Math Real Analysis

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HW6Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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HW6Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 HW 6

(a): Discuss the uniform convergence of the following sequence of functions on different intervals:

1. fn(x) = x n

1+xn on the following intervals respectively

I1 = {x ∈ R : 0 ≤ x ≤ b}, with b < 1;

I2 = {x ∈ R : 0 ≤ x ≤ 1};

I3 = {x ∈ R : a ≤ x ≤ ∞}, with 1 < a.

ANS: For I1: first of all, x n ≤ bn and bn → 0 as n → 0. Hence fn(x) =

x n

1+xn ≤ bn approaches to 0

as n → ∞. By M-test, it is uniformly convergent.

For I2: Let x 1 n = 1 − 1/n and x2

n = 1 − 2/n. Both sequence converges to 1 and in [0, 1]. However,

fn(x 1 n ) → e−1/(1 + e−1) which is different from fn(x

2 n ) → e−2/(1 + e−2). Hence it is not uniformly

convergent on I2.

For I3: First of all, we change fn(x) to be 1/(1 + x −n). For any x ≥ a > 1, one has xn → ∞,

that is, x−n → 0 as n → ∞. Therefore, fn(x) → 1 for all x ∈ [a, ∞) with a > 1. Consider gn(x) = 1 − fn(x) = x

−n/(1 + x−n). By letting y = x−1, and hence x ∈ [a, ∞] implies y ∈ [0, 1/a]. a > 1 implies 1/a < 1, and gn(y) = y

n/(1 + yn). By I1, we get gn(y) is uniformly convergent on [0, 1/a], which is equivalent to fn(x) uniformly converges to 1 on [a, ∞].

2. fn(x) = 1

1+nx on the following intervals respectively

I1 = {x ∈ R : 0 < x < ∞};

I2 = {x ∈ R : a ≤ x ≤ ∞}, with 0 < a.

ANS: For I1, we let x 1 n = 1/n and x2

n = 2/n. Both sequence converges to 0. However, fn(x

1 n ) → 1/2

which is different from fn(x 2 n ) → 1/3. Hence it is not uniformly convergent on I1.

For I2: we have fn(x) ≤ 1/(nx) ≤ 1/(na) → 0 as n → ∞. Hence fn(x) is uniformly convergent on I2, by M-test.

3. fn(x) = n 2 x 2

1+n3x3 on the following intervals respectively

I1 = {x ∈ R : 0 < x < ∞};

I2 = {x ∈ R : a ≤ x ≤ ∞}, with 0 < a.

ANS: For I1, we let x 1 n = 1/n and x2

n = 2/n. Both sequence converges to 0. However, fn(x

1 n ) → 1/2

which is different from fn(x 2 n ) → 4/9. Hence it is not uniformly convergent on I1.

For I2: we have fn(x) ≤ 1/(nx) ≤ 1/(na) → 0 as n → ∞. Hence fn(x) is uniformly convergent on I2, by M-test.

4. fn(x) = x n sin(nx) on the following intervals respectively

I1 = {x ∈ R : a ≤ x ≤ b} with − 1 < a < b < 1;

I2 = {x ∈ R : −1 < x < 1}.

1

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Math 3001, Winter 2020 HW 6

ANS: For I2, we let x 1 n = 1−1/n which converges to 1. However, lim fn(x

1 n ) = lim(1−1/n)n sin(n−1)

which does not exist as lim sin(n − 1) does not exist. (See HW 1 for the proof). Hence it is not uniformly convergent on I2.

For I1: we have |fn(x)| ≤ x n ≤ (max{|a|, |b|})n → 0 as n → ∞. Hence fn(x) is uniformly convergent

on I1, by M-test.

(b): Prove the following statements.

1. Show that if {fn} converges uniformly on a set S, then {fn} also uniformly converges on any every subset of S;

Proof. By definition, fn uniformly converges to f on S implies that

sup x∈S

|fn(x) − f(x)| → 0

as n → ∞. For any subset T of S, one has

0 ≤ sup x∈T

|fn(x) − f(x)| ≤ sup x∈S

|fn(x) − f(x)| → 0

as n → ∞. Hence, sup x∈T

|fn(x) − f(x)| → 0

which implies that fn uniformly converges to f on T .

2. Show that if {fn} converges uniformly on sets S1, S2, · · · , Sm with m < ∞, then {fn} also uniformly converges on S = ∪m

i=1 Si;

Proof. By definition, fn uniformly converges to f on Si implies that

Ai = sup x∈Si

|fn(x) − f(x)| → 0

as n → ∞. Let A = ∑

1≤i≤m {Ai}. Then A → 0 as n → 0, since the sum

∑ 1≤i≤m

{Ai} is involved with finite many terms. Hence,

0 ≤ sup x∈S

|fn(x) − f(x)| ≤ A → 0

which implies that fn uniformly converges to f on S.

3. Give an example where {fn} converges uniformly on sets S1, S2, · · · but {fn} is not uniformly convergent on S = ∪∞

i=1 Si.

Proof. Consider Si = [0, 1 − 1/i] and fn as in problem (a)-1. Then by problem (a)-1-I1, fn uniformly converges on each Si with i = 1, 2, · · · . However, by problem (a)-1-I2, fn is not uniformly convergent on S = ∪∞

i=1 [0, 1 − 1/n] = [0, 1). (Although, in (a)-1-I2, I2 = [0, 1], my solution in fact

works also for [0, 1) since all sequence are in [0, 1)).

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