[Problem Books in Mathematics] Problems in Real and Complex Analysis || Real- and Complex-valued Functions

3 Real- and Complex-valued Functions

3.1. Real-valued Functions

3.1. For each n in D and some index m greater than n, Sm > Sn-l. Let n' be the least such m. If n 0,

whereas Sp-l - Sn-l :s; 0, and so Sn' - Sp-l > 0. In other words, each p in (n, n') is distinguished, i.e., the numbers n, n + 1, ... ,n' - 1 constitute a block or a part of a block.

Thus a monotonely increasing enumeration of the elements of D begins

with a block: nl ~f nl, nl + 1, ... , n/ -1. Let n2 be the first distinguished index after nl/-1 and then continue the enumeration: n2, n2+1, ... , n2' -1,

etc. In this way D is completely enumerated and if n# is the last member

of a block to which n belongs and if n# > n, then Sn# > Sn-l· 0

3.2. If D = 0 then D is open and no further discussion is required. If

D ::f. 0, xED, limy=xf(y) ~ Lx, x' > x, and f (x') > Lx, for some N(x), x E N(x), x' f{. N(x) and SUPYEN(x) f(y) < f (x'). Hence N(x) c D, D is open, and uniquely expressible as follows: D = Un (an, bn).

Assume x E (an, bn ) and f(x) > Lbn ' Because bn f{. D, if x" > bn then Lbn ~ f (x") whence Lx ~ f(x) > f (x"). Thus if f (x') > Lx then x' E (x,bn ]. Let c be the supremum of all such x'. Then x < c :s; bn . If c = bn then f(c) ~ Lx as claimed. If c < bn there is a c' such that c' > c

and f(c') > Lc k f(c) ~ Lx) and so c' E (x,bn ], c:S; c', a contradiction. Hence c = bn . 0

192

[ Note s3.1: Figure s3.1, showing the graph of y = f(x) for a continuous f, suggests the origin of the name running water lemma for the result in 3.2. The situations in 3.1 and 3.2 are analogous. Both discoveries are due to F. Riesz who used them to give perspicuous proofs of a) the differentiability a.e. of a monotone function and b) the Birkhoff pointwise ergodic theorem, (cf. 5.160, 5.161).]

B. R. Gelbaum, Problems in Real and Complex Analysis© Springer-Verlag New York, Inc. 1992

Solutions 3.3-3.4 193

y-axis

y = [(x)

o an+ l x-axis

Figure s3.1.

3.3. If E ~f {p : p E lR[x) , deg(p) ::; R + 8 - I} then E is isomorphic to CR+S. For T : E 3 P 1-+ (p(l), ... ,p(R-I)(I),p(2), ... ,p(S-I)(2)) E CR+S,

if T(p) = 0 and p(x) = E~;-I Ak (x - l)k then

Ao = Al = ... = AR - I = 0.

If 1 $ s ::; 8 then

R+S-I 1 whence Ek=R (k _ s + l)!k!Ak = 0, 1 ::; s $ 8, a system of 8 homo-

geneous linear equations in the 8 variables, k!Ak , R ::; k ::; R + 8 - 1. def 8 For N = R + - 1, the matrix of coefficients is, modulo column/row

permutations,

1 1

1 1 N! (N -1)! (N - 8 + I)!

M(N, S) 'Jf ( ; i i

(N - 8 + I)! (N - 8)! (N - 28+2)!

Mathematical induction and the row operations for reducing M(N, 8) to a manageable form show there are positive constants K(N, 8) such that det(M(N, S)) = K(N, S)det(M(N -1, S - 1)). Hence det(M(N, 8)) i= 0, Ak == 0, i.e., T is surjective. 0

3.4. If def{O ifx<bn

8n : 1R 3 x 1-+ = 1 if bn ::; x ' n EN,

f : 1R 3 x I-+~f E~=I an 8n (x), and E > 0, there is an N such that N> n and E:=N +1 am < Eo There is a positive 8 such that if y < bn ::; x and

194 3. Real- and Complex-valued Functions: Solutions

x - y < 8 then, among bb"" bN , only bn lies in (y, x]. Hence

an + E ;::: f(x) - fey) = L am = an + L am;::: an, bmE(Y,x] m#-n

bmE(Y,x)

whence f (bn + 0) - f (bn - 0) = an and so ContU) C S. Furthermore, if xES and E > 0 for some N, L::=N+l an < E. There

is a positive 8 such that

(x - 8, x + 8) n {bn} nEN C {bn} N+1:::;n<oo ,

whence if y E (x-8, x+8) then If(x) - f(y)1 < E. Thus S C ContU)· 0

3.5. Direct calculations reveal the truth of each of the following statements.

i. On each of the intervals (-00, Xl) , (Xb X2) , ... , (Xn-b xn) , (xn' 00), 8 is a linear function.

ii. On JR., 8 is continuous.

iii. If Xk ~ Xl + 15k, 1 ::; k ::; n, then 0 = 81 < 82 < ... < 8n and

8(0) =

! L:~=l Xk - nO

(2p - n): + (n - 2p)Xl - L:1=1 15k + L:~=P+1 15k

nO - L:k=l Xk

iv. On (-OO,Xl], 8 is monotonely decreasing, on [Xn' 00) 8 is monotonely increasing, and on [Xk' Xk+l]

{ monotonely decreasing if 2k - n < 0

8 is constant if 2k - n = 0 . monotonely increasing if 2k - n > 0

Hence if n is odd, 8(0) reaches its minimal value when 0 = X n+l . 2

If n is even, [x1j,x1j+d·

8( 0) reaches its minimal value throughout the interval o

3.6. The Heine-Borel theorem implies there is a finite set {Xdl<k<K in JR. and a set {8k}1:::;k:5K of positive numbers such that: - -

Xl - 81 < p < Xl < X2 - 82 < Xl + 81 < X2 < ... < XK - 15K < XK < XK-l + 8K, < q < XK + 15K;

U (Xk - 15k, Xk + 8x ) :) [p, q];

{Xk-l - 8k- l < ak < Xk-l < bk < Xk-l + 15k- I }

=} {f (ak) ::; f (bk)} ,k = 2,3, ... ,K + 1.

Solutions 3.7-3.8 195

Hence if X2 - 02 < Y2 < Xl + 01, ... ,XK - OK < YK < XK-I + OK-I then f(p) :S f (Y2) :S f (Y3) :S ... :S f (YK-I) :S f(q)· D

3.7. If ¢ is convex,

def b - C def C - a u < a < c 0, a + (3 = 1, and c = aa + (3b. Hence

¢(b) - ¢(a) = ¢(b) - a¢(a) - (3¢(b) < ..:.....::¢(--:-b)_-...:....;¢(,-,-c) b - a b - aa - (3b - b - c

and, by a similar calculation, ¢(c) - ¢(a) :S ¢(b~ - ¢(a). The inequalities c-a -a

u < X :S x' < Y :S y' < V can be applied first when x = a, x' = c, and Y = b and then when x' = a, Y = c, and y' = b. The argument is reversible.

The sketch in Figure s3.2 shows the geometry of the situation. D

t-axis

o x x' y

Figure 83.2.

s-axis

3.8. If ¢ is neither monotonely decreasing nor monotonely decreasing, there are numbers a, b, c such that a max{¢(a) , ¢(c)} or b) ¢(b) < min{¢(a) < ¢(c)}. The convexity of ¢ rules out a) and implies for b) that if a" < a' < a then ¢ (a') ~ ¢(a) [whence ¢(a") ~ ¢(a' )] andifb" > b' > bthen¢(b') ~ ¢(b) [whence¢(b") ~ ¢(b')].

