M ATH 140 B - HW 5 S OLUTIONS

Problem 1 (WR Ch 7 #8). If ( I (x) =

0

(x ≤ 0),

1

(x > 0),

if {x n } is a sequence of distinct points of (a, b), and if series f (x) =

∞ X

c n I (x − x n )

P

|c n | converges, prove that the

(a ≤ x ≤ b)

n=1

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

Solution. Let f k (x) =

k X

c n I (x − x n ).

n=1

By the Weierstrass M -test (Theorem 7.10) with M n = |c n |, { f k (x)} converges uniformly to f (x). Let E = [a, b] \ {x n : n ∈ N}. Since each f k (x) is continuous on E , then by Theorem 7.12 we know that f is continuous on E .

Problem 2 (WR Ch 7 #9). Let { f n } be a sequence of continuous functions which converge uniformly to a function f on a set E . Prove that lim f n (x n ) = f (x)

n→∞

for every sequence of points x n ∈ E such that x n → x, and x ∈ E . Is the converse of this true?

Solution. Since f is the uniform limit of continuous functions, it is continuous (Theorem 7.12). Since f is continuous and x n → x, we know that f (x n ) → f (x) (Theorem 4.2). Set ² > 0. Then there is some N1 ∈ N such that | f (x n ) − f (x)|

0, we choose N >

M ² , so that for n

| f n (x n ) − f (x)| = | xnn − 0| =

|x n | n

≤

M N

≥ N we have

< ².

This proves that f n (x n ) → f (x).

Problem 3 (WR Ch 7 #10). Letting (x) denote the fractional part of the real number x, consider the function f (x) =

∞ (nx) X 2 n=1 n

(x real ).

Find all discontinuities of f , and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.

Solution. First notice that the function g (x) = (x) is discontinuous on Z and continuous on R \ Z. This means that g n (x) = (nx) is discontinuous at all x such that nx ∈ Z, which means only at rational numbers of the form

m n

(where m, n ∈ Z and gcd(m, n) = 1). Let E = R \ Q. As

n ranges over all positive integers, we see that g n (x) only has discontinuities at points which lie outside E , so that if we let f k (x) =

k g (x) k (nx) X X n = , 2 2 n=1 n n=1 n

we see that each f k (x) is continuous on E . By the Weierstrass M -test (Theorem 7.10) with Mn =

1 , n2

we know that { f k (x)} converges uniformly to f (x), and thus by Theorem 7.12 we

know that f is continuous on E since each f k (x) is continuous on E . What remains is to prove that f is discontinuous at all rational points. Let x =

a b

for a, b ∈ Z,

gcd(a, b) = 1. Then g b (x) is discontinuous at x, or more specifically lim y→x− g b (x) = 1 and lim y→x+ = 0. More generally, at any discontinuity of g n , we will have that the limit from the left will be larger than the limit from the right, meaning that lim f k (y) − lim f k (y) ≥

y→x−

y→x+

1 b2

for k ≥ b.

Since f k (x) → f (x) pointwise (the series is bounded by P (nx) 1 (N > b) such that ∞ n=N +1 n 2 < 3b 2 , so that µ ¶ Ã lim f (y) − lim f (y) =

y→x−

y→x+

P∞

n=1 1/n

lim f N (y) − lim f N (y) + lim f

y→x−

y→x+

y→x−

1 1 1 > 2 − 2 2 = 2 > 0. b 3b 3b 2

2

), there exists some N ∈ N

∞ X n=N +1

(nx) n2

− lim f y→x+

∞ X n=N +1

! (nx) n2

Therefore f is discontinuous at every rational point. The fact that f is Riemann integrable follows directly from Theorem 7.16.

Problem 4 (WR Ch 7 #11). Suppose { f n }, {g n } are defined on E , and (a)

P

f n has uniformly bounded partial sums;

(b) g n → 0 uniformly on E ; (c) g 1 (x) ≥ g 2 (x) ≥ g 3 (x) ≥ · · · for every x ∈ E . Prove that

P

f n g n converges uniformly on E .

