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21-355: Real Analysis 1
Carnegie Mellon University
Professor Ian Tice - Fall 2013
Project LATEXd by Ivan Wang
Last Updated: November 15, 2013
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CONTENTS 21-355 Notes
Contents
1 The Number Systems 3
1.1 The Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Positivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.2 Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.3 Multiplication. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 The Integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Properties of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.2 Algebraic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 The Rationals and Ordered Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3.1 Fields and Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4 Problems with Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.4.1 Bounds (Infimum and Supremum) . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.5.1 Defining the Real Numbers: Dedekind Cuts . . . . . . . . . . . . . . . . . . . 9
1.5.2 Defining the Real Numbers: The Least Upper Bound Property . . . . . . . . 10
1.5.3 Defining the Real Numbers: Addition . . . . . . . . . . . . . . . . . . . . . . 10
1.5.4 Defining the Real Numbers: Multiplication . . . . . . . . . . . . . . . . . . . 11
1.5.5 Defining the Real Numbers: Distributivity. . . . . . . . . . . . . . . . . . . . 11
1.5.6 Defining the Real Numbers: Archimedean . . . . . . . . . . . . . . . . . . . . 12
1.6 Properties ofR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.6.1 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Sequences 13
2.1 Convergence and Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.1.1 Squeeze Lemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.2 Monotonicity and limsup, liminf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3.1 Limsup Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.4 Special Sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3 Series 17
3.1 Convergence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.1 Cauchy Criterion Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.2 Logarithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 The numbere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
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3.3 More Convergence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.4 Algebra of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.5 Absolute Convergence and Rearrangements . . . . . . . . . . . . . . . . . . . . . . . 22
4 Topology ofR 24
4.1 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.1.1 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.1.2 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.1.3 Limit Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.1.4 Closure, Interior, and Boundary Sets . . . . . . . . . . . . . . . . . . . . . . . 26
4.2 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2.1 Heine-Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3 Connected Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
5 Continuity 305.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.1.1 Divergence Criteria. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5.3 Compactness and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5.4 Continuity and Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.5 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.6 Monotone Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
6 Differentiation 35
6.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
6.2 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.3 Darbouxs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6.4 LHopitals Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
6.5 Higher Derivatives and Taylors Theorem . . . . . . . . . . . . . . . . . . . . . . . . 38
7 Riemann-Stieltjes Integration 39
7.1 The R-S Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397.2 Integrability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
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1 The Number Systems
1.1 The Natural Numbers
Theorem (existence ofN): There exists a set N satisfying the following properties, known as thePeano Axioms:
PA1 0 N
PA2 There exists a function S : N N called the successor function. In particular,S(n) N.
PA3 n N. S(n)= 0
PA4 S(n) =S(m) = n= m (S is injective, one-to-one)
PA5 [Axiom of Induction] Let P(n) be a property associated to each n N.IfP(0) is true, and P(n) = P(S(n)), then P(n) is true n N.
Definition: PA1 = 0 N. PA2 = S(0) N.
Define 1 =S(0), 2 =S(1), 3 =S(2), etc.PA2guarantees that {0, 1, 2, } N.
PA3prevents wraparound: no successor can map to a negative number.
PA4prevents stagnation: the cycle does not terminate.
Theorem: N ={0, 1, 2, }
Proof: We know that {0, 1, 2, } N, so it suffices to prove that N {0, 1, 2, }.
LetP(n) denote the proposition that n {0, 1, 2, }. Clearly P(0) is true.
Suppose P(n) is true; then n {0, 1, 2, } = S(n) {0, 1, 2, }by construction.
Hence,P(S(n)) is true. By induction, PA5 guarantees that P(n) is true n N.It follows that N {0, 1, 2, }.
Definition: For any m N, we define 0 +m= m.Then ifn+m is defined for n N, we set S(n) +m= S(n+m).
Proposition (Properties of Addition):
1. n N. n+ 0 =n (0 is the additive identity)
2. m, n N. n+S(m) =S(n+m)
3. m, n N. m+n= n+m (commutativity)
4. k,m,n N. k+ (m+n) = (k+m) +n (associativity)5. k,m,n N. n+k = n+m = k= m (cancelation)
Proof:
1. LetP(n) be n+ 0 =n.P(0) is true because 0 + 0 = 0 by definition.Note P(n) = S(n) + 0 = S(n+ 0) = S(n), so P(S(n)) is true. By induction,(1) is true.
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2. Fixm N. LetP(n) denote n +S(m) =S(n+m).P(0) is true because 0 +S(m) =S(m) =S(0 +m).P(n) = S(n)+ S(m) =S(n + S(m)) =S(S(n + m)) =S(S(n) + m), soP(S(n))is true. By induction, since m N was arbitrary, (2) is true.
3. Letm be fixed and P(n) denote n+m= m+n.P(0) is true since 0 + m= m by definition, and m + 0 =m by 1, so 0 + m= m =m+ 0.Suppose P(n); then S(n) + m= S(n + m) =S(m + n) =m + S(n), soP(S(n)) istrue. By induction and arbitrary choice ofm, (3) is true.
4. Fix k, m N and let P(n) denote k+ (m+n) = (k+m) +n.P(0) is true as k+ (m+ 0) =k +m= (k+m) + 0.SupposeP(n); thenk+(m+S(n)) =k+S(m+n) =S(k+(m+n)) =S(k+m)+n=(k+m) +S(n) by (2). By induction and arbitrary choice, (4) is true.
5. Fix m, n N and let P(k) denote proposition 5.P(0) is true because n + 0 =n = n+m = m= 0 = k= m.Suppose P(k); also, supposem + S(k) =n + S(k). ThenS(m + k) =m + S(k) =n+S(k) = S(n+k) = m+k = n+k = m = n (by 4). By the axiom of
induction, (5) is true.
1.1.1 Positivity
Definition: We say that n N is positive ifn = 0.
Proposition (Properties of Positivity):
1. n, m N, ifm is positive, then m +n is positive.
2. n, m N, ifm +n= 0, then m = n = 0.
3. n N, ifn is positive, then there exists a unique m N such that n= S(m).
1.1.2 Order
Definition: For allm, n N, m n or n m iffn = m+p for some p N.m < nor n > m iffm n m=n. The relation provides what is called an order on N.
Proposition (Properties of Order):
Letj, k,m,n N. Then:
1. n n (reflexitivity)
2. m n k m = k n (transitivity)
3. m n m n = m= n (anti-symmetry)
4. jk m n = j+m k +n (order preservation)
5. m < n S(m) n
6. m < n n= m+p for some positive p N.
7. n m S(n)> m
8. n= 0 0< n
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Theorem (Trichotomy of Order): Let m, n N. Then exactly one of the following is true:
m < n m= n m > n
Proof: Show that no two can be true simultaneously (by definition of< and>), and then at leastone must be true (by induction on n).
1.1.3 Multiplication
Definition: Fix m N. Define 0 m = 0. Now, if n m is defined for some n N, we defineS(n) m= n m+m.
Proposition (Properties of Multiplication):
Fix k ,m, n N. Then:
1. m n= n m (commutativity)
2. m, n are positive = mn is positive
3. m n= 0 m= 0 n= 0 (no zero divisors)
4. k (m n) = (k m) n(associativity)
5. k m= k n k is positive = m= n (cancelation)
6. k (m+n) = (m+n) k= k m+k n (distributivity)
7. m < n k l k, l are positive = m k < n l
1.2 The Integers
Consider the following relation on the set N N:
(m, n) (m, n) m+n = m+n
Lemma: is an equivalence relation.
Proof:
Reflexivity: m+n= m+n = (m, n) (m, n)
Symmetry: (m, n) (m, n) = m+ n = m +n = m +n = m+ n =(m, n) (m, n)
Transitivity: Suppose (m, n) (m, n) (m, n) (m, n). Then:
m+n = m+n m+n = m+n
=m+n = m+n=(m, n) (m, n)
Definition: Write the equivalence class of (m, n) as [(m, n)] ={(p, q)| (p, q) (m, n)}.Define the integersZ ={[(m, n)]}.
Lemma: Suppose (m, n) (m, n), (p, q) (p, q). Then:
1. (m+p, n+q) (m+p, n+q)
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2. (mp+nq,mq+np) (mp+nq, mq+np)
Proof: Consider equalities (a) :m+n = m+n and (b) :p+q=p+q(by definition of).
Using linear combinations of (a) and (b), we derive the two rules of the lemma:
1. (a) + (b)
2. (a)(p+q) + (b)(m+n)
Definition: Let [(m, n)], [(p, q)] Z. Then:
1. [(m, n)] + [(p, q)] = [(m+p, n+q)] (addition of integers)
2. [(m, n)] [(p, q)] = [(mp+nq,mq+np)] (multiplication of integers)
By the lemma, these are well-defined operations.
Note that for all m, n N:
[(m, 0)] = [(n, 0)] m+ 0 =n+ 0 m= n
[(m, 0)] + [(n, 0)] = [(m+n, 0)]
[(m, 0)] [(n, 0)] = [(mn, 0)]
As such, the set {[(n, 0)]| n N} Z behaves exactly like a copy ofN.
Definition: For n N we set n Z to be n := [(n, 0)].
For x = [(m, n)] Z we define x= [(n, m)].
1.2.1 Properties of Integers
(We can see that every integer x Z can be represented as x := m nwhere x = [(m, n)].)
