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

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!

• I think there are proofs of this in references 13, 14, and 15 of this Wikipedia article. May 16, 2019 at 6:05

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 $$a_i>0$$.

$$d=\sum_{i has the signum of $$(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, $$f(t)=tr(e^L)+t.tr(e^LH)$$.