ANS: The trouble point is x = 0, which is not inside the interval. So, other points are “normal” and one gets the pointwise limit by treating x as a constant, which is:

f(x) = lim n→∞

n ln

(

)

= lim y→0

y ln y = 0.

Now, we determine the maximal value of g(y) = y ln y whose derivative is g′(y) = ln y + 1. One can get the critical point is y = e−1 and this point is a minimal point. Moreover, if y < 1/e, g′(y) < 0 and hence g(y) is decreasing on (0, 1/e). Note that g(y) ≤ 0 on D = (0, 1). Hence, |g(y)| obtains its maximal at y = 1/e, and |g(y)| is increasing on (0, 1/e).

Note our functions fn(x) = x2

n ln (

)

. So for n big enough, say n ≥ 3, then x2/n ∈ (0, 1/e) for all

x ∈ (0, 1). Hence, one can see that |fn(x)| ≤ |fn(1)| = | 1 n ln (

1 n

)

| = 0 as n → ∞.

By definition (or M-test), one has fn is uniformly convergent to 0 on D = (0, 1).

2. fn(x) = sin(nx)

n2 on D = (−∞, ∞).

ANS: Clearly, we have |fn(x) − 0| ≤ 1/n

2 → 0

as n → ∞. Hence by definition (or M-test), one has fn is uniformly convergent to 0 on D = (−∞, ∞).

3. fn(x) = 1

1+n3x on D = [0, ∞).

ANS: The trouble point is x = 0, and fn(0) = 1 which implies f(0) = limn→∞ fn(0) = 1.

So, other points are “normal” and one gets the pointwise limit by treating x as a constant, which is:

f(x) = lim n→∞

1 + n3x = 0.

Note that, each fn is continuous. If fn is uniformly convergent to f(x), then f(x) should be continuous. However, f(x) is not continuous at x = 0. This implies that fn is not uniformly convergent to f(x) on D = [0, ∞).

Another method is: choosing x1n = 1/n 3 and hence fn(x

1 n) = 1/2 → 1/2 as n → ∞. Choosing

x2n = 2/n 3 and hence fn(x

2 n) = 1/3 → 1/3 as n → ∞. Hence fn is not uniformly convergent to f(x)

on D = [0, ∞).

4. fn(x) = x 2e−2nx on D = [0, ∞).

ANS: The trouble point is x = 0, and fn(0) = 0 which implies f(0) = limn→∞ fn(0) = 0.

Other points are “normal” and one gets the pointwise limit by treating x as a constant, which is:

f(x) = lim n→∞

x2e−2nx = 0.

So, f(x) = 0 for all x ∈ D.

Clearly, the derivative of f ′n(x) = 2x(1 − nx)e −2nx. This implies that x = 1/n is a local maximal

point for fn(x). Moreover, one can have x = 1/n is a global maximal point for fn(x), namely

|fn(x) − 0| ≤ fn(1/n) = e −2/n2 → 0

as n → ∞. Hence, by definition (or M-test), one has fn is uniformly convergent to 0 on D = [0, ∞).

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Math 3001, Winter 2020 Homework 7

(b): Prove or disprove the following problems.

1. Suppose fn(x) is uniformly convergent to f on D = [a, b]. Let c ∈ [a, b]. Is fn uniformly convergent to f on D1 = [a, c] and/or D2 = [c, b]?

2. Let a < c < b. Suppose fn(x) is uniformly convergent to f on D1 = [a, c] and D2 = [c, b]. Is fn uniformly convergent to f on D = [a, b].

3. Suppose that fn(x) is uniformly convergent to f on Ii, i = 1, 2, · · · Is fn uniformly convergent to f on ∪mi=1Ii with finite m. How about the uniform convergence of fn on ∪

∞ i=1Ii.

Here is the solution for much more general cases, and our problems are the special case,

namely on intervals.

(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 = ∪mi=1Si;

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=1Si.

Proof. Consider fn(x) = xn

1+xn on the following intervals respectively

I1 = {x ∈ R : 0 ≤ x ≤ b}, with b < 1;

I2 = {x ∈ R : 0 ≤ x ≤ 1};

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Math 3001, Winter 2020 Homework 7

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

1+xn ≤ xn approaches to 0 as

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

For I2: Let x 1 n = 1 − 1/n and x

2 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.

Consider Si = [0, 1 − 1/i]. Then fn uniformly converges on each Si with i = 1, 2, · · · . However, fn is not uniformly convergent on S = ∪∞i=1[0, 1 − 1/n] = [0, 1).

(c): Let f(x) be a continuous function in [0, 1] with f(1) = 0. Prove that gn(x) = f(x)x 2n is uniformly

convergent in D = [0, 1]. Proof. First of all, as f is a continuous function in [0, 1] , there is a M > 0, such that |f(x)| < M for all x ∈ [0, 1].

Second, gn(1) = 0 and hence g(1) = limn→∞ gn(1) = 0. For all x ∈ [0, 1), as |f(x)| < M and 0 ≤ limn→∞ |gn(x)| ≤ limn→∞ Mx

2n = 0, which means that g(x) = 0 for all x ∈ [0, 1]. To have gn(x) = f(x)x

2n is uniformly convergent g(x) = 0 in D = [0, 1], one needs to prove that, for all ǫ > 0, there is N(ǫ) > 0, such that

|gn(x)| < ǫ, ∀x ∈ [0, 1], ∀n > N(ǫ).

Note that f(1) = 0 and the continuity of f implies that, for all ǫ > 0, there is δ(ǫ) > 0, such that,

|f(x)| < ǫ, ∀x ∈ (1 − δ(ǫ), 1].

This implies that

|gn(x)| = |f(x)x 2n| ≤ |f(x)| < ǫ, ∀x ∈ (1 − δ(ǫ), 1], ∀n ≥ 1.

On the other hand, x2n is uniformly convergent to 0 in [0, 1−δ(ǫ)], that is, for all ǫ > 0, there is N(ǫ) > 0, such that

|x2n| < ǫ/M, ∀x ∈ [0, 1 − δ(ǫ)], ∀n > N(ǫ).

Therefore, |gn(x)| = |f(x)x

2n| < Mx2n < ǫ, ∀x ∈ [0, 1 − δ(ǫ)], ∀n > N(ǫ).

In conclusion, one has for all ǫ > 0, there is N(ǫ) > 0, such that

|gn(x)| < ǫ, ∀x ∈ [0, 1], ∀n > N(ǫ).

That is, gn(x) is uniformly convergent to 0 on D = [0, 1].

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