## Arzelà-Ascoli

"A great many propositions in analysis are trivial for finite sets, true and reasonably simple for infinite compact sets; and either false or extremely difficult to prove for noncompact sets." - Edwin Hewitt

The Arzelà-Ascoli theorem is an important result in mathematical analysis.

In this post we will be stating and proving the general form of the theorem. We will do this by establishing three lemmas and combining them to obtain the result.

#### Arzelà-Ascoli Theorem

Let \(\mathcal{F}\) be an equicontinuous family from a separable metric space \((X, d_X)\) to a complete metric space \((Y, d_Y)\) such that \(\overline{\{f_n(x) : 0 \leq n < \infty \}}\) is compact for all \(x\).

Then there is a subsequence \(\{f_{n_j}\}_{j=1}^\infty\) converging pointwise to a continuous function and the convergence is uniform on compact subsets.

##### Corollary

Let \(\{f_n : [0,1] \to \mathbb{R} \}\) be uniformly bounded and equicontinuous. Then some subsequence converges in \(C[0,1]\).

The Arezlà Ascoli theorem is so important because it allows us to characterise compact sets of continuous functions.

It has applications in the theory of ordinary and partial differential equations, in probability theory, in complex analysis, and in harmonic analysis.

One of the key assumptions in the hypothesis is that of equicontinuity.

This is a concept I hadn't seen before learning about the Arzelà-Ascoli theorem.

#### Equicontinuity

Let \((X,d_x)\), \((Y,d_Y)\) be metric spaces, \(C(X,Y)\) be the space of continuous functions from \(X\) to \(Y\).

A family \(\mathcal{F} \subseteq C(X,Y)\) of functions is equicontinuous if

\(\forall \epsilon > 0, \forall x,y \in X, \exists \delta > 0\) such that \(d_X(x,y) < \delta \implies d_Y(f(x), f(y)) < \epsilon\) \(\forall f \in \mathcal{F}\)

In words, this means that for an equicontinuous family, the value of \(\delta\) depends on \(\epsilon\) and \(x\) but not on \(f \in \mathcal{F}\).

Another way to explain equicontinuity is that the same choice of \(\delta\) will work on all \(f \in \mathcal{F}\).

Now we will state and prove some lemmas required for proving the theorem.

#### Lemma I

Let \(\{f_n: X \to Y\}_{n=1}^\infty\) where \(Y\) is complete be equicontinuous and assume that for some \(D \subseteq X\) dense that \(\{f_n(x)\}_{n=1}^\infty\) converges for \(x \in D\). Then \(f_n(x)\) converges \(\forall x \in X\) and \(f(x) = \lim f_n(x)\) is continuous.

##### Proof

It suffices to show that \(\{f_n(x)\}_{n=1}^\infty\) is Cauchy for all \(x \in X\).

Well, what do we know?

By equicontinuity we have \(\forall \epsilon > 0\) \(\exists \delta \) such that \(d_x(x,y) < \delta \implies d_Y(f_n(x), f_n(y)) < \frac{\epsilon}{3}\)

We also know

\(\forall x\) \(\exists y \in D\) such that \(d_X(x,y) < \delta\) so we have

\(d_Y(f_n(x), f_m(x)) < d_Y(f_n(x), f_n(y)) + d_Y(f_n(y), f_m(y)) + d_Y(f_m(y), f_m(x))\)

The first and third terms are bounded by \(\frac{\epsilon}{3}\) by equicontinuity, and the second term is Cauchy in \(D\), so with a large enough \(N\) we can also bound it by \(\frac{\epsilon}{3}\).

Thus we have \(d_Y(f_n(x), f_m(x)) < \epsilon\), and hence that \(\{f_n(x)\}_{n=1}^\infty\) is Cauchy in \(Y\) for all \(x \in X\).

We now know that \(f = \lim f_n\) exists, we must now show that it is continuous.

Let \(\epsilon > 0, x \in X\). We have a \(\delta>0\) such that \(d_X(x,y) < \delta \implies (d_Y(f_n(x), f_n(y)) < \frac{\epsilon}{2} \forall n)\)

We also know that \(d_Y\) is continuous, so we have

\(\frac{\epsilon}{2} \geq \lim_{n\to \infty}d_y(f_n(x), f_n(y)) = d_Y(f(x),f(y))\)

So \(f\) is continuous.\(\blacksquare\)

#### Lemma II

Let \(\{f_n: X \to Y\}_{n=1}^\infty\) be an equicontinuous family, and assume \(\exists f : X \to Y\) such that \(f_n(x) \to f(x)\) \(\forall x \in X\). If \((X, d_X)\) is compact then \(f_n \to f\) uniformly.

##### Proof

Since \(X\) is compact \(\{f_n\} \cup \{f\}\) is an equicontinuous family.