Theorem: Every x Z satisfies exactly one of the following:
1. x= n for some n N\{0}
2. x= 0
3. x= nfor some n N\{0}
Proof: Writex = [(p, q)] for some p, q N. By trichotomy of order on N we know that p < qorp= qor p > q. Each of these correlates to one of the three properties.
Corollary: Z ={0, 1, 2, . . .} { 1, 2, 3, . . .}
1.2.2 Algebraic Properties
Proposition: Let x, y, z Z. Then the following hold:
1. x+y = y +x
2. x+ (y+z) = (x+y) +z
3. x+ 0 = 0 +x= x
4. x+ (x) = (x) +x= 0
5. xy= yx
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1.4 Problems withQ 21-355 Notes
1. x, y Eexactly one of the following is true: x < yor x = y or y < x (trichotomy)
2. x,y,z E, x < y y < z = x < z (transitivity)
Definition: Let F be a field. Then we define x y= x+ (y) and xy =xy1 (for y = 0).
Theorem: Q is an ordered field with order 0. Thenthe well-ordering principle implies that !n S(q). n= min S(q).
Since n S(q), we know that q = mn for some m Z. Then q2 = (mn)
2 = m2
n2 =m2 = 2n2 = m2 is even. We claim that m is also even (proof is exercise to reader).
Then l Z. m = 2l. Then 4l2
= (2l)2
= m2
= 2n2
= n2
= 2l2
= n2
is even= n is even = n= 2p for some p N+.
Hence q= mn = 2l2p =
lp = p S(q). But clearly p < n, which contradicts the fact
that n is the minimal element. By contradiction, the theorem must be true.
1.4.1 Bounds (Infimum and Supremum)
Informally, Q has holes:
Definition: Let Ebe an ordered set with order
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Theorem: Q does not satisfy the least upper bound property.
Proof : Consider the set A= {x Q |x >0, x2 2}.
Note that 0 < 1 = 12 2 = 1 A, so A is non-empty. Also, 2 4 = 22 implies(x A = 0< x2 2 and using the same logic as
before, we can choose n large enough such that pq 1n is an upper bound ofA. But
pq
1n 0 = (n N). x < n.
Lemma: IfQ is Archimedean, then (p < q Q).(r Q). p < r < q.(Proofs in HW 2.)
1.5.1 Defining the Real Numbers: Dedekind Cuts
Definition: We say that C P(Q) is a cut(Dedekind cut) iff the following hold:
(C1) =C, C = Q
(C2) Ifp Cand q Q with q < p, then q C.
(C3) Ifp C, (r Q). p < r r C.
Lemma: Suppose C is a cut. Then:
1. p C, q / C = p < q
2. r / C, r < s = s / C
3. Cis bounded above
Lemma: Let q Q. Then {p Q |p < q}is a cut.
Proof: Call the set C. We prove the 3 properties of a cut:
(C1) q 1 C = C = ; q+ 1 / C = C = Q.
(C2) Ifp Cand r Q such that r < p, then r < p < q = r < q = r C.
(C3) Letp Cwhere p < q. Since Q is Archimedean,(r Q). p < r < q = r C.
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Definition: Given q Q we write Cq ={p Q |p < q}. By the above lemma, Cq is a cut.
Definition: We write R ={C P(Q)| C is a cut} = .
Lemma: The following hold:
1. A, B R, exactly one of the following holds: A B, A= B, B A.
2. A, B, C R,A B B C = A C.
Definition: IfA, B R we say that A< B A B, and A B A B. This definesan order on R by the above lemma.
1.5.2 Defining the Real Numbers: The Least Upper Bound Property
Lemma: Suppose =E R is bounded above. Then B:=AEA R.
Theorem: R satisfies the least upper bound property.
Proof: Let =E R be bounded above and set B=AEA R. We claim B= sup E.
First, we show that Bis an upper bound ofE. Let A E. Then A B = A B
(by definition). This is true for all A E, so B is an upper bound.We claim that for C R. C< B = C is not an upper bound ofE. IfC < B, thenC B. This impliesb B. b / C = (A E). b A b / C. ThenA > C sinceotherwiseA C = b C, b / C. Hence C q). p / A}. ThenR, +, 0R = C0 = {p Q | p < 0}satisfy the field axioms.
Proof:
(A1) A + B R by previous lemma.
(A2) A + B= {a+b}= {b+a}= B+ A.
(A3) A + (B+ C) ={a+ (b+c)}= {(a+b) +c}= (A + B) + C.
(A4) ShowA R.0R+ A= A.
(A5) Show thatA R, then A + (A) = 0R using Archimedean property.
Theorem (Ordered Field): Let A, B, C R. IfA< Bthen A + C
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1.5.4 Defining the Real Numbers: Multiplication
Lemma: Let A, B R,A, B> 0R. Then C={q Q |q 0} {a b| a A, b B, a, b >0} R.
Proof:
(C1) 0 C = C = . A, Bare bounded above by, say M1, M2, so M1 M2+ 1 / Cand C = Q.
(C2) Letp Cand q < p. Ifq 0 then q C by definition. Ifq >0 then 0 < q < p,but then 0 < p = p = a b for a A, b B,a ,b > 0. Then 0< q < a b =qa < b = 00 satisfies p < a b C,so r = a b is the desired element of C. However, if p > 0, then p = a b fora A, b B, a, b >0. Choose s A such that a < s, t Bsuch that t > b. Thenp= a b < s t S, sor = s tproves the claim.
Definition of Multiplication: Let A, B R.
1. IfA > 0, B> 0 we set A B= {q Q |q 0} {a b| a A, b B, a, b >0} R.
2. IfA = 0 or B= 0, we set A B= 0R.3. IfA > 0 and B< 0, let A B= (A (B)).
4. IfA < 0 and B> 0, let A B= ((A) B).
5. IfA < 0 and B< 0, let A B= (A) (B).
Theorem: R, satisfies (M1-M5) with 1R = C1, andA> 0 = A1 ={q Q |q 0} {q Q |q >0, p > q. p1 / A} R;A< 0 = A1 =(A)1.
Proof : HW3 (similar to addition).
Theorem: IfA, B> 0, thenA B> 0.
Proof : By definition C0 A B = 0 A B . Equality is impossible since A, B> 0.
1.5.5 Defining the Real Numbers: Distributivity
Theorem: LetA, B, C R. Then A (B+ C) =A B+ A C.
Proof : We prove the case where all are positive. The other cases are in HW.
Letp A(B+ C). Ifp 0 then p A B + A Bis trivial (both products contain theinterval less than 0).
Ifp >0, p = a(b+c) for a A, b B, c C fora >0, b+c >0.
Regardless of sign ofb or c,a b A B , a c A C . Hence p = a(b + c) =a b + a cA B+ A C. SoA(B+ C) A B + A C.
Finally, we show the converse is true; let p A B+ A C = p = r +s forr A B , s A C . Case on positivity ofp,r,s to show p A(B+ C).
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1.5.6 Defining the Real Numbers: Archimedean
Theorem: For p, q Q, the following are true:
1. Cp+q =Cp+ Cq
2. Cp= Cp3. Cpq =CpCq
4. Ifp = 0 then Cp1 = (Cp)15. p < q Q Cp < Cq R
Proof: HW.
Definition: For q Q we sayCq R. Then Q R.
Theorem: There exists an ordered field satisfying the least upper bound property; R is unique (forany ordered field F satisfying these properties, F = R up to isomorphism; and R is Archimedean.
Proof : The basic assertion is Steps (0)-(4). Step (5) proves 1, Step (6) proves 3.
1.6 Properties ofR
Notation: think ofR as numbers, not cut notation.
Proposition: R satisfies the following:
Theorem: For p, q Q, the following are true:
1. R is Archimedean: x R, x >0.n N. x < n
2. N R is not bounded above
3. inf{ 1n |n N, n 1}= 0
4. x R the set B(x) ={m Z |x < m}has a minimum in Z.
5. x, y R, x < y.q Q. x < q < y
Remarks:
1. (5) is interpreted as the density ofQ R. Any element x R can be approxi-mated to arbitrary accuracy by elements ofQ.
2. (4) allows us to define the integer part of anyx R. We can setx= min B(x) 1 Z. Then x x 0 and n N, n 1. Then !y R. y >0 yn =x.
Proof: The case n= 1 is trivial so assume n 2.
SetE= {z R |z >0 zn < x}. We want to showE= and is bounded above. Sett= x1+x ; then 0< t 0.We claim that yn < xand yn > xare both impossible (proof is exercise), so yn =x.
Definition: Let n 1; for x R, x >0, we write x1n =y where yn =x. We set 0
1n = 0.
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1.6.1 Absolute Value
For x R, we define the function | |: R {r R |r 0}:
|x|=
x ifx >0
0 ifx = 0
x ifx 0.|an|< M (n l).Lemma: If a sequence converges, then it is bounded.
Definition: Given{an}, {bn} Rwe define{an+ bn} Rto be the sequence whose elements arean+bn. We similary define {can}for a fixed c R, {anbn}, and {an/bn}where bn= 0, n l.
Theorem (algebra of convergence): Let{an}, {bn} R, c R, and assume that an a, bn bas n . Then the following hold:
1. an+bna+b as n
2. can ca as n
3. anbn ab as n
4. Ifbn= 0 and b= 0, then an/bna/b as n .
Proof: (1), (2) are in next weeks HW.
