Some Notes on Rudin’s book: Principles of Mathematical ... · Some Notes on Rudin’s book: Principles of Mathematical Analysis, 3/e Written by Meng-Gen, Tsai email: [email protected]

Some Notes on Rudin’s book:

Principles of Mathematical Analysis, 3/e

Written by Meng-Gen, Tsai

email: [email protected]

Exercise 1.1. If r is rational (r 6= 0) and x is irrational, prove that r + x

and rx are irrational.

Proof. If r + x is rational, then x = (r + x) − r is also rational, a contra-

diction. Also, if rx is rational, then x = rxr

is also rational since r 6= 0, a

contradiction. �

Exercise 1.2. Prove that there is no rational number whose square is 12.

Proof. Let f(x) = x2 − 12 ∈ Q[x]. Hence all possible rational roots are

1,−1, 2,−2, 3,−3, 4,−4, 6,−6, 12,−12.

But none of them satisfy f(x) = 0. Hence no rational number whose square

is 12. (If so, then my proof implies that f(x) = x2−12 has a rational root.) �

Exercise 1.3. Prove Proposition 1.15: The axioms for multiplication imply

the following statements.

(a) If x 6= 0 and xy = xz then y = z.

(b) If x 6= 0 and xy = x then y = 1.

(c) If x 6= 0 and xy = 1 then y = 1/x.

(d) If x 6= 0 then 1/(1/x) = x.

1

Proof. We prove (a). By (M3), we have y(M4)= 1y

(M5)= x(1/x)y

(M2)=

(1/x)xy = (1/x)xz(M5)= 1z

(M4)= z.

We prove (b). xy = x(M4)= 1x

(M2)= x1. By part (a), y = 1.

We prove (c). xy = 1(M5)= x(1/x). By part (a), y = 1/x.

We prove (d). x(1/x)(M5)= 1. By part (c), 1/(1/x) = x. �

Exercise 1.4. α is a lower bound of E ⇒ x ≥ α for all x ∈ E. Since E

is nonempty, we can pick an element x0 ∈ E. β is a upper bound of E ⇒x0 ≤ β. Hence

α ≤ x0 ≤ β.

So α ≤ β. �

Exercise 1.5. If A is a set of real numbers which is bounded below, then

inf A is a lower bound, i.e.,

inf A ≤ x

for all x ∈ A and if inf A ≥ b for any other lower bound b of A. Now if A is

bounded below, then −A is bounded above. In fact, if b ≤ x for all x ∈ A,

then −b ≥ −x for all x ∈ A which means −b ≥ y for all y ∈ −A. Now if

sup(−A) is the least upper bound of −A it follows that − sup(−A) is a lower

bound of A. In fact,

x ∈ A ⇒ −x ∈ −A ⇒ sup(−A) ≥ −x ⇒ − sup(−A) ≤ x.

As noted above, if b is any lower bound for A then −b is an upper bound for

−A so −b ≥ sup(−A) and b ≤ − sup(−A). This is the definition of inf A so

inf A = − sup(−A).

�

Exercise 1.6. We prove (a). mq = np since m/n = p/q. Thus bmq = bnp.

By Theorem 1.21 we know that (bmq)1/(mn) = (bnp)1/(mn), that is, (bm)1/n =

(bp)1/q, that is, br is well-defined.

2

We prove (b). Let r = m/n and s = p/q where m, n, p, q are integers, and

n > 0, q > 0. Hence (br+s)nq = (bm/n+p/q)nq = (b(mq+np)/(nq))nq = bmq+np =

bmqbnp = (bm/n)nq(bp/q)nq = (bm/nbp/q)nq. By Theorem 1.21 we know that

((br+s)nq)1/(nq) = ((bm/nbp/q)nq)1/(nq), that is br+s = bm/nbp/q = brbs.

We prove (c). Note that br ∈ B(r). For all bt ∈ B(r) where t is rational

and t ≤ r. Hence, br = btbr−t ≥ bt1r−t since b > 1 and r − t ≥ 0. Hence br is

an upper bound of B(r). Hence br = sup B(r).

We prove (d). bxby = sup B(x) sup B(y) ≥ btxbty = btx+ty for all rational

tx ≤ x and ty ≤ y. Note that tx + ty ≤ x + y and tx + ty is rational.

Therefore, sup B(x) sup B(y) is a upper bound of B(x + y), that is, bxby ≥sup B(x + y) = bx+y. Conversely, we claim that bxbr = bx+r if x ∈ R and

r ∈ Q. The following is my proof.

bx+r = sup B(x + r) = sup{bs|s ≤ x + r, s ∈ Q}= sup{bs−rbr|s − r ≤ x, s − r ∈ Q}= br sup{bs−r|s − r ≤ x, s − r ∈ Q}= br sup B(x) = brbx.

And we also claim that bx+y ≥ bx if y ≥ 0. The following is my proof: (r ∈ Q)

B(x) = {br|r ≤ x} ⊂ {br|r ≤ x + y} = B(x + y),

Therefore, sup B(x + y) ≥ sup B(x), that is, bx+y ≥ bx. Hence,

bx+y = sup B(x + y)

= sup{br|r ≤ x + y, r ∈ Q}= sup{bsbr−s|r ≤ x + y, s ≤ x, r, s ∈ Q}≥ sup{sup B(x)br−s|r ≤ x + y, s ≤ x, r, s ∈ Q}= sup B(x) sup{br−s|r ≤ x + y, s ≤ x, r, s ∈ Q}= sup B(x) sup{br−s|r − s ≤ x + y − s, s ≤ x, r − s ∈ Q}= sup B(x) sup B(x + y − s)

≥ sup B(x) sup B(y) = bxby.

Therefore, bx+y = bxby. �

3

Exercise 1.8. If we can define an other on the complex field, then one and

only one of the statements

0 < i, 0 = i, 0 > i.

is true. First, 0 = i is impossible. (0 = i ⇒ 02 = i2, 0 = 1, a contradiction).

Second, suppose 0 < i. Then

0 < i, and 0 i. Hence 0 < −i. Then

0 < −i, and 0 < −i ⇒ 0 < (−i)2 = i2 = −1,

a contradiction again. Hence no order can be defined in the complex field

that turns it into an ordered field. �

Exercise 1.13. By Theorem 1.33, |x|+ |x− y| ≥ |y| and |y|+ |x− y| ≥ |x|.Thus |x − y| ≥ ||x| − |y||. �

Exercise 1.14.

|1 + z|2 + |1 − z|2 = (1 + z)(1 + z) + (1 − z)(1 − z)

= (1 + z)(1 + z) + (1 − z)(1 − z)

= (1 + z + z + zz) + (1 − z − z + zz)

= 2 + 2zz = 2 + 2 = 4.

�

Exercise 1.15. In the proof of Theorem 1.35, equality holds if and only if

B = 0 or∑ |Baj − Cbj |2 = 0, that is,

b1 = · · · = bn = 0

or

aj = rbj

4

for all 1 ≤ j ≤ n, and for some r ∈ R. �

Exercise 1.17. |x + y|2 + |x− y|2 = (x + y) · (x + y) + (x− y) · (x− y) =

(x ·x+2x ·y +y ·y)+ (x ·x− 2x ·y +y ·y) = 2x ·x+2y ·y = 2|x|2 +2|y|2.�

Exercise 1.18. Write x = (x1, · · · , xk) ∈ Rk. If the i-th coordinate xi of x

is zero for some 1 ≤ i ≤ k, we can pick

y = (0, · · · , 1︸︷︷︸

i -th position

, · · · , 0) ∈ Rk.

If all coordinates of x are nonzero, we can pick

y = (1,−x1/x2, 0, · · ·0).

It is possible since k ≥ 2. Thus we have x · y = 0. It is not true if k = 1

since 1 is not a zero-divisor in R1. �

Exercise 2.1. For any element x of the empty set, x is also an element of ev-

ery set since x does not exist. Hence, the empty set is a subset of every set. �

Exercise 2.2. For every positive integer N there are only finitely many

equations with

n + |a0| + |a1| + ... + |an| = N.

(since 1 ≤ n ≤ N and 0 ≤ |a0| ≤ N). We collect those equations as CN .

Hence⋃

CN is countable. For each algebraic number, we can form an equa-

tion and this equation lies in CM for some M and thus the set of all algebraic

numbers is countable. �

Exercise 2.3. If not, R = { all algebraic numbers } is countable, a contra-

diction. �

Exercise 2.4. If R − Q is countable, then R = (R − Q)⋃

Q is countable, a

contradiction. Thus R − Q is uncountable. �

5

Exercise 2.5. Put

A = {1/n : n ∈ N}⋃

{1 + 1/n : n ∈ N}⋃

{2 + 1/n : n ∈ N}.

A is bounded by 3, and A contains three limit points −0, 1, 2. �

Exercise 2.6. For any point p of X − E ′, that is, p is not a limit point E,

there exists a neighborhood of p such that q is not in E with q 6= p for every

q in that neighborhood.

Hence, p is an interior point of X − E ′, that is, X − E ′ is open, that is,

E ′ is closed.

Next, if p is a limit point of E, then p is also a limit point of E since

E = E⋃

E ′. If p is a limit point of E, then every neighborhood Nr(p) of p

contains a point q 6= p such that q ∈ E. If q ∈ E, we completed the proof. So

we suppose that q ∈ E − E = E ′ − E. Then q is a limit point of E. Hence,

Nr′(q)

where r′ = 12min(r − d(p, q), d(p, q)) is a neighborhood of q

and contains a point x 6= q such that x ∈ E. Note that Nr′(q) contains

in Nr(p)− {p}. That is, x 6= p and x is in Nr(p). Hence, q also a limit point

of E. Hence, E and E have the same limit points.

Last, the answer of the final sub-problem is no. Put

E = {1/n : n ∈ N},

and E ′ = {0} and (E ′)′ = ∅. �

Exercise 2.7. We proof (a). Method 1. Bn is the smallest closed subset of

X that contains Bn. Note that⋃

Ai is a closed subset of X that contains

Bn, thus

Bn ⊃n⋃

i=1

Ai.

If p ∈ Bn − Bn, then every neighborhood of p contained a point q 6= p

such that q ∈ Bn. If p is not in Ai for all i, then there exists some

6

neighborhood Nri(p) of p such that (Nri

(p) − p)⋂

Ai = φ for all i. Take

r = min{r1, r2, · · · , rn}, and we have Nr(p)⋂

Bn = φ, a contradiction. Thus

p ∈ Ai for some i. Hence

Bn ⊂n⋃

i=1

Ai.

that is,

Bn =

n⋃

i=1

Ai.

Method 2. Since⋃n

i=1 Ai is closed and Bn =⋃n

i=1 Ai ⊂ ⋃ni=1 Ai, Bn ⊂

⋃ni=1 Ai.

We prove (b). Since B is closed and B ⊃ B ⊃ Ai, B ⊃ Ai for all i. Hence

B ⊃ ⋃Ai.

Note: My example is Ai = (1/i,∞) for all i. Thus, Ai = [1/i,∞), and

B = (0,∞), B = [0,∞). Note that 0 is not in Ai for all i. Thus this inclusion

can be proper. �

Exercise 2.8. For the first part of this problem, the answer is yes.

Reason. For every point p of E, p is an interior point of E. That is, there

is a neighborhood Nr(p) of p such that Nr(p) is a subset of E. Then for every

real r′, we can choose a point q such that d(p, q) = 1/2 min(r, r′). Note that

q 6= p, q ∈ Nr′(p), and q ∈ Nr(p). Hence, every neighborhood Nr′(p) contains

a point q 6= p such that q ∈ Nr(p) ⊂ E, that is, p is a limit points of E.

For the last part of this problem, the answer is no. Consider A = {(0, 0)}.A′ = φ and thus (0, 0) is not a limit point of E. �

Exercise 2.9. We prove (a). If E is non-empty, take p ∈ E◦. We need to

show that p ∈ (E◦)◦. Since p ∈ E◦, there is a neighborhood Nr of p such

that Nr is contained in E. For each q ∈ Nr, note that Ns(q) is contained in

Nr(p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interior point

of E, that is, Nr is contained in E◦. Hence E◦ is always open.

We prove (b). (⇒) It is clear that E◦ is contained in E. Since E is

open, every point of E is an interior point of E, that is, E is contained in

7

E◦. Therefore E◦ = E. (⇐) Since every point of E is an interior point of E

(E◦(E) = E), E is open.

We prove (c). If p ∈ G, p is an interior point of G since G is open. Note

that E contains G, and thus p is also an interior point of E. Hence p ∈ E◦.

Therefore G is contained in E◦. (Thus E◦ is the biggest open set contained

in E. Similarly, E is the smallest closed set containing E.)

We prove (d). Suppose p ∈ X − E◦. If p ∈ X − E, then p ∈ X − E

clearly. If p ∈ E, then N is not contained in E for any neighborhood N of p.

Thus N contains an point q ∈ X − E. Note that q 6= p, that is, p is a limit

point of X − E. Hence X − E◦ is contained in X − E.

Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − E◦ clearly.

If p ∈ E, then every neighborhood of p contains a point q 6= p such that

q ∈ X −E. Hence p is not an interior point of E. Hence X − E is contained

in X − E◦. Therefore X − E◦ = X − E.

For (e), the answer is no. Take X = R and E = Q. Thus E◦ = ∅ and

E◦

= (R)◦ = R 6= ∅.For (f), the answer is no. Take X = R and E = Q. Thus E = R, and

E◦ = ∅ = ∅. �

Exercise 2.10. (a) d(p, q) = 1 > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) = d(q, p)

since p = q implies q = p and p 6= q implies q 6= p. (c) d(p, q) ≤ d(p, r)+d(r, q)

for any r ∈ X if p = q. If p 6= q, then either r = p or r = q, that is, r 6= p or

r 6= q. Thus, d(p, q) = 1 ≤ d(p, r) + d(r, q). By (a)-(c) we know that d is a

metric.

Every subset of X is open and closed. We claim that for any p ∈ X,

p is not a limit point. Since d(p, q) = 1 > 1/2 if q 6= p, there exists an

neighborhood N1/2(p) of p contains no points of q 6= p such that q ∈ X.

Hence every subset of X contains no limit points and thus it is closed. Since

X − S is closed for every subset S of X, S = X − (X − S) is open. Hence

every subset of X is open.

Every finite subset of X is compact. Let S = {p1, ..., pn} be finite. Con-

sider an open cover {Gα} of S. Since S is covered by Gα, pi is covered by

Gαi, thus {Gα1

, ..., Gαn} is finite subcover of S. Hence S is compact. Next,

8

suppose S is infinite. Consider an open cover {Gp} of S, where

Gp = N 1

2

(p)

for every p ∈ S. Note that q is not in Gp if q 6= p. If S is compact, then S

can be covered by finite subcover, say

Gp1, ..., Gpn.

Then there exists q such that q 6= pi for all i since S is infinite, a contradic-

tion. Hence only every finite subset of X is compact. �

Exercise 2.11. (1) d1(x, y) is not a metric. Since d1(0, 2) = 4, d1(0, 1) = 1,

and d1(1, 2) = 1, d1(0, 2) > d1(0, 1) + d1(1, 2). Thus d1(x, y) is not a metric.

(2) d2(x, y) is a metric. (a) d(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) =√

|p − q| =√

|q − p| = d(q, p). (c) |p − q| ≤ |p − r| + |r − q|,√

|p − q| ≤√

|p − r| + |r − q| ≤√

|p − r|+√

|r − q|. That is, d(p, q) ≤ d(p, r)+ d(r, q).

(3) d3(x, y) is not a metric since d3(1,−1) = 0.

(4) d4(x, y) is not a metric since d4(1, 1) = 1 6= 0.

(5) d5(x, y) is a metric since |x − y| is a metric.

Claim: d(x, y) is a metric, then d′(x, y) = d(x,y)1+d(x,y)

is also a metric.

(a) d′(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d′(p, q) = d′(q, p). (c) Let

x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y + z.

d′(p, q) ≤ d′(p, r) + d′(r, q)

⇔ x

1 + x≤ y

1 + y+

z

1 + z

⇔ x(1 + y)(1 + z) ≤ y(1 + z)(1 + x) + z(1 + x)(1 + y)

⇔ x + xy + xz + xyz ≤ (y + xy + yz + xyz) + (z + xz + yz + xyz)

⇔ x ≤ y + z + 2yz + xyz

⇐ x ≤ y + z

Thus, d′ is also a metric. �

9

Exercise 2.12. Suppose that {Oα} is an arbitrary open covering of K. Let

E ∈ {Oα} consists 0. Since E is open and 0 ∈ E, 0 is an interior point of

E. Thus there is a neighborhood N = Nr(0) of 0 such that N ⊂ E. Thus N

contains1

[1/r] + 1,

1

[1/r] + 2, · · ·

Next, we take finitely many open sets En ∈ {Oα} such that 1/n ∈ En

for n = 1, 2, · · · , [1/r]. Hence {E, E1, ..., E[1/r]} is a finite subcover of K.

Therefore, K is compact.

Note: The unique limit point of K is 0. Suppose p 6= 0 is a limit point of

K. Clearly, 0 < p < 1. Thus there exists n ∈ Z+ such that

1

n + 1< p <

1

n.

Hence Nr(p) where r = min{

1n− p, p − 1

n+1

}

contains no points of K, a

contradiction. �

Exercise 2.13. Let K be consist of 0 and the numbers 1/n for n = 1, 2, 3, · · ·Let xK = {xk : k ∈ K} and x + K = {x + k|k ∈ K} for x ∈ R. We take

Sn = (1 − 1

2n) +

K

2n+1,

S =

∞⋃

n=1

Sn

⋃

{1}.

Claim: S is compact and the set of all limit points of S is K⋃{1}. Clearly,

S lies in [0, 1], that is, S is bounded in R. Note that Sn ⊂[

1− 12n , 1− 1

2n+1

]

.

By Exercise 12 and its note, we have that all limit points of S⋂

[0, 1) is

0,1

2, · · · ,

1

2n, · · ·

Clearly, 1 is also a limit point of S. Therefore, the set of all limit points of

S is K⋃{1}. Note that K

⋃{1} ⊂ S, that is, K is compact. We completed

the proof of our claim. �

10

Exercise 2.14. Take {On} = {(1/n, 1)} for n = 1, 2, 3, · · · . For every

x ∈ (0, 1),

x ∈(

1

[1/x] + 1, 0

)

∈ {On}

Hence {On} is an open covering of (0, 1). Suppose there exists a finite sub-

cover (1

n1

, 1

)

, · · · ,

(1

nk

, 1

)

where n1 < n2 < · · · < nk, respectively. Clearly 12np

∈ (0, 1) is not in any

elements of that subcover, a contradiction.

Note: By the above we know that (0, 1) is not compact. �

Exercise 2.15. For closed: [n,∞). For bounded: (−1/n, 1/n) − {0}. �

Exercise 2.16. Let S = (√

2,√

3)⋃

(−√

3,−√

2). Then E = {p ∈ Q : p ∈S}. Clearly, E is bounded in Q. Since Q is dense in R, every limit point of

Q is in Q. (We regard Q as a metric space). Hence, E is closed in Q.

To prove that E is not compact, we form a open covering of E as follows:

{Gα} = {Nr(p)|p ∈ E and (p − r, p + r) ⊂ S}.

Surely, {Gα} is a open covering of E. If E is compact, then there are finitely

many indices α1, · · · , αn such that

E ⊂ Gα1

⋃

· · ·⋃

Gαn .

For every Gαi= Nri

(pi), take p = max1≤i≤n pi. Thus, p is the nearest point

to√

3. But Nr(p) lies in E, thus [p + r,√

3) cannot be covered since Q is

dense in R, a contradiction. Hence E is not compact.

Finally, the answer is yes. Take any p ∈ Q, then there exists a neighbor-

hood N(p) of p contained in E. (Take r small enough where Nr(p) = N(p),

and Q is dense in R.) Thus every point in N(p) is also in Q. Hence E is also

open. �

11

Exercise 2.17. Let

E ={ ∞∑

n=1

an

10n|an = 4 or an = 7

}

.

Claim 1. E is uncountable. If not, we list E as follows:

x1 = 0.a11a12 · · ·a1n · · ·x2 = 0.a21a22 · · ·a2n · · ·· · · · · ·xk = 0.ak1ak2 · · ·akn · · ·· · · · · ·

(Prevent ending with all digits 9) Let x = 0.x1x2 · · ·xn · · · where

xn =

{

4 if ann = 7

7 if ann = 4

By my construction, x /∈ E, a contradiction. Thus E is uncountable.

Claim 2. E is not dense in [0, 1]. E⋂

(0.47, 0.74) = ∅.Claim 3. E is compact. Clearly, E is bounded. For every limit point p

of E, we will show that p ∈ E. If not, write the decimal expansion of p as

follows

p = 0.p1p2 · · · pn · · ·

Since p /∈ E, there exists the smallest k such that pk 6= 4 and pk 6= 7. When

pk = 0, 1, 2, 3, select the smallest l such that pl = 7 if possible. (If l does not

exist, then p < 0.4. Thus there is a neighborhood of p such that contains no

points of E, a contradiction.) Thus

0.p1 · · · pl−14pl+1 · · · pk−17 < p < 0.p1 · · ·pk−14.

Thus there is a neighborhood of p such that contains no points of E, a

contradiction. When pk = 5, 6,

0.p1 · · · pk−147 < p < 0.p1 · · · pk−174.

12

Thus there is a neighborhood of p such that contains no points of E, a

contradiction. When pk = 8, 9, it is similar. Hence E is closed. Therefore E

is compact.

Claim. E is perfect. Take any p ∈ E, and I claim that p is a limit point

of E. Write p = 0.p1p2 · · · pn · · · Let

xk = 0.y1y2 · · · yn · · ·

where

yn =

pk if k 6= n

4 if pn = 7

7 if pn = 4

Thus, |xk −p| → 0 as k → ∞. Also, xk 6= p for all k. Hence p is a limit point

of E. Therefore E is perfect. �

Exercise 2.18. Yes. The following claim will show the reason.

Claim. Given a measure zero set S, we have a perfect set P contains no

elements in S.

Since S has measure zero, there exists a collection of open intervals {In}such that

S ⊂⋃

In and∑

|In| < 1.

Consider E = R −⋃In. E is nonempty since E has positive measure. Thus

E is uncountable and E is closed. Therefore there exists a nonempty perfect

set P contained in E by Exercise 28. P⋂

S = ∅. Thus P is our required

perfect set. �

Exercise 2.19. For (a), we recall the definition of separated: A and B

are separated if A⋂

B and A⋂

B are empty. Since A and B are closed sets,

A = A and B = B. Hence A⋂

B = A⋂

B = A⋂

B = ∅. Hence A and B

are separated.

