Rudin Chap9 Some Solutions

PRINCIPLES OF MATHEMATICAL ANALYSIS.

WALTER RUDIN

Disclaimer: these solutions were typed at warp speed, often with little orno preparation. And very little proofreading. Consequently, there are bound to beloads of mistakes. Consider it part of the challenge of the course to find these errors.

(And when you do find some, please let me know so I can fix them!)

7. Sequences and Series of Functions

1. Prove that every uniformly convergent sequence of bounded functions is uni-formly bounded.{fn} is uniformly Cauchy, so |fn(x)−fm(x)| < 1, for n,m bigger than some

large N . Then∣∣∣|fn(x)| − |fN(x)|

∣∣∣ ≤ |fn(x)− fN(x)| < 1, so

|fn(x)| < |fN(x)|+ 1.

Let M be a bound for fN . Then

|fn(x)| < M + 1,

so {fn}∞n=N is uniformly bounded by M + 1. Define

K := max{sup |f1|, sup |f2|, . . . , sup |fN−1|,M + 1}.Then {fn}∞n=1 is uniformly bounded by K.

2. Show that {fn}, {gn} converge uniformly on E implies {fn + gn} convergesuniformly on E. If, in addition, {fn}, {gn} are sequences of bounded functions,prove that {fngn} converges uniformly.

supx∈E

{|(fn + gn)(x)− (f + g)(x)|} = supx∈E

{|fn(x)− f(x) + gn(x)− g(x)|}≤ sup

x∈E{|fn(x)− f(x)|+ |gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)|+ supx∈E

|gn(x)− g(x)|n→∞−−−−−→ 0 + 0.

When {fn}, {gn} are bounded sequences,

supx∈E

{|(fngn)(x)− (fg)(x)|}

March 23, 2006. Solutions by Erin P. J. Pearse.

1

Principles of Mathematical Analysis — Walter Rudin

= supx∈E

{|fngn(x)− fgn(x) + fgn(x)− fg(x)|}≤ sup

x∈E{|fn(x)− f(x)||gn(x)|+ |f(x)||gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)||gn(x)|+ supx∈E

|f(x)||gn(x)− g(x)|≤ sup

x∈E|fn(x)− f(x)| ·Mg + sup

x∈EMf · |gn(x)− g(x)|.

We can introduce the uniform bounds Mf and Mg by problem 1 and theadditional hypothesis. Then it is clear that the last line goes to 0 as n →∞.

3. Construct sequences {fn}, {gn} which converge uniformly on a set E, but suchthat {fngn} does not converge uniformly.

Work on I = (0, 1). Let fn(x) = 1x

+ xn

and gn(x) = − 11−x

− 1−xn

so thatgn is the horizontal reflection of fn, translated 1 to the right. (Graph them tosee it.)

fn(x) = 1x

+ xn

n→∞−−−−−→ 1x

= f(x) sup0≤x≤1

|fn − f | = supx∈I

xn

= 1n→ 0

gn(x) = − 11−x

− 1−xn

n→∞−−−−−→ − 11−x

= g(x) sup0≤x≤1

|gn − g| = supx∈I

1−xn

= 1n→ 0

fngn(x) = −(n + (x− 1)2)(n + x2)

n2x(x− 1)

n→∞−−−−−→ 1

x− x2= fg(x)

sup0≤x≤1

|fngn − fg| = supx∈I

∣∣∣∣x2(x− 1)2 + n + 2nx(x− 1)

n2x(x− 1)

∣∣∣∣ = ∞, for any n.

To see the sup is infinite, check x = 0, 1.

4. Consider the sum

f(x) =∞∑

n=1

1

1 + n2x.

For what x does the series converge absolutely? On what intervals does itconverge uniformly? On what intervals does it fail to converge uniformly? Isf bounded?

For xk = − 1k2 , we get

f(xk) =∞∑

n=1

1

1− (nk

)2 .

The kth term of the sum is undefined, so f is undefined at xk = − 1k2 , k =

1, 2, . . .To be completed.

2

Solutions by Erin P. J. Pearse

5. Define a sequence of functions by

fn(x) =

0(x < 1

n+1

),

sin2 πx

(1

n+1< x ≤ 1

n

),

0(

1n

< x).

Show that the series {fn} converges to a continuous function, but not uni-formly. Use the series

∑fn to show that absolute convergence, even for all x,

does not imply uniform convergence.For x ≤ 0, fn(x) = 0 for every n, so lim fn(x) = 0. For x > 0,

n > N :=[

1x

]=⇒ 1

n< x =⇒ fn(x) = 0.

Thus fn(x)pw−−−→ f(x) := 0 for x ∈ R.

To see that the convergence is not uniform, consider

f ′n(x) = 2(sin π

x

) (cos π

x

) (− πx2

).

f ′n(x) = 0 when• sin π

x= 0, in which case x = 1

kfor some k ∈ Z, or

• cos πx

= 0, in which case x = 22k+1

for some k ∈ Z.For each fn, only a few of these values occur where fn is not defined to be 0,so checking these values of x,

fn

(1n

)= sin(nπ) = 0

fn

(1

n+1

)= sin((n + 1)π) = 0

fn

(2

2n+1

)= sin

(2n+1

2π)

= 1.

