The Converse of the Arzelà-Ascoli Theorem

# The Converse of the Arzelà-Ascoli Theorem

Recall from The Arzelà–Ascoli Theorem page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of continuous functions defined on a compact set $D \subseteq \mathbb{R}^n$ and if $\mathcal F \{ f_n : n \in \mathbb{N} \}$ is uniformly bounded and equicontinuous on $D$ then there exists a subsequence $(f_{n_k}(x))_{k=1}^{\infty}$ that converges uniformly to a function $f \in C(D, \mathbb{R})$.

We will now prove that the converse of the Arzelà-Ascoli theorem is also true particularly for when we consider functions defined on a compact interval (a closed and bounded interval) $J$.

Theorem 1 (Converse of the Arzelà-Ascoli Theorem): Let $J \subseteq \mathbb{R}$ be a compact interval and let $\mathcal F$ be a collection of continuous functions defined on $J$. If every sequence of functions $(f_n(x))_{n=1}^{\infty}$ in $\mathcal F$ has a uniformly convergent subsequence then $\mathcal F$ is uniformly bounded and equicontinuous. |

**Proof:**We break this proof into two parts. In the first part we show that $\mathcal F$ is uniformly bounded. In the second part we show that $\mathcal F$ is equicontinuous.

**Part 1 (Uniform Boundedness of $\mathcal F$ on $J$):**Assume instead that $\mathcal F$ is not uniformly bounded on $J$. Then there exists a sequence $(f_n(x))_{n=1}^{\infty}$ in $\mathcal F$ and a sequence of real numbers $(x_n)_{n=1}^{\infty}$ in $J$ such that for all $n \in \mathbb{N}$ we have that:

\begin{align} \quad | f_n(x_n) | > n \quad (\dagger) \end{align}

- Since every sequence in $\mathcal F$ contains a uniformly convergent subsequence, in particular, we have that $(f_n(x))_{n=1}^{\infty}$ has a subsequence $(f_{n_k}(x))_{k=1}^{\infty}$ that converges uniformly to a function $f \in C(J, \mathbb{R})$. Now, since $f$ is continuous on $J$ and $J$ is compact we have that $f$ is uniformly continuous on $J$. But every continuous function defined on a compact interval $J$ is bounded. So there exists a $B \in \mathbb{R}$, $B > 0$ such that:

\begin{align} \quad | f(x) | \leq B \quad (*) \end{align}

- Furthermore, since $(f_{n_k}(x))_{k=1}^{\infty}$ converges uniformly to $f(x)$ on $J$ we have that for $\displaystyle{\epsilon^* = B > 0}$ there exists an $N \in \mathbb{N}$ such that if $k \geq N$ then for all $x \in J$:

\begin{align} \quad | f_{n_k}(x) - f(x) | < \epsilon^* = B \quad (**) \end{align}

- Therefore, if $n \geq N$ then from $(*)$ and $(**)$ we have that:

\begin{align} \quad | f_{n_k}(x_{n_k}) | = | f_{n_k}(x_{n_k}) - f(x_{n_k}) + f(x_{n_k}) | \leq | f_{n_k}(x_{n_k}) - f(x_{n_k}) | + | f(x_{n_k}) | < B + B = 2B \end{align}

- But if $k > 2B$ then from $(\dagger)$ we have that:

\begin{align} \quad | f_{n_k}(x_{n_k}) | > n_k \geq k > 2B \end{align}

- Hence $2B < | f_{n_k}(x_{n_k}) | < 2B$ which is a contradiction. So the assumption that $\mathcal F$ was not uniformly bounded on $J$ was false. Hence $\mathcal F$ is uniformly bounded on $J$.

**Part 2 (Equicontinuity of $\mathcal F$ on $J$):**Assume that $\mathcal F$ is not equicontinuous on $J$. Then for some $\epsilon_0 > 0$ there exists a sequence of functions $(f_n(x))_{n=1}^{\infty}$ and sequences $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ of real numbers in $J$ such that if $\displaystyle{| x_n - y_n | < \frac{1}{n}}$ then:

\begin{align} \quad | f_n(x_n) - f_n(y_n) | \geq \epsilon_0 \end{align}

- Since $(f_n(x))_{n=1}^{\infty}$ is a sequence of functions in $\mathcal F$ there exists a uniformly convergent subsequence $(f_{n_k}(x))_{k=1}^{\infty}$ that converges to $f \in C(J, \mathbb{R})$. So for $\displaystyle{\frac{\epsilon_0}{3} > 0}$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then for all $x \in J$:

\begin{align} \quad | f_{n_k}(x) - f(x) | < \frac{\epsilon_0}{3} \quad (*) \end{align}

- Furthermore, since $f$ is continuous on the compact interval $J$ we have that $f$ is uniformly continuous on $J$. So for $\displaystyle{\frac{\epsilon_0}{3} > 0}$ there exists a $\delta > 0$ such that for all $x, y \in J$ with $| x - y | < \delta$ we have that:

\begin{align} \quad | f(x) - f(y) | < \frac{\epsilon_0}{3} \quad (**) \end{align}

- Let $N_2 \in \mathbb{N}$ be such that $\displaystyle{N_2 > \frac{1}{\delta} > 0}$. Then for all $k \geq N_2$ we have that:

\begin{align} \quad | x_{n_k} - y_{n_k} | \leq \frac{1}{n_k} \leq \frac{1}{k} < \frac{1}{M} < \delta \end{align}

- So let $N = \max \{ N_1, N_2 \}$. Then if $n \geq N$ we have that $(*)$ and $(**)$ hold so:

\begin{align} \quad | f_{n_k}(x_{n_k}) - f_{n_k}(y_{n_k}) | = | f_{n_K}(x_{n_k}) - f(x_{n_k}) + f(x_{n_k}) - f(y_{n_k}) + f(y_{n_k}) - f_{n_k}(y_{n_k}) | < \frac{\epsilon_0}{3} + \frac{\epsilon_0}{3} + \frac{\epsilon_0}{3} = \epsilon_0 \end{align}

- So $\epsilon_0 \leq | f_{n_k}(x_{n_k}) - f_{n_k}(y_{n_k}) | < \epsilon_0$, which is a contradiction. So the assumption that $\mathcal F$ is not equicontinuous on $J$ is false. So $\mathcal F$ is equicontinuous on $J$. $\blacksquare$