For (b), Suppose A⋂

B is not empty. Thus we can pick p ∈ A⋂

B. For

p ∈ A, there exists a neighborhood Nr(p) of p contained in A since A is open.

For p ∈ B = B⋃

B′, if p ∈ B, then p ∈ A⋂

B. Note that A and B are

disjoint, and it’s a contradiction. If p ∈ B′, then p is a limit point of B. Thus

13

every neighborhood of p contains a point q 6= p such that q ∈ B. Take an

neighborhood Nr(p) of p containing a point q 6= p such that q ∈ B. Note that

Nr(p) ⊂ A, thus q ∈ A. With A and B are disjoint, we get a contradiction.

Hence A⋂

B is empty. Similarly, A⋂

B is also empty. Thus A and B are

separated.

For (c), suppose A⋂

B is not empty. Thus there exists x such that x ∈ A

and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = B⋃

B′, thus if x ∈ B,

then d(p, x) > δ, a contradiction. The only possible is x is a limit point of

B. Hence we take a neighborhood Nr(x) of x contains y with y ∈ B where

r = δ−d(x,p)2

. Clearly, d(y, p) > δ. But,

d(y, p) ≤ d(y, x) + d(x, p) < r + d(x, p) =δ − d(x, p)

2+ d(x, p)

=δ + d(x, p)

2<

δ + δ

2= δ,

and it is absurd. Hence A⋂

B is empty. Similarly, A⋂

B is also empty.

Thus A and B are separated.

Note: Take care of δ > 0. Think a while and you can prove the next

sub-exercise.

For (d), let X be a connected metric space. Take p ∈ X, q ∈ X with

p 6= q, thus d(p, q) > 0 is fixed. Let

A = {x ∈ X|d(x, p) < δ} and B = {x ∈ X|d(x, p) > δ}.

Take δ = δt = td(p, q) where t ∈ (0, 1). Thus 0 < δ < d(p, q). p ∈ A since

d(p, p) = 0 < δ, and q ∈ B since d(p, q) > δ. Thus A and B are non-empty.

By (c), A and B are separated. If X = A⋃

B, then X is not connected,

a contradiction. Thus there exists yt ∈ X such that y /∈ A⋃

B. Let

E = Et = {x ∈ X|d(x, p) = δt} ∋ yt.

For any real t ∈ (0, 1), Et is nonempty. Next, Et and Es are disjoint if

t 6= s (since a metric is well-defined). Thus X contains a uncountable set

{yt|t ∈ (0, 1)} since (0, 1) is uncountable. Therefore, X is uncountable. Note:

It is a good exercise. If that metric space contains only one point, then it

must be separated.

14

Supplement exercises given by Lee Sin-Yi. (a) Let A = {x|d(p, x) < r}and B = {x|d(p, x) > r} for some p in a metric space X. Show that A, B

are separated.

(b) Show that a connected metric space with at least two points must be

uncountable. [Hint: Use (a)] For (a), by definition of separated sets, we want

to show A⋂

B = φ, and B⋂

A = φ. In order to do these, it is sufficient to

show A⋂

B = φ. Let x ∈ A⋂

B = φ, then we have:

(1) x ∈ A ⇒ d(x, p) ≤ r; (2) x ∈ B ⇒ d(x, p) > r

It is impossible. So, A⋂

B = φ.

For (b), suppose that C is countable, say C = {a, b, x3, · · · }. We want to

show C is disconnected. So, if C is a connected metric space with at least

two points, it must be uncountable. Consider the set S = {d(a, xi)|xi ∈ C},and thus let r ∈ R − S and inf S < r < sup S. And construct A and B

as in (a), we have C = A⋃

B, where A and B are separated. That is C

is disconnected. Another proof of (b). Let a ∈ C, b ∈ C, consider the con-

tinuous function f from C into R defined by f(x) = d(x, a). So, f(C) is

connected and f(a) = 0, f(b) > 0. That is, f(C) is an interval. Therefore,

C is uncountable. �

Exercise 2.20. Closures of connected sets is always connected, but interiors

of those is not. The counterexample is

S = N1(2, 0)⋃

N1(−2, 0)⋃

{x-axis} ⊂ R2.

Since S is path-connected, S is connect. But S◦ = N1(2)⋃

N1(−2) is dis-

connected clearly.

Claim. If S is a connected subset of a metric space, then S is connected.

If not, then S is a union of two nonempty separated set A and B. Thus

A⋂

B = A⋂

B = ∅. Note that

S = S − T = A ∪ B − T = (A ∪ B) ∩ T c = (A ∩ T c) ∪ (B ∩ T c)

where T = S − S. Thus

(A ∩ T c) ∩ B ∩ T c ⊂ (A ∩ T c) ∩ B ∩ T c ⊂ A ∩ B = ∅.

15

Hence (A ∩ T c) ∩ B ∩ T c = ∅. Similarly, A ∩ T c ∩ (B ∩ T c) = ∅.Now we claim that both A∩ T c and B ∩ T c are nonempty. Suppose that

B ∩ T c = ∅. Thus

A ∩ T c = S ⇔ A ∩ (S − S)c = S

⇔ A ∩ (A ∪ B − S)c = S

⇔ A ∩ ((A ∪ B) ∩ Sc)c = S

⇔ A ∩ ((Ac ∩ Bc) ∪ S) = S

⇔ (A ∩ S) ∪ (A ∩ Ac ∩ Bc) = S

⇔ A ∩ S = S.

Thus B is empty, and it is absurd. Thus B ∩ T c is nonempty. Similarly,

A∩ T c nonempty. Therefore S is a union of two nonempty separated sets, a

contradiction. Hence S is connected. �

Exercise 2.21. For (a), we claim that A0

⋂B0 is empty. (B0

⋂A0 is sim-

ilar). If not, take x ∈ A0

⋂B0. x ∈ A0 and x ∈ B0. x ∈ B0 or x is a

limit point of B0. x ∈ B0 will make x ∈ A0

⋂B0, that is, p(x) ∈ A

⋂B, a

contradiction since A and B are separated.

Claim: x is a limit point of B0 ⇒ p(x) is a limit point of B. Take

any neighborhood Nr of p(x), and p(t) lies in B for small enough t. More

precisely,

x − r

|b− a| < t < x +r

|b− a| .

Since x is a limit point of B0, and (x−r/|b−a|, x+r/|b−a|) is a neighborhood

N of x, thus N contains a point y 6= x such that y ∈ B0, that is, p(y) ∈ B.

Also, p(y) ∈ Nr. Therefore, p(x) is a limit point of B. Hence p(x) ∈ A⋂

B,

a contradiction since A and B are separated. Hence A0

⋂B0 is empty, that

is, A0 and B0 are separated subsets of R.

We prove (b). Suppose not. For every t0 ∈ (0, 1), neither p(t0) ∈ A

nor p(t0) ∈ B (since A and B are separated.) Also, p(t0) ∈ A⋃

B for all

t0 ∈ (0, 1). Hence (0, 1) = A0

⋃B0, a contradiction since (0, 1) is connected.

For (c), let S be a convex subset of Rk. If S is not connected, then S

is a union of two nonempty separated sets A and B. By (b), there exists

16

t0 ∈ (0, 1) such that p(t0) /∈ A∪B. But S is convex, p(t0) must lie in A∪B,

a contradiction. Hence S is connected. �

Exercise 2.22. Consider S = the set of points which have only rational

coordinates. For any point x = (x1, x2, · · · , xk) ∈ Rk, we can find a rational

sequence {rij} → xj for j = 1, · · · , k since Q is dense in R. Thus,

ri = (ri1, ri2 , · · · , rik) → x

and ri ∈ S for all i. Hence S is dense in Rk. Also, S is countable, that is, S

is a countable dense subset in Rk, Rk is separable. �

Exercise 2.23. Let X be a separable metric space, and S be a countable

dense subset of X. Let a collection {Vα} = { all neighborhoods with rational

radius and center in S }. We claim that {Vα} is a base for X.

For every x ∈ X and every open set G ⊂ X such that x ∈ G, there exists

a neighborhood Nr(p) of p such that Nr(p) ⊂ G since x is an interior point

of G. Since S is dense in X, there exists {sn} → x. Take a rational number

rn such that rn < r2, and {Vα} ∋ Nrn(sn) ⊂ Nr(p) for enough large n. Hence

we have x ∈ Vα ⊂ G for some α. Hence {Vα} is a base for X. �

Exercise 2.24. Fix δ > 0, and pick x1 ∈ X. Having chosen x1, · · · , xj ∈ X,

choose xj+1, if possible, so that d(xi, xj+1) ≥ δ for i = 1, · · · , j. If this process

cannot stop, then consider the set A = {x1, x2, · · · , xk}. If p is a limit point

of A, then a neighborhood Nδ/3(p) of p contains a point q 6= p such that

q ∈ A. q = xk for only one k ∈ N. If not, d(xi, xj) ≤ d(xi, p) + d(xj, p) ≤δ/3+ δ/3 < δ, and it contradicts the fact that d(xi, xj) ≥ δ for i 6= j. Hence,

this process must stop after finite number of steps.

Suppose this process stop after k steps, and X is covered by Nδ(x1),

Nδ(x2), · · · , Nδ(xk), that is, X can therefore be covered by finite many neigh-

borhoods of radius δ.

Take δ = 1/n(n = 1, 2, 3, · · · ), and consider the set A of the centers of

the corresponding neighborhoods.

Fix p ∈ X. Suppose that p is not in A, and every neighborhood Nr(p).

17

Note that Nr/2(p) can be covered by finite many neighborhoods Ns(x1), · · · , Ns(xk)

of radius s = 1/n where n = [2/r] + 1 and xi ∈ A for i = 1, · · · , k. Hence,

d(x1, p) ≤ d(x1, q)+d(q, p) ≤ r/2+s < r where q ∈ Nr/2(p)∩Ns(x1). There-

fore, x1 ∈ Nr(p) and x1 6= p since p is not in A. Hence, p is a limit point of A if

p is not in A, that is, A is a countable dense subset, that is, X is separable. �

Exercise 2.25. For every positive integer n, there are finitely many neigh-

borhood of radius 1/n whose union covers K (since K is compact). Collect

all of them, say {Vα}, and it forms a countable collection. We claim that

{Vα} is a base.

For every x ∈ X and every open set G ⊂ X, there exists Nr(x) such that

Nr(x) ⊂ G since x is an interior point of G. Hence x ∈ Nm(p) ∈ {Vα} for

some p where m = [2/r] + 1. For every y ∈ Nm(p), we have

d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.

Hence Nm(p) ⊂ G, that is, Vα ⊂ G for some α, and therefore {Vα} is a

countable base of K. Next, collect all of the center of Vα, say D, and we

claim that D is dense in K. (D is countable since Vα is countable.) For all

p ∈ K and any ǫ > 0 we can find Nn(xn) ∈ {Vα} where n = [1/ǫ] + 1. Note

that xn ∈ D for all n and d(p, xn) → 0 as n → ∞. Hence D is dense in K.

�

Exercise 2.26. By Exercises 2.23 and 2.24, X has a countable base. It

follows that every open cover of X has a countable subcover {Gn}, n =

1, 2, 3, · · · . If no finite subcollection of {Gn} covers X, then the complement

Fn of G1

⋃ · · ·⋃ Gn is nonempty for each n, but⋂

Fn is empty. If E is a set

contains a point from each Fn, consider a limit point of E.

Note that Fk ⊃ Fk+1 ⊃ · · · and Fn is closed for all n, thus p lies in Fk for

all k. Hence p lies in⋂

Fn, but⋂

Fn is empty, a contradiction. �

Exercise 2.27. Let {Vn} be a countable base of Rk, let W be the union

of those Vn for which E⋂

Vn is at most countable, and we will show that

P = W c. Suppose x ∈ P . (x is a condensation point of E). If x ∈ Vn for

18

some n, then E⋂

Vn is uncountable since Vn is open. Thus x ∈ W c. (If

x ∈ W , then there exists Vn such that x ∈ Vn and E⋂

Vn is uncountable, a

contradiction.) Therefore P ⊂ W c.

Conversely, suppose x ∈ W c. x /∈ Vn for any n such that E⋂

Vn is count-

able. Take any neighborhood N(x) of x. Take x ∈ Vn ⊂ N(x), and E⋂

Vn is

uncountable. Thus E⋂

N(x) is also uncountable, x is a condensation point

of E. Thus W c ⊂ P . Therefore P = W c. Note that W is countable, and

thus W ⊂ W⋂

E = P c⋂

E is at most countable.

To show that P is perfect, it is enough to show that P contains no isolated

point. (since P is closed). If p is an isolated point of P , then there exists a

neighborhood N of p such that N⋂

E = φ. p is not a condensation point of

E, a contradiction. Therefore P is perfect. �

Exercise 2.28. Let X be a separable metric space, let E be a closed set on

X. Suppose E is uncountable. (If E is countable, there is nothing to prove.)

Let P be the set of all condensation points of E. Since X has a countable

base, P is perfect, and P c⋂

E is at most countable by Exercise 27. Since E

is closed, P ⊂ E. Also, P c⋂

E = E − P . Hence E = P⋃

(E − P ).

For corollary: if there is no isolated point in E, then E is perfect. Thus

E is uncountable, a contradiction.

Note: It’s also called Cauchy-Bendixon Theorem. �

Exercise 2.29. Since O is open, for each x in O, there is a y > x such that

(x, y) ⊂ O. Let b = sup{y | (x, y) ⊂ O}. Let a = inf{z | (z, x) ⊂ O}. Then

a < x < b, and Ix = (a, b) is an open interval containing x.

Now Ix ⊂ O, for if w ∈ Ix, say x < w < b, we have by the definition of b

a number y > w such that (x, y) ⊂ O, and so w ∈ O).

Moreover, b /∈ O, for if b ∈ O, then for some ǫ > 0 we have (b−ǫ, b+ǫ) ⊂ O,

whence (x, b + ǫ) ⊂ O, contradicting the definition of b. Similarly, a /∈ O.

Consider the collection of open intervals {Ix}, x ∈ O. Since each x ∈ O

is contained in Ix, and each Ix ⊂ O, we have O =⋃

Ix.

Let (a, b) and (c, d) be two intervals in this collection with a point in

common. Then we must have c < b and a < d. Since c /∈ O, it does not

19

belong to (a, b) and we have c ≤ a. Thus a = c. Similarly, b = d, and

(a, b) = (c, d). Thus two different intervals in the collection {Ix} must be

disjoint. Thus O is the union of the disjoint collection {Ix} of open inter-

vals, and it remains only to show that this collection is countable. But each

open interval contains a rational number since Q is dense in R. Since we

have a collection of disjoint open intervals, each open interval contains a

different rational number, and the collection can be put in one-to-one corre-

spondence with a subset of the rationals. Thus it is a countable collection. �

Exercise 2.30. I prove Baire’s theorem directly. Let Gn be a dense open

subset of Rk for n = 1, 2, 3, · · · . I need to prove that⋂∞

1 Gn intersects any

nonempty open subset of Rk is not empty.

Let G0 is a nonempty open subset of Rk. Since G1 is dense and G0 is

nonempty, G0

⋂G1 6= ∅. Suppose x1 ∈ G0

⋂G1. Since G0 and G1 are

open, G0

⋂G1 is also open, that is, there exist a neighborhood V1 such that

V1 ⊂ G0

⋂G1. Next, since G2 is a dense open set and V1 is a nonempty

open set, V1

⋂G2 6= ∅. Thus, I can find a nonempty open set V2 such that

V2 ⊂ V1

⋂G2. Suppose I have get n nonempty open sets V1, V2, · · · , Vn

such that V1 ⊂ G0

⋂G1 and Vi+1 ⊂ Vi

⋂Gn+1 for all i = 1, 2, · · · , n − 1.

Since Gn+1 is a dense open set and Vn is a nonempty open set, Vn

⋂Gn+1

is a nonempty open set. Thus we can find a nonempty open set Vn+1 such

that Vn+1 ⊂ Vn

⋂Gn+1. By induction, I can form a sequence of open sets

{Vn|n ∈ Z+} such that V1 ⊂ G0

⋂G1 and Vi+1 ⊂ Vi

⋂Gi+1 for all n ∈ Z+.

Since V1 is bounded and V1 ⊃ V2 ⊃ · · · ⊃ Vn ⊃ · · · , by Theorem 2.39 I know

that ∞⋂

n=1

Vn 6= ∅.

Since V1 ⊂ G0

⋂G1 and Vn+1 ⊂ Gn+1, G0

⋂(⋂∞

n=1 Gn) 6= ∅.Note: By Baire’s theorem, I’ve proved the equivalent statement. Next,

Fn has a empty interior if and only if Gn = Rk − Fn is dense in Rk. Hence

we completed all proof. �

Exercise 3.1. Since {sn} is convergent, for any ǫ > 0, there exists N

20

such that |sn − s| < ǫ whenever n ≥ N . By Exercise 1.13 I know that

||sn| − |s|| ≤ |sn − s|. Thus, ||sn| − |s|| < ǫ, that is, {sn} is convergent.

The converse is not true for sn = (−1)n. �

Exercise 3.2.

√n2 + n − n =

n√n2 + n + n

=1

√

1/n + 1 + 1→ 1

2

as n → ∞. �

Exercise 3.3. First, I show that {sn} is strictly increasing. It is trivial that

s2 =√

2 +√

s1 =

√

2 +√√

2 >√

2 = s1. Suppose sk > sk−1 when k < n.

By the induction hypothesis,

sn =√

2 +√

sn−1 >√

2 +√

sn−2 = sn−1

By the induction, {sn} is strictly increasing. Next, we will show that {sn}is bounded by 2. Similarly, we apply the induction again. Hence {sn} is

strictly increasing and bounded, that is, {sn} converges. �

Exercise 3.4. By direct computing s1 = 0, s2 = 0, s3 = 12, s4 = 1

4, s5 =

34, s6 = 3

8, s7 = 7

8, · · · So lim inf sn = 1

2, lim sup sn = 1. �

Exercise 3.5. Let a = lim sup an, b = lim sup bn, and c = lim sup(an + bn).

Suppose {an} and {bn} is bounded below. If a = +∞ or b = +∞, then the

conclusion always holds. Hence we assume that a and b are finite. Hence give

ǫ > 0, there is an integer N such that an + bn < a + b + ǫ whenever n ≥ N .

It follows that c is finite. Since c = lim sup(an + bn), for the same ǫ there

is an integer m ≥ N such that c − ǫ < am + bm. Put n = m and combine

them, we have c− ǫ < a + b + ǫ, or c < a + b + 2ǫ. Since ǫ is arbitrary small,

c ≤ a + b, that is,

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn.

If a = −∞ and b = −∞. Then given any real M , an > M and bm > M

for at most a finite number of values of n and m, so (an + bn) > 2M for at

21

most a finite number of values of n. Hence c = −∞. So we also have

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn.

�

Exercise 3.6. For (a). The partial sum sn =∑n

k=1 ak is

sn =

n∑

k=1

ak =

n∑

k=1

(√

k + 1 −√

k) =√

n + 1 − 1.

Hence∑

an = lim sn = ∞ is divergent.

For (b).

an =

√n + 1 −√

n

n=

1

n(√

n + 1 +√

n).

So

an ≤ 1

2n3/2.

By Theorem 3.25 and 3.28,∑

an converges.

For (c).

lim supn→∞

n√

|an| = lim supn→∞

n

√

( n√

n − 1)n = limn→∞

( n√

n − 1) = 0

by Theorem 3.20(d). The root test indicates that∑

an converges.

For (d). an → 0 implies that |z| > 1 (an → 0, ⇒ 1 + zn → ∞, ⇒zn → ∞). Hence

∑an diverges when |z| ≤ 1. When |z| > 1, there is N such

that |z|n − 1 > 12|z|n whenever n ≥ N . Hence

|an| =1

|1 + zn| ≤1

|z|n − 1<

2

|z|n

whenever n ≥ N . By Theorem 3.25 and 3.26,∑ |an| converges. So

∑an

converges absolutely if |z| > 1. �

Exercise 3.7. By Cauchy’s inequality,

k∑

n=1

an

k∑

n=1

1

n2≥

k∑

n=1

an

√an

n

22

for all n ∈ N . Also, both∑

an and∑

1n2 are convergent; thus

∑kn=1 an

√an

n

is bounded. Besides,√

an

n≥ 0 for all n. Hence

∑ √an

nis convergent. �

Exercise 3.8. We prove a more general exercise.

Dirichlet’s Convergence Test: Let {an} and {bn} be sequences of real

numbers such that {∑nk=0 ak} is bounded and {bn} decreases with 0 as its

limit. Then∑∞

n=0 anbn converges.

Let An =∑n

k=0 an and let M be an upper bound for {bn}. By partial

summation formula, we have

∣∣∣∣

q∑

n=p

anbn

∣∣∣∣

≤ |q−1∑

n=p

An(bn − bn+1)| + |Aqbq| + |Ap−1bp|

≤ M

q−1∑

n=p

(bn − bn+1) + |Aqbq| + |Ap−1bp|)

≤ M(bp − bq) + Mbq + Mbp

= 2Mbp → 0

as p, q → ∞. Hence∑∞

n=0 anbn is a Cauchy sequence, that is,∑∞

n=0 anbn

converges.

Return to this exercise, suppose {bn} is decreasing and bn → b, then the

new sequence {bn − b} decreases with 0 as its limit. Also,∑

an converges

implies that {∑nk=0 ak} is bounded. By Dirichlet’s Convergence Test,

∞∑

n=0

an(bn − b)

converges. Also,∑

an converges implies that∑

ban converges. Add them,

and we have∑

anbn converges. Finally, suppose {bn} is increasing and

bn → b. Thus the neq sequence {b − bn} decreases with 0 as its limit. Using

the same argument, we also get the same conclusion. �

Exercise 3.9. (a) limn→∞ αn = limn→∞(n3)1/n = 1. R = 1/α = 1.

(b) limn→∞ αn = limn→∞(2n/n!)1/n = limn→∞ 2/(n!)1/n = 0. R = +∞.

(c) limn→∞ αn = limn→∞(2n/n2)1/n = 2/1 = 2. R = 1/α = 1/2.