So Mn = supx{|fn(x)− f(x)|} = 1 for each n, and clearly Mn → 1 6= 0.The series

∑∞n=1 fn(x) converges absolutely for all x ∈ R: for any fixed x,

there is only one nonzero term in the sum.The series

∑∞n=1 fn(x) does not converge uniformly: check partial sums.

supx

∣∣∣∣∣N∑

n=1

fn(x)−N+1∑n=1

fn(x)

∣∣∣∣∣ = supx|fN+1(x)| = 1,

so the sequence of partial sums is not Cauchy (in the topology of uniformconvergence), hence cannot converge.

6. Prove that the series ∞∑n=1

(−1)n x2 + n

n2

converges uniformly in every bounded interval, but does not converge abso-lutely for any value of x.

To see uniform convergence on a bounded interval,

supa<x<b

{|fN(x)− f(x)|} = supa<x<b

{∣∣∣∑∞

n=N+1(−1)n x2+n

n2

∣∣∣}

3


={∣∣∣

∑∞n=N+1

(−1)n c2+nn2

∣∣∣}

(7.1)

where c := max{|a|, |b|}. We will use the alternating series test to show∑∞n=1(−1)n c2+n

n2 converges. From this, it will follow that (7.1) goes to 0 asN →∞.(i) For all x ∈ R, x2+n

n2 > 0. So the sum alternates.

(ii) For fixed x, lim x2+nn2 = lim 1

2n= 0. (L’Hop)

(iii) To check monotonicity, prove

x2 + n + 1

(n + 1)2≤ x2 + n

n2

by cross-multiplying.

7. For n = 1, 2, 3, . . . , and x ∈ R, put fn(x) = x1+nx2 . Show that {fn} converges

uniformly to a function f , and that the equation f ′(x) = limn→∞ f ′n(x) iscorrect if x 6= 0 but false if x = 0.

8. If

I(x) =

{0 (x ≤ 0),

1 (x > 0),

if {xn} is a sequence of distinct points in (a, b), and if∑ |cn| converges, then

prove that the series

f(x) =∞∑

n=1

cnI(x− xn) (a ≤ x ≤ b)

converges uniformly, and that f is continuous for every x 6= xn.

4


9. {fn} are continuous and fnunif−−−→ f on E. Show lim fn(xn) = f(x) for every

sequence {xn} ⊆ E with xn → x.Pick N1 such that

n ≥ N1 =⇒ supx∈E

|fn(x)− f(x)| < ε/2.

Then surely |fn(xn) − f(xn)| < ε/2 holds for each n ≥ N1. By Thm. 7.12, fis continuous, so pick N2 such that

n ≥ N2 =⇒ |f(xn)− f(x)| < ε/2.

Then we are done because

|fn(xn)− f(x)| ≤ |fn(xn)− f(xn)|+ |f(xn)− f(x)|.

11. {fn}, {gn} are defined on E with:

(a) {∑Nn=1 fn} uniformly bounded, (b) gn

unif−−−→ g on E, and (c) gn(x) ≤gn−1(x) ∀x ∈ E, ∀n. Prove

∑fngn converges uniformly on E.

Define a := supx∈E |fn(x)| and bn := supx∈E |gn(x)|. Then

supx∈E

∣∣∣∣∣∞∑

N+1

fn(x)gn(x)

∣∣∣∣∣ ≤∞∑

N+1

supx∈E

|fn(x)gn(x)| ≤∞∑

N+1

anbn → 0,

by Thm. 3.42.

13. {fn} is monotonically increasing on R, and 0 ≤ fn(x) ≤ 1

(a) Show ∃f, {nk} such that f(x) = limk→∞ fnk(x),∀x ∈ R.

By Thm. 7.23, we can find a subsequence {fni} such that {fni

(r)} con-verges for every rational r. Thus we may define f(x) := supr≤x f(r),where the supremum is taken over r ∈ Q. It is clear that f is monotone,because

x < y =⇒ {r ≤ x} ⊆ {r ≤ y}and the supremum can only increase on a larger set. Thus, f has at mosta countable set of discontinuities, by Thm. 4.30, pick x such that f iscontinuous at x.We want to show fni

(x) → f(x). Fix ε > 0. Since f is continuous at x,choose δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Now pick arational number r ∈ [x− δ

3, x]. Then

|fni(x)− f(x)| ≤ |fni

(x)− fni(r)|+ |fni

(r)− f(r)|+ |f(r)− f(x)|. (7.2)

On the RHS of (7.2): the last term is less than ε/3 by the choice of δ;and the middle term is less than ε/3 whenever i ≥ N1 for some large N1,because the subsequence converges on the rationals. It remains to showthe first term gets small.

5


Pick some rational s ∈ [x, x + δ/3]. Then r ≤ x ≤ s and the continuityof f at x shows

|r − s| < δ =⇒ |f(r)− f(s)| < ε/3.

Also, since the fniare monotone,

fni(r) ≤ fni

(x) ≤ fni(s). (7.3)

With

|fni(r)− fni

(s)| ≤ |fni(r)− f(r)|+ |f(r)− f(s)|+ |f(s)− fni

(s)|,some large N2, i ≥ N2 gives |fni

(r) − fni(s)| < ε. By (7.3), this shows

|fni(r)− fni

(x)| < ε.

