n !1nx

1+ n 2 x 2 = 0 for all x 2R:———————————————————————————————————————————————–Proof. If x = 0 ; then nx

1+ n 2 x 2 = 0 for all n 2N and so limn !1

nx1+ n 2 x 2 = 0 :

If x > 0; then 0 < nx1+ n 2 x 2 1

nx and since limn !1

1nx = 1

x limn !1

1n = 0 ; the squeeze theorem implies that

Thus, limn !1

nx1+ n 2 x 2 = 0 .

If x < 0; then 1nx nx

1+ n 2 x 2 < 0 and since limn !1

1nx = 1

x limn !1

1n = 0 ; the squeeze theorem implies that

Thus, limn !1

nx1+ n 2 x 2 = 0 .

Therefore, limn !1

nx1+ n 2 x 2 = 0 for all x 2R:

———————————————————————————————————————————————–283 . Evaluate lim

n !1nx

1+ nx for x 2R; x 0:———————————————————————————————————————————————–Proof. If x = 0 ; then nx

1+ nx = 0 for all n 2N and so limn !1

nx1+ nx = 0 :

If x > 0; thennx

1+ nx 1 = 11+ nx ! 0 as n ! 1 :

Hence, limn !1

nx1+ nx = 0 if x = 0

1 if x > 0 :

———————————————————————————————————————————————–284 . Evaluate lim

n !1x n

1+ x n for x 2R; x 0:———————————————————————————————————————————————–Proof. If x = 0 ; then xn

1+ x n = 0 for all n 2N and so limn !1

x n

1+ x n = 0 :

If 0 < x < 1; then xn

1+ x n ! 0 as n ! 1 (since xn ! 0 as n ! 1 ): If x = 1 ; then xn

1+ x n = 11+1 = 1

2 for all n 2N and so limn !1

x n

1+ x n = 12 :

If x > 1; then x n

1+ x n 1 = 11+ x n ! 0 as n ! 1 (since xn ! 1 as n ! 1 ):

Hence, limn !1x n

1+ x n = 8<:0 if 0 x < 112 if x = 11 if x > 1 9=;

:

———————————————————————————————————————————————–285 . Evaluate lim

n !1sin( nx )1+ nx for x 2R; x 0:

———————————————————————————————————————————————–Proof. If x = 0 ; then sin( nx )

1+ nx = 0 for all n 2N and so limn !1

sin( nx )1+ nx = 0 :

If x > 0; then 0 sin( nx )1+ nx

11+ nx < 1

nx ! 0 as n ! 1 .

Hence, by the squeeze theorem, limn !1

sin( nx )1+ nx = 0 :

105



Thus, limn !1

sin( nx )1+ nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–286 . Evaluate lim

n !1e nx for x 2R; x 0:

———————————————————————————————————————————————–Proof. If x = 0 ; then e nx = 1 for all n 2N and so lim

n !1e nx = 1 :

If x > 0; then 0 < ex

< 1 and so limn !1 (ex

)n

= 0 :Thus, lim

n !1e nx = 1 if x = 0

0 if x > 0 :

———————————————————————————————————————————————–287 . Show that lim

n !1xe nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–Proof. If x = 0 ; then xe nx = 0 for all n 2N and so lim

n !1xe nx = 0 :

If x > 0; then 0 < e x < 1 and so limn !1

x (e x )n = x limn !1

(e x )n = 0 :Thus, lim

n !1xe nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–288 . Show that lim

n !1x2e nx = 0 and lim

n !1n2x2e nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–

Proof. If x = 0 ; then x2

enx

= 0 for all n 2N and so limn !1 x2

enx

= 0 : If x > 0; then 0 < e x < 1 and so lim

n !1x2 (e x )n = x2 lim

n !1(e x )n = 0 :

Thus, limn !1

x2e nx = 0 for x 2R; x 0:

If x = 0 ; then n 2x2e nx = 0 for all n 2N and so limn !1

n2x2e nx = 0 :

If x > 0; then recall that enx = 1

Pk=0

n k x k

k! and so, since x > 0; enx n 3 x 3

3! :

Hence, e nx 6n 3 x 3 and so n2x2e nx 6n 2 x 2

n 3 x 3 = 6nx ! 0 as n ! 1 :

Therefore, limn !1

n2x2e nx = 0 for x 2R; x 0:

———————————————————————————————————————————————–289 . Show that lim

n !1(cos x)2n exists for all x 2R. What is its limit?

———————————————————————————————————————————————–Proof. If x =2Z; then 1 < cos x < 1 and so, 0 (cos x)2 < 1 =) limn !1 (cos x)2n = 0 : If x 2Z; then cos x = 1 and so, (cos x)2 = 1 = ) (cos x)2n = 1 for all n 2N=) lim

n !1(cos x)2n = 1 :

Hence, limn !1

(cos x)2n = 0 if x =2Z1 if x 2Z :

———————————————————————————————————————————————–290 . Show that if a > 0; then the convergence of the sequence n x

x + n o1n =1

is uniform on [0; a]; but notuniform on [0;1 ):———————————————————————————————————————————————–Proof. Let f n (x) = x

x + n for n = 1 ; 2; 3;::: and x 0:Then jf n (x)j< 1 for all n and for all x 0:We showed in an exercise above that lim

n !1f n (x) = 0 for all x 0:

Also, kf n (x) 0k[0;a ] = supx2[0;a ]

xx + n =

aa + n ! 0 as n ! 1 : (Note that if 0 x < y; then

xx + n

yy+ n ):

Therefore, the convergence of ff n (x)g1n =1 is uniform on [0; a]:

To show that the convergence is not uniform on [0;1 ); let 0 = 14 and note that f n (n) = n

n + n = 12 > 0 :

———————————————————————————————————————————————–291 . Show that if a > 0; then the convergence of the sequence n nx

1+ n 2 x 2 o1n =1

is uniform on [a; 1 ); but notuniform on [0;1 ):———————————————————————————————————————————————–Proof. Let f n (x) = nx

1+ n 2 x 2 for n = 1 ; 2; 3;::: and x 0: Let > 0 be given.

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Then 9n 2N : n > 1a and so 1

an < :Hence, for all x a; we have that jf n (x)j 1

nx 1an < :

Therefore, the convergence of ff n (x)g1n =1 is uniform on [a; 1 ):

To show that the convergence is not uniform on [0;1 ); let 0 = 14 and note that f n ( 1

n ) = 12 > 0 :

———————————————————————————————————————————————–292 . Show that if 0 < b < 1; then the convergence of the sequence n

xn1+ x n o1n =1

is uniform on [0; b]; but notuniform on [0; 1]:———————————————————————————————————————————————–Proof. Let f n (x) = xn

1+ x n for n = 1 ; 2; 3;::: and x 2[0; b]:Then jf n (x)j 1 for all n and for all x 2[0; b]:We showed in an exercise above that lim

n !1f n (x) = 0 for all x 2[0; b]:

Also, kf n (x) 0k[0;b] = supx2[0;b]

x n

1+ x n = bn

1+ bn ! 0 as n ! 1 :

Therefore, the convergence of ff n (x)g1n =1 is uniform on [0; b]:

To show that the convergence is not uniform on [0; 1]; let 0 = 14 and note that f n ( 1

21 =n ) = 13 > 0 :

———————————————————————————————————————————————–293 . Show that if a > 0; then the convergence of the sequence n

sin( nx )1+ nx o1n =1is uniform on [a; 1 ); but not

uniform on [0;1 ):———————————————————————————————————————————————–Proof. Let f n (x) = sin( nx )

1+ nx for n = 1 ; 2; 3;::: and x 2[a; 1 ): Let > 0 be given. Then there is an N 2Nsuch that 1

Na < : Hence, for all n N; we have that

jf n (x)j= sin( nx )1+ nx

11+ nx 1

nx 1na 1

Na < for all x 2[a; 1 ):Therefore, the convergence of ff n (x)g1n =1 is uniform on [a; 1 ):

To show that the convergence is not uniform on [0; 1]; let 0 = 14 and note that f n (

2n ) = 22+ > 0 :

———————————————————————————————————————————————–294 . Show that the sequence x2e nx converges uniformly on [0;1 ):———————————————————————————————————————————————–Proof. Let f n (x) = x2e nx for n = 1 ; 2; 3;::: and x 2[0;1 ):We showed in an exercise above that lim

n !1f n (x) = 0 for all x 2[0;1 ):

To calculate the uniform norm of f n (x) 0 = f n (x) on [0;1 ); we …nd the derivative and solve

f 0n (x) = e nx (2x nx 2) = 0

to obtain the points yn = 0 and xn = 2n :

Using the …rst derivative test, we get that f n (x) attains its maximum on [0;1 ) at xn = 2n :

Therefore, we obtain

kf n k[0;1 ) = f n (xn ) = f n ( 2n ) = 4

n 2 e 2 ! 0 as n ! 1 :

Thus we see that convergence is uniform on [0;1 ):

———————————————————————————————————————————————–295 . Show that if a > 0; then the sequence n2x2e nx 1

n =1 converges uniformly on [a; 1 ); but it does notconverge uniformly on [0;1 ):———————————————————————————————————————————————–Proof. Let f n (x) = n2x2e nx for n = 1 ; 2; 3;::: and x 2[a; 1 ):We showed in an exercise above that lim

n !1f n (x) = 0 for all x 2[0;1 ):

Now, we can …nd N 2N large enough so that 2N < a and so 2

n < a for all n N:Let n N: To calculate the uniform norm of f n (x) 0 = f n (x) on [a; 1 ); we …nd the derivative and solve

107





Thus, jf n (x)gn (x) f (x)g(x)j= jf n (x)gn (x) f n (x)g(x) + f n (x)g(x) f (x)g(x)j= jf n (x) [gn (x) g(x)] + g(x) [f n (x) f (x)]jjf n (x)j jgn (x) g(x)j+ jg(x)j jf n (x) f (x)j< M

2M + L 2L = ; for all x 2A and all n N:

Therefore, f n gn

A

fg:

———————————————————————————————————————————————–299 . Let (f n ) be a sequence of functions that converges uniformly to f on A and that satis…es jf n (x)j M for all n 2N and all x 2A: If g is continuous on the interval [ M; M ]; show that the sequence (g f n )converges uniformly to g f on A:———————————————————————————————————————————————–Proof. We …rst show that jf (x)j M for all x 2A:Let > 0 be given. Since f n

Af; we can …nd N 2N : jf n (x) f (x)j< for all x 2A and all n N:

Hence, jf (x)j + jf n (x)j + M:Since our choice of > 0 was arbitrary, we conclude that jf (x)j M for all x 2A:

Now, we go back to the exercise. Suppose that g is continuous on the interval [ M; M ]:Then g is uniformly continuous on the interval [ M; M ]:Let > 0 be given. There exists > 0 : whenever x; y 2[ M; M ] with jx yj< ; we have thatjg(x) g(y)j< :For this ; since f n

Af; we can …nd N 2N : jf n (x) f (x)j< for all x 2A and all n N:

Thus, for all x 2A and all n N; we have that

j(g f n ) (x) (g f ) (x)j= jg (f n (x)) g(f (x))j< since f n (x) and f (x) are in [ M; M ] for all x 2Aand jf n (x) f (x)j< for all x 2A and all n N:

Therefore, (g f n )A

(g f ) :

———————————————————————————————————————————————–300 . Show that the sequence xn

1+ x n does not converge uniformly on [0; 2] by showing that the limitfunction is not continuous on [0; 2]:———————————————————————————————————————————————–Proof. If x = 0 ; then x

n

1+ x n = 0 for all n 2N and so limn !1x

n

1+ x n = 0 :

If 0 < x < 1; then limn !1

x n

1+ x n =lim

n !1x n

1+ limn !1

x n = 01+0 = 0 :

If x = 1 ; then xn

1+ x n = 12 for all n 2N and so lim

n !1x n

1+ x n = 12 :

If 1 < x 2; then limn !1

x n

1+ x n = 1 . (note that x n

1+ x n 1 = 11+ x n ! 0 as n ! 1 ):

Hence, the sequence of continuous functions xn

1+ x n converges to the function f de…ned by:

f (x) = 8<:

0 if 0 x < 11=2 if x = 11 if 1 < x 2

9=;

:

Since f is not continuous on [0; 2]; the convergence cannot be uniform on [0; 2]: ———————————————————————————————————————————————–301 . Construct a sequence of functions on [0; 1] each of which is discontinuous at each point of [0; 1]and which converges uniformly to a function that is continuous at every point.———————————————————————————————————————————————–Solution. For n = 1 ; 2; :::; de…ne f n : [0; 1] ! R as follows:

f n (x) = 1

n if x 2Q\ [0; 1]0 if x 2[0; 1]nQ :

109



Note that each f n is discontinuous at each point of [0; 1].Also, jf n (x) 0j 1

n for all n 2N and all x 2[0; 1]:Thus, the sequence (f n ) converges uniformly to the zero function on [0; 1] which is continuouseverywhere on [0; 1]:

———————————————————————————————————————————————–302 . Suppose (f n ) is a sequence of continuous functions on an interval I that converges uniformly on I toa function f : If (xn ) I converges to x0 2I; show that limn !1 (f n (xn )) = f (x0) :———————————————————————————————————————————————–Proof. Let > 0 be given. Since each f n is continuous and f n

I f; we conclude that f is continuous on I :

Thus, we can …nd N 2N such that:

(i) jf n (x) f (x)j< 2 for all x 2I and all n N: In particular, jf n (xn ) f (xn )j<

2 for all n N:(This is because f n

I f ):

(ii) jf (xn ) f (x0)j< 2 for all n N: (This is by continuity of f ):

Hence, for all n N; jf n (xn ) f (x0)j= jf n (xn ) f (xn ) + f (xn ) f (x0)j jf n (xn ) f (xn )j+ jf (xn ) f (x0)j< 2 +

2 = :

This is exactly what it means for limn !1 (f n (xn )) to be f (x0) : ———————————————————————————————————————————————–303 . Let f : R! R be uniformly continuous on R and let f n (x) = f (x + 1

n ) for x 2R: Show that (f n )converges uniformly on R to f :———————————————————————————————————————————————–Proof. Let > 0 be given. Then 9 > 0 : whenever x; y 2R with jx yj< ; we have jf (x) f (y)j< :Let N 2N be large enough so that 1

N < : Then 1n 1

N < for all n N:Let x 2R: Then x (x + 1

n ) = 1n < for all n N:

Hence, f (x) f (x + 1n ) < for all n N:

Thus, since our choice of x 2R was arbitrary, we have proved that:

8 > 0 9N 2N : 8x 2R; jf (x) f n (x)j= f (x) f (x + 1n ) < for all n N:

This is exactly what it means for f n (x) to converge to f (x) uniformly on R. ———————————————————————————————————————————————–304 . Let f n (x) = 1

(1+ x ) n for x 2[0; 1]: Find the pointwise limit f of the sequence (f n ) on [0; 1]:Does (f n ) converge uniformly to f on [0; 1]?———————————————————————————————————————————————–Proof. For x = 0 ; f n (x) = 1

(1+0) n = 1 for all n and so limn !1

f n (0) = 1 :

If 0 < x < 1; then 1 + x > 1 =) (1 + x)n! 1 as n ! 1 =) 1

(1+ x ) n ! 0 as n ! 1 :If x = 1 ; then f n (x) = 1

2n ! 0 as n ! 1 :

Thus, the pointwise limit f (x) = 1 if x = 00 if 0 < x 1 :

Since each f n is continuous on [0; 1]; but f (x) is not, we conclude that the convergence cannot beuniform.

———————————————————————————————————————————————–305 . Suppose that the sequence (f n ) converges uniformly to f on the set A; and suppose that each f nis bounded on A: (That is, for each n there is a constant M n such that jf n (x)j M n for all x 2A):Show that the function f is bounded on A:———————————————————————————————————————————————–Proof. Since (f n ) converges uniformly to f on the set A; 9N 2N :

jf n (x) f (x)j< 1 for all x 2A and all n N

110





Proof. It is easy to check, as we did many times before, that f (x) = 0 if x = 01 if 0 < x 1 :

Note that each f n is continuous but f is not continuous, and so the convergence cannot be uniform.Since f is continuous except at x = 1 ; we conclude that f is integrable and R

10 f (x)dx = R

10 1dx = 1 :

Now,

R 1

0 f n (x)dx =

R 1

0nx

1+ nx dx =

R 1

0nx +1 1

1+ nx dx =

R 1

0 1 11+ nx dx = 1

R 1

01

1+ nx dx = 1 1n ln j1 + nx j

10

= 1 ln(1+ n )

n ! 1 as n ! 1 :Hence, R 1

0 f (x)dx = 1 = limn !1 R 1

0 f n (x)dx:

———————————————————————————————————————————————–311 . Let gn (x) = nx (1 x)n for x 2[0; 1]; n 2N: Discuss the convergence of (gn ) and R

10 gn (x)dx :

———————————————————————————————————————————————–Proof. It is clear that gn (x) converges pointwise to the function g(x) = 0 on [0; 1]:This convergence is not uniform because gn ( 1

n ) = n 1n (1 1

n )n = (1 1n )n ! e 1 as n ! 1 :

By taking the …rst derivative, it can be easily checked that gn (x) attains its maximum at the point xn = 11+ n :

Hence, kgn (x)k[0;1] = gn ( 11+ n ) = n

1+ n (1 11+ n )n 1 1 = 1 for all n 2N:

Note that each gn 2 R[0; 1] and g 2 R[0; 1]: Also, we showed that jgn (x)j 1 for all x 2[0; 1]; n 2N:Thus, by the bounded convergence theorem, we conclude that lim

n !1 R 1

0 gn (x)dx = R 1

0 g(x)dx = 0 :

———————————————————————————————————————————————–

312 . Let fr 1 ; r 2 ; ; rn ; g be an enumeration of the rational numbers in [0; 1]; and let

f n (x) = 1 if x 2 fr 1 ; r 2 ; ; rn g0 otherwise :

Show that f n 2 R[0; 1] for each n 2N; that f 1(x) f 2(x) f n (x) ; and that f (x) = limn !1

(f n (x))is the Dirichlet function which is not Riemann integrable on [0; 1]:———————————————————————————————————————————————–Proof. Each f n is continuous on [0; 1] except on a …nite set fr 1; r 2 ; ; rn g: Hence, each f n 2 R[0; 1]:By de…nition, f n (x) = f n +1 (x) for all x 2[0; 1]nfr n +1 g and f n (r n +1 ) = 0 but f n +1 (r n +1 ) = 1 :Hence, f n (x) f n +1 (x) for all x 2[0; 1] and for all n 2N: That is, f 1(x) f 2(x) f n (x) :Note that if x 2[0; 1]nQ then f n (x) = 0 for all n 2N and so lim

n !1(f n (x)) = 0 :

Also, if x

2[0; 1]

\Q; then x = rN for some N

2N and so f n (x) = f n (r N ) = 1 for all n N:

Hence, limn !1(f n (x)) = 1 :

Thus, the pointwise limit of (f n ) is the function f de…ned by:

f (x) = 1 if x 2Q\ [0; 1]0 if x 2[0; 1]nQ

:

This is precisely the Dirichlet function.

