Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / E / E4 / c

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

Continuity and nonvanishing on the compact interval implyHence $m\Vert f-g{\Vert }_{\infty }\le d_{\alpha }(f,g)\le M\Vert f-g{\Vert }_{\infty }$. A $d_{\alpha }$-Cauchy sequence is therefore uniformly Cauchy, so (i) gives a continuous uniform limit; the upper bound then gives convergence in $d_{\alpha }$. Thus $\boxed{(C([0,1]),d_{\alpha })\text{ is complete}}$. The positive lower bound is what makes the argument work.