(b) If f is continuous, show fnk→ f uniformly on compact sets.

Let K be compact. Fix ε > 0. Since f is uniformly continuous onK, pick δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Since K

is compact, we can find {x1, . . . xJ} such that K ⊆ ⋃Jj=1 Bδ(xj), where

Bδ(xj) := (xj − δ, xj + δ).

16. {fn} is equicontinuous on a compact set K, fnpw−−−→ f on K. Prove {fn}

converges uniformly on K.Define f by f(x) := lim fn(x). Fix ε. From equicontinuity, find δ such that

|x− y| < δ =⇒ |fn(x)− fn(y)| < ε/3, ∀n, x, y.

Letting n →∞, this gives

|x− y| < δ =⇒ |f(x)− f(y)| < ε/3, ∀x, y.

Since K is compact, we can choose a finite set {x1, . . . , xJ} such that

K ⊆⋃J

j=1Bδ(xj)

where Bδ(xj) := (xj − δ, xj + δ). For each xj, we know fn(xj) → f(xj), sopick N big enough that

n ≥ N =⇒ |fn(xj)− f(xj)| < ε/3, ∀j = 1, . . . , J.

For any x ∈ K, x ∈ Bδ(xj) for some j. Thus for all x ∈ K,

|fn(x)− f(x)| ≤ |fn(x)− fn(xj)|+ |fn(xj)− f(xj)|+ |f(xj)− f(x)|.

6


18. Let {fn} be uniformly bounded and Fn(x) :=∫ x

af(t) dt for x ∈ [a, b]. Prove

∃{Fnk} which converges uniformly on [a, b]

We need to show {Fn} is equicontinuous. Then by Thm. 6.20, each Fn iscontinuous; and by Thm. 7.25(b), we’re done. So fix ε > 0, let x < y, and letM be the uniform bound on the {fn}.

|Fn(x)− Fn(y)| =∣∣∣∣∫ y

x

fn(t) dt

∣∣∣∣ ≤∫ y

x

|fn(t)| dt ≤ M(y − x)

Then pick any δ < ε/M and

|x− y| < δ =⇒ |Fn(x)− Fn(y)| ∀n,

so {Fn} is equicontinuous.

7


20. f is continuous on [0, 1] and∫ 1

0f(x)xn dx = 0, n = 0, 1, 2, . . . . Prove that

f(x) ≡ 0.Let g be any polynomial. Then g(x) = a0 + a1x + a2x

2 + · · · + aKxK . Bylinearity of the integral,

∫ 1

0

f(x)g(x) dx =K∑

k=0

ak

∫ 1

0

f(x)xk dx =K∑

k=0

0 = 0.

By the Weierstrass theorem, let {fn}∞n=1 be a sequence of polynomials whichconverge uniformly to f on [0, 1]. Then

∫ 1

0

f 2(x) dx = limn→∞

∫ 1

0

f(x)fn(x) dx = limn→∞

0 = 0.

Then by Chap. 6, Exercise 2, f 2(x) ≡ 0. Thus f(x) ≡ 0.

21. Let K be the unit circle in C and define

A :=

{f(eiθ) =

N∑n=0

cneinθ ... cn ∈ C, θ ∈ R}

.

To see that A separates points and vanishes at no point, note that A con-tains the identity function f(eiθ) = eiθ.

To see that there are continuous functions on K that are not in the uniformclosure of A, note that

∫ 2π

0

f(eiθ)eiθ dθ = 0 ∀f ∈ A, (7.4)

and hence for g = lim gn (uniform limit) with gn ∈ A,

∫ 2π

0

g(eiθ)eiθ dθ = limn→∞

∫ 2π

0

gn(eiθ)eiθ dθ = 0,

by Thm. 7.16. Thus, all functions in the closure of A satisfy (7.4). However,if we choose an h which is not in A, like

h(eiθ) = e−iθ,

then h is clearly continuous on K, and

∫ 2π

0

h(eiθ)eiθ dθ =

∫ 2π

0

1 dθ = 2π.

Thus h is not in the uniform closure of A.

8


22. Assume f ∈ R(α) on [a, b] and prove that there are polynomials Pn such that

limn→∞

∫ b

a

|f − Pn|2dα = 0.

We need to find {Pn} such that ‖f−Pn‖2n→∞−−−−−→ 0. Fix ε > 0. By Chap. 6,

Exercise 12, we can find g ∈ C[a, b] such that ‖f − g‖2 < ε/2. Note that∫ b

a

|g − P |2 ≤∫ b

a

sup |g − P |2 = sup |g − P |2(b− a).

Then by the Weierstrass theorem, we can find a polynomial P such that

‖g − P‖2 ≤ sup |g − P |(b− a) < ε/2.

By Chap. 6, Exercise 11, this gives ‖f − P‖2 < ε.

