Proof. sketch [br724A]

搬砖题

$\{ f_{n} \}$ converges uniformly on $K$ $\implies$ for all $\epsilon > 0$, there exists an integer $N > 0$ such that for any $n, m > N$ we have $\lVert f_{n} - f_{m} \rVert < \epsilon$.

Pick any $k > N$. $f_{k} \in \mathscr{C}(K) \implies$ $f_{k}$ continuous on $K$. Now $K$ is compact $\implies$ $f_{k}$ is uniformly continuous on $K$. So there exists $\delta > 0$ such that for any $x, y \in K$ with $d(x, y) < \delta$, we have $\lvert f_{k}(x) - f_{k}(y) \rvert < \epsilon$. Then for any $n > N$, $$ \lvert f_{n}(x) - f_{n}(y) \rvert = \lvert (f_{n}(x) - f_{k}(x)) + (f_{k}(x) - f_{k}(y)) + (f_{k}(y) - f_{n}(y)) \rvert < 3\epsilon. $$ For $1 \leq n \leq N$, $f_{n}$ is uniformly continuous, just take $\delta' = \min_{1 \leq i \leq N}\delta_{i} > 0$, and our final choice of $\delta^*$ is $\min \{ \delta, \delta' \}$, which is positive.