(3): Note that |anbn ab| = |anbn abn +abn ab| |anbn abn|+ |abn ab| =|bn||an a| + |a||bn b|. Since bn b we know that M >0.|bn|< M(n l).
Let > 0. Since an a and bn b we may choose N1 such that n N1 =|an a| 0. Since an a as n , we knowNl. n N = |an a|< . We claim Kl. k K = (k) N.
If not, then (k) < N(k l); but k (k) < N(k l) is a contradiction. Thenthe claim is true, and k K = (k) N = |a(k) a| < . Since > 0 wasarbitrary, we deduce {a(k)} a as k .
Remark: Converse fails. Example: an = (1)n;a2n= +1 +1, but a2n+1=1 1.
2.3.1 Limsup Theorem
Theorem: Let{an} R be bounded. The following hold:
1. Every subsequence of{an} is bounded.
2. If{ank} is a subsequence, then lim supk ank lim supn an.
3. If{ank} is a subsequence, then lim infn an lim infk ank .
4. There exists a subsequence{ank}such that limk ank = lim supn an.
5. There exists a subsequence{ank}such that limk ank = lim infn an (= (4)).Proof:
1. Trivial.
2. Sincek (k),{a(n)|n k} {an | n k}for every order-preserving. HenceSk = sup{a(n)} |n k} sup{an | n k}= Tk. But:limsupn a(n) = limsupk{a(n) | n k} limsupk{an | n k} =limsupn an.
3. Similar to (2); exercise to reader.
4. Too lazy to LATEX; exercise to reader.
5. Exercise to reader.
Theorem: Suppose {an} R; the following are equivalent:
1. an a as n
2. {an} is bounded, and every convergent subsequence converges to a.
3. {an} is bounded, and lim supn an = lim infn an.
Proof: (1) = (2) proven already.
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(2) = (3)Limsup theorem (4,5) = {a(k)}, {a(k)} subsequences such thata(k) lim supn an, a(k)liminfn an as k . By (2) the limits must agree.
(3) = (1)Limsup theorem (1-3) = {a(k)}. lim infn anlim infk a(k) lim supk a(k)limsupn an. As the first and last are equal, by transitivity it follows all subsequences
satisfy lim infk a(k)= lim supk a(k). Asanis a subsequence of itself, it thereforeconverges to some a as n .
Theorem (Bolzano-Weierstrass): If{an} R is bounded then there exists a convergent subse-quence. Proof from (4) or (5) of Limsup Theorem.
2.4 Special Sequences
Definition: Given an R for 0 k n, n N we definen
k=0an = a0+a1+ +an.
Lemma (Binomial Theorem): Let x, y R and n N. Then (x+ y)n =n
k=0 nk
xkynk, wherenk
:= n!k!(nk)! N.Theorem: In the following assuming thatn 1:
1. Letx R, x >0. Then an = 1nx 0 as n .
2. Letx R, x >0. Then an = x1/n 1 as n .
3. Letan = n1/n; then an 1 as n .
4. Leta, x R, x >0. Then na
(1+x)a 0 as n .
5. Letx R, |x|< 1. Then an = xn 0 as n .
3 Series
Definition: Let{an}n=l R; for p < qwe write
qn=pan = (ap+ +aq).
1. We define, for each n l, Sn =n
k=l
ak R to be the nth partial sum of{an}
n=l.
2. If s R. Sn s as n , then
n=lan = s. We say the infinite seriesn=lan converges.
3. If the series does not converge, it diverges.
Examples
1. Let an = xn for n 0, x R. Then Sn =
nk=0xk. Notice that (1 x)Sn =
nk=0xk n
k=0xk+1 =
nk=0x
k n+1
k=1xk = 1 xn+1.
SoSn =n
k=0xk = (1x
n+1
1x ). If|x|< 1 then Sn 11x by special seq (5).
2. Suppose{bn}n=0 Rwherebn b as n . Setan = bn+1 bn forn 0. Then the series
n=0an converges and in fact
n=0= b b0.
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3.1 Convergence Results
We develop tools that will let us deduce the convergence of a series without knowing its value.
Theorem: Suppose
n=lan converges. Then an 0 as n .
Proof: Notice thatan = Sn Sn1 and so limn an = limn(Sn Sn1) =S S= 0.
Corollary:n=0(1)n and n=0n diverge, as neither sequences converge to 0.Corollary: The series
n=0x
n converges |x|< 1.
Proof: |x| 1 = |xn|= |x|n 1(n N). The converse was proved last time.
Next, we provide a characterization of convergence in terms of the size of the tails of the series.
Theorem:
n=lan converges >0.Nl. m k N = |m
n=kan|< .
Proof:
n=lan converges Sk =k
n=lan converges {Sk} is Cauchy.
This is useful in practice because we can guarantee a series converges without knowing its value.
Theorem:1. Ifn k.|an| bn for some k l, and
n=l
bn converges, thenn=l
an converges.
2. Ifn k.0 an bn for some k l, andn=l
an diverges, thenn=l
bn diverges.
Proof: (1) Let > 0 and prove with previous theorem and induction on triangle inequality. (2)follows from contrapositive.
Examples:
1.
n=0
(1)n2n converges because |
(1)n2n |=
12n and
n=0
12n converges (
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On the other hand, ifm 2k,Sm a1 + + a2k =a1 +a2 + (a3 + a4) + + (a2k11 + +a2k)
12a1+a2+ + 2
k1a2k = 12Tk.
Now, if
n=02na2n converges, then Tn T as n and so Sm limn Tm =T,
which means {Sm}is bounded and
n=1an converges.
Similarly, ifn=1an converges, then Tk 2limn Sn = {Tk} is bounded =n=02na2n converges.
Theorem: Letp R. Then
n=11np converges p >1.
Proof:
Ifp 0 the result is trivial since 1np 1 (the sequences converges to 0). Assume thatp >0. Then 1(n+1)p
1np , so we can apply the Cauchy criterion:
n=1
1
np converges
n=0
2n
(2n)p converges.
But n=0 2n
(2n)p = n=0 1(2p1)n , and this series converges 12p1 1.Notice
n=1
1n is divergent, but
n=1
1n1+r converges r > 0. To try to find intermediate series,
we need the logarithm.
3.1.2 Logarithm
Definition: From Supplemental Reading 3, for every 1 < b R, we define a functionlogb : {x R |x >0} R such that
1. blogbx =x (x >0)
2. logb(1) = 0, logbb= 1
3. 0< x < y logbx 0, z R)
5. logb is a bijection
6. limn
logbnnr = 0 (r R, r >0)
Then from (6), for large nand p >0 we know:n n(logbn)
p n np =n1+p = 1n1+p 1
n(logbn)p
1n .
So 1n(logbn)pis such an intermediate series.
Theorem: Letb >1.
n=2
1n(logbn)
p converges p >1. (n 2 = logbn >0)
Proof:n=2
1n(logbn)
p converges n=1
2n
2n(logb2n)p converges by Cauchy criterion, but
n=1
1(logb2)
pnp = 1(logb2)
p
n=1
1np converges p > 1.
In particular,n=2
1n logbn
is divergent.
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3.2 The number e
Lemma:
n=01n! converges.
Proof: Ifn 2 then:
Sn=
n
k=0
1
k!= 1 + 1 +
1
2 1+ +
1
n(n 1) 2 1
1 + 1 +1
2+
1
2 2+ +
1
2n1
1 +k=0
1
2k = 1 + 2 = 3
Since Sn is increasing and bounded, we know that
n=01n! converges.
Definition: We set e =n=0
1n! . Note that e >1.
Theorem: e= limn(1 + 1n)n.
Proof: Let Sn=n
k=01k! , Tn = (1 +
1n)
n
. Then by the Binomial Theorem:
Tn = (1 +1
n)n =
nk=0
n!
k!(n k)!
1
nk
= 1 + 1 + 1
2!
n(n 1)
n2 + +
1
n!
n(n 1) 1
nn
= 1 + 1 + 1
2!(1
1
n) +
1
3!(1
1
n)(1
2
n) + +
1
n!(1
1
n) (1
n 1
n )
1 + 1 + 1
2!+ +
1
n!=Sn
Hence, lim supn Tn lim supn Sn = limn Sn = e.
OTOH, fix m N. Then for n m:
Tn 1 + 1 + 1
2!(1
1
n) + +
1
m!(1
1
n) (1
m 1
n )
= lim infn Tn lim infn RHS 1 + 1 +
1
2!lim infn (1
1
n) + +
1
m!lim infn (1
1
n (1
m 1
n ) =
Then, lettingm , e = limm Sm lim infn Tn.
Thus, e lim infn Tn lim supn Tne = limn Tn = e.
Theorem: n 1.0 < e Sn < 1nn! . Also,e R\Q is irrational.
Proof: Since Sn is increasing, 0< e Sn is clear. The other side can be seen from algebra.
Now, suppose e Q; then e = pq forp, q N, p , q 1.
Then 0< q!(e Sq)< 1q (q 1). Notice that q!e= q!
pq = (q 1)!p N and
q!(1 + 12! + + 1q!) N.
Henceq!(e Sq) Z; but this yields an integer between 0 and 1, a contradiction. So eis irrational.
Remark: In fact, e is transcendental.
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3.3 More Convergence Results
Theorem (Root Test): Suppose {an}n=l R and {|an|
1/n} is bounded. Let 0 = lim supn |an|1/n.Then the following holds:
1. If 1, then n=lan diverges.
3. if= 1, both convergence and divergence are possible.