Let \(\epsilon > 0\), choose \(\delta\) such that:

\(d_X(x,y) < \delta\) implies both \(d_Y(f_n(x), f_n(y)) < \frac{\epsilon}{3}\) and \(d_Y(f(x), f(y)) < \frac{\epsilon}{3}\)

Let \(\{B_X(y, \delta)\}_{y\in X}\) be an open cover of \(X\) that admits a finite sub cover \(\bigcup_{i=1}^m B_X(y_i, \delta)\).

Because we have a finite sub cover, every \(x\in X\) is within \(\delta\) of some \(y_i\).

For each \(i\) there exists an \(N_i\) such that for all \(n\) greater than \(N_i\) we have \(d_Y(f_{n}(y_i), f(y_i)) < \frac{\epsilon}{3}\)

Let \(N \geq \max_{i=1,...,m}N_i\) then we have

\(d_Y(f_n(x), f(x)) \leq d_Y(f_n(x), f_n(y_i)) + d_Y(f_n(y_i), f(y_i)) + d_Y(f(y_i), f(x)) < \epsilon\)

Thus the sequence is uniformly convergent.\(\blacksquare\)

#### Lemma III

Let \(D\) be a countable set, \(\{f_n : D \to Y\}\) where \(Y\) is complete and assume that \(\forall x \in D\) the set \(\overline{\{f_n(x) : 1 \leq n < \infty \}}\) is compact. Then there exists a subsequence \(\{f_{n_j}\}_{j=1}^\infty\) convergent on all \(D\)

##### Proof

Since \(D\) is countable, let \(q_1, q_2, q_3, ... \) be an enumeration of the elements of \(D\).

\(\overline{\{f_n(x) : 1 \leq n < \infty \}}\) is compact for each \(q_i\), in particular for \(q_1\).

Sequential and 'regular' compactness are equivalent in a metric space, so we know that there exists a sequence of functions \(\{f_{n,1}\}_{n=1}^\infty\), such that \(\{f_{n,1}(q_1)\}_{n=1}^\infty\) converges.

Since \(\overline{\{f_{n,1}(x) : 1 \leq n < \infty \}}\) is a closed subset of a compact set, it is compact. Hence it also admits convergent sequences, in particular it admits a convergent sequence for \(q_2\).

That is, there exists a sequence of elements \(\{f_{n,2}\}_{n=1}^\infty\), such that \(\{f_{n,2}(q_2)\}_{n=1}^\infty\) converges.

The crucial thing to note, is that \(\{f_{n,2}\}_{n=1}^\infty\) is a subsequence of \(\{f_{n,1}\}_{n=1}^\infty\).

This holds precisely because of the aforementioned fact that \(\overline{\{f_{n,1}(x) : 1 \leq n < \infty \}}\) is compact.

Thus we get the following diagram.

\(f_{1,1}\) \(f_{2,1}\) \(f_{3, 1}\) \(f_{4, 1}\) \(f_{5,1}\) \(\cdots\)

\(f_{1,2}\) \(f_{2,2}\) \(f_{3, 2}\) \(f_{4, 2}\) \(f_{5,2}\) \(\cdots\)

\(f_{1,3}\) \(f_{2,3}\) \(f_{3, 3}\) \(f_{4, 3}\) \(f_{5,3}\) \(\cdots\)

\(f_{1,4}\) \(f_{2,4}\) \(f_{3, 4}\) \(f_{4, 4}\) \(f_{5,4}\) \(\cdots\)

\(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

The important things to note about this diagram is that the \(i\)th row converges when evaluated at any point \(q_1, q_1, ..., q_i \in D\), and that each row is a subsequence of the row above.

Remembering the goal is to show that there exists a subsequence that converges on all of \(D\) we take the sequence formed by the diagonal entries of the above diagram, that is, take \(\{f_{n,n}\}_{n=1}^\infty\).

This subsequence converges for \(q_1\), since \(\{f_{n,n}\}_{n=1}^\infty\) is a subsequence of \(\{f_{n,1}\}_{n=1}^\infty\) by definition.

\(\{f_{n,n}\}_{n=1}^\infty\) converges for \(q_i\) because all but a finite number of elements of \(\{f_{n,n}\}_{n=1}^\infty\) are also in \(\{f_{i,n}\}_{n=1}^\infty\), and a convergent sequence with finitely many elements added to the beginning still converges.

Thus it is exactly this diagonal subsequence that is convergent on all of \(D\). \(\blacksquare\)

##### Proof (Arzelà-Ascoli)

Let \(C\) be a countable dense subset of \((X, d_x)\).

By Lemma III there exists a subsequence \(\{f_{n_j}\}_{j=1}^\infty\) that is convergent on all of \(C\).

By Lemma I we can extend the convergence of \(\{f_{n_j}\}_{j=1}^\infty\) to all of \(X\).

By Lemma II we get uniformity on compact subsets.\(\blacksquare\)

This completes the proof of the general statement of the Arzelà-Ascoli Theorem. I hope you've enjoyed learning about it!