# Point-wise bounded and equicontinuous sequence of functions has a uniformly convergent subsequence

Problem

We have a sequence $(f_n)$ of continuous functions on a compact metric space K. It is also given that $(f_n)$ is point-wise bounded and equicontinuous. Now show that $(f_n)$ has a sub-sequence that converges uniformly on K.

• It means the domain is $K$. Usually, if nothing about the codomain is said, that is assumed to be $\mathbb{R}$ (or $\mathbb{C}$). Not that it matters much here, any metric space in which bounded subsets are relatively compact will do. – Daniel Fischer Jun 15 '15 at 15:38
• @DanielFischer Even in that case they are compact but taking a set like {'a','b','c','d'} will not work for my logic. So should I say in general my logic will not work? – User Not Found Jun 15 '15 at 15:39

First of all what is meant by the phrase "functions on a compact metric space". Does it mean that the domain is $K$ or does it mean that the range is also $K$.

It means that the domain of the functions is $K$. Nothing is said about the codomain. Usually, when nothing is said about the codomain, and it is not clear from the context, the codomain is assumed to be $\mathbb{R}$ or $\mathbb{C}$. But the conclusion holds when the codomain is any metric space in which bounded subsets are relatively compact, and when in the premises "$(f_n)$ is point-wise bounded" is replaced with "$\{ f_n(x) : n \in \mathbb{N}\}$ is relatively compact for all $x\in K$", the codomain can be any metric space.

The assertion is more or less one direction of the Ascoli-Arzelà theorem, which characterises relatively compact sets in some function spaces. Two key points are

1. on a compact (metric) space, every equicontinuous family is uniformly equicontinuous, and
2. a compact metric space is separable.

There is a good chance that your text/lecture does not distinguish between equicontinuity and uniform equicontinuity and calls plain equicontinuity what is in fact uniform equicontinuity, so let's first make clear what the first point means.

Let $T$ a topological space, and $M$ a metric space. Further, let $\mathscr{F}$ be a family (set) of functions $T\to M$. The family $\mathscr{F}$ is equicontinuous at $t_0 \in T$, if for every $\varepsilon > 0$ there is a neighbourhood $U_\varepsilon$ of $t_0$ such that

$$\bigl(\forall f\in \mathscr{F}\bigr)\bigl(\forall t\in U_\varepsilon\bigr)\bigl(d(f(t),f(t_0)) \leqslant \varepsilon\bigr).$$

Note that equicontinuity at $t_0$ does not imply continuity of any $f\in \mathscr{F}$ at points $t\neq t_0$. The family $\mathscr{F}$ is equicontinuous if it is equicontinuous at $t$ for all $t\in T$.

If $(M_1,d_1)$ and $(M_2,d_2)$ are metric spaces, a family $\mathscr{F}$ of functions $M_1 \to M_2$ is uniformly equicontinuous if for every $\varepsilon > 0$ there is a $\delta > 0$ such that

$$\bigl(\forall f\in \mathscr{F}\bigr)\bigl(\forall x,y \in M_1\bigr)\bigl(d_1(x,y) \leqslant \delta \implies d_2(f(x),f(y)) \leqslant \varepsilon\bigr).$$

In particular, every member of $\mathscr{F}$ is then uniformly continuous. A uniformly equicontinuous family is equicontinuous, but in general not vice versa, e.g. if $f\colon M_1 \to M_2$ is continuous, then the singleton family $\{ f\}$ is equicontinuous, but it is uniformly equicontinuous (if and) only if $f$ is uniformly continuous. However, if $M_1$ is compact, then every equicontinuous family is uniformly equicontinuous: Let $\varepsilon > 0$ be given. By equicontinuity, for every $x\in M_1$ there is a $\rho(x) > 0$ such that $d_1(x,y) \leqslant \rho(x) \implies d_2(f(x),f(y)) \leqslant \varepsilon/2$ for all $f\in \mathscr{F}$. Set $\delta(x) = \rho(x)/2$. Since $M_1$ is compact, the open cover $\{ B_{\delta(x)}(x) : x \in M_1\}$ of $M_1$ has a finite subcover, i.e. there is an $n\in\mathbb{N}$ and points $x_0,\dotsc, x_n \in M_1$ with

$$M_1 = \bigcup_{\nu = 0}^n B_{\delta(x_\nu)}(x_\nu).\tag{\ast}$$

Set $\delta = \min \{ \delta(x_\nu) : 0 \leqslant \nu \leqslant n\}$. For $x,y$ with $d_1(x,y) \leqslant \delta$, by $(\ast)$, there is an $x_\nu$ with $x \in B_{\delta(x_\nu)}(x_\nu)$. Since $\delta \leqslant \delta(x_\nu)$, it follows that $d_1(y,x_\nu) \leqslant d_1(y,x) + d_1(x,x_\nu) < 2\delta(x_\nu) = \rho(x_\nu)$, and hence

$$d_2(f(x),f(y)) \leqslant d_2(f(x),f(x_\nu)) + d_2(f(x_\nu),f(y)) \leqslant \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

Now to your problem: Let $D = \{ x_n : n \in \mathbb{N}\}$ a countable dense subset of $K$. Assuming the codomain of the $f_n$ is $\mathbb{R}$, the point-wise boundedness implies - by the Bolzano-Weierstraß theorem - that there is a subsequence $(f_{0,m})_{m\in\mathbb{N}}$ of $(f_n)$ such that the sequence $\bigl(f_{0,m}(x_0)\bigr)_{m\in \mathbb{N}}$ is convergent - say to $y_0$. The sequence $\bigl(f_{0,m}(x_1)\bigr)_{m\in \mathbb{N}}$ is bounded, hence by Bolzano-Weierstraß, there is a subsequence $\bigl(f_{1,m}\bigr)_{m\in \mathbb{N}}$ such that $\bigl(f_{1,m}(x_1)\bigr)_{m\in \mathbb{N}}$ also converges - say to $y_1$. Since $\bigl(f_{1,m}\bigr)$ is a subsequence of $\bigl(f_{0,m}\bigr)$, we still have $f_{1,m}(x_0) \to y_0$. Continuing in this way, having found a subsequence $\bigl(f_{k,m}\bigr)$ such that $f_{k,m}(x_j) \to y_j$ for $0 \leqslant j \leqslant k$, we find a subsequence $\bigl(f_{k+1,m}\bigr)$ of $\bigl(f_{k,m}\bigr)$ such that $f_{k+1,m}(x_{k+1}) \to y_{k+1}$. Taking the diagonal sequence $(g_k)_{k\in \mathbb{N}}$, where $g_k = f_{k,k}$, we obtain a subsequence of $(f_n)$ such that $g_k(x_j) \to y_j$ for all $j\in \mathbb{N}$.

Then you need to show that the sequence $(g_k)$ is uniformly convergent. For that, we use the uniform equicontinuity of the sequence [which is implied by the equicontinuity of $(f_n)$, since $(g_k)$ is a subsequence of that]. Given $\varepsilon > 0$, choose $\delta > 0$ according to the definition of uniform equicontinuity. Show that there is an $m$ such that

$$K = \bigcup_{\mu = 0}^m B_\delta(x_\mu).$$

Use the convergence of $g_k(x_\mu)$ to deduce the existence of an $n_\varepsilon \in \mathbb{N}$ such that

$$\max \{ \lvert g_k(x) - g_n(x)\rvert : x \in K\} \leqslant 3\varepsilon$$

for all $k,n \geqslant n_\varepsilon$.

• +1 For your majestic effort +1 For such a great explanation. But unfortunately I can only give +1... – User Not Found Jun 16 '15 at 15:48