Equicontinuity on a compact metric space turns pointwise to uniform convergence

Solution 1:

Let $\varepsilon>0$, we shall show that there exists an $n_0=n_0(\varepsilon)$, such that, $$ n\ge n_0\quad\Longrightarrow\quad \lvert\, f_n(x)-f(x)\rvert<\varepsilon, $$ for all $x\in K$.

1. As $\{f_n\}$ is equicontinuous, there exists a $\delta>0$, such that for all $x,y\in K$: $$ d(x,y)<\delta\quad\Longrightarrow\quad \lvert\, f_n(x)-f_n(y)\rvert<\frac{\varepsilon}{3}, \tag{1} $$ for all $n\in\mathbb N$. If we let $n\to\infty$, then $(1)$ implies that $$ d(x,y)<\delta\quad\Longrightarrow\quad \lvert\, f(x)-f(y)\rvert\le \frac{\varepsilon}{3}, $$

2. Since $K$ is compact, it can be covered by finitely many balls of radius $\delta$, i.e., there exist $k\in\mathbb N$ and $z_1,\ldots,z_k\in K$, such that $$ K\subset B(z_1,\delta)\cup\cdots\cup B(z_k,\delta). $$

3. As $f_n(z_j)\to f(z_j)$, for $j=1,\ldots,k$, we can find $n_0$, such that $$ n\ge n_0\quad\Longrightarrow\quad \lvert\, f_n(z_j)-f(z_j)\rvert<\frac\varepsilon 3, $$ for all $j=1,\ldots,k$.

4. Finally, if $x\in K$ and $n\ge n_0$, then there exists a $j\in\{1,\ldots,k\}$, for which $x\in B(z_j,\delta)$, and hence $$ \lvert\, f_n(x)-f(x)\rvert\le \lvert\, f_n(x)-f_n(z_j)\rvert +\lvert\, f_n(z_j)-f(z_j)\rvert +\lvert\, f(z_j)-f(x)\rvert< \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}={\varepsilon}. $$ Ὅπερ ἔδει δεῖξαι.