———————————————————————————————————————————————–313 . Show that if a convergent series contains only a …nite number of negative terms, then it is absolutelyconvergent.———————————————————————————————————————————————–Proof. Let 1

Pn =1

xn be such a series. Let N 2N be large enough so that xn 0 for all n N:

Since 1Pn =1xn < 1 ; we get that 1Pn = N

xn < 1 and so 1Pn = N jxn j= 1Pn = N xn is absolutely convergent.

Say, 1

Pn = N xn = a < 1 and let b =

N 1

Pn =0 jxn j:

Then 1

Pn =1 jxn j= b + a < 1 and so 1

Pn =1xn is absolutely convergent.

———————————————————————————————————————————————–314 . Show that if a series is conditionally convergent, then the series obtained from its positive terms isdivergent and the series obtained from its negative terms is divergent.———————————————————————————————————————————————–

112



Proof. Let 1

Pn =1xn be such a series. For n = 1 ; 2; 3; :::; de…ne the sequences (an ) and (bn ) as follows:

an = xn if xn > 00 if xn 0 and bn = 0 if xn 0

xn if xn < 0 :

Then 1Pn =1an is the series of positive terms of 1Pn =1

xn and 1Pn =1bn is the series of negative terms of 1Pn =1

xn :

Suppose to the contrary that 1

Pn =1an < 1 :

Then, since 1

Pn =1xn < 1 ; we get that 1

Pn =1xn

1

Pn =1an = 1

Pn =1(xn an ) = 1

Pn =1bn < 1 :

Since 1

Pn =1 jan j= 1

Pn =1an < 1 and 1

Pn =1 jbn j= 1

Pn =1bn = 1

Pn =1bn < 1 ; we conclude that

1

Pn =1 jxn j= 1

Pn =1(jan j+ jbn j) =

1

Pn =1 jan j+ 1

Pn =1 jbn j< 1 :

Hence, 1

Pn =1

xn is absolutely convergent. A contradiction.

Therefore, 1

Pn =1an = 1 : Similarly, 1

Pn =1bn = 1 :

———————————————————————————————————————————————–315 . If 1

Pn =1xn is conditionally convergent, give an argument to show that there exists a rearrangement

whose partial sums diverge to 1:———————————————————————————————————————————————–Proof. By the previous problem, the series 1

Pn =1an and 1

Pn =1bn of positive and negative terms of 1

Pn =1xn are

both divergent. That is, 1

Pn =1an = + 1 and 1

Pn =1bn = 1 ( ) :

Consider the rearrangement de…ned as follows: (we can do this because of ( ))

Take poisitve terms from 1Pn =1an until the partial sum exceeds 1, then take negative terms from 1Pn =1

bn

until the partial sum is less than 1, then take positive terms until the partial sum exceeds 2, then negativeterms untill the partial sum is less than 2, ... and so on.

———————————————————————————————————————————————–316 . Find an explicit expression for the n -th partial sum of 1

Pn =2ln 1 1

n 2 and show that this series

converges to ln 2: Is this convergence absolute?———————————————————————————————————————————————–Solution. sn =

n

Pk=2ln 1 1

k 2 = ln n

Qk =21 1

k 2 = ln n

Qk=2

k 2 1k 2 = ln

n

Qk=2

(k 1)( k +1)k 2

= ln (n 1)!( n +1)!2( n !)2 = ln n +1

2n :

Thus, limn !1

sn = limn !1

ln n +12n = ln lim

n !1

n +12n = ln(1 =2) = ln 2: (The second equality follows because

ln x is continuous at x = 1 =2 and n +12n is a sequence that converges to 1=2).

Now, note that 0 < 1 1n 2 < 1 for all n 2 and so ln 1 1

n 2 < 0 for all n 2:Since all the terms of the series are negative, the convergence is indeed absolute.

———————————————————————————————————————————————–317 . (a) If 1

Pn =1an is absolutely convergent and (bn ) is a bounded sequence, show that 1

Pn =1an bn is absolutely

convergent.

(b) Give an example to show that if the convergence of 1

Pn =1an is conditional and (bn ) is a bounded

113



sequence, then 1

Pn =1an bn may diverge.

———————————————————————————————————————————————–Proof. (a) Suppose that jbn j M for all n 2N: Then jan bn j M jan j for all n 2N; and so1

Pn =1 jan bn j

1

Pn =1

M jan j= M 1

Pn =1 jan j< 1 since 1

Pn =1

an is absolutely convergent.

Hence, 1Pn =1an bn is absolutely convergent.

(b) Let an = ( 1) n

n and bn = ( 1)n for n = 1 ; 2; 3; ::::

Then 1

Pn =1an is conditionally convergent and (bn ) is a bounded sequence.

Also, 1

Pn =1an bn = 1

Pn =1

( 1) n

n ( 1)n = 1

Pn =1

1n = 1 :

———————————————————————————————————————————————–318 . Give an example of a convergent series 1

Pn =1an such that 1

Pn =1a2

n is not convergent.———————————————————————————————————————————————–Solution. Let an = ( 1) n

p n for n 2N: Then 1

Pn =1an is convergent, but 1

Pn =1a2

n = 1

Pn =1

1n = 1 :

———————————————————————————————————————————————–319 . Give an example of a divergent series 1Pn =1

an with (an ) decreasing and such that limn !1(na n ) = 0 :

———————————————————————————————————————————————–Solution. Let an = 1

n ln n for n 2N: Then 1

Pn =1an = 1 , and (an ) is decreasing.

Also, limn !1

(na n ) = limn !1

1ln n = 0 :

———————————————————————————————————————————————–320 . If (an ) is a sequence and if lim

n !1n 2an exists in R; show that 1


———————————————————————————————————————————————–Proof. Since lim

n !1n2an exists in R, the sequence n2an must be bounded, say by M .

That is, n2an M for all n 2N and so jan j M n 2 for all n 2N:

Hence, 1

Pn =1

jan j1

Pn =1

M n 2 < 1 :

Therefore, 1Pn =1an is absolutely convergent.

———————————————————————————————————————————————–321 . If (an k ) is a subsequence of (an ), then the series 1

Pn =1an k is called a subseries of 1

Pn =1an : Show that

1

Pn =1an is absolutely convergent if and only if every subseries of it is convergent.

———————————————————————————————————————————————–Proof. (=) ) Suppose that 1

Pn =1an is absolutely convergent, and let (an k ) be a subsequence of (an ) :

We want to show that 1

Pn =1an k is convergent.

For n 2N; de…ne

bn = an k if n = nk0 otherwise :

Note that 1

Pn =1an k = 1

Pn =1bn . Also, jbn j jan j for all n 2N and so 1

Pn =1 jbn j1

Pn =1 jan j< 1 :

Hence, 1

Pn =1 jan k j< 1 and so 1

Pn =1an k < 1 :

(( =) Suppose that every subseries of 1

Pn =1an is convergent. Then, in particular, the subseries of positive

114



terms and the subseries of negative terms are convergent and hence are absolutely convergent as we did

in a previous problem. Hence, 1


———————————————————————————————————————————————–322 . Let a > 0: Show that the series 1

Pn =1


———————————————————————————————————————————————–Proof. If 0 < a < 1; then lim

n !1an = 0 and so lim

n !11

1+ a n = 11+0 = 1 : Thus, 1Pn =1

11+ a n is divergent.

If a = 1 ; then 1

Pn =1

11+ a n = 1

Pn =1

12 = 1 :

If a > 1; then 1a < 1 and so 1

Pn =1

11+ a n < 1

Pn =1

1a n = 1

Pn =1

1a

n = 1=a1 (1 =a ) = 1

a 1 < 1 :

Hence, the series 1

Pn =1


———————————————————————————————————————————————–323 . Establish the convergence or divergence of the series whose n th term is:(a) (n (n + 1)) 1=2 ;(b) n2 (n + 1) 1=2 ;(c) n!=nn ;(d) ( 1)n n= (n + 1) :———————————————————————————————————————————————–Solution. (a) an = 1p n (n +1)

: Let bn = 1n for n 2N: Then lim

n !1a nbn

= limn !1

np n (n +1)= 1 :

Thus, by the Limit Comparison Test and since 1

Pn =1bn is divergent, we get that 1

Pn =1

1p n (n +1)is divergent .

(b) an = 1p n 2 (n +1): Let bn = 1

n 3 = 2 for n 2N: Then limn !1

a nbn

= limn !1

n 3 = 2

p n 2 (n +1)= lim

n !1 q n 3

n 2 (n +1) = 1 :

Thus, by the Limit Comparison Test and since 1

Pn =1bn is absolutely convergent, we get that 1

Pn =1

1p n 2 (n +1)

is absolutely convergent .

(c) We have a n +1a n

= (n +1)! n n

n !(n +1) n +1 = nn +1

n= 1 1

n +1

n:

Hence, limn !1

a n +1a n

= limn !1

1 1n +1

n = e 1 < 1:

Thus, by the Ratio Test and since limn !1

a n +1a n

< 1; we get that 1

Pn =1

n !n n is absolutely convergent .

(d) Note that limn !1

( 1) n nn +1 does not exist and so 1

Pn =1

( 1) n nn +1 is divergent .

———————————————————————————————————————————————–324 . If a;b > 0; show that 1

Pn =1(an + b) p converges if p > 1 and diverges if p 1:

———————————————————————————————————————————————–Proof. If p < 0; then lim

n !1(an + b) p = 1 and so 1

Pn =1(an + b) p diverges.

If p = 0 ; then 1

Pn =1

(an + b) p = 1

Pn =1

1 = 1 and so 1

Pn =1

(an + b) p diverges.

If 0 1; let cn = 1n p : Then (an + b) p

cn= n p

(an + b) p ! 1a p as n ! 1 :

Since 1

Pn =1

1n p < 1 for p > 1; we get by the comparison test that 1

Pn =1(an + b) p converges.

———————————————————————————————————————————————–325 . Discuss the series whose nth term is:

115



(a) n !3 5 7 (2 n +1) ;

(b) (n !)2

(2 n )! ;

(c) 2 4 (2 n )3 5 (2 n +1) ;

(d) 2 4 (2 n )5 7 (2 n +3) :

———————————————————————————————————————————————–Solution. (a) a n +1

a n= (n +1)!(3 5 7 (2 n +1))

(3 5 7 (2 n +1)(2 n +3)) n ! = n +12n +3 ! 1

2 < 1 as n ! 1 : Hence, by the Ratio Test, we

get that the series 1

Pn =1an converges.

(b) a n +1a n

= (( n +1)!) 2 (2 n )!(2 n +2)!( n !)2 = (n +1) 2

(2 n +1)(2 n +2) = n 2 +2 n +14n 2 +6 n +2 ! 1

4 < 1 as n ! 1 : Hence, by the Ratio Test, we

get that the series 1

Pn =1an converges.

(c) a n +1a n

= [2 4 (2 n )(2 n +2)][3 5 (2 n +1)][3 5 (2 n +1)(2 n +3)][2 4 (2 n )] = 2n +2

2n +3 ! 1 as n ! 1 : Hence, the Ratio Test is not applicable.So, we try the Raabe’s Test. Note that 2n +2

2n +3 = 1 12n +3 1 1

n for all n 2N:

Hence, 1

Pn =1

an does not converge absolutely and since the terms of 1

Pn =1

an are all positive, we conclude

that1

Pn =1 an does not converge.

(d) It can be checked that in this case, a n +1a n

= 2n +22n +5 = 1 3

2n +5 1 an for any 1 < a < 3=2 and

for all n 2N: Let a = 4 =5: Then by Raabe’s test, 1

Pn =1an is convergent.

———————————————————————————————————————————————–326 . Show that the series 1 + 1

2 13 + 1

4 + 15 1

6 + + is divergent.———————————————————————————————————————————————–Proof. Note that the sum 1

3n 2 + 13n 1 1

3n = 9n 2 2(3 n 2)(3 n 1)(3 n ) :

Let (sn ) be the sequence of partial sums of our series.

Then s3n =n

Pk=1

13k 2 + 1

3k 1 13k =

n

Pk =1

9k 2 2(3 k 2)(3 k 1)(3 k) and so lim

n !1s3n = 1

Pk =1

9k 2 2(3 k 2)(3 k 1)(3 k) :

But the series whose nth term is 9n 2 2

(3 n 2)(3 n 1)(3 n ) is divergent. (compare it with 1

n):

Hence, limn !1

s3n = 1 :Since (s3n ) is a divergent subsequence of (sn ) ; we conclude that (sn ) is divergent.Hence, the series 1 + 1

2 13 + 1

4 + 15 1

6 + + is divergent.

———————————————————————————————————————————————–327 . For n 2N; let cn = 1

1 + 12 + + 1

n ln n: Show that (cn ) is a decreasing sequence of positivenumbers. The limit C of this sequence is called Euler’s Constant and is approximately equal to 0:577.Show that if we put

bn = 11 1

2 + 13 1

2n

then the sequence (bn ) converges to ln 2: [Hint: bn = c2n cn + ln 2 :]———————————————————————————————————————————————–

Proof. Note that ln n = R n

11x dx and that

11 +

12 + +

1n 1 is the Riemann sum of the function f (x) =

1x

with respect to the tagged partition whose intervals are of the form [k 1; k] for 2 k n and whose tagsare the left endpoints of the intervals.Since f (x) = 1

x is decreasing on [1;1 ); we get that

ln n = R n

11x dx 1

1 + 12 + + 1

n 1 < 11 + 1

2 + + 1n 1 + 1

n :

Hence, cn = 11 + 1

2 + + 1n ln n > 0 for all n 2N (1) :

To show that the sequence (cn ) is decreasing, note that R n +1

n1x dx > 1

n +1 for all n 2N:

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Hence, for all n 2N; ln(n + 1) ln(n) > 1n +1 =) ln n > 1

n +1 ln(n + 1)=) cn = 1

1 + 12 + + 1

n ln n > 11 + 1

2 + + 1n + 1

n +1 ln(n + 1) = cn +1 :

Thus, the sequence (cn ) is strictly decreasing (2) .

By (1) and (2) , we get that the sequence (cn ) is convergent indeed.

Let bn = 11 1

2 + 13 1

2n : If we show that bn = c2n cn + ln 2 for all n 2N; then we wouldget that lim

n !1bn = lim

n !1(c2n cn + ln2) = lim

n !1c2n lim

n !1cn + ln 2 = C C + ln 2 = ln 2 and so we will

be done.Note that c2n cn + ln 2 = 1

n +1 + + 12n + ln ( n) ln(2n) + ln 2 = 1

n +1 + + 12n = dn :

Thus, we need to check that bn = dn for all n 2N:We will prove this by induction.For n = 1 ; we have b1 = 1 1=2 = 1 =2 and d1 = 1 =2:Now, suppose the result is true for n: That is, bn = dn : We need to show that the result holds for n + 1 ;i.e., bn +1 = dn +1 = 1

n +2 + + 12n +2 :

Note that dn +1 = 1n +2 + + 1

2n +2 = 1n +1 + 1

n +2 + + 12n +2 1

n +1 = dn + 12n +1 + 1

2n +2 1n +1

= dn + 12n +1 1

2( n +1) = bn + 12n +1 1

2( n +1) = bn +1 :

Hence, by the induction theorem, bn = dn = c2n cn + ln 2 for all n 2N: ———————————————————————————————————————————————–328 . Test the following series for convergence and for absolute convergence:

(a) 1

Pn =1

( 1) n +1

n 2 +1 ; (b) 1

Pn =1

( 1) n +1

n +1 ; (c) 1

Pn =1

( 1) n +1 nn +2 ; (d) 1

Pn =1( 1)n +1 ln n

n :———————————————————————————————————————————————–Solution. (a) ( 1) n +1

n 2 +1 = 1n 2 +1 1

n 2 for all n 2N. Since 1

Pn =1

1n 2 < 1 ; the series 1

Pn =1

( 1) n +1

n 2 +1 converges

absolutely and so it converges.

(b) ( 1) n +1

n +1 = 1n +1 and since P 1

n +1 = 1 (comparing it with 1n ); the series does not converge

absolutely. Since xn = 1n +1 is a decreasing sequence the converges to 0, and since the sequence of partial

sums of the series 1

Pn =1

( 1)n +1 , which is

f1; 0; 1; 0; 1; 0;:::

g; is bounded, the Dirichlet’s test implies that the

series 1

Pn =1

( 1) n +1

n +1 converges.

(c) Note that limn !1

( 1) n +1 nn +2 does not exist and so the series 1

Pn =1

( 1) n +1 nn +2 diverges.

(d) Note that xn = ln nn is decreasing for n 4 and xn ! 0 as n ! 1 by L’Hospital’s Rule.

Also, the sequence of partial sums of the series 1

Pn =1( 1)n +1 ; which is f1; 0; 1; 0; 1; 0;:::g; is bounded.

Hence, the Dirichlet’s test implies that the series 1

Pn =1( 1)n +1 ln n

n converges.

Now,

R 1

1ln x

x dx = 12 (ln x)2]11 = 1 and so, by the integral test, we get that the series 1

Pn =1

( 1)n +1 ln nn does

not converge absolutely. ———————————————————————————————————————————————–329 . Give an example to show that the Alternating Series Test may fail if (an ) is not a decreasing sequence.———————————————————————————————————————————————–Solution. For n 2N; de…ne the sequence (an ) as follows:

an = 1=n2 if n is odd1=n if n is even

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Then, clearly, (an ) is a sequence of strictly positive numbers and (an ) converges to 0.

Also, 1

Pn =1( 1)n an = 1 + 1 =2 1=9 + 1 =4 1=25 + 1 =6 1=49 + 1 =8 is divergent because if

it were conditionally convergent, then the series obtained from its negative terms must be divergent.It is clear, though, that the series of negative terms is absolutely convergent.

Hence, 1

Pn =1

( 1)n an is divergent.

Note that all of this happened because the sequence (an ) was not decreasing to 0.

———————————————————————————————————————————————–330 . Consider the series

1 12 1

3 + 14 + 1

5 16 1

7 + + ;

where the signs come in pair. Does it converge?———————————————————————————————————————————————–Proof. Let (xn ) = 1

n and (yn ) = f1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; g:

Then the sequence (xn ) is decreasing to 0 and the sequence of partial sums of the series 1

Pn =1yn is given

by f1; 0; 1; 0; 1; 0; 1; 0; 1; 0; 1; 0; g is bounded by 1. Hence, by the Dirichlet’s Test, we deduce that

the series 1 12

13 +

14 +

15

16

17 + + converges.

———————————————————————————————————————————————–331 . Let an 2R for n 2N and let p < q: If the series 1

Pn =1

a nn p is convergent, show that the series 1

Pn =1

a nn q is

also convergent.———————————————————————————————————————————————–Proof. Note that 1

Pn =1

a nn q = 1

Pn =1

a nn q p + p = 1

Pn =1

1n q p

a nn p = 1

Pn =1xn yn ; where xn = 1

n q p and yn = a nn p for n 2N:

Note that, since q p > 0; the sequence (xn ) is monotonically decreasing and xn ! 0 as n ! 1 :

Moreover, the series 1

Pn =1yn = 1

Pn =1

a nn p is given to be convergent.

