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

Take $f_n(t)=\sin (2\pi nt)$. Its domain norm is $1$, whereas its image norm is $1+2\pi n$. Hence $T$ is unbounded, so it has no finite operator norm and is not a compact operator.

For the inclusion, $\Vert Tf{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, with equality for a nonzero constant. Thus $\Vert T\Vert =1$. The domain unit ball is uniformly bounded and has $\Vert f^{\prime }{\Vert }_{\infty }\le 1$, so it is equicontinuous. The Arzela-Ascoli theorem makes its image relatively compact in $C[0,1]$. Therefore $T$ is compact.

Differentiation satisfies $\Vert Tf{\Vert }_{\infty }\le \Vert f\Vert$, so its operator norm is at most $1$. With $f_n(t)=\sin (2\pi nt)/(2\pi n)$, the ratio $\Vert T{f}_n{\Vert }_{\infty }/\Vert f_n\Vert$ tends to $1$. Hence $\Vert T\Vert =1$.

Normalize $g_n=f_n/(1+(2\pi n{)}^{-1})$, so $\Vert g_n\Vert =1$. Their derivatives are $\cos (2\pi nt)/(1+(2\pi n{)}^{-1})$. Distinct cosine functions are orthogonal in $L^2(0,1)$, and these normalizing factors stay bounded away from zero. Consequently the derivative images have pairwise $L^2$ distances bounded away from zero. Since uniform convergence would imply $L^2$ convergence, no subsequence converges uniformly. Thus $T$ is not compact.

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 $h=0$.