= Solution
\b[Yes.] <Continuity> and nonvanishing on the compact <interval> give constants
$$
0<m=\min_{[0,1]}|\alpha|\leq\max_{[0,1]}|\alpha|=M<\infty.
$$
Consequently
$$
\boxed{m\|f-g\|_\infty\leq d_\alpha(f,g)\leq M\|f-g\|_\infty.}
$$
A $d_\alpha$-<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>.
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