Hence, by Abel’s test, we get that the series 1

Pn =1

xn yn = 1

Pn =1

a nn q is convergent.

———————————————————————————————————————————————–332 . If p and q are positive numbers, show that 1Pn =2

( 1)n (ln n ) pn q is a convergent series.

———————————————————————————————————————————————–Proof. If we de…ne f (x) = (ln x ) p

x q ; then we get that f 0(x) < 0 for x su¢ciently large. Hence, the functionf (x) is decreasing for x su¢ciently large, say for x M: Let N 2N be such that N M:Then we get that the sequence (ln n ) p

n q is decreasing for all n N:Moreover, lim

n !1(ln n ) p

n q = 0 by applying L’Hospital’s Rule.

Hence, by the Alternating Series Test, we deduce that the series 1

Pn =2( 1)n (ln n ) p

n q is a convergent series.

———————————————————————————————————————————————–333 . Discuss the series whose nth term is:(a) ( 1)n n n

(n +1) n +1 ; (b) n n

(n +1) n +1 ; (c) ( 1)n (n +1) n

n n ; (d) (n +1) n

n n +1 :———————————————————————————————————————————————–Solution. (a) ( 1)n n n

(n +1) n +1 = xn yn ; where xn = n n(n +1) n = 1 1

n +1

nand yn = ( 1) n

n +1 :

Note that the sequence (xn ) is monotone and converges to e 1 : Also, 1

Pn =1yn = 1

Pn =1

( 1) n

n +1 is convergent.

Hence, by Abel’s test, we get that the series 1

Pn =1xn yn = 1

Pn =1( 1)n n n

(n +1) n +1 is convergent.

(b) Let an = n n

(n +1) n +1 and bn = 1n for n 2N: Then a n

bn= n

n +1

n +1

! e 1 as n ! 1 :

Hence, by the Comarison Test and since 1

Pn =1

1n = 1 ; we deduce that the series 1

Pn =1an diverges.

118



(c) Note that limn !1

( 1)n (n +1) n

n n does not exist and so the series 1

Pn =1( 1)n (n +1) n

n n diverges.

(d) Let an = (n +1) n

n n +1 and bn = 1n +1 for n 2N: Then a n

bn= n +1

nn +1

! e as n ! 1 :

Hence, by the Comarison Test and since 1

Pn =1

1n +1 = 1 ; we deduce that the series 1

Pn =1

an diverges.

———————————————————————————————————————————————–334 . If the partial sums of 1Pn =1

an are bounded, show that the series 1Pn =1an e nt converges for t > 0:

———————————————————————————————————————————————–Proof. Fix t > 0; and let xn = e nt = 1

e tn for n 2N: Then the sequence (xn ) is decreasing and

converges to 0. (Note that for t > 0; e t < 1 and so e nt ! 0 as n ! 1 ):

Hence, by the Dirichlet’s Test, we deduce that 1

Pn =1an e nt converges.

———————————————————————————————————————————————–335 . If the partial sums sn of 1


Pn =1

a nn converges to 1

Pn =1

s nn (n +1) :

———————————————————————————————————————————————–Proof. Recall the Abel’s Lemma which says that if (an ) and (bn ) are sequences in R and if the sequenceof partial sums of (an ) is denoted by sn ; with s0 = 0 ; and if m > n; then

m

Pk = n +1ak bk = ( bm sm bn +1 sn ) + m 1

Pk = n +1(bk bk +1 ) sk

In our case, bn = 1n : Also, note that

m

Pk=1ak bk = ( bm sm b1s0) +

m 1

Pk =1(bk bk+1 ) sk = bm sm +

m 1

Pk=1(bk bk+1 ) sk

In other words,m

Pk =1

a kk = sm

m +m 1

Pk=1

1k 1

k+1 sk = smm +

m 1

Pk=1

1k(k +1) sk :

Note that limm !1

s mm = 0 since (sm ) is a bounded sequence.

Hence, 1

Pk =1

a kk = lim

m !1

m

Pk =1

a kk = lim

m !1

s mm + lim

m !1

m 1

Pk=1

1k(k +1) sk = 0 + 1

Pk=1

1k(k +1) sk = 1

Pk =1

1k(k +1) sk :

———————————————————————————————————————————————–336. Show that the hypothesis that the sequence (xn ) is decreasing in the Dirichlet’s Test can be replaced

by the hypothesis that 1

Pn =1 jxn xn +1 j is convergent.———————————————————————————————————————————————–Proof. We recall the Dirichlet’s Test:

Dirichlet’s Test : If (xn ) is a decreasing sequence with limn !1

xn = 0 ; and if the partial sums (sn ) of the series1

Pn =1yn are bounded, then the series 1

Pn =1xn yn is convergent.

What we need to prove, thus, is the following modi…ed version:

Modi…ed Dirichlet’s Test : If (xn ) is a sequence such that 1Pn =1 jxn xn +1 j is convergent, with limn !1

xn = 0 ;and

if the partial sums (sn ) of the series 1

Pn =1yn are bounded, then the series 1


Proof . Let jsn j B for all n 2N: If m > n; then it follows by Abel’s Lemma (see the previous problem)

and the fact that 1

Pn =1 jxn xn +1 j is convergent, that: (Here, we use Cauchy’s Criterion for Convergence)

m

Pk = n +1xk yk jxm xn +1 jB +

m 1

Pk= n +1 jxk xk +1 jB = B "jxm xn +1 j+m 1

Pk= n +1 jxk xk +1 j# n: [ > 0 was chosen so that things would work!].

Hence, by Cauchy’s Criterion for Convergence of Series, we deduce that 1


———————————————————————————————————————————————–337. Show that if the partial sums sn of the series 1

Pk =1

ak satisfy jsn j Mn r for some r < 1; then the

series 1Pk =1

a kk converges.

———————————————————————————————————————————————–Proof. Recall the Abel’s Lemma which says that if (an ) and (bn ) are sequences in R and if the sequenceof partial sums of (an ) is denoted by sn ; with s0 = 0 ; and if m > n; then

m

Pk = n +1ak bk = ( bm sm bn +1 sn ) +

m 1

Pk = n +1(bk bk +1 ) sk

In our case, bn = 1n : Thus,

m

Pk = n +1

a kk = 1

m sm 1n +1 sn +

m 1

Pk= n +1

s kk(k +1)

Hence,m

Pk= n +1

a kk

1m jsm j+ 1

n +1 jsn j+m 1

Pk= n +1js k jk (k+1) 1

m Mm r + 1n +1 Mn r +

m 1

Pk = n +1

Mk r

k(k +1)

= M 1m m r + 1

n +1 n r +m 1

Pk= n +1

k r

k(k +1) !:

Note that, since r < 1; 1m m r ! 0 as m ! 1 ; and 1

n +1 n r ! 0 as n ! 1 .

Also, 1

Pk=1

k r

k (k+1) < 1 since we can compare it with 1k 2 r and 1

Pk=1

1k 2 r < 1 since 2 r > 1:

Hence, as m; n ! 1 ; we get thatm 1

Pk = n +1

k r

k(k+1) ! 0:

Gathering all the information that we got, we deduce that

m

Pk = n +1

a k

k ! 0 as m; n ! 1 and so, by the

Cauchy’s Criterion for Convergence, we get that the series 1

Pk =1

a kk converges.

———————————————————————————————————————————————–338. Suppose that 1

Pn =1an is a convergent series of real numbers. Either prove that 1

Pn =1bn converges or

give a counter-example, where we de…ne bn by:(a) an

n ; (b) p a nn (an 0) (c) p a n

n (an 0); (d) n1=n an :———————————————————————————————————————————————–Solution. (a) Since the sequence 1

n is monotone and converges to 0, and since 1

Pn =1an is a convergent

series, Abel’s Test implies that 1

Pn =1

a nn is convergent.

(b) Note that, by Cauchy-Schwartz’ inequality,

1

Pn =1

p a nn

1

Pn =1p an

21=2 1

Pn =1

1n

21=2

= 1

Pn =1an

1=2 1

Pn =1

1n 2

1=2

This is a product of two convergent series and so it converges. Hence, by the Comparison Test, we deduce

that the series 1

Pn =1

p a nn is convergent.

(c) Let an = 1n (ln n ) 2 (Note that the series 1

Pn =2

1n (ln n ) 2 is convergent by the Integral Test).

120



Then 1

Pn =2 p a nn = 1

Pn =2 q 1n 2 (ln n ) 2 = 1

Pn =2

1n ln n which is divergent by the Integral Test again.

(d) The sequence n1=n is monotonically decreasing for n 3 with limn !1

n1=n = 1 ; and since 1

Pn =1an

is a convergent series, Abel’s Test implies that 1

Pn =1

n1=n an is convergent.

———————————————————————————————————————————————–339. Discuss the convergence and uniform convergence of the series 1

Pn =1f n where f n is given by:

(a) x2 + n2 1 (b) (nx ) 2 (x 6= 0) (c) sin(x=n2) (d) (xn + 1) 1 (x 0)———————————————————————————————————————————————–Solution. (a) We have 1

x 2 + n 2 1n 2 for all x 2R and all n 2N: Also, 1

Pn =1

1n 2 < 1 :

Hence, by the Weierstrass M-Test, we conclude that 1

Pn =1x2 + n2 1 converges uniformly on R.

(b) For any x 6= 0 ; 1

Pn =1

1n 2 x 2 = 1

x 21

Pn =1

1n 2 < 1 : Hence, the series converges for all x 6= 0 :

For a …xed a > 0; we claim that 1

Pn =1

1n 2 x 2 converges uniformly on the set jxj a:

This is because jxj a =) x2 a2 =) 1x 2 1

a 2 =) 1n 2 x 2 1

n 2 a 2 ; and since 1Pn =1

1n 2 a 2 < 1 ;

the Weierstrass M-Test implies that 1

Pn =1(nx ) 2 converges uniformly on Rn( a; a ).

The series does not converge uniformly on Rnf0g because if we take the sequence (xn ) = 1n ; then

1

Pn =1f n (xn ) = 1

Pn =11 = 1 :

(c) We have sin xn 2 jx jn 2 for all x 2R and so 1

Pn =1sin x

n 2 converges for all x 2R:

The convergence is not uniform on R because if (xn ) = n2 ; then 1

Pn =1sin x n

n 2 = 1

Pn =1sin(1) = 1 :

Yet, on any interval of the form [ a; a ] ; with a > 0; we claim that the convergence is uniform.

This is because sin xn 2 j

x

jn 2 an 2 for all x 2[ a; a ] and the Weierstrass M-Test impliesuniform convergence.

(d) If x = 0 ; then f n (x) = 1 for all n 2N and so the series diverges.If 0 < x < 1; then f n (x) = 1

1+ x n ! 1 as n ! 1 and so the series diverges.If x = 1 ; then f n (x) = 1

2 for all n 2N and so the series diverges.

If x > 1; then f n (x) = 11+ x n 1

xn and 1

Pn =1

1x

n < 1 : Hence, the series converges.

The convergence is not uniform on (1;1 ) since if we choose (xn ) = n1=n then xn ! 1+ but

f n (xn ) = 11+ n and 1

Pn =1

11+ n = 1 :

We claim that for any a > 1; the series converges uniformly on [a; 1 ):

This is because for all x a; jf n (x)j= 11+ x n

1x n

1a n and

1

Pn =1

1a n < 1 :

Hence, by the Weierstrass M-Test, we conclude that the convergence is uniform on [a; 1 ):

———————————————————————————————————————————————–340. If 1

Pn =1an is an absolutely convergent series, show that the series 1

Pn =1an sin nx is absolutely and

uniformly convergent.———————————————————————————————————————————————–Proof. We have jan sin nx j jan j for all x 2R and all n 2N: Also, 1

Pn =1 jan j< 1 :

Hence, by the Weierstrass M-Test, we conclude that the convergence is uniform on R:

121



Clearly, by the same inequality, we see that the convergence is absolute.

———————————————————————————————————————————————–341. Determine the radius of convergence of the power series 1

Pn =1an xn , where an is given by:

(a) 1=nn ; (b) n =n!; (c) n n =n!; (d) (ln n) 1 (n 2); (e) (n!)2 =(2n)!;(f) n p n :———————————————————————————————————————————————–Solution. (a) We have limsup jan j

1=n = lim 1 =n = 0 ; and so R = + 1 :(b) We have limsup jan j

1=n = limsup n =n =(n!)1=n = lim n =n

(n !)1 =n = 0 ; and so R = + 1 :

(c) We have lim a na n +1

= lim n n (n +1)!n !(n +1) n +1 = lim n n

(n +1) n = lim 1 1n +1

n= e 1 ; and so R = e 1 :

(d) We have lim a na n +1

= lim ln( n +1)ln n

L’Hospital’s Rule= 1 ; and so R = 1 :

(e) We have lim a na n +1

= lim (n !)2 (2 n +2)!(2 n )!(( n +1)!) 2 = lim (2n +1)(2 n +2)

(n +1) 2 = 4 ; and so R = 4 :

(f) We have limsup jan j1=n = limsup n 1=p n = lim n 1=p n = 1 ; and so R = 1=1 = 1 :

———————————————————————————————————————————————–342. If an = 1 if n = k2

0 otherwise and bn = 1 if n = k!0 otherwise for k 2N, …nd the radius od convergence

of the power series 1Pn =1an xn and of 1Pn =1

bn xn :———————————————————————————————————————————————–Solution. We have limsup jan j

1=n = limsup f1; 0; 0; 1; 0; 0; 0; 0; 1; 0; g= 1 ; and so R = 1 =1 = 1 :Also, lim sup jbn j

1=n = limsup f1; 1; 0; 0; 0; 1; 0; 0; g= 1 ; and so R = 1 =1 = 1 :

———————————————————————————————————————————————–343. If 0 < p jan j q for all n 2N; …nd the radius of convergence of 1

Pn =1an xn :

———————————————————————————————————————————————–Solution. We have 0 < p 1=n jan j

1=n q 1=n and so, since lim p1=n = lim q 1=n = 1 ; the Squeeze Theoremimplies that lim jan j

1=n = 1 : Hence, limsup jan j1=n = 1 and so, R = 1 =1 = 1 :

———————————————————————————————————————————————–344. Let f (x) = 1

Pn =1

an xn for jxj< R: If f (x) = f ( x) for all jxj< R; show that an = 0 for all n odd.

———————————————————————————————————————————————–Proof. We have an = f ( n ) (0)n ! for all n 2N: Note that f (n ) (x) = ( 1)n f (n ) ( x) for all n 2N:

Hence, f (n ) (0) = f (n ) (0) for all n odd and so f (n ) (0) = 0 for all n odd.Consequently, an = 0 for all n odd.

———————————————————————————————————————————————–345. Prove that if f is de…ned for jxj< r and if there exists a constant B such that f (n ) (x) B for all

jxj< r and n 2N; then the Taylor series expansion

1

Pn =1

f ( n ) (0)n ! xn

converges to f (x) for jxj< r:———————————————————————————————————————————————–

Proof. We have to show that the sequence (Rn (x)) of remainders converges to 0 for each x with jxj< r:Recall that

Rn (x) = f ( n +1) (c)(n +1)! xn +1

for some c between 0 and x:Hence, jRn (x)j= f ( n +1) (c)

(n +1)! xn +1 B(n +1)! r n +1 ! 0 as n ! 1 :

Hence, the Taylor series expansion 1

Pn =1

f ( n ) (0)n ! xn converges to f (x) for jxj< r:

———————————————————————————————————————————————–

122



346. Find a series expansion for R x

0 e t 2dt for x 2R:

———————————————————————————————————————————————–Solution. For each t 2R; we have e t 2

= 1

Pn =0

( t 2 )n

n ! = 1

Pn =0

( 1) n t 2 n

n ! :

Hence,

R x

0 e t 2dt =

R x

0 1

Pn =0

( 1) n t 2 n

n ! dt = 1

Pn =0

( 1) n

n !

R x

0 t2n dt = 1

Pn =0

( 1) n

n !x 2 n +1

2n +1 :

———————————————————————————————————————————————–347. Let f : R! R be a Lebesgue integrable function, i.e., R R fdm < 1 .(a) Prove that m(fx 2R : f (x) = 1g) = 0 :(b) Prove that 8 > 0; m(fx 2R : jf (x)j g) < 1 :(c) Prove that the Lebesgue integral is absolutely continuous with respect to the Lebesgue measure, i.e.,if f is Lebesgue integrable on A , then 8 > 0 9 > 0 :R B jf jdm < whenever B A with m(B ) < :(d) Prove that 8 > 0; 9 a compact set K R : R RnK jf jdm < :(e) Prove that 8 > 0; 9 M > 0 and a measurable set A R : jf (x)j M on A ; and R RnA jf jdm < :———————————————————————————————————————————————–Proof. (a) If m(fx 2R : f (x) = 1g) = a > 0; then R R fdm R fx2R :f (x )= 1g fdm = 1 :

(b) Let > 0 be given. If m(fx 2R : jf (x)j g) = 1 ; then R R jf jdm R fx2R :jf (x )j gjf jdmm(fx 2R : jf (x)j g) = 1 :

(c) Let > 0 be given. f is Lebesgue integrable on A

, jf

j is Lebesgue integrable on A .

jf j 0 =) 9 a sequence of simple functions ' n %f on A , ' n (x) 0 on A for each n = 1 ; 2; 3;::::Thus, by the Lebesgue monotone convergence theorem, lim

n !1 R A ' n dm = R A jf jdm:Hence, 9N :R A (jf j ' n ) dm <

2 8n N:

Write ' N =k

Xi =1

a i (A i ); where A =k

[i =1

A i : Let a = max1 i k

a i :

Then for = 2a ; we have that R B j' N jdm < a m (B ) = a =

2 whenever B A with m(B ) < :Hence, R B jf jdm = R B (jf j ' N + ' N ) dm = R B (jf j ' N ) dm + R B ' N dm <

2 + 2 = :

(d) Let > 0 be given and for n 2Z; set A n = [n; n + 1] . Then R R jf jdm =1

Xn = 1R A n jf jdm < 1 :

Thus,

9N :

N

Xn = 1R

An j

f

jdm +

1

Xn = N R A

n jf

jdm < :

Let K = [ N; N ]: Then K is compact, and R RnK jf jdm =N

Xn = 1R A n jf jdm +1

Xn = N R A n jf jdm < :

(e) Let > 0 be given. By (c), 9 > 0 :R B jf jdm < 2 whenever B R with m(B ) < :

By (d), 9 a compact set K R :R RnK jf jdm < 2 .

