Theorems

Arzelà–Ascoli

Version 1 (215)

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

In other words

If \mathcal{F} is an equicontinuous and uniformly bounded family of continuous functions on \mathbb{R}, then any sequence drawn from \mathcal{F} 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, be a compact metric space, and let (f_\alpha)_{\alpha \in A} {(f_\alpha)_{\alpha \in A}} be a family of bounded continuous functions from X to Y. The following are equivalent:

  • (i) \{f_\alpha \mid \alpha \in A\} is a precompact subset of the bounded continuous functions from X to Y.
  • (ii) (f_\alpha)_{\alpha \in A} is pointwise precompact (i.e. for every x \in X, the set \{f_\alpha(x) \mid \alpha \in A\} is precompact in Y) and equicontinuous.
  • (iii) (f_\alpha)_{\alpha \in A} 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 \epsilon > 0, by equicontinuity of \mathcal{Y} there is a \delta > 0 such that \left| f(x) - f(y) \right| < \epsilon whenever $d_X(x,y) < \delta$.

Key observation: since K \subset \cup_{k \in K} B(k, \delta) is compact, there exists a finite subcover \cup_{j=1}^n B(p_j, \delta) \supset K.

Let M_j = \sup_{f \in \mathcal{Y}} \left| f(p_j) \right| < \infty and set M = \max \{M_j \:|\: j=1, \cdots, N\} < \infty.

Given x \in K, $x \in B(p_j, \delta)$ 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.

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