When are the operator norm and the Hilbert-Schmidt norm equal?

Question

Let $H$ be Hilbert space and let $\|\cdot\|$ and $\|\cdot\|_2$ denote the usual operator norm and the Hilbert-Schmidt norm respectively. I know that if $T \in \mathscr{B}(H)$ is a Hilbert-Schmidt operator, then $\|T\| \leq \|T\|_2$. Is there a particular type of operator for which $\|T\| = \|T\|_2$?

Answer 1

Equality occurs if and only if $T$ is a rank one operator. One proof is as follows.

If $T$ is bounded then $\|T\|$ is finite, so we can exclude the case that $\|T\|_2$ fails to be finite. Suppose then that $\|T\|_2$ is finite. Then $T$ is compact and there exist countable orthonormal sets $(u_j),(v_j)$ such that $$ T(x) = \sum_{j} \sigma_j \langle x,v_j \rangle u_j $$ where $\sigma_1 \geq \sigma_2 \geq \sigma_3 \geq \cdots$ are non-negative. It follows that $$ \|T\|^2 = \sigma_1^2, \quad \|T\|_2^2 = \sum_{j} \sigma_j^2. $$ These two can only be equal if $\sigma_j = 0$ for all $j \geq 2$, which occurs if and only if $T$ has rank $1$.