Theorem (Ratio Test): Let{an}n=l R. Then
n=lan:
1. converges if{|an+1an |}n=l is bounded and lim supn
|an+1||an|
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Definition: The seriesn=0
cn, where cn=n
k=0
akbnk, is called the Cauchy product
of the seriesn=0
an,n=0
bn.
Remark: If
an,
bnconverge,
cndoes not necessarily converge if neither series has convergentabsolute values.
3.5 Absolute Convergence and Rearrangements
Proposition: If
n=l |an|converges, then
n=lan converges. Proof is trivial.
Definition: Suppose
n=lan converges. If
n=l |an|converges, the series convergesabsolutely. If|an|diverges, the series is conditionally convergent.
Example:
n=1(1)n
n is conditionally convergent, while
n=1(1)nn2 is absolutely convergent.
Lets try to manipulate the series without being careful.
=
n=1
(1)
n+1
n = 1 12+ 13 14+
= limk
(Sk =k
n=0
(1)n+1
n ) = lim
k(S2k =
2kn=0
(1)n+1
n )
but: S2k = (1 1
2) + (
1
3
1
4+ + (
1
2k 1
1
2k)> 0
Hence, >0. But the next step is questionable:
2=
n=1(2)(1)n+1
n?
=
k=02
2k+ 1
k=12
2k
?=
k=0
2
2k+ 1
k=1
1
k =
k=0
1
2k+ 1
k=1
1
2k =
= 2= >0 a contradiction!
Problem: rearrangement is a delicate issue.
Definition: Let :{m Z| m l} {m Z| m l} be a bijection. The series
n=la(n) iscalled a rearrangement of
n=lan.
Theorem: If
n=lan is absolutely convergent, then every rearrangement converges to
n=lan.
Proof: Let >0.
Since
n=lan converges absolutely, Nl. k m N =k
n=m |an|< 2 .
Letk :
n=m |an| 2 < .
Now choose M N such that {l, l + 1, . . . , N } {(l), (l+ 1), . . . , (M)}. Thenm M = |
mn=lan
mn=la(n)|
n=N|an|< .
Hence limm(m
n=lan
n=la(n)) = 0 and from this we deducelimm
mn=la(n)= limm
mn=lan =
n=lan.
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When a series is only conditionally convergent, the situation is vastly worse.
Theorem: Suppose
n=0an is conditionally convergent. Let c R.There exists a rearrangement (bijection) : N N such that
n=0a(n)= c.
Lemma: Suppose
n=0an is conditionally convergent and set:
bn
= an ifan > 00 ifan0
cn
= an ifan < 00 ifan 0
Then
n=0bn and
n=0cn both diverge.
Proof: Suppose not; one of the series is convergent. If
bn converges, then cn = bn an =cn =
bn
an; but |an|= bn+cnand so
|an|=
bn+
cnis convergent, a contradiction.
A similar argument holds if
cn converges.
Rearrangement Theorem Proof:
Let{a+n }n=0 denote the subsequence of{bn |bn > 0 or bn = 0 an = 0}. Let{a
n }
n=0
denote the subsequence of{cn | cn>0}(from last lemma). Note:
1. a+n 0, an 0 since an0 = bn 0, cn 0.
2.
a+n and
an both diverge because they differ by 0 from
bn,
cnrespectively.
Setm0= n0= 1. Since
a+n diverges we may use the well-ordering principle: m1=min{k N |
kn=0a
+n > c}. Similarly, n1= min{k N |
m1n=0a
+n
kn=0a
n < c}.
Next, ifmp and np are known, we set:
mp+1= min
k N |
p1l=0
mlj=1+ml
a+j
p1l=0
nlj=1+nl
aj +k
j=1+mp
a+j > c
np+1= mink N |
p1l=0
mlj=1+ml
a+j
p1l=0
nlj=1+nl
aj +
mp+1j=1+mp
a+j
kj=1+np
aj < c
Consider the series (a+1 + + a+m1) (a
1 + + a
n+1) + (a
+1+m1
+ + a+m+2) (a1+n1
+ +an2) + . This is clearly a rearrangement of
n=0an.
Write Ap =mp+1
l=1+mpa+l , A
p =
np+1l=1+np
al , and let Sj denote the jth partial sum of
the rearrangement.By construction, lim supj Sj = lim supp(
p+1l=0 A
+l
pl=0A
l ) and
liminfj Sj = liminfp(p
l=0A+l +
pl=0A
l ).
Also,c < p+1l=0 A
+l
pl=0A
l < c+a
+mp+1 and c a
np+1 <
p+1l=0 A
+l
p+1l=0 A
l < c.
Thus, by the squeeze lemma, limp(
p+1l=0 A+l
pl=0Al ) = limp(
pl=0A+l pl=0A
l ) =c, and so limj Sj =c =
n=0a(n) = c.
Remark: One can also rearrange such that
a(n) = .
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21-355 Notes
4 Topology ofR
Our goal in Section 4 is to develop some tools for understanding the topology ofR, which is asort of generalized qualitative geometry.
4.1 Open and Closed Sets
4.1.1 Open Sets
Definition:
1. Fora, b R with a b, we define:
(a, b) ={x R |a < x < b} [a, b) ={x R |a x < b}
(a, b] ={x R |a < x b} [a, b] ={x R |a x b}
2. Forx R and >0, we set B(x, ) = (x , x+) and B[x, ] = [x , x+].We call the set B(x, ) a neighborhood ofxor a ball of radius centered at x.
3. A set E R is open ifx E. >0. B(x, ) E.In other words, every point in Ehas a neighborhood contained in E.
Examples:
1. is vacuously open.
2. R is open because x R. B(x, 1) R.
3. Ifa < b then (a, b) is open.Proof : Fix x (a, b) and let = min{x a, b x} > 0. Thena x < x 0. a /[a, b) and hence B (a, )[a, b).5. [a, b] is not open, nor is (a, b] by previous argument.
6. E= {a} is not open.
7. E= { 1n |n N, n 1} is not open: >0. B(1, )E.
Lemma: IfE R is open A (some index set), then
AE is open.
Proof : Let x
AE. Thenx E0 for some 0 A. Since E0 is open, > 0. B(x, ) E0
AE.
Lemma: IfEi R is open for i [n], n N, then
ni=1Ei is open.
Remark: Infinite intersections of open sets need not be open. Let En = (1
n ,
1
n), n 1.Then
n=1En = {0} which is closed.
4.1.2 Closed Sets
Definition: We say E R is closed iffEc = R\E is open.
Lemma: E is open Ec is closed (by definition).
Examples:
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1. is closed because c = R is open.
2. R is closed because Rc = is open.
3. [a, b] is closed because [a, b]c = (, a) (b, ) is the union of open sets, and thusopen.
4. [a, b) and (a, b] are not closed because [a, b)c = (, a) [b, ) and B(b, ) [a, b)c ( >0).
5. {a}is closed since {a}c = (, a) (a, ), both open sets.
6. SupposeE R is finite. Write E= {ai| i [n]} where a1< a2< .. . < an. ThenEc = (, a1) (a1, a2) (an1, an) (an, ), all of which are open.
7. E= { 1n |n N, n 1}is not closed. Ec = (, 0]
n=1(
1n+1 ,
1n) (1, ) is not
open because B(0, ) E= { 1n | 1 < n} = = B(0, ) /E
c.
8. E= {0} { 1n |n 1}is closed, asEc = (, 0)
n=1(
1n+1 ,
1n) (1, ) is open.
Lemma:
1. IfE R is closed A, then
AE is closed.
2. IfEi R is closed i [n] then ni=1Ei is closed.Proof : The complement is the union ofEc (open by claim), which is open by previous lemma.
Remark: Example (7) shows that infinite unions of closed sets need not be closed.
4.1.3 Limit Points
Definition: Let E R.
1. A pointx R is a limit point ofE iff. >0.(B(x, ) E)\{x} = .
2. A pointx E is called isolatedif it is not a limit point.
Example: E={ 1n |n 1}. 0 is a limit point, but 1n Eis isolated, since B( 1n , 1n(n+1)) E={ 1n}.
Theorem: LetE R. E is closed every limit point ofEis contained in E.
Proof:
=:AssumeE is closed and x R is a limit point ofE. Ifx Ec then, since Ec is open, > 0. B(x, ) Ec = B(x, ) E= . But this contradicts the fact that x is alimit point ofE; thus x E.
=:Suppose Eis not closed; then Ec is not open and so >0.x Ec. B(x, ) E= .
Since x Ec, (B(x, ) E)\{x}= B (x, ) E= and hence x is a limit point ofE.Then x E Ec, a contradiction; and Eis closed.
Definition: Let{xn}n=l Sfor some set S. We say {xn} is eventually constant if
Nl. xn = xN (n N).
Proposition: LetE R. Thenx is a limit point ofE {xn}n=1 Esuch that the sequence
is not eventually constant and xn x as n .
Proof:
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4.1 Open and Closed Sets 21-355 Notes
=:Suppose x is a limit point ofE, i.e. > 0. (B(x, ) E)\{x} = . Set r1 = 1 andchoose x1 Esuch that x1(B(x, r) E)\{x}.Setrn = min(
1n , |x xn1|) and choose xn (B(x1, rn) E)\{x}.
