If the domain is a compact set, pointwise equicontinuity implies uniform equicontinuity.

Question

Conversely, is it true that if every sequence of pointwise equicontinuous functions from $M$ to $\mathbb{R}$ is uniformly equicontinuous, them $M$ is compact?

Answer 1

Such a space need not be compact. Take any metric space with the discrete metric, $d(x,y) = 1$ for $x\neq y$. Then every family of functions $M \to \mathbb{R}$ is uniformly equicontinuous. But if the space contains infinitely many points, it is not compact.

A connected metric space on which every pointwise equicontinuous sequence of functions is uniformly equicontinuous, however, must be compact.

More, a connected metric space on which every continuous function is uniformly continuous must be compact.

First, if $(X,d)$ is an incomplete metric space, let $(\tilde{X},\tilde{d})$ be its completion. For every $p \in \tilde{X}\setminus X$, the function

$$f\colon X\to \mathbb{R}; \quad f\colon x \mapsto \frac{1}{\tilde{d}(x,p)}$$

is continuous but not uniformly continuous. It is continuous because $x\mapsto \tilde{d}(x,p)$ is continuous and nonzero on $X$, and for a sequence $(x_n)$ in $X$ converging to $p$, the sequence $\bigl(f(x_n)\bigr)$ is not a Cauchy sequence, hence $f$ is not uniformly continuous.

If a metric space $(X,d)$ is not totally bounded, it contains a sequence $(x_n)$ of points with $d(x_k,x_n) \geqslant 4\varepsilon$ for $k\neq n$ and some $\varepsilon > 0$. For $n\in\mathbb{Z}^+$, let

$$\varphi_n(x) = \max \left\{0, 1- \frac{d(x_n,x)}{\varepsilon} \right\}.$$

Then $\varphi_n$ is continuous, and since $B_\varepsilon(x)$ intersects the support of at most one $\varphi_n$, the function

$$f(x) = \sum_{n=1}^\infty n\cdot \varphi_n(x)$$

is well-defined and continuous. If $X$ is connected, $f$ is not uniformly continuous: then for every $n$ there is an $y_n \in X$ with $d(x_n,y_n) = \frac{\varepsilon}{2n}$ and hence $\lvert f(y_n) - f(x_n)\rvert = \frac{1}{2}$, but $\lim\limits_{n\to\infty} d(x_n,y_n) = 0$.

The connectedness is a stronger assumption than is needed, it suffices that for infinitely many of the $x_n$ there is a point $y_n$ with $\frac{c}{n} \leqslant d(x_n,y_n) \leqslant \delta_n$ where $c$ is some positive constant, and $\delta_n \to 0$. But an exact criterion is not obvious.

Answer 2

This is not necessary. To see why this is the case, let us construct a counterexample as follows: Consider a family of constant functions $\{f_n|n\in \mathbb{N}\}$. Here, $f_n(x)=n$ for all $x\in M$. This family is uniformly equicontinuous. However, the sequence $(f_n)_n$ doesn't have a convergent sub-sequence in every (usual) norm or metric.