Math 104: Midterm 2 sample solutions

1. To show that f is uniformly continuous, choose ϵ > 0. Since f_{n} → f uniformly, there exists an N such that n > N implies that

|f_{n}(x) − f(x)| < \frac{ϵ}{3}

for all x ∈ (a, b). Now consider f_{N+1} : since this is uniformly continuous, there exists a δ > 0 such that if x, y ∈ (a, b) and |x − y| < δ, then

|f_{N+1} (x) − f_{N+1} (y)| < \frac{ϵ}{3}.

Now, for any x, y ∈ (a, b) with |x − y| < δ,

|f(x) − f(y)| ≤ |f(x) − f_{N+1} (x)| + |f_{N+1} (x) − f_{N+1} (y)| + |f_{N+1} (y) − f(y)|
< \frac{ϵ}{3} + \frac{ϵ}{3} + \frac{ϵ}{3} = ϵ

and hence f is uniformly continuous.

2. For three values 0, 1, and 2,

d_{1}(0, 1) + d_{1}(1, 2) = 1^{4} + 1^{4} = 2

but

d_{1}(0, 2) = 2^{4} = 16

and hence the triangle inequality is violated, so d_{1} is not a metric.

Since d_{2}(0, 0) = 1, it does not satisfy the property that d(x, x) = 0 for all x ∈ , and hence d_{2} is not a metric. Since d_{3}(0, 1) = 2, and d_{3}(1, 0) = 1 it is not symmetric, and hence it is not a metric.

3. First, consider the series

∑_{n=0}^{∞} \frac{x^{n}}{n^{\sqrt{n}}}

so that the coefficients are a_{n} = n^{−\sqrt{n}} . Then

β = lim sup |a_{n}|^{1/n}
= lim sup |n^{−\sqrt{n}} |^{1/n}
= lim sup n^{−1/\sqrt{n}} .

Since the power will always be negative, all of the terms in this sequence must be less than or equal to 1, so β ≤ 1. Consider the subsequence of terms for n = 2^{k}. Then

n^{−1/\sqrt{n}} = 2^{−k/2^{k/2}} .

It is known that k/2^{k/2} → 0 as k → ∞. Hence 2^{−k/2^{−k}} → 1 as k → ∞. Hence, since the subsequence tends to 1 as n → ∞, then β ≥ 1. Combining with the result above, β = 1, and the radius convergence is R = 1.

For the second series

∑_{n=0}^{∞} 4^{n} x^{2n+1}

then

a_{n} = \{ \begin{matrix}2^{n−1} & if n is odd \\ 0 & if n is even. \end{matrix}

Consider the subsequence of odd terms:

|a_{2k+1} |^{1/(2k+1)} = |2^{2k+1−1} |^{1/(2k+1)} = 2^{2k/(2k+1)}

which converges to 2 as k → ∞. Hence

β = lim sup |a_{n}|^{1/n} = 2

and therefore the radius of convergence is R = ^{1}/2.

For the third series,

∑_{n=0}^{∞} x^{n^{2}},

the coefficients are a_{n} = 1 if n is a square number and zero otherwise. Hence

|a_{n}|^{1/n} = \{ \begin{matrix}1 & if n is a square number \\ 0 & otherwise. \end{matrix}

Hence

β = lim sup |a_{n}|^{1/n} = 1

since there are infinitely many terms which are 1 in the sequence. Hence the radius of convergence is R = 1.

4. If g is bounded on S, then there exists an M > 0 such that |g(x)| < M for all x ∈ S. Now consider any ϵ > 0. If f_{n} converges uniformly to f, then there exists an N such that n > N implies that

|f_{n}(x) − f(x)| < \frac{ϵ}{M}

for all x ∈ S. Now consider g · f_{n}:

|g(x)f_{n}(x) − g(x)f(x)| = |g(x)| · |f_{n}(x) − f(x)| < M\frac{ϵ}{M} = ϵ

and hence it uniformly converges to g · f.

5. If f_{n} converges uniformly to f, then there exists an N such that n > N implies that

|f_{n}(x) − f(x)| < 1

for all x ∈ S. By using the triangle inequality,

|f(x)| < |f_{N+1} (x)| + 1.

Since f_{N+1} is bounded, |f_{N+1} (x)| < M for all x and for some M ≥ 0, and thus |f(x)| < M + 1 for all x. Hence f is bounded.

6. Since a continuous function on a closed interval is bounded, then for each f_{n}, there exists M_{n} such that |f_{n}(x)| < M_{n} for all x ∈ [0, 1]. Since f_{n} converges uniformly to f, there exists an N ∈ such that for all n > N,

|f_{n}(x) − f(x)| ≤ 1

Since the f_{n} converge uniformly and are continuous, the limit f is continuous also, and therefore bounded, so that |f(x)| < M′for all x ∈ [0, 1], for some M′ > 0. By using the triangle inequality,

|f_{n}(x)| < |f(x)| + 1 < M′+ 1

for all n > N. Now define

L = max{M_{1}, M_{2}, …, M_{N}, M′+ 1}

Consider any x ∈ [0, 1] and any n ∈ . If n ≤ N, then |f_{n}(x)| < M_{n} ≤ L. If n > N, then |f_{n}(x)| < M′+ 1 ≤ L. Thus L is an upper bound for the set A = {|f_{n}(x)| : n ∈ , x ∈ [0, 1]}. Hence 0 ≤ sup A ≤ L, and thus sup A must be finite.