Are Hilbert-Schmidt operators on separable Hilbert spaces "Hilbert Schmidt" on the space of Hilbert Schmidt Operators? 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!
 A: 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.
A: No it's not.
If $s:H\to H$ is a rank one projection, then
$s\otimes\mathrm{id}:H\otimes \overline H\to H\otimes \overline H$
is a projection of rank $\dim(H)$.
In particular, if $H$ is infinite dimensional, then
$s\otimes\mathrm{id}:H\otimes \overline H\to H\otimes \overline H$
is a projection of infinite rank, and thus not Hilbert-Schmidt.
Here, I've identified the space of Hilbert-Schmidt operators with the Hilbert space tensor product $H\otimes \overline H$.