Let D n = fx 2K : jf (x)j> n g: Then m(D n ) n !1! 0: Thus, 9M : m(D n ) < for all n M:Let A = K nD M and note that RnA = ( RnK ) [ (D M ) :Then A is measurable, jf (x)j M on A , and R RnA jf jdm = R RnK jf jdm + R D M jf jdm <

2 + 2 = :

———————————————————————————————————————————————–348. Show that if f is a Lebesgue integrable function on A and A n = fx 2A : jf (x)j ng; thenlim

n !1n m (A n ) = 0 :

———————————————————————————————————————————————–Proof. Since f is a Lebesgue integrable function on A , m(A n ) ! 0 as n ! 1 :Thus, by problem 1(c), lim

n !1 R A n jf jdm = 0 : Note that R A n jf jdm n m (A n ) 0:Hence, lim

n !1n m (A n ) = 0 :

———————————————————————————————————————————————–349. Show that if f 0 on a set A , m(A ) > 0; and R A fdm = 0 ; then f = 0 a.e. on A .———————————————————————————————————————————————–

Proof. A = A 0 [1

[n =1

B n ; where A 0 = fx 2A : f (x) = 0 g and B n = fx 2A : f (x) > 1n g:

123



Note that B =1

[n =1

B n = fx 2A : f (x) 6= 0 g:

If m(B ) > 0; then since m(B )1

Xn =1m(B n ); there is N : m(B N ) = a > 0:

Then

R B N

fdm > 1N m(B N ) > 0: But

R B N

fdm

R A fdm = 0 ; a contradiction.

Thus, m(B ) = 0 and so f = 0 a.e. on A . ———————————————————————————————————————————————–350. Show that if R B fdm = 0 for every measurable subset B of A ; m(A ) > 0; then f = 0 a.e. on A .———————————————————————————————————————————————–Proof. Let B 1 = fx 2A : f (x) 0g and B 2 = fx 2A : f (x) 0g:Both B 1 and B 2 are measurable subsets of A and so R B 1

fdm = 0 and R B 2fdm = 0 :

But R B 1fdm = R A f + dm and R B 2

fdm = R A f dm: Also, f + 0 and f 0:Hence, by problem 3, f + = 0 a.e. on A , and f = 0 a.e. on A .But f = f + f : Thus, f = 0 a.e. on A .

———————————————————————————————————————————————–351. Show that if f is Lebesgue integrable on A and R A fdm = R A jf jdm; then either f 0 a.e. on A ,or f 0 a.e. on A .

———————————————————————————————————————————————–Proof. We …rst note that if a;b > 0 and ja bj= a + b; then either a = 0 or b = 0 :Now, let a = R A f + dm and b = R A f dm:Then R A fdm = a b and R A jf jdm = a + b:Thus, either a = 0 or b = 0 = ) either R A f + dm = 0 or R A f dm = 0f + ;f 0=) either f + = 0 a.e. on A or f = 0 a.e. on A =) either f 0 a.e. on A , or f 0 a.e. on A .

———————————————————————————————————————————————–352. Let ff n g be a sequence of nonnegative and measurable functions on A such that lim

n !1 R A f n dm = 0 :

Show that f nm

! 0: ( f nm

! 0 means f n converges to 0 in measure).———————————————————————————————————————————————–Proof. Let > 0 be given and set A n = fx 2A : f n (x) g: Then 0 m(A n ) R A n

f n dm R A f n dm:Thus, since

R A f n dm n !1

! 0; m(A n ) n !1

! 0 and so f n

m

!0:

———————————————————————————————————————————————–353. Show that in problem 6, convergence in measure cannot be replaced by a.e. convergence.———————————————————————————————————————————————–Proof. Let A = [0 ; 1]: Let f 1 = [0;1] ; f 2 = [0;1=2]; f 3 = [1=2;1]; f 4 = [0;1=4]; f 5 = [1=4;1=2] ;f 6 = [1=2;3=4] ; f 7 = [3=4;1] ; f 8 = [0;1=8] ; :::; and so on. Clearly, lim

n !1 R A f n dm = 0 :But f n 9 0 a.e. on A .

———————————————————————————————————————————————–354. Let ff n gbe a sequence of measurable functions on a set A , m(A ) < 1 : Show that lim

n !1 R A jf n j1+ jf n jdm = 0

if and only if f nm

! 0:———————————————————————————————————————————————–Proof. (=) ) Suppose that lim

n !1

R A jf n j1+ jf n jdm = 0 , let > 0; and set A n = fx 2A : jf n (x)j g:

Note that jf n (x)j () jf n

j1+ jf n j 1+ : Hence, 0

1+ m(

An ) R

A n jf n

j1+ jf n jdm R A j

f n

j1+ jf n jdm n

!1! 0:Thus, m(A n ) n!1! 0 and consequently, f n

m

! 0:(( =) Suppose that f n

m

! 0, let > 0; and set A n = fx 2A : jf n (x)j g:Note that jf n (x)j< () jf n j1+ jf n j <

1+ :

Now, R A jf n j1+ jf n jdm = R A njf n j1+ jf n jdm + R A nA

njf n j1+ jf n jdm R A n

1dm + R A nAn 1+ dm = m(A n ) +

1+ m(A nA n )

m(A n ) + 1+ m(A ) n !1!

1+ m(A ):Hence, lim

n !1 R A jf n j1+ jf n jdm 1+ m(A ): But m(A ) < 1 and our choice of > 0 was arbitrary.

Thus, limn !1 R A jf n j1+ jf n jdm = 0 :

124



———————————————————————————————————————————————–355. Show that the assumption m(A ) < 1 is essential in problem 8.———————————————————————————————————————————————–Proof. For n = 1 ; 2; 3;:::; let f n (x) = 1

n : Then f nm

! 0 on A = R:But

R A jf n j1+ jf n jdm =

R R

1=n1+1 =n dm =

R R

1n +1 dm = 1

n +1

R R dm = 1 :

———————————————————————————————————————————————–356. Suppose that f is nonnegative and measurable on A , m(A ) < 1 : Let A k = fx 2A : k f (x) < k +1 g:


Xk=0

km (A k ) < 1 :

———————————————————————————————————————————————–

Proof. We …rst note that km(A k ) R A kfdm < (k + 1) m(A k ) for k = 0 ; 1; 2;:::: Also, A =

1

[k=0

A k and

the A k ’s are mutually disjoint and so m(A ) =1

Xk =0

m(A k ) and R A fdm =1

Xk=0 R A kfdm:

Hence,1

Xk =0

km (A k )1

Xk =0 R A kfdm <

1

Xk=0

(k + 1) m(A k ) = m(A ) +1

Xk =0

km (A k ):

Thus,1

Xk=0km (A k ) R

A fdm < m (A ) +1

Xk=0km (A k ):

Therefore, f is Lebesgue integrable on A if and only if 1

Xk=0

km (A k ) < 1 :

———————————————————————————————————————————————–357. Suppose that f is nonnegative and measurable on A , m(A ) < 1 : Let B k = fx 2A : f (x) kg:


Xk=0

m(B k ) < 1 :

———————————————————————————————————————————————–

Proof. We …rst note that1

Xk=0

m(B k ) =1

Xk =0

(k + 1) m(A k ), where A k = fx 2A : k f (x) < k + 1g: Thus,

1

Xk=0

m(B k ) <

1 ,1

Xk =0

(k + 1) m(A k ) <

1 ,1

Xk =0

m(A k ) +1

Xk =0

km (A k ) <

1 ,m(A ) +

1

Xk=0

km (A k ) <

1,

1

Xk =0

km (A k ) < 1Problem 10

, f is Lebesgue integrable on A .

———————————————————————————————————————————————–

358. Suppose that f is nonnegative and integrable on A , m(A ) < 1 : For > 0; de…ne S ( ) =1

Xk =0

k m(A k );

where A k = fx 2A : k f (x) < (k + 1) g: Prove that lim! 0

S ( ) = R A fdm:———————————————————————————————————————————————–

Proof. We …rst note that k m(A k ) R A kfdm < (k + 1) m(A k ) for k = 0 ; 1; 2; :::: Also, A =

1

[k=0

A k and

the A k ’s are mutually disjoint and so m(A ) =1

Xk =0

m(A k ) and

R A fdm =

1

Xk=0

R A k

fdm:

Hence, S ( ) =1

Xk =0

k m(A k )1

Xk=0 R A kfdm <

1

Xk=0

(k + 1) m(A k ) = m(A ) +1

Xk=0

k m(A k ) = m(A ) + S ( ):

Thus, S ( ) R A fdm < m (A ) + S ( ): Letting ! 0; we get that lim! 0

S ( ) R A fdm lim! 0

S ( ):

Thus, lim! 0

S ( ) = R A fdm:

———————————————————————————————————————————————–359. Let ff n g be a sequence of nonnegative functions converging to a function f on R; and suppose thatlim

n !1 R R f n dm = R R fdm < 1 : Show that for each measurable set A R; limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–

125



Proof. By Fatou’s Theorem, R A fdm = R A limf n dm limR A f n dm limR A f n dmlim(R R f n dm R RnA f n dm) = lim R R f n dm limR RnA f n dm = R R fdm limR RnA f n dm

Fatou’s Theorem

R R fdm R RnA limf n dm = R R fdm R RnA fdm = R A fdm:Hence,

R A fdm lim

R A f n dm lim

R A f n dm

R A fdm and so lim

R A f n dm = lim

R A f n dm =

R A fdm:

Thus, limn !1 R

A f n dm =

R A fdm:

———————————————————————————————————————————————–360. Suppose that ff n g is a sequence of measurable functions on A such that jf n j g; where g is integrableon A . Show that R A limf n dm limR A f n dm limR A f n dm R A limf n dm:———————————————————————————————————————————————–Proof. Let gn = g f n ; n = 1 ; 2; 3;:::: Then each gn is nonnegative and measurable on A .Hence, by Fatou’s Theorem, R A limgn dm limR A gn dm:But limgn = lim( g f n ) = g limf n ; and so R A limgn dm = R A (g limf n )dm = R A gdm R A limf n dm:Also, limR A gn dm = lim R A (g f n )dm = R A gdm limR A f n dm:

Thus, R A gdm R A limf n dm R A gdm limR A f n dm and so limR A f n dm R A limf n dm :Now, let hn = g + f n ; n = 1 ; 2; 3; :::: Then each hn is nonnegative and measurable on A .Hence, by Fatou’s Theorem,

R A limhn dm lim

R A hn dm:

But limhn = lim( g + f n ) = g + lim f n ; and so

R A limhn dm =

R A (g + lim f n )dm =

R A gdm +

R A limf n dm:

Also, limR A hn dm = lim R A (g + f n )dm = R A gdm + lim R A f n dm:Thus, R A gdm + R A limf n dm R A gdm + lim R A f n dm and so R A limf n dm limR A f n dm :

It is always true that limR A f n dm limR A f n dm :

Thus, R A limf n dm limR A f n dm limR A f n dm R A limf n dm:

———————————————————————————————————————————————–361. Suppose that ff n g is a sequence of real-valued functions integrable on A , m(A ) < 1 : Show thatif f n

Af , then lim

n !1 R A f n dm = R A fdm: ( f n A

f means f n converges uniformly to f on A ).———————————————————————————————————————————————–Proof. Let > 0 be given. Since f n

Af; there exists N : jf n f j<

m ( A ) for all x 2A and for all n N:

Hence, for n N; we have

R A f n dm

R A fdm =

R A (f n f )dm

R A jf n f jdm <

R A m ( A ) dm = :

Thus, limn !1 R A f n dm = R

A fdm: ———————————————————————————————————————————————–362. Let f n (x) = nx n 1 (n + 1) xn ; x 2(0; 1):

(a) Show that R (0 ;1)

1

Xn =1f n dm 6=

1

Xn =1 R (0 ;1) f n dm:

(b) Show that1

Xn =1 R (0 ;1) jf n jdm = 1 :

———————————————————————————————————————————————–

Proof. We …rst note that for x 2(0; 1);1

Xn =1nx n 1 (n + 1) xn =

1

Xn =1

ddx xn xn +1 = d

dx

1

Xn =1xn xn +1

= ddx ( x x 2

1 x ) = ddx (x) = 1 :

(a) We have that R (0 ;1)

1

Xn =1f n dm = R (0 ;1)

1

Xn =1nx n 1 (n + 1) xn dm = R (0 ;1) 1 dm = m ((0; 1)) = 1 :

On the other hand,1

Xn =1 R (0 ;1) f n dm =1

Xn =1 R (0 ;1) nx n 1 (n + 1) xn dm =1

Xn =1[xn (1 x)]1

0 =1

Xn =10 = 0 :

Hence, R (0 ;1)

1

Xn =1

f n dm = 1 6= 0 =1

Xn =1 R (0 ;1) f n dm:

(b) We …rst note that jf n j= nx n 1 (n + 1) xn for x 2(0; n

n +1 )(n + 1) xn nx n 1 for x 2( n

n +1 ; 1) :

126





(b) Find limn !1 R 1

0 f n (x)dx:

(c) Is there a function g(x) 2L1(0; 1) : g(x) f n (x) for all n?———————————————————————————————————————————————–Solution. (a) Let x 2(0; 1): Then there is N large so that 1

N x; and so f n (x) = 0 for all n N:Thus, lim

n !1f n (x) = 0 :

(b) limn !1 R 10 f n (x)dx = limn !1 R

1n

0 ndx = limn !1 [nx ]1n0 = limn !1 1 = 1 :

(c) Since limn !1 R

10 f n (x)dx = 1 6= 0 = R

10 lim

n !1f n (x)dx; the Lebesgue dominated convergence theorem implies

that no such g exists.

———————————————————————————————————————————————–367. (a) Let (a; b) (0; 2 ) be an open interval. Show that lim

n !1 R b

a cos(nx )dx = 0 :(b) Let A (0; 2 ) be a measurable set. Show that lim

n !1 R A cos(nx )dx = 0 :———————————————————————————————————————————————–Proof. (a) R b

a cos(nx )dx = hsin( nx )n i

b

a= sin( nb ) sin( na )

n ; so R ba cos(nx )dx 2

n ; and so

limn !1 R

ba cos(nx )dx = 0 : Hence, lim

n !1 R b

a cos(nx )dx = 0 :

(b) Let A (0; 2 ) be a measurable set and > 0 be given. Then there exists and open set G such that

A G and m(G nA ) < : Thus, G = A [ (G nA ): Write G = 1[k =1(ak ; bk ); a disjoint union of open intervals.

Then limn !1 R G cos(nx )dx = lim

n !11

Xk=1 R bk

a kcos(nx )dx =

1

Xk=1

limn !1 R

bk

a kcos(nx )dx

(a )=

1

Xk =1

0 = 0 :

Let N be large enough so that R G cos(nx )dx < 2 for all n N:

Since m(G nA ) < ; R G nA cos(nx )dx R G nA jcos(nx )jdx R G nA dx = m(G nA ) < 2 :

Thus, for n N; R A cos(nx )dx = R G cos(nx )dx R G nA cos(nx )dx R G cos(nx )dx + R G nA cos(nx )dx<

2 + 2 = : Therefore, lim

n !1 R A cos(nx )dx = 0 :

———————————————————————————————————————————————–368. Let f n (x) = n

1+ n 2 x 2 :(a) Does f n

!0 a.e. on [0; 1]?

(b) Does f n ! 0 in L1[0; 1]?(c) Find lim

n !1 R 1

0sin x

x f n (x)dx and justify your answer.———————————————————————————————————————————————–Solution. (a) For x 2(0; 1]; lim

n !1f n (x) = lim

n !1n

1+ n 2 x 2 = limn !1

11n + nx 2 = 1

0+ 1 = 0 :Thus, f n ! 0 a.e. on [0; 1]:(b) lim

n !1 R 1

0 f n (x)dx = limn !1 R

10

n1+ n 2 x 2 dx u= nx= lim

n !1 R n

01

1+ u 2 du = limn !1

[arctan u]n0 = limn !1

(arctan n) = 2 :

Thus, f n 9 0 in L1[0; 1]:

(c) R 1

0sin x

x f n (x)dx = R 1

0n sin x

x (1+ n 2 x 2 ) dx u = nx= R n

0sin ( u

n )( u

n )(1+ u 2 )du:

For n = 1 ; 2; 3;:::; let gn (u) = ( sin ( un )

( un )(1+ u 2 )

for 0 u n

0 for u > n ): (Remember that limx! 0+

sin xx = 1) :

Then limn !1 R

10

sin xx f n (x)dx = lim

n !1 R n

0sin (

un )( u

n )(1+ u 2 ) du = limn !1 R 10 gn (u)du:

Note that jgn (u)j=sin ( u

n )( u

n )(1+ u 2 ) 1

(1+ u 2 ) for all u 2[0;1 ):

Also, R 10

1(1+ u 2 ) du =

2 < 1 and so by the Lebesgue dominated convergence theorem, we get that

limn !1 R 1

0sin x

x f n (x)dx = limn !1 R 1

0 gn (u)du = R 10 lim

n !1gn (u) du = R 1

01

(1+ u 2 ) du = 2 :

———————————————————————————————————————————————–369. Compute lim

n !1 R 11

sin x1+ nx 2 dx and justify your answer.

———————————————————————————————————————————————–

128



Solution. Note that sin x1+ nx 2

11+ nx 2 1

1+ x 2 for all n 2N and for all x 2R:

Also, R 11

11+ x 2 dx = < 1 : Thus, by the Lebesgue dominated convergence theorem, we get that

limn !1 R 1

1sin x

1+ nx 2 dx = R 11 lim

n !1sin x

1+ nx 2 dx = R 11 0dx = 0 :

———————————————————————————————————————————————–370. Let f be a nonnegative Lebesgue integrable function on (0;

1): Prove that lim

n

!1

1n

R n

0 xf (x)dx = 0 :———————————————————————————————————————————————–Proof. For n = 1 ; 2; 3;:::; let f n (x) =

xn f (x) for 0 x n

0 for x > n :

Then limn !1

1n R

n0 xf (x)dx = lim

n !1 R n

0xn f (x)dx = lim

n !1 R 10 f n (x)dx:

Note that jf n (x)j= xn f (x) f (x) for all x 2[0;1 ) and for all n 2N: Also, R 1

0 f (x)dx < 1 :Thus, by the Lebesgue dominated convergence theorem, we get thatlim

n !11n R

n0 xf (x)dx = lim

n !1 R 10 f n (x)dx = R 1

0 limn !1

f n (x) dx = R 10 0dx = 0 :

———————————————————————————————————————————————–371. Let f be nonnegative and measurable on [0; 1]: Let A = fx 2[0; 1] : f (x) > 1g:Show that lim

n !1 R 1

0 (f (x))n dx exists if and only if m(A ) = 0 :———————————————————————————————————————————————–Proof. (=) ) Suppose that m(A) > 0: Then 9r > 1 : m(fx 2[0; 1] : f (x) rg) > 0:

Let B = fx 2[0; 1] : f (x) rg: Then R B (f (x))

ndx r

nm(B ):

Thus, limn !1 R

10 (f (x))n dx lim

n !1 R B (f (x))n dx = limn !1

r n m(B ) = 1 :

(( =) Suppose that m(A ) = 0 : Then R 1

0 (f (x))n dx = R A (f (x))n dx + R [0;1]nA (f (x))n dx = R [0;1]nA (f (x))n dx:

Note that on [0; 1]nA ; (f (x))n 1 for all n 2N and R [0;1]nA 1dx = m([0; 1]nA ) m ( A )=0

= 1 < 1 :

Thus, by the Lebesgue dominated convergence theorem, we get that limn !1 R

10 (f (x))n dx =

limn !1 R [0;1]nA (f (x))n dx = R [0;1]nA lim

n !1(f (x))n dx = m(C ) 1; where C = fx 2[0; 1] : f (x) = 1 g:

———————————————————————————————————————————————–372. Show that if f is Riemann integrable on [a; b] and f (x) = 0 for x 2[a; b] \ Q; then R

ba f (x)dx = 0 :

———————————————————————————————————————————————–Proof. We have that f is Riemann integrable on [a; b] if and only if f is continuous a.e. on [a; b]:Let D =

fx

2[a; b] : f is discontinuous at x

g: Then m(D ) = 0 and f is continuous on [a; b]

nD :

Since [a; b] \ Q is dense in [a; b]; we get that f (x) = 0 for all x 2[a; b]nD :

Thus, R b

a f (x)dx = R D f (x)dx + R [a;b ]nD f (x)dx m ( D )=0

= 0 + R [a;b ]nD 0dx = 0 :

———————————————————————————————————————————————–373. Prove the following Schwartz inequality :

If f and g are Lebesgue integrable on (a; b); then R b

a f (x)g(x)dx2

R b

a (f (x))2 dx R b

a (g(x))2 dx :———————————————————————————————————————————————–Proof. 0 R

ba R

ba (f (x)g(y) f (y)g(x))2 dxdy = R

ba R

ba f (x)g(y))2 2f (x)f (y)g(x)g(y) + ( f (y)g(x) 2 dxdy

= R b

a R b

a (f (x))2(g(y)) 2dxdy 2 R b

a R b

a f (x)f (y)g(x)g(y)dxdy + R b

a R b

a (f (y))2(g(x))2dxdy

= R b

a (f (x))2dx R b

a (g(y))2dy 2 R b

a f (x)g(x)dx R b

a f (y)g(y)dy + R b

a (f (y))2dy R b

a (g(x))2dx

=

R b

a (f (x))2dx

R b

a (g(x))2dx 2

R b

a f (x)g(x)dx

R b

a f (x)g(x)dx +

R b

a (f (x))2dx

R b

a (g(x))2dx

= 2 R b

a (f (x))2dx R b

a (g(x))2dx 2 R b

a f (x)g(x)dx 2 :

Hence, R b

a f (x)g(x)dx2

R b

a (f (x))2 dx R b

a (g(x))2 dx :

———————————————————————————————————————————————–374. Let f : [a; b] ! R be nonnegative and Lebesgue integrable.

