Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 22G (Linear Analysis)

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.

For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.