23

(d) limn→∞ αn = limn→∞(n3/3n)1/n = 1/3. R = 1/α = 3. �

Exercise 3.10. Recall α = lim supn→∞n√

|an| and R = 1α. Since an are

integers, infinitely many of which are distinct from zero, n√

|an| ≥ 1 for all

n. Hence

α = lim supn→∞

n√

|an| ≥ 1,

R =1

α≤ 1,

that is, the radius of convergence is at most 1. �

Exercise 3.11. For (a). Note that an

1+an→ 0 iff 1

1

an+1

→ 0 iff 1an

→ ∞ iff

an → 0 as n → ∞. If∑

an

1+anconverges, then an → 0 as n → ∞. Thus for

some ǫ′ = 1 there is an N1 such that an < 1 whenever n ≥ N1. Since∑

an

1+an

converges, for any ǫ > 0 there is an N2 such that am

1+am+ · · · + an

1+an< ǫ for

all n > m ≥ N2. Take N = max(N1, N2). Thus ǫ > am

1+am+ · · · + an

1+an>

am

1+1+ · · ·+ an

1+1= am+···+an

2for all n > m ≥ N . Thus

am + · · · + an < 2ǫ

for all n > m ≥ N . It is absurd. Hence∑

an

1+andiverges.

For (b). aN+1

sN+1+· · ·+ aN+k

sN+k≥ aN+1

sN+k+· · ·+ aN+k

sN+k=

aN+1+···+aN+k

sN+k=

sN+k−sN

sN+k=

1− sN

sN+k. If

∑an

snconverges, for any ǫ > 0 there exists N such that am

sm+ · · ·+

an

sn< ǫ for all m, n whenever n > m ≥ N . Fix m = N and let n = N + k.

Thus

ǫ >am

sm

+ · · · + an

sn

=aN

sN

+ · · ·+ aN+k

sN+k

≥ 1 − sN

sN+k

for all k ∈ N . But sN+k → ∞ as k → ∞ since∑

an diverges and an > 0.

Take ǫ = 1/2 and we obtain a contradiction. Hence∑

an

sndiverges.

For (c). sn−1 ≤ sn iff 1s2n≤ 1

snsn−1iff an

s2n

≤ an

snsn−1= sn−sn−1

snsn−1iff an

s2n

≤1

sn−1− 1

snfor all n. Hence

k∑

n=2

an

s2n

≤k∑

n=2

(1

sn−1− 1

sn) =

1

s1− 1

sn.

Note that 1sn

→ 0 as n → ∞ since∑

an diverges. Hence∑

an

s2n

converges.

24

For (d).∑

an

1+nanmay converge or diverge, and

∑an

1+n2anconverges. To

see this, we put an = 1/n. an

1+nan= 1

2n, that is,

∑an

1+nan= 2

∑1/n diverges.

Besides, if we put

an =1

n(log n)p

where p > 1 and n ≥ 2, then

an

1 + nan

=1

n(log n)2p((log n)p + 1)<

1

2n(log n)3p

for large enough n. By Theorem 3.25 and Theorem 3.29,∑

an

1+nanconverges.

Next,∑ an

1 + n2an=

∑ 1

1/an + n2<

∑ 1

n2.

for all an. Note that∑

1n2 converges, and thus

∑an

1+n2anconverges. �

Exercise 3.12. For (a).

am

rm+ · · ·+ an

rn>

am + · · · + an

rm=

rm − rn

rm= 1 − rn

rm

if m < n. If∑

an

rnconverges, for any ǫ > 0 there exists N such that

am

rm

+ · · ·+ an

rn

< ǫ

for all m, n whenever n > m ≥ N . Fix m = N . Thus

am

rm+ · · · + an

rn> 1 − rn

rm= 1 − rn

rN

for all n > N . But rn → 0 as n → ∞; thus am

rm+ · · · + an

rn→ 1 as n → ∞. If

we take ǫ = 1/2, we will get a contradiction.

For (b). Note that rn+1 < rn iff√

rn+1 <√

rn iff√

rn +√

rn+1 < 2√

rn iff√

rn+√

rn+1√rn

< 2 iff (√

rn − √rn+1)

√rn+

√rn+1√

rn< 2(

√rn − √

rn+1) iff rn−rn+1√rn

<

2(√

rn −√rn+1) iff an√

rn< 2(

√rn −√

rn+1) since an > 0 for all n. Hence,

k∑

n=1

an√rn

<

k∑

n=1

2(√

rn −√rn+1) = 2(

√r1 −

√rk+1).

Note that rn → 0 as n → ∞. Thus∑

an√rn

is bounded. Hence∑

an√rn

converges. �

25

Note: We say that∑

an converges faster than∑

bn if

limn→∞

an

bn= 0.

According the above exercise, we can construct a faster convergent series

from a known convergent one easily. It implies that there is no perfect tests

to test all convergences of the series from a known convergent one.

Exercise 3.13. Put An =∑n

k=0 |ak|, Bn =∑n

k=0 |bk|, Cn =∑n

k=0 |ck|. Then

Cn = |a0b0| + |a0b1 + a1b0| + · · · + |a0bn + a1bn−1 + · · ·+ anb0|≤ |a0||b0| + (|a0||b1| + |a1||b0|) + · · ·

+(|a0||bn| + |a1||bn−1| + · · · + |an||b0|)= |a0|Bn + |a1|Bn−1 + · · · + |an|B0

≤ |a0|Bn + |a1|Bn + · · ·+ |an|Bn

= (|a0| + |a1| + · · ·+ |an|)Bn = AnBn ≤ AB

where A = lim An and B = lim Bn. Hence {Cn} is bounded. Note that

{Cn} is increasing, and thus Cn is a convergent sequence, that is, the Cauchy

product of two absolutely convergent series converges absolutely. �

Exercise 3.14. For (a). The proof is straightforward. Let tn = sn − s,

τn = σn − s. (Or you may suppose that s = 0.) Then

τn =t0 + t1 + · · ·+ tn

n + 1.

Choose M > 0 such that |tn| ≤ M for all n. Given ǫ > 0, choose N so that

n > N implies |tn| < ǫ. Taking n > N in τn = (t0 + t1 + · · · + tn)/(n + 1),

and then

|τn| ≤|t0| + · · ·+ |tN |

n + 1+

|tN+1| + · · · + |tn|n + 1

<(N + 1)M

n + 1+ ǫ.

Hence, lim supn→∞ |τn| ≤ ǫ. Since ǫ is arbitrary, it follows that limn→∞ |τn| =

0, that is, lim σn = s.

26

Note: If you know O.Stolz Theorem, the proof will be very simple. A

similar exercise is stated here: If lim an = a, lim bn = b, then

limn→∞

a1bn + a2bn−1 + · · ·+ anb1

n= ab.

Another proof of (a). Given ǫ > 0, there exists N1 such that |sn − s| < ǫ

whenever n > N1. Thus

σn − s ≤ |s0 − s| + |s1 − s| + · · · + |sn − s|n + 1

≤ |s0 − s| + · · ·+ |sN1− s|

n + 1+

n + 1 − N1

n + 1ǫ.

Note that |s0 − s| + · · ·+ |sN1− s| is fixed. Take N2 > 0 such that

|s0 − s| + · · · + |sN1− s|

n + 1< ǫ

whenever n > N2. Let N = max{N1, N2}. Hence,

|σn − s| < ǫ +n + 1 − N1

n + 1ǫ < 2ǫ.

For (b). Let sn = (−1)n. Hence |σn| ≤ 1/(n + 1), that is, lim σn = 0.

However, lim sn does not exists.

For (c). Let

sn =

1 , n = 0,

n1/4 + n−1 , n = k2 for some integer k,

n−1 , otherwise.

It is obvious that sn > 0 and lim sup sn = ∞. Also,

s0 + · · ·+ sn = 1 + nn−1 +⌊√

n⌋n1/4 = 2 +

⌊√n⌋n1/4.

That is,

σn =2

n + 1+

⌊√n⌋n1/4

n + 1.

The first term 2/(n + 1) → 0 as n → ∞. Note that

0 ≤ ⌊√n⌋n1/4

n + 1< n1/2n1/4n−1 = n−1/4.

27

It follows that the last term → 0. Hence, lim σn = 0.

For (d).

n∑

k=1

kak =

n∑

k=1

k(sk − sk−1) =

n∑

k=1

ksk −n∑

k=1

ksk−1

=

n∑

k=1

ksk −n−1∑

k=0

(k + 1)sk

= nsn +

n−1∑

k=1

ksk −n−1∑

k=1

(k + 1)sk − s0

= nsn −n−1∑

k=1

sk − s0 = (n + 1)sn −n∑

k=0

sk

= (n + 1)(sn − σn).

That is,

sn − σn =1

n + 1

n∑

k=1

kak.

Note that {nan} is a complex sequence. By (a),

limn→∞

(1

n + 1

n∑

k=1

kak

)

= limn→∞

nan = 0.

Also, lim σn = σ. Hence by the previous equation, lim s = σ.

For (e). If m < n, then

n∑

i=m+1

(sn − si) + (m + 1)(σn − σm)

= (n − m)sn −n∑

i=m+1

si + (m + 1)(σn − σm)

= (n − m)sn −( n∑

i=0

si −m∑

i=0

si

)

+ (m + 1)(σn − σm)

= (n − m)sn − (n + 1)σn + (m + 1)σm + (m + 1)(σn − σm)

= (n − m)sn − (n − m)σn.

Hence,

sn − σn =m + 1

n − m(σn − σm) +

1

n − m

n∑

i=m+1

(sn − si).

28

For these i, recall an = sn − sn−1 and |nan| ≤ M for all n,

|sn − si| =

∣∣∣∣

n∑

k=i+1

ak

∣∣∣∣≤

n∑

k=i+1

|ak| ≤n∑

k=i+1

M

i + 1=

(n − i)M

i + 1

≤ (n − (m + 1))M

(m + 1) + 1=

(n − m − 1)M

m + 2.

Fix ǫ > 0 and associate with each n the integer m that satisfies

m ≤ n − ǫ

1 + ǫ< m + 1.

Thusn − m

m + 1≥ ǫ and

n − m − 1

m + 2< ǫ,

orm + 1

n − m≤ 1

ǫand |sn − si| < Mǫ.

Hence,

|sn − σ| ≤ |σn − σ| + 1

ǫ(|σn − σ| + |σm − σ|) + Mǫ.

Let n → ∞ and thus m → ∞ too, and thus

lim supn→∞

|sn − σ| ≤ Mǫ.

Since ǫ was arbitrary, lim sn = σ. �

Exercise 3.15. Theorem 3.22.∑

an converges if and only if for every ǫ > 0

there is an integer N such that∣∣∣∣

m∑

k=n

an

∣∣∣∣≤ ǫ

if m ≥ n ≥ N .

Theorem 3.23. If∑

an converges, then limn→∞ an = 0.

Theorem 3.25(a). If |an| ≤ cn for n ≥ N0, where N0 is some fixed integer,

and if∑

cn converges, then∑

an converges.

Theorem 3.33 (Root Test). Given∑

an, put α = lim supn→∞n√

|an|.Then

29

(a) if α < 1,∑

an converges;

(b) if α > 1,∑

an diverges;

(c) if α = 1, the test gives no information.

Theorem 3.34 (Ratio Test). The series∑

an

(a) converges if lim supn→∞∣∣an+1

an

∣∣ < 1,

(b) diverges if∣∣an+1

an

∣∣ ≥ 1 for n ≥ n0, where n0 is some fixed in-

teger.

Theorem 3.42. Suppose

(a) the partial sums An of∑

an form a bounded sequence;

(b) b0 ≥ b1 ≥ b2 ≥ · · · ;(c) limn→∞ = 0.

Then∑

bnan converges.

Theorem 3.45. If∑

an converges absolutely, then∑

an converges.

Theorem 3.47. If∑

an = A, and∑

bn = B, then∑

(an +bn) = A+B,

and∑

can = cA for any fixed c ∈ R.

Theorem 3.55. If∑

an is a series in Ck which converges absolutely, then

every arrangement of∑

an converges, and they all converge to the same sum.

Proof of Theorem 3.22. Theorem 3.11(c) implies it.

Proof of Theorem 3.23. Put m = n + 1 in Theorem 3.22, we get the

conclusion.

Proof of Theorem 3.25(a). Given ǫ > 0, there exists N ≥ N0 such that

m ≥ n ≥ N impliesm∑

k=n

ck ≤ ǫ,

30

by the Cauchy criterion. Hence

∣∣∣∣

m∑

k=n

ak

∣∣∣∣≤

m∑

k=n

|ak| ≤m∑

k=n

ck ≤ ǫ,

and (a) follows.

Exercise 3.16. For (a).

xn − xn+1 = xn − 1

2

(

xn +α

xn

)

=1

2

(

xn − α

xn

)

=1

2

(x2

n − α

xn

)

> 0

since xn > α. Hence {xn} decreases monotonically. Also, {xn} is bounded

by 0; thus {xn} converges. Let lim xn = x. Hence lim xn+1 = lim 12(xn + α

xn)

iff x = 12(x + α

x) iff x2 = α. Note that xn > 0 for all n. Thus x =

√α.

lim xn =√

α.

For (b).

xn+1 =1

2

(

xn +α

xn

)

⇒ xn+1 −√

α =1

2

(

xn +α

xn

)

−√α

⇒ xn+1 −√

α =1

2

x2n − 2xn

√α + α

xn

⇒ xn+1 −√

α =(xn −√

α)2

2xn

⇒ ǫn+1 =ǫ2n

2xn<

ǫ2n

2√

α.

Hence ǫn+1 < β( ǫ1β)2n

where β = 2√

α by induction.

For (c).

ǫ1

β=

2 −√

3

2√

3=

1

2√

3(2 +√

3)=

1

6 + 4√

3<

1

10.

Thus

ǫ5 < β(ǫ1

β)24

< 2√

3 · 10−16 < 4 · 10−16,

ǫ6 < β(ǫ1

β)25

< 2√

3 · 10−32 < 4 · 10−32.

31

�

Note. It is an application of Newton’s method. Let f(x) = x2 − α in

Exercise 5.25.

Exercise 3.20. Since {pn} is a Cauchy sequence, ∀ǫ > 0 there exists N1

such that |pn − pm| < ǫ whenever m, n ≥ N1. Also, since {pnj} converges to

p ∈ X, there exists N2 such that |pnj− p| < ǫ whenever j ≥ N2. Thus

|pn − p| ≤ |pn − pnj| + |pnj

− p|

for each n ≥ N1. Choose j such that j ≥ N2 and nj ≥ N1. Thus |pn−pnj| < ǫ

and |pnj− p| < ǫ. Hence |pn − p| < 2ǫ whenever n ≥ N1. Therefore the full

sequence {pn} converges to p. �

Exercise 3.21. En is bounded ∀n ⇒ En is nonempty ∀n. So we pick pn ∈ En

for each n. With the relation En ⊃ En+1 and Page 53, we know that {pn}is a Cauchy sequence in X. Since X is complete, pn converges to a point

p ∈ X. Since En is closed and En ⊃ En+1, the limit point p lies in each En.

So

p ∈∞⋂

n=1

En.

If there is q ∈ ⋂∞n=1 En such that p 6= q, then diam(

⋂∞n=1 En) ≥ d(p, q) > 0,

where d is the metric of X. But∞⋂

n=1

En ⊂ Ek

for all k. Hence

diam

( ∞⋂

n=1

En

)

≤ diamEk

for all k. Take limit and limn→∞ diamEn ≥ d(p, q), where d(p, q) is a fixed

positive number, contradicts with limn→∞ diamEn = 0. Hence⋂∞

n=1 En con-

sists of exactly one point. �

Exercise 3.22. See 2 Exercise 2.30. �

32

Exercise 3.23. For any ǫ > 0, there exists N such that d(pn, pm) < ǫ and

d(qm, qn) < ǫ whenever m, n ≥ N . Note that

d(pn, qn) ≤ d(pn, pm) + d(pm, qm) + d(qm, qn).

It follows that

|d(pn, qn) − d(pm, qm)| ≤ d(pn, pm) + d(qm, qn) < 2ǫ.

Thus {d(pn, qn)} is a Cauchy sequence in X. Hence {d(pn, qn)} converges. �

Exercise 3.24. For (a). Suppose there are three Cauchy sequences {pn},{qn}, and {rn}. First, d(pn, pn) = 0 for all n. Hence, d(pn, pn) = 0 as n → ∞.

Thus it is reflexive. Next, d(qn, pn) = d(pn, qn) → 0 as n → ∞. Thus it is

symmetric. Finally, if d(pn, qn) → 0 as n → ∞ and if d(qn, rn) → 0 as

n → ∞, d(pn, rn) ≤ d(pn, qn) + d(qn, rn) → 0 + 0 = 0 as n → ∞. Thus it is

transitive. Hence this is an equivalence relation.

For (b). To show △(P, Q) is well-defined. Suppose {pn} and {p′n} are

equivalent; {qn} and {q′n} are equivalent. Let △′(P, Q) = lim d(p′n, q′n). Since

d(pn, qn) ≤ d(pn, p′n) + d(p′n, q

′n) + d(q′n, qn),

so we take the limit:

limn→∞

d(pn, qn) ≤ limn→∞

d(pn, p′n) + lim

n→∞d(p′n, q′n) + lim

n→∞d(q′n, qn).

Note that the equivalence of sequences, and we have

△(P, Q) ≤ △′(P, Q).

Similarly,

d(p′n, q′n) ≤ d(p′n, pn) + d(pn, qn) + d(qn, q′n),

we have

△′(P, Q) ≤ △(P, Q).

Hence

△(P, Q) = △′(P, Q).

33

So △(P, Q) is well-defined (unchanged). Hence △ is a distance function in

X∗.

For (c). Let {Pn} be a Cauchy sequence in (X∗,△). We wish to show that

there is a point P ∈ X∗ such that △(Pn, P ) → 0 as n → ∞. For each Pn,

there is a Cauchy sequence in X, denoted {Qkn}, such that △(Pn, Qkn) → 0

as k → ∞. Let ǫn > 0 be a sequence tending to 0 as n → ∞. From the double

sequence {Qkn} we can extract a subsequence Q′n such that △(Pn, Q

′n) < ǫn

for all n. From the triangle inequality, it follows that

△(Q′n, Q

′m) ≤ △(Q′

n, Pn) + △(Pn, Pm) + △(Pm, Q′m). (*)

Since {Pn} is a Cauchy sequence, given ǫ > 0, there is an N > 0 such that

△(Pn, Pm) < ǫ for m, n > N . We choose m and n so large that ǫm < ǫ,

ǫn < ǫ. Thus the inequality (*) shows that {Q′n} is a Cauchy sequence in X.

Let P be the corresponding equivalence class in S. Since

△(P, Pn) ≤ △(P, Q′n) + △(Q′

n, Pn) < 2ǫ

for n > N , we conclude that Pn → P as n → ∞. That is, X∗ is complete.

For (d). {p} ∈ Pp and {q} ∈ Pq. By part (b),

△(Pp, Pq) = limn→∞

d(p, q) = d(p, q).

So the mapping ϕ is an isometry of X into X∗.

For (e). Take any element P ∈ X∗. Take {pn} ∈ P , where pn ∈ X. Then

Ppn converges to P . In fact,

limn→∞

△(Ppn, P ) = limn→∞

limm→∞

d(pn, pm) = 0

(since {pn} is a Cauchy sequence). Hence ϕ(X) is dense in X∗. If X is

complete, then we take any element P ∈ X∗. Take {pn} ∈ P , where pn ∈ X.

Since X is complete, pn converges to p ∈ X. So {p} ∈ P , P = Pp ∈ ϕ(X).

Hence ϕ(X) = X∗ if X is complete. �

Exercise 3.25. The completion of X is the real number system with the

metric d(x, y) = |x − y|. �

34

Exercise 4.1. No. Take f(x) = 1 if x ∈ Z; f(x) = 0 otherwise. �

Exercise 4.2. If f(E) is empty, the conclusion holds trivially. If f(E) is

non-empty, then we take an arbitrary point y ∈ f(E). Thus, there exists

p ∈ E such that y = f(p). Thus p ∈ E or p ∈ E ′. Also, note that f(E) =

f(E)⋃

(f(E))′. Now we consider the following two cases:

Case 1: If p ∈ E, then y ∈ f(E) ⊂ f(E).

Case 2: Suppose p ∈ E ′. Since f is continuous at x = p, given ǫ > 0,

there exists δ > 0 such that dY (f(x), f(p)) < ǫ whenever dX(x, p) < δ for all

x ∈ X. Since p is a limit point of E, then for some δ > 0 there exists x ∈ E.

Thus f(x) ∈ Nǫ(f(p)) for some f(x) ∈ f(E). Since ǫ is arbitrary, f(p) is an

adhere point of f(E) in Y . Thus f(p) ∈ f(E).

By case 1 and 2, we proved that f(E) ⊂ f(E).

Now we show that f(E) can be a proper subset of f(E). Define a function

f(x) from X = (0, +∞) to Y = R1 by

f(x) =1

x.

Let E = Z+. Thus

f(E) = f(E) = {1/n : n ∈ Z+},f(E) = {1/n : n ∈ Z+} = {0}

⋃

{1/n : n ∈ Z+}.

�

Exercise 4.3. To prove Z(f) is closed, it suffice to show that every limit

point of Z(f) is contained in Z(f) itself. Let x ∈ X be a limit point of Z(f).

Thus, there is a convergent sequence {xn} in Z(f) converging to x. Since f

is continuous, limn→∞ f(xn) = f(limn→∞ xn). It follows that 0 = f(x) and

thus x ∈ Z(f) exactly. �

Exercise 4.4. First we need to show f(E) is dense in f(X), that is, every

point of f(X) is a limit point of f(E), or a point of f(E) (or both).

Take any y ∈ f(X), and then there exists a point p ∈ X such that

y = f(p). Since E is dense in X, thus p is a limit point of E or p ∈ E. If

35

p ∈ E, then y = f(p) ∈ f(E). Thus y is a point of f(E), done. If p is a

limit point of E and p /∈ E. Since f is continuous on X, for every ǫ > 0

there exists δ > 0 such that dY (f(x), f(p)) < ǫ for all points x ∈ X for which

dX(x, p) < δ. Since p is a limit point of E, there exists q ∈ Nδ(p) such that

q 6= p and q ∈ E. Hence

f(q) ∈ Nǫ(f(p)) = Nǫ(y).

and f(q) ∈ f(E). Since p /∈ E, f(p) /∈ f(E), and f(q) 6= f(p). Hence f(p) is

a limit point of f(E). Thus f(E) is dense in f(X).