23. Put P0 = 0 and define Pn+1(x) := Pn(x) + (x2 − P 2n(x)) /2 for n = 0, 1, 2, . . . .

Prove that limn→∞ Pn(x) = |x| uniformly on [−1, 1].Note that if Pn is even, the definition will force Pn+1 to be even, also. Now

P1 = x2

2, so assume 0 ≤ Pn−1 ≤ 1 for |x| ≤ 1. Then

Pn = Pn−1 +x2

2− P 2

n−1

2=

x2

2+ Pn−1

(1− Pn−1

2

).

By elementary calculus,

0 ≤ y ≤ 1 =⇒ f(y) = y(1− y2) takes values in [0, 1

2]. (7.5)

Since |x| ≤ 1 also implies x2

2∈ [0, 1], this gives 0 ≤ Pn ≤ 1. Then

0 ≤ |x|+ Pn(x)

2≤ 1

0 ≤ 1− |x|+ Pn(x)

2≤ 1. (7.6)

To see Pn(x) ≤ |x|, consider that for x ≥ 0, the inequality

x− P1(x) = x− x2/2 ≥ 0

holds in virtue of the positivity of f in (7.5). Then by the symmetry of evenfunctions, this is true for |x| ≤ 1. Now suppose Pn−1 ≤ |x|, i.e., |x|−Pn−1(x) ≥0. Then the given identity, and (7.6), give

|x| − Pn = (|x| − Pn−1)(1− 1

2(|x|+ Pn−1)

) ≥ 0.

We have established

0 ≤ Pn(x) ≤ Pn+1(x) ≤ |x| for− 1 ≤ x ≤ 1. (7.7)

Now, for n = 1, it is clear that

|x| − P1(x) = |x| − |x|22≤ |x|

(1− |x|

2

)1

.

9


So suppose |x| − Pn−1(x) ≤ |x|(1− |x|

2

)n−1

. Then, multiplying each side by(1− |x|

2− Pn−1(x)

2

)and using the identity, we get

(|x| − Pn−1(x))(1− |x|

2− Pn−1(x)

2

)≤ |x|

(1− |x|

2

)n−1 (1− |x|

2− Pn−1(x)

2

)

|x| − Pn(x) ≤ |x|(1− |x|

2

)n−1 (1− |x|

2

)

|x| − Pn(x) ≤ |x|(1− |x|

2

)n

.

Now consider gn(x) = x(1− x

2

)non [0, 1]. gn(x) =

(1− x

2

)n−1(1− (n+1)x

2

)

shows that gn has extrema at 2 and 2n+1

; only the latter is in [0, 1]. Then for

fn(x) = |x|(1− |x|

2

)n

, fn has extrema

fn

(± 2n+1

)= 2

n+1

(n

n+1

)n< 2

n+1.

We have established

|x| − Pn(x) ≤ |x|(1− |x|2

x

)n

< 2n+1

. (7.8)

Now sup∣∣∣|x| − Pn(x)

∣∣∣ < 2n+1

n→∞−−−−−→ 0 and Pnunif−−−→ |x|.

10


8. Some Special Functions

7. If 0 < x < π2, prove that 2

π< sin x

x< 1.

To see the first inequality, suppose ∃x0 ∈ (0, π2) with 2

π≥ sin x0

x 0. Since

limx→0sin x

x= 1, IVT gives an y with 2

π= sin y

y. Define g(x) = sin x − 2

πx so

g(y) = 0. Then g′(x) = cos(x) − 2π

and g′′(x) = − sin x < 0 for x in theinterval. By IVT again, there is a point z ∈ (0, y) with g′(z) = 0, so g′(x) < 0

for x ∈ (y, π2). Then g(y) = 0 implies that g(π

2) < 0. <↙

To see the second inequality, put f(x) = x− sin x. Then f ′(x) = 1− cos xshows f ′(0) = 0 and f ′(x) > 0 for 0 < x < π

2.

8. For n = 0, 1, 2, . . . and x ∈ R, prove | sin nx| ≤ n| sin x|.This is clearly true (with equality) for n = 0, 1, so induct on n; suppose

| sin nx| ≤ n| sin x|. We could use this assumption, to say

| sin(n + 1)x| = | sin(nx + x)| ≤ | sin nx|+ | sin x| = (n + 1)| sin x|,if we could prove the central inequality. From the definitions of C(x) and S(x)given in (46),

sin(x + y) = sin x cos y + cos x sin y.

Applying this,

| sin(nx + x)| ≤ | sin nx cos x|+ | cos nx sin x|≤ | sin nx|+ | sin x|,

since | cos x| ≤ 1 for all x.

9. (a) Put sN = 1 + 12

+ · · ·+ 1N

. Prove that γ = limN→∞(sN − log N) exists.Note that

γ = limN→∞

N∑

k=1

1

k− log N =

∫ ∞

1

(1

[x]− 1

x

)dx.

The following sum telescopes:

N∑

k=1

[1

k− (log(k + 1)− log k)

]= sN − log(N + 1),

so rewrite the summand as

ak :=1

k− log

(1 +

1

k

).

We will bound ak by 12x−2. Note that for

f(x) =x2

2− x + log(1 + x),

11


we have f ′(x) = x− 1+ 11+x

and f ′′(x) = 1 = 1(1+x)2

. This gives f ′(0) = 0

and f ′′(x) > 0 for all x > 0, so that f is always positive for x > 0 and

0 ≤ x− log(1 + x) ≤ x2

2.