Hence ¢ is monotonely decreasing on (-00, a) and ¢ is monotonely increasing on (b,oo).


Ifp~sup{a: ¢! on (-00, a)} , q~finf{b: ¢ion(b,oo)}then p :::; q and [p, q] is the required interval.

Examples: a) ¢(x) = e"; b) ¢(x) = x+ Ixl; c) ¢(x) == 0; d) ¢(x) = e-"; e)

{(x + a)2 if -00 < x :::; -a < 0

¢(x)= 0 if-a<x:::;a (x-a)2 if a < X,oo

f) ¢(x) = Ixl - x.

3.9. If u < c :::; x < y :::; d < v then

¢(e) - ¢(a) < ¢(y) - ¢(x) < ¢(b) - ¢(d) e-a - y-x - b-d

and so ¢ E Lip(l). The criterional inequality of 3.7 shows that the difference quotient

¢(x + h) - ¢(x) h

o

is a monotonely increasing function of h. It follows that the right- and left-hand derivatives D±¢ of ¢ exist everywhere.

In particular, at each point a, D_¢(a) :::; D+¢(a). [Any line through (a, ¢(a)) and with a slope m satisfying D_¢(a) :::; m :::; D+¢(a) is a supporting line.]

From 3.9 and the differentiability a.e. of a monotone function it follows that ¢ is differentiable a.e. 0

[ Note 83.2: The fact that (u, v) is open is essential to the continuity and differentiability properties of ¢. For example, if (u,v) is replaced by (0,1] and

¢(x)={~ ifO<x<l if x = 1

then ¢ is convex but is discontinuous when x = 1. Similar observations apply when (u, v) is replaced by any nonopen interval.]

3.10. The functions 'if; : x f--+ x 2 and ¢ : x f--+ e-" are convex, but ¢ 0 'if; : x f--+ e-,,2 is not convex. 0

3.11. a) If ¢" exists, ¢" > 0, and ¢ is not convex then there is a triple r-q q-p

p, q, r such that u --¢(p) + --¢(r). The r-p r-p

map f : x f--+ ¢(x) - r - x ¢(p) + x - p ¢(r) is continuous and positive at r-p r-p

q, whence f has a positive maximum on [p, r]. Because f(p) = fer) = 0,

Solution 3.12 197

this maximum occurs at some S in (p, r) and thus 0 < ¢" (s) = f" (s) ::; 0, a contradiction.

b) If p < rand ¢'(p) > ¢'(r), the convexity of ¢ implies

¢(r) > ¢(p) + ¢'(p)(r - p),

whence

¢(r) + (p - r)¢'(r) > ¢(r) + (p - r)¢'(p)

> ¢(p) + ¢'(p)(r - p) + ¢'(p)(p - r) = ¢(p).

In geometric terms, the inequalities above say that the point (p, ¢(P)) does not lie above the supporting line through (r, ¢(r)), a contradiction. Hence

{p ::; r} =* {¢' (P) ::; ¢' (r)} ,

i.e., ¢' is monotonely increasing and so ¢" ::::: o. An alternative proof uses a truncated Taylor series. If x E (s, t) then

for suitable ~ and 1],

(s x)2 ¢(s) = ¢(x) + ¢'(x)(s - x) + ¢"(~) ~! '

¢(t) = ¢(x) + ¢'(x)(t - x) + ¢"(1]) (t _,X)2. 2.

For some A in (0,1), x = AS + (1 - A)t and, owing to the nonnegativity of ¢", for some nonnegative P,

A¢(S) + (1 - A)¢(t) = ¢(x) + ¢'(x) (AS - AX + (1- A)t - (1- A)X) + P

::::: ¢(x) + 0 = ¢(AS + (1 - A)t).

Note that if ¢(x) = x then ¢ is convex and yet ¢"(X) == 0: the implication {¢" exists and ¢ is convex} =* {¢" > O} is invalid. 0

3.12. a) Ifln(¢) is convex, x, y E JR,O ::; a, (3, a + (3 = 1 then

In(¢)(ax + (3y) ::; a In(¢)(x) + (3ln(¢)(y)

¢(ax + (3y) ::; (¢(x))a. (¢(y))~.

However for u, v positive, uav~ ::; au + (3v. (PROOF. Because

(ta )' It=c = {~ca-l ::; a aca - 1 ::::: a

(at + (3)' == a,

talt=l= at + (3lt=l= 1,

if c = 1 if c > 1 ifO<c<l

198 3. Real- and Complex-valued Functions:

it follows that for all positive t, ta ~ at + (3. Hence if t = !!:. then v

Ua

V ~ (a!!:. + (3) v, va v

ua v/3 ~ au + (3v.]

By induction it follows that if

K

ak, Uk 2 0, 1 ~ k ~ K, and L ak = 1 k=l

then, ambiguous cases excluded,

K K

II U~k ~ Lakuk, k=l k=l

as required.

Solutions

b) Graphical considerations show that the map ¢ x 1--+ 1 + Ixl is

convex. However, for 'l/J ~ In¢,

'l/J'(x) = {~~: 1-x

if x> 0

if x < 0

whence

{x =I O} '* {'l/J"(x) = - (1 ~ x)2 < o} . Hence 3.11 implies'l/J is not convex. D

3.13. i. True: There is a sequence {xn}nEN such that Xn < Xn2+l and

g(x) 2 n if x ~ xn. If ¢(xn) = n - 1,n E N, and if ¢ is piecewise . linear , continuous, and nonnegative then ¢ is also convex and monotonely decreasing, ¢ ~ g, and ¢(x) --t 00 as x --t o.

ii. False: For 9 : [0,00) '3 x 1--+ In(1 + x), if the convex function ¢ is such that ¢ ~ 9 then (cf. Figure s3.3), since -g is convex, 9 is concave and

en In (1 + e-n ) As n --t 00, --t 00 and 1 + --t 1 whence ¢(x) -f 00 as

n n x --t 00. D

Solution 3.14 199

y-axis

o

y = In(1 + xl

x-axis

Figure s3.3. The graph of y = In(1 + x)

[ Note s3.3: Associated with the notion of convexity is that of midpoint convexity, viz.:

A function f in JR(a,b) is midpoint convex iff for all x, y

in (a,b), f (~+~) ~ ~f(X) + ~f(Y). Hence if a, f3 E {t : t = k2-n , k, n EN}, and a + f3 = 1,

then f(ax + f3y) ~ af(x) + f3f(y). If, to boot, f is continuous, then f is convex. The existence of a discontinuous linear (hence convex) function ¢ in JRIR (cf. [GeO] and, for generalizations, 6.115) shows that the converse is false.]

3.14. For the maps

k: [0,00) :3 s I-t foB b(r) dr,

g: [0,00) :3 s I-t k(s) + c(s),

T: C([O,oo),JR):3 f I-t (F: [0,00):3 s I-t foB a(r)f(r)dr) ,


the following obtain:

d (T~:(S)) _ a(s)T(y)(s) ::; a(s)g(s);

d (e- I: a(r)drT(y)(s)) t

ds ::; ef. a(r)dr a(s)g(s);

T(y)(t) ::; lot a(s)g(s)eI: a(r)dr ds;

lot a(s) [yes) - g(s)eI: a(r) dr] ds::; 0;

from which the result follows. Owing to the importance of fixed-point theory, the alternative solution

below is of interest. If Mo is an endomorphism of a Banach space X and if, for all x in X,

:E~=1 M(j(x) converges then for each y in X, the map

My : X 3 x 1--+ Y + Mo(x)

has a unique fixed point, namely P ~f Y + :E~=1 M(j(x). If, to boot, X is a function space and Mo preserves positivity then

{x::; y + Mo(x)} ~ {x::; P}.

