Past exam of the options II course of the University of Oxford / 2010 / E / E4 / c

Yes. A Cauchy sequence in the supremum norm converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-norm convergence is uniform convergence. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $ϵ$, choose a continuous $f_n$ within $ϵ/3$ of $f$ uniformly, then use its continuity at $x_0$ and the triangle inequality. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.

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.