= Completeness of a continuous weighted uniform norm
Let $K$ be a nonempty <compact space> and let $\alpha:K\to\mathbb R$ be continuous and nowhere zero. On $C(K)$, define $\|f\|_\alpha=\sup_{x\in K}|\alpha(x)f(x)|$. Multiplication by $\alpha$ is an isometry onto $C(K)$ with its ordinary <uniform norm>, because division by the continuous nonvanishing $\alpha$ provides the inverse. Thus the weighted space is a <Banach space>. Equivalently, the <extreme value theorem> gives $0<m=\min|\alpha|\leq\max|\alpha|=M<\infty$, and $m\|f\|_\infty\leq\|f\|_\alpha\leq M\|f\|_\infty$. A weighted <Cauchy sequence> is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.
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