Uniform convergence and sup norm

Let $X$ be a set and let $f_n,f:X\to\mathbb{R}$ (or $\mathbb{C}$). Define the sup norm (when finite) by $\|h\|_\infty=\sup_{x\in X} |h(x)|.$

Uniform convergence and sup norm: The sequence $f_n$ converges uniformly to $f$ on $X$ if and only if $\|f_n-f\|_\infty \to 0.$ Moreover, whenever the sup norms are finite, $\bigl|\|f_n\|_\infty-\|f\|_\infty\bigr|\le \|f_n-f\|_\infty.$

This inequality shows the sup norm is continuous with respect to uniform convergence and is used constantly in approximation arguments on compact sets .

Proof sketch: The equivalence is immediate from the definitions. For the inequality, note that for every $x$, $|f_n(x)|\le |f(x)|+|f_n(x)-f(x)|,$ take suprema to get $\|f_n\|_\infty\le \|f\|_\infty+\|f_n-f\|_\infty$, and reverse the roles of $f_n$ and $f$.