**Version 1** **(215)**

Let *K* be a compact subset of and let be an equicontinuous family of functions. Assume that *f(x)* is bounded for each fixed as *f* varies over *Y*. Then *Y* is pre-compact, i.e. every sequence in *Y* has a subsequence converging to some (equivalently, the closure of *Y* in *C(K)* is compact.)

**In other words**

If is an equicontinuous and uniformly bounded family of continuous functions on , then any sequence drawn from has a subsequence which converges (uniformly on any compact set) to some continuous function.

**In complex analytic terms**

Any equicontinuous and uniformly bounded collection of continuous functions forms a normal family: this is the second half of Montel’s theorem (the first half states that a uniformly bounded family of continuous functions is equicontinuous.)

**Version 2 (Tao)**

Let *Y* be a metric space, *X *be a compact metric space, and let {(f_\alpha)_{\alpha \in A}} be a family of bounded continuous functions from *X* to *Y*. The following are equivalent:

- (i) is a precompact subset of the bounded continuous functions from
*X*to*Y*. - (ii) is pointwise precompact (i.e. for every , the set is precompact in
*Y)*and equicontinuous. - (iii) is pointwise precompact and uniformly equicontinuous (= equicontinuous + uniform bound on size of neighborhoods (radius of balls))

**Proof (Arzelà–Ascoli diagonalisation)**

Now a standard technique in analysis, as first shown to you by Sarnak

Given , by equicontinuity of there is a such that whenever $d_X(x,y) < \delta$.

Key observation: since is compact, there exists a *finite* subcover .

Let and set .

Given , $x \in B(p_j, \delta)$ for some $j = 1, \cdots, N$; then $\left| f(x) – f(p_j) \right| < \epsilon$. Then

\[ \left|f(x)\right| < M_j + \epsilon \leq M + \epsilon \]

for all $x \in K$; hence $f$ is uniformly bounded as $f$ varies over $\mathcal{Y}$ and $x$ varies over $K$.

Next, let $E \subset K$ be a countable dense subset of $K$. We claim that any sequence $\{f_n\}\stdseq \subset \mathcal{Y}$ has a subsequence $\{f_{n_j}\}_{j=1}^{\infty}$ for which $\{f_{n_j}(e)\}_{j=1}^{\infty}$ converges for each $e \in E$.

This we prove using a diagonal argument (not quite Cantor’s.) Enumerate $E = \{e_1, e_2, \cdots \}$. Write out a table

\[ \begin{array}{l l l l l}

& f_1 & f_2 & f_3 & \cdots \\

e_1 & f_1(e_1) & f_2(e_1) & f_3(e_1) & \cdots \\

e_2 & f_1(e_2) & f_2(e_2) & f_3(e_2) & \cdots \\

e_3 & f_1(e_3) & f_2(e_3) & f_3(e_3) & \cdots \\

\vdots & \vdots & \vdots & \vdots & \ddots

\end{array} \]

First construct a family of sequences as follows: let $A_1$ be a convergent subsequence of the elements of the first row.

For $n > 1$, let $A’_n$ be the subsequence of the $n\thc$ row of elements picked out by the indices of $A_{n-1}$, and let $A_n$ be a convergent subsequence of $A’_n$.

At each step a convergent subsequence exists because $f \in \mathcal{Y}$ is uniformly bounded, and hence $f_i(e_j) \in S$ for all $i,j \in \nats$, where $S$ is some closed, bounded (hence compact) interval of reals.

Now define the sequence $\{f_{n_j}\}_{j=1}^{\infty}$ as follows: for each $j$, look at the $j\thc$ element of $A_j$, which is of the form $f_k(x_l)$ for some natural numbers $k,l$. Then $n_j := k$, i.e. $f_{n_j} = f_k$. By recursion, this subsequence converges pointwise for every $e_i \in E$.

Finally, we will show that the subsequence we have just constructed is convergent in $C(K)$, by showing that it is uniformly Cauchy, i.e. given any $\epsilon > 0$, there exists a $N \in \nats$ such that $\left| f_n(x) – f_m(x) \right| < \epsilon$ for all $x\in K$ whenever $n,m \geq N$.

We have the finite open cover $\cup_{j=1}^n B(p_j, \delta) \supset K$. For each $j = 1, \cdots, N$, $B(p_j, \delta)$ contains some element of $E$—call it $e’_j$—since $E$ is dense.

Note $\{f_n(e’_j)\}\stdseq$ is a Cauchy sequence for each fixed $j$, since we have already proven that $\{f_n(e)\}\stdseq$ converges for any $e \in E$. Hence there exists a natural number $R_j$ such that whenever $n,m \geq R_j$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$.

Pick $R = \max\{R_1, \cdots, R_N\}$. Then if $n,m \geq R$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$ for all $j=1, \cdots, N$.

Now, given any $x \in K$, $X \in B(p_j, \delta)$ for some natural number $j$. Then

\[ \left| f_n(x) – f_m(x) \right| \leq \left| f_n(x) – f_n(e’_j) \right| + \left| f_n(e’_j) – f_m(e’_j) \right| + \left| f_m(e’_j) – f_m(x) \right| < \epsilon \]

for all $n,m \geq R$.

There are also other proofs (e.g. using functional analysis, which I shall perhaps explore one day.)

**In short**

A way of obtaining convergence / compactness in function space from stronger continuity / boundedness assumptions.