# Strong Concavity in Lieb's inequality

> Question 1: Can one get strong concavity in one of Lieb's inequalities (the p'th power one or the exp-log one) under some conditions on the commutators of the matrices?
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Concretely, let us look at Lieb's concavity theorem for $p=1$: For all PD $A$, and Hermitian $K$, $f(A) = Tr(\sqrt{A}\, K \sqrt{A}\, K^\dagger)$ is a concave function of $A$.

> **Conjecture/Question**: For some function $\alpha:\mathbb{Z}_+ \rightarrow (0,\infty)$, for all $K$, $f(A) = Tr(\sqrt{A}\, K \sqrt{A}\, K^\dagger)$ is strongly-concave in $A$ with $\nabla^2 f(A) \succeq \alpha_d Tr([A^{-1/2},K] [K^\dagger, A^{-1/2}]) \cdot I_{d \otimes d}$.
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### Sketch for derivative along identity:

Fix $K$. Let us compute the Hessian of $f(A) = Tr(\sqrt{A}\, K \sqrt{A}\, K^\dagger)$ along $I_d$. That is, let $g(x) = f(A + x I_d)$. By Lieb's concavity theorem, $g$ is a concave function, and we want to say that $g''(0)$ is positive if $A,K$ do *commute* in a quantifiable way.

>Claim: Fix $A, K$ and let $g(x) = f(A + x I_d)$. Then, $g''(0) \geq (1/2) Tr([A^{-1/2},K] [K^\dagger, A^{-1/2}])$.

**Proof**: The first idea is to use the expansion of $\sqrt{A}$ as done implicitly in the Winger-Yasse paper. Suppose we write $\sqrt{A + x B} = S + x N - x^2 T + \cdots$. Then, $S^2 = A$, $S N + N S = B$, and $S T + T S = N^2$. In our case, as $ B = I_d$, we get that $N = S^{-1}/2$ and $T = S^{-3}/8$. Thus, Hessian of $g(x) = f(A + x I_d)$ is given by
$$\begin{align*}
Tr(\sqrt{A + x B} K \sqrt{A + x B} K^\dagger) &\approx Tr( (S + x N - x^2 T) K (S + x N - x^2 T) K^\dagger)\\
&\approx g(0) + x (\;) - x^2 \left( Tr(S K T K^\dagger) + Tr(T K S K^\dagger) - Tr(N K N K^\dagger)\right).
\end{align*}$$

Now, define an operator on $d\times d$ matrices by $Q \equiv S \otimes T + T \otimes S - N \otimes N$. Then, it is easy to check that $\langle Q(K), K^\dagger \rangle = \left( Tr(S K T K^\dagger) + Tr(T K S K^\dagger) - Tr(N K N K^\dagger)\right)$.

Thus, Lieb's concavity theorem for instance becomes equivalent to $Q$ being PSD (as operator on matrices). In our case, since we want a lower bound for all $K$, let us define another operator $P \equiv (S^{-1} \otimes I_d - I_d \otimes S^{-1})^2 \succeq 0$. We will show that $Q \succeq P/2$ as operators. The claim would follow from this as we'd have
$$\langle Q(K), K^\dagger \rangle \geq (1/2) \langle P(K), K^\dagger \rangle = (1/2) Tr([A^{-1/2},K] [K^\dagger, A^{-1/2}]).$$

Note that in our case $Q \equiv S \otimes S^{-3}/8 + S^{-3} \otimes S/8 - S^{-1} \otimes S^{-1}/4$. Clearly, for any PD $X,Y$, $X + Y - \sqrt{X} \sqrt{Y} - \sqrt{Y} \sqrt{X} = (\sqrt{X} - \sqrt{Y})^2$. Applying this to $X = S \otimes S^{-3}$ and $Y = S^{-3} \otimes S^{-1}$, we first get

$$ S \otimes S^{-3} + S^{-3} \otimes S - 2 S^{-1} \otimes S^{-1} = \left(S^{1/2} \otimes S^{-3/2} - S^{-3/2} \otimes S^{1/2}\right)^2.$$

Therefore, to finish our claim, we need to show that
$$(1/8) \left(S^{1/2} \otimes S^{-3/2} - S^{-3/2} \otimes S^{1/2}\right)^2 \succeq (1/2) (S^{-1} \otimes I_d - I_d \otimes S^{-1})^2.$$

Since the above only involve $S$, it suffices to show it for the case when $S$ is diagonal. In which case, the above inequality is equivalent to showing that for positive numbers $x, y$, $(x y^{-3} - x^{-3} y)^2 \geq 4 (x^{-2} - y^{-2})^2$. The latter is equivalent to $(x^2 + y^2)^2 \geq 4 x^2 y^2$ which is true.