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!

  • $\begingroup$ Could anyone please give me some hint? $\endgroup$
    – RKC
    Aug 29, 2021 at 16:51
  • $\begingroup$ Is it correct that the $\eta$-term in that inequality is independent of $A,B$? $\endgroup$
    – daw
    Sep 4, 2021 at 12:21
  • $\begingroup$ @daw yes that's correct. $\endgroup$
    – RKC
    Sep 5, 2021 at 4:21
  • $\begingroup$ @daw the hint was given as follows $:$ Use some integral expression for $\log (X + Y) - \log (X)$ for positive definite matrices $X$ and $Y.$ $\endgroup$
    – RKC
    Sep 5, 2021 at 4:26
  • $\begingroup$ This looks suspiciously like an information-theoretic value. Perhaps there's a proof that can be offered by calculating a relevant entropy? $\endgroup$ Sep 15, 2021 at 17:43

1 Answer 1


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).

  • $\begingroup$ I get $$\frac {1} {\lambda} \varphi (\lambda P) - \varphi (P) = -\log (\lambda) \cdot \text {Tr} (P).$$ $\endgroup$
    – RKC
    Sep 10, 2021 at 6:34
  • 1
    $\begingroup$ Also you have missed a $\lambda$ in the denominator of $\eta (t,1-t)$ in the last inequality you have obtained. Although these doesn't make any difference at last. The last equality will become an inequality as a result. But I have to confess that it's a wonderful answer. Thanks a lot. $\endgroup$
    – RKC
    Sep 10, 2021 at 6:49
  • $\begingroup$ You are right about the errors, however upon reading I discovered another error: The necessary condition in the beginning is not $\mathrm{Tr}(A)≥1$ but $≤1$. This means the second step starts with a false assumption and the proof only goes through if both $A$ and $B$ have trace $≤1$. Thinking more about it one can construct counterexamples to the equation if one allows either $A$ or $B$ to have trace $≥1$, take for example: $$A=e^{100}\ \Bbb 1, \quad B = \begin{pmatrix}1&0\\0&\frac12\end{pmatrix}$$ Then: $$\varphi(M_{1/2})-1/2\varphi(A)-1/2\varphi(B) \approx 10^{43}$$ $\endgroup$
    – s.harp
    Sep 12, 2021 at 19:03
  • $\begingroup$ which is not compatible with the inequality. $\endgroup$
    – s.harp
    Sep 12, 2021 at 19:03
  • 1
    $\begingroup$ The inequality is true for trace $≤1$ - don't forget that $\log(t)$ is negative as $0<t<1$. $\endgroup$
    – s.harp
    Sep 19, 2021 at 10:28

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