Solution. Let A n (x) =

Pn

f . k=1 n

Choose M such that |A n (x)| ≤ M for all n. Given ² > 0, by

uniform continuity there is an integer N such that g N (x) ≤ (²/2M ) for all x ∈ E . For N ≤ p ≤ q, we have ¯ ¯ ¯ ¯ q ¯X ¯ ¯ q−1 ¯ ¯ ¯ ¯X ¯ f n (x)g n (x)¯ = ¯ A n (x)(g n (x) − g n+1 (x)) + A q (x)g q (x) − A p−1 (x)g p (x)¯ ¯ ¯n=p ¯ ¯n=p ¯ ¯ ¯ ¯ q−1 ¯ ¯X ¯ ≤M¯ (g n (x) − g n+1 (x)) + g q (x) + g p (x)¯ ¯n=p ¯ = 2M g p (x) ≤ 2M g N (x) ≤ ². Convergence follows from the Cauchy criterion for uniform convergence.

Problem 5 (WR Ch 7 #15). Suppose f is a real continuous function on R1 , f n (t ) = f (nt ) for n = 1, 2, 3, . . ., and { f n } is equicontinuous on [0, 1]. What conclusion can you draw about f ?

Solution. We can conclude f is constant on [0, ∞). The fact that { f n } is equicontinuous means that for every ² > 0, there exists a δ > 0 such that |s − t | < δ

=⇒

| f n (s) − f n (t )| < ²

for all n ∈ N.

For any x ∈ [0, ∞), set ² > 0 and find a δ > 0 so that the above inequality holds. Then choose N to be the smallest integer such that N > x/δ (so that x/N < δ). Then if we set s = 0 and t = x/N , we have |s − t | = x/N < δ, so by the inequality above we have | f (0) − f (x)| = | f n (s) − f n (t )| < ². But our choice of ² was arbitrary, so that means f (0) = f (x) for all x ∈ [0, ∞), proving our claim.

3

Problem 6 (Supp. HW2 #4). Given an example of a metric space X and a sequence of functions { f n } on X such that { f n } is equicontinuous but not uniformly bounded.

Solution. Let X = R and f n (x) = n. Then for any ² > 0, choose any δ > 0 and we have | f n (x) − f n (y)| = |n − n| = 0 < ² whenever |x − y| < δ, so { f n } is equicontinuous. If it were uniformly bounded then there would be some M > 0 such that | f n (x)| < M for all n ∈ N and x ∈ R, but this is clearly not possible by taking n > M .

Problem 7 (Supp. HW2 #5). Give an example of a uniformly bounded and equicontinuous sequence of functions on R which does not have any uniformly convergent subsequences.

Solution. Let

n ≤ x ≤ n + 12

2(x − n) f n (x) = 2(n + 1 − x) 0

n + 12 < x ≤ n + 1 . otherwise

For example, we graph f 3 (x) below. In loose terms, f n (x) is zero everywhere except for a “triangle” of height 1 on the interval [n, n + 1].

From this definition it’s clear that | f n (x)| ≤ 1 for all n ∈ N and x ∈ R, so the sequence is uniformly bounded. To prove equicontinuity, set some 1 > ² > 0 and choose δ = ²/2, so that if

4

|x − y| < δ and x < y we have ¯ ¯ ¯0 − 2(y − n)¯ , ¯ ¯ ¯2(x − n) − 2(y − n)¯ , ¯ ¯ | f n (x) − f n (y)| ≤ max ¯2(x − n) − 2(n + 1 − y)¯ , ¯ ¯ ¯2(n + 1 − x) − 2(n + 1 − y)¯ , |2(n + 1 − x) − 0|

= 2|x − y| < 2δ < ².

The reason this sequence doesn’t have any uniformly convergent subsequences is that the sequence converges pointwise to 0, so any subsequence must converge pointwise to 0, but f n (n + 21 ) = 1, so if we have some subsequence { f nk } and we set ² < 1, then sup | f nk (x) − f (x)| ≥ 1 > ² x∈R

5

for all k.

