If $K \subset \mathbb{R}^{n}$ is compact, then every equicontinuous set $\chi \subset C(K;\mathbb{R}^{m})$ is uniformly equicontinuous

Question

Prove that if $K \subset \mathbb{R}^{n}$ is compact, then every equicontinuous set $X \subset C(K;\mathbb{R}^{m})$ is uniformly equicontinuous.

I don't know any criteria to show that a given set is uniformely equicontinuous so, I tried to show this only by definition, but I could not. Would anyone have any suggestions?

Is it necessary to write down the definition of equicontinuous and uniformly equicontinuous set?

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Edit. Using Daniel Fischer hint:

Proof. By hypothesis, for each $x \in K$, given $\epsilon > 0$, there is $\delta_{x} > 0$ such that if $x,y_{x} \in K$ $$\Vert x - y_{x} \Vert < \delta_{x} \Longrightarrow \Vert f(x) - f(y_{x}) \Vert < \epsilon,\quad \forall f \in X.$$ Note that $\bigcup_{x \in K} B_{\delta_{x}}(x)$ is an open cover of $K$ and this, has a Lebesgue Number, that is, there is $\delta > 0$ such that for every $x \in K$, exist an open $U$ of the cover such that $B_{\delta}(x) \subset U$. Thus, for every $x,y \in K$, if $\Vert x - y \Vert < \delta$ then $x,y \in U$ for some $U$ of the cover. Therefore $$\Vert f(x) - f(y) \Vert \leq \Vert f(x) - f(y_{x}) \Vert + \Vert f(y_{x}) - f(y) \Vert < 2\epsilon,\quad \forall f \in X.$$ Then, $X$ is uniformly equicontinuous.

Is correct?

Answer 1

Consider the set $C(X,\mathbb{R}^m)$ of all continuous functions from $X$ to $\mathbb{R}^m$. On this set, consider the distance:$$d(f,g)=\begin{cases}\sup\left\{d\bigl(f(x),g(x)\bigr)\,\middle|\,x\in X\right\}&\text{ if }(\forall x\in X):d\bigl(f(x),g(x)\bigr)\leqslant1\\1&\text{ otherwise.}\end{cases}$$Now, consider the map$$\begin{array}{rccc}h\colon&K&\longrightarrow&C(X,\mathbb{R}^m)\\&k&\mapsto&\left(\begin{array}{ccc}X&\longrightarrow&\mathbb{R}^m\\f&\mapsto&f(k)\end{array}\right).\end{array}$$It is not hard to prove that $h$ is continuous if and only if $X$ is equicontinuous and that $h$ is uniformly continuous if and only if $X$ is uniformly equicontinuous. Now, use the compacity of $K$ and the fact that a continuous function from a compact metric space into a matric space is always uniformly continuous.

Answer 2

With pointwise equicontinuity, for any $\epsilon > 0$ and for each $x \in K$, there exists $\delta(\epsilon,x) > 0$ such that $|f(y) - f(x)| < \epsilon/2$ for every $f$ when $|y-x| < \delta(\epsilon,x)$.

If $K$ is compact then there is a finite set $\{x_1 , \ldots, x_n\}$ such that

$$K \subset \bigcup_{j=1}^n B(x_j,\delta(\epsilon,x_j)/2)$$

Take $\delta(\epsilon) = \frac{1}{2} \min(\delta(\epsilon,x_1) \ldots, \delta(\epsilon,x_n))$. For any $x,y \in K$, we have $|x - x_j| < \frac{1}{2} \delta(\epsilon,x_j) < \delta(\epsilon,x_j)$ for some $j \in \{1,\ldots,n\}$. If $|y - x| < \delta(\epsilon)$, we also have $|y - x_j| \leqslant |y-x| + |x - x_j| < \delta(\epsilon,x_j)$

Now uniform equicontinuity follows , since $|x-y| < \delta(\epsilon)$ implies that for every $f$,

$$|f(y) - f(x)| \leqslant |f(y) - f(x_j)| + |f(x) - f(x_j)| < \epsilon$$