Prove that R b

a f (x)cos xdx2

+ R b

a f (x)sin xdx2

R b

a f (x)dx2

:———————————————————————————————————————————————–Proof. R

ba f (x)cos xdx

2+ R

ba f (x)sin xdx

2= R

ba p f (x)p f (x)cos xdx

2+ R

ba p f (x)p f (x)sin xdx

2

129



Schwartz

R b

a f (x)dx R b

a f (x)cos2 xdx + R b

a f (x)dx R b

a f (x)sin 2 xdx

= R b

a f (x)dx hR b

a f (x)cos2 xdx + R b

a f (x)sin 2 xdx i= R

ba f (x)dx hR

ba f (x) cos2 x + sin 2 x dxi= R

ba f (x)dx

2:

———————————————————————————————————————————————–

375. Find R [0;1] fdm; where f (x) = x2

for x 2[0; 1]nQ1 for x 2[0; 1] \ Q : Is f Riemann integrable on [0; 1]?———————————————————————————————————————————————–Solution. Since [0; 1] \ Q is dense in [0; 1] and f (x) = x2 for x 2[0; 1]nQ; we have that f is notcontinuous at any point in (0; 1): Thus, f is not Riemann integrable.Now, m([0; 1] \ Q) = 0 and so m([0; 1]nQ) = 1 :Thus, R [0;1] fdm = R [0;1]nQ x2dm + R [0;1]\ Q 1dm = R [0;1]nQ x2dm = R 1

0 x2dx = 13 :

———————————————————————————————————————————————–376. Let C be the Cantor set .

(a) Find R [0;1] fdm; where f (x) = 8<:sin( x) for x 2[0; 1=2]nC

cos( x) for x 2[1=2; 1]nC

x2 for x 2C9=;

:

(b) Let f be de…nd on [0; 1] as follows: f (x) = 8><>:0 for x 2C

n for x 22n 1

[k=1

C n;k 9>=>;;where C n;k is a removed

interval of length 13n : Find R [0;1] fdm:

———————————————————————————————————————————————–Solution. (a) We …rst note that m(C ) = 0 and so m([0; 1=2]nC ) = 1 =2 and m([1=2; 1]nC ) = 1 =2:Now, R [0;1] fdm = R [0;1=2]nC sin( x)dm + R [1=2;1]nC cos( x)dm + R C x2dm

Thus, R [0;1] fdm = R 1=20 sin( x)dx + R 1

1=2 cos( x)dx + 0 = 0 :

(b) R [0;1] fdm =1

Xn =1

2n 1

Xk =1 R C n;kfdm =

1

Xn =1

2n 1

Xk=1 R C n;kndm =

1

Xn =1

2n 1

Xk =1

n m (C n;k ) =1

Xn =1n0@

2n 1

Xk=1

13n 1A

=1

Xn =1

n 2n 1

3n = 12

1

Xn =1

n( 23 )n = 3 :

———————————————————————————————————————————————–377. De…ne the Cantor-Lebesgue function ' : [0; 1] ! [0; 1] as follows:

If x is an element of the Cantor set C and x =1

Xn =1

a n3n with an = 0 or 2; then we put

' (x) = ' (1

Xn =1

a n3n ) =

1

Xn =1

a n2

12n ;

that is, if an is the n -th ternary digit for x; then the n-th binary digit for ' (x) is a n2 :

We extend ' to [0; 1] by setting:

' (x) = sup f' (y) : y 2C ; y < xg(a) Show that the Cantor function maps the Cantor set C onto [0; 1].(b) Show that ' is increasing and continuous on [0; 1]:(c) Show that ' 0(x) = 0 a.e. on [0; 1]:(d) Compute R [0;1] 'dm and justify your answer.———————————————————————————————————————————————–Proof. See my notes on Cantor Lebesgue Function: http://www.adwan.net/RealAnalysis/Cantor.pdf .———————————————————————————————————————————————–378. Let A [0; 1] be a perfect nowhere dense set with m(A ) > 0: Show that A (x) = 1 for x 2A

0 for x =2A

130



is not Riemann integrable on [0; 1] but it is Lebesgue integrable on [0; 1].———————————————————————————————————————————————–Solution. Since A [0; 1]; we have that 0 < m (A ) 1:Now, R [0;1] A dm = R A 1dm + R [0;1]nA 0dm = m(A ): Thus, A (x) is Lebesgue integrable on [0; 1].Now, B = [0 ; 1]nA is dense in [0; 1] and A (x) 0 on B :Let a

2A : Then

9 a sequence

fbn

g in B such that lim

n !1bn = a:

Note that A (bn ) = 0 for all n 2N; and A (a) = 1 : This implies that A is not continuous at a:Thus, A D = fx 2[0; 1] : A is discontinuous at xg and so m(D ) m(A ) > 0:Hence, A (x) is not Riemann integrable on [0; 1]:

———————————————————————————————————————————————–379. Let fr n g be the sequence of rational numbers and de…ne f (x) = Xfn :r n <x g

12n :

(a) Show that f is continuous on the irrationals.(b) Show that f is discontinuous on the rationals.(c) Compute R [0;1] fdm and justify your answer.———————————————————————————————————————————————–Proof. (a) Let x0 be an irrational number, and let > 0 be given.

Since1

Xn =1

1

2n <

1; we can …nd N :

1

Xn = N

1

2n < :

Now, let > 0 be small enough so that (x0 ; x0 + ) \ f r 1 ; r 2 ;:::;r N 1g= ? :

Then 8x 2(x0 ; x0); f (x0) f (x) = Xfn :r n <x 0 g

12n Xfn :r n <x g

12n = Xfn :x r n <x 0 g

12n <

1

Xn = N

12n < :

Also, 8x 2(x0 ; x0 + ); f (x) f (x0) = Xfn :r n <x g

12n Xfn :r n <x 0 g

12n = Xfn :x 0 <r n <x g

12n <

1

Xn = N

12n < :

Thus, 8x 2(x0 ; x0 + ); jf (x) f (x0)j< :Hence, f is continuous at x0 :

(b) Let x0 be a rational number. Then x0 = rm for some m 2N:

Thus, if x > x 0 ; then f (x) f (x0) =

Xfn :x 0 r n <x g

12n = 0

@ 12m +

Xfn :x 0 <r n <x g

12n 1A

> 12m :

Thus, f is discontinuous at x0 :

(c) R [0;1] fdm = R [0;1] 0@Xfn :r n <x g

12n 1A

dm = R [0;1]

1

Xn =1(r n ;1 ) (x) 1

2n !dm =1

Xn =1

12n R [0;1] ( r n ;1 ) (x) dm

=1

Xn =1

12n (m (( r n ;1 ) \ [0; 1])) :

———————————————————————————————————————————————–

380. Let fr n g be the sequence of rational numbers in [a; b] and de…ne f (x) =1

Xn =1

12n

1p jx r n j:

(a) Show that R b

a f (x)dx < 1 :(b) Show that

R b

a (f (x))2 dx =

1:

———————————————————————————————————————————————–Proof. (a) R b

a f (x)dx = R ba

1

Xn =1

12n

1p jx r n jdx Monotone Convergence Theorem=

1

Xn =1 R ba

12n

1p jx r n jdx

=1

Xn =1

12n hR

r n

a1p r n x dx + R

br n

1p x r ndxi=

1

Xn =1

12n h[ 2p r n x]r n

a + [2p x r n ]br n i=

1

Xn =1

12n 2p r n a + 2 p b r n

1

Xn =1

p b a + p b a2n 1 = 2 p b a

1

Xn =1

12n 1 = 4 p b a < 1 :

131





Then f is Lebesgue integrable on R and R R fdm = 1 :Then for such a continuous function g; we have R R jf gjdm = 0 Problem 3=) j f gj= 0 a.e. on R=) f = g a.e. on R=) m(fx 2R : 0 < g (x) < 1g) = 0 = ) m(g 1(0; 1)) = 0 :But g is continuous =) g 1(0; 1) is open =) m(g 1(0; 1)) > 0: A contradiction.Hence, there is no such g and the statement is False.———————————————————————————————————————————————–385. We say that functions f n ; n = 1 ; 2; 3; :::; integrable on A are equi-integrable if 8 > 0 9 > 0 such thatfor every measurable subset B of A, R B jf n jdm < for n = 1 ; 2; 3; :::; whenever m(B ) < :Show that if ff n g is a convergent sequence, say f n ! f; of equi-integrable functions on a set A , m(A ) < 1 ;then lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–Proof. Let > 0 be given. By equi-integrability of ff n g; 9 > 0 such that for every measurablesubset B of A, R B jf n jdm <

3 for n = 1 ; 2; 3; :::; whenever m(B ) < :By Fatou’s Theorem, R B jf jdm = R B lim jf n jdm limR B jf n jdm <

3 :By Egorov’s Theorem, 9 a measurable subset C A ; m(A nC ) < such that f n

Cf:

Thus, 9N : jf n (x) f (x)j< 3m ( C ) for all n N and for all x 2C :

Therefore, for n N;

R A f n dm

R A fdm =

R A (f n f )dm

R A jf n f jdm =

R C jf n f jdm+

R A nC jf n f jdm

R C jf n f jdm + R A nC jf n jdm + R A nC jf jdmm ( A

nC )<

< R C 3m ( C ) dm + 3 + 3 = :Thus, lim

n !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–386. (Fatou in measure) Show that if f n 0 and f n

m

! f on A , then R A fdm limR A f n dm:———————————————————————————————————————————————–Proof. Let an = R A f n dm and a = lim an : Then there is a subsequence an k ! a as k ! 1 :Since f n

m

! f; we get that f n km

! f and so 9 a subsequence f n k m

a.e.

! f on A as m ! 1 :

Thus, Fatou’s Theorem =) R A fdm = R A limm !1

f n k m dm limm !1 R A f n k m dm = lim

m !1an k m = a:

Hence, R A fdm limR A f n dm:

———————————————————————————————————————————————–387. (Monotone Convergence in measure) Show that if f n

0; f n

f n +1 ; and f n

m

!f on A ,

then limn !1 R A f n dm = R

A fdm:———————————————————————————————————————————————–Proof. Since f n

m

! f , 9 a subsequence f n k

a.e.

! f on A as k ! 1 :Thus, by the Lebesgue Monotone Convergence Theorem, lim

k!1 R A f n k dm = R A fdm:

Since f n f n +1 ; we conclude that limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–388. (Lebesgue DCT in measure) Show that if f n

m

! f on A , and there exists a function g integrableon A such that jf n (x)j g(x) for all n 2N and for all x 2A ; then lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–Proof. Let an = R A f n dm and a = R A fdm: If lim

n !1an 6= a; then 9 a subsequence fan k g: an k ! b 6= a:

Since f nm



a.e.

! f on A as m ! 1 :Now, f n k m (x) g(x) for all m

2N and for all x

2A :

Thus, by the Lebesgue Dominated Convergence Theorem, we get that limm !1 R A f n k m dm = R A fdm:In other words, lim

m !1an k m = a: But lim

m !1an k = b =) lim

m !1an k m = b 6= a: A contradiction.

Hence, limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–389. (Equi-integrability in measure) Show that if ff n g is a sequence of equi-integrable functions ona set A , m(A ) < 1 ; and f n

m

! f on A , then limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–Proof. Let an = R A f n dm and a = R A fdm: If lim


133



Since f nm

! f; we get that f n k

m


a.e.

! f on A as m ! 1 :Since ff n k m g are equi-integrable, we can use problem 39 to get that lim

m !1 R A f n k m dm = R A fdm:In other words, lim


m !1an k = b =) lim



———————————————————————————————————————————————–

390. Let f ; f n : R ! R be integrable functions such that f na.e.

! f on R:Show that limn !1 R R jf n f jdm = 0 if and only if lim

n !1 R R jf n jdm = R R jf jdm:———————————————————————————————————————————————–Proof. (=) ) 0 R R jf n jdm R R jf jdm = R R (jf n j jf j) dm R R jf n f jdm ! 0 as n ! 1 :Thus, lim

n !1 R R jf n jdm = R R jf jdm:(( =) We prove it …rst for sets of …nite measure. So let A R be a measurable set, m(A ) < 1 and let

> 0 be given. Since jf j is integrable on A ; 9 > 0 : whenever C A is measurable and m(C ) < ;we have R C jf jdm <

4 : Since limn !1 R A jf n jdm = R A jf jdm; we can …nd N 1 such that R C jf n jdm <

2 8n N 1 :

Since f na.e.

! f on A , Egorov’s Theorem =) 9 a measurable set B A , m(A nB ) < such that f n B

f:

Thus, 9N 2 : jf n (x) f (x)j< 4m ( B ) for all n N 2 and for all x 2B :

Let N = max fN 1 ; N 2g: Then, for n N;

R A jf n f jdm =

R B jf n f jdm +

R A nB jf n f jdm

R B 4m ( B ) dm + R A nB jf n jdm + R A nB jf jdmm ( A

nB )<

4m ( B ) R B dm +

2 + 4 = :

Therefore, limn !1 R A jf n f jdm = 0 :

Now, Let A m = [m; m + 1] : Then R =1

[m = 1A m and so R R jf n f jdm =

1

Xm = 1R A m jf n f jdm:

Hence, limn !1 R R jf n f jdm = lim

n !11

Xm = 1R A m jf n f jdm =1

Xm = 1lim

n !1 R A m jf n f jdm =1

Xm = 10 = 0 :

———————————————————————————————————————————————–391. Use Fubini’s theorem to prove that R R n e jxj2 dx = n= 2 :———————————————————————————————————————————————–Proof.

R R n e jxj2

dx =

R R

R R

R R e x 2

1 x 22 x 2

n dx1dx2 dxn =

R R

R R

R R e x 2

1 e x 22 e x 2

n dx1dx2 dxnFubini=

R R e x 2

1 dx1

R R e x 2

2 dx2

R R e x 2

n dxn :

Now, R R e x 2dx

2= R R e x 2

dx R R e x 2dx Fubini= R R R R e x 2 y2

dxdy Polar Coordinates= R 2

0 R 10 e r 2

(rdrd )Fubini= R

20 d R 1

0 re r 2dr = (2 ) (h1

2 e r 2i10 ) = : Thus, R R e x 2dx = 1=2 :

Therefore, R R n e jxj2 dx = R R e x 21 dx1 R R e x 2

2 dx2 R R e x 2n dxn = 1=2 1=2 1=2 = n= 2 :

———————————————————————————————————————————————–392. Let a > 1: Prove that R

10

x sin x1+( nx ) a dx = o( 1

n ):———————————————————————————————————————————————–Proof. We need to show that lim

n !1n R

10

x sin x1+( nx ) a dx = 0 : Let f n (x) = nx sin x

1+( nx ) a ; n = 1 ; 2; 3; ::::Then lim

n !1f n (x) = 0 since a > 1:

Now, if x 2[0; 1n ]; then nx 1 and so nx 1 + ( nx )a which imples that jf n (x)j 1:

If x

2[1

n ; 1]; then nx(nx ) a = 1

(nx ) a 1

1 since (nx )a 1 1: Hence, nx 1 + ( nx )a and so

jf n (x)

j1:

Therefore, for all n and for all x 2[0; 1]; jf n (x)j 1:Moreover, R

10 1dx = 1 < 1 and so, by the Lebesgue DCT, we get that

limn !1

nR 10

x sin x1+( nx ) a dx = R 1

0 limn !1

nx sin x1+( nx ) a dx = R 1

0 0dx = 0 :

———————————————————————————————————————————————–393. Compute lim

n !1 R 10

dx(1+ x

n ) n x 1 =n and justify your answer.———————————————————————————————————————————————–Solution. For x 2(0;1 ); let f n (x) = 1

(1+ xn ) n x 1 =n ; n = 1 ; 2; 3; :::: Then lim

n !1f n (x) = e x :

For x 2 (0; 1); we have f n (x) 1x 1 =n for n = 1 ; 2; 3;::::

134



For x 2[1;1 ); we have f n (x) 1(1+ x

n ) n 1(1+ x

2 ) 2 for n 2:

Let gn (x) = 1x 1 =n for x 2(0; 1)

1(1+ x

2 ) 2 for x 2[1;1 ) for n 2: Then f n (x) gn (x) on (0;1 ) for n 2:

Then limn !1

R 1

0 gn (x)dx = limn !1 h

R

10

1x 1 =n dx +

R

10

1(1+ x

2 ) 2 dxi= limn !1 h n

n 1 + 23i= 5

3 :

Also, limn !1 gn (x) = g(x); where g(x) = 1 for x

2(0; 1)

1(1+ x

2 ) 2 for x 2[1;1 ) ; and R 10 g(x)dx = 53 :Thus, gn (x) ! g(x) in L1((0 ;1 )) :Therefore, by the variant of Lebesgue DCT, (see problem 56), we get thatlim

n !1 R 10

dx(1+ x

n ) n x 1 =n = limn !1 R 1

0 f n (x)dx = R 10 lim

n !1f n (x)dx = R 1

0 e x dx = 1 :

———————————————————————————————————————————————–394. Compute lim

n !1 R 1

0log( x + n )

n e x cos x dx and justify your answer.———————————————————————————————————————————————–Solution. Let f n (x) = log( x + n )

n e x cos x; n = 1 ; 2; 3; ::::: Then limn !1

f n (x) = 0 on (0; 1):

Now, jf n (x)j= log( x + n )n e x cos x e x for n = 1 ; 2; 3;:::: Since R

10 e x dx = 1 e 1 < 1 ;

the Lebesgue DCT implies that limn !1

R 1

0log( x + n )

n e x cos xdx =

R 1

0 limn !1

log( x + n )n e x cos x dx =

R 1

0 0dx = 0 :———————————————————————————————————————————————–395. (a) Compute limn !1 R

10 nx1+ n 2 x 2 dx and justify your answer.(b) Compute lim

n !1 R 1

0n 3 = 2 x

1+ n 2 x 2 dx and justify your answer.———————————————————————————————————————————————–Solution. (a) lim

n !1 R 1

0nx

1+ n 2 x 2 dx = limn !1

12n R

10

2n 2 x1+ n 2 x 2 dx = lim

n !11

2n log 1 + n2x2 10 = lim

n !1log (1+ n 2 )

2n = 0 :

(b) limn !1 R

10

n 3 = 2 x1+ n 2 x 2 dx = lim

n !11

2p n R 1

02n 2 x

1+ n 2 x 2 dx = limn !1

12p n log 1 + n2x2 1

0 = limn !1

log (1+ n 2 )2p n = 0 :

———————————————————————————————————————————————–

396. Compute1

Xn =0 R =2

0 1 p sin xn

cos xdx and justify your answer.