Suppose p ∈ X − E. Since E is dense in X, p is a limit point of E and

p /∈ E. Hence we can take a sequence {qn} → p such that qn ∈ E and qn 6= p

for all n. (More precisely, since p is a limit point, every neighborhood Nr(p)

of p contains a point q 6= p such that q ∈ E. Take r = rn = 1/n, and thus

rn → 0 as n → ∞. At this time we can get q = qn → p as n → ∞.) Hence

g(p) = g(

limn→∞

qn

)

= limn→∞

g(qn)

= limn→∞

f(qn) = f(

limn→∞

qn

)

= f(p).

Thus g(p) = f(p) for all p ∈ X. �

Exercise 4.5. Note that the following fact: Every open set of real numbers

is the union of a countable collection of disjoint open intervals.

Thus, consider Ec =⋃

(ai, bi), where i ∈ Z, and ai < bi < ai+1 < bi+1.

We extend g on (ai, bi) as following:

g(x) = f(ai) + (x − ai)f(bi) − f(ai)

bi − ai

(g(x) = f(x) for x ∈ E). Thus g is well-defined on R1, and g is continuous

on R1 clearly.

Next, consider f(x) = 1/x on a open set E = R−0. f is continuous on E,

but we cannot redefine f(0) = any real number to make new f(x) continue

at x = 0.

Next, consider a vector valued function

f(x) = (f1(x), · · · , fn(x)),

36

where fi(x) is a real valued function. Since f is continuous on E, each

component of f , fi, is also continuous on E, thus we can extend fi, say gi,

for each i = 1, · · · , n. Thus,

g(x) = (g1(x), · · · , gn(x))

is a extension of f(x) since each component of g, gi, is continuous on R1

implies g is continuous on Rn. �

Exercise 4.6. (⇒) Let G = {(x, f(x))|x ∈ E}. Since f is a continuous

mapping of a compact set E into f(E), by Theorem 4.14 f(E) is also com-

pact. We claim that the product of finitely many compact sets is compact.

Thus G = E × f(E) is also compact.

(⇐) Define

g(x) = (x, f(x))

from E to G for x ∈ E. We claim that g(x) is continuous on E. Consider

h(x, f(x)) = x from G to E. Thus h is injective, continuous on a compact set

G. Hence its inverse function g(x) is injective and continuous on a compact

set E.

Since g(x) is continuous on E, the component of g(x), say f(x), is con-

tinuous on a compact E.

Proof of claim: We prove that the product of two compact spaces is

compact; the claim follows by induction for any finite product.

Step 1. Suppose that we are given sets X and Y , with Y is compact.

Suppose that x0 ∈ X, and N is an open set of X × Y containing the “slice”

x0 × Y of X × Y . We prove the following: There is a neighborhood W of x0

in X such that N contains the entire set W × Y .

The set W × Y is often called a tube about x0 × Y .

First let us cover x0 × Y by basis elements U × V (for the topology of

X × Y ) lying in N . The set x0 × Y is compact, being homeomorphic to Y .

Therefore, we can cover x0 × Y by finitely many such basis elements

U1 × V1, · · · , Un × Vn.

37

(We assume that each of the basis elements Ui×Vi actually intersects x0×Y ,

since otherwise that basis element would be superfluous; we could discard it

from the finite collection and still have a covering of x0 × Y .) Define

W = U1

⋂

· · ·⋂

Un.

The set W is open, and it contains x0 because each set Ui × Vi intersects

x0 × Y . We assert that the sets Ui × Vi, which were chosen to cover the slice

x0 × Y , actually cover the tube W × Y . Let x × y ∈ W × Y . Consider the

point x0 × y of the slice x0 × Y having the same y-coordinate as this point.

Now x0 × y ∈ Ui × Vi for some i, so that y ∈ Vi. But x ∈ Uj for every j

(because x ∈ W ). Therefore, we have x × y ∈ Ui × Vi, as desired.

Since all the sets Ui × Vi ⊂ N , and since they cover W × Y , the tube

W × Y ⊂ N also.

Step 2. Now we prove the claim. Let X and Y be compact sets. Let Abe an open covering of X × Y . Given x0 ∈ X, the slice x0 × Y is compact

and may therefore be covered by finitely many elements A1, · · · , Am of A.

Their union N = A1

⋃ · · ·⋃ Am is an open set containing x0 ×Y ; by Step 1,

the open set N contains a tube W × Y about x0 × Y , where W is open in

X. Then W × Y is covered by finitely many elements A1, · · · , Am of A.

Thus, for each x ∈ X, we can choose a neighborhood Wx of x such

that the tube Wx × Y can be covered by finitely many element of A. The

collection of all the neighborhoods Wx is an open covering of X; therefore by

compactness of X, there exists a finite subcollection

{W1, · · · , Wk}

covering X. The union of the tubes

W1 × Y, · · · , Wk × Y

is all of X × Y ; since each may be covered by finitely many elements of A,

so may X × Y be covered. �

38

Exercise 4.7. Since x2 + y4 ≥ 2xy2, f(x, y) ≤ 2 for all (x, y) ∈ R2. That is,

f is bounded (by 2). Next, select

(xn, yn) =

(1

n3,1

n

)

.

(xn, yn) → (0, 0) as n → ∞, and g(xn, yn) = n/2 → ∞ as n → ∞, that

is, g(x, y) is unbounded in every neighborhood of (0, 0) by choosing large

enough n.

Next, select

(xn, yn) =

(1

n2,1

n

)

.

(xn, yn) → (0, 0) as n → ∞, and f(xn, yn) = 1/2 for all n. Thus,

limn→∞

f(xn, yn) =1

26= 0 = f(0, 0)

for some sequence {(xn, yn)} ⊂ R2. Thus, f is not continuous at (0, 0).

Finally, we consider two cases of straight lines in R2: (1) x = c and (2)

y = ax + b. (equation of straight lines).

(1) x = c: If c 6= 0, f(x, y) = cy2/(c2 + y4) and g(x, y) = cy2/(c2 + y6) are

continuous since cy2, c2 + y4, and c2 + y6 are continuous on R1 respect to y,

and c2 + y4, c2 + y6 are nonzero. If c = 0, then f(x, y) = g(x, y) = 0, and it

is continuous trivially.

(2) y = ax+b: If b 6= 0, then this line dose not pass (0, 0). Then f(x, y) =

x(ax+b)2/(x2+(ax+b)4) and g(x, y) = x(ax+b)2/(x2+(ax+b)6). By previous

method we conclude that f(x, y) and g(x, y) are continuous. If b = 0, then

f(x, y) = 0 if (x, y) = (0, 0); f(x, y) = a2x/(1 + a4x2), and g(x, y) = 0 if

(x, y) = (0, 0); g(x, y) = a2x/(1 + a6x4). Thus, f(x, y) → 0/1 = 0 = f(0, 0)

and g(x, y) → 0/1 = 0 = f(0, 0) as x → 0. Thus, f and g are continuous.

Both of two cases imply that the restriction of both f and g to every

straight line in R2 are continuous. �

Exercise 4.8. Let E is bounded by M > 0, i.e., |x| ≤ M for all x ∈ E.

Since f is uniformly continuous, take ǫ = 1 there exists δ > 0 such that

|f(x) − f(y)| < ǫ

39

whenever |x−y| < δ where x, y ∈ E. For every x ∈ E, there exists an integer

n = nx such that

nδ ≤ x < (n + 1)δ.

Since E is bounded, the collection of S = {nx|x ∈ E} is finite. Suppose

x ∈ E and x is the only one element satisfying nδ ≤ x < (n + 1)δ for some

n. Let x = xn, and thus

|f(x)| ≤ |f(xn)|

for all x ∈ E⋂

[nδ, (n + 1)δ). If there are more than two or equal to two

element satisfying that condition, then take some one as xn. Since |x−xn| < δ

for all x ∈ E⋂

[nδ, (n + 1)δ),

|f(x) − f(xn)| < 1

for all x ∈ E⋂

[nδ, (n + 1)δ), that is,

|f(x)| < 1 + |f(xn)|.

So

|f(x)| < maxn∈S

(1 + f(xn)).

(The maximum is meaningful because S is finite.) Thus f(x) is bounded. �

Note: If boundedness of E is omitted from the hypothesis, then we define

f(x) = x for x ∈ R1. Hence f is uniformly continuous on R1, but f(R1) = R1

is unbounded.

Exercise 4.9. Recall the original definition of uniformly continuity: for

every ǫ > 0 there exists δ > 0 such that dY (f(p), f(q)) < ǫ for all p and q in

X for which dX(p, q) < δ.

(⇐) Given ǫ > 0. ∀p, q ∈ X for which dX(p, q) < δ. Take

E = {p, q},

and thus diamE = supp,q∈E d(p, q) = d(p, q) < δ. Hence diamf(E) < ǫ. Note

that diamf(E) ≥ d(f(p), f(q)) since p, q ∈ E. Hence d(f(p), f(q)) < ǫ. Thus

for every ǫ > 0 there exists δ > 0 such that dY (f(p), f(q)) < ǫ for all p and

q in X for which dX(p, q) < δ.

40

(⇒) ∀E ⊂ X with diamE < δ. ∀p, q ∈ R, d(p, q) ≤ diamE < δ. Thus we

have

d(f(p), f(q)) <ǫ

2

for all p, q ∈ E. Hence diamf(E) ≤ ǫ/2 < ǫ. Thus to every ǫ > 0 there exists

a δ > 0 such that diamf(E) < ǫ for all E ⊂ X with diamE < δ. �

Exercise 4.10. If f is not uniformly continuous, then for some ǫ > 0

there are sequences {pn}, {qn} in X such that dX(pn, qn) → 0 but dY (f(pn),

f(qn)) > ǫ.

Since E = {pn : n ∈ Z+} is an infinite subset of a compact set X, then

E has a limit point in X; that is, there is a subsequence of {pn}, which we

denote {pkn}, converges to some point x ∈ X. Then, since dX(pn, qn) → 0,

we see that qkn → x as n → ∞. Using the fact that f is continuous on X,

we have

f(pkn) → f(x), f(qkn) → f(x).

That is, there is a positive integer N such that

dY (f(pkn), f(x)) ≤ ǫ

2,

dY (f(qkn), f(x)) ≤ ǫ

2

for all n ≥ N . Hence for all n ≥ N , we find

dY (f(pkn), f(qkn)) ≤ dY (f(pkn), f(x)) + dY (f(x), f(qkn))

<ǫ

2+

ǫ

2= ǫ.

The last inequality contradicts our hypotheses, and the result is established.

�

Exercise 4.11. Let {xn} be a Cauchy sequence in X. ∀ǫ′ > 0, ∃N such

that dX(xn, xm) < ǫ′ whenever m, n ≥ N . Since f is a uniformly continuous,

∀ǫ > 0, ∃δ > 0 such that dY (f(x), f(y)) < ǫ whenever d(x, y) < δ. Take

ǫ′ = δ. Thus

dY (f(xn), f(xm)) < ǫ

41

whenever m, n ≥ N for some N ; that is, {f(xn)} is a Cauchy sequence. �

Exercise 4.12. Theorem. Let f be a function defined on a set S in X and

assume that f(S) ⊂ Y . Let g be defined on f(S) with value in Z, and let h

denote the composite function defined by

h(x) = g(f(x))

if x ∈ S. If f is uniformly continuous on S and if g is uniformly continuous

on f(S), then h is uniformly continuous on S.

Proof. Since f is uniformly continuous on S, ∀ǫ′ > 0, ∃δ > 0 such that

dY (f(x), f(y)) < ǫ′

whenever for all x, y ∈ S for which dX(x, y) < δ. Also, since g is uniformly

continuous on f(S), ∀ǫ > 0, ∃δ′ > 0 such that

dZ(g(x), g(y)) < ǫ

whenever for all x, y ∈ f(S) for which dY (x, y) < δ′. Take ǫ′ = δ′; that is,

∀ǫ > 0, ∃δ > 0 such that

dZ(g(f(x)), g(f(y))) < ǫ

whenever for all x, y ∈ S for which dX(x, y) < δ. Thus h is uniformly con-

tinuous on S. �

Exercise 4.13. Here I give a natural proof. Take x ∈ X −E. E is dense in

X, ⇒ ∃{xn} in E such that xn → x, ⇒ {xn}: Cauchy. Since f is a uniformly

continuous mapping, by Exercise 4.11, {f(xn)} is a Cauchy sequence in E.

R: complete ⇒ f(xn) → L ∈ R. Define

g(x) =

{

f(x) if x ∈ E

limn→∞

f(xn) if x ∈ X − E, xn → x.

g(x) is well-defined. Suppose there is another sequence x′n → x, and f(x′

n) →L′. Then

{x1, x′1, x2, x

′2, · · · , xn, x′

n, · · · }

42

is still a Cauchy sequence. Hence

|L − L′| ≤ |L − f(xn)| + |f(xn) − f(x′n)| + |f(x′

n) − L′| → 0,

that is, L = L′. Hence g(x) is an extension of f(x). We prove that g(x) is also

uniformly continuous. Take x, y ∈ X, then there exist xn → x, and yn → y,

where {xn}, {yn} in E. Given ǫ > 0. Since f is uniformly continuous, there

is δ > 0 such that |f(xn) − f(yn)| < ǫ whenever d(xn, yn) < δ. If we take

n large enough (d(xn, x) < δ/3, d(yn, y) < δ/3 whenever n ≥ N ′ for some

integer N .) and d(x, y) small enough (d(x, y) < δ/3), we have

d(xn, yn) ≤ d(xn, x) + d(x, y) + d(y, yn) < δ.

So

|f(xn) − f(yn)| < ǫ.

Also, g(x) = limn→∞

f(xn), g(y) = limn→∞

f(yn), there is an integer N ′′ such that

|g(x) − f(xn)| < ǫ, and |g(y)− f(yn)| < ǫ

whenever n ≥ N ′′. Take N = max{N ′, N ′′}. Hence

|g(x) − g(y)| ≤ |g(x) − f(xn)| + |f(xn) − f(yn)| + |f(yn) − g(y)| < 3ǫ

whenever d(x, y) < δ/3 and n ≥ N . Hence g is uniformly continuous. To

prove the uniqueness of g: Suppose h is another such function. On E, g = h.

On X − E,

h(x) = limn→∞

h(xn) = limn→∞

f(xn) = g(x).

Hence g = h on X. So the g is unique.

In the above proof, we know that the conclusion is still true if we replace

the range space R by any complete metric space (including Rk surely). Also,

a metric space is compact iff it is complete and totally bounded. So it is also

true if we replace R by any compact metric space.

But if we replace R by any metric space, the conclusion may be wrong.

Consider a subset S of Q ∩ [0, 1]:

S =

{n

am

|n ≤ am, n ∈ Z+ ∪ {0}}

43

where am =∑m

k=0 10k. For example, 311

, 67111

, · · · ∈ S. Hence S is dense in

Q ∩ [0, 1]. Define f on S by

f

(n

am

)

=

{

1 if 2n ≥ am

0 if 2n < am.

For example, f( 311

) = 0 and f( 67111

) = 1. So f is a uniformly continuous

function. But f cannot extend to Q∩ [0, 1]. Take 12∈ Q∩ [0, 1], we have two

sequences {pn} ={

⌊an/2⌋an

}

and {qn} ={

⌈an/2⌉an

}

such that lim f(pn) = 0 and

lim f(qn) = 1. �

Exercise 4.14. Let g(x) = f(x) − x. If g(1) = 0 or g(0) = 0, then the

conclusion holds trivially. Now suppose g(1) 6= 0 and g(0) 6= 0. Since f is

from I to I, 0 6= f(x) 6= 1. Thus,

g(1) = f(1) − 1 < 0,

g(0) = f(0) − 0 > 0.

Since g is continuous on [0, 1], by Intermediate Value Theorem (Theorem

4.23)

g(c) = 0

for some c ∈ (0, 1). Hence f(c) = c for some c ∈ (0, 1). �

Exercise 4.15. If not, there exist three points x1 < x2 < x3 ∈ R1 such that

f(x2) > f(x1), f(x2) > f(x3)

or

f(x2) < f(x1), f(x2) < f(x3).

WLOG, we only consider the case that f(x2) > f(x1), f(x2) > f(x3) for

some x1 < x2 < x3. Since f is continuous on R1, for ǫ = f(x2)−f(x1)2

> 0 there

exists δ1 > 0 such that

|f(x) − f(x1)| < ǫ

whenever |x − x1| < δ1. That is,

f(x) <f(x1) + f(x2)

2< f(x2)

44

whenever x < x1 + δ1. Note that δ1 < x2 − x1. Hence we can take y1 ∈(x1, x1 + δ1). Similarly, for ǫ = f(x2)−f(x3)

2> 0 there exists δ2 > 0 such that

|f(x) − f(x3)| < ǫ

whenever |x − x3| < δ2. That is,

f(x) <f(x2) + f(x3)

2< f(x2)

whenever x > x3 − δ2. Note that δ2 < x3 − x2. Hence we can take y2 ∈(x3 − δ2, x3). Note that y2 > y1. Since f is continuous on a closed set [y1, y2],

f take a maximum value at p ∈ [y1, y2]. Note that

supx∈(x1,x3)

f(x) ≤ supx∈[y1,y2]

f(x)

by previous inequations. Also,

supx∈(x1,x3)

f(x) ≥ supx∈[y1,y2]

f(x)

Hence supx∈(x1,x3) f(x) = supx∈[y1,y2] f(x). Since (x1, x3) is an open set,

f((x1, x3)) is also open. Note that f(p) ∈ f((x1, x3)) but f(p) is not an

interior point of f((x1, x3)). (Otherwise f(p) + ǫ ∈ f((x1, x3)) for some

ǫ > 0. That is, f(p) + ǫ > f(p) which is a contradiction with the maximum

of f(p).) �

Exercise 4.16. Since [x]+(x) = x is a continuous function, [x] and (x) have

the same discontinuities.

For [x], its discontinuities are integers. So (x)’s discontinuities are also

integers. �

Exercise 4.17. Recall the definition of simple discontinities: both f(x+)

and f(x−) exist. First, let E be the set on which f(x−) < f(x+). With

each point x ∈ E, associate a rational triple (p, q, r) such that

(a) f(x−) p.

45

This association is possible. Q is dense in R ⇒ the existence of p. f(x−)

exists ⇒ for any ǫ > 0, there exists x−a > δ > 0 such that |f(t)−f(x−)| < ǫ

whenever t ∈ (x− δ, x)∩ (a, b). Choose ǫ = p−f(x−)2

> 0, and take δ = q ∈ Q.

So a < q < t < x implies f(t) < p, that is, the existence of (b) is OK.

Similarly, the existence of (c) is OK too.

The set of all such triples is countable. Claim: each triple is associated

with at most one point of E. If not, then there are at least two points x and

y with x < y such that

(a) f(x−) p.

(a′) f(y−) p.

for some rational triple (p, q, r). Hence q < x < y < r. However, if we take

t0 ∈ (x, y), then f(t0) > p by (c) and f(t0) < p by (b′), a contradiction.

Hence our claim holds. So E is countable.

Similarly, let F be the set on which f(x−) = f(x+). With each point

x ∈ F , associate a rational triple (p, q) such that

(a) a f(x−).

As the above argument, we conclude that F is also countable. Combine E

and F . Hence E ∪ F , the set of point at which f has a simple discontinuity,

is at most conutable. �

Exercise 4.18. WLOG, we only consider f on [0, 1]. Given ǫ > 0, for any

x0 ∈ [0, 1], there are finitely many positive integers n satisfying that

n <1

ǫ;

that is, there are finitely many rational numbers m/n such that

f

(m

n

)

≥ ǫ.

46

Thus we can take δ > 0 such that the neighborhood Nδ(x0) of x0 contains

no such rational numbers. (If x0 ∈ Q, we may remove x0.) Hence for all x

the following

|f(x)| < ǫ

holds whenever 0 < |x − x0| < δ. Thus we proved that

f(x0+) = f(x0−) = 0

for all x0 ∈ [0, 1].

If x0 is rational, then f(x0) = 0; that is, f(x) is continuous at x = x0.

If x0 is not rational, then f(x0) 6= 0; that is, has a simple discontinuity at

x = x0. � Note. f is Riemann-integrable on [0, 1].

Exercise 4.19. Let S = {x|f(x) = r}. If xn → x0, but f(xn) > r > f(x0)

for some r and all n since Q is dense in R, then f(tn) = r for some tn

between x0 and xn; thus tn → x0. Hence x0 is a limit point of S. Since S is

closed, f(x0) = r, a contradiction. Hence, lim sup f(xn) ≤ f(x0). Similarly,

lim inf f(xn) ≥ f(x0). Hence, lim f(xn) = f(x0), and f is continuous at x0.

�

Note: Original problem is stated as follows: Let f be a function from the

reals to the reals, differentiable at every point. Suppose that, for every r, the

set of points x, where f ′(x) = r, is closed. Prove that f ′ is continuous. If

we replace Q into any dense subsets of R, the conclusion also holds.

Exercise 4.20. We prove (a). (⇐) If x ∈ E ⊂ E, then

infz∈E

d(x, z) ≤ d(x, x) = 0

since we take z = x ∈ E. Hence ρE(x) = 0 if x ∈ E. Suppose x ∈ E − E,

that is, x is a limit point of E. Thus for every neighborhood of x contains

a point y 6= x such that q ∈ E. It implies that d(x, y) → 0 for some y ∈ E,

that is, ρE(x) = infz∈E d(x, z) = 0 exactly.

(⇒) Suppose ρE(x) = infz∈E d(x, z) = 0. Fixed some x ∈ X. If d(x, z) =

0 for some z ∈ E, then x = z, that is x ∈ E ⊂ E. If d(x, z) > 0 for all z ∈ E,

47

then by infz∈E d(x, z) = 0, for any ǫ > 0 there exists z ∈ E such that

d(x, z) < ǫ,

that is,

z ∈ Nǫ(x).

Since ǫ is arbitrary and z ∈ E, x is a limit point of E. Thus x ∈ E ′ ⊂ E.

We prove (b). For all x ∈ X, y ∈ X, z ∈ E,

ρE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z).

Take infimum on both sides, and we get that

ρE(x) ≤ d(x, y) + ρE(y).