More particularly, if X = C ([0, al, R), 0 ::; t ::; a, and Mo = T, the results just cited are applicable: for y read g, for x read y, and for P read

o

3.15. The solution involves two steps: a) showing that for a, bin R, the differential equation problem:

y" + (1 + q)y = O,y(O) = a, y'(O) = b

has a unique solution; b) the solution is bounded. a) Because q E L1 ([0,00)), the map

T: C ([0, Rl, R) 3 f 1--+ ([0,00) 3 t 1--+ lot sin(t - s)q(s)f(s) dS)

(s3.1)

is one for which E:'l IITnlloo converges if 0 ::; t ::; R. Hence (cf. 3.14) if w E C([O,oo),R), the map Mw : C([O,oo),R) 3 f 1--+ W - T(f) has a unique fixed point v. If w" is continuous and w(O) = a, w'(O) = b then

Solution 3.16 201

v(O) = a,v'(O) = b and direct calculation shows v" + (1 + q)v = 0: (s3.1) has a solution.

Since the differential equation is homogeneous, the question of uniqueness of solution for the problem is resolved by examining the case in which a = b = O. The original equation is equivalent to the system

(s3.2)

Any solutions Yl, Y2 of the system must satisfy

Because

the second proof in Solution 3.14 applied to (s3.2) shows Yl = 0 (whence Y2 = 0): the problem has at most one solution.

b) If w" + w = 0, w(O) = a,and w'(O) = b then

Iv(t)1 :::; Ilwll oo + lot Iq(s)llv(s)1 ds;

3.14 applied again yields

whence v is bounded. o 3.16. a) Since I is continuous, E-I(f,O) is the union of a null set Nand an open subset U of (0, 1). Furthermore,

and U± are uniquely of the form lJnEN (an ±, bn ±). On U±, III = ±I and III", as the limit of second order difference quotients for ±I, is measurable. Because I is in C2 ([0, 1], lR),

{ 0 if x E [0,1] \ U h : [0,1] :3 x ,........ III" h ot erwise

is in LR'([O, 1],,\).


b) Since

r1 00 l bn± io g(x)h(x)dx= ~ g(x)h(x)dx,

o n=1 a n ±

it suffices to consider a single term of the sum. Then, (a, b) denoting any one of (an ±, bn ±), integration by parts twice yields

lb g(x)h(x) dx = g(x)III'(x)l~ -g"(x)ll(x)ll~ + lb g"(x)ll(x)1 dx.

Since 9 (b) III' (b) ::; ° and 9 (a) III' (a) ~ 0, the first summand in the right member above is nonpositive. Because III (a) = III (b) = 0, the second summand is zero. Hence

lb g(x)h(x) dx ::; lb g"(x)ll(x)1 dx. o

3.17. i. The map

I: (0,1) 3 x ~ ~ sin (~) is continuous and

is a Cauchy sequence, whereas its I-image is not. ii. If I is not continuous then for some Xo in (0,1) and some sequence

{xn}nEN'

lim Xn = Xo and I (xn ) -1+ I (xo) n--->oo

as n -+ 00. If Y2n = xn , Y2n-1 = xo, n E N,

then {Yn} nEN is a Cauchy sequence and its I-image is not, a contradiction. o

3.18. The hypothesis implies that if Xn -+ ° as n -+ 00 then {f (xn )} nEN

is a Cauchy sequence. Thus the argument in Soluition 3.17ii implies limx--->o I(x) exists. 0

3.19. The map I : JR. 3 r ~ A {x : sinx > r, x E [0,471']} is monotonely decreasing, 1([-1,1]) = I (JR.) = [0,471'], and on [-1,1], I is injective. Thus

9 ~f 1-1 1 [0,41fJ is monotonely decreasing and g(x) > r iff x < I(r). 0

3.20. a) If Ik is the function for which the graph of Y = !hex) is that in Figure 83.4 then limk--->oo Ik(X) == 1 and I~ (~) = k, kEN.

Solution 3.21 203

y-axis

-l-----l-------<j __ --*-----< __ --::----- X - axis o (2,0) (1,0)

Figure s3.4.

b) Because each fk is monotone, each f~ exists a.e.,

E ~ {x : f£{x) exists and is finite, kEN}

is measurable, and )..(E) = 1. The Fundamental Theorem of Calculus for Lebesgue integration implies fk(l)- fk(O) ~ IE fHx) dx, kEN, and Fatou's lemma (cf. 4.3) permits the conclusion:

However, f~(x) ~ 0 if x E E. D

3.21. Since AIR is countable and contains Q, AIR is a dense Fu' If, for some f, AIR = ContU) it is also a (dense) Go (cf. 2.97) and therefore a set of the second category, a contradiction.

As an Fu, AIR is the union of count ably many closed sets Fn such that

Fn c Fn+1. If Bn ~ (Fn \ Fn- 1) \ (Fn \ Fn-1t and

{ 2-n

f(x) = 0 if x E Bn if x (j UnEN Bn


then Discont(f) = AIR whence Cont(f) = R \ AIR [GeO]. 0

3.22. Because f~(x) - f~(y) = J: f::(t)dt, the sequence {f~}nEN is equicontinuous. If M' as described does not exist then, via passage to

subsequences as needed, it may be assumed that an ~f IIfnlloo 10 and that

M~ ~ Ilf~lIoo i 00. Hence for some positive E and for each n in N, there is

an Xn such that {Ix - xnl < E} :::} {lf~(x)1 > ~M~}. For some Cn,

fn(x) = 1x f~(t) dt + en, n E N, Xn

and since fn (xn) = Cn, it follows that len I :::; an, n E N. If Ix - xnl < E 1 1

then Ifn(x)1 2:: 2M~E - an· Hence Ilfnlloo 2:: 4M~E - an -+ 00 as n -+ 00, a

contradiction. 0

3.23. If U is open in R, f-l(U) is the union of two disjoint sets:

Cu ~ Cont(f) n f-l(U) and Du ~f Discont(f) n f-l(U).

By hypothesis, oX (Du) = O. If x E Cu, for some positive E(X), {y : If(x) - yl < E(X)} C U and

there is a positive b(x) such that

f({x': Ix'-xl<b(x)})c{y: If(x)-yl<E(X)}CU.

Hence f-l(U) = (UXECu {x' : lx' - xl < b(x)}) UDu, a union of an open (hence measurable) set and a null set. 0

3.24. If f'(x) = X(-oo,Oj 0 f(x) is valid on an open set U containing {O} then

1'( ) = {O if f(x) > 0 x 1 if f(x) :::; 0 '

(s3.3)

and, by hypothesis, f(O) = O.

i. If E > 0 and f(x) = 0 on [0, E) then f'(x) = 0 on (0, E) while (s3.3) requires that f' (x) = l.

ii. If f(x) > 0 throughout (0, E) then f'(x) = 0 throughout (0, E) whence for some positive constant p, f(x) = p throughout (0, E) and hence f is not continuous at zero.

iii. If f(x) < 0 throughout (O,E) then f'(x) = 1 throughout (O,E) and for some constant k, f(x) = x+k throughout (0, E). Because f(O) = 0, k = 0; the hypothesis implies f' (x) = 0 throughout (0, E), a contradiction.

Hence if E > 0 then f:

iv. cannot be ze .. o throughout [0, E);

Solution 3.25 205

v. can be neither only positive nor only negative throughout (0, e).

