Exercise 6.2.15

Question

[Arzela-Ascoli Theorem] For each $n \in \mathbf{N}$, let $f_{n}$ be a function defined on $[0,1]$. If $\left(f_{n} ight)$ is bounded on $[0,1]$ extemdash that is, there exists an $M>0$ such that $\left|f_{n}(x) ight| \leq M$ for all $n \in \mathbf{N}$ and $x \in[0,1]$ extemdash and if the collection of functions $\left(f_{n} ight)$ is equicontinuous (Exercise 6.2.14), follow these steps to show that $\left(f_{n} ight)$ contains a uniformly convergent subsequence. \begin{enumerate}[label=(\alph*)] \item Use Exercise 6.2.13 to produce a subsequence $\left(f_{n_{k}} ight)$ that converges at every rational point in $[0,1]$. To simplify the notation, set $g_{k}=f_{n_{k}}$. It remains to show that $\left(g_{k} ight)$ converges uniformly on all of $[0,1]$. \item Let $\epsilon>0$. By equicontinuity, there exists a $\delta>0$ such that $$ \left|g_{k}(x)-g_{k}(y) ight|<\frac{\epsilon}{3} $$ for all $|x-y|<\delta$ and $k \in \mathbf{N}$. Using this $\delta$, let $r_{1}, r_{2}, \ldots, r_{m}$ be a finite collection of rational points with the property that the union of the neighborhoods $V_{\delta}\left(r_{i} ight)$ contains $[0,1]$. Explain why there must exist an $N \in \mathbf{N}$ such that $$ \left|g_{s}\left(r_{i} ight)-g_{t}\left(r_{i} ight) ight|<\frac{\epsilon}{3} $$ for all $s, t \geq N$ and $r_{i}$ in the finite subset of $[0,1]$ just described. Why does having the set $\left\{r_{1}, r_{2}, \ldots, r_{m} ight\}$ be finite matter? \item Finish the argument by showing that, for an arbitrary $x \in[0,1]$, $$ \left|g_{s}(x)-g_{t}(x) ight|<\epsilon $$ for all $s, t \geq N$. \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item ...is this actually a question? The rational numbers in $[0,1]$ are countable, so the results from Exercise 6.2.13 can be applied.

\item $r_i$ is a rational number, so $g_n(r_i)$ is a Cauchy sequence, and we can find $N_i$ for each $r_i$ where $s,t>N_i$ ensures $\left|g_s(r_i) - g_t(r_i)\right| <\epsilon/3$. Then just have $N = \max \{N_1,N_2,\dots, N_m\}$.
 We need $\left\{r_{1}, r_{2}, \ldots, r_{m}\right\}$ to be finite so that the final operation of taking the maximum of all $N_i$ is valid.

\item We have
 $$
\left|g_n(x) - g_m(x)\right| \leq \left|g_n(x) - g_n(a)\right| + \left|g_n(a) - g_m(a)\right| + \left|g_m(a) - g_m(x)\right|
$$
where $a$ is some rational number.
The first and last terms can be made less than $\epsilon/3$ by continuity of $g_n$ and $g_m$ and by choosing $a$ close enough to $x$, while the middle term can be made less than $\epsilon/3$ from part (b).

\end{enumerate}

Exercise 6.2.15

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Exercise 6.2.15

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