Problem 1 (WR Ch 7 #8). If ( I (x) =

0

(x ≤ 0),

1

(x > 0),

if {x n } is a sequence of distinct points of (a, b), and if series f (x) =

∞ X

c n I (x − x n )

P

|c n | converges, prove that the

(a ≤ x ≤ b)

n=1

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

Solution. Let f k (x) =

k X

c n I (x − x n ).

n=1

By the Weierstrass M -test (Theorem 7.10) with M n = |c n |, { f k (x)} converges uniformly to f (x). Let E = [a, b] \ {x n : n ∈ N}. Since each f k (x) is continuous on E , then by Theorem 7.12 we know that f is continuous on E .

Problem 2 (WR Ch 7 #9). Let { f n } be a sequence of continuous functions which converge uniformly to a function f on a set E . Prove that lim f n (x n ) = f (x)

n→∞

for every sequence of points x n ∈ E such that x n → x, and x ∈ E . Is the converse of this true?

Solution. Since f is the uniform limit of continuous functions, it is continuous (Theorem 7.12). Since f is continuous and x n → x, we know that f (x n ) → f (x) (Theorem 4.2). Set ² > 0. Then there is some N1 ∈ N such that | f (x n ) − f (x)|

0, we choose N >

M ² , so that for n

| f n (x n ) − f (x)| = | xnn − 0| =

|x n | n

≤

M N

≥ N we have

< ².

This proves that f n (x n ) → f (x).

Problem 3 (WR Ch 7 #10). Letting (x) denote the fractional part of the real number x, consider the function f (x) =

∞ (nx) X 2 n=1 n

(x real ).

Find all discontinuities of f , and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.

Solution. First notice that the function g (x) = (x) is discontinuous on Z and continuous on R \ Z. This means that g n (x) = (nx) is discontinuous at all x such that nx ∈ Z, which means only at rational numbers of the form

m n

(where m, n ∈ Z and gcd(m, n) = 1). Let E = R \ Q. As

n ranges over all positive integers, we see that g n (x) only has discontinuities at points which lie outside E , so that if we let f k (x) =

k g (x) k (nx) X X n = , 2 2 n=1 n n=1 n

we see that each f k (x) is continuous on E . By the Weierstrass M -test (Theorem 7.10) with Mn =

1 , n2

we know that { f k (x)} converges uniformly to f (x), and thus by Theorem 7.12 we

know that f is continuous on E since each f k (x) is continuous on E . What remains is to prove that f is discontinuous at all rational points. Let x =

a b

for a, b ∈ Z,

gcd(a, b) = 1. Then g b (x) is discontinuous at x, or more specifically lim y→x− g b (x) = 1 and lim y→x+ = 0. More generally, at any discontinuity of g n , we will have that the limit from the left will be larger than the limit from the right, meaning that lim f k (y) − lim f k (y) ≥

y→x−

y→x+

1 b2

for k ≥ b.

Since f k (x) → f (x) pointwise (the series is bounded by P (nx) 1 (N > b) such that ∞ n=N +1 n 2 < 3b 2 , so that µ ¶ Ã lim f (y) − lim f (y) =

y→x−

y→x+

P∞

n=1 1/n

lim f N (y) − lim f N (y) + lim f

y→x−

y→x+

y→x−

1 1 1 > 2 − 2 2 = 2 > 0. b 3b 3b 2

2

), there exists some N ∈ N

∞ X n=N +1

(nx) n2

− lim f y→x+

∞ X n=N +1

! (nx) n2

Therefore f is discontinuous at every rational point. The fact that f is Riemann integrable follows directly from Theorem 7.16.

Problem 4 (WR Ch 7 #11). Suppose { f n }, {g n } are defined on E , and (a)

P

f n has uniformly bounded partial sums;

(b) g n → 0 uniformly on E ; (c) g 1 (x) ≥ g 2 (x) ≥ g 3 (x) ≥ · · · for every x ∈ E . Prove that

P

f n g n converges uniformly on E .