Thenn 1.{xn}n=1 Eand|x xn1|< |x xn|and |x xn| 0. N 1. n N = |x xn| < . Then {xn | n N} B(x, ) E.If {xn | n N} = {x} then {xn} is eventually constant, a contradiction. Hence={xn |n N}\{x} (B(x, ) E)\{x} = (B(x, ) E)\{x} = , and hence xis a limit point.
Corollary: Let E R. The following are equivalent (proof follows from last theorem):
1. Eis closed.
2. Ifx R is a limit point ofE,x E.
3. If{xn}n=lEis such that xn x as n , then x E.
Corollary: Let E R and E= . Suppose Eis closed.
1. IfEis bounded above, then sup E E, i.e. sup E= max E.
2. IfE is bounded below, then infEE, i.e. infE= min E.
4.1.4 Closure, Interior, and Boundary Sets
Definition: Let E R.
1. LetO(E) ={V R |V Eand V is open} P(R)
C(E) ={C R |E Cand C is closed} P(R).
Note that O(E) and R C(E).2. We defineE0 =
VO(E)V, and call this set the interior ofE.
We define E=
CC(E)C, and call this set the closure ofE.
3. We defineE= E\E0 to be the boundary ofE.
Theorem: LetE R. The following hold:
1. E0 E E
2. E0 is open and E,Eare closed.
3. For every x E,x E0 x E.
4. E= {x R | >0. B(x, ) E= and B (x, ) Ec
= }.5. E is open E=E0, E is closed E= E.
Proof:
1. Trivial.
2. E0 is an arbitrary union of open sets and thus open; E is an arbitrary intersectionof closed sets, so its closed. E= E\E0 = E (R\E0) is the intersection of twoclosed sets, so its closed.
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3. Trivial.
4. Supposex E. Show the two properties of the set are satisfied via contradiction.Next, assume x in the set, and show that x E.
5. Trivial.
Corollary: Let E R. Then Eis closed EE.
Proof: E is closed = E = E = E E E. On the other hand, if E E thenE E=E0 EE, soE= E.
Theorem (Bolzano-Weierstass, Part 2): Let E R be infinite and bounded. ThenEhas alimit point.
Proof: Since E is infinite we may construct a non-eventually-constant sequence {xn}n=0E. We
do so by choosing x0 E arbitrarily, and xn E\{x0, . . . , xn1} for any n N+. Since E isbounded, the sequence is too, so B-W implies there exists a convergent subsequence{xnk}
k=0E.
This subsequence is not eventually constant by construction, so its limit is a limit point.
4.2 Compact Sets
Definition:
1. Let A be some index set and assume A. V R. We writeV={V}A forthe collection of all of these subsets.
2. IfE R andE
AV, then we say V is a cover ofE.
3. IfV R is open A and V is a cover ofE, we say V is an open cover.
4. LetVbe a cover ofE. We sayW={V}A is asubcover ofE ifA A and Wis a cover ofE.
5. LetVbe a cover ofE. IfA is finite, then W={V}A is afinite subcover ofE,ifW is a subcover ofE.
Examples:
1. EveryE R admits a cover: E=
xE{x}.
2. EveryE R admits an open cover: E
xEB(x, ) for >0.
3. If E is finite and V is an open cover, we claim there is a finite open subcover.Indeed, write E = {ai | 1 i n} and choose Vi such that ai Vi . ThenE
ni=1Vi and {Vi}
ni=1 {V}A. Hence every open cover of a finite set
admits a finite open subcover.
4. E = { 1n | n 1}. V = {B(1n ,
1n(n+1)}
n=1 is an open cover of E. Note that
B(
1
n ,
1
n(n+1)) E=
1
n , so there does not exist a finite subcover.5. E= {0}{ 1n |n 1}. Suppose Vis an open cover ofE. Since 0 E,0 A.0
V0. Since V0 is open, >0. B(0, ) V0. Then B(0, ) E= {1n ||; nN}
where N = min{n N | n 1 }. Hence E\B(0, ) = {1n | 1 n N}. There
existVn for n [N] such that 1n Vn . Then E
Nn=0Vn and Ehas a finite
subcover.
6. Let a < b and E= (a, b). Then V={(a+ 1n+1 , b 1n+1)}nN is an open cover of
E. Since these intervals are nested, there cannot be a finite subcover.
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Definition: Let E R. We say that E is compact if every open cover of E admits a finitesubcover.
Examples:
1. is trivially compact.
2. Ris not compact becauseV={B(0, n)}nN is an open cover that clearly does notadmit a finite subcover ofR.
3. Any finite setE R is compact.
4. (a, b) for a < b is not compact.
5. { 1n |n 1} is not compact.
6. {0} { 1n |n 1} is compact.
Notice in each of our examples of compact sets that the set is closed and bounded.
4.2.1 Heine-Borel Theorem
Theorem: LetK R. Then Kis compact Kis closed and bounded.
Proof:
= Suppose Kis compact.
Notice that
n=1B(0, n) = R (since R is Archimedean) and so K R =
n=1B(0, n).Then {B(0, n)}n=1 is an open cover ofK. Since K is compact, a finite subcover :K
mi=1B(0, ni) for some m N.
Setr= maxi[m]ni. Then Km
i=1B(0, ni) B(0, r) = Kis bounded.
Now we showK is closed. Let x KC. For eachy Kwe setry = 12 |x y|> 0. Then
B(y, ry) B(x, ry) = (y K). Also, {B(y, ry)}yK is an open cover.
K compact = a finite subcover: K n
i=1B(yi, ryi). Set r = mini[n]ri > 0and notice that B(yi, ryi) B(y, r) = . Hence
ni=1B(yi, ryi) B(x, r) = =
K B(x, r) = = B(x, r) KC. This means that KC is open and so Kis closed.
= (Heine-Borel) Suppose K is closed and bounded. IfK= were done, so suppose K= .
Notice that K bounded = infK, sup K R, and K closed = infK, sup K K.In particular, sup K= max K, inf = min K. Let Vbe an open cover ofK.
Let E = {x K | Vadmits a finite subcover ofK[infK, x]} K. Notice thatK [infK, infK] = {infK} is a finite set and hence compact; thus V admits a finitesubcover ofK [infK, infK]. Hence infKE, and so E=. Clearly E is boundedabove by sup K. By LUB property, sup E R and sup E sup K.
We want to show sup E= sup K = max E. Notice thatn 1. xn E K suchthat sup E 1n < xn sup E. Then xn sup E as n , and so sup E K (sinceK is closed).
Write V = {V}A. Since sup E K, 0 A such that sup E V0 . But V0 isopen so > 0. B(sup E, ) V0. By definition, x E. sup E < x sup E.HenceVadmits a finite subcoverofK [infK, x], i.e. K [infK, x]
ni=1Vi . Then
K [infK, sup E]n
i=0Vi = sup EE = sup E= max E.
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Assume for sake of contradiction that max E < max K. Let K = K\n
i=0Vi . K
is closed since its the intersection of closed sets. K = since otherwise K ni=0Vi = max E= max K.
Let y = infK = min K (since K is closed) and note that y > max E. Then K[infK, y] =K[infK, min K]
ni=0Vi{y}. But sincey K
K, Vn+1 Vsuchthaty Vn+1. HenceK [infK, y]
n+1i=0 Vi = y E = max E < y max E,
a contradiction. We then deduce that max E= max K = K=K [min K, max K]is covered by a finite subcover ofV; thus, Kis compact.
Corollary:
1. IfK R is compact and E R is closed, then E K is compact.
2. IfK R is compact and E Kis closed, then Eis compact.
3. IfKi R is compact for i [n], thenn
i=1Ki is compact.
4. IfK R is compact A, then
AK is compact.
4.3 Connected Sets
Definition: We say two sets A, B R are separated ifA B= A B = .A setE R is disconnected ifE= A B such that a= , B= andA, B are separated.If a set is E R is not disconnected, we say its connected.
Examples:
1. (0, 1) and [1, 2) are not separated, though they are disjoint, since (0, 1) [1, 2) =[0, 1] [1, 2) ={1} = .
2. (a, b) and (b, c) fora < b < c are separated, since (a, b) (b, c) = = (a, b) (b, c).Then (a, c)\{b} is disconnected, since (a, c)\{b}= (a, b) (b, c).
3. Similarly, a R.(, a) an (a, ) are separated.
Then R\{a}= (, a) (a, ) is disconnected.
Theorem: LetE R. Then E is connected (x, y Eandx < z < y = zE).
Proof:2 = 1:
If (2) is false then x, y E and z (x, y) such that z / E. ThenE = Lz Rz forLz =E (, z) and Rz =E (z, ). Since x Lz, y Rz, and Lz (, z) andRz (z, ), it follows thatLz and Rz are separated. Hence E is disconnected.
1 = 2
Suppose E is disconnected. Write E= A B with A, B= and A B= A B= .Letx A andy B. Without loss of generality, we assume x < y.
Let z = sup(A [x, y]). Clearly z A and so z / B = z = y = x z y . Ifz /A thenz =x = x < z < y andz /A B=E. Otherwise, ifz A, thenz / B.B is closed, so BC is open; and hence we can find w such that z < w < y, w /B, andw /A. Then x < w < y and w /A B=E. In all cases, then, 2 is true.
Corollary: R, (, a), (, a], (a, ), [a, ), (a, b), (a, b], [a, b),and[a, b] are all connected.
