(6). Define $f_n : [0, 1] \rightarrow \mathbb{R}$ by the equation $f_n(x) = x^n$. Show that the sequence $(f_n(x))$ converges for each $x \in [0, 1]$ but the sequence $(f_n)$ does not converge uniformly.

First we show that for each $x \in [0, 1]$ the sequence of points $(f_n(x))$ of $\mathbb{R}$ converges.

Let $x \in [0, 1]$. If $x = 0$ or $x = 1$, then the sequence $(f_n(x))$ is constant (either $(0, 0, \ldots)$ or $(1, 1, \ldots)$ respectively), so it clearly converges. Otherwise, we show that it converges to zero. Given $\epsilon >0$, set $N$ so that

$$ N > \frac{\ln \epsilon}{\ln x} $$

Then if $n \geq N$, we have $x^n < \epsilon$ by some simple algebra. So we have chosen $N$ so that for all $n \geq N$, $f_n(x) \in B(-\epsilon, \epsilon)$ so $(f_n(x))$ converges to $0$.

Now we show that $(f_n)$ does not converge uniformly. Let $f: [0, 1] \rightarrow \mathbb{R}$ be an arbitrary function; we will show that no matter what $f$ is, the sequence $(f_n)$ does not converge to it. Note that since each $f_n \geq 0$ everywhere, we can assume $f \geq 0$.

First we show that if $f(x) > 0$ at any $x \in (0, 1)$, then $(f_n)$ does not converge uniformly to $f$. Then we will show that even if $f(x) = 0$ for all $x \in (0, 1)$, $(f_n)$ still does not converge uniformly to $f$.

Suppose $f(x) > 0$ for some $x\in (0, 1)$. Then set $\epsilon = \frac{1}{2} f(x)$. Then for any integer $N$, there is always some $n \geq N$ such that for the $x$ given, $x^n < \frac{1}{2}f(x)$. In particular, for any $n$ that satisfies

$$ n > \frac {\ln \left(\frac{1}{2}f(x)\right)} {\ln x} $$

we have that $x^n < \frac{1}{2} f(x)$ and hence $x^n < f(x) - \epsilon$. So we have found some $\epsilon > 0$ so that whatever our choice of $N$, we have an $n \geq N$ and an $x \in (0, 1)$ for which $f_n(x) \notin B(f(x), \epsilon)$. So $(f_n)$ cannot converge uniformly to $f$.

Then suppose that $f(x) = 0$ for all $x \in (0, 1)$. Then set $\epsilon = \frac{1}{2}$. For any $N$, there is always some $n \geq N$ and $x \in (0, 1)$ such that

$$ x^n > \frac{1}{2} \iff x > \frac{1}{\sqrt[n]{2}} $$

This is the case because $\sqrt[n]{2} > 1$ for all $n$, so we can just choose $n = N + 1$ and find some $x < 1$ that satisfies the above. But then for the chosen $n$ and $x$, $f_n(x) \notin B(f(x), \epsilon)$. Hence $(f_n)$ cannot converge uniformly to $f$.

Thus there is no function to which $(f_n)$ converges uniformly.

(7). Let $X$ be a set, and let $f_n: X \rightarrow \mathbb{R}$ be a sequence of functions. Let $\overline{\rho}$ be the uniform metric on the space $\mathbb{R}^X$. Show that the sequence $(f_n)$ converges uniformly to the function $f: X \rightarrow \mathbb{R}$ if and only if the sequence $(f_n)$ converges to $f$ as elements of the metric space $(\mathbb{R}^X, \overline{\rho})$.

First suppose $(f_n)$ converges uniformly to $f$. We want to show that the sequence $(f_n)$ converges via the $\overline{\rho}$-metric. Let $\epsilon > 0$. Then set $\epsilon' = \min\{\frac{\epsilon}{2}, \frac{1}{2} \}$. By uniform convergence, there is some $N$ such that for all $x \in X$ and $n \geq N$,

$$ d(f_n(x), f(x)) < \epsilon' \implies \vert f_n(x) - f(x) \vert < \epsilon' $$

where $d$ is the standard metric over $\mathbb{R}$. Then for all $n > N$,

$$ \begin{array}{rlr} \overline{\rho}(f_n, f)

&=\sup\left\{\overline{d} \left(f_n(x) - f(x)\right) \mid x \in X \right\} &\text{ (by definition of } \overline{\rho} \text{)} \\

&=\sup\left\{\min \left( \vert f_n(x) - f(x) \vert, 1\right) \mid x \in X \right\}

&\text{ (by definition of } \overline{d} \text{)} \\

&= \sup\{ \vert f_n(x) - f(x) \vert \mid x \in X \}

& \text{(since } \epsilon' \leq \frac{1}{2} < 1 \text{)} \\

& \leq \epsilon' \leq \frac{\epsilon}{2} < \epsilon

& \text{(since } \vert f_n(x) - f(x) \vert < \epsilon' \text{)}

\end{array} $$