In particular, this is true for x = 1k. Since

∑1k2 converges, this shows

limN→∞

N∑

k=1

ak = limN→∞

(sN − log(N + 1)

)

exists. Finally,

log(N + 1)− log N = log

(1 +

1

N

)N→∞−−−−−→ log 1 = 0

shows that limN→∞ (sN − log N) = 0.

(b) Roughly how large must m be such that N = 10m satisfies sN > 100?From the above, and problem #13, we have

0 ≤ sN − log N ≤ π2

6=⇒ log N ≤ sN ≤ log N +

π2

6.

Then m ≈ 43.43 will ensure log 10m > 100.

12. Suppose that f is periodic with f(x) = f(x + 2π), and for δ ∈ (0, π),

f(x) =

{1, |x| ≤ δ,

0, δ < |x| < π.

(a) Compute the Fourier coefficients of f .

cn =1

2π

∫ δ

−δ

e−inx dx =−1

2πin

(e−inδ − einδ

)=

sin nδ

πn.

(b) Conclude that

∞∑n=1

sin nδ

n=

π − δ

2, (0 < δ < π).

∑

n∈Z=

∑

n∈Z

sin nδ

πn

2∞∑

n=1

cn =1

π

∑

n∈Z

sin nδ

πn.

12


(c) Deduce from Parseval’s theorem that∞∑

n=1

sin2 nδ

n2δ=

π − δ

2.

Parseval’s theorem gives equality between

1

2π

∫ π

−π

|f(x)|2 dx =1

2π

∫ δ

−δ

1 dx =δ

π

and ∑

n∈Z|cn|2 =

∑

n∈Z

sin2 nδ

π2n2=

δ

π2

∑

n∈Z

sin2 nδ

n2δ.

Then ∞∑n=1

|cn|2 =1

2

(δ

π− c0

)=

Note that c0 = δπ

also.

(d) Let δ → 0 and prove that∫ ∞

0

(sin x

x

)2

dx =π

2.

(e) Put δ = π/2 in (c). What do you get?

2

π

∞∑n=0

sin2(π2n)

n2=

π

4=⇒

∞∑

k=0

1

(2k + 1)2=

π2

8.

13. Put f(x) = x for 0 ≤ x < 2π, and apply Parseval’s Theorem to conclude∞∑

n=1

1

n2=

π2

6.

Apply it to the 2π-periodic function f(x) = x on (−π, π) instead. Integra-tion by parts gives

cn =1

2π

∫ π

−π

xeinx dx =(−1)n

in,

so with Parseval’s theorem,

∑

n∈Z

1

n2=

∑

n∈Z|cn|2 =

1

2π

∫ π

−π

|f(x)|2 dx =π2

3.

Since we have |cn| = |c−n|, we find the desired series by subtracting

c0 =1

2π

∫ π

−π

x dx = 12[x2]π−π = 0

13


from each side, and dividing the remainder by 2.

∞∑n=1

1

n2=

1

2

(∑

n∈Z|cn|2 − c0

)=

1

2

(π2

3− 0

)=

π2

6.

9. Functions of Several Variables

5. Prove that to every A ∈ L(Rn,R) there corresponds a unique y ∈ Rn suchthat Ax = x · y.

For N = {x ∈ Rn ... Ax = 0}, N is a closed subspace by Problem 4. IfN = Rn, then Ax = 0 · x ∀x, and we’re done. So suppose N 6= Rn. Thenthere is a nonzero vector

x0 ∈ N⊥ = {u ∈ Rn ... u · v = 0,∀v ∈ N}.We check Ax = yA · x for yA defined by

yA =Ax0

‖x0‖2x0.

First, if x ∈ N , then Ax = 0 = yA · x. Next, if x = αx0, then

Ax = A(αx0) = αAx0 =Ax0

‖x0‖2x0 · αx0 = yA · αx0.

Since the functions A(x) and yA · x are linear functions of x and agree on Nand x0, they must agree on the space spanned by N and x0. But N and x0

span Rn, since every element y ∈ Rn can be written

y =

(y − Ay

Ax0

x0

)+

Ay

Ax0

x0.

Thus Ax = yA · x for all x ∈ Rn. If Ax = y′ · x also, then

‖y′ − yA‖2 = A(y′ − yA)− A(y′ − yA) = 0.

So y′ = yA is unique. Equality of the norms comes from the inequalities:

‖A‖ = sup‖x‖≤1

|Ax| = sup‖x‖≤1

|yA · x| ≤ sup‖x‖≤1

‖yA‖‖x‖ = ‖yA‖, and

‖A‖ = sup‖x‖≤1

|Ax| ≥∣∣∣∣A

(yA

‖yA‖)∣∣∣∣ = yA · yA

‖yA‖ = ‖yA‖.

6. If f(0, 0) = 0 and

f(x1, x2) =x1x2

x21 + x2

2

, for (x1, x2) 6= 0,

prove that D1f(x1, x2) and D2f(x1, x2) exist for every point (x1, x2) ∈ R2,although f is not continuous at (0, 0).