Thus, owing to (s3.3), if e > 0, f' assumes both values zero and one on (0, e) and no others. As a derivative, f' enjoys the intermediate value property and a contradiction emerges. 0

3.25. If, for some x,y, x < y and f(x) > fey) then for some positive p

and all e in (O,p), fey) < f(x) - e(y - x).

u ·axis

(y, j(x) - E(y - x»

(y,j(y»

L-... ___ .....I.-____ ....L __ .....I.-_--L. _____ x-axis o x x~ y

Figure s3.5. The situation: x < y, f(x) > fey), and Xoo < y.

As indicated in Figure s3.5, by hypothesis, for some Xl in (x, y),


If y > Xoo ~f sup {Xl : X < Xl < y,J (Xl) ~ f(x) - f (Xl - X)}

then f (xoo) ~ f(x) - f (xoo - x) and, by hypothesis, for some z in (xoo, y),

fez) ~ f (xoo) - f (z - xoo ) ~ f(x) - f (xoo - x) - f (z - xoo) = f(x) - feZ - x),

a contradiction. Thus

Xoo = y and fey) ~ f(x) - fey - x).

Since the last inequality obtains for all f in (O,p), fey) ~ f(x), another contradiction. 0

3.26. For the Cantor function <Po, let h be <Po.

i. On the closure of each interval Jln , n E N, deleted in the construction of the Cantor set Co, define the homolog hn of h.

ii. On the closure of each interval J2n , n E N, deleted in the construction of the sequence {hn}nEN' define the homolog!2n of h·

iii. By induction, for each k in N, there can be defined a sequence {fkn} nEN such that: a) the domain of fkn is the closure of one of the intervals deleted in the construction of {fk-l,n} nEN; b) each /kn on its domain is the homolog of h.

The sequence {hn}nEN' {!2n}nEN' ... can be rearranged so that there emerges a single sequence {hn}nEN' each member of which, on its original domain, is homologous to h on [0,1]. Furthermore, each hn may be extended to be constantly zero to the left of its domain and constantly one to

the right of its domain. Then h ~f 2::=1 ~: is continuous and monotonely

increasing while 1 - h ~ f is monotonely decreasing. Owing to Fubini's

theorem on derivatives of sequences of functions, h' ~ 2::=1 ~~ and, since

h~ ~ 0, n E N, f' ~ 0. In sum, the function f meets all the requirements: f is a monotonely decreasing function, hence monotonely increasing on no interval and yet f' ~ ° ~ o. 0

3.27. For x, y in (0,1), the formula

d: (0,1)23 {x,y} 1-+ max {If(x) - f(y)l, Ix - yl}

Solutions 3.28-3.30 207

defines a metric and furthermore, d(x, y) 2:: Ix - yl. Because I is continuous, if limn -+oo IXn - xl = 0 then limn -+oo II (xn) - l(x)1 = 0, whence limn -+oo d (Xn' x) = 0: d and I I endow (0, 1) with the same topology.

Since {d(x,y) < f} ::::} {1/(x) - l(y)1 < f}, I is uniformly continuous on ((0,1),d). 0

3.28. If n = 1, I may be regarded as the restriction to (-r, r) of a function F holomorphic in D(O,r)o. If I ~ 0 then F- 1 (0) is countable and thus A (1-1(0)) = O.

Assume the result is valid when n = 1,2, ... , N - 1. When n = N, let A be 1-1(0) n B(O, r)o. Fubini's theorem for product measures implies

that if A"N ~ {(Xl'"'' XN-l) : (XI,"" XN-l, XN) E A} then

For fixed XN in (-r,r), AXN in JRN - 1 is the possibly empty subset on which I, regarded as a function on JRN - l when Xn is fixed, is zero. By inductive assumption, AN -1 (AXN) = 0 whence AN (A) = o. 0

[ Note 83.4: The last result implies, when Matnn is regarded 2

as a subset of JRn , Nondiagnn is the set of nondiagonable n x n matrices, and Singnn is the set of singular n x n matrices, that

Thus "almost all" n x n matrices are invertible and diagonable. In the same vein, every algebraic or analytic variety in JRn is

a null set.]

3.29. For 1,9 in V, Mf V Mg resp. kf V kg serves for Mfvg resp. kfvg and Mf 1\ Mg resp. kf 1\ kg serves for Mf!\g resp. kf!\g. The vector space character of V flows from the definitions.

The Daniell construction in Chapter 1 shows that there is a measure v such that P(f) = JlRn I(x) dv(x) whence v = fL· Each nonnegative I in V is the limit of a monotonely increasing sequence in Co (JRn , JR) and the monotone convergence theorem implies the validity of (3.1) for f. For an arbitrary I in V, 1= 1+ + 1-· 0

3.30. For to fixed in JR, the map 9 : ~ 3 x 1--+ (f (x, to))2 + (O/~' to)) 2

is in C (~, JR) and 9 > O. Because ~ is compact, m ~f minxEI; 9 (x, to) > 0 and for some Xto in ~, 9 (Xto' to) = m.

If, arbitrarily close to to, there is a tk and in ~ a corresponding Xtk

such that 9 (Xtk' tk) = 0 then it may be assumed that for some Xoo in ~,


limk-+oo IIXtk -xooll = 0 and thus g(xoo,to) = 0, a contradiction. Hence, for some open set N (to), if tEN (to) then minxEE 9 (x, t) > O.

If, for each k in N, there are in N (to) two points, tkl, tk2 such that

Itkl - tol+l tk2 - tol < ~ and for some Xk in~, 1 (Xk, tki) = 0, i = 1,2, then

Rolle's theorem implies for some tk3 between tkl and tk2, of (~t' tk3) = o. It may be assumed that for some x in ~, limk-+oo IIXk - xii = O. Hence 01 (x, to) .. _ at = O. SInce mInxEE 9 (x, t) > 0, 1 (x, to) =I- O. Thus for large k,

1 (Xk, tkl) . 1 (Xk, tk2) =I- 0, a contradiction. It follows that for h, t2 close enough to to and different from each other (one may be to), and any x in ~, (f (x, tl))2 + (f (x, t2))2 > O. 0

3.31. For ~ as in 3.30, let K be ~ n 1-1(0). Then K is compact and g(x) > 0 on K. Hence for some positive € and some open U containing K, g(x) ;::: € on U. For some positive 8, I(x) ;::: 8 on the compact set ~ \ U. Because U may be chosen so that 0 fJ: U and since if x =I- 0,

it follows that

{x} { I(x) } { x } { g(x) } IIxll E ~ \ U * IIxll m ;::: 8 and IIxll E U * Ilxll m ;::: € •

The required inequality obtains if C = ~ and D = ~. o

3.32. For a ~f (al,.'" an) in Bl (JRn) let f(a) be (It (a), ... , In (a)). If

b ~f (bl , ... ,bn ) E Bl (JRn ) then for some Cij between aj and bj ,

There is a positive {) such that

{II;~; 1100 < 8,i,j = 1, ... ,n} * {lIf(a) - f(b) II < ~lla - b U}.