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5 Continuity
5.1 Limits of Functions
Definition: Let E R, f :E R, and p R be a limit point. Let q R.We say limxpf(x) = q or f(x) q as x p iff > 0. > 0. x E 0 < |x p| < =|f(x) q|< .
Examples:
1. E= [0, 1], f(x) =x. Let p = 12 . limx 12 f(x) = 12 .
Proof: Let > 0; choose = > 0. Thenx [0, 1] and 0 < |x 12 | < S =|f(x) 12 |< .
2. E= [0, 1], f(x) =x (for x = 12), f(x) = 37 (for x= 12).
By the proof of (1), the claim still holds.
3. f(x) = xn on E = (0, 1) for 2 n N. 0 is a limit point of E; we claimlimx0xn = 0.Proof : Let >0; choose =1/n >0. Thenx(0, 1) and 0 < |x 0|< =
0< x < = 0< xn < n = = |f(x) 0|= xn < .
4. limxpx= p whenever p is a limit point ofE.
5. Ifx E. f(x) = 1 then limxpf(x) = 1 whenever p is a limit point ofE.
6. Let E = R and f(x) = cos(x). From HW6, | cos(x) 1| x2ex2
. We claimlimx0cos(x) = 1.Proof : Let > 0. Choose= min(1,
/e) > 0. Then for x R, 0 < |x 0| 0. x E, 0 < |x 0| < =|f(x) q|< 1. But x E, |x|< = x= 1n ,
1 < n, and|f(x) q|= |
11/n q|=
|n q|< 1, which is a contradiction.
Definition: Let f : E R for some E R. IfA Ewe define f(A) = {f(x)|x A} R asthe image ofA under f. IfB R we define f1(B) ={x E|f(x) B} as the pre-image ofBunderf.
Lemma: Suppose f :E R. Then A B E = f(A) f(B), andA B R = f1(A) f1(B) E.
5.1.1 Divergence Criteria
Theorem (Divergence Criteria): Let E R, f :E R, p be a limit point ofE, q R. Thefollowing are equivalent:
1. limxpf(x) =q
2. For every open set V R such thatq V, an open set U R with p U suchthat f(U E\{p}) V. (Topological characterization)
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3. If{xn}n=l E satisfies xn = p (n l) and xn p as n , the sequence
{f(xn)}n=l R converges andf(xn) qas n . (Sequential characterization)
Proof:
(1) = (2) :Assume (1) and let V R be open with qV . Since V is open, >0. B(q, ) V.Since limx
pf(x) = q, > 0. x E 0 < |x p| < = |f(x) q| < . Let
U = B(p, ) (an open set). Then x U E\{p} = x E |x p| < =|f(x) q|< = f(x) B(q, ) V. So f(U E\{p}) V as desired.
(2) = (3):Assume (2) and let {xn}
n=l E satisfyxn=p, xn p. Let >0 and set V =B(q, )
(open). From (2), open U such that f(U E\{p}) V and p U. Since U is open, > 0. B(p, ) U. Since xn p as n , N l. n N = ) < |xn p| < = xn U E\{p} = f(xn) V =B(q, ). Hence n N = |f(xn) q|< ,and f(x) qas n .
(1) = (3):Suppose (1) is false; then > 0. > 0. x E with 0 < |x p| < such that|f(x) q| . For n N, n 1, set = 1n to find xn Esuch that 0< |xn p|0. >0. x E |x p|< = |f(x) f(p)|<
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Iff :E R is continuous at each p Ewe say f is continuous on E.
Remarks:
1. In order to be continuous at p E, f must be defined at p. Contrast this tolimxpf(x), in which case p need only be a limit point ofE.
2. Informally one can think of continuous functions as those approximated well nearp byf(p), i.e. f(x) f(p) when x p.
3. In the definition, the value of may depend on the point p. If a function iscontinuous on Ethen for a given >0 the = (p) may vary greatly as p varies.
4. Ifp E is isolated (not a limit point ofE), then f is vacuously continuous at p:x E, |x p|< for small enough = x= p.
Example:
We saw last time that limxpP(x) = P(p) for all polynomials P : R R. Hence >0. >0. x R, 0< |x p|< = |P(x) P(p)|< . Hence P is continuousat p.
Theorem: LetE R, f :E R, p Ebe a limit point ofE. Then:
f is continuous at p limxp f(x) =f(p)
Corollary (Algebra of Continuity): LetE R, f , g : E R, andp E. Assume thatf , g arecontinuous atp. Then the following hold:
1. If, R then f+ g is continuous at p.
2. f g is continuous at p.
3. Ifg (p)= 0 then fg :E\g1({0}) R is well-defined and continuous at p.
Proof: Ifp is isolated, the claim is vacuously true. Assume p is not isolated, i.e. p is a limit pointofE. Then the last theorem and algebra of limits gives the result.
Corollary: LetE R, f , g : E R. Iff , g are continuous on E, then:
1. If, R then f+ g is continuous on E.
2. f g is continuous on E.
3. Ifg (x)= 0 (x E), then fg is continuous on E.
Theorem: Let E, F R, f : E R, g : F R. Assumef(E) F, f is continuous at p E,andg is continuous at f(p) F. Theng f :E R(where (g f)(x) =g(f(x))) is continuous atp. Moreover, iff is continuous on Eand g is continuous on F, then g f is continuous on E.
Proof: Let >0.
Sinceg is continuous atf(p), >0. y Fand |y f(p)|< = |g(y)g(f(p))|< .Since f is continuous at p, >0. x E , |x p|< = |f(x) f(p)|< .
Since f(E) Fwe know that x E, |x p|< = f(x) F, |f(x) f(p)|< =|g(f(x)) g(f(p))|< . Hence, g f is continuous by definition.
Examples:
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1. exp, cos, sin : R Rare continuous onR (proof in HW). YAlso, log : (0, ) Ris continuous on (0, ).
2. Let R and set f : (0, ) R via f(x) =x. Notice thatf(x) = exp( log x).Since log and exp are continuous, f(x) =x is continuous.
Definition: LetE Rand A E. We say A is relatively open in E iffA = U Efor some opensetU R. Similarly, we say A is relatively closed in E iffA = C Efor some closed C R.
Proposition: Let A E R. The following hold:
1. A is relatively open in E x A. >0. B(x, ) A E.
2. A is relatively closed in E A= BC Efor some relatively open B E.
Theorem (Continuity Criteria): Let E R, f :E R. The following are equivalent:
1. f is continuous on E.
2. Ifp Eis a limit point ofE, then limxpf(x) =f(p).3. Ifp E is a limit point ofE and {xn}
n=l E satisfies xn p as n , then
f(xn) f(p) as n .
4. IfV R is open, then f1(V) Eis relatively open in E.5. IfC R is closed, then f1(C) Eis relatively closed in E.
Proof:
(1) (2) (3) follows from the sequential criterion of limits, previous theorem.(4) (5) follows since f1(VC) = (f1(V))C E.
(1) = (4):Let V R be open and choose pf1(V). SinceV is open, >0. B(f(p), ) V.It suffices to show, via previous proposition, that >0. B(p, ) Ef1(V). Sincef is continuous on E, > 0. x E, |x p| < = |f(x) f(p)| < . Thatis, x B(p, ) E = |f(x) f(p)| < = f(x) B(f(p), ) V. HenceB(p, ) E f1(V).
(4) = (1):Let p E, >0, and V =B (f(p), ). Thenf1(B(f(p), ))E is relatively open inE = (by previous proposition) >0. B(p, ) Ef1(B(f(p), )). Then x Eand|x p|< = f(x) B(f(p), ) = |f(x) f(p)|< . Since, pwere arbitrary,we deduce fis continuous on E.
5.3 Compactness and Continuity
Theorem: SupposeK R is compact andf :K R is continuous onK. Thenf(K) is compact.
Proof:
Note that for E R, f(f1(E)) E and E f1(f(E)). Let {V}A be an opencover of f(K). Since f is continuous and V is open, f
1(V) is relatively open inK = f1(V) =U K for some open U R.
Since{V}A coverf(K), we see that{f1(V)}Ais a cover ofK. Then{U}A isan open cover ofK. Since Kis compact, there exists a finite subcover: K
ni=1Ui .
Then Kn
i=1Ui K=n
i=1f1(Vi) = f(K)
ni=1f(f
1(Vi))n
i=1Vi .As we have extracted a finite open subcover off(K), f(K) is compact.
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Extreme Value Theorem: Let K R be compact and f : K R be continuous. Thenx0, x1K such that f(x0) f(x) f(x1) (x K). That is, f(x0) = minxKf(x) = min f(K)and f(x1) = maxxKf(x) = max f(K).
Proof: From last theorem, we know f(K) is compact, so its closed and bounded. From a pre-vious theorem, closed and bounded sets contain their infimum and supremum (and thus min, max).
Definition: Let E R and f :E R. We sayf is uniformly continuous on E iff:
>0. >0. x, y E |x y|< = |f(x) f(y)|<
Remarks:
1. f is uniformly continuous on E = f is continuous on E.
2. The key difference is that for uniform continuity, >0 works for all points in E.
Examples:
1. Let E= (0, 1) and f(x) = 1x . Its trivial thatf is continuous on E, but it is notuniformly continuous.
Proof: Suppose it is; then for = 12 , > 0. x,y (0, 1) |x y | < =|f(x) f(y)|< 12 . ChooseN N such that
1
< N. Thenx = 1n , y= 1n+1 satisfy
|x y|= 1n(n+1) 1n2 < ifn N. Then
12 >|f(x) f(y)|= |n (n + 1)|= 1, a
contradiction.