14


First, the derivatives are

D1f(x1, x2) =x2(x

22 − x2

1)

(x21 + x2

2)2

, and D2f(x1, x2) =x1(x

21 − x2

2)

(x21 + x2

2)2

,

and so clearly exist wherever (x, y) 6= (0, 0). To check the origin, consider thatalong the axes,

D1f(x1, 0) =0

x41

= 0, and D2f(0, x2) =0

x42

= 0.

However, for f to be continuous at (0, 0), we must have

f(0, 0) = lim(x,y)→(0,0)

f(x, y),

no matter how (x, y) → (0, 0)! Define γ(t) : R→ R2 by

γ(t) = (x(t), y(t)) = (t, t),

so that we approach the origin along the diagonal. Then

limt→0

f(x(t), y(t)) = limt→0

f(t, t) = limt→0

t2

2t2= lim

t→0

1

2=

1

26= 0.

7. Suppose that f is a R-valued defined in an open set E ⊆ Rn, and that thepartial derivatives D1f, . . . , Dnf are bounded in E. Prove that f is continuousin E.

Take m = 1 as in 9.21. Pick x ∈ E and consider the ball of radius rB(x, r) ⊆ E. Choose h ∈ Rn such that ‖h‖ < r. Define vk ∈ Rn by

vk = (h1, . . . , hk, 0, . . . , 0) for k = 1, . . . , n.

Then as in (42),

f(x + h)− f(x) =n∑

j=1

[f(x + vj)− f(x + vj−1)] .

Since |vk| < r for 1 ≤ k ≤ n, and since B = B(x, r) is convex, the segmentswith endpoints x + vj−1 and x + vj lie in B. Then vj = vj−1 + hjej, where

hjej = (0, . . . , 0, hj, 0, . . . , 0).

Then the Mean Value Theorem applies to the partials and shows that the jth

summand ishj(Djf)(x + vj−1 + tjhjej)

for some tj ∈ (0, 1). By hypothesis, we have M such that

|(Djf)(x)| ≤ M j = 1, . . . , n, ∀x ∈ E.

Applying this to the absolute value of the above difference,

|f(x + h)− f(x)| ≤n∑

j=1

|hj| ·M ≤ nM max{|hj|} ≤ nM‖h‖.

15


8. Suppose that f is a differentiable real function in an open set E ⊆ Rn, andthat f has a local maximum at a point x ∈ E. Prove that f ′(x) = 0.

Let {ej} be the standard basis vectors of Rn, and define

ϕj(t) = f(x0 + tej), for t ∈ R.

Then ϕj : R→ R is differentiable, so by Thm. 5.8,

ϕ′j(0) = (Djf)(x0) = 0.

But the partial derivatives (Djf)(x0) are precisely the columns of the matrix(Df)(x0) = f ′(x).

9. If f is a differentiable mapping of a connected open set E ⊆ Rn into Rm, andin f ′(x) = 0 for every x ∈ E, prove f is constant in E.

First, suppose m = 1. Pick some x0 ∈ E and define

C = {x ∈ E ... f(x) = f(x0)},D = {x ∈ E ... f(x) 6= f(x0)} = E \ C.

Clearly, C ∪ D = E and C ∩ D = ∅. If we can show each of C, D is open,then D must be empty, or else we’d have a disconnection of the connected setE.

E is open, so for any x ∈ C, pick ε > 0 such that B(x, ε) ⊆ E. Thenfor any y ∈ B(x, ε), the segment [x, y] ⊆ B(x, ε). Since f ′(x) = 0, we havef(x) = f(y) by Thm. 5.11(b). Also, we chose x ∈ C, so f(x) = f(y) = f(x0).This puts B(x, ε) ⊆ C and shows C is open.

Now suppose we have some x ∈ E \C. Since E is open, we can again chooseε > 0 such that B(x, ε) ⊆ E. By the exact same argument, we get

y ∈ B(x, ε) =⇒ f(x) = f(y) 6= f(x0),

so that B(x, ε) ⊆ D and D is open.Finally, if m > 1, then apply this argument to each component of f .

10. Suppose f is a differentiable mapping of R into R3 such that |f(t)| = 1 forevery t. Prove f ′(t) · f(t) = 0. Interpret this geometrically.

Since ‖f(t)‖ = 1, we have

(f1(t))2 + (f2(t))

2 + (f3(t))2 = 1.

Differentiating both sides,

2f1(t)f′1(t) + 2f2(t)f

′2(t) + 2f3(t)f

′3(t) = 0

f(t) · f ′(t) = 0.

This means that for any curve on the unit sphere, the tangent at p ∈ S1 isorthogonal to p, i.e., the surface of a sphere is orthogonal to the radius at anypoint.

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14. Define f(0, 0) = 0 and f(x, y) = x3/(x2 + y2) for (x, y) 6= (0, 0).

-2-1 0 1 2

-2-1

012

-2

-1

0

1

2

-1 0 1 2

-2-1

012

(a) Prove that D1f and D2f are bounded functions in R2 so that f is con-tinuous.We have the partial derivatives

D1f =x4 + 3x2y2

x4 + 2x2y2 + y4, and

D2f =−2x3y

x4 + 2x2y2 + y4.