Solution. Let A n (x) =

Pn

f . k=1 n

Choose M such that |A n (x)| ≤ M for all n. Given ² > 0, by

uniform continuity there is an integer N such that g N (x) ≤ (²/2M ) for all x ∈ E . For N ≤ p ≤ q, we have ¯ ¯ ¯ ¯ q ¯X ¯ ¯ q−1 ¯ ¯ ¯ ¯X ¯ f n (x)g n (x)¯ = ¯ A n (x)(g n (x) − g n+1 (x)) + A q (x)g q (x) − A p−1 (x)g p (x)¯ ¯ ¯n=p ¯ ¯n=p ¯ ¯ ¯ ¯ q−1 ¯ ¯X ¯ ≤M¯ (g n (x) − g n+1 (x)) + g q (x) + g p (x)¯ ¯n=p ¯ = 2M g p (x) ≤ 2M g N (x) ≤ ². Convergence follows from the Cauchy criterion for uniform convergence.

Problem 5 (WR Ch 7 #15). Suppose f is a real continuous function on R1 , f n (t ) = f (nt ) for n = 1, 2, 3, . . ., and { f n } is equicontinuous on [0, 1]. What conclusion can you draw about f ?

Solution. We can conclude f is constant on [0, ∞). The fact that { f n } is equicontinuous means that for every ² > 0, there exists a δ > 0 such that |s − t | < δ

=⇒

| f n (s) − f n (t )| < ²

for all n ∈ N.

For any x ∈ [0, ∞), set ² > 0 and find a δ > 0 so that the above inequality holds. Then choose N to be the smallest integer such that N > x/δ (so that x/N < δ). Then if we set s = 0 and t = x/N , we have |s − t | = x/N < δ, so by the inequality above we have | f (0) − f (x)| = | f n (s) − f n (t )| < ². But our choice of ² was arbitrary, so that means f (0) = f (x) for all x ∈ [0, ∞), proving our claim.

3

Problem 6 (Supp. HW2 #4). Given an example of a metric space X and a sequence of functions { f n } on X such that { f n } is equicontinuous but not uniformly bounded.

Solution. Let X = R and f n (x) = n. Then for any ² > 0, choose any δ > 0 and we have | f n (x) − f n (y)| = |n − n| = 0 < ² whenever |x − y| < δ, so { f n } is equicontinuous. If it were uniformly bounded then there would be some M > 0 such that | f n (x)| < M for all n ∈ N and x ∈ R, but this is clearly not possible by taking n > M .

Problem 7 (Supp. HW2 #5). Give an example of a uniformly bounded and equicontinuous sequence of functions on R which does not have any uniformly convergent subsequences.

Solution. Let

n ≤ x ≤ n + 12

2(x − n) f n (x) = 2(n + 1 − x) 0

n + 12 < x ≤ n + 1 . otherwise

For example, we graph f 3 (x) below. In loose terms, f n (x) is zero everywhere except for a “triangle” of height 1 on the interval [n, n + 1].

From this definition it’s clear that | f n (x)| ≤ 1 for all n ∈ N and x ∈ R, so the sequence is uniformly bounded. To prove equicontinuity, set some 1 > ² > 0 and choose δ = ²/2, so that if

4

|x − y| < δ and x < y we have ¯ ¯ ¯0 − 2(y − n)¯ , ¯ ¯ ¯2(x − n) − 2(y − n)¯ , ¯ ¯ | f n (x) − f n (y)| ≤ max ¯2(x − n) − 2(n + 1 − y)¯ , ¯ ¯ ¯2(n + 1 − x) − 2(n + 1 − y)¯ , |2(n + 1 − x) − 0|

= 2|x − y| < 2δ < ².

The reason this sequence doesn’t have any uniformly convergent subsequences is that the sequence converges pointwise to 0, so any subsequence must converge pointwise to 0, but f n (n + 21 ) = 1, so if we have some subsequence { f nk } and we set ² < 1, then sup | f nk (x) − f (x)| ≥ 1 > ² x∈R

5

for all k.