Definition: A function f :E R is Lipschitz ifx, y E.k >0.|f(x) f(y)| k|x y|.
Claim: Iff is Lipschitz, it is uniformly continuous. Proof: let = k .
Theorem: LetK Rbe compact and f :K R be continuous. Then f is uniformly continuouson K.
5.4 Continuity and Connectedness
Theorem: Let E R be connected and f :E R be continuous on E. IfXE is connected,then f(X) is connected.
Intermediate Value Theorem: Let a < b R. Suppose f : [a, b] R is continuous. Iff(a)< c < f(b) or f(b)> c > f(a) for some c R, then x (a, b). f(x) =c.
5.5 Discontinuities
Lemma: If p is a limit point of E R then p is a limit point of E+
p
= E(p, ) or Ep
=E (, p).
Definition: Let E R, f :E R,pbe a limit point ofE,q R.
1. Ifp is a limit point ofEp, we say limxpf(x) =q >0. >0. x Ep,0< p = |f(x) q|< .
2. Ifp is a limit point ofE+p, then limxp+f(x) =q >0. >0. x E+p,0< x p < = |f(x) q|< .
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Proposition: Ifp is not a limit point ofE+p then limxpf(x) = limxpf(x). Ifp is not a limitpoint ofEp then limxpf(x) = limxp+f(x).
Proposition: Ifp is both a limit point of either E+p or Ep, then
limxp f(x) =q limxp
f(x) = limxp+
f(x) =q
Definition: SupposeE R,f :E R,p Eis a limit point ofE. Suppose further that p is nota point of continuity off.
1. We sayfhas a simple discontinuity ofp ifp is not a limit point ofE+p and limxpf(x) exists,p is not a limit point ofEp and limxp+f(x) exists, orp is a limit point ofE+p andE
p and limxp+f(x), limxpf(x) both exist.
2. Otherwise, we sayfhas an essential discontinuity ofp.
5.6 Monotone Functions
Definition: Let E R and f :E R. We say:
f is non-decreasing (increasing) ifx, y Eand x < y = f(x) f(y) (f(x)< f(y)), andf is non-increasing (decreasing) ifx, y Eandx < y = f(y) f(x) (f(y)< f(x)).Iff is non-increasing or non-decreasing, f is monotone.
Theorem: Suppose f : (a, b) R is monotone, and let p (a, b). Then limxpf(x) andlimxp+f(x) both exist. Moreover, iffis non-decreasing, then
limxp
f(x) = sup f((a, p)) f(p) inff((p, b)) = limxp+
f(x)
Corollary: Iff : (a, b) R is monotone, then fhas no essential discontinuities.
Example: f(x) =x is non-decreasing and fhas countably many simple discontinuities.
Theorem: Iff : (a, b) R is monotone, thenfhas at most countably many simple discontinuities.
6 Differentiation
6.1 The Derivative
Definition: Assumef : [a, b] R for a < b R. For allx [a, b], the function : (a, b)\{x} R
via (t) = f(t)f(x)
tx is well-defined, and x is a limit point of (a, b)\{x}. If limtx(t) exists wewrite f(x) = limtx(t) and say that f is differentiable at x.
We define f :{x [a, b]| x is differentiable at x} R to be the derivative off. Iff is differen-
tiable x E[a, b], we say fis differentiable on E.Definition (General): Let E R, f : E R, and x E be a limit point of E. Define
: E\{x} R by(t) = f(t)f(x)tx . If limtx(t) exists we say fis differentiable at x, and writef(x) = limtx(t).
Proposition (locality of derivative): Supposef : E R, g : F R, x E F is a limiitpoint ofE F, and that f and g are differentiable at x. Iff = g on E F then f(x) = g(x).This shows that f(x) only depends on the value off near x.
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Proposition (Newtonian approximation): Let f : E R, xEbe a limit point ofE, andL R. Then the following are equivalent:
1. f is differentiable at x and f(x) =L
2. >0. >0. t E |x t|< = |f(t) (f(x) +L(t x))|< |t x|
Proof follows from definition of limtx(t). Newtons approximation says differentiablefunctions are those that can be well-approximated by affine functions +x. Contin-
uous functions are those well-approximated by constants, while differentiable functionsare well-approximated by the next simplest function.
Theorem: Supposef :E R,x Eis a limit point ofE, andf is differentiable at x. Then f iscontinuous atx.
Proof: By definition, ift E\{x} thenf(t)f(x) =(t)(tx). Thenf(t) =f(x)+(t)(tx) andhence limtxf(t) =f(x)+limtx(t)(t x) =f(x) + f(x)0 =f(x). By the limit chracterizationof continuity, we deduce that fis continuous at x.
Remark: The converse fails. Let f(x) = |x| on R. Since ||x| |y| | |x y |, f is Lipschitz
and hence uniformly continuous. However, for x = 0, t > 0 = (t) = |t|0t0 = 1 and
t < 0 = (t) = t0t0 = 1. Then limt0(t) = 1 = limt0+(t) = 1, so f(0) doesnot exist.
Theorem (Algebra of Derivatives): Letf, g: E R be differentiable at x E. Then:
1. f+g : E R is differentiable at x and (f+ g)(x) =f(x) +g(x)
2. f g: E R is differentiable at xand (f g)(x) =f(x)g(x) +f(x)g(x)
3. If g(x) = 0 then fg : E\g1({0}) R is differentiable at x and ( fg )
(x) =g(x)f(x)f(x)g(x)
g(x)2
Examples:
1. f(x) =+x on R = f(x) = limtxf(t)f(x)
tx =(x R).
2. f(x) =xn forn N = f(x) =nxn1. Proof by induction.
3. Every polynomialP(x) =N
n=0anxn is differentiable, andP(x) =
Nn=0nanx
n1.
4. R(x) = P(x)Q(x) is dfiferentiable when P, Q are polynomials at points p R where
Q(p)= 0.
Theorem (Chain Rule): Supposef :E R is differentiable atx E,f(E) F, andg : F Ris differentiable atf(x) F. Thengf :E R is differentiable atxand (gf)(x) =g (f(x))f(x).
6.2 Mean Value Theorems
Definition: Let f : E R. We say thatf has a local maximum at x E if > 0. t E and|x t|< = f(t) f(x). We say fhas a local minimumatx E iffhas a local maximum.
Iffhas either a local max or min at x E, we sayfhas a local extremumat x.
Theorem: Supposef : E R is differentiable at x E and x is a limit point of both E+x andEx. Iffhas a local extremum at x, then f(x) = 0.
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Proof: It suffices to assume that f has a local max at x. Let > 0 such that t E and|x t|< = f(t) f(x). Then t E , 0< x t < = f(t)f(x)tx 0 and 0< t x < =f(t)f(x)
tx 0. So limtxf(t)f(x)
tx =f(x) 0, limtx
f(t)f(x)tx =f
(x) 0, and thus f(x) = 0.
Remark: The result is false ifx is not a limit point of either E+x or Ex. Consider f(x) = x on
E= [0, 1]; fhas a local min at x = 0, local max at x = 1, but f(x) = 1x [0, 1].
Theorem (Monotonicity part 1): Letf :E R be differentiable at x E.
1. Iff is non-decreasing on E, then f(x) 0.
2. Iffis non-increasing on E, then f(x) 0.
Cauchys Mean Value Theorem: Suppose thatf, g: [a, b] R are continuous on [a, b], differ-entiable on (a, b). Then x (a, b).(g(b) g(a))f(x) = (f(b) f(a))g(x).
Proof:
Consider h: [a, b] R via h(x) = (g(b) g(a))f(x) (f(b) f(a))g(x). To prove theresult, it suffices to findx (a, b) such that h(x) = 0 (since h is cont, diff on [a, b] and(a, b)).
Notice thath(a) =g(b)f(a) g(a)f(b) =h(b).
Ifh is constant, thenh(x) = 0 trivially and were done. Assumeh is not constant; thent (a, b). h(t)> h(a) or h(t)< h(a).
If h(t) h(a), then Extreme Value Theorem guarantees that x (a, b). h(x) =max h([a, b]) and Local Extremum Theorem = h(x) = 0.
If h(t) < h(a) then EVT guarantees x (a, b). h(x) = min h([a, b]) and LET =h(x) = 0.
Corollary (Mean Value Theorem): If f : [a, b] R is cont and diff on [a, b], (a, b) thenx (a, b). f(b) f(a) =f
(x)(b a). Proof: Set g(x) =x.
Corollary (Monotonicity part 2): Suppose f : (a, b) R is differentiable on (a, b). Thefollowing hold:
1. x (a, b). f(x)> 0 = f is increasing.
2. x (a, b). f(x) 0 = f is non-decreasing.
3. x (a, b). f(x) = 0 = f is constant.
4. x (a, b). f(x) 0 = f is non-increasing.
5. x (a, b). f(x)< 0 = f is decreasing.
Proof: by MVT, ifa < x1 < x2 < b, then f(x2) f(x1) =f(x)(x2 x1).
6.3 Darbouxs Theorem
Definition: We say a function g : R R is periodic with periodp >0 ifg(x +p) =g(x) (x R).
Theorem (Darboux): Supposef : [a, b] R is differentiable on [a, b] and f(a) < < f (b).Thenx (a, b). f(x) =.
