Let $\mathbb P_n$ be the space of all $n \times n$ self-adjoint positive definite matrices. Consider the function $\varphi: \mathbb P_n \longrightarrow \mathbb R$ defined by $$\varphi (A) = -\text {tr}\ (A \log A).$$ Show that for all $t \in (0,1)$ $$\varphi ((1 - t) A + t B) \leq (1 - t) \varphi (A) + t \varphi (B) - \eta (t,1-t)$$ where $\eta (t,1 - t) = t \log (t) + (1 - t) \log (1 - t).$
I know that $\varphi$ is operator concave. But I don't have any idea as to how to bound $\varphi$ from above. Could anyone please give me some hint?
Thanks a bunch!
For notational simplicity denote with
$$M_t := (1-t) A+tB$$
now note that by positivity of $A$ and $B$
$$M_t ≥ tB, \qquad M_t ≥(1-t)A$$
and that $\log$ is monotone, so also $$\log(M_t)≥ \log(tB)=\log(t)\Bbb 1+\log(B), \quad \log(M_t) ≥\log(1-t)A = \log(1-t)\Bbb1 + \log(A)$$
This allows you do do the following simplification with the trace, for $P$ positive you have:
\begin{align}\mathrm{Tr}(P\log(M_t) ) &= \mathrm{Tr}(\sqrt{P}\log(M_t)\sqrt{P})\\ &≥ \mathrm{Tr}(\sqrt{P}(\log(t)+\log(B))\sqrt{P}) \\&= \log(t)\mathrm{Tr}(P) +\mathrm{Tr}(P\log(B))\end{align}
Same with $A$. Then:
$$\varphi(M_t) ≤ -(1-t)\mathrm{Tr}(A\log(A))-t\mathrm{Tr}(B\log(B)) -(1-t)\mathrm{Tr}(A)\cdot \log(1-t) - t\mathrm{Tr}(B)\cdot \log(t) $$
This yields the inequality:
$$\varphi((1-t)A+tB) ≤ (1-t) \varphi(A) + t\varphi(B) -(1-t)\log(1-t)\mathrm{Tr}(A)-t\log(t)\mathrm{Tr}(B)$$
In the event that both $\mathrm{Tr}(A), \mathrm{Tr}(B)$ are $≤1$ you recover the desired inequality. However the inequality may fail if either $A$ or $B$ has trace larger than $1$ - and this may happen in any dimension. A very simple example in dimension $1$, let $t=\frac12$, $B=1$ and $A=e^{10}$, then:
$$\varphi(\frac12e^{10}+\frac12) = -\frac12 (e^{10}+1)(-\ln(2)+\ln(e^{10}+1))\approx -103\,000$$ $$\frac12\varphi(e^{10})+\frac12\varphi(1)=-5 e^{10}\approx -110\,000,\qquad \eta(\frac12,\frac12) = \ln(2)\approx 0.69$$ So $$\varphi(\frac12 e^{10}+\frac12) \not≤ \frac12 \varphi(e^{10})+\frac12 \varphi(1) + \eta(\frac12,\frac12) $$
(the number $e^{10}$ was chosen arbitrarily - of course you can get much better smaller numbers for which it fails).