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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!

RKC
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1 Answers1

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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).

s.harp
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  • I get $$\frac {1} {\lambda} \varphi (\lambda P) - \varphi (P) = -\log (\lambda) \cdot \text {Tr} (P).$$ – RKC Sep 10 '21 at 06:34
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    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. – RKC Sep 10 '21 at 06:49
  • 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: $$\varphi(M_{1/2})-1/2\varphi(A)-1/2\varphi(B) \approx 10^{43}$$ – s.harp Sep 12 '21 at 19:03
  • which is not compatible with the inequality. – s.harp Sep 12 '21 at 19:03
  • Do you mean that the inequality is false? – RKC Sep 12 '21 at 20:08
  • Yes, the inequality is false as stated. It works for $A,B$ so that $\mathrm{Tr}(A)≤1, \mathrm{Tr}(B)≤1$ (this is what the proof in the answer actually shows if you correct my mistake). Thinking more about it there is no need even for matrices, the positive reals alone admit counter-examples. $A=e^{100}$, $B=1$, $t=\frac12$ is one such counter example. – s.harp Sep 13 '21 at 07:57
  • @s.harp Nice answer. Can we correct the inequality, e.g. what is the correct term of $\eta(t, A)$? – River Li Sep 15 '21 at 00:03
  • @RiverLi I edited the answer to be more clear about what inequality holds and what inequality fails - the way to correct $\eta$ would be: $$\eta(t; \mathrm{Tr}(A), \mathrm{Tr}(B) ) = - (1-t)\ln(t) \mathrm{Tr}(A) - t\ln(t)\mathrm{Tr}(B)$$ – s.harp Sep 15 '21 at 17:07
  • @s.harp It is very nice to give $\eta$. – River Li Sep 16 '21 at 01:41
  • If $\text {Tr} (A), \text {Tr} (B) \leq 1$ will the inequality not be reversed? At least I think so. – RKC Sep 18 '21 at 08:03
  • I think the inequality may fail hold if both the traces are lesser than $1.$ Could you please check your argument again and confirm? – RKC Sep 18 '21 at 08:15
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    The inequality is true for trace $≤1$ - don't forget that $\log(t)$ is negative as $0<t<1$. – s.harp Sep 19 '21 at 10:28