Solutions 3.33-3.34 209

Because d(id) = id, it follows that if g = f - id then d(g) = d(f) - id. There is a positive"., such that

{s~p IId(f) (x) - idll <".,} => {II~~; -8iilloo < 8,i,j = 1, ... ,n}, 1

whence Ila- bll-Ilf(a) -f(b)11 ~ IIg(a) -g(b)1I < "2l1a- bll. Consequently,

IIf( a) - f(b) II ~ ~ Iia - bll and thus f is injective. D

3.33. For x in lRn, d(f)(x) E [lRn, lR] and, since [lRn, lR] and lRn are naturally isomorphic as vector spaces, if f E C2 (lRn , lR) then d 2 (f) is in [lRn , [lRn ll. [For each x, d 2 (f) (x) may be represented by the Hessian matrix

( ::~~:~ ) :i=l'] By definition, there is a map f; : [-1,1] 3 t 1--+ (0,00) such that

limt_O f;(t) = 0 and for some positive 8,

{Ilull < 8} => {lldf (xo + tu) - df (xo) - d 2 f (xo) (tu) II = f;(t)ltl·llull}·

Since df (xo) = 0, Ildf (xo + tu) - d 2 f (xo) tull = f;(t)ltl·llull· If df (xo + tu) = 0 for some nonzero t in [-1,1] and some nonzero

u, then, since d 2 f (xO)-l exists, IId2f (xo)-lll ~f K > O. If Y ~f d2f (xo)

then tu = d 2 f (XO)-l (y) ~ Kllyll, whence y i= 0 and lIyll ~ f;(t)KIlYII, i.e., 1 ~ f;(t)K.

1 1 8 If IIxn - xoll < -, df (xn ) = 0, n E N, and - < -2 then when n > no,

n no

there is a Un such that II Un II = ~ and a tn such that Xn = Xo + tn Un and

Itnl < .!.. But then 1 ~ f; (tn) K, a contradiction, since limn_oo f; (tn) = O. n

Hence for some open set N (xo), df(y) i= 0 for all y in N (xo) \ {xo}. D

3.34. If W = var[O,l] (f) - ~, there is a partition Po determined by n + 2

partition points for which Var[O,l],Po(f) > W -~. There is a positive 8 such

that {Ix - yl < 8} => {If(x) - f(y)1 < 8:}' If IFI < min{8, IPol} ~f a

and PI is the common refinement of Po and P, then f; f;

Var[O,l],P(J) + 2n 8n ~ Var[O,l],P! (J) ~ Var[O,I],Po (f) > W - "2

3f; Var[O,l],P(f) > W - "4 = Var[O,l] (f) - f;.

When f = XU} then Var[O,I](J) = 2. However, for any partition P

determined by points in [0,1] \ {~}, Var[O,l],P(J) = O. D


3.35. There are functions Ii, 1 :s; i :s; 4, each monotonely increasing, nonnegative, and bounded for which I = It - h + i (13 - 14). For each Ii, fro,l] II(x) dx :s; li(l) - li(O) < 00, whence!, E Ll ([0, 1), .\). 0

3.36. Only the situation for left-continuity is discussed. The other situations are treated, mutatis mutandis.

If I is of bounded variation and left-continuous at a, if E > 0, and if

def o = Xo < Xl < ... < Xn < Xn+1 < a n

L II (Xk+1) - I (xk)1 + I/(a) - I (xn+l)1 > var[O,a](I) - E k=O

then var[O,Xn+l](I) > var[O,a](I) - E - I/(a) - I (xn+dl· As Xn+l i a, limxn+l ja var[O,Xn +l] (I) ~ var[O,a] (I)-E. Because var[O,x] (I) increases monotonely, var[O,a] (I) ~ limxn+1 ja var[O,Xn +l] (I) ~ var[O,a] (I) - E.

Conversely, if var[O,x] (I) is left-continuous at a, then

and as Xn+1 i a the right member above descends to zero. o 3.37. If, for some a, I/(a + 0) - I(a - 0)1 = 6. > 0 then for all small

positive 8, II(a + 8) - I(a - 8)1 > ~ a~ so on [a - 8, a + 8], I assum: no

values between (I(a - 0) /\ I(a + 0)) + 3 and (I(a - 0) V I(a + 0)) - 3' a

contradiction. 0 [ Note s3.5: Hence if a function I is of bounded variation and is a derivative then I is continuous.)

3.38. There is a sequence {/mn}m,nEN of piecewise linear, continuous, even functions such that

{I if Ix - al < ~ or Ix + al < ~

I mn (x) = (m 2 2 ) m ( 2 2 ) o if x ~ -a - m' -a + m U a - m' a + m

Imn 1 X[-a-.l. -a+.l.] + X[a-.l. a+.l.] as n -+ 00. tn' 1'n Tn' 'Tn

Then 0 = limn-->oo J~l Imn(x)g(x) dx = J~:~r + J:~r g(x) dx. Because g is continuous at a and -a, if E > 0 and m is large, '"

o

Solutions 3.39-3.43 211

3.39. The assertion holds if I(x) = X{l}(x). o 3.40. Direct calculation shows that on [0,1] if a ::; 1 then la is not of bounded variation and that if a> 1 then la is absolutely continuous. 0

3.41. If

then I has the required properties.

ifO::;x<1

if x = 1

o 3.42. If 0< a < 1, Co< is a Cantor subset of [0,1], and >. (Co<) = a, then

I(x) ci;1 1x (1- XC.,(t)) dt

is absolutely continuous.

If [a, b] c [0, 1] then (a, b) is either a subinterval of some interval deleted in the construction of Co or some such interval is a subinterval of (a, b). It follows that

I(b) - I(a) ~ b - a.

Furthermore f' ~ 1 - Xc., whence I' ~ ° on Co. o 3.43. The map () in Solution 2.57ii resp. (}-l ~f ~ is a continuous strictly monotonely increasing function on Co resp. S. The intervals deleted from [0,1] in the construction of Co are in bijective order-preserving correspondence with intervals deleted from [0,1] to yield S.

An analog CPs of CPo should be monotonely increasing, continuous, and constant on each interval deleted in the construction of S. Owing to the denseness of [0,1] \ S, the values of CPs on the deleted intervals then determine the values of CPs throughout [0,1]. Thus there is some arbitrariness about the definition of CPs.

A straightforward analogy with CPo is

CPs : S 3 x = f: f n (x)(1 - r)rn - 1 t---+ f: fn2~) n=l n=l

which is a continuous and monotonely increasing map of S onto [0,1]. Consequently, at the ends of each interval deleted in the construction of S, the values of CPs are the same and CPs may be extended in one and only one way to a continuous and monotonely increasing map of [0,1] onto itself. Figure s3.6 depicts the graph of the second approximant to CPs.


y-axis

-O-f"---r-('--I -_ -'-2r-)--"-1 --2r"'---r-'-(1---2'-r)------ x-axis

Figure s3.6.

a) The complement in the unit square of each approximant to 98 con

sists of two congruent figures. Hence frO,1] 4>8 (X) dx = ~. b) The length of 98 may be approximated by the lengths of polygons

that represent the successive approximants to 98. Induction shows that, f3n denoting the length of each interval in the nth group of deleted intervals, at the Nth stage of approximation, the length of the polygonal approximant

is 2::=12n-1f3n + VI + (1- 2::=12n-1f3n /. Hence the length of 98 is 2

(I). 1

c) Symmetry arguments show J[0,1]4>a{X) dx = '2. On the other hand,

the discussion in b) now leads to 1- a + VI + a 2 as the length of r a. The length of r a is less than 2 if 0 < a < 1. 0

[ Note s3.6: When 0 < r < ~, a more interesting analogy with

4>0 (for which r = ~) follows.

Dual to an ~ {1- r)rn-l, n E N, is bn ~ r{l- r)n-1, n E N. Furthermore, 2::'=1 bn = 1 and bn ::; 2:~n+1 bk, whence the set T of all subsums of 2:::1 bn is [0,1].

Solutions 3.44-3.45

Thus the formula

00 00

<Ps : S 3 x ~f L En(x)(l - r)rn- 1 ~ L En(x)r(l - rt- 1

n=1 n=1

defines a continuous monotonely increasing map of S onto [0,1]. Hence the values of <P s at the endpoints of each interval deleted in the construction of S are the same and <P s may be extended in one and only one way to a continuous and monotonely increasing function mapping [0,1] onto itself.