———————————————————————————————————————————————–Solution. For x 2[0;

2 ]; let f n (x) = 1 p sin xn

cos x: Then f n (x) 0 for all n and for all x 2[0; 2 ]:

Now, if one sets gn (x) =

n

Xk=0f k (x); then one gets that gn (x) %g(x) =

1

Xk =0f k (x) as n ! 1 :

Thus, By the Lebesgue monotone convergence theorem, we get that

limn !1 R

=20 gn (x)dx = R

= 20 lim

n !1gn (x) dx = R

=20

1

Xk=0

f k (x)!dx = R =2

0

1

Xk=0

1 p sin xk

cos x!dx

= R =2

0 cos x1

Xk =0

1 p sin xk!dx

(1 p sin x ) 1= R

=20

cos x1 (1 p sin x ) dx = R

= 20

cos xp sin x dx = 2 :

———————————————————————————————————————————————–397. Compute lim

n !1 R 1

0 (1 + xn ) n sin( x

n )dx and justify your answer.———————————————————————————————————————————————–Solution. Let f n (x) = (1 + x

n ) n sin( xn ); n = 1 ; 2; 3; ::::: Then lim

n !1f n (x) = 0 on (0; 1):

Now,

jf n (x)

j= (1 + x

n ) n sin( xn ) (1 + x

n ) n (1 + x2 ) 2 for all x

2[0; 1] and for all n 2:

Since R 10 (1 + x2 ) 2dx = 23 < 1 ; the Lebesgue DCT implies that limn !1 R

10 (1 + xn ) n sin( xn )dx =

R 1

0 limn !1

(1 + xn ) n sin( x

n ) dx = R 1

0 0dx = 0 :———————————————————————————————————————————————–398. Compute lim

n !1 R 10

1+ nx(1+ x ) n dx and justify your answer.

———————————————————————————————————————————————–Solution. Let f n (x) = 1+ nx

(1+ x ) n ; n = 1 ; 2; 3; ::::: Then limn !1

f n (x) = 0 on (0; 1):

Now, jf n (x)j= 1+ nx(1+ x ) n 1 for all n:Since R

10 1dx = 1 < 1 ; the Lebesgue DCT implies

135



that limn !1 R

10

1+ nx(1+ x ) n dx = R

10 lim

n !11+ nx

(1+ x ) n dx = R 1

0 0dx = 0 :———————————————————————————————————————————————–399. Show that f (x) = 1 for x = 1

n0 otherwise is Riemann integrable on [0; 1] and compute R

10 f (x)dx:

———————————————————————————————————————————————–Proof. Note that f (x) is continuous on A = (0 ; 1]nf1n : n 2Ng and m(f1

n : n 2Ng [ f0g) = 0 :

Thus, f is continuous a.e. on [0; 1] and so f is Riemann integrable on [0; 1]:Now, R

10 f (x)dx = R A f (x)dx + R [0;1]nA f (x)dx = R A 0dx + R [0;1]nA f (x)dx = 0 + 0 = 0 :

———————————————————————————————————————————————–400. (Variant of Fatou’s Theorem) Let fgn g be a sequence of integrable functions on A such thatgn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such that f n

a.e.

! f on A ,and f n gn a.e. for each n 2N: Prove that lim R A f n dm R A limf n dm = R A fdm:———————————————————————————————————————————————–Proof. Let hn (x) = gn (x) f n (x): Then hn (x) 0 a.e. for each n 2N:Now, Fatou’s theorem =) R A (limhn ) dm lim R A hn dm=) R A (lim( gn f n )) dm lim R A (gn f n ) dm=) R A (limgn ) limf n dm lim R A gn dm R A f n dm=)

R A (lim gn ) dm

R A limf n dm lim

R A gn dm lim

R A f n dm

Now, since gn ! g in L1

(A

); we get that limn !1 R A gn dm = R

A gdm and so=) R A gdm R A limf n dm R A gdm lim R A f n dm=) lim R A f n dm R A limf n dm = R A fdm:

———————————————————————————————————————————————–401. (Variant of Fatou’s Theorem in measure) Let fgn g be a sequence of integrable functions on Asuch that gn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such thatf n

m

! f on A , and f n gn a.e. for each n 2N: Prove that lim R A f n dm R A limf n dm:———————————————————————————————————————————————–Proof. Let an = R A f n dm and a = lim an : Then there is a subsequence an k ! a as k ! 1 :Since f n

m



a.e.

! f on A as m ! 1 :Thus, Variant of Fatou’s Theorem =) a = lim R A f n k m dm R A limf n k m dm = R A fdm:Hence, liman = lim

R A f n dm

R A limf n dm =

R A fdm:

———————————————————————————————————————————————–402. (Variant of Lebesgue’s DCT) Let fgn g be a sequence of integrable functions on A such thatgn ! g in L1(A ): Suppose that ff n g is a sequence of measurable functions on A such that f n

a.e.

! f on A ,and jf n j gn a.e. for each n 2N: Prove that lim

n !1 R A f n dm = R A fdm:———————————————————————————————————————————————–Proof. Applying Fatou’s theorem and the variant of Fatou’s theorem, we get that:

R A jf jdm = R A (lim jf n j) dmFatou

lim R A jf n jdm lim R A jf n jdmVariant of Fatou

R A lim jf n j dm= R A jf jdm: Hence, R A jf jdm = lim R A jf n jdm = lim R A jf n jdm and so lim

n !1 R A jf n jdm = R A jf jdm:

Now, to get the result, we note that f na.e.

! f on A =) f +na.e.

! f + and f na.e.

! f on A .

———————————————————————————————————————————————–403. (Variant of Lebesgue’s DCT in measure) Let fgn g be a sequence of integrable functions on Asuch that gn

!g in L1(A ): Suppose that

ff n

g is a sequence of measurable functions on A such that

f n m! f on A , and jf n j gn a.e. for each n 2N: Prove that limn !1 R A f n dm = R A fdm:

———————————————————————————————————————————————–Proof. Let an = R A f n dm and a = R A fdm: If lim


Since f nm



a.e.

! f on A as m ! 1 :Now, f n k m (x) gn (x) for all m 2N and for a.e. x 2A :Thus, by the variant of Lebesgue DCT, we get that lim

m !1 R A f n k m dm = R A fdm:In other words, lim


m !1an k = b =) lim



136



———————————————————————————————————————————————–404. Let 1 p < 1 ; f n 2L p(R) such that f n

a.e.

! f on R: Suppose that(i) 9n1 > 0 and A R with m(A ) < 1 : R RnA jf n j

p dm 1 8n n1 ; and(ii) 9n0 > 0 and 0 < < 1 : whenever m(F ) < ; R F jf n j

p dm 1 8n n0 :Show that f 2L p(R):———————————————————————————————————————————————–Proof. f n a.e.! f on R =) j f n j p a.e.! jf j p on R:By Fatou’s Theorem, R R jf j

p dm = R R lim jf n j p dm limR R jf n j

p dm:Thus, R RnA jf j

p dm limR RnA jf n j p dm 1; and whenever m(F ) < ; R F jf j

p dm limR F jf n j p dm 1.

Now, since m(A ) < 1 ; Egorov’s Theorem =) 9B A : m(A nB ) < and jf n j p

B jf j p :

Thus, 9N 1 : jjf n j p

jf j pj 1 8x 2B and 8n N 1 =) on B , jf j

p 1 + jf N 1 j p.

Hence, R R jf j p dm = R RnA jf j

p dm + R B jf j p dm + R A nB jf j

p dm 1 + R B (1 + jf N 1 j p)dm + 1

= 2 + m(B ) + R B jf N 1 j p dm < 1 :

Therefore, f 2L p(R):

———————————————————————————————————————————————–405. Prove the following Tchebyshev’s Inequality : Let f be nonnegative and measurable on A :If > 0; then m(fx 2A : f (x) > g) 1

R A fdm:

———————————————————————————————————————————————–Proof. Let B = fx 2A : f (x) > g: Then R A fdm R B fdm R B dm = m (B ):Thus, m(B ) 1 R A fdm:

———————————————————————————————————————————————–406. Let f n (x) = p ne nx on [0; 1]: Prove that:(a) f n (x) ! 0 pointwise in (0; 1]:(b) R 1

0 (f n (x))2dx C for all n 2N:(c) f n does not converge in L2(0; 1):(d) R 1

0 f n (x)g(x)dx ! 0 for each g 2L2(0; 1):———————————————————————————————————————————————–Proof. (a) Let x 2(0; 1]: Then lim

n !1f n (x) = lim

n !1p ne nx = lim

n !1p nenx

L’Hopital= lim

n !11

2p nxe nx = 0 :

(b)

R 1

0 (f n (x))2dx =

R 1

0 (p ne nx )2dx =

R 1

0 ne 2nx dx = 12 1 e 2n 1 for all n 2N:

(c) limn !1 R 1

0 (f n (x))2dx = limn !112 1 e 2n =

12 ; whereas R

10 limn !1 (f n (x))2dx = 0 :

(d) Let ' 2L2(0; 1) be a simple function. Write ' (x) =k

Xi =1

a i A i (x); where [0; 1] =k

[i =1

A i :

Then limn !1 R

10 f n (x)' (x)dx = lim

n !1 R 1

0 f n (x) k

Xi =1

a i A i (x)!dx!= limn !1

k

Xi =1

a i R 1

0 f n (x) A i (x)dx != lim

n !1

k

Xi =1

a i R A if n (x)dx !:

Now, R 1

0 f n (x)dx = R 1

0p ne nx dx = 1p n R

10 ne nx dx = 1p n (1 e n ) n !1! 0: Hence, f n converges in L1(0; 1):

Thus, limn !1

R A i

f n (x)dx =

R A i

limn !1

f n (x)dx =

R A i

0dx = 0 ; and so limn !1

R 1

0 f n (x)' (x)dx = 0 :Now, let g

2L2(0; 1) and let > 0 be given: Then, by density of simple functions in L2(0; 1);

9 a simple function ' 2L2(0; 1) : R 1

0 jg(x) ' (x)j2 dx < 2

4 :By what we did above, 9N :R

10 jf n (x)' (x)jdx <

2 for all n N:Thus, R

10 jf n (x)g(x)jdx = R

10 jf n (x)g(x) f n (x)' (x) + f n (x)' (x)jdx

R 1

0 jf n (x)g(x) f n (x)' (x)jdx + R 1

0 jf n (x)' (x)jdx= R

10 jf n (x)j jg(x) ' (x)jdx + R

10 jf n (x)' (x)jdx

137



Schwartz Z 1

0 jf n (x)j2 dx

1=2

| {z } 1 by (b)R

10 jg(x) ' (x)j

2 dx1=2

+ R 1

0 jf n (x)' (x)jdx

2

4

1=2+

2 = for all n N:

Thus, R 1

0 f n (x)g(x)dx ! 0 for each g 2L2(0; 1):

———————————————————————————————————————————————–407. Let f n (x) = 1

jx 1n j

1 = 2 on (0; 1): Prove that:

(a) f n (x) converges pointwise on (0; 1):(b) f n (x) converges in measure on (0; 1):(c) f n converges in L1(0; 1):(d) @ g 2L1(0; 1) : f n (x) g(x) for a.e. x 2(0; 1) and for all n 2N:———————————————————————————————————————————————–Proof. (a) Let x 2(0; 1)nf1

m : m 2Ng: Then limn !1

f n (x) = limn !1

1

jx 1n j

1 = 2 = 1p x :

If x = 1m for some m 2N; then for n m + 1 ; we also get that lim

n !1f n ( 1

m ) = limn !1

1

j1m

1n j

1 = 2 = 1p 1m :

(b) Since m ((0; 1)) = 1 < 1 ; and f n (x) ! 1p x pointwise in (0; 1); f n (x) m

! 1p x :

(c) We have R 1

0 limn !1f n (x) dx = R

10

1p x dx = 2 :

Now, limn !1 R

10 f n (x)dx = lim

n !1 R 1=n

01p 1n x

dx + R 1

1=n1p x 1

ndx = lim

n !1 h2q 1n xi

1=n

0+ h2q x 1

n i1

1=n

= limn !1

2q 1n + 2q 1 1

n = 2 :

Thus, R 10 lim

n !1f n (x) dx = lim

n !1 R 10 f n (x)dx and so, f n converges in L1(0; 1):

———————————————————————————————————————————————–408. Let p 1: Suppose f n 2L p(R), sup n jf n j 2L p(R); and f n

a.e.

! f in R. Prove that f 2L p(R) andthat f n ! f in L p(R):———————————————————————————————————————————————–Proof. Let g(x) = sup n jf n (x)j: Then jf n (x)j

p g(x) p for all x 2R and for all n 2N:Now, f n

a.e.

!f in R =

) jf n

j p a.e.

! jf

j p in R:

Fatou’s theorem =) R R jf (x)j p dx = R R lim jf n (x)j p dx limR R jf n (x)j p dx R R g(x) pdx < 1 :Now, Lebesgue’s DCT =) lim

n !1 R R jf n (x)j p dx = R R jf (x)j

p dx:

———————————————————————————————————————————————–409. Let A R be measurable with m(A ) < 1 : Let ff n g be a sequence of functions in L1(A ) such thatf n

m

! 0 on A , and kf n kL 1 ( A ) M for all n 2N. Prove that limn !1 R A jf n gj

1=2 dm = 0 for all g 2L1(A ):———————————————————————————————————————————————–Proof. Let g 2L1(A ) and suppose that R A jf n gj

1=2 dm 9 0:Then 9 a subsequence R A jf n k gj

1=2 dm r > 0 for all k = 1 ; 2; 3; ::::

Now, R A f n k m g 1=2 dm = R A f n k m

1=2

jgj1=2 dm

Schwartz

R A f n k m dm 1=2

R A jgjdm 1=2 :Since f n k

m

! 0; 9 a subsequence f n k m

a.e.

! 0 as m ! 1 :Let > 0 be given. Since m(A ) < 1 ; Egorov’s theorem =) 9B A : m(A nB ) <

2M and f n k m

B0:

Thus, 9N : f n k m (x) < 2m ( B ) 8x 2B and 8m N:Now, R A nB f n k m dm f n k m L 1 ( A ) m(A nB ) M

2M = 2 :

Thus, 8m N; R A f n k m dm = R B f n k m dm + R A nB f n k m dm < R B 2m ( B ) dm + R A nB f n k m dm < 2 +

2 = :

Hence, limm !1 R A f n k m dm 1=2 = 0 and since g 2L1(A ); lim

m !1 R A f n k m dm 1=2

R A jgjdm 1=2 = 0 :

Thus, limm !1 R A f n k m g 1=2 dm = 0 ; a contradiction to R A jf n k gj

1=2 dm r > 0 for all k = 1 ; 2; 3; ::::

Therefore, limn !1 R A jf n gj

1=2 dm = 0 for all g 2L1(A ):

———————————————————————————————————————————————–

138



410. Let f be a measurable function on [0; 1] and let A = fx 2[0; 1] : f (x) 2Zg:Prove that A is measurable and that lim

n !1 R 1

0 [cos( f (x))]2n dx = m(A ):———————————————————————————————————————————————–Proof. A = [n 2Z

f 1fng ; and f 1fng= f 1[n; 1 ) n f 1(n; 1 ) is measurable since f is measurable.

