Let $(X,d)$ be a compact metric space.
I would like to prove that if $(f_n)_{n \in \mathbb{N}}$ is a sequence of continuos functions $f_n:X \to Y$ that converge uniformly in $X$, then $(f_n)_{n \in \mathbb{N}}$ is a uniformly equicontinuous family of functions.
I know that if $X$ is compact uniformly equicontinuous is equivalent to equicontinuous. But I don't know how to prove this statement either!
Any help?
I am assuming that $Y$ is a metric space, whose metric I'm also writing as $d$. I'll also use without proof the fact that a continuous function on a compact metric space is uniformly continuous.
Let $f$ denote the limit of the sequence $(f_n)_{n\in\mathbb{N}}$. Let $\epsilon > 0$ - we wish to find $\delta >0$ so that $d(x,y)<\delta\implies d(f_n (x),f_n (y))<\epsilon$ for all $x,y\in X$ and all $n$.
By uniform convergence of $(f_n)$ to $f$, pick $N$ large enough so that $n>N$ implies $\Vert f-f_n \Vert_\infty<\epsilon /3$.
Now, for each $n\le N$, pick $\delta_n>0$ so that $d(x,y)<\delta_n$ implies $d(f_n(x),f_n (y))<\epsilon$. Additionally, pick $\delta_f>0$ so that $d(x,y)<\delta_f$ implies $d(f(x),f (y))<\epsilon/3$ (possible by the uniform continuity of $f$ which is in part given by the Uniform Limit Theorem).
Finally, take $\delta = \min\{\delta _f , \delta_1,\dots ,\delta_N\}>0$ (possible because this is a finite set).
Now let $x,y\in X$ with $d(x,y)<\delta$. For $n\le N$, we are done, because $d(x,y)<\delta\le \delta_n$. Then for $n>N$, we use the triangle inequality: $$d(f_n(x),f_n(y))\le d(f_n(x),f(x))+d(f(x),f(y))+d(f(y),f_n(y))\\ <\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$ and we're done.
Let $(X,d_X)$ be a compact metric space and let $(Y, d_Y)$ be a metric space. Suppose that $\{f_n\}_{n=1}^\infty$ is a sequence of continuous functions on $X$ with values in $Y$, and suppose $f_n \to f$ uniformly on $X$. Let $\varepsilon>0$ be given. By the definition of uniform convergence, we know there exists $N \in \mathbb{N}$ so that \begin{equation} \sup_{x \in X} \,d_Y\left(\,f_n(x),f(x)\right)<\frac{\varepsilon}{6} \text{ whenever } n \geq N. \end{equation} Since a $(d_X, d_Y)$-continuous function will be uniformly continuous on the compact metric space $(X,d_X)$, we know there exists $\delta_N>0$ so that \begin{equation} d_Y\left(\,f_N(x),f_N(y)\right)<\frac{\varepsilon}{3} \text{ whenever } d_X(x,y)<\delta_N \text{ and } x,y \in X. \end{equation} Hence we have shown that \begin{align} \ \,d_Y\left(\,f(x),f(y)\right) & \leq d_Y\left(\,f(x),f_N(x)\right)+d_Y\left(\,f_N(x),f_N(y)\right)+d_Y\left(\,f_N(y),f(y)\right) \\& < \frac{\varepsilon}{6}+\frac{\varepsilon}{3}+\frac{\varepsilon}{6}=\frac{4\varepsilon}{6} \; \text{ whenever } d_X(x,y)<\delta_N \text{ and } x,y \in X. \end{align}
Notice we have now shown that if $x,y \in X$, $d_X(x,y)<\delta_N$, and $n \geq N$, then \begin{align} \ \,d_Y\left(\,f_n(x),f_n(y)\right) & \leq d_Y\left(\,f_n(x),f(x)\right)+d_Y\left(\,f(x),f(y)\right)+d_Y\left(\,f(y),f_n(y)\right) \\& < \varepsilon. \end{align} Since every function in the finite collection $\{ f_1, \ldots, f_{N-1}\}$ is uniformly continuous on $X$, we know that for $k=1,\ldots, N-1$ there exists a number $\delta_k>0$ so that \begin{equation} d_Y\left(\,f_k(x),f_k(y)\right)<\varepsilon \; \text{ whenever } d_X(x,y)<\delta_k \text{ and } x,y \in X. \end{equation} Setting $\delta = \min \{\delta_1, \ldots, \delta_{N-1}, \delta_N \}$ shows that the collection of functions $\{f_n: n \in \mathbb{N}\}$ satisfies the definition of a uniformly equicontinuous family on $X$.
