Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 19G / b / iv / Solution

The integration functional satisfies $|Tf|\le \Vert f{\Vert }_{\infty }{\int }_0^1|h|$. If $h$ never vanishes, its sign is constant and $f=\mathrm{sgn}h$ attains equality. In general use the continuous functions $f_{\epsilon }=h/\sqrt{h^2+\epsilon }$, of supremum norm at most $1$. Thenby the dominated convergence theorem. Therefore $\Vert T\Vert ={\int }_0^1|h|$. Every bounded subset of its finite-dimensional target $\mathbb{R}$ is relatively compact, so $T$ is compact, including when $h=0$.