Uniformly continuous implies: the sequence is bounded in some way

Question

If $(f_n)$ is a sequence of bounded Borel-measurable functions $f_n:K\to\mathbb{K}$ ( $K\subseteq \mathbb{R}$ compact) such that $\|f_n-f\|_{\infty}\to 0$ (uniform convergence). This implies: $\exists C>0$ such that $|f_n(x)|<C$ $\forall n\in \mathbb{N}, x\in K$. My question is, is this the same as $\sup_{n\in\mathbb{N}}\|f_n\|_{\infty}<\infty$? Does uniform convergence imply: $\sup_{n\in\mathbb{N}}\|f_n\|_{\infty}<\infty$?

Answer 1

Yes. Choose $N$ so that $n \ge N$ implies $\|f_n - f\|_\infty \le 1$. For all such $n$, $\|f_n\|_\infty \le 1 + \|f\|_\infty$. Then $$\sup_{n \in \mathbb N} \|f_n\|_\infty \le \|f_1\|_\infty + \cdots + \|f_N\|_\infty + \|f\|_\infty + 1.$$