Let's consider a separable Hilbert space $(\mathcal H, \langle\cdot, \cdot\rangle\_{\mathcal H})$ with Norm $||\cdot||\_{\mathcal H} := \langle\cdot, \cdot\rangle^{1/2}\_{\mathcal H},$ orthonomal basis $(e\_j)$ of $\mathcal H$ and let $s\colon \mathcal H \rightarrow \mathcal H$ be a Hilbert Schmidt operator, denoted by $s \in \mathcal{S\_H}.$ It's known that $(\mathcal{S\_H}, \langle\cdot, \cdot\rangle\_{\mathcal{S\_H}})$ is a (separable ??) Hilbert space and for $s\_1, s\_2 \in \mathcal{S\_H}\colon$ $\langle s\_1, s\_2\rangle\_{\mathcal{S\_H}}=\sum\_{j=1}^\infty\langle s\_1(e\_j), s\_2(e\_j)\rangle\_{\mathcal H},$ such that $||s\_1||^2\_{\mathcal{S\_H}} = \sum\_{j=1}^\infty||s\_1(e\_j)||^2\_{\mathcal H}.$ Now, assuming that $\mathcal{ S\_H}$ indeed is separable, let $(\phi\_j)$ be an orthonormal basis of $\mathcal S\_{\mathcal H}.$ It easily can be shown, since $||\cdot||\_{\mathcal{S\_H}}$ is sub-multiplicative, that for $s \in \mathcal{S\_H}\colon s\colon \mathcal{S\_H} \rightarrow \mathcal{S\_H}$ is a bounded operator as well, denoted by $s \in \mathcal{L\_{S\_H}}.$


Is $s\colon\mathcal{S\_H} \rightarrow \mathcal{S\_H}$ "Hilbert-Schmidt" as well ($s \in \mathcal{S\_{S\_H}}$)? 


Thank you in advance!



Are Hilbert-Schmidt operators on separable Hilbert spaces "Hilbert Schmidt" on the space of Hilbert Schmidt Operators?
The answer is No, assuming of course that for you $s$ acts on $\mathcal{S}\_\mathcal{H}$ by left-multiplication. There is no need to assume that $\mathcal{S}\_\mathcal{H}$ is separable, since it actually is, with an explicit countable orthonormal basis given by the elementary operators $e\_{ij}$ that act as $e\_{ij}(e\_k) = e\_i \delta\_{jk}$ on the orthonormal basis of $\mathcal{H}$.


The Hilbert-Schmidt norm of $s$ on $\mathcal{S}\_\mathcal{H}$ is $\operatorname{tr}(s' s)$, where $s'$ is the adjoint of $s$ on $\mathcal{S}\_\mathcal{H}$. It is not hard to tell that $s' = s^\*$, that is, left-multiplication by $s^\*$, where $s^\*$ is the adjoint of $s$ on $\mathcal{H}$. Let $r = s^\* s$ on $\mathcal{H}$ and denote by $r$ also the left-multiplication by $r$ on $\mathcal{S}\_\mathcal{H}$. Its trace is $\operatorname{tr}(r) = \sum\_{ij} (e\_{ij}, r e\_{ij})\_{\mathcal{S}\_\mathcal{H}}$. A quick calculation gives $(e\_{ij}, r e\_{ij})\_{\mathcal{S}\_\mathcal{H}} = (e\_i, r e\_i)\_\mathcal{H}$. So, each of the diagonal elements $(e\_i, r e\_i)\_\mathcal{H}$ occurs infinitely many times in the summation for $\operatorname{tr}(r)$, which hence cannot be finite unless all diagonal elements of $r$ vanish. But we know that that's impossible from $r = s^\* s$.


This answer is essentially the same as the one by André Henriques, which appeared as I was typing, but goes into more explicit detail.


