Suppose that $A$, $B$ are deterministic $n\times n$ matrices and $G$ a Gaussian matrix of i.i.d. entries $N(0,1)$.


I'd like to establish an upper bound of the trace norm of $AGB$ as
$$
\mathbb{E}\|AGB\|\_\ast \leq \max\{\|A\|\_\ast \|B\|\_F, \|A\|\_F\|B\|\_\ast \},
$$
where $\|\cdot\|\_\ast$ denotes the trace norm and $\|\cdot\|\_F$ the Frobenius norm. I can show the inequality with an additional $\log n$ factor on the right-hand side but I do not think the $\log n$ factor is necessary...


(**First question**: is it true to replace $\max$ with $\min$?)


[Update] As answered by Mikael de la Salle below, it is possible to replace $\max$ with $\min$ for $p\in [1,2]$, to obtain that
$$
\mathbb{E}\|AGB\|\_p \leq \min\{\|A\|\_p \|B\|\_F, \|A\|\_F \|B\|\_p\}.
$$


For $p>2$, it is necessary to have $\max$. Take $A$ to be identity matrix and $B = e\_1e\_1^T$ (zero matrix except the top-left entry being $1$), then $\|AGB\|\_p \sim \sqrt{n}$ while $\|A\|\_p \|B\|\_F = n^{1/p} < \sqrt{n}$.


**Second question**: Does the following hold for $p>2$?
$$
\mathbb{E}\|AGB\|\_p \leq C\max\{\|A\|\_p \|B\|\_F, \|A\|\_F \|B\|\_p\}.
$$
for some constant $C$ (which can depend on $p$)? I could get the inequality with an extra $\log n$ factor on the right-hand side, but I am not sure if it is necessary. For $p=\infty$, it is too easy to get $\sqrt{n}\|A\|\_{op}\|B\|\_{op}$, which is bigger than the bound I want.


**Third question**: How about the lower bound on $\mathbb{E}\|AGB\|\_p$? Do we have
$$
\mathbb{E}\|AGB\|\_{p} \geq c\max\{\|A\|\_p \|B\|\_F, \|A\|\_F \|B\|\_p\}
$$
for some constant $c$?


I can show that
$$
\mathbb{E}\|AGB\|\_{op} \geq \max\{\|A\|\_{op} \|B\|\_F, \|A\|\_F \|B\|\_{op}\}
$$
So I think the lower bound inequality holds for $p\geq 2$. For $p = 1$, if we take $A$ to be identity matrix and $B=e\_1e\_1^T$ then $\mathbb{E}\|AGB\|\_{p}\approx \sqrt{n}$, so the inequality cannot hold with $\max$. So the question is, will it hold
$$
\mathbb{E}\|AGB\|\_{p} \geq c\min\{\|A\|\_p \|B\|\_F, \|A\|\_F \|B\|\_p\}
$$
for $p\in [1,2)$? Again I can show it with an extra $1/\log n$ factor on the right but I am not sure if it is necessary.



trace norm of AGB, where G is Gaussian random matrix
[Edit: Now I answer all questions.]


The answer to the first question is yes, the answer to the second question is no, and the answer to the third question is *if and only if $p \geq 2$* (only a guess in the case $p<2$).


* First question


The inequality
$$ \mathbb E \|A G B\|\_\* \leq \min( \|A\|\_\* \|B\|\_F,\|A\|\_F \|B\|\_\*)$$
always holds. By symmetry, it is enough to prove the inequality $\mathbb E \|A G B\|\_\* \leq \|A\|\_\* \|B\|\_F$. And by writing $A$ as a sum of rank one operators ("the unit ball of the trace class is the convex hull of the norm $1$ rank $1$ matrices"), we can assume that $A$ has rank one.


In that case, $A G B$ has rank one for every $G$. Using (1) that for a rank $1$ matrix, the trace norm and Frobenius norm coincide, and (2) that the $L^1$ norm is less than the $L^2$ norm, we get
$$ \mathbb E \|A G B\|\_\* = \mathbb E \|A G B\|\_F \leq (\mathbb E \|A G B\|\_F^{2})^{\frac 1 2} = \|A\|\_F \|B\|\_F=\|A\|\_\* \|B\|\_F.$$
The second equality is a straighforward computation, at least when $A$ and $B$ are diagonal.


* Second question, and third question when $p \geq 2$.


It follows from the non-commutative Khintchine inequalities of Françoise Lust-Picard that, for every $2 \leq p < \infty$,
$$ \mathbb E \|A G B\|\_p \simeq \max( \|A\|\_p \|B\|\_2,\|A\|\_2 \|B\|\_p)\ if\ p \geq 2$$
(up to constants depending on $p$, whose growth rate is known as $p \to \infty$).


Indeed, the non-commutative Khintchine inequalities states that if $(C\_k)$ is a family of matrices and $g\_k$ are iid $N(0,1)$ random variables,
$$ \mathbb E \| \sum\_k g\_k C\_k\|\_p \simeq \max ( \|(\sum C\_k C\_k^\*)^{\frac 1 2}\|\_p, \|(\sum C\_k^\* C\_k)^{\frac 1 2}\|\_p).$$
(the inequality is sometimes stated in terms of Bernoulli random variables instead of gaussians, and with $L^p$-norm, of $L^2$ norm instead of $L^1$-norm on the left-hand side, but all versions are equivalent by Kahane's inequalities and standard probabilistic arguments). Applying this to $C\_{i,j}$, the product of the $i$-th column of $A$ and and $j$-th row of $B$, one gets the answer because $\sum\_{i,j} C\_{i,j} C\_{i,j}^\* = \|B\|\_2^2 A A^\*$ and $\sum\_{i,j} C\_{i,j}^\* C\_{i,j}=\|A\|\_2^2 B^\*B$.


* Third question if $p \leq 2$.


Here also, the non-commutative Khintchine inequalities (due to Lust-Picard and Pisier when $p=1$) allow, in principle, to give an explicit equivalent of $\mathbb E\|A G B\|\_p$. Indeed, when $p \leq 2$, the inequality reads
$$ \mathbb E \| \sum\_k g\_k C\_k\|\_p \simeq \inf \|(\sum D\_k D\_k^\*)^{\frac 1 2}\|\_p+ \|(\sum E\_k^\* E\_k)^{\frac 1 2}\|\_p),$$
(up to universal constants)
where the infimum is over all families $(D\_k,E\_k)$ such that $C\_k=D\_k+E\_k$ for all $k$.


This gives a very involved proof of the inequality of the first question. I believe that it can be deduced from this inequality that the inequality 
$$ \mathbb E \|A G B\|\_p \geq \min( \|A\|\_p \|B\|\_2,\|A\|\_2 \|B\|\_p)$$
does not hold if $p<2$, but this deserves to be checked.


