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 $\Vert f{\Vert }_{\alpha }={\sup }_{x\in K}|\alpha (x)f(x)|$. 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 $0<m=\min |\alpha |\le \max |\alpha |=M<\infty$, and $m\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_{\alpha }\le M\Vert f{\Vert }_{\infty }$. A weighted Cauchy sequence is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.