Similarly, we also have ρE(y) ≤ d(x, y) + ρE(x). Hence

|ρE(x) − ρE(y)| ≤ d(x, y)

for all x ∈ X, y ∈ X. Thus ρE is a uniformly continuous function on X. �

Exercise 4.21. Let

ρF (x) = infz∈F

d(x, z)

for all x ∈ K. By Exercise 4.20(a), we know that

ρF (x) = 0 ⇔ x ∈ F = F

(since F is closed). That is, ρF (x) = 0 if and only if x ∈ F . Since K and

F are disjoint, ρF (x) is a positive function. Also ρF (x) is continuous by

Exercise 4.20(b). Thus ρF (x) is a continuous positive function. Since K is

compact, ρF (x) takes minimum m > 0 for some x0 ∈ K. Take δ = m/2 > 0

as desired.

Next, let X = R, A = Z+ − {2}, and B = {n + 1/n|n ∈ Z+}. Hence

A and B are disjoint, and they are not compact. Suppose there exists such

δ > 0. Take

x =

[1

δ

]

+ 1 ∈ A, y = x +1

x∈ B.

48

However,

d(x, y) =1

x<

1

1/δ= δ,

a contradiction. �

Exercise 4.22. Note that ρA(p) and ρB(p) are (uniformly) continuous on

X, and ρA(p) + ρB(p) > 0. (Clearly, ρA(p) + ρB(p) ≥ 0 by the definition. If

ρA(p) + ρB(p) = 0, then p ∈ A⋂

B by Exercise 20, a contradiction). Thus

f(p) = ρA(p)/(ρA(p) + ρB(p)) is continuous on X. Next, f(p) ≥ 0, and

f(p) ≤ 1 since ρA(p) ≤ ρA(p) + ρB(p). Thus f(X) lies in [0, 1].

Next, f(p) = 0 ⇔ ρA(p) = 0 ⇔ p ∈ A, and f(p) = 1 ⇔ ρB(p) = 0 ⇔ p ∈B by Exercise 20.

Now we prove a converse of Exercise 3: Every closed set A ⊂ X is Z(f)

for some continuous real f on X. If Z(f) = φ, then f(x) = 1 for all x ∈ X

satisfies our requirement. If Z(f) 6= φ, we consider two possible cases: (1)

Z(f) = X; (2) Z(f) 6= X. If Z(f) = X, then f(x) = 0 for all x ∈ X. If

Z(f) 6= X, we can choose p ∈ X such that f(p) 6= 0. Note that Z(f) and

{p} are one pair of disjoint closed sets. Hence we let

f(x) =ρZ(f)(x)

ρZ(f)(x) + ρ{p}(x).

By the previous result, we know that f(x) satisfies our requirement. Hence

we complete the whole proof.

Note that [0, 1/2) and (1/2, 1] are two open sets of f(X). Since f is

continuous, V = f−1([0, 1/2)) and W = f−1((1/2, 1]) are two open sets.

f−1({0}) ⊂ f−1([0, 1/2)), and f−1({1}) ⊂ f−1((1/2, 1]). Thus, A ⊂ V and

B ⊂ W . Thus a metric space X is normal. �

Exercise 4.23. First, we prove that if f is convex in (a, b) and if a < s <

t 0 such that t = λs + (1 − λ)u. Since f is convex in

(a, b),

f(t) ≤ λf(s) + (1 − λ)f(u).

49

Let

m =f(u) − f(s)

u − s(t − s) + f(s)

= (f(u) − f(s))(1 − λ) + f(s)

= λf(s) + (1 − λ)f(u).

Thusm − f(s)

t − s=

f(u) − f(s)

u − s.

Hence

f(t) − f(s)

t − s≤ (λf(s) + (1 − λ)f(u)) − f(s)

t − s

=m − f(s)

t − s=

f(u) − f(s)

u − s.

Similarly, f(u)−f(s)u−s

≤ f(u)−f(t)u−t

.

Lemma. If f is convex on (a, b) and if x, y, x′, y′ are points of (a, b) with

x ≤ x′ < y′ and x < y ≤ y′, then the chord over (x′, y′) has larger slope than

the chord over (x, y); that is,

f(y)− f(x)

y − x≤ f(y′) − f(x′)

y′ − x′ .

The above lemma is very easy by our previous result. Now we turn to

show that f is continuous on (a, b). Choose p ∈ (a, b). Since p is an interior

point of (a, b), there exists α > 0 such that (p − α, p + α) ⊂ (a, b). Hence

[c, d] ⊂ (a, b) where c = p − α/2, d = p + α/2. Then by lemma, we have

f(c) − f(a)

c − a≤ f(y) − f(x)

y − x≤ f(b) − f(d)

b − d

for x, y ∈ [c, d]. Thus |f(y)−f(x)| ≤ M |x−y| in [c, d]. Hence f is continuous

at x = p.

Finally, we show that every increasing convex function of a convex func-

tion is convex. Let f(x) be a convex function on (a, b), g(x) be an increasing

convex function on f((a, b)). For any x, y ∈ (a, b), 0 < λ < 1,

g(f(λx + (1 − λ)y)) ≤ g(λf(x) + (1 − λ)f(y))

≤ λg(f(x)) + (1 − λ)g(f(y))

50

since g is increasing. Hence g(f(x)) is convex. �

Exercise 4.24. Claim 1.

f

(x1 + · · · + x2n

2n

)

≤ f(x1) + · · ·+ f(x2n)

2n.

for all n.

Induction on n. When n = 1, it is OK. Assume n = k the conclusion

holds, then as n = k + 1,

f

(x1 + · · ·+ x2k+1

2k+1

)

= f

(1

2(x1 + · · ·+ x2k

2k+

x2k+1 + · · ·+ x2k+1

2k)

)

≤ 1

2

(

f

(x1 + · · ·+ x2k

2k

)

+ f

(x2k+1 + · · · + x2k+1

2k

))

≤ 1

2

(f(x1) + · · ·+ f(x2k)

2k+

f(x2k+1) + · · ·+ f(x2k+1)

2k

)

=f(x1) + · · ·+ f(x2k+1)

2k+1.

by induction hypothesis. Hence by the induction, claim 1 holds.

Claim 2.

f

(x1 + · · · + xn

n

)

≤ f(x1) + · · ·+ f(xn)

n.

for all n.

Reverse induction on n. Put

xn =x1 + · · ·+ xn−1

n − 1.

Then

f(xn) = f

(x1 + · · ·+ xn−1

n+

xn

n

)

= f

(x1 + · · ·+ xn

n

)

≤ f(x1) + · · · + f(xn)

n.

51

So

f

(x1 + · · · + xn−1

n − 1

)

≤ f(x1) + · · ·+ f(xn−1)

n − 1.

Hence claim 2 holds.

Back to the proof. Given x, y ∈ (a, b), λ ∈ (0, 1). irst suppose λ = p/q ∈(0, 1) is rational, where p, q ∈ N. Let xi = x for 1 ≤ i ≤ p, and xj = y for

p + 1 ≤ j ≤ q. Hence

f

(p

qx +

q − p

qy

)

≤ p

qf(x) +

(

1 − p

q

)

f(y).

Finally we take λ ∈ (0, 1) and choose a rational sequence λn → λ. Since

f is continuous,

f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y).

�

Exercise 4.25. We prove (a). Take z /∈ K + C. (If we cannot take such z,

then K + C = Rk. K + C is closed trivially.) Put F = z − C, the set of all

z − y with y ∈ C. Then K and F are disjoint. Note that F is closed, and

we choose δ as in Exercise 21.

We will show that the open ball with center z and radius δ does not

intersect K + C. If not, there exists y ∈ K + C such that d(z,y) ≤ δ. Since

y ∈ K + C, ∃k ∈ K, c ∈ C such that y = k + c. Let f = z − c ∈ F . Thus

d(k, f) = d(y − c, z − c) = d(y, z) > δ,

a contradiction.

We prove (b). Let A = C1 +C2. Claim: 0 is a limit point of A. Let ǫ > 0.

Take arbitrary n ∈ N such that 1/n < ǫ, and we partition the interval [0, 1)

into n subintervals:[

0,1

n

)

,

[1

n,2

n

)

, · · · ,

[n − 1

n, 1

)

.

Since n + 1 numbers,

0 · α − [0 · α], 1 · α − [1 · α], · · · , nα − [nα],

52

are all in [0, 1), there exist two different integer k, l with 0 ≤ k, l ≤ n such

that kα − [kα] and lα − [lα] in the same subinterval. If the front number is

larger, then

0 < (kα − [kα]) − (lα − [lα]) <1

n< ǫ.

Let u = [lα] − [kα], v = k − l; thus

0 < u + vα < ǫ

where u + vα ∈ Nǫ(0)⋂

(A − {0}). Thus 0 is a limit point of A.

For any x ∈ R and a neighborhood N of x, we can take two reals a, b

such that x ∈ (a, b) ⊂ N . According the previous conclusion, we can find

two integers u, v such that

0 1, there exists an integer m such that

a

u + vα< m <

b

u + vα,

a < m(u + vα) < b.

Thus m(u + vα) ∈ N⋂

A, that is, N⋂

A is not empty. Thus x is a limit

point of A.

Note. Let C1 = {n + 1/3n|n ∈ N} and C2 = {−n + 1/3n|n ∈ N}. Then

C1 and C2 are closed but C1 + C2 isn’t. �

Exercise 4.26. Let Z ′ = g(Y ) be a metric space induced from Z. Then

g : Y → Z ′ is injective and surjective. Since Y is compact, so is Z ′. Also, the

inverse of g, denoted by g−1 is also continuous. Since Y and Z ′ are compact,

by Theorem 4.19 both g and g−1 are uniformly continuous. By Exercise 4.12,

we know that

f(x) = g−1(h(x))

is uniformly continuous if h is uniformly continuous. Also, by Theorem 4,9,

f(x) = g−1(h(x))

53

is continuous if h is continuous.

For a counterexample, let X = Z = [0, 1], Y = [0, 1/2) ∪ [1, 3/2],

f(x) =

{

x if x ∈ [0, 1/2),

x + 1/2 if x ∈ [1/2, 1],

g(x) =

{

x if x ∈ [0, 1/2),

x − 1/2 if x ∈ [1, 3/2].

Observe that g is uniformly continuous on Y , and h(x) = x is uniformly

continuous on Z, but f is not (uniformly) continuous on X. �

Exercise 5.1. |f(x)−f(y)| ≤ (x−y)2 for all real x and y. Fix y, | f(x)−f(y)x−y

| ≤|x − y|. Let x → y, therefore,

0 ≤ limx→y

f(x) − f(y)

x − y≤ lim

x→y|x − y| = 0.

It implies that (f(x) − f(y))/(x − y) → 0 as x → y. Hence f ′(y) = 0, or

f = constant. �

Exercise 5.2. For every pair x > y in (a, b), f(x) − f(y) = f ′(c)(x − y)

where y < c < x by Mean Value Theorem. Note that c ∈ (a, b) and f ′(x) > 0

in (a, b), hence f ′(c) > 0. f(x) − f(y) > 0 or f(x) > f(y) if x > y, i.e., f is

strictly increasing in (a, b).

Let ∆g = g(x0 + h) − g(x0). Note that x0 = f(g(x0)). So

(x0 + h) − x0 = f(g(x0 + h)) − f(g(x0)),

and

h = f(g(x0) + ∆g) − f(g(x0)) = f(g + ∆g) − f(g).

Apply the fundamental lemma of differentiation,

h = [f ′(g) + η(∆g)]∆g,

1

f ′(g) + η(∆g)=

∆g

h

54

Note that f ′(g(x)) > 0 for all x ∈ (a, b) and η(∆g) → 0 as h → 0, thus,

limh→0

∆g/h = limh→0

1

f ′(g) + η(∆g)=

1

f ′(g(x)).

Thus g′(x) = 1f ′(g(x))

, g′(f(x)) = 1f ′(x)

. �

Exercise 5.3. For every x < y with x, y ∈ R, we will show that f(x) 6= f(y).

By Mean Value Theorem, g(x) − g(y) = g′(c)(x − y) for some x < c < y.

(x − y) + ǫ((x) − g(y)) = (ǫg′(c) + 1)(x − y),

that is,

f(x) − f(y) = (ǫg′(c) + 1)(x − y). (*)

Since |g′(x)| ≤ M , −M ≤ g′(x) ≤ M for all x ∈ R. Thus 1 − ǫM ≤ǫg′(c) + 1 ≤ 1 + ǫM , where x < c < y. Take c = 1

2M, and ǫg′(c) + 1 > 0

where x < c < y for all x, y. Take into equation (*), and f(x) − f(y) < 0

since x − y < 0, that is, f(x) 6= f(y), that is, f is injective. �

Exercise 5.4. Let f(x) = C0x + · · ·+ Cn

n+1xn+1. f is differentiable in R and

f(0) = f(1) = 0. Thus, f(1) − f(0) = f ′(c) for some c ∈ (0, 1) by Mean

Value Theorem. Also,

f ′(x) = C0 + C1x + · · ·+ Cn−1xn−1 + Cnx

n.

Thus, c ∈ (0, 1) is one real root between 0 and 1 of that equation. �

Exercise 5.5. f(x + 1) − f(x) = f ′(c)(x + 1 − x) where x < c < x + 1 by

Mean Value Theorem. Thus, g(x) = f ′(c) where x < c < x + 1, that is,

limx→+∞

g(x) = limx→+∞

f ′(c) = limc→+∞

f ′(c) = 0.

�

Exercise 5.6. Our goal is to show g′(x) > 0 for all x > 0 ⇔ g′(x) =xf ′(x)−f(x)

x2 > 0 ⇔ f ′(x) > f(x)x

. Since f ′(x) exists, f(x) − f(0) = f ′(c)(x− 0)

55

where 0 < c < x by Mean Value Theorem. ⇒ f ′(c) = f(x)x

where 0 < c < x.

Since f ′ is monotonically increasing, f ′(x) > f ′(c), that is, f ′(x) > f(x)x

for

all x > 0. �

Exercise 5.7.

f ′(t)

g′(t)=

limt→xf(t)−f(x)

t−x

limt→xg(t)−g(x)

t−x

=limt→x f(t)

limt→x g(t)= lim

t→x

f(t)

g(t)

Surely, this holds also for complex functions. �

Exercise 5.8. Since f ′(x) is continuous on a compact set [a, b], f ′(x) is

uniformly continuous on [a, b]. Hence, for any ǫ > 0 there exists δ > 0 such

that

|f ′(t) − f ′(x)| < ǫ

whenever 0 < |t−x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Thus, f ′(c) = f(t)−f(x)t−x

where

c between t and x by Mean Value Theorem. Note that 0 < |c − x| < δ and

thus |f ′(c) − f ′(x)| < ǫ, thus,

|f(t) − f(x)

t − x− f ′(x)| < ǫ

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.

It does not hold for vector-valued functions. Consider f(x) = (cos x, sin x),

[a, b] = [0, 2π], and x = 0. Hence f ′(x) = (− sin x, cos x). Take any 1 > ǫ > 0,

there exists δ > 0 such that∣∣∣∣

f(t) − f(0)

t − 0− f ′(0)

∣∣∣∣< ǫ

whenever 0 < |t| < δ by our hypothesis. Thus∣∣∣∣

(cos t − 1

t,sin t

t

)

− (0, 1)

∣∣∣∣< ǫ,

∣∣∣∣

(cos t − 1

t,sin t

t− 1

)∣∣∣∣< ǫ,

(cos t − 1

t)2 + (

sin t

t− 1

)2

< ǫ2 < ǫ,

56

2

t2+ 1 − 2(cos t + sin t)

t< ǫ

since 1 > ǫ > 0. Note that

2

t2+ 1 − 4

t<

2

t2+ 1 − 2(cos t + sin t)

t

But 2t2

+ 1 − 4t→ +∞ as t → 0. It is absurd. �

Exercise 5.8. Note. We prove a more general exercise as following. Suppose

that f is continuous on an open interval I containing x0, suppose that f ′ is

defined on I except possibly at x0, and suppose that f ′(x) → L as x → x0.

Prove that f ′(x0) = L.

By L’Hospital’s rule,

limh→0

f(x0 + h) − f(x0)

h= lim

h→0f ′(x0 + h)

By our hypothesis, f ′(x) → L as x → x0. Thus,

limh→0

f(x0 + h) − f(x0)

h= L,

Thus f ′(x0) exists and f ′(x0) = L. �

Exercise 5.10. Write f(x) = f1(x) + if2(x), where f1(x), f2(x) are real-

valued functions. Thus,

df(x)

dx=

df1(x)

dx+ i

df2(x)

dx,

Apply L’Hospital’s rule to f1(x)x

and f2(x)x

, we have

limx→0

f1(x)

x= lim

x→0f ′

1(x) and limx→0

f2(x)

x= lim

x→0f ′

2(x)

Combine f1(x) and f2(x), we have

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0

f1(x)

x+ i

f2(x)

x= lim

x→0

f(x)

x,

or

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0f ′

1(x) + i limx→0

f ′2(x) = lim

x→0f ′(x)

57

Thus, limx→0f(x)

x= limx→0 f ′(x). Similarly, limx→0

g(x)x

= limx→0 g′(x). Note

that B 6= 0. Thus,

limx→0

f(x)

g(x)= lim

x→0

(f(x)

x− A

)x

g(x)+ A

x

g(x)

= (A − A)1

B+

A

B=

A

B.

In Theorem 5.13, we know g(x) → +∞ as x → 0. (f(x) = x, and

g(x) = x + x2ei

x2 ). �

Exercise 5.11. By L’Hospital’s rule with respect to h,

limh→0

f(x + h) + f(x − h) − 2f(x)

h2= lim

h→0

f ′(x + h) − f ′(x − h)

2h.

Note that

f ′′(x) =1

2(f ′′(x) + f ′′(x))

=1

2(limh→0

f ′(x + h) − f ′(x)

h+ lim

h→0

f ′(x − h) − f ′(x)

−h)

=1

2limh→0

f ′(x + h) − f ′(x − h)

h

= limh→0

f ′(x + h) − f ′(x − h)

2h.

Thus,f(x + h) + f(x − h) − 2f(x)

h2→ f ′′(x)

as h → 0.

Counterexample: f(x) = x|x| on R. �

Exercise 5.12. f ′(x) = 3|x|2 if x 6= 0. Consider

f(h) − f(0)

h=

|h|3h

.

Note that |h|/h is bounded and |h|2 → 0 as h → 0. Thus,

f ′(0) = limh→0

f(h) − f(0)

h= 0.

58

Hence, f ′(x) = 3|x|2 for all x. Similarly,

f ′′(x) = 6|x|.

Thus,f ′′(h) − f(0)

h= 6

|h|h

Since |h|h

= 1 if h > 0 and = −1 if h < 0, f ′′′(0) does not exist. �

Exercise 5.13. For (a). (⇒) f is continuous iff for any sequence {xn} → 0

with xn 6= 0, xan sin x−c

n → 0 as n → ∞. In particular, we take

xn =

(1

2nπ + π/2

) 1

c

> 0

and thus xan → 0 as n → ∞. Hence a > 0. If not, then a = 0 or a < 0. When

a = 0, xan = 1. When a < 0, xa

n = 1/x−an → ∞ as n → ∞. It contradicts.

(⇐) f is continuous on [−1, 1] − {0} clearly. Note that

−|xa| ≤ xa sin (x−c) ≤ |xa|,

and |xa| → 0 as x → 0 since a > 0. Thus f is continuous at x = 0. Hence f

is continuous.

For (b). f ′(0) exists iff xa−1 sin (x−c) → 0 as x → 0. In the previous proof

we know that f ′(0) exists if and only if a − 1 > 0. Also, f ′(0) = 0.

Exercise 5.15. Suppose h > 0. By using Taylor’s theorem, f(x + h) =

f(x) + hf ′(x) + h2

2f ′′(ξ) for some x < ξ < x + 2h. Thus h|f ′(x)| ≤ |f(x +

h)| + |f(x)| + h2

2|f ′′(ξ)|, h|f ′(x)| ≤ 2M0 + h2

2M2, or

h2M2 − 2h|f ′(x)| + 4M0 ≥ 0. (*)

Since equation (*) holds for all h > 0, its determinant must be non-positive.

Thus 4|f ′(x)|2 − 4M2(4M0) ≤ 0, |f ′(x)|2 ≤ 4M0M2, or

(M1)2 ≤ 2M0M2

�

59

Note: There is a similar exercise. Suppose f(x) defined on R is a twice-

differentiable real-valued function, and

Mk = sup−∞<x<+∞

|f (k)(x)| < +∞

for k = 0, 1, 2. Prove that M21 ≤ 2M0M2. Write

f(x + h) = f(x) + f ′(x)h +f ′′(ξ1)

2h2 (*)

for x < ξ1 < x + h or x > ξ1 > x + h, and

f(x − h) = f(x) − f ′(x)h +f ′′(ξ2)

2h2 (**)

for x− h < ξ2 < x or x− h > ξ2 > x. (*) minus (**) implies that f(x + h)−f(x − h) = 2f ′(x)h + h2

2(f ′′(ξ1) − f ′′(ξ2)). 2h|f ′(x)| ≤ |2hf ′(x)| implies that

2h|f ′(x)| ≤ |f(x + h)| + |f(x − h)| + h2

2(|f ′′(ξ1)| + |f ′′(ξ2)|), or 2h|f ′(x)| ≤

2M0 + h2M2. Thus

M2h2 − 2|f ′(x)|h + 2M0 ≥ 0.

Since this equation holds for all h, its determinant must be non-positive.

Thus 4|f ′(x)|2 − 4M2(2M0) ≤ 0, |f ′(x)|2 ≤ 2M0M2, or

M21 ≤ 2M0M2

Exercise 5.16. Suppose a ∈ (0,∞), and M0, M1, M2 are the least up-

per bounds of |f(x)|, |f ′(x)|, |f ′′(x)| on (a,∞). Hence, M21 ≤ 4M0M2. Let

a → ∞, M0 = sup |f(x)| → 0. Since M2 is bounded, therefore M21 → 0 as

a → ∞. It follows that sup |f ′(x)| → 0 as x → ∞. �

Exercise 5.17. We take α = 0andβ = 1 in Theorem 5.15, and we obtain

that

f(1) = f(0) + f ′(0) +f ′′(0)

2+

f (3)(s)

6

60

where s ∈ (0, 1). Take α = 0andβ = −1, and

f(−1) = f(0) − f ′(0) +f ′′(0)

2− f (3)(t)

6

where t ∈ (−1, 0). Thus

1 =f ′′(0)

2+

f (3)(s)

6where s ∈ (0, 1), (*)

0 =f ′′(0)

2− f (3)(s)

6where s ∈ (−1, 0). (**)

(*) minus (**) and thus

f (3)(s)

6+

f (3)(t)

6where s ∈ (0, 1) and t ∈ (−1, 0),

or

f (3)(s) + f (3)(t) = 6 where s, t ∈ (−1, 1).

