How to prove concavity of $\textrm{Tr}(\exp(H+\log X))$?

Question

I am studying random matrices and in the notes that I am reading, the teacher uses Lieb's inequality without proving it, that is $$X \to \textrm{Tr}(\exp(H+\log X))$$ is concave on the set of symmetric matrices, where $H$ is symmetric too. How to prove this? What would be a first step?

I can only understand that to define $\log X$ one has to choose a base where $X$ is diagonal, but that's pretty much how far I get. Thanks in advance for any clue!

Answer 1

Note that the considered inequality is valid on the set of hermitian $>0$ matrices. The proof of Lieb's inequality (dated 1973) is not simple. It's easier to prove the result in the particular case when $X=I_n$; we also assume that the matrices are real symmetric ( $L ,H=[h_{i,j}]\in S$) or real symm. $>0$ (in $S^+$).

Let $f:t\in (-\epsilon,\epsilon)\mapsto tr(\exp (L+\log(I+tH))$. We show that $f"(0)\leq 0$. Note that $\log(I+tH)=tH-(t^2/2)H^2+\cdots$.

$f'(t)=tr((H-tH^2+\cdots)\exp(L+tH-(t^2/2)H^2+\cdots))$.

In particular, $f'(0)=tr(He^L)$. More complicated

$f"(0)=tr(-H^2e^L)+tr(HDg_{L}(H))$ where $g(Z)=e^Z$; that is

$f"(0)=tr(-H^2e^L+H\int_0^1 e^{uL}He^{(1-u)L}du)$ or

$f"(0)=\int_0^1 tr(He^{uL}He^{(1-u)L}-H^2e^L)du$.

It suffices to show that, for every $u$,

$d=tr(HA^uHA^{1-u}-H^2A)\leq 0$ where $A=e^L>0$.

We may assume that $A=diag(a_i)$ where $a_i>0$.

$d=\sum_{i<j}{h_{i,j}}^2({a_i}^u-{a_j}^u)(-{a_i}^{1-u}+{a_j}^{1-u})$ has the signum of $(a_i-a_j)(a_j-a_i)$, that is, $d\leq 0$, and we are done.

EDIT. Moreover, it is not difficult to see that $d=0$ iff $HA=AH$ iff $HL=LH$. In this case, $f(t)=tr(e^L)+t.tr(e^LH)$.