Thus, A is a countable union of measurable sets and so is measurable.Now, let f n (x) = [cos( f (x))]2n : Then for x 2A ; f (x) = m 2Z; and so f n (x) = [cos( f (x))]2n

= [cos( m)]2n = 1 ; and for x =2A ; f (x) =2Z and so 0 f n (x) < 1:

Hence, limn !1

f n (x) = 1 x 2A

0 x =2A = A (x):

Now, jf n (x)j 1 for all n 2N and for all x 2[0; 1]: Also, R 1

0 1dx = 1 < 1 ; and so by the Lebesgue DCTwe get that lim

n !1 R 1

0 [cos( f (x))]2n dx = R 1

0 limn !1

[cos( f (x))]2n dx = R 1

0 A (x)dx = m(A ):

———————————————————————————————————————————————–411. Show that if m(A ) < 1 ; and 0 1 and let p0 be such that 1 p + 1

p0 = 1 : Then 1 p0 = 1 p1

p2:

Then by Hölder’s inequality: R A jf j p1 dm R A jf j

p1 p dm1p R A 1 p0

dm1

p 0:

Thus, R A jf j p1 dm R A jf j

p2 dm1p (m(A ))

1p 0

=) R A jf j p1 dm

1p 1 R A jf j

p2 dm 1pp 1 (m(A ))

1p 0p 1 = R A jf j

p2 dm 1=p2 (m(A ))1

p 1 1p 2 :

Thus, if f 2L p2 (A ); then kf k p1 kf k p2(m(A ))

1p 1

1p 2 and since m(A ) < 1 ; f 2L p1 (A ):

———————————————————————————————————————————————–412. Given f 2L2([0; 1]) de…ne K (x) = 1

x 4 = 3 R x

0 f (t)dt:Show that kK k1 C kf k2 where C is a constant independent of f :———————————————————————————————————————————————–Proof.

kK

k1 =

R 1

0 1x 4 = 3

R x

0 f (t)dt dx =

R 1

0 1x 1 = 3

1x

R x

0 f (t)dt dx

R 1

0 1x 1 = 3

1x R

x0 jf (t)jdt dx

Schwartz

R 1

0 1x 1 = 3

2 dx12

R 1

01x R

x0 jf (t)jdt

2dx

12

Mean Value Theorem= R 1

01

x 2 = 3 dx12

R 1

0 jf (cx )j2 dx

12

; for some 0 < cx < x

Now, R 10 jf (cx )j

2 dx12

R 10 jf (x)j

2 dx12

= kf k2 :

Also, R 1

01

x 2 = 3 dx12

= p 3:

Hence, kK k1 R 1

01

x 2 = 3 dx12

R 1

0 jf (cx )j2 dx

12 p 3 R

10 jf (x)j

2 dx12

= p 3kf k2 :

———————————————————————————————————————————————–413. Recall that the convolution of two integrable functions f and g is de…ned by:

(f g) (x) = R R f (x y)g(y)dy:

Let f 0 such that R R f (x)dx = A < 1: De…ne the sequence f n = f f f

| {z } n times

:

Prove that f n 2L1(R) for each n 2N, and f n ! 0 in L1(R):———————————————————————————————————————————————–Proof. Note that R R (f f ) (x)dx = R R R R f (x y)f (y)dydx Fubini= R R f (y) R R f (x y)dx dyu = x y

= R R f (y) R R f (u)du dy = R R f (u)du R R f (y)dy = A2 :

139



We claim that R R f n (x)dx = An : To prove it, we proceed by induction.Suppose that R R f n 1(x)dx = An 1 :

Then R R f n (x)dx = R R (f n 1 f )(x)dx as above= R R f n 1(x)dx R R f (x)dx = An 1A = An < 1 < 1 :Thus, f n 2L1(R) for each n 2N:Now, lim

n

!1 R R jf n (x)jdx f 0= lim

n

!1 R R f n (x)dx = lim

n

!1

An 0 A< 1= 0 : Hence, f n ! 0 in L1(R):

———————————————————————————————————————————————–414. Let ff n g be a sequence of measurable functions on A , m(A ) < 1 ; such that f nm

! f on A ,and jf n (x)j C for x 2A , and n = 1 ; 2; 3; :::: Show that if g is continuous on [ C; C ]; thenlim

n !1 R A g(f n )dm = R A g(f )dm:———————————————————————————————————————————————–Proof. Let > 0 be given. Since g is continuous on a compact set [ C; C ]; g is uniformly continuouson [ C; C ]: Thus, 9 > 0 : whenever x; y 2[ C; C ]; and jx yj< =) j g(x) g(y)j< :We now show that g(f n ) m

! g(f ) on A .Note that fx 2A : jg(f n ) g(f )j g fx 2A : jf n f j g; and consequently,0 m(fx 2A : jg(f n ) g(f )j g) m(fx 2A : jf n f j g) ! 0 as n ! 1 :Thus, lim

n !1m(fx 2A : jg(f n ) g(f )j g) = 0 and so g(f n ) m

! g(f ) on A .Now, jg(f n )j M = max

x2[ C;C ]g(x) for all n = 1 ; 2; 3;::: and for all x 2A :

Also, since m(A ) < 1 ; we see that R A Mdm = M m (A ) < 1 :Thus, by the Lebesgue DCT in measure, problem 42, we get that lim

n !1 R A g(f n )dm = R A g(f )dm:

———————————————————————————————————————————————–415. Let ff n g be a sequence of measurable functions on A , m(A ) < 1 ; such that f n

m

! f on A .Then lim

n !1 R A sin( f n )dm = R A sin(f )dm:———————————————————————————————————————————————–Proof. As in the previous problem, we get that sin(f n ) m

! sin( f ) on A .Now, jsin(f n )j 1 for all n and for all x 2A : Note that R A 1dm = m(A ) < 1 :Thus, by the Lebesgue DCT in measure, problem 42, we get that lim

n !1 R A sin( f n )dm = R A sin( f )dm:

———————————————————————————————————————————————–416. Suppose that g satis…es a Lipschitz condition on R; i.e., 9C > 0 : 8x; y 2R; jg(x) g(y)j C jx yj:Show that if f is Lebesgue integrable on [a; b]; then g(f ) is Lebesgue integrable on [a; b]:———————————————————————————————————————————————–Proof. Let > 0 be given. Since f 2L1([a; b]); 9 a continuous function ' such that R [a;b ] jf ' jdm < :Thus, R [a;b ] jg(f )jdm = R [a;b ] jg(f ) g(' ) + g(' )jdm R [a;b ] jg(f ) g(' )jdm + R [a;b ] jg(' )jdm

C R [a;b ] jf ' jdm + R [a;b ] jg(' )jdmg ' is continuous

< C (b a) + M (b a) < 1 ; where M = max[a;b ] jg(' )j:

Hence, R [a;b ] jg(f )jdm < 1 and g(f ) is Lebesgue integrable on [a; b]:

———————————————————————————————————————————————–417. Suppose that g satis…es a Lipschitz condition on R; i.e., 9C > 0 : 8x; y 2R; jg(x) g(y)j C jx yj:Show that if ff n g is a sequence of measurable functions on [a; b] such that f n

m

! f and there is a Lebesgueintegrable function G such that jf n (x)j G(x); then lim

n !1 R [a;b ] g(f n )dm = R [a;b ] g(f )dm:———————————————————————————————————————————————–Proof. Let > 0 be given. Then fx 2[a; b] : jg(f n ) g(f )j g fx 2[a; b] : jf n f j

C g; andconsequently, 0 m(

fx

2[a; b] :

jg(f n ) g(f )

j g) m(

fx

2[a; b] :

jf n f

j

C g)

!0 as n

! 1:

Thus, g(f n ) m

! g(f ) on [a; b].Now, jg(f n ) g(G)j C jf n Gj 2C jGj and so jg(f n )j 2C jGj+ jg(G)j for all n and for all x 2[a; b]:We have shown, in problem 70, that g(G) is Lebesgue integrable on [a; b]:Hence, 2C jGj+ jg(G)j is Lebesgue integrable on [a; b]:Thus, by the Lebesgue DCT in measure, problem 42, we get that lim

n !1 R [a;b ] g(f n )dm = R [a;b ] g(f )dm:

———————————————————————————————————————————————–418. Let ff n gbe a sequence of functions converging to f on A such that R A jf n j

p dm < 1 and R A jf j p dm < 1 ;

1 p < 1 : Show that limn !1 R A jf n f j

p dm = 0 if and only if limn !1 R A jf n j

p dm = R A jf j p dm:

———————————————————————————————————————————————–

140



Proof. (=) ) Suppose that limn !1 R A jf n f j

p dm = 0 : Then by Minkowski’s inequality, we get that

kf n k p = kf n f + f k p kf n f k p + kf k p and kf k p = kf f n + f n k p kf n f k p + kf n k p :

Thus, 0 kf n k p kf k p kf n f k p ! 0 as n ! 1 :

Hence, limn !1 kf n k p kf k p = 0 and so lim

n !1

R A jf n j

p dm =

R A jf j

p dm:

(( =) We prove it …rst for m(A ) < 1 : Suppose that limn !1 R A jf n j p dm = R A jf j p dm and let > 0 be given.

We can …nd N 1 large so that R A jf n j p dm < R A jf j

p dm + 2p for all n N 1 :

Since jf j p is Lebesgue integrable on A , we can use problem 1c to …nd > 0 such that whenever B A is

measurable and m(B ) < ; we have that R B jf j p dm <

2p +1 :Now, Egorov’s theorem implies that we can …nd a measurable set C A with m(A nC ) < such that f n

Cf:

Thus, we can …nd N 2 large so that for all x 2C ; we have that jf n (x) f (x)j< 1 =p

m ( C ) 1 =p for all n N 2 :Hence, for n N = max fN 1 ; N 2g; we have

R A jf n f j p dm = R C jf n f j

p dm+R A nC jf n f j p dm R C jf n f j

p dm+2 p R A nC jf j p dm + R A nC jf n j

p dm

R C jf n f j p dm + 2 p R A nC jf j

p dm + R A nC jf j p dm +

2p = R C jf n f j p dm + 2 p 2R A nC jf j

p dm + 2p

R C

jf n f

j p dm + 2 p+1

R A

nC

jf

j p dm +

R C

jf n f

j p dm + 2 p+1

2p +1 +

R C

jf n f

j p dm + 2

R C 1 =p

m ( C ) 1 =p

pdm + 2 + 2 = 3 : Thus, lim

n !1 R A jf n f j p dm = 0 :

If m(A ) = 1 ; then, by problem 1, 9a measurable set B A ; m(B ) < 1 so that R A jf j p dm < R B jf j

p dm + :Then we proceed as above to show that lim

n !1 R A jf n f j p dm = 0 :

———————————————————————————————————————————————–419. Let ff n g be a sequence of functions in L p[a; b]; 1 0 : kf n k p C for n = 1 ; 2; 3; :::: Prove that 8g 2Lq[a; b]; with 1

p + 1q = 1 ;

limn !1 R [a;b ] f n gdm = R [a;b ] fgdm:———————————————————————————————————————————————–

Proof. By Fatou’s theorem, R [a;b ] jf j p dm limR [a;b ] jf n j

p dmkf n kp C

C p: Thus, f 2L p[a; b]

Let us show that limn !1 R

[a;b ]

jf n

j p dm =

R [a;b ]

jf

j p dm:

This is equivalent, by problem 72, to proving that limn !1 R [a;b ] jf n f j

p dm = 0 :

So let > 0 be given. Then for 1 = q

C q > 0; we have the following:

whenever B [a; b]; and m(B ) < 1 =) 8 n 2N, R B jf n jdmHölder

R B jf n j p dm 1=p

R B 1qdm 1=q

= kf n k p (m(B ))1=q C ( q

C q )1=q = :Also, we can …nd 2 > 0 :whenever B [a; b]; and m(B ) < 2; we have that R B jf jdm < :Let = min f 1 ; 2g: Then whenever B [a; b]; and m(B ) < ; we have that 8n; R B jf n jdm; R B jf jdm < :Now, Egorov’s theorem =) 9A [a; b]; m([a; b]nA ) < : f pn

Af p:

Thus, 9N : jf n (x) f (x)j p < 1 =p

(m ( A )) 1 =p for all x 2A ; and for all n N:Hence, for n N , we have

R [a;b ]

jf n f

j p dm =

R A

jf n f

j p dm +

R [a;b ]

nA

jf n f

j p dm

R A

1 =p

(m ( A )) 1 =p p dm + 2 p R [a;b ]nA jf n j

p dm + R [a;b ]nA jf j p dm

m ([a;b ]nA )<

R A m ( A ) dm + 2 p ( + )(2 p+1 + 1) :

Thus, limn !1 R [a;b ] jf n f j

p dm = 0 and so, by problem 72, limn !1 R [a;b ] jf n j

p dm = R [a;b ] jf j p dm:

Similarly, to show that limn !1 R [a;b ] f n gdm = R [a;b ] fgdm 8g 2Lq[a; b]; we show that

limn !1 R [a;b ] jf n g fg jdm = 0 and then use the result in problem 72.

141



(Note that, by Hölder’s inequality, R [a;b ] f n gdm < 1 and R [a;b ] fgdm < 1 ):

Now, 0 R [a;b ] jf n g fg jdm = R [a;b ] jf n f j jgjdmHölder

R [a;b ] jf n f j p dm

1=p

R [a;b ] jgjq dm

1=q n !1! 0:Thus, lim

n !1 R [a;b ] jf n g fg jdm = 0 and we are done by problem 72.

———————————————————————————————————————————————–420. Let

ff n

g be a sequence of functions in L2(R): Suppose that

kf n

kL 2 (R )

M for all n

2N; and

f n a.e.! f on R: Prove that limn !1 R R f n gdm = R R fgdm for all g 2L2(R):

———————————————————————————————————————————————–Proof. Let us do it …rst when A R is such that m(A ) < 1 : Note that this would be exactly problem 73with p = q = 2 : Thus, lim

n !1 R A f n gdm = R A fgdm for all g 2L2(R): (Here, g 2L2(R) is not a typo, why?).

Now, write R= [m 2Z

[m; m + 1] ; and let g 2L2(R):

Then limn !1 R R f n gdm = lim

n !1 Xm 2Z R [m;m +1] f n gdm!m a nd n are independent= Xm 2Z

limn !1 R [m;m +1] f n gdm

= Xm 2Z R [m;m +1] fgdm = R R fgdm:

———————————————————————————————————————————————–

421. Let ff n g be a sequence of functions in L p

[a; b]; 1 p < 1 ; which converges in norm to f 2L p

[a; b];and let fgn g be a sequence of measurable functions such that jgn j C for n = 1 ; 2; 3;:::; and gna.e.

! g:Show that f n gn ! fg in L p[a; b]:———————————————————————————————————————————————–Proof. Note that for all n; jfg n j

p C p jf j p and R [a;b ] C p jf j

p dm = C pR [a;b ] jf j p dm < 1 :

Moreover, f pg pn

a.e.

! f pg p on [a; b] and so, by the Lebesgue DCT, limn !1 R [a;b ] f pg p

n dm = R [a;b ] f pg pdm:Hence, by problem 72, we have that lim

n !1 R [a;b ] jfg n fg j p dm = 0 .

Thus, we have R [a;b ] jf n gn fg j p dm

1=p= R [a;b ] jf n gn fg n + fg n fg j

p dm1=p

Hölder

R [a;b ] jf n gn fg n j p dm

1=p+ R [a;b ] jfg n fg j

p dm1=p

=

R [a;b ]

jgn

j p

jf n f

j p dm

1=p+

R [a;b ]

jf

j p

jgn g

j p dm

1=p

C pR [a;b ] jf n f j p dm

1=p+ R [a;b ] jf j

pjgn gj

p dm1=p

! 0 as n ! 1 :

Thus, limn !1 R [a;b ] jf n gn fg j

p dm = 0 ; and so, by problem 72, limn !1 R [a;b ] f pn g p

n dm = R [a;b ] f pg pdm:Therefore, f n gn ! fg in L p[a; b]:

———————————————————————————————————————————————–422. Let f ; f n : R ! R be measurable functions such that f n

a.e.

! f on R and there exists a function gintegrable on R such that jf n (x)j g(x) a.e. for all n 2N. Show that f n ! f almost uniformly on R:(That is, 8 > 0; 9 a measurable set A R : m(RnA ) < and f n

Af ):

———————————————————————————————————————————————–Proof. Since jf n (x)j g(x) a.e. for all n 2N; R R jf n jdm R R gdm < 1 a.e. for all n 2N. Thus, each f nis Lebesgue integrable on R: Now, Fatou’s theorem =)

R R jf jdm lim

R R jf n jdm

R R gdm < 1 .

Thus, f is Lebesgue integrable on R: By the Lebesgue DCT, we get that limn !1 R

R jf n

jdm =

R R j

f

jdm:

Thus, by problem 72, we get that limn !1 R R jf n f jdm = 0 :

Let > 0 be given. Then 9N :R R jf n f jdm < 2 for all n N:Now, by Tchebyshev’s inequality, problem 59, we get that if B = fx 2R : jf n (x) f (x)j> g; thenm(B ) < 1 R R jf n f jdm < 2

= ; and so if A = RnB ; then m(RnA ) = m(B ) < ; and on A , we have that

jf n (x) f (x)j for all n N: This means that f n A

f and so we are done.

———————————————————————————————————————————————–423. Suppose that f is nonnegative and measurable on a set A , where 0 < m (A ) < 1 : Suppose also thatthere are positive and such that 1

m ( A ) R A fdm and 1m ( A ) R A f 2dm :

142



Show that if 0 < < 1 and A = fx 2A : f (x) g; then m(A ) m(A )(1 )2 2 :

———————————————————————————————————————————————–Proof. R A fdm 2 Schwartz

R A f 2dm R A 12dm = R A f 2dm m(A ) ( m(A )) m(A ) = (m(A ))2 < 1 :Thus, f is Lebesgue integrable on A :

Now,

R A fdm

2 Schwartz

R A f 2dm

R A 12dm

R A f 2dm (m(A )) m(A )m(A ):

On the other hand, R A nA fdm m(A nA ) m(A ):Thus, m (A ) R A fdm = R A nA fdm + R A fdm m(A ) + R A fdm:

Hence, R A fdm m (A ) m(A ) = (1 )m(A ) and so R A fdm2

2(1 )2 (m(A ))2 :

Therefore, 2(1 )2 (m(A )) 2 R A fdm2

m(A )m(A ):

This immediately implies that m(A ) m(A )(1 )2 2 :

———————————————————————————————————————————————–424. We say that a measurable function f on a set A is essentially bounded if m(fx 2A : jf (x)j> r g) = 0for some real number r: In this case, we de…ne the essential supremum of f by:

kf k1 = inf fr : m(fx 2A : jf (x)j> r g) = 0 g:

Show that if f is essentially bounded on [a; b]; then lim p!1 R [a;b ] jf j

p dm1=p

= kf k1 :———————————————————————————————————————————————–Proof. If kf k1 = 0 ; then f = 0 a.e. in [a; b] and so lim

p!1 R [a;b ] jf j p dm

1=p= 0 = kf k1 :

If kf k1 = r > 0; then let B = fx 2[a; b] : jf (x)j> r g: Then m(B ) = 0 and on [a; b]nB ; jf j r:

Thus, R [a;b ] jf j p dm

1=p= R [a;b ]nB jf j

p dm + R B jf j p dm

1=p m ( B )=0= R [a;b ]nB jf j

p dm1=p

R [a;b ]nB r pdm1=p

= r R [a;b ]nB dm1=p

= kf k1 (m([a; b]nB ))1=p m ( B )=0= kf k1 (m([a; b]))1=p = kf k1 (b a)1=p :

Thus, lim p!1 R [a;b ] jf j

p dm1=p

lim p!1 kf k1 (b a)1=p = kf k1 :

Now, given any 0 < < r; we have, by de…nition of r; that m(fx 2A : jf (x)j> r g) = > 0:

Thus, R [a;b ] jf j

p

dm1=p

R fx2A :jf (x )j>r gjf j p

dm1=p

R fx2A :jf (x )j>r g (r ) p

dm1=p

= ( r ) 1=p

;and so lim

p!1 R [a;b ] jf j p dm

1=pr = kf k1 :

Therefore, Letting ! 0; we get that lim p!1 R [a;b ] jf j

p dm1=p

kf k1 :

Thus, kf k1 lim p!1 R [a;b ] jf j

p dm1=p

lim p!1 R [a;b ] jf j

p dm1=p

kf k1 :

Hence, lim p!1 R [a;b ] jf j

p dm1=p

= kf k1 :

———————————————————————————————————————————————–425. Prove the following Hölder’s Inequality : If 1 p 1 and 1

p + 1 p0 = 1 ; then kfgk1 kf k p kgk p0 ;

that is,

R A

jfg

jdm

R A

jf

j p dm 1=p

R A

jg

j p0

dm1=p 0

; and

R A

jfg

jdm

kf

k1 R A

jg

jdm:

———————————————————————————————————————————————–Proof. Let f 1(x) = jf j

p (x) and f 2(x) = jgj p0

(x): Then Hölder’s inequality, problem 92, implies that

R A f 1=p1 (x)f 1=p 0

2 (x)dx R A f 1(x)dx 1=p

R A f 2(x)dx 1=p0

: But, f 1=p1 = jf j; and f 1=p 0

2 = jgj:

Thus, R A jfg jdm R A jf j p dm 1=p

R A jgj p0

dm1=p0

:Now, a.e. on A , we have jfg j kf k1 jgj; and so R A jfg jdm kf k1 R A jgjdm:

———————————————————————————————————————————————–426. (a) Prove the following Minkowski’s Inequality : If 1 p 1; then kf + gk p kf k p + kgk p ;

that is, R A jf + gj p dm 1=p

R A jf j p dm 1=p + R A jgj

p dm 1=p ; and kf + gk1 kf k1 + kgk1 :

143



(b) Show that Minkowski’s Inequality fails for 0 < p < 1:———————————————————————————————————————————————–Proof. (a) We consider 3 cases:(Case i) p = 1 :Then

R A jf + gjdm

R A (jf j+ jgj) dm =

R A jf jdm +

R A jgjdm:

(Case ii) p =

1:

Then we have jf j kf k1 and jgj kgk1 a.e. on A . Thus, jf + gj jf j+ jgj kf k1 + kgk1 a.e. on A,and consequently, kf + gk1 kf k1 + kgk1 :(Case iii) 1 < p < 1 :Then we have kf + gk

p p = R A jf + gj

p dm = R A jf + gj p 1

jf + gjdm R A jf + gj p 1

jf jdm+R A jf + gj p 1

jgjdm:In the last integral, apply Hölder’s inequality to jf + gj

p 1 and jgjwith exponents p0 = p p 1 and p; respectively.