Hence

f (3)(x) ≥ 3 for some x ∈ (−1, 1).

�

Exercise 5.19. For (a). Write Dn = f(βn)−f(0)βn

βn

βn−αn+ f(αn)−f(0)

αn

−αn

βn−αn.

Note that f ′(0) = limn→∞f(αn)−f(0)

αn= limn→∞

f(βn)−f(0)βn

. Thus for any

ǫ > 0, there exists N such that L − ǫ < f(αn)−f(0)αn

< L + ǫ, and L − ǫ <f(βn)−f(0)

βn< L + ǫ whenever n > N where L = f ′(0) respectively. Note that

βn/(βn − αn) and −αn/(βn − αn) are positive. Hence,

βn

βn − αn

(L − ǫ) <f(βn) − f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ǫ),

and−αn

βn − αn(L − ǫ) <

f(αn) − f(0)

αn

−αn

βn − αn<

−αn

βn − αn(L + ǫ).

Combine two inequations, and we get L− ǫ < Dn < L + ǫ. Hence, lim Dn =

L = f ′(0).

For (b). We process as above prove. Note that −αn/(βn − αn) < 0 only

implies the following two inequations

βn

βn − αn

(L − ǫ) <f(βn) − f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ǫ),

61

−αn

βn − αn

(L + ǫ) <f(αn) − f(0)

αn

−αn

βn − αn

<−αn

βn − αn

(L − ǫ).

Combine them, and we have

L − βn + αn

βn − αnǫ < Dn < L +

βn + αn

βn − αnǫ.

Note that {βn/(βn −αn)} is bounded, ie,∣∣∣

βn

βn−αn

∣∣∣ ≤ M for some constant M .

Thus ∣∣∣∣

βn + αn

βn − αn

∣∣∣∣=

∣∣∣∣

2βn

βn − αn− 1

∣∣∣∣≤ 2M + 1.

Hence, L − (2M + 1)ǫ < Dn < L + (2M + 1)ǫ, or lim Dn = L = f ′(0).

For (c). By Mean Value Theorem, Dn = f ′(tn) for some tn between

αn and βn. Note that min{αn, βn} < tn < max{αn, βn}, max{αn, βn} =12(αn + βn + |αn − βn|), and min{αn, βn} = 1

2(αn + βn − |αn − βn|). Thus,

max{αn, βn} → 0 and min{αn, βn} → 0 as αn → 0 and βn → 0. By squeezing

principle for limits, tn → 0. With the continuity of f ′, we have

lim Dn = lim f ′(tn) = f ′(lim tn) = f ′(0).

Example: Let f be defined by

f(x) =

{

x2 sin(1/x) (x 6= 0),

0 (x = 0).

Thus f ′(x) is not continuous at x = 0, and f ′(0) = 0. Take αn = 1π/2+2nπ

and βn = 12nπ

. It is clear that αn → 0, and βn → 0 as n → ∞. Also,

Dn =−4nπ

π(π/2 + 2nπ)→ −2

π

as n → ∞. Thus, lim Dn = −2/π exists and is different from 0 = f ′(0). �

Exercise 5.22. For (a). If not, then there exists two distinct fixed points,

say x and y, of f . Thus f(x) = x and f(y) = y. Since f is differentiable,

by applying Mean Value Theorem we know that f(x) − f(y) = f ′(t)(x − y)

where t is between x and y. Since x 6= y, f ′(t) = 1 and it is absurd.

For (b). We show that 0 < f ′(t) < 1 for all real t. First, f ′(t) =

1 + (−1)(1 + et)−2et = 1− et

(1+et)2. Since et > 0, (1 + et)2 = (1 + et)(1 + et) >

62

1(1 + et) = 1 + et > et > 0 for all real t. Thus 1 > (1 + et)−2et > 0 for all

real t. Hence 0 < f ′(t) < 1 for all real t.

Next, since f(t)− t = (1− et)−1 > 0 for all real t, f(t) has no fixed point.

For (c). Suppose xn+1 6= xn for all n. (If xn+1 = xn, then xn = xn+1 = · · ·and xn is a fixed point of f).

By Mean Value Theorem, f(xn+1) − f(xn) = f ′(tn)(xn+1 − xn) where tn

is between xn and xn+1. Thus,

|f(xn+1) − f(xn)| = |f ′(tn)||(xn+1 − xn)|.

Note that |f ′(tn)| is bounded by A < 1, f(xn) = xn+1, and f(xn+1) = xn+2.

Thus |xn+2 − xn+1| ≤ A|xn+1 − xn| or

|xn+1 − xn| ≤ CAn−1

where C = |x2−x1|. For two positive integers p > q, |xp−xq| ≤ |xp−xp−1|+· · · + |xq+1 − xq| = C(Aq−1 + Aq + · · ·+ Ap−2) ≤ CAq−1

1−A. Hence

|xp − xq| ≤CAq−1

1 − A.

Therefore, for any ǫ > 0, there exists N = [logAǫ(1−A)

C] + 2 such that |xp −

xq| < ǫ whenever p > q ≥ N . By Cauchy criterion we know that {xn}converges to x. Thus,

limn→∞

xn+1 = f( limn→∞

xn)

since f is continuous. Thus, x = f(x), i.e., x is a fixed point of f .

For (d). Since xn+1 = f(xn), it is trivial. �

Exercise 5.25. For (b). We show that xn ≥ xn+1 ≥ ξ by induction. By

Mean Value Theorem, f(xn)−f(ξ) = f ′(cn)(xn−ξ) where cn ∈ (ξ, xn). Since

f ′′ ≥ 0, f ′ is increasing and thus f(xn)xn−ξ

= f ′(cn) ≤ f ′(xn) = f(xn)xn−xn+1

, or

f(xn)(xn − ξ) ≤ f(xn)(xn − xn+1).

Note that f(xn) > f(ξ) = 0 since f ′ ≥ δ > 0 and f is strictly increasing.

Thus, xn − ξ ≤ xn − xn+1 or ξ ≤ xn+1.

63

Note that f(xn) > 0 and f ′(xn) > 0. Thus xn+1 < xn. Hence, xn >

xn+1 ≥ ξ. Thus, {xn} converges to a real number ζ . Suppose ζ 6= ξ, then

xn+1 = xn − f(xn)

f ′(xn).

Note that f(xn)f ′(xn)

> f(ζ)δ

. Let α = f(ζ)δ

> 0, be a constant. Thus,

xn+1 < xn − α

for all n. Thus, xn < x1 − (n − 1)α, that is, xn → −∞ as n → ∞. It

contradicts. Thus, {xn} converges to ξ.

For (c). By Taylor’s theorem, f(ξ) = f(xn)+ f ′(xn)(ξ−xn)+ f ′′(tn)2(xn−ξ)2

, or

0 = f(xn)+ f ′(xn)(ξ−xn)+ f ′′(tn)2(xn−ξ)2

, or 0 = f(xn)f ′(xn)

−xn + ξ + f ′′(tn)2f ′(xn)

(xn − ξ)2.

Thus

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

where tn ∈ (ξ, xn).

For (d). By (b) we know that 0 ≤ xn+1 − ξ for all n. Again by (c) we

know that xn+1 − ξ = f ′′(tn)2f ′(xn)

(xn − ξ)2. Note that f ′′ ≤ M and f ′ ≥ δ > 0.

Thus xn+1 − ξ ≤ A(xn − ξ)2 ≤ 1A(A(x1 − ξ))2n

by the induction. Thus,

0 ≤ xn+1 − ξ ≤ 1

A[A(x1 − ξ)]2

n

.

For (e). If x0 is a fixed point of g(x), then g(x0) = x0, that is, x0− f(x0)f ′(x0)

=

x0 or f(x0) = 0. It implies that x0 = ξ and x0 is unique since f is strictly

increasing. Thus, we choose x1 ∈ (ξ, b) and apply Newton’s method, we can

find out ξ. Hence we can find out x0. Next, by calculating g′(x) = f(x)f ′′(x)f ′(x)2

,

we have

0 ≤ g′(x) ≤ f(x)M

δ2.

As x near ξ from right hand side, g′(x) near f(ξ) = 0.

For (f). xn+1 = xn − f(xn)f ′(xn)

= −2xn by calculating. Thus, xn = (−2)n−1x1

for all n, thus {xn} does not converges for any choice of x1, and we cannot

find ξ such that f(ξ) = 0 in this case. �

64

Exercise 5.26. Suppose A > 0. (If not, then f = 0 on [a, b] clearly.) Fix

x0 ∈ [a, b], let M0 = sup |f(x)|, M1 = sup |f ′(x)| for a ≤ x ≤ x0. For any such

x, f(x)−f(a) = f ′(c)(x−a) where c is between x and a by using Mean Value

Theorem. Thus |f(x)| ≤ M1(x − a) ≤ M1(x0 − a) ≤ A(x0 − a)M0. Hence

M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0] by taking x0 = a + 12A

.

Repeat the above argument by replacing a with x0, and note that 12A

is a

constant. Hence, f = 0 on [a, b]. �

Exercise 5.27. Suppose y1 and y2 are solutions of that problem. Since

|φ(x, y2) − φ(x, y1)| ≤ A|y2 − y1|, y(a) = c, y′1 = φ(x, y1), and y′

2 = φ(x, y2),

by Exercise 26 we know that y1 − y2 = 0, y1 = y2. Hence, such a problem

has at most one solution.

Note: Suppose there is initial-value problem

y′ = y1/2 with y(0) = 0.

If y1/2 6= 0, then y1/2dy = dx. By integrating each side and noting that

y(0) = 0, we know that f(x) = x2/4. With y1/2 = 0, or y = 0. All solutions

of that problem are

f(x) = 0 and f(x) = x2/4.

Why the uniqueness theorem does not hold for this problem? One reason is

that there does not exist a constant A satisfying

|y′1 − y′

2| ≤ A|y1 − y2|if y1 and y2 are solutions of that problem. (since 2/x → ∞ as x → 0 and

thus A does not exist.) �

Theorem. Let φj(j = 1, · · · , k) be real functions defined on a rectangle Rj

in the plane given by a ≤ x ≤ b, αj ≤ yj ≤ βj.

A solution of the initial-value problem

y′j = φ(x, yj), yj(a) = cj (αj ≤ cj ≤ βj)

is, by definition, a differentiable function fj on [a, b] such that fj(a) = cj,

αj ≤ fj(x) ≤ βj, and

f ′j(x) = φj(x, fj(x)). (a ≤ x ≤ b)

65

Then this problem has at most one solution if there is a constant A such that

|φj(x, yj2) − φj(x, yj1)| ≤ A|yj2 − yj1|

whenever (x, yj1) ∈ Rj and (x, yj2) ∈ Rj.

Proof. Suppose y1 and y2 are solutions of that problem. For each compo-

nents of y1 and y2, say y1j and y2j respectively, y1j = y2j by using Exercise

26. Thus, y1 = y2 �

Exercise 5.29. Theorem. Let Rj be a rectangle in the plain, given by

a ≤ x ≤ b, min yj ≤ yj ≤ max yj. (since yj is continuous on the compact

set, say [a, b], we know that yj attains minimal and maximal.) If there is a

constant A such that{

|yj+1,1 − yj+1,2| ≤ A|yj,1 − yj,2| (j < k)

|∑kj=1 gj(x)(yj,1 − yj,2)| ≤ A|yk,1 − yk,2|

whenever (x, yj,1) ∈ Rj and (x, yj,2) ∈ Rj.

Proof. Since the system y′1, · · · , y′

k with initial conditions satisfies a fact

that there is a constant A such that |y′1−y′

2| ≤ A|y1 −y2|, that system has

at most one solution. Hence,

y(k) + gk(x)y(k−1) + · · ·+ g2(x)y′ + g1(x)y = f(x),

with initial conditions has at most one solution. �

Exercise 6.1. Note that L(P, f, α) = 0 for all partition P of [a, b]. Thus∫ a

b

fdα = 0.

Take a partition P such that

P =

{

a, a +1

n(b − a), · · · , a +

k

n(b − a), · · · , a +

n − 1

n(b − a), b

}

for all positive integer n > 1. Thus

U(P, f, α) =

n∑

i=1

Mi∆αi ≤2(b − a)

n

66

for all positive integer n > 1. Thus

0 ≤ inf U(P, f, α) ≤ 2(b − a)

n

for all positive integer n. Thus inf U(P, f, α) = 0. Hence∫ a

bfdα = 0; thus,

∫ a

bfdα = 0. �

Exercise 6.2. If not, then there is p ∈ [a, b] such that f(p) > 0. Since

f is continuous at x = p, for ǫ = f(p)/2, there exist δ > 0 such that

|f(x) − f(p)| < ǫ whenever x ∈ (x − δ, x + δ)⋂

[a, b], that is,

0 <1

2f(p) < f(x) <

3

2f(p)

for x ∈ Br(p) ⊂ [a, b] where r is small enough. Next, consider a partition P

of [a, b] such that

P =

{

a, p − r

2, p +

r

2, b

}

.

So

L(P, f) ≥ r · 1

2f(p) =

rf(p)

2,

or

sup L(P, f) ≥ L(P, f) ≥ rf(p)

2> 0.

It is absurd since∫ b

af(x)dx = sup L(P, f) = 0. Hence f = 0 for all x ∈ [a, b].

� Note: The above conclusion holds under the condition that f is continu-

ous. If f is not necessary continuous, then we cannot get this conclusion. (A

counter-example is shown in Exercise 6.1).

Exercise 6.4. If f(x) = 0 for all irrational x, f(x) = 1 for all rational x,

prove that f /∈ R on [a, b] for any a < b.

Proof. Take any partition P of [a, b], say

a = x0 ≤ x1 ≤ · · · ≤ xn−1 ≤ xn = b.

By P we can construct the new partition P ′ without repeated points, and

U(P, f) = U(P ′f), L(P, f) = L(P ′, f). Say P ′

a = y0 < y1 < · · · < ym−1 < ym = b.

67

Note that Q is dense in R, and R \ Q is also dense in R. Hence,

Mi = supyi−1≤x≤yi

f(x) = 1,

mi = infyi−1≤x≤yi

f(x) = 0.

Hence,

U(P, f) = U(P ′, f) =

m∑

i=1

Mi∆yi = b − a,

L(P, f) = L(P ′, f) =m∑

i=1

mi∆yi = 0.

for any partition of [a, b]. So,

∫ b

a

fdx = inf U(P, f) = b − a,

∫ b

a

fdx = sup L(P, f) = 0.

Thus, f /∈ R on [a, b]. �

Exercise 6.5. The first answer is NO. Define f(x) = −1 for all irrational

x ∈ [a, b], f(x) = 1 for all rational x ∈ [a, b]. Similarly, f /∈ R by Exercise

6.4.

However, if we assume that f 3 ∈ R, the answer is YES. Let Φ = x1/3,

we apply Theorem 6.11 and thus get f ∈ R. �

Exercise 6.7. (a) Take any c > 0, and note that f is bounded, say |f | ≤ M

for some M . Hence,

∫ c

0

−Mdx ≤∫ c

0

f(x)dx ≤∫ c

0

Mdx,

−cM ≤∫ c

0

f(x)dx ≤ cM,

−cM ≤∫ 1

c

f(x)dx −∫ 1

0

f(x)dx ≤ cM.

68

Letting c → 0 and thus∫ 1

cf(x)dx −

∫ 1

0f(x)dx → 0, that is,

∫ 1

cf(x)dx →

∫ 1

0f(x)dx as c → 0. Thus this definition of the integral agrees with the old

one.

(b) Define

f(x) = n(−1)n, where1

n + 1< x ≤ 1

n

or

f(x) =sin(1/x)

x.

�

Exercise 6.8. Since f is nonnegative and decreases monotonically on [1,∞),

we have for n ≥ 2,

k∑

n=2

f(n) ≤∫ k

1

f(x)dx ≤k−1∑

n=1

f(n).

We define

g(x) =

∫ y

1

f(t)dt.

If this integral converges, then g(x) is a non-decreasing function which tends

to a limit, and so g(k) is a bounded non-decreasing sequence. Thus denote

sk =∑k

n=2 f(n), we see that sk ≤ g(k), and sk tends to a limit. The fact that∑∞

n=1 f(n) converges if∫ ∞1

f(x)dx converges is now established. If the inte-

gral diverges, then g(x) → +∞ as x → +∞. Therefore g(k) → +∞. Since

g(k) ≤ ∑k−1n=1 f(n), we conclude that the series diverges. Hence

∫ ∞1

f(x)dx

converges if∑∞

n=1 f(n). Thus,∫ ∞

1f(x)dx converges if and only if

∑∞n=1 f(n)

converges. �

Exercise 6.9. Integration by parts gives us

∫ M

0

cos x

1 + xdx =

sin x

1 + x

∣∣∣∣

x=M

x=0

+

∫ M

0

cos x

(1 + x)2dx

=sin M

1 + M+

∫ M

0

sin x

(1 + x)2dx

69

for M > 0. By letting M → ∞, we have∫ ∞

0

cos x

1 + xdx =

∫ ∞

0

sin x

(1 + x)2dx.

Also,∫ ∞

0cos x1+x

dx does not converges abolutely, but∫ ∞0

sin x(1+x)2

dx does.

Exercise 6.10. We prove (a). Claim: if a and b are positive and 0 < α < 1,

then

aαb1−α ≤ αa + (1 − α)b

and that the equality holds if and only if a = b. To prove it, we let f(x) =

αx + (1 − α) − xα, where x = a/b. Thus

f ′(x) = α − αxα−1 = α(1 − xα−1).

Hence f(x) attains its minimum at x = 1 (, a = b,) and f(1) = 0. Hence

f(x) = αx + (1 − α) − xα ≥ 0. Put x = a/b,

α

(a

b

)

− (1 − α) −(

a

b

)α

≥ 0

or

aαb1−α ≤ αa + (1 − α)b.

Next, let α = 1/p, a = u1/α, and b = v1/(1−α). Hence

uv ≤ up

p+

vq

q.

Equality holds if and only if a = b, that is, up = vq.

We prove (b). If 0 ≤ f ∈ R(α) and 0 ≤ g ∈ R(α) then f p and gq ∈ R(α)

by Theorem 6.11. It also follows that fg ∈ R(α) and using the part (a):

∫ b

a

fgdα ≤ 1

p

∫ b

a

f pdα +1

q

∫ b

a

gqdα = 1.

We prove (c). If f and g are complex-valued in R(α) then |f | and |g| are

nonnegative elements of R(α) and fg ∈ R(α). Moreover

∣∣∣∣

∫ b

a

fgdα

∣∣∣∣≤

∫ b

a

|f ||g|dα.

70

If I =∫ b

a|f |p 6= 0 and J =

∫ b

a|g|q 6= 0 then apply the conclusion of the part

(b) to |f |/c and |g|/d where cp = I and dq = J . Hence

∣∣∣∣

∫ b

a

fgdα

∣∣∣∣≤ cd =

{∫ b

a

|f |pdα

}1/p{ ∫ b

a

|g|qdα

}1/q

If I = 0 (J = 0 is similar), we can reverse the roles of p and q, then

∫ b

a

|f |(c|g|)dα ≤ cq 1

p

∫ b

a

|g|qdα

for any c > 0. Since c is arbitrary,

∫ b

a

|f ||g|dα = 0.

Hence the inequality still holds.

Exercise 6.11. Put p = q = 2 in Exercise 4.10, we have

( ∫ b

a

|uv|dα

)2

≤∫ b

a

|u|2dα +

∫ b

a

|v|2dα. (*)

Hence,

‖u + v‖22 =

∫ b

a

|u + v|2dα

=

∫ b

a

|u|2dα +

∫ b

a

(uv + vu)dα +

∫ b

a

|v|2dα

=

∫ b

a

|u|2dα +

∫ b

a

|uv + vu|dα +

∫ b

a

|v|2dα

≤∫ b

a

|u|2dα + 2

∫ b

a

|uv|dα +

∫ b

a

|v|2dα (by (*))

≤({∫ b

a

|u|2dα

}1/2

+

{∫ b

a

|v|2dα

}1/2)2

= (‖u‖2 + ‖v‖2)2.

Therefore, ‖u + v‖2 ≤ ‖u‖2 + ‖v‖2. Putting u = f − g and v = h − g gives

‖f − h‖2 ≤ ‖f − g‖2 + ‖g − h‖2.

71

�

Exercise 6.13. (a) Put t2 = u,

f(x) =

∫ x+1

x

sin(t2)dt

=

∫ (x+1)2

x2

sin u

2u1/2du

=− cos u

2u1/2

∣∣∣∣

u=(x+1)2

u=x2

−∫ (x+1)2

x2

cos u

4u3/2du

=cos(x2)

2x− cos[(x + 1)2]

2(x + 1)−

∫ (x+1)2

x2

cos u

4u3/2du.

To get a bound of∫ (x+1)2

x2

cos u4u3/2 du, we replace cos u by −1:

∣∣∣∣

∫ (x+1)2

x2

cos u

4u3/2du

∣∣∣∣<

∣∣∣∣

∫ (x+1)2

x2

−1

4u3/2du

∣∣∣∣=

1

2x(x + 1)

for x > 0. Hence, for x > 0,

|f(x)| ≤∣∣∣∣

cos(x2)

2x

∣∣∣∣+

∣∣∣∣

cos[(x + 1)2]

2(x + 1)

∣∣∣∣+

∣∣∣∣

∫ (x+1)2

x2

cos u

4u3/2du

∣∣∣∣

<1

2x+

1

2(x + 1)+

1

2x(x + 1)=

1

x.

(b) By (a),

f(x) =cos(x2)

2x− cos[(x + 1)2]

2(x + 1)−

∫ (x+1)2

x2

cos u

4u3/2du,

2xf(x) = cos(x2) − cos[(x + 1)2] +1

x + 1cos[(x + 1)2]

− 2x

∫ (x+1)2

x2

cos u

4u3/2du.

Let

r(x) =1

x + 1cos[(x + 1)2] − 2x

∫ (x+1)2

x2

cos u

4u3/2du.

Note that ∣∣∣∣

1

x + 1cos[(x + 1)2]

∣∣∣∣≤ 1

x + 1<

1

x,

72

2x

∫ (x+1)2

x2

cos u

4u3/2du < 2x

1

2x(x + 1)<

1

x

if x > 0. Thus |r(x)| < 1/x + 1/x = c/x and c = 2 is a constant if x > 0.

Exercise 7.1. Prove that every uniformly convergent sequence of bounded

functions is uniformly bounded.