The reader might wish to calculate f[o,IJ <ps(x) dx and the length of the graph of y = <ps(x).]

213

3.44. If p == 0, E=(p) = JR, E>(p) = E<Cp) = 0. The definitions imply

that if p(x) ~f aoxn + ... + an and ao =f. 0 then

Thus it suffices to discuss only E> (p) and to assume that ao > o. When

( -a1 ) n=l,E>(p)= ~,oo. Ifn=Nand,whenn=1, ... ,N-1,for

some a in JR, E>(p) = (a, 00), then for some {3 in JR, E> (p') = ((3,00) and E>,o(p) is a union of a finite number of pairwise disjoint open intervals.

If E>,o(p) = JR then for some {3 in JR, E> (p) = ({3, 00). If rk < Sk < rk+1, E>,o(p) = Uf=1 (rk, Sk), and {3 ~ SK-l, then for

some 'Y in (SK-b 00), p'("() = 0 in contradiction of the definition of {3.

Hence for a ~f max{{3,rK}, E>(p) = (a, 00) (and SK = 00). 0

3.45. If J is Riemann integrable, it must be bounded since otherwise the set of approximating Riemann sums is unbounded in JR.

Since

Discont(f) = U { x : limy=xJ(y) -limy=xJ(y) ~ ~ } ~ U E::" nEN nEN

if >. (Discont(f)) = {j > 0 then for some no, >. (E n1J ~f E > O. For each partition used to calculate the upper and lower Riemann sums, the sum of


the lengths of the partition intervals containing E..L is at least € and thus "0

the difference between the upper and lower Riemann sums is at least ~, whence I is not Riemann integrable, a contradiction.

no

Conversely, if I is bounded and >. (Discont(f)) = 0 then for positive €, E. is compact and >. (E.) = O. Hence E. can be covered by finitely many pairwise disjoint open intervals of small length-sum. For a partition consisting of such intervals and others (on each of which I is uniformly continuous), the difference between the upper and lower Riemann sums is small if the mesh of the partition is small. 0

3.46. For Qn [O,lJ ~f {rdkEN,ln ~ L;=lXrk,limn-oo/n = XlQln[O,lj,

each In is Riemann integrable while Discont (XlQln[O,lj) = [0, 1J and XlQln[O,lj

is not Riemann integrable (cf. 3.45). 0

3.47. Let <Po be the Cantor function and let I be <Po . X[O,lj\co' Because I is bounded and >. (Discont(f)) = 0, I is Riemann integrable.

For O! positive, there is a homeomorphism H : Co. ho~eo Co [GeOJ and the Tietze extension theorem implies that H has a continuous extension g: [0, 1J I-----t [O,lJ. If h ~ log then >.(Discont(h)) = >. (Co.) > 0 and h is not Riemann integrable. 0

3.48. If I is strictly increasing, all its difference quotients are nonnegative and thus f' 2: 0 everywhere. If D contains a connected set containing two points, then f'(x) = 0 on a nonempty open interval (a, b), whence I is constant on (a, b), a contradiction.

Conversely if f' 2: 0 everywhere, Rolle's theorem implies I is monotonely increasing. If a < band I(a) = I(b) then I'(x) = 0 on (a,b), an impossibility if D is totally disconnected. 0

3.49. If m is a strict local maximum value of I, if Am ~f 8 n 1-1(m), and if # (Am) > No then by virtue of 2.27, for some a in Am, and every N(a), # (N(a) n Am) > No, whence arbitrarily close to a there are points where the value of I is m, i.e., a is not the site of a strict local maximum: # (Am) ::; No·

Let M be the set of strict local maxima of f. If #(M) > No then for some m in M, and all pin (0,00), # ((m,m + p) n M) > No whence

# (1-1 ((m,m + p) n M)) > No. Thus in E ~f 1-1 ((m, m + p) n M) there is a b such that for every N(b), # (N(b) n E) > No. Because # (Am) ::; No for some c in N(b)nE, I(c) i:- m and so I(c) is in (m, m+p) whence m ~ 8: #(M) ::; No and #(8) = LmEM # (Am) ::; No· 0

3.50. If a < b, there is a sequence {[an' bn)} nEN such that

an ~ PnTq" , bn ~ (Pn + 1) Tq", Pn, qn EN, (a, b) = U [an' bn). nEN

Solutions 3.51-3.52 215

For some no, a* ~f infn~no an and b* ~f sUPn~no bn are such that

If(a) - f (a*)1 + If(b) - f (b*)1 < b - a no

If(b) - f(a)1 ::; If(b) - f (b*)1 + L If (bn) - f (an)1 + If (a*) - f(a)1 n=l

no

< b - a + LTqn . M < (M + 1)(b - a). n=l

Hence f E Lip(1) and so f E AC whence for all x,

The Fundamental Theorem of Calculus for Lebesgue integration implies that f'(x) ~ O. Hence for each n in N and each kin Z, f(O) = f (k2-n). Since {k2-nhEZ is dense in lR and f is continuous, f(x) == f(O). 0

nEN

3.51. If no such a exists then, for all x, f(x) - g(x) =f:. O. Since f and 9

are continuous, f - 9 is of one sign, say f - 9 > 0 and on [0,1], for some positive 8, f(x) > g(x)+8. Thus fof > gof+8 = fog+8 > gog+28. By induction, it follows that {n E N} => {fin} > gin} + n8}, an impossibility ifn8> 1.

For f: [0,00) 3t---+ In(1 +x) and g: [0,00) 3 x t---+ eHx , it follows that f 0 g(x) = 1 + x = go f(x) and yet, for all nonnegative x, In(1 + x) < eHx .

o 3.52. Because f is continuous, if f, 181 > 0 th~p for some h6 in Q \ {O}, Ih6 - 81 ::; 82 and If(c + 8) - f (c + h6)1 < f181. Hence

1 fCc + 8~ - f(c) _ LI ::; 1 fCc + 8) -/ (c + h6) 1

+ 1 f (c +~] - f(c) . h; - LI.

For small 181, the second term in the right member above is small since ~ is near one and the first member is small by virtue of the choice of h6.

. f(c+h)-f(c) If f = XiQi and c = 0 then hmhEiQi\O,h->O h = 0 although,

since f is nowhere continuous, f'(O) does not exist. 0


3.53. If supp(h) c [a, bJ then for F(x) ct;bf f: f(t) dt, integration by parts

implies fJR f(t)h(t) dt = - f: F(t)h'(t) dt = - f: g(t)h'(t) dt whence F = g. Because F' = f, g' exists and is f. 0

3.54. If limn -+oo f (xn) = y then limn -+oo Xn ct;bf x exists and f(x) = y. Hence f (ll~n) is closed.

The inequality (3.2) implies that f is injective and hence, confined to a closed ball in jRn, is a homeomorphism. From Brouwer's invariance of domain theorem it follows that f is open whence f (jRn) is both open and closed. Since jRn is connected, f is surjective: f is bijective. Finally, (3.2) also implies f- 1 , which exists because f is bijective, is continuous. 0

3.55. The proof is an application of the average principle: an average lies between the supremum and infimum of the numbers averaged.

If If(O)1 = SUPxEB(O,I) If(x)1 ct;bf M, Ilxoll = r < 1, and If (xo)1 < M then

If(O)1 = M = 1-21 1 f(y) dyl ::; -21 1 If(y)1 dy. (s3.4) nr 1Ilyll=r nr 1Ilyll=r

Because f is continuous, If(y)1 < M near Xo whence the value of the last member in (s3.4) is less than M, a contradiction.

