 Solution

Yes. Continuity and nonvanishing on the compact interval give constantsConsequentlyA $d_{\alpha }$-Cauchy sequence is therefore supremum-norm Cauchy and has a continuous uniform limit $f$ by part (i). The upper bound then makes it converge to $f$ also in $d_{\alpha }$. Equivalently, multiplication by $\alpha$ is an isometry from this weighted space onto ordinary $C([0,1])$, since division by the continuous nonvanishing $\alpha$ gives its inverse.

This is completeness of a continuous weighted uniform norm.