Proof. Let fn → f uniformly on S in a metric space. fn is bounded for all

n. Given ǫ = 1, there is an integer N such that |fn(x)−fm(x)| < 1 whenever

n, m ≥ N and x ∈ S. Put m = N , and we have

|fn(x)| ≤ |fN(x)| + 1

whenever n ≥ N and x ∈ S. Let

M = max

{

supx∈S

|f1(x)|, · · · , supx∈S

|fN−1(x)|, supx∈S

|fN(x)| + 1

}

.

(fn is bounded ⇒ the existence of supx∈S |fn(x)|.) Therefore |fn(x)| ≤ M

for all n and x ∈ S. Hence {fn} is uniformly bounded on S. �

Exercise 7.2. If {fn} and {gn} converge uniformly on a set E, prove that

{fn + gn} converges uniformly on E. If, in addition, {fn} and {gn} are

sequences of bounded functions, prove that {fngn} converges uniformly on

E.

Proof. Let hn(x) = fn(x) + gn(x) for each n. Since {fn} (resp. {gn})converges uniformly on E, given ǫ > 0, there exists N such that |fn(x) −fm(x)| < ǫ/2 (resp. |gn(x)−gm(x)| < ǫ/2) for all x ∈ E whenever n, m ≥ N .

Thus |hn(x) − hm(x)| ≤ |fn(x) − fm(x)| + |gn(x) − gm(x)| < ǫ for all x ∈ E,

i.e., {fn + gn} converges uniformly on E.

By exercise 7.1, {fn} and {gn} are uniformly bounded on E. Let M be the

bound of them, i.e., |fn(x)| ≤ M and |gn(x)| ≤ M for all x and all n. Again

we let hn(x) = fn(x)gn(x) for each n. Since {fn} (resp. {gn}) converges

uniformly on E, given ǫ > 0, there exists N such that |fn(x) − fm(x)| <

ǫ/(4M) (resp. |gn(x) − gm(x)| < ǫ/(4M)) for all x ∈ E whenever n, m ≥ N .

73

Thus,

|hn(x) − hm(x)| = |fn(x)(gn(x) − gm(x)) + gm(x)(fn(x) − fm(x))|≤ |fn(x)||gn(x) − gm(x)| + |gm(x)||fn(x) − fm(x)|< ǫ

whenever n, m ≥ N and x ∈ E, i.e, {fngn} converges uniformly on E. �

Exercise 7.3. Construct sequences {fn}, {gn} which converge uniformly on

some set E, but such that {fngn} does not converge uniformly on E. (of

course, {fngn} must converge on E).

Proof. Let E = [1, 2] ⊂ R. Let

fn(x) = x(1 +1

n),

gn(x) =

{1n

if x ∈ R \ Q

b + 1n

if x = ab

with (a, b) = 1 and b > 0

on E. ({gn(x)} is a sequence of unbounded functions clearly.)

Claim 1. {fn} and {gn} converge uniformly on E.

limn→∞

fn(x) = f(x) = x

limn→∞

gn(x) = g(x) =

{

0 if x ∈ R \ Q

b if x = ab∈ Q with (a, b) = 1 and b > 0.

Given ǫ > 0, there exists N = [2ǫ] + 1 > 2

ǫsuch that |fn(x) − f(x)| =

| xn| ≤ 2

n≤ 2

N< ǫ whenever n ≥ N , x ∈ E. Therefore, fn → f uniformly on

E. Next, given ǫ > 0, there exists N = [1ǫ]+1 > 1

ǫsuch that |gn(x)−g(x)| ≤

1n≤ 1

N< ǫ whenever n ≥ N , x ∈ E. Hence, gn → g uniformly on E.

Claim 2. {fngn} does not converge uniformly on E.

limn→∞

fn(x)gn(x) = f(x)g(x)

=

{

0 if x ∈ R \ Q

a if x = ab∈ Q with (a, b) = 1 and b > 0.

It suffices to show that {fngn} does not converge uniformly on F = E ∩ Q.

If not, given ǫ = 1 and consequently there exists N such that |fn(x)gn(x) −

74

f(x)g(x)| < 1 whenever n ≥ N and x ∈ F . Now we let x = N+1N

∈ F ,

n = N . Hence,

1 > |fn(x)gn(x) − f(x)g(x)|

=

∣∣∣∣

N + 1

N

(

1 +1

N

)(

N +1

N

)

− (N + 1)

∣∣∣∣

=

∣∣∣∣

N3 + 2N2 + 2N + 1

N3

∣∣∣∣

> 1,

and thus we reach a contradiction. �

Exercise 7.4. Consider

f(x) =

∞∑

n=1

1

1 + n2x.

For what values of x does the series converge absolutely? On what inter-

vals does it converge uniformly? On what intervals does it fail to converge

uniformly? Is f continuous whenever the series converges? Is f bounded?

Proof. Let sm(x) =∑n=m

n=11

1+n2xbe the partial sum of f .

First, we will find values of x such that the series converge abso-

lutely. x /∈ {− 1k2 | k ∈ N} clearly. When x = 0, the series

∑1 diverges.

When x > 0, the series∑

(1+n2x)−1 = x−1∑

1x−1+n2 < x−1

∑n−2 converges

(absolutely) by the comparison test. To complete our calculation, we consider

the rest case x ∈ R− \{− 1k2 | k ∈ N}. Recall:

∑(1+n2x)−1 = x−1

∑1

x−1+n2 .

For each fixed x, there exists N such that x−1 + n2 > 0 whenever n ≥ N .

Hence, it suffices to determine whether the series

∞∑

n=N

1

1 + n2x= x−1

∞∑

n=N

1

x−1 + n2

converges absolutely. It is similar to the case x > 0, and consequently the

new series from N to infinity converges absolutely too. Therefore, the series∑

(1 + n2x)−1 converges absolutely for all real x except 0 ∪ {− 1k2 | k ∈ N}.

Second, we will find intervals such that the series converge uni-

formly or does not converge uniformly. By the above deduction and

75

Weierstrass M-test (theorem 7.10), f(x) converges uniformly on [A, +∞)

with A > 0. Does f converge uniformly on (0, +∞)? The answer is no. If

not, sm → f uniformly on (0, +∞).

sm(x) =

n=m∑

n=1

1

1 + n2x<

n=m∑

n=1

1 = m

is bounded (by m) for all x ∈ (0, +∞). By exercise 7.1, f(x) is uniformly

bounded. Since

sm(m−3) =

n=m∑

n=1

1

1 + n2 · m−3>

n=m∑

n=1

1

1 + m2 · m−3

=m

1 + m2 · m−3=

m2

1 + m> m − 1

for each m, we obtain a contradiction. It also implies that f is unbounded.

Next, we consider other possible intervals that f converge uniformly on. We

list two possible cases: (i) (−∞,−1) and (ii) Ek = (−k−2,−(k + 1)−2).

Case (i). Note that 0 > (1+n2x)−1 > (1−n2)−1 for every x ∈ (−∞,−1).

Thus,

0 ≤∞∑

n=2

∣∣∣∣

1

1 + n2x

∣∣∣∣<

∞∑

n=2

∣∣∣∣

1

1 − n2

∣∣∣∣.

By Weierstrass M-test,∑∞

n=2(1 + n2x)−1 (and so f) converges uniformly on

(−∞,−1).

Case (ii). If x ∈ Ek (or −k−2 < x < −(k + 1)−2), then∣∣∣∣

1

1 + n2x

∣∣∣∣<

∣∣∣∣

1

1 − n2/k2

∣∣∣∣

for n ≥ k + 2. Anain by Weierstrass M-test,∑∞

n=k+1(1 + n2x)−1 (and so

f) converges uniformly on Ek. In sum, f converges uniformly on all real

numbers except (0, A) ∪ {−k−2 | k ∈ N}), and does not converges uniformly

on (0, A) where A > 0.

Third, we will show that f is continuous whenever f converges. It

is clear. For any x ∈ R \ (0 ∪ {−k−2 | k ∈ N}), x must lie in some closed

interval [a, b] that f converges uniformly on. Since sm are continuous for all

m, f is continuous whenever f converges. �

76

Exercise 7.5. Let

fn(x) =

0(x < 1

n+1

),

sin2 πx

(1

n+1≤ x ≤ 1

n

),

0(

1n

< x).

Show that {fn} converges to a continuous function, but not uniformly. Use

the series∑

fn to show that absolute convergence, even for all x, does not

imply uniform convergence.

Proof. By the definition of fn(x), it suffices to consider that fn(x) lie on

(0, 1) only. We will show that the limit function

limn→∞

fn(x) = 0

and thus {fn} converges to a continuous function 0. To show this, we fix

any x ∈ (0, 1). There exists N such that x > 1n

whenever n ≥ N . Hence

limn→∞ fn(x) = 0. However, {fn} does not converge uniformly. If not, then

given ǫ = 1 > 0 and consequently there is N such that |fn(x) − 0| < 1

whenever n ≥ N and x ∈ R. In particular, |fN( 1N+1/2

)| = 1 < 1, which is

absurd.

Let s(x) =∑

fn(x) and sm(x) =∑n=m

n=1 fn(x) be the partial sums of∑

fn(x). To show the absolute absolute convergence of s(x) =∑

fn(x), we

fix x ∈ (0, 1) first. Thus, there exists an unique N such that 1N+1

< x ≤ 1N

.

Hence,∑

|fn(x)| = s(x) = fN (x) = sin2 π

x,

and consequently s(x) converges absolutely on (0, 1) (and on R).

To show that {sn} is not uniformly convergent, we suppose {sn} is uni-

formly convergent and get a contradiction. By our assumption, sn → s

uniformly on R. Given ǫ = 1 > 0, there is N such that |sn(x) − s(x)| < 1

whenever n ≥ N and x ∈ R. Hence, for n = N and x = 12N+ 1

2

,

∣∣∣∣sN

(1

2N + 12

)

− s

(1

2N + 12

)∣∣∣∣= |0 − 1| = 1,

which is absurd. �

77

Exercise 7.6. 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.

Proof. Let

fn(x) = (−1)n, gn(x) =x2 + n

n2.

Fixed a bounded interval [a, b]. Note: (i)∑

fn(x) is uniformly bouned, (ii)

gn(x) → 0 uniformly on [a, b], and (iii) gn+1(x) ≤ gn(x) on [a, b]. Hence,

by Dirichlet test for uniform convergences, the series∑

(−1)n x2+nn2 converges

uniformly on every bounded interval [a, b].

Next, notice that

∞∑

n=1

|(−1)n x2 + n

n2| =

∞∑

n=1

x2 + n

n2>

∞∑

n=1

1

n

diverges for any x ∈ R, and thus∑

(−1)n x2+nn2 does not converge absolutely

for any value of x. �

Exercise 7.7. For n = 1, 2, 3, · · · , x real. Put

fn(x) =x

1 + nx2.

Show that {fn} converges uniformly to a function f , and that the equation

f ′(x) = limn→∞

f ′n(x)

is correct if x 6= 0, but false if x = 0.

Proof. For a fixed x ∈ R, limn→∞ fn(x) = 0. Also,

|fn(x) − f(x)| =

∣∣∣∣

x

1 + nx2

∣∣∣∣≤ 2√

n

for all x ∈ R. The last inequality is due to the inequality for arithmetic and

geometric means. Thus, fn → f uniformly on R.

78

f ′(x) = 0 and f ′n(x) = 1−nx2

(1+nx2)2. For x = 0, f ′(0) = 0 and limn→∞ f ′

n(0) =

limn→∞ 1 = 1. For x 6= 0, f ′(x) = 0 and

limn→∞

f ′n(x) = lim

n→∞

−nx2 + 1

n2x4 + 2nx2 + 1= 0,

i.e., f ′(x) = limn→∞ f ′n(x) if x 6= 0. �

Exercise 7.8. If

I(x) =

{

0 (x ≤ 0),

1 (x > 0),

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

that the series

f(x) =

∞∑

n=1

cnI(x − xn)

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

Proof. Note that |cn| = |cnI(x− xn)| and∑ |cn| converges. By Weierstrass

M-test, f(x) converges uniformly on (a, b).

Next, let

sm(x) =n=m∑

n=1

cnI(x − xn) = c1I(x − x1) + · · · cmI(x − xm)

for each m. sm(x) are continuous on (a, b) \ {x1, · · · , xm} and thus on

(a, b) \ {xk | k ∈ N}. Hence, f is continuous on (a, b) \ {xk | k ∈ N}.�

Exercise 7.9. Let {fn} be a sequence of continuous functions which con-

verges uniformly to a function f on a set E. Prove that

limn→∞

fn(xn) = f(x),

for every sequence of points xn ∈ E such that xn → x, and x ∈ E. Is the

converse of this true?

Proof. Given ǫ > 0. By our assumption, there exists N1 such that

|fn(x) − f(x)| <ǫ

2

79

whenever n ≥ N1 and all x ∈ E. Also, since f is continuous on E by theorem

7.12, limn→∞ f(xn) = f(x), that is, there exists N2 such that

|f(xn) − f(x)| <ǫ

2

whenever n ≥ N2. Now, if n ≥ N = max{N1, N2}, we have

|fn(xn) − f(x)| ≤ |fn(xn) − f(xn)| + |f(xn) − f(x)| <ǫ

2+

ǫ

2= ǫ,

i.e.,

limn→∞

fn(xn) = f(x).

The converse is not true. Here is our example:

fn(x) = xn (0 < x < 1).

limn→∞ fn(x) = 0 but {fn} is not convergent uniformly. �

Exercise 7.16. Suppose that {fn} is an equicontinuous sequence of functions

on a compact set K, and {fn} converges pointwise on K. Prove that {fn}converges uniformly on K.

Proof. First, by the hypothesis that {fn} is equicontinuous on K, given

ǫ > 0 and thus there exists δ > 0 such that

|fn(x) − fn(y)| <ǫ

3(*)

for all n whenever d(x, y) < δ with x, y ∈ K. For such δ > 0, we form an

open covering {Ui} of K where Ui = {x ∈ K | d(x, xi) < δ} for each xi ∈ K.

By the compactness of K, there are finitely many indices 1, · · · , r such that

{U1, · · · , Ur} also covers K.

Second, since {fn} converges pointwise on K, for a fixed x ∈ K and given

ǫ > 0, there is N such that

|fn(x) − fm(x)| <ǫ

3(**)

whenever n, m ≥ N .

80

Back to the proof. Now we fix any x ∈ K. there exists Ui such that

x ∈ Ui, i.e., d(x, xi) < δ. Hence,

|fn(x) − fn(xi)| <ǫ

3(by (*)) and

|fn(xi) − fm(xi)| <ǫ

3(by (**)) and

|fm(xi) − fm(x)| <ǫ

3(by (*)).

Three inequalities implies that

|fn(x) − fm(x)| <ǫ

3+

ǫ

3+

ǫ

3= ǫ

for all x ∈ K. Therefore, {fn} converges uniformly on K by Cauchy crite-

rion. �

Exercise 8.1. Define

f(x) =

{

e−1/x2

(x 6= 0),

0 (x = 0).

Prove that f has derivatives of all orders at x = 0, and that f (n)(0) = 0 for

n = 1, 2, 3, · · · .Proof. Induction.

f ′(0) = limx→0

e−1/x2 − 0

x − 0= lim

x→0

1/x

e1/x2= lim

y→±∞

y

ey2= 0

where y = 1/x. By induction hypothesis, f (n−1)(0) = 0. Clearly,

f (n−1)(x) = Pn−1

(1

x

)

e−1

x2

where Pn−1(1x) is a polynomial of an indeterminate variable 1

x. Hence,

f (n)(0) = limx→0

Pn−1(1/x)e−1/x2 − 0

x − 0= lim

y→±∞

yPn−1(y)

ey2= 0

where y = 1/x. �

81

Exercise 8.2. For a fixed i,

∑

j

aij =∑

ji

aij

=∑

jj

aij

= −1 +∑

i>j

2j−i = −1 + 1 = 0.

Hence∑

i

∑

j aij =∑

i(−21−i) = −2 and∑

j

∑

i aij =∑

j 0 = 0. �

Exercise 8.4. (a) By L’Hospital’s rule,

limx→0

bx − 1

x= lim

x→0

log(b)bx

1= log b

if b > 0.

(b) By L’Hospital’s rule,

limx→0

log(1 + x)

x= lim

x→0

(1 + x)−1

1= 1.

(c) By part (b),

e = exp

(

limx→0

log(1 + x)

x

)

= limx→0

exp

(log(1 + x)

x

)

= limx→0

(1 + x)1

x .

(d) By part (c),

ex =

(

limy→0

(1 + y)1

y

)x

= limy→0

(1 + y)xy = lim

n→∞

(

1 +x

n

)n

.

�

Exercise 8.7. Let f(x) = sinxx

for all x > 0. Thus f ′(x) = x cos x−sinxx2 . Let

g(x) = x cos x − sin x for all real x. Thus g′(x) = −x sin x. g′(x) < 0 on

(0, π/2), and g(x) is decreasing strictly on (0, π/2). Also, g(π/2) = −1 < 0,

82

and g(x) < 0 for all 0 < x < π/2. f ′(x) < 1 on (0, π/2), and thus f(x) is

decreasing strictly on (0, π/2).

Note that limx→0sin x

x= 1 and f(π/2) = 2/π. Hence 2

π< f(x) < 1, that

is,2

π<

sin x

x< 1.

�

Exercise 8.8. For n = 0, 1, we have | sin 0x| ≤ 0| sinx| and | sinx| ≤ | sin x|.Assume that | sin kx| ≤ k| sin x|. We have

| sin(k + 1)x| = | sin(kx + x)| = | sin kx cos x + sin x cos kx|≤ | sin kx|| cos x| + | sin x|| cos kx| ≤ k| sin x| + | sin x|= (k + 1)| sin x|.

�

Exercise 8.9. (a) Let γn = sn − log n = 1 + 12

+ · · ·+ 1n− log n. Thus

γn+1 − γn =1

n + 1− log(n + 1) + log n

=1

n + 1−

∫ n+1

n

1

xdx

=

∫ n+1

n

(1

n + 1− 1

x

)

dx

≤ 0

for all n ≥ 0. Hence {γn} is a decreasing sequence. Also,

γn =n∑

k=1

1

k−

n−1∑

k=1

∫ k+1

k

1

xdx

=1

n+

n−1∑

k=1

∫ k+1

k

(1

k− 1

x

)

dx

≥ 0,

that is, {γn} is bounded. Therefore, γ = limn→∞ γn exists.

83

(b) Note that sN − log N ≥ 0 for all N , and thus we can choose m >

100/ log 10. �

Exercise 8.10. Given N , let p1, · · · , pk be those primes that divide at least

one integer ≤ N . Then

N∑

n=1

1

n≤

k∏

j=1

(

1 +1

pj+

1

p2j

+ · · ·)

=

k∏

j=1

(

1 − 1

pj

)−1

≤ exp

( k∑

j=1

2

pj

)

.

The last inequality holds because (1−x)−1 ≤ e2x for 0 ≤ x ≤ 12. Now letting

N → ∞, and thus∑∞

n=11n≤ exp

(∑

p2p

). Since the left hand side,

∑∞n=1

1n,

diverges,∑

p 1/p is divergent. �

Exercise 8.11. Given ǫ > 0. Let

g(t) = t

∫ ∞

0

e−txf(x)dx − 1 = t

∫ ∞

0

e−tx[f(x) − 1]dx

on [0, 1]. Since f(x) → 1 as x → +∞, there exists M > 0 such that

|f(x) − 1| < ǫ/2 whenever x ≥ M . Thus

|g(t)| ≤ t

∫ ∞

0

e−tx|f(x) − 1|dx

= t

∫ M

0

e−tx|f(x) − 1|dx + t

∫ ∞

M

e−tx|f(x) − 1|dx

≤ t

∫ M

0

e−txLdx + t

∫ ∞

M

e−tx ǫ

2dx

≤ t

∫ M

0

Ldx +ǫ

2e−tM

≤ tML +ǫ

2

where L > 0 is a bound of |f(x) − 1| on [0, M ]. (f ∈ R([0, M ]) and thus f

is bounded.) Choose t = min(1, ǫ2ML

) and thus |g(t)| < ǫ, i.e.,

limt→0+

t

∫ ∞

0

e−txf(x)dx = 1.

�

84

Exercise 8.12. (a) c0 = δ/π. For m 6= 0,

cm =1

2π

∫ π

−π

f(x)e−imxdx =1

2π

∫ δ

−δ

e−imxdx

=1

2π· e−imx

−im

∣∣∣∣

x=δ

x=−δ

=sin(mδ)

mπ.

(b) Thus,

f(x) ∼ δ

π+ 2

∞∑

n=1

sin(nδ)

nπcos(nx).

For x = 0, we can apply theorem 8.14 to get

f(0) =δ

π+ 2

∞∑

n=1

sin(nδ)

nπ,

that is,∞∑

n=1

sin(nδ)

n=

π − δ

2.

(c) By Parseval’s theorem,

1

2π

∫ π

−π

|f(x)|2dx =∞∑

−∞|cn|2 =

δ2

π2+ 2

∞∑

n=1

sin2(nδ)

n2π2.

1

2π

∫ π

−π

|f(x)|2dx =1

2π

∫ δ

−δ

12dx =δ

π.

Hence ∞∑

n=1

sin2(nδ)

n2δ=

π − δ

2.

(d)

∫ ∞

0

(sin x

x

)2

dx = limN→∞

1

N

∞∑

n=1

(sin n

NnN

)2

= limN→∞

∞∑

n=1

sin2 nN

n2

N

= limδ→0

∞∑

n=1

sin2(nδ)

n2δ= lim

δ→0

π − δ

2=

π

2.

(e) Put δ = π2

in (c), we have

∞∑

n=1

sin2(nπ2

)n2π2

=π − π

2

2.

85

π

4=

2

π

∞∑

n=1

sin2(nπ2

)

n2=

2

π

∞∑

n=1

1

(2n − 1)2,

that is,∞∑

n=1

1

(2n − 1)2=

π2

8.

�

Exercise 8.13. c0 = 12π

∫ 2π

0xdx = π. For m 6= 0,

cm =1

2π

∫ 2π

0

xe−imxdx

=1

2π· 1

−imxe−imx +

1

2π· 1

m2e−imx

∣∣∣∣

x=2π

x=0

=1

−im.

By Parseval’s theorem,

1

2π

∫ 2π

0

|f(x)|2dx =∞∑

−∞|cn|2.