If f is constant then If(O)1 = SUPXEB(O,I) IIf(x)ll· 0

3.56. Because f is uniformly continuous on K, if f > ° there is a positive 8 such that on K, {Ix - yl < 8} =? {If(x) - f(y)1 < fl. Consequently,

{Ix -yl < f} =? {If[tj(x) - f[tj(y)1 = If(x+t) - f(y+t)1 < f},

whence {f[tnj} nEN is uniformly bounded and equicontinuous. The ArzelaAscoli theorem implies the result. 0

3.57. For any x in [O,IJ and some ~ in (0, x),

f(x) ::; If(x)1 ::; fox f(t) dt = f(~)x ::; Ilflloox.

It follows by repeated integration that for each n in N and each x in [0,1), xn

If(x)1 ::; IIflloo,. 0 n.

3.58. For each real a, there are coefficients bn(a), n E N, and a positive r(a) such that for x in (a-r(a), a+r(a», 1'(x) = E~=1 bn(a)(x-a)n. The series

(x - a)n+1 "''''=--1 bn (a) converges in (a - r( a), a + r( a» and represents a L.Jn_ n + 1 function gao

On the other hand, foX l' (t) dt = ga (x) for x near a and, since l' is continuous, g~ = 1', f = ga + constant, whence f is real analytic. 0

Solutions 3.59-3.60 217

3.59. If t E (0,1], f(n+l)(tx) is a monotonely increasing function of x. Taylor's formula

f(x) = f(O) + f'(O)x + ... + __ xn + (1 - tt f(n+l) (tx) dt -fen) (0) (11 ) xn+1

n! 0 n!

~f f(O) + j'(O)x + ... + f(n) (0) xn + R,,(x) n!

leads to the conclusion that if 0 S; x < c then

xn+1 J/(l-t)n+1f(n+1)(tc)dt OS;Rn(x)S; 0 ,

n.

( fen) (O)cn ) Xn+1 f(c) - f(O) - j'(O)c - ... - n!

whence for all x, limn --+oo Rn(x) = o. 3.60. a) Abel summation implies

Ii an sin nx sin ~I = ~ lam cos (n-~) x

+ i (an+1 - an) cos (n +~) x - aM cos (M +~) xl

S; ~ [am + f: (an - an+1) + aM 1 = am·

def [1] If x E (0,71") and n = -; then when m S; n

m

18m (x)1 S; L nan x S; Kmx S; Knx S; K. n=1

If m > n then, since

1.1 Ix 111 --Slll-X> --- > - >-an+1 2 - an+171" - an+1(n+l)7I" - K7I"'

o


it follows that

18m (x)1 ~ 18n (x)1 + I f aksinkxl ~ K + ~n+~ ~ K +K1r. k=n+! sm"2

Hence if x E (0,1rJ, 18N(x)1 ~ K(l + 1r). b) When 0 E (0, 21r), Euler's formula eiO = cosO+sinO and the formula

for the sum of a geometric progression imply that there is a constant C1

such that It, sin nol ~f IKN(O)I ~ Isin ~~/2)1· Abel summation implies that if N 2 M and 0 < a ~ 0 ~ b < 21r then

for some constant C2 ,

N

18N(O) - 8M(O)1 ~ L lak - ak+11·IK k(O)1 k=M+1

+ laNI·IKN(O)1 + laM+11·IKM+!(O)1

~ C2 (laNI + laM+11)

whence convergence is uniform on [a, b]. c) If 2:;;=1 sin nO converges uniformly on [0,21r] as N -4 00, 0 = 2~'

and f > 0 then for large M and all N in (M,oo),

f N-M (M) "2 > 8N(O) - 8M(O) 2 (N - M)aN = N NaN = 1- N NaN·

Conversely, if an 10 and limn-;co nan = 0 the conclusion in a) implies

that it suffices to prove uniform convergence on [O,~]. Owing to point

wise convergence everywhere on [0,21rJ, consideration can be focussed on ,",co . 0 def R d def {} L.m=M+1 anSm; = [~]a:ef fM = maxn2':M nan ..

If 0 < 0 ~ 4" then (j = lI(O) EN and for M m N,

RM = {E~~L-+1 + 2:~=v(O)+l an sin nO ~f R'u + R'J.r if M + 1 ~ lI(O) .

2:~=M+1 an sin nO ~f 8M if M + 1 > lI(O)

If M + 1 ~ lI(O), the inequality I sin nOI ~ InOI implies

v(O) 1

/R'u/ ~ 0 L nan ~ lI(O) fM+1 (lI(O) - M) ~ fM+!· n=M+1

Solution 3.61 219

In sum, R~ ~ 0. 7r • 0 7rO •

If ° < 0 < - then sm - > -. Smce - -4 2-8

if M + 1 :::; v(O) Abel summation reveals that for some constant C3 ,

00

IR~I:::; L (an - an+d IKn(O)1 + av(O) IKv(o)-l(O)1 n=v(O)

8 16C2 :::; 2C1aV (O)-O :::; -- (v(O) + l)av(o) :::; C3EM.

7r 7r

If M + 1 > v(O) the same kind of calculation reveals that for some

constant C4 , ISMI :::; C4 EM. Hence if C ~f max{Cl,C2 ,C3 ,C4 } then

I f an sin nol :::; CEm· n=M+l

o

3.2. Complex-valued Functions

3.61. For n in Nand j in [0, n - 1],

(j+1) (j) 1 ( (j)) Xn ----;:;- = Xn ;: +;: f Xn ;:

and induction shows, since f is defined on [0,1]' that

Direct calculation shows that

whence {xn} nEN is uniformly bounded and equicontinuous. The ArzelaAscoli theorem implies that there is a subsequence {Xnk hEN and a contin-

uous function x such that x nk ~ x. The equicontinuity of {xnk } nEN and the uniform continuity of f imply

that if "1 > 0, there is an no and a positive {j such that if nk > no then


. {I} j j+l Hence If nk > max ~, no and - ::; s < t ::; -- then u nk nk

Xnk(t) - xnk(s) = I (xnk (L)) t - s nk

I (Xnk (s)) - 'fJ ::; xnk (t~ = :nk (s) ::; I (Xnk (s)) + 'fJ

x(t) - x(s) I (x(s)) - 2'fJ ::; t _ s ::; I(x(s)) + 2'fJ.

It follows that

I(x(s)) - 2'fJ ::; limt=s X(t~ = :(s) ::; limt=s X(t~ = :(s) ::; I(x(s)) + 2'fJ

whence x'(s) exists and x'(s) = I(x(s)).

Because X(s) ~f J; I(x(t)) dt and x(s) are both solutions of

z' = I(z), z(O) = 0, (s3.5)

the uniqueness theorem for differential equations implies X = x. If the entire sequence {xn}nEN fails to converge to x, some subsequence converges uniformly to a different limit, say y. However both x and y are solutions of (s3.5). 0

3.62. If {AhEN C N n and limk~oo Il/k - 11100 = 0, then for each k and

. . Ilk (Xk + h) - Ik (Xk) I some Xk In [0,1], If h 1= 0, h ::; n. It may be assumed

that for some x in [O,IJ, x - Xk ~f Ok --+ 0 as k --+ 00. Thus for large k,

10k : h I is near one and

I I (x + hh - I (x) I ::; I I(x + h) ~ A(x + h) I

I Ik (Xk + Ok + h) - A (Xk) I· 10k + h I + Ok + h h

+ I Ik (Xk) : I (Xk) I + I I (Xk) : I (x) I def I+II+III+IV.