Corollary: Iff : [a, b] R is differentiable on [a, b], then f has no simple discontinuities.
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6.4 LHopitals Rule 21-355 Notes
6.4 LHopitals Rule
Theorem: Suppose f , g: [a, b] R are continuous on [a, b], differentiable on (a, b), and g(x)= 0(x (a, b)). Assume that limxa
f(x)g(x) =L. Iff(a) =g(a) = 0, then limxa
f(x)g(x) =L.
Proof:
We claim first that g(x)= 0 for x (a, b]. Otherwise, g(x) = 0 for some x (a, b] =
0 = g(x)
g(a)
xa = g(z) for some z (a, x), a contradiction. So fg : (a, b] R is well-
defined.
Let{xn}n=l (a, b] satisfyxn a as n . We claim that limn
f(xn)g(xn)
=L. Oncethis is established, the sequential characterization of limits yields the desired result.
To prove the claim, we apply Cauchys Mean Value Theorem on [ a, xn]: yn (a, xn)such that f(yn)g(xn) = f(yn)(g(xn) g(a)) = g(xn)(f(xn) f(a)) = g(xn)f(xn).Then n l.
f(xn)g(xn)
= f(xn)g(xn)
. Since a < yn < xn, the squeeze lemma implies yn a.
Hence limnf(xn)g(xn)
= limnf(xn)g(xn)
= limxaf(x)g(x) =L.
Remarks:
1. The theorem is also true if we take limits att.2. If f, g : (a, b] R and limxaf(x) = limxag(x) = 0, then the theorem still
works.
6.5 Higher Derivatives and Taylors Theorem
Definition: Suppose f : E R is differentiable at x E, and x is a limit point of {y E | f(x) exists}. We say f is twice differentiable at x if f : {y E | f(y) exists} R isdifferentiable atx; and f(x) =f(2)(x) = (f)(x). Similarly, for n N with n >2, we say f is n-times differentiable atx ifx is a limit point of{y E|f(n1)(y) exists}and f(n1) is differentiableat x, in which case f(n)(x) = (f(n1))(x).
Iff(n) exists n N,n 1 we say fis infinitely differentiable at x.
Theorem (Taylor): Supposef : [a, b] R. Assumef(n1) is continuous on [a, b] andf(n) existson (a, b). Let x, y [a, b] with x=y. Thenz (min{x, y}, max{x, y}) such that
f(y) =n1k=0
f(k)(x)
k! (y x)k +
f(n)(z)
n! (y x)n
(called the Taylor polynomial or Taylor approximation).
Proof:
Supposex < y(y < xis handled without loss of generality). LetP(t) =
n1k=0
f(k)(x)k! (t
x)k
, and set M= f(y)
P(y)
(yx)n . It suffices to prove thatM= f(n)(z)
n! for somez (x, y).Defineg(t) =f(t) P(t) M(t x)n, and notice thatg(n)(t) =f(n)(t) n!M. As such,it suffices to show that g (n)(z) = 0 for some z (x, y).
By construction, g(k)(x) = 0 (k = 0, . . . , n 1), and g(y) = 0 (by choice of M).
By Mean Value Theorem, xi (x, y). g(x1) =
g(y)g(x)yx = 0. Similarly, x2
(x, x1). g(x2) =
g(x1)g(x)x1x = 0. Iterating, we eventually findxn1 (x, y). g
(n1)(xn1) =
0. Then 0 =g(n)(z) = g(n1)(xn1)g(n1)(x)
xn1x = 0 for some z (x, xn1).
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21-355 Notes
7 Riemann-Stieltjes Integration
7.1 The R-S Integral
Definition: Let a, b R with a b. A partition of [a, b] is a finite ordered set P ={x0, . . . , xn}such that a= x0 x1 xn =b. Write [a, b] ={P |Pis a partition of [a, b]}. For brevitywell write = [a, b].
Universal Assumptions: Throughout7 we will always assume that:
1. f : [a, b] R is bounded: x [a, b]. m f(x) M, wherem = inff([a, b]), M=sup f([a, b])
2. : [a, b] R (the integrator or weight function) is non-decreasing (in particular, is also bounded)
Definition: For each P [a, b] we associate to fthe following quantities (P ={x0, . . . , xn}):
1. mi= inf{f(x)| x [xi1, xi]} for i [n]2. Mi= sup{f(x)| x [xi1, xi]}fori [n]
3. i = (xi) (xi1) 0 fori [n]We write U(P,f ,) =
ni=1Mii, L(P,f ,) =
ni=1mii.
Uis the upper Riemann-Stieltjes sum, and L is the lower R-S sum.
Remark: Clearly m((b) (a)) =n
i=1m((xi) (xi1)) =n
i=1mi n
i=1Mii Mn
i=1i =M((b) (a)). HenceP [a, b]. m((b) (a))L(P,f ,)U(P,f ,)M((b) (a)).
Definition of Integral: We define
ba f d = sup{L(P,f ,) | P [a, b]}, and
ba f d = inf{U(P,f ,) | P [a, b]}.
Both are well-defined by the remark.Ifba f d=
ba f d then we say fis R-S integrable with respect to , and writeb
af d=ba f d=
ba f d.
We write R([a, b]; ) ={f : [a, b] R |f is bounded,fis R-S integrable with respect to }.
When(x) =x,baf dx is the Riemann integral and we write R([a, b]).
Heuristics: The function assigns different weights to different points in [a, b]. The
intuition is thatbaf d is a weighted Riemann integral. If is continuous then we
have a geometric interpretation of
baf d: consider the curve in R
2 parameterized by
(x(t), y(t)) = ((t), f(t)); ba
f d= area under this curve.
Lemma: Let f(x) =C (x [a, b]). Then f R([a, b]; ) andba f d= C((b) (a)).
Proof: For any P [a, b] we have mi=Mi = C. Hence U(P,f ,) =L(P,f ,) =n
i=1Ci =
C((b) (a)). Soba f d= sup{(L(P,f ,)}= C((b) (a)) = inf{U(P,f ,}=
ba f d.
Definition: IfP, P [a, b] and every point in P is in P, we say P is a refinement ofP.IfP1, P2 [a, b] we define the common refinement P1#P2[a, b] byP1#P2= P1 P2, orderedappropriately.
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7.2 Integrability Criteria 21-355 Notes
Proposition: If P, P [a, b] and P is a refinement of P, then L(P,f ,) L(P, f ) U(P, f , ) U(P,f ,).
Theorem:ba f d
ba f d.
Proof: Let P1, P2 [a, b]. Then L(P1, f , ) L(P1#P2, f , ) U(P1#P2, f , ) U(P 2, f , )
by last proposition. Hence ba f d= sup{L(P1, f , )| P1 [a, b]} U(P2, f , ). Then ba f dinf{U(P2, f , )| P2 [a, b]}=
ba f d.
7.2 Integrability Criteria
Theorem (Riemann)*: f R([a, b]; ) >0.P [a, b]. U(P,f ,) L(P,f ,)< .
Proof:
(1) = (2):
Let >0. By definition,P1, P2 [a, b].ba f d
2 < L(P1, f , ), and U(P2, f , )
ba f d+ 2 . Let P = P1#P2. Then U(P1, f , ) U(P2, f , ) < ba f d+ 2 . HenceU(P,f ,) L(P,f ,)< .
(2) = (1):
For any partition P [a, b] we knowba f d U(P1, f , ) and L(P,f ,)
ba f d.
We also knowba f d
ba f d. Then (2) implies that for > 0 we have P
[a, b]. U(P1, f , ) L(P1, f , ) < . But then 0 ba f d
ba f d < ( > 0)
and soba f d=
ba f d = f R([a, b]; ).
Lemma: Let P [a, b]. The following are true:
1. IfP is a refinement ofP, then U(P, f , ) L(P, f , ) U(P,f ,) L(P,f ,).2. If si, ti [xi1, xi] (i [n]), then 0
ni=1 |f(si) f(ti)|i U(P,f ,)
L(P,f ,).
3. Iff R([a, b]; ), thenti[xi1, xi] = |n
i=1f(ti)ibaf d| U(P,f ,)
L(P,f ,).
Theorem: Iffis continuous on [a, b], then f R([a, b]; ).
Proof:
Note that EVT impliesf is bounded. Let >0 and choose k >0 sok((b) (a))< .Since fis cont on compact [a, b], f is uniformly continuous. Then >0. x, y [a, b]and |x y|< = |f(x) f(y)|< k.
Choose a partition P [a, b] such that xi xi1< (i [n]). Then by EVT,si, ti [xi1, xi] such that mi = f(si), Mi = f(ti). Then U(P,f ,) L(P,f ,) =n
i=1(Mi mi)i=n
i=1(f(ti) f(si))in
i=1ki= kn
k=1i=k((b) (a)) < . Since > 0 was arbitrary, we deduce Riemanns Theorem thatf R([a, b]; ).
Theorem: Suppose that f is monotone and is continuous on [a, b]; then f R([a, b]; ).
Remarks:
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7.2 Integrability Criteria 21-355 Notes
1. If = x, then all monotone functionsfare inR([a, b]), i.e. all monotone functionsare Riemann integrable.
2. Monotone functions dont have to be continuous, sofwith simple discontinuitiesin R([a, b]; ) when is continuous.
3. Monotone functions can have countably infinite sets of discontinuity; R-S integralscan handle infinite discontinuities.
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