Thus,4π2

3= π2 + 2

∞∑

n=1

1

n2

or ∞∑

n=1

1

n2=

π2

6.

�

Exercise 8.14. We deduce 12π

∫ π

−π|x|e−imxdx and 1

2π

∫ π

−πx2e−imxdx with

m 6= 0 for further computing.∫

xe−imxdx =1

−imxe−imx −

∫1

−ime−imxdx

=1

−imxe−imx +

1

m2e−imx + constant;

∫

x2e−imxdx =1

−imx2e−imx −

∫2

−imxe−imxdx

=1

−imx2e−imx − 2

−im

∫

xe−imxdx.

86

Thus∫ π

−π

|x|e−imxdx =

∫ 0

−π

−xe−imxdx +

∫ π

0

xe−imxdx

= −(

xe−imx

−im+

e−imx

m2

)∣∣∣∣

0

−π

+

(xe−imx

−im+

e−imx

m2

)∣∣∣∣

π

0

=

{πim

+ π−im

if m is even

(− 2m2 − π

im) + (− 2

m2 + πim

) if m is odd

=

{

0 if m is even

− 4m2 if m is odd;

∫ π

−π

xe−imxdx =

{π

−im+ π

−imif m is even

( 2m2 + π

im) + (− 2

m2 + πim

) if m is odd

=

{

− 2πim

if m is even2πim

if m is odd

= (−1)m+1 2π

im,

and∫ π

−π

x2e−imxdx =

(1

−imx2e−imx

)∣∣∣∣

π

−π

− 2

−im

∫ π

−π

xe−imxdx

= (−1)m 4π

m2.

Hence,

c0 =1

2π

∫ π

−π

(π − |x|)2dx =π2

3,

cm =1

2π

∫ π

−π

(π − |x|)2e−imxdx

=1

2π

∫ π

−π

(π2 − 2π|x| + x2)e−imxdx

= −∫ π

−π

|x|e−imxdx +1

2π

∫ π

−π

x2e−imxdx

=

{

0 + 2m2 if m is even

4m2 + −2

m2 if m is odd

=2

m2.

87

Therefore,

f(x) = (π − |x|)2 ∼ π2

3+

∞∑

n=1

4

n2cos nx.

Given δ = 1. Note that

|f(x + t) − f(x)| =∣∣(π − |x + t|)2 − (π − |x|)2

∣∣

=∣∣2π − |x + t| − |x|

∣∣ ·

∣∣|x + t| − |x|

∣∣

≤ (2π + 1)|t|

for all t ∈ (−δ, δ) = (−1, 1). By theorem 8.14,

f(x) = (π − |x|)2 =π2

3+

∞∑

n=1

4

n2cos nx

on [−π, π]. Let x = 0, π2 = π2

3+

∑∞n=1

4n2 , i.e.,

∑∞n=1

1n2 = π2

6.

By Parseval’s theorem,

1

2π

∫ π

−π

|f(x)|2dx =

∞∑

−∞|cn|2 =

π4

9+

∞∑

n=1

8

n4.

1

2π

∫ π

−π

|f(x)|2dx =1

2π

∫ π

−π

(π − |x|)4dx

=1

2π

∫ 0

−π

(π + x)4dx +1

2π

∫ π

0

(π − x)4dx

=1

2π

(π + x)5

5

∣∣∣∣

0

−π

+1

2π

−(π − x)5

5

∣∣∣∣

π

0

=π4

5.

Hence∑∞

n=11n4 = π4

90. �

Exercise 8.15. For KN(x) = 1N+1

· 1−cos(N+1)x1−cos x

. Recall

Dn(x) =sin(n + 1

2)x

sin x2

,

88

or

sinx

2Dn(x) = sin(n +

1

2)x.

Hence

sinx

2

N∑

n=0

Dn(x) =

n∑

n=0

sin(n +1

2)x.

Multiply sin x2

on both sides,

(sinx

2)2

N∑

n=0

Dn(x) =N∑

n=0

sin(n +1

2)x · sin x

2

=1

2

N∑

n=0

(cos nx − cos(n + 1)x)

=1

2(1 − cos(N + 1)x).

Note that 2(sin x2)2 = 1 − cos x, and thus

KN (x) =1

N + 1

N∑

n=0

Dn(x) =1

N + 1· 1 − cos(N + 1)x

1 − cos x.

(a) By KN(x) = 1N+1

· 1−cos(N+1)x1−cos x

, and | cosx| ≤ 1, KN (x) ≥ 0.

(b) In Theorem 8.14, we know

1

2π

∫ π

−π

Dn(x)dx = 1.

for all 0 ≤ n ≤ N . Hence

1

2π

∫ π

−π

KN(x)dx =1

2π

∫ π

−π

1

N + 1

N∑

n=0

Dn(x)dx

=1

N + 1

N∑

n=0

1

2π

∫ π

−π

Dn(x)dx

=1

N + 1(N + 1) = 1.

(c) cos x < cos δ whenever 0 < δ ≤ |x| ≤ π. Thus

KN(x) =1 − cos(N + 1)x

(N + 1)(1 − cos x)≤ 2

(N + 1)(1 − cos δ).

89

By equation (78) on page 189,

sN (f ; x) =1

2π

∫ π

−π

f(x − t)DN(t)dt.

Hence

σN(f ; x) =1

2π

∫ π

−π

f(x − t)KN (t)dt.

For Fejer’s theorem. By (a) and (b),

|σN(f ; x) − f(x)| =

∣∣∣∣

1

2π

∫ π

−π

f(x − t)KN(t)dt − f(x)

∣∣∣∣

≤ 1

2π

∫ π

−π

|f(x − t) − f(x)|KN(t)dt.

Given ǫ > 0, we choose π > δ > 0 such that |y−x| < δ implies |f(y)−f(x)| <

ǫ/2. Let M = sup |f(x)|. Hence,

1

2π

∫ −δ

−π

|f(x− t) − f(x)|KN(t)dt ≤ 1

2π

∫ −δ

−π

2M · 2

(N + 1)(1 − cos δ)dt

=4M(π − δ)

2π(N + 1)(1 − cos δ)

<2M

(N + 1)(1 − cos δ),

1

2π

∫ π

δ

|f(x− t) − f(x)|KN(t)dt <2M

(N + 1)(1 − cos δ),

and

1

2π

∫ δ

−δ

|f(x − t) − f(x)|KN(t)dt ≤ ǫ

4π

∫ δ

−δ

KN(t)dt

≤ ǫ

4π

∫ π

−π

KN(t)dt =ǫ

2.

Therefore,

|σN (f ; x) − f(x)| ≤ 1

2π

∫ π

−π

|f(x − t) − f(x)|KN(t)dt

≤ 4M

(N + 1)(1 − cos δ)+

ǫ

2

< ǫ

90

for all large enough N , which proves the conclusion that σN(f ; x) → f(x)

uniformly on [−π, π]. �

Exercise 8.17. (a) Suppose f increases monotonically on [−π, π]. By Ex-

ercise 17 of Chapter 6,

ncn =1

2π

∫ π

−π

f(x)ne−inxdx

=ie−inπ

2πf(π) − ie−inπ

2πf(−π) − 1

2π

∫ π

−π

ie−inxdf

for n 6= 0. Hence,

|ncn| =

∣∣∣∣

ie−inπ

2πf(π) − ie−inπ

2πf(−π) − 1

2π

∫ π

−π

ie−inxdf

∣∣∣∣

≤∣∣∣∣

ieinπ

2πf(π)

∣∣∣∣+

∣∣∣∣

ie−inπ

2πf(−π)

∣∣∣∣+

1

2π

∣∣∣∣

∫ π

−π

ie−inxdf

∣∣∣∣

≤ 1

2π(|f(π)| + |f(−π)|) +

1

2π(f(π) − f(−π))

are bounded for n 6= 0, that is, {ncn} is a bounded sequence.

Exercise 8.29. Prove that every continuous mapping f of D into D has a

fixed point in D. (This is the 2-dimensional case of Brouwer’s fixed-point

theorem.) Hint: Assume that f(z) 6= z for every z ∈ D. Associate to each

z ∈ D the point g(z) ∈ T which lies on the ray that starts at f(z) and passes

through z. Then g maps D into T , g(z) = z if z ∈ T , and g is continuous,

because

g(z) = z − s(z)[f(z) − z],

where s(z) is the unique nonnegative root of a certain quadratic equation

whose coefficients are continuous function of f and z. Apply Exercise 28.

Exercise 8.30. Use Stirling’s formula to prove that

limx→∞

Γ(x + c)

xcΓ(x)= 1

for every real constant c.

91

Proof. By Stirling’s formula,

limx→∞

Γ(x + c)

xcΓ(x)= lim

x→∞

(x+c−1e

)x+c−1√

2π(x + c − 1)

xc(x−1e

)x−1√

2π(x − 1)

=1

eclim

x→∞

√

x + c − 1

x − 1

(x + c − 1

x − 1

)x+c−1

.

Note that

limx→∞

(x + c − 1

x − 1

)x+c−1

= limx→∞

[(

1 +c

x − 1

)x−1]x+c−1

x−1

= ec.

and limx→∞

√x+c−1x−1

= 1. Hence

limx→∞

Γ(x + c)

xcΓ(x)= 1

for every real c. �

Exercise 8.31. In the proof of Theorem 7.26 it was shown that

∫ 1

−1

(1 − x2)ndx ≥ 4

3√

n

for n = 1, 2, 3, · · · . Use Theorem 8.20 and Exercise 30 to show the more

precise result

limn→∞

√n

∫ 1

−1

(1 − x2)ndx =√

π.

Proof. By change of variable, letting t = x2,

∫ 1

0

(1 − x2)ndx =

∫ 1

0

1

2t−

1

2 (1 − t)ndt =Γ(1

2)Γ(n + 1)

2Γ(n + 32)

.

Note that∫ 0

−1(1 − x2)ndx =

∫ 1

0(1 − x2)ndx. Hence

limn→∞

√n

∫ 1

−1

(1 − x2)ndx =Γ(√

n12)Γ(n + 1)

Γ(n + 32)

= Γ

(1

2

)

=√

π.

�

92

Exercise 9.1. Given x = c1x1 + · · ·+ckxk, y = d1y1 + · · ·+dhyh ∈ span(S),

where xi,yj ∈ S and ci, dj are scalars. Thus x + y = c1x1 + · · · + ckxk +

d1y1 + · · · + dhyh is a linear combinations of elements of S. Also, for every

scalar c, cx = c · (c1x1 + · · ·+ ckxk) = (cc1)x1 + · · ·+(cck)xk again is a linear

combinations of elements of S. Thus span(S) is a vector space. �

Exercise 9.2. To show BA is linear, it suffices to show that (BA)(x1+x2) =

(BA)(x1) + (BA)(x2) and (BA)(cx) = c(BA)(x) for all x,x1,x2 ∈ X and

all scalars c. In fact, (BA)(x1 + x2) = B(A(x1 + x2)) = B(Ax1 + Ax2) =

B(Ax1) + B(Ax2) = (BA)(x1) + (BA)(x2), and (BA)(cx) = B(A(cx)) =

B(cAx) = cB(Ax) = c(BA)(x). Thus BA is linear.

To show A−1 is linear, it suffices to show that A−1(x1 + x2) = A−1(x1) +

A−1(x2) and A−1(cx) = cA−1(x) for all x,x1,x2 ∈ X and all scalars c.

Since A is invertible, A maps X onto X and thus x1 = Ay1, x2 = Ay2,

x = Ay for some y1,y2,y ∈ X. Hence, A−1(x1 + x2) = A−1(Ay1 + Ay2) =

A−1(A(y1 + y2)) = y1 + y2 = A−1(x1) + A−1(x2); A−1(cx) = A−1(cAy) =

A−1(A(cy)) = cy = cA−1(x). Thus A−1 is linear.

To show A−1 is invertible, it needs to show that A−1 is a linear opera-

tor on X which (i) is one-to-one and (ii) maps X onto X. For (i), given

A−1(x1) = A−1(x2). Since A is invertible, A maps X onto X and thus

x1 = Ay1, x2 = Ay2 for some y1,y2 ∈ X. Thus, A−1(Ay1) = A−1(Ay2) or

y1 = y2, that is, x1 = x2. For (ii), given x ∈ X we have Ax ∈ X such that

A−1(Ax) = x. Thus A−1 is invertible. �

Exercise 9.3. To show A is 1-1, it suffices to show that Ax1 = Ax2 implies

x1 = x2. Since A ∈ L(X, Y ), 0 = Ax1 − Ax2 = A(x1 − x2). By assumption,

x1 − x2 = 0, that is, x1 = x2. �

Exercise 9.4. For all x,y ∈ N (A), and for all scalars c, we need to show

that x+y ∈ N (A) and cx ∈ N (A). It follows by A(x+y) = A(x)+A(y) =

0 + 0 = 0 and A(cx) = cA(x) = c0 = 0. Hence N (A) is a vector space.

For all x,y ∈ R(A), and for all scalars c, we need to show that x + y ∈R(A) and cx ∈ R(A). x and y are all in R(A) implies that there exist

93

x1,y1 ∈ X such that Ax1 = x and Ay1 = y. Thus, x + y = Ax1 + Ay1 =

A(x1 + y1), i.e., x + y ∈ R(A). Also, cx = cAx1 = A(cx1), i.e., cx ∈ R(A).

Hence R(A) is a vector space. �

Exercise 9.5. By Schwarz inequality,

‖A‖ = sup|x|≤1

|Ax| = sup|x|≤1

|x · y| ≤ sup|x|≤1

|x||y| = |y|.

Now we pick x = |y|−1y if y 6= 0. (The case y = 0 is trivial.) |x| = 1 and

|Ax| = |x · y| = |y| ≤ ‖A‖. Hence, ‖A‖ = |y|. �

Exercise 9.6. For (x, y) 6= (0, 0), (D1f)(x, y) = 3x2y+y3

(x2+y2)2and (D2f)(x, y) =

3xy2+x3

(x2+y2)2. For (x, y) = (0, 0), (D1f)(0, 0) = (D2f)(0, 0) = 0. But f is not con-

tinuous at (0, 0). In fact, to calculate the limit lim(x,y)→(0,0) f(x, y), we make

changes of variables by x = r cos θ and y = r sin θ. (x, y) → (0, 0) if and only

if r = 0, and f(x, y) = cos θ sin θ = 12sin(2θ). Hence lim(x,y)→(0,0) f(x, y) =

limr→012sin(2θ) = 1

2sin(2θ) dose not exist. �

Exercise 9.7. Fix x ∈ E and ǫ > 0. Since E is open, there is an open ball

S ⊂ E, with center at x and radius r. Suppose h =∑

hjej , h < r, put

v0 = 0, and vk = h1e1 + · · ·+ hkek, for 1 ≤ k ≤ n. Then

f(x + h) − f(x) =

n∑

j=1

(f(x + vj) − f(x + vj−1)). (*)

Since |vk| < r for all k and since S is convex, the segments with end points

x+vj−1 and x+vj lie in S. Since vj = vj−1 +hjej, the mean value theorem

shows that the j-th summand in (*) is equal to

hj(Djf)(x + vj−1 + θjhjej)

for some θj ∈ (0, 1). Since Djf are bounded in E, there is M > 0 such that

94

|Djf | ≤ M . Thus

|f(x + h) − f(x)| ≤n∑

j=1

|f(x + vj) − f(x + vj−1))|

=n∑

j=1

|hj(Djf)(x + vj−1 + θjhjej)|

≤ M

n∑

j=1

|hj | ≤ Mnr < ǫ

if we choose r < ǫ/(Mn). Hence f is continuous in E. �

Exercise 9.8. By definition 5.7, there exists δ > 0 such that f(x+h) ≤ f(x)

for all h ∈ E with |h| < δ. To show f ′(x) = 0, it suffices to show that

(Djf)(x) = 0 for 1 ≤ j ≤ n. If −δ < t < 0, then |tej | < δ and

f(x + tej) − f(x)

t≥ 0.

Letting t → 0, we see that (Djf)(x) ≥ 0. If 0 < t < δ, then |tej | < δ and

f(x + tej) − f(x)

t≤ 0,

which shows that (Djf)(x) ≤ 0. Hence (Djf)(x) = 0. � (The idea of this

proof comes from theorem 5.8.)

Exercise 9.11. For each x ∈ Rn, the gradient of fg at x is

▽(fg)(x) =

n∑

i=1

(Di(fg))(x)ei

=n∑

i=1

(gDi(f) + fDi(g))(x)ei

=

n∑

i=1

(gDi(f))(x)ei +

n∑

i=1

(fDi(g)(x)ei

= (g ▽ f + f ▽ g)(x).

Thus ▽(fg) = f ▽ g + g ▽ f . In particular, 0 = ▽1 = ▽(f · 1/f) =

f ▽ (1/f) + 1/f ▽ f , i.e., ▽(1/f) = −f−2 ▽ f whenever f 6= 0. �

95

Exercise 9.13. Let g(x, y, z) = (x, y, z) · (x, y, z) = x2 + y2 + z2 of R3 into

R, and F (t) = g(f(t)) = f(t) · f(t) = 1 of R into R. Note that g′(x) = 2x

since (D1g)(x, y, z) = 2x, (D2g)(x, y, z) = 2y, and (D3g)(x, y, z) = 2z. By

theorem 9.15,

F ′(t) = g′(f(t))f ′(t) = 2f(t) · f ′(t).

Since 1 = |f(t)|, 1 = f(t) · f(t) = F (t) and thus 0 = f(t) · f ′(t). We interpret

this result geometrically. Given a particle on the unit sphere with a smooth

movement f(t) by the time t. The result f(t) ·f ′(t) = 0 says that the direction

of velocity of this particle point at the center of this unit sphere. �

Exercise 9.15. (a) (x4 + y2)2 − 4x4y2 = (x4 − y2) ≥ 0 implies that 4x4y2 ≤(x4 + y2)2. To show f is continuous on R2, it suffices to show f is continuous

at (0, 0). Note that

f(x, y) = x2 + y2 − 2x2y − x2 4x4y2

(x4 + y2)2

and thus lim(x,y)→(0,0) f(x, y) = 0 = f(0, 0).

Exercise 9.16.

f ′(0) = limt→0

f(t) − f(0)

t= lim

t→0

[

1 + 2t sin

(1

t

)]

= 1.

For t 6= 0,

f ′(t) = 1 + 4t sin

(1

t

)

− cos

(1

t

)

.

Hence, f ′(0) = 1 and f ′(t) is bounded (by 6). But f is not one-to-one in any

neighborhood of 0 since f ′ is not monotonic in any neighborhood of 0. �

Exercise 11.1. If f ≥ 0 and∫

Efdµ = 0, prove that f(x) = 0 almost

everywhere on E. Hint: Let En be the subset of E on which f(x) > 1/n.

Write A = ∪En. Then µ(A) = 0 if and only if µ(En) = 0 for every n.

Proof. Let En be the subset of E on which f(x) > 1/n. Write A = ∪En. In

fact, A = {x ∈ E | f(x) > 0}. Claim: µ(A) = 0 if and only if µ(En) = 0 for

96

every n. One side is trivial. Conversely, if µ(En) = 0 for all n, then we have

µ(En) → µ(A) by theorem 11.3, that is, µ(A) = 0. Therefore, for every n

1

nµ(En) ≤

∫

En

fdµ ≤∫

E

fdµ = 0

implies µ(En) = 0. Since {x ∈ E | f(x) > 0} = A = ∪En, we have µ(A) = 0.

�

Exercise 11.2. If∫

Afdµ = 0 for every measurable subset A of a measurable

set E, then f(x) = 0 almost everywhere on E.

Proof. Let A = {x ∈ E | f(x) ≥ 0} be a measurable subset of E. Thus,∫

Afdµ =

∫

Af+dµ = 0. By exercise 11.1, f+ = 0 almost everywhere on

A and thus on E. Similarly, f− = 0 almost everywhere on E. Hence,

f = f+ − f− = 0 almost everywhere on E. �

Exercise 11.3. If {fn} is a sequence of measurable functions, prove that

the set of points x at which {fn(x)} converges is measurable.

Proof. Let E be the set of points x at which {fn(x)} converges. x ∈ E if

and only if for every ǫ > 0 there exists N such that n ≥ N implies

|fn(x) − f(x)| < ǫ

for some f(x), which can be restate as that

x ∈∞⋂

k=1

∞⋂

n=N

{x ∈ X | |fn(x) − f(x)| < 1/k}

for some integer N = N(x, k). For simplicity, we define f(x) = 0 if x ∈ X\E.

Hence,

E =

∞⋂

k=1

∞⋃

m=1

∞⋂

n=m

{x ∈ X | |fn(x) − f(x)| < 1/k}

is measurable since every fn is a measurable function. �

97

Exercise 11.5. Put

g(x) =

{

0 (0 ≤ x ≤ 12),

1 (12

< x ≤ 1),

f2k(x) = g(x) (0 ≤ x ≤ 1),

f2k+1(x) = g(1 − x) (0 ≤ x ≤ 1).

Show that

lim infn→∞

fn(x) = 0 (0 ≤ x ≤ 1),

but ∫ 1

0

fn(x)dx =1

2.

[Compare with (77).]

Proof. Since

f2k(x) =

{

0 (0 ≤ x ≤ 12),

1 (12

< x ≤ 1),

f2k+1(x) =

{

1 (0 ≤ x < 12),

0 (12≤ x ≤ 1),

lim infn→∞ fn(x) = 0, but∫ 1

0fn(x)dx = 1

2for all n. �

Exercise 11.6. Let

fn(x) =

{1n

(|x| ≤ n),

0 (|x| > n).

Then fn(x) → 0 uniformly on R1, but∫ ∞

−∞fn(x)dx = 2 (n = 1, 2, 3, · · · ).

(We write∫ ∞−∞ in place of

∫

R1 .) Thus uniform convergence does not imply

dominated convergence in the sense of Theorem 11.32. However, on sets

of finite measure, uniformly convergent sequences of bounded functions do

satisfy Theorem 11.32.

Proof. Clearly, limn→∞ fn(x) = 0. Put

Mn = supx∈R1

|fn(x) − 0| =1

n.

98

By theorem 7.9, fn → 0 uniformly on R1 if and only if Mn = 1n→ 0 as

n → ∞. But∫ ∞−∞ fn(x)dx = 1

n· 2n = 2 for all n.

99

Documents

Some Notes on Rudin’s book: Principles of Mathematical ... · Some Notes on Rudin’s book: Principles of Mathematical Analysis, 3/e Written by Meng-Gen, Tsai email: [email protected]