Past exam of the functional analysis II course of the University of Oxford / 2023 / 3 / b

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

As in the preceding part, take $T$ to be a bounded linear operator. For $x\in X$, the Hilbertian basis expansion and continuity of $T$ giveThe triangle inequality, the Cauchy-Schwarz inequality for finite sums, and the Parseval identity for a Hilbertian basis givePassing to the limit and taking the supremum over the unit ball provesThus the Hilbert-Schmidt norm controls the operator norm. Without the boundedness interpretation, the nonzero $R$ constructed in the preceding part has zero printed column sum and makes the requested inequality false.

Use the intended space of Hilbert-Schmidt operators, with the given Hilbert-Schmidt inner product linear in its first argument. It remains to prove completeness in its Hilbert-Schmidt norm. Let $(T_n)$ be a Cauchy sequence in that norm. The bound $\Vert T_n-T_m\Vert \le \Vert T_n-T_m{\Vert }_2$ makes it a Cauchy sequence in operator norm. The space $\mathcal{B}(X)$ is a Banach space, so $T_n\to T$ in operator norm for some bounded linear operator $T$.

For completeness, that last standard fact follows by taking $Tx={\lim }_n{T}_{n}x$ for each $x$: the Cauchy sequence $(T_n)$ is uniformly bounded in operator norm, so its limit is linear and bounded, and the uniform Cauchy estimate on the unit ball gives operator norm convergence.

Given $\epsilon >0$, choose $N$ so that $\Vert T_n-T_m{\Vert }_2<\epsilon$ for $n,m\ge N$. Fix $n\ge N$ and any positive integer $K$. Since $T_m{e}_k\to T{e}_k$ for each $k$, passage to the limit in a finite sum givesTaking the supremum over $K$ gives ${\sum }_k\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$. Also,Thus $T\in S_2(X)$, and $\Vert T_n-T{\Vert }_2\le \epsilon$ for $n\ge N$. This proves Hilbert-Schmidt operators form a Hilbert space:Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $⟨R,R{⟩}_{S_2(X)}=0$. The form would fail positive definiteness and would not be an inner product; the boundedness correction is needed here too.