Boundedness at ±∞:

|D1f | ≤ 3(x4 + 3x2y2)

x4 + 2x2y2= 3, as |x| → ∞, and

|D2f | = |2x3||x4/y + 2x2y + y3|

|y|→∞−−−−−→ 0.

Boundedness away from ∞: only need to check zeroes of the denomina-tors, and there is only (0, 0).

D1f(0, 0) = limx→0

f(x, 0)− f(0, 0)

xlimx→0

1 = 1, and

D2f(0, 0) = limy→0

f(0, y)− f(0, 0)

ylimy→0

0 = 0.

Thus f is continuous by Prob. 7.

(b) Let u ∈ R2, |u| = 1. Show (Duf)(0, 0) exists and |(Duf)(0, 0)| ≤ 1.The directional derivative is

Duf(0, 0) = limt→0

f [(0, 0) + tu]− f(0, 0)

t= lim

t→0

1

t· t3x3

t2(x2 + y2)

=x3

x2 + y2,

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for u = (x, y). Since |u| = x2 + y2 = 1, this implies |x| ≤ 1. Hence wehave

Duf(0, 0) = x3 and |Duf(0, 0)| = |x3| ≤ 1.

(c) Let γ be a differentiable mapping of R into R2 with γ(0) = (0, 0) and|γ′(0)| > 0. Put g(t) = f(γ(t)) and prove that g is differentiable for everyt ∈ R.Put γ(t) = (x(t), y(t)) so that

g(t) =x(t)3

x(t)2 + y(t)2, for t 6= 0.

Since γ is differentiable for t 6= 0, we have x(t), y(t) differentiable for t 6= 0,and hence g(t) is differentiable for t 6= 0, by the chain rule, Thm. 9.15.1

It remains to check that g is differentiable at the origin.

g′(0) = limh→0

g(h)− g(0)

h

= limh→0

1

h· x(h)3

x(h)2 + y(h)2

= limh→0

x(h)3

h3· h2

x(h)2 + y(h)2

=

(limh→0

x(h)

h

)3 (limh→0

h2

x(h)2 + y(h)2

)

= (x′(0))3 · 1

|γ′(0)|2 .

The last two lines are easiest to see by working backwards.

Also show γ ∈ C ′ =⇒ g ∈ C ′.Note that g′(t) = f ′(γ(t))γ′(t). Since the additional hypothesis is thatγ′(t) is continuous (which is equivalent to saying x(t), y(t) are continuous,by Thm. 4.10), we just need that f ′(γ(t)) is continuous. Since the chainrule gives

g′(t) =x(t)2 (x(t)2x′(t) + 3y(t)2x′(t)− 2x(t)y(t)y′(t))

(x(t)2 + y(t)2)2 ,

it is clear that g′(t) is continuous whenever x, y are not simultaneously 0.For t = 0 (or other t0 s.t. x(t0) = y(t0) = 0), replace x(t) by x′(0)t + o(t)(using Taylor’s Thm.) and similarly replace y(t) by y′(0)t + o(t). Thus

g′(t) ≈ (x′(0)t+o(t))2((x′(0)t+o(t))2x′(t)+3(y′(0)t+o(t))2x′(t)−2(x′(0)t+o(t))(y′(0)t+o(t))y′(t))((x′(0)t+o(t))2+(y′(0)t+o(t))2)2

,

1By (a), the partials of f exist and are continuous in an open nbd of (0, 0). Hence, f is continuouslydifferentiable at (0, 0) by Thm. 9.21.

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so that g′(t) t→0−−−−→ g′(0).

(d) In spite of this, f is not differentiable at (0, 0).Let u = (x, y) be a unit vector. By Rudin’s (40),

Du = D1f(0, 0)x + D2f(0, 0)y = 1 · x + 0 · y = x.

But from (b) we have Du = x3 6= x for any x other than −1, 0, 1.

(Bonus problem) This problem isn’t in Rudin.Define f : R2 → R by

f(x, y) =

{0, (x, y) = (0, 0),

x2yx4+y2 , otherwise.

Let γa be any straight line through the origin with slope a and γa(0) = (0, 0). Thismeans that (up to parametrization) γa(t) = (t, at) for some a ∈ R, or γ∞(t) = (0, t).Show that

limt→0

f(γa(t)) = f(0, 0) = 0,

but that f is not continuous at the origin.

-1

-0.5

0

0.5

1-1

0

1-0.5-0.2500.250.5

-1

-0.5

0

0.5

1

-0.5-0.2500.25

Continuity of the restriction follows by a basic computation:

limt→0

f(γa(t)) = limt→0

at3

t4 + (at)2= lim

t→0

at

t2 + a2= 0, and

limt→0

f(γ∞(t)) = limt→0

0

0 + t2= 0,

To see the other part, approach the origin along a parabolic curve: let γ(t) =(x(t), y(t)) = (t, t2). Then

limt→0

f(γ(t)) = limt→0

t2t2

t4 + (t2)2= lim

t→0

t4

2t4= lim

t→0

1

2=

1

26= 0.

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15. Define f(x, y) = x2 + y2 − 2x2y − 4x6y2

(x4+y2)2.

