# Harrison Chapman

## Math 317: Homework 6

Due: Friday, March 15, 2019
1. (19.4)

1. Prove that if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on $S$. (Hint: Try proof by contradiction.)

Proof. Assume that $f$ is uniformly continuous on $S$ but suppose to the contrary that $f$ is unbounded on $S$. Then for all $n \in \mathbb N$, there exists as number $x_n$ so that $f(x_n) > n$. Use these to define the sequence $(x_n)$.

$S$ is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence $(x_{n_k})$. As $f$ is assumed to be uniformly continuous, this means that $(f(x_{n_k}))$ is itself a Cauchy, hence convergent, sequence. Say it converges to the number $y$.

But for all $n_k > y$, we have that $f(x_{n_k}) > n_k > y$. This implies that the sequence cannot possibly converge to $y$, a contradiction. Hence $f$ must be bounded on $S$. $\Box$

2. Explain why (a) gives a proof that $1/x^2$ is not uniformly continuous on $(0, 1)$.

$1/x^2$ is unbounded on $(0,1)$, so it cannot be uniformly continuous.

2. (23.1, 23.2) For each of the following power series, find the radius of convergence and determine the exact interval of convergence.

1. $\displaystyle\sum \left(\dfrac xn\right)^n$

The interval of convergence is $(-\infty, \infty)$.

2. $\displaystyle\sum \left(\dfrac {n^3}{3^n}\right)x^n$

The interval of convergence is $(-3, 3)$.

3. $\displaystyle\sum \left( \dfrac{3^n}{n 4^n} \right)x^n$

The interval of convergence is $[-4/3, 4/3)$.

4. $\displaystyle\sum x^{n!}$

Notice that the sequence of terms is $a_n = 1$ if $n = k!$ for some $k \in \mathbb N$, or $a_n = 0$ otherwise (if the series index starts at 0 instead of 1, then $a_1 = 2$, but this does not change the tail of the sequence). Hence the limit superior of the roots of terms is going to be $1$.

The interval of convergence is $(-1, 1)$.

3. (24.4) For $x \in [0, \infty)$, let $f_n(x) = \dfrac{x^n}{1 + x^n}$.

1. Find $f(x) = \lim f_n(x)$, the pointwise limit of $(f_n)$.

The pointwise limit is the function

To see this, take $x$ to be in any of these cases and evaluate the limit $\lim x^n/(1 + x^n)$.

1. Determine whether $f_n \to f$ uniformly on $[0, 1]$.

It cannot, as the pointwise limit is not continuous on $[0, 1]$, while each of the functions $f_n$ are.

1. Determine whether $f_n \to f$ uniformly on $[0, \infty)$.

It cannot, as the pointwise limit is not continuous on $[0, \infty)$, while each of the functions $f_n$ are.