For any nonzero h and large k, I, III, and IV are small while II does not exceed a number near n. Hence I E N n and so N n is closed.

There is a piecewise linear continuous "sawtooth" function G for which

at each point x, I G(x + hh - G(x) I > n if h is sufficiently small while IIGlloo

Solutions 3.63-3.64 221

is arbitrarily small [GeO]. For any neighborhood N(I) of an f in Nn, there is a polynomial p such that p + G E N (I) \ N n. Hence the interior N~ of N n is empty and so, as a closed set with no interior, N n is nowhere dense. 0

[Note 83.7: The set A ~f 0([0, l],q \ (UnENNn) is of the second category and consists entirely of (continuous) nowhere differentiable functions.]

3.63. Write fn = !R (In) + i<;J< (In) ~ Un + ivn· If Ilfnlloo +0 as n -+ 00,

it may be assumed that for some positive 8, some xoo , and some sequence {xn}nEN' Un (xn) ~ 8 and limn-+oo Xn = Xoo. The mean value theorem and the hypothesis imply

whence for large n and all x in [xoo - ~'Xoo +~], un(x) ~ ~. If 9 is the piecewise linear function such that

_{1 on [xoo-~,xoo+~] g(x)- [8 8]

o off Xoo - 2'xoo + 2

f1 82 then for large n, io un(x )g(x) dx ~ 16' a contradiction. o 3.64. Because f(O) = 0, for some F in 0 1 ([-11",11"], q,

F(x) = {f(X) if x E [0,11"] - f( -x) if x E [-11",0].

Integration by parts and the Schwarz inequality show that if

00

F(x) = I>nsinnx n=l

[since F E 0 1 ([-11",11"], q, the right member converges everywhere to F(x)] then

bn = :n (i:I'(x)cosnxdx + 1'" f'(x) cos nx dX)

Ib 1 < 111'112 n - .j1rn

21'" If(x)12 dx = J~ IF(x)12 dx = 11" ~ Ibn l2 ~ (~:2 ) 1'" II'(xW dx

as required. 0


3.65. The Fundamental Theorem of Calculus implies that if E is a bounded subset of C1 ru., C) then E is equicontinuous. The ArzelaAscoli theorem implies T(E) is compact. 0

3.66. a) Since II 11(1) ;::: II 1100, X is closed in C 1 ([0, 1], C). b) Because II II 00 ~ II II (1), the inclusion map

T: c1 ([0,1], C) '-+ C ([0,1], C)

restricted to X is a continuous bijection of Banach spaces and thus T is open, whence T- 1 is continuous. Hence for some M, II 11(1) ~ Mil 1100' Let

1 . k be M and let K be 1.

c) The mean value theorem for derivatives implies that a bounded subset of C1 ([0,1], C) is equicontinuous. By virtue of 3.65, the inclusion map T of b) above is compact, whence T(E) is compact. In particular,

T (B (0, It) is compact while, since T is open, T (B (0, It) is open. Since a Banach space is finite-dimensional iff it is locally compact, it follows that X is finite-dimensional. 0

3.67. Since I E C1 ('][', C), I is represented by its Fourier series:

I(z) = L anzn. nEZ\{O}

Furthermore, the Fourier series for f' is 2:nEZ\{O} nan z n - 1 , whence by virtue of Parseval's equation,

1If'11~ = L n2 lanl2 , II/II~ = L lanl2 ~ II!'II~· nEZ\{O} nEZ\{O}

Equality obtains iff for n other than ±1, an = O. o 3.68. Note that E1 = L2 ([0,1], ),), D (C1 ([0, 1], C)) eEl, and for all I in C1([0,1],C), IID(f)II' ~ 11/11". Because C 1 ([0, 1],C) is dense in Ell D may be extended to a linear continuous map D : E2 ~ L2 ([0, 1], ),).

If D(f) = 0 then for some sequence {In} nEN in C1 ([0,1]' C),

lim Il/n - III" = lim liD (fn)II' = O. n---+OCl n-+OCl

The Schwarz inequality and the choice of the sequence {fn}nEN imply

lim {1 I/n(x)1 dx = 0 n-oo}o

lim (1 (1/m(x) _ In(x)12 + 1/:n(x) - 1~(x)12) dx = O. m,n--+oo io

Solutions 3.69-3.71 223

It may be assumed that limn --+oo J n ~ J and limn --+oo J~ ~ O. The Fundamental Theorem of Calculus implies that for all x in [0,1],

Hence limn --+oo In(O) ~f C exists and, since x may be chosen off the null set on which {fn}nEN fails to converge, J(x) ~ c as required. D

3.69. a) The triangle inequality la + bl ~ lal + Ibl for complex numbers shows that D(O, 1) is convex.

b) If 0 ~ tn ~ tn+1 and L:;;=I ti = 1, the Euclidean structure of ]R2 yields the result if N = 2. If N > 2 and 0 < h < 1 then the formulre

and mathematical induction reduce the argument to the case when N = 2. D

3.70. If L:;;=I aneinx == 0 on [a, b] then J : e 3 z 1---+ L:;;=I aneinz is an entire function vanishing on a compact infinite subset of e. Hence J(z) == 0 and thus J(x) == 0 on [0,271']. The orthogonality relations

1271"" {O e·nx e -.mx dx = o 271'

ifm #n ifm =n

lead to the conclusion that an == o. D

3.71. If A ~f (aij)~j:l and det(A) = 0, the result follows. Hence it may be assumed that det(A) # 0, whence the rows of A are linearly independent vectors al, ... ,~ in en. The Gram-Schmidt orthonormalization

process applied to {ai} 1 '-I' The 1,,)- 2,3-

diagonal entries of rare 'Yii ~f Ilai - L:~:i (~, Uj)rI 2: Ilaill-i. Because

(Uij)~j:I is a unitary matrix, Idet [(Uij)~j:I] 1= 1, whence

D

224 3. Real- and Complex-valued Functions:

3.72. For m in Nand z in D(O, 1), the functions

def J (~z) Fp(z) = (p _ 1 )' 1 ~ p ~ m,

J --z m

Solutions

are continuous and IFp(z)1 == 1, 1 ~ p ~ m. For large m, and all z, because the numerator and denominator of the fraction defining Fp are near each other, Fp(z) f:. -1, 1 ~ p ~ m. Hence Fp (D(O, 1)) is a connected compact subarc Ap of '][" \ {-I}, i.e., there is a continuous function

'TIp: D(O, 1) ;:, z 1-+ 'TIp(z) E (-11",11")

such that Fp(z) = ei'1/p(z) E Ap. Since

J (E-z) J(z) = J(O)· II (p ~ 1 )'

l$p$mJ --z m

it follows that if J(O) ~f ei '1/0,-11" < 'TIo ~ 11", and ¢(z) ~f 'TIo + EZ'=o'TIk(Z) then J(z) = eiq,(z). 0

3.73. If G(O) ~ 9 (ei(l)+1r)) - 9 (eilJ ) then G(11") = g(l) - g(-l) = -G(O). Hence either G(O) = 0, p = 1, and p = -lor for some 0 near 11", G(O) and G(O) are of opposite sign, whence at some 'IjJ in [0,11"), G('IjJ) = O. Thus if p = ei'lj! and p = ei (1/!+1r) then g(p) = 9 (P). 0

3.74. Because hiT maps '][" into itself, 3.73 implies that at least one of some pair of antipodal points is not left fixed. 0

3.75. If no point is left fixed by J then for z in D(O, 1), the closed half-line determined by z and J(z) and of which J(z) is the endpoint meets '][" in a point h(z). The map h is continuous and h leaves each point of'][" fixed, in contradiction of 3.74. 0

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