-2-101

2

-2-1

01

2

-10

0

10

20

-10

0

10

(a) Prove that 4x4y2 ≤ (x4 + y2)2, so that f ∈ C0.

(x4 + y2)2 − 4x4y2 = x8 + 2x4y2 + y4 − 4x4y2

= (x4 − y2)2 ≥ 0

Now f(x, y) = x2 + y2 − 2x2y − x2ϕ(x, y), where 0 ≤ ϕ(x, y) ≤ 1. Thisgives f continuous at (0, 0).

(b) For 0 ≤ θ < π, t ∈ R, let gθ(t) = f(t cos θ, t sin θ). Show that gθ(0) = 0,g′θ(0) = 0, g′′θ (0) = 2. Each gθ thus has a strict local minimum at t = 0,i.e., on each line through (0, 0), f has a strict local minimum at (0, 0).

Since sin θ, cos θ are bounded, gθ(0) = f(0, 0) = 0 by (a). By somenightmarishly tortuous (but elementary) calculation, get the other tworesults.

(c) Show that (0, 0) is not a local minimum of f , since f(x, x2) = −x4.

Substituting in y = x2:

f(x, x2) = x2 + x4 − 2x4 − 4x10

(2x4)2

= x2 − x4 − x2

= −x4.

16. Define f(t) = t+2t2 sin(

1t

), f(0) = 0. Then f ′(0) = 1, f ′ bounded in (−1, 1),

but f is not injective in any nbd of 0.

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-0.2 -0.1 0.1 0.2

-0.2

-0.1

0.1

0.2

First note that f ′(t) = 1 − 2 cos(

1t

)+ 4t sin

(1t

)for t 6= 0. Since sin t, cos t

are bounded by 1, we have

|f ′(t)| ≤ 1 + 4 + 2 = 7, for t ∈ (−1, 1) \ {0}.At t = 0, we have that

f(t)− f(0)

t=

t + 2t2 sin 1t

t= 1 + 2t sin 1

t

shows f ′(0) = 1 by the Sandwich theorem applied to t + 2t2 and t − 2t2. Sof ′ is bounded in all of (−1, 1).

Define sn = 12nπ

for n = 1, 2, . . . , so that

f ′(sn) = 1− 2 cos(2nπ) + 2sin 2nπ

nπ= 1− 2 = −1.

Then define tn = 2(4n+3)π

for n = 1, 2, . . . , so that

f ′(tn) = 1− 2 cos (4n+3)π2

+ 8(4n+3)π

sin (4n+3)π2

= 1 + 8(4n+3)π

> 0.

Since f ′ changes sign between each successive sn and tn, and since sn, tn → 0,f fails to be injective in every neighbourhood of 0.

17. Let f = (f1, f2) : R2 → R2 be given by

f1(x, y) = ex cos y, f2(x, y) = ex sin y.

(a) What is the range of f?

Note that if we identify (0, 1) =√−1 = i, then

f(z) = f(x + iy) = ez = ex+iy = ex cos y + iex sin y.

But you only really need to see that (ex cos y, ex sin y) is polar coordinatesfor a point with radius ex and argument y, to see that the range is anypoint of R2 with radius r > 0, i.e., R2 \ (0, 0).

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(b) Show that the Jacobian of f is nonzero. Thus, every point of R2 has aneighbourhood in which f is injective. However, f is not injective.

The Jacobian is the determinant of partials:∣∣∣∣ex cos y −ex sin yex sin y ex cos x

∣∣∣∣ = e2x(cos2 x + sin2 x) = e2x 6= 0.

But f is not injective, since f(x, y) = f(x, y + 2nπ), for n ∈ Z.

(c) Put a = (0, π3) and b = f(a). Let g be the continuous inverse of f ,

defined in a neighbourhood of b, such that g(b) = a. Find an explicitformula for g, compute f ′(a) and g′(b), and verify Rudin’s (52).

For u = ex cos y, v = ex sin y, one verifies that

x = 12log(u2 + v2) = log r, for r = ex, and

y = tan−1 vu

= arg θ, for θ = vu.

For the derivatives,

f ′(a) =

[12

−√

32√

32

12

],

and

g′(u, v) = 1u2+v2

[u v

−v u

]=⇒ g′(b) =

[12

√3

2

−√

32

12

],

Finally,

f ′(g(u, v)) = f ′(log

√u2 + v2, tan−1 v

u

)

=√

u2 + v2

[cos

(tan−1 v

u

) − sin(tan−1 v

u

)

sin(tan−1 v

u

)cos

(tan−1 v

u

)]

=√

u2 + v2

[ u√u2+v2 − v√

u2+v2

v√u2+v2 − u√

u2+v2

]

=

[u −vv u

]

= [g′(u, v)]−1

(d) What are the images under f of lines parallel to the coordinate axes?

Lines parallel to the x-axis are mapped to straight lines through theorigin, parameterized exponentially. Lines parallel to the y-axis are cir-cles about the origin of radius ex, parameterized with constant speed.Diagonal lines γ(t) = (at + b, btc) will get mapped to (et cos t, et sin t),counterclockwise logarithmic spirals emanating from the origin.

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Rudin Chap9 Some Solutions

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Transcript of Rudin Chap9 Some Solutions