= Solution
The integration functional satisfies $|Tf|\leq\|f\|_\infty\int_0^1|h|$. If $h$ never vanishes, its sign is constant and $f=\operatorname{sgn}h$ attains equality. In general use the <continuous functions> $f_\varepsilon=h/\sqrt{h^2+\varepsilon}$, of supremum <norm> at most $1$. Then
$$
Tf_\varepsilon=\int_0^1\frac{h^2}{\sqrt{h^2+\varepsilon}}\,dt\longrightarrow\int_0^1|h(t)|\,dt
$$
by the <dominated convergence theorem>. Therefore \b[$\|T\|=\int_0^1|h|$]. Every bounded subset of its finite-dimensional target $\mathbb R$ is relatively compact, so \b[$T$ is compact], including when $h=0$.
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