We know that the class of Hilbert-Schmidt operators $\mathcal{L}_2 \mathcal{(H)}$ on a Hilbert Space $\mathcal{H}$ forms a Hilbert Space with the inner product $\langle u,v\rangle =\operatorname{tr}(v^*u)$, where the trace function is defined for Trace-Class Operators by
$$\operatorname{tr}(u)=\sum_{x\in E}\langle ux,x \rangle$$
where $E$ is an orthonormal basis of $\mathcal{H}$. The norm for this topology is the Hilbert-Schmidt norm $\Vert u\Vert_2=\sum_{x\in E} \Vert u(x)\Vert ^2$. We also know that the operator norm $\Vert \cdot \Vert \leq \Vert \cdot \Vert_2$, which implies the Hilbert-Schmidt norm topology is weaker than the operator norm topology.
My question is whether the Hilbert-Schmidt norm topology is strictly weaker than the operator norm topology or whether they are equivalent on $\mathcal{L}_2 \mathcal{(H)}$?
Best Answer
In the separable case (ie $E$ countable), it’s not. Consider an explicit bijection $E \rightarrow \mathbb{N}$, and let $V_n$ be the vector subspace generated by the first $n$ vectors of $E$. Let $T_n$ be $n^{-1/2}$ times the orthogonal projection on $V_n$. Then $\|T_n\|_{HS}=1$ but $\|T_n\|_{op}=n^{-1/2}$.