Holder inequality for matrices I am interested in the following version of the Holder inequality. Let $D \in M_n(\mathbb{C})$ be a positive semi-definite matrix of trace $1$ and $A, B \in M_n(\mathbb{C}).$ Does it follow that 
$$ |\mbox{Tr } DAB| \leq (\mbox{Tr } D|A|^p)^{1/p} (\mbox{Tr } D|B|^q)^{1/q}, $$
where $1/p + 1/q = 1$?
I found the case $D = I$ with proof (for the tracial state) in the literature but nothing more. 
 A: Recall that $|U|=(U^*U)^{1/2}$. If $D=I$, then, in general, the proposed inequality does not work; the correct inequality is $|tr(A^*B)|\leq (tr(|A|^p)^{1/p}(tr(|B|^q)^{1/q}$.
Of course, we consider $tr(DA^*B)$ where $D$ is symmetric $\geq 0$ (the condition $tr(D)=1$ is useless). Using a reasoning by continuity, we may assume that $D>0$ and , moreover, that $D$ is diagonal $>0$. One has $|tr(DA^*B)|=|tr(D^{1/p}A^*BD^{1/q})|=|<AD^{1/p},BD^{1/q}>|\leq (tr(|AD^{1/p}|)^p)^{1/p}(tr(|BD^{1/q}|)^q)^{1/q}$.
The question: is $tr((|AD^{1/p}|)^p)=tr(|A|^pD)$ true ?
If $p=2$, then the answer is yes; indeed the LHS is $tr(D^{1/2}A^*AD^{1/2})$ and the RHS is $tr(A^*AD)$. If $p\not= 2$, then the equality does not work (at least , I think so).
A: This is true.
You need to view $\mathrm{tr}$ as a dual pairing now. And write it as 
$$\mathrm{tr}(DAB)=\mathrm{tr}(D^{\frac{1}{p}}ABD^{\frac{1}{q}})=\langle D^{\frac{1}{p}}A,BD^{\frac{1}{q}}\rangle,$$
Now you view them as from Schatten-p and Schatten-q classes respectively, and apply the usual Hölder inequality.