This gives R A jf + gj p 1

jgjdm kf + gk p 1 p kgk p : Similarly, R A jf + gj

p 1jf jdm kf + gk

p 1 p kf k p :

Thus, kf + gk p p kf + gk

p 1 p kf k p + kf + gk

p 1 p kgk p = kf + gk

p 1 p (kf k p + kgk p):

Dividing both sides by kf + gk p 1 p ; which we can assume to be 6= 0 and 6= 1 ; we get that

kf + gk p kf k p + kgk p :

(b) Let A = (0 ; 1); f (x) = (0 ;1=2) (x); and g(x) = (1 =2;1) (x):Then kf + gk p = 1 ; but kf k p + kgk p = 2 1=p + 2 1=p = 2 1 (1 =p) < 1:

———————————————————————————————————————————————–427. Let f : [0; 1] ! R be positive and continuous. For p 6= 0 ; let N ( p) = R

10 f p(x)dx

1=p:

(a) Compute lim p!1

N ( p):

(b) Compute lim p! 1

N ( p):

(c) Compute lim p! 0

N ( p):———————————————————————————————————————————————–Solution. (a) By problem 78, lim

p!1N ( p) = kf k1

f is continuous= maxx2[0;1] jf (x)j:

(b) For p < 0; p = j pj: Thus, lim p! 1

N ( p) = lim p! 1 R

10 f p(x)dx

1=p

= lim

j p

j!1 R 1

0 ( 1f )j pjdx

1=j pj = 1lim

j p j!1(

R 1

0 ( 1f ) j p j dx )1 = j p j = 1

k1f

k1

f is continuous= 1max

x 2 [0 ; 1]

j 1

f ( x )

j = min

x

2[0;1] jf (x)j:

(c) lim p! 0

ln(N ( p)) = lim p! 0

ln R 10 f p(x)dx

1=p= lim

p! 0

ln (R 10 f p (x )dx )

pL’Hopital

= lim p! 0

ddp ln R 1

0 f p(x)dx

= lim p! 0

ddp (R 1

0 f p (x )dx )

R 10 f p (x )dx

why?= lim

p! 0

(R 10

ddp f p (x )dx )

R 10 f p (x )dx = lim

p! 0

(R 10 f p (x ) ln( f (x )) dx )

R 10 f p (x )dx = R

10 ln(f (x))dx;

because f p(x)ln( f (x)) p! 0

[0;1]ln( f (x)) ; and f p(x)

p! 0

[0;1]1:

Hence, lim p! 0

N ( p) = lim p! 0

eln N ( p) = elim

p ! 0(ln N ( p))

= eR 10 ln( f (x )) dx :

———————————————————————————————————————————————–428. Suppose f 2C 1([a; b]); f (a) = f (b) = 0 ; and R

ba f 2(x)dx = 1 :

(a) Show that

R b

a xf (x)f 0(x)dx = 12 :

(b) Show that 14 R

ba (f 0(x))2 dx R

ba x2f 2(x)dx :

———————————————————————————————————————————————–Proof. (a) R

ba xf (x)f 0(x)dx = R

ba xf (x)d(f (x))

integration by parts= xf 2(x) b

a R b

a f (x)d(xf (x))f (a )= f (b)=0

= 0 R b

a f (x) ( f (x) + xf 0(x)) dx = R b

a f 2(x)dx R b

a xf (x)f 0(x)dx = 1 R b

a xf (x)f 0(x)dx:Thus, 2R

ba xf (x)f 0(x)dx = 1 and so R

ba xf (x)f 0(x)dx = 1

2 :

(b) 14 = ( 1

2 )2 = R b

a xf (x)f 0(x)dx2 Schwartz

R b

a (xf (x))2 dx R b

a (f 0(x))2 dx :

———————————————————————————————————————————————–

144










a f (x)g(x)dx R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:

(b) If 0 < p < 1; and f and g are positive, then R b

a f (x)g(x)dx R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:———————————————————————————————————————————————–Proof. (a) Let f 1(x) = f p(x) and f 2(x) = gq(x): Then Hölder’s inequality, problem 92, implies that

R b

a f 1=p1 (x)f

1=q2 (x)dx R

ba f 1(x)dx

1=p

R b

a f 2(x)dx1=q

: But, f 1=p1 (x) = f (x); and f

1=q2 (x) = g(x):

Thus, R b

a f (x)g(x)dx R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:

(b) Let r = 1 p > 1; and s be such that 1

r + 1s = 1 : Also, let f 1(x) = ( fg ) p (x) and f 2(x) = 1

gp (x ) :Then f (x)g(x) = f r1 ; f p(x) = f 1(x)f 2(x); and gq(x) = f s2 (x); and so, by part (a),

R b

a f p(x)dx = R b

a f 1(x)f 2(x)dx R b

a f r1 (x)dx1=r

R b

a f s2 (x)dx1=s

= R b

a f (x)g(x)dx1=r

R b

a gq(x)dx1=s

= R b

a f (x)g(x)dx p

R b

a gq(x)dx1 p

:

Thus, R b

a f (x)g(x)dx p

R b

a f p(x)dx R b

a gq(x)dx p 1

=)

R b

a f (x)g(x)dx

R b

a f p(x)dx1=p

R b

a gq(x)dxp 1

p=

R b

a f p(x)dx1=p

R b

a gq(x)dx1=q

:

———————————————————————————————————————————————–440. Suppose that f is continuous on [0; 1] and there is a > 0 such that 0 f (x) a2=3 8x 2[0; 1]:Show that if R

10 f (x)dx = a; then R

10 p f (x)dx a2=3 :

———————————————————————————————————————————————–Proof. By Hölder’s inequality, with p > 1; a = R

10 f (x)dx = R

10 (f (x))

12 p +1 1

2 p dx R 1

0 p f (x)dx1=p

a2 p 1

3 p :

Thus, a1 ( 2 p 13 p ) R

10 p f (x)dx

1=p=) a

p +13 p R

10 p f (x)dx

1=p=) a

p +13 R

10 p f (x)dx:

Since this holds for any p > 1; lim p! 1+

ap +1

3 R 10 p f (x)dx =) a2=3 R 1

0 p f (x)dx:

———————————————————————————————————————————————–441. Prove the following Minkowski inequality for integrals : If f 1 ; f 2;:::;f n are nonnegative and Riemannintegrable on [a; b]; then

(a) If p > 1; then

R b

a (f 1(x) + + f n (x)) p dx1=p

R b

a f p1 (x)dx1=p

+ +

R b

a f pn (x)dx1=p

:

(b) If 0 < p < 1; and f i 0, then R b

a (f 1(x) + + f n (x)) p dx 1=p

R b

a f p1 (x)dx 1=p + + R b

a f pn (x)dx 1=p :———————————————————————————————————————————————–Proof. (a) If one sets S n (x) = f 1(x) + f 2(x) + + f n (x); then one gets that

R b

a S pn (x)dx = R b

a (f 1(x) + f 2(x) + + f n (x)) S p 1n (x)dx

= R ba f 1(x)S p 1

n (x)dx + R ba f 2(x)S p 1

n (x)dx + + R ba f n (x)S p 1

n (x)dxHölder’s inequality

1p + 1

q =1 R b

a f p1 (x)dx1=p

R b

a S p 1n

q (x)dx1=q

+ + R b

a f pn (x)dx1=p

R b

a S p 1n

q (x)dx1=q

( p 1) q= p= R

ba f p1 (x)dx

1=p

R b

a S pn (x)dx1=q

+ + R b

a f pn (x)dx1=p

R b

a S pn (x)dx1=q

=

R b

a S pn (x)dx1=q

R b

a f p1 (x)dx1=p

+ +

R b

a f pn (x)dx1=p

:

Now, R b

a S pn (x )dx

(R ba S pn (x )dx )1 =q = R

ba S pn (x)dx1 1

q = R ba S pn (x)dx

1=p:

Thus, R b

a S pn (x)dx1=p

R b

a f p1 (x)dx1=p

+ + R b

a f pn (x)dx1=p

:

(b) It su¢ces to apply the inequality in problem 93(b).

———————————————————————————————————————————————–442. Assume that f 1 ; f 2 ;:::;f n are nonnegative and Riemann integrable on [a; b]:(a) If p > 1; then R

ba (f 1(x) + + f n (x)) p dx R

ba f p1 (x)dx + + R

ba f pn (x)dx:

(b) If 0 0, and for p > 1; x p1 + + x p

n (x1 + + xn ) p ; then the resultfollows immediately. Let us show that for x1 ; x2 > 0; and p > 1; we have x p

1 + x p2 (x1 + x2) p :

Dividing both sides by x p2 ; we get that x 1

x 2

p+1 x 1

x 2+ 1

p; and so it su¢ces to prove that t p +1 (t + 1) p

for all t > 0 and p > 1: Let f (t) = ( t + 1) p t p 1: Then f (0) = 0 ; and f 0(t) = p (t + 1) p 1 t p 1 > 0:

(This is because for p > 1; and t > 0; we have that 1 + 1t

p 1

> 1).Thus, f is increasing and so f (t) 0 for all t > 0: Hence, t p + 1 (t + 1) p for all t > 0 and p > 1:Therefore, if p > 1; then R

ba (f 1(x) + + f n (x)) p dx R

ba f p1 (x)dx + + R

ba f pn (x)dx:

(b) If we show that for x1 ;:::;x n 0, and for 0 0; and 0 0 and 0 0; we have that 1 + 1t

p 1 < 1).Thus, f is decreasing and so f (t) 0 for all t > 0: Hence, (t + 1) p t p + 1 for all t > 0 and 0 0; is strictly increasing on [0; c] and f (0) = 0 : Let f 1(x) denote

the inverse of f : Show that for x 2[0; c]; R x

0 f (t)dt + R f (x )

0 f 1(t)dt = xf (x):———————————————————————————————————————————————–Proof. For x 2[0; c]; let g(x) = R

x0 f (t)dt + R

f (x )0 f 1(t)dt:

Then g0(x) = R x0 f (t)dt 0 + R f (x )

0 f 1(t)dt 0= f (x) + f 0(x)f 1(f (x)) = f (x) + xf 0(x):Now, g(x) = R

x0 g0(t)dt = R

x0 (f (t) + tf 0(t)) dt = R

x0 f (t)dt + R

x0 tf 0(t)dt

by parts= R

x0 f (t)dt + [tf (t)]x

0 R x

0 f (t)dt = R x

0 f (t)dt + xf (x) R x

0 f (t)dt = xf (x):

———————————————————————————————————————————————–444. Prove the following Young inequality : Suppose that f 2C ([0; c]); c > 0; is strictly increasing on [0; c]and f (0) = 0 : Let f 1(x) denote the inverse of f :(a) Show that for any a 2[0; c] and for any b 2[0; f (c)]; we have

R a

0 f (t)dt +

R b

0 f 1(t)dt ab:(b) Show that equality holds in part (a) if and only if b = f (a):———————————————————————————————————————————————–Proof. (a) WLOG, say f (a) b: (otherwise, b < f (a) and so f 1(b) < a and we apply the same method).Now, R

a0 f (t)dt + R

b0 f 1(t)dt = R

a0 f (t)dt + R

f (a )0 f 1(t)dt + R

bf (a ) f 1(t)dt

problem 97= af (a) + R

bf (a ) f 1(t)dt:

Also, since f is strictly increasing on [0; c]; we get that R b

f (a ) f 1(t)dt R b

f (a ) adt = a(b f (a) = ab af (a):

Hence, R a

0 f (t)dt + R b

0 f 1(t)dt = af (a) + R b

f (a ) f 1(t)dt af (a) + ab af (a) = ab:

(b) (( =) If b = f (a); then this is exactly problem 97.(=) ) If R

a0 f (t)dt + R

b0 f 1(t)dt = ab; and WLOG f (a) b; then ab = af (a) + R

bf (a ) f 1(t)dt:

If f (a) < b; then R b

f (a ) f 1(t)dt > ab af (a); and we would get that ab > ab a contradiction.

———————————————————————————————————————————————–445. Show that for a; b 0; (1 + a)ln(1 + a) (1 + a) + ( eb b) ab:———————————————————————————————————————————————–Proof. Let f (x) = ln(1 + x) for x 2[0; c]; c > 0: Then f is strictly increasing on [0; c] and f (0) = ln 1 = 0 :Also, f 1(x) = ex 1: Thus, by problem 98, we have that for any a; b 0; R a

0 f (t)dt + R b0 f 1(t)dt ab:

(Of course, we will have a and b …rst and then we choose c > maxfa; bg):But R

a0 f (t)dt + R

b0 f 1(t)dt = R

a0 ln(1 + t)dt + R

b0 (et 1) dt = eb b 1 + (( a + 1) ln ( a + 1) a) :

———————————————————————————————————————————————–446. Suppose that f is positive and Riemann integrable on [a; b]:(a) Show that (b a)2 R

ba f (x)dx R

ba

1f (x ) dx :

(b) Show that if 0 < m f (x) M; then R b

a f (x)dx R b

a1

f (x ) dx (m + M ) 2

4mM (b a)2 :

149





———————————————————————————————————————————————449. Let f be continuous on [0; 1]; and suppose that g is a nonnegative continuous function on R with

g(x + 1) = g(x): Show that limn !1

1

Z 0

f (x)g(nx )dx = 0@1

Z 0

f (x)dx1A0@1

Z 0

g(x)dx1A:

———————————————————————————————————————————————

Proof. Let a =1

Z 0

g(x)dx and M = supx2[0;1] jf (x)j:

Then, since g(x) 0; we have that g(x) a for all x 2[0; 1]:

Now,1

Z 0

f (x)g(nx )dx 0@1

Z 0

f (x)dx1A0@1

Z 0

g(x)dx1A=

1

Z 0

f (x)g(nx )dx1

Z 0

af (x)dx

=1

Z 0

f (x) (g(nx ) a) dx1

Z 0

jf (x)j jg(nx ) ajdx a g(x ) 8x2[0;1]

=1

Z 0

jf (x)j(a g(nx )) dx

M 1

Z 0

(a g(nx )) dx:

We will be done if we show that limn !1

1

Z 0

(a g(nx )) dx = 0 :

Note that1

Z 0

(a g(nx )) dx = a1

Z 0

g(nx )dx and so we have to show that limn !1

1

Z 0

g(nx )dx = a:

Now, limn !1

1

Z 0

g(nx )dx y= nx=dy = ndx

limn !1

1n

n

Z 0

g(y)dy g(y)= g(y+1)

= limn !1

1n 0@

n1

Z 0

g(y)dy1A=

1

Z 0

g(y)dy = a:

———————————————————————————————————————————————450. Let f 2L1(R). Show that there exists a sequence xn ! 1 such that xn f (xn ) ! 0:———————————————————————————————————————————————

Proof. For x 2R; let g(x) = xf (x): Suppose that @ a sequence xn ! 1 such that g(xn ) ! 0:Then we can …nd ;M > 0 : for all x M; we have that jg(x)j :Now, f (x) = g(x )

x and so jf (x)j= g(x )x

jxj for all x M:But R jxj M jxjdx = 1 and so f =2L1(R); a contradiction.Hence, there is a sequence xn ! 1 such that xn f (xn ) ! 0:

———————————————————————————————————————————————451. Let g : Rd ! R be nonnegative with R R d g(y)dy = 1 ; let g (x) = d g(x= ) for > 0; and de…nef (x) = ( g f ) (x) = R R d g (y)f (x y)dy: If f : Rd ! R is continuous and bounded, then show thatfor any compact subset K Rd , f

K f as ! 0:

———————————————————————————————————————————————Proof. We …rst note that for any > 0;

R R d g (x)dx = d

R R d g(x= )dx

y= x==

dy = d dx

R R d g(y)dy = 1 :

Let K Rd be a compact subset. Then for all x 2K; we have:jf (x) f (x)j= R R d g (y)f (x y)dy R R d g (y)f (x)dy = R R d g (y) [f (x y) f (x)] dy

= R R d d g(y= ) [f (x y) f (x)] dyz = y=

=dz = d dy R R d g(z) [f (x z) f (x)] dz R R d g(z) jf (x z) f (x)jdz

(1):

Since f is bounded, 9M > 0 : jf (y) f (x)j M for all x; y 2Rd :Since f is continuous on K , f is uniformly continuous on K:We are now ready to show that f

K f as ! 0:

Let > 0 be given. Since g 2L1(R); 9N 2N :R R d nB N (0) g(y)dy < 2M :

Now, (1) can be written as: (Remember that f is bounded and so R R d g(z) jf (x z) f (x)jdz M )

151



R R d g(z) jf (x z) f (x)jdz = Z R d nB N (0)g(z) jf (x z) f (x)jdz

| {z } I 1

+ Z B N (0)g(z) jf (x z) f (x)jdz

| {z } I 2

:

For I 1 ; we have that I 1 R R d nB N (0) g(z)Mdz = M R R d nB N (0) g(z)dz < M 2M =

2 ; 8x 2Rd :

For I 2 ; we go as follows: (our aim is to make I 2 < 2 )Since f is uniformly continuous on K; we can …nd 0 > 0 : whenever jx yj< 0; we have jf (x) f (y)j<

2 :Choose small enough so that N < 0: Then j zj< 0 for all z 2BN (0):Hence, I 2 = R B N (0) g(z) jf (x z) f (x)jdz R B N (0) g(z) 2 dz =

2 R B N (0) g(z)dz 2 R R d g(z)dz =

2 ; 8x 2K:

Hence, we have shown that for small enough, and for all x 2K; we have that jf (x) f (x)j< :Thus, for any compact subset K Rd , f

K f as ! 0:

———————————————————————————————————————————————

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