Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 19G (Linear Analysis)

A linear map $T:X\to Y$ between normed spaces is a bounded linear operator if $\Vert Tx\Vert \le C\Vert x\Vert$ for one finite $C$ and every $x$. This implies $\Vert Tx-Ty\Vert \le C\Vert x-y\Vert$, hence continuity. Conversely, continuity at zero supplies $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$ for nonzero $x$. Thus boundedness and continuity are equivalent for linear maps.

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$, equivalently ${\sup }_{x\ne 0}\Vert Tx\Vert /\Vert x\Vert$. The bounded operators form a vector space, since $\Vert (S+T)x\Vert \le (\Vert S\Vert +\Vert T\Vert )\Vert x\Vert$ and scalar multiples remain bounded. Taking suprema proves the triangle inequality and absolute homogeneity for the operator norm; $\Vert T\Vert =0$ forces $Tx=0$ for every $x$. Thus $B(X,Y)$ is a normed space, whether or not $X,Y$ are complete.

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 $h=0$.