Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 10A / Solution

Assume $a<b$. On continuous functions on this compact interval, both proposed distances are finite and symmetric, and vanish on identical functions. For the supremum norm, zero distance plainly implies pointwise equality. For the L2 norm, if a continuous difference is nonzero at any point, it stays bounded away from zero on a subinterval of positive length. Its squared integral is then positive, so zero distance also implies equality.

For the first metric, the pointwise triangle inequality followed by the supremum gives $d_1(x,z)\le d_1(x,y)+d_1(y,z)$. For the second, put $u=x-y$, $v=y-z$. The Cauchy-Schwarz inequality givesTaking square roots proves the required triangle inequality. Thus both are metrics.

The continuous-function space is complete in the uniform metric. If $(x_n)$ is a Cauchy sequence in $d_1$, at each $t$ its scalar values have a limit $x(t)$, by completeness of the scalar field. Given $\epsilon >0$, choose $N$ with $\Vert x_n-x_m{\Vert }_{\infty }<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ gives $|x_n(t)-x(t)|\le \epsilon$ uniformly in $t$. Hence there is uniform convergence. To see that the limit is continuous at $t_0$, approximate it uniformly by one continuous $x_n$ and useThe outer terms can be made small uniformly, and the middle term is small near $t_0$. This proves the needed uniform limit theorem and completeness.

For the second metric, let $H$ be zero on $(-1,0)$ and one on $(0,1)$; its value at zero is immaterial to the integral. The supplied ramp functions are continuous and satisfyAlthough $H$ is not in the continuous-function space, the integral triangle inequality yields $d_2(x_m,x_n)\le (3m{)}^{-1/2}+(3n{)}^{-1/2}\to 0$. Thus the sequence is Cauchy. If it converged in $d_2$ to a continuous $x$, the same inequality would give $\Vert x-H{\Vert }_2=0$. Continuity would then force $x=0$ on the negative half-interval and $x=1$ on the positive half-interval, contradicting continuity at zero. Therefore the continuous-function space is not complete in $d_2$; its limit in the larger L2 space is the discontinuous step.