Let $B$ be an $n\times n$ matrix, and define $f$ to be the function that maps positive semidefinite (PSD) $n\times n$ matrices $A$ to real numbers by 


$$
f(A) = \mathrm{trace}( (B^\*A^2B)^{1/3}).
$$


In other words, $f$ maps $A$ to the sum of $1/3$-powers of the eigenvalues of the PSD matrix $B^\*A^2B$. 



> 
> Is the function $f$ concave over the PSD cone? I.e. is it true that for any two PSD matrices $X$ and $Y$, $f((X + Y)/2) \ge f(X)/2 + f(Y)/2$?
> 
> 
> 


A more general question is whether the function $f(A) = \mathrm{trace}( (B^\*A^2B)^{p})$ is concave over the PSD cone for $0< p < 1/2$. 


[Carlen and Lieb](https://arxiv.org/pdf/0710.4167.pdf) have some closely related results, but I could not find the particular combination of matrix powers in their paper or in other related work on trace inequalities. 



Concavity of the trace of a matrix power
Unfortunately, the conjectured function is not concave. Here is a ~~simple~~ *simpler* counterexample.


\begin{equation\*}
 B = \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},\quad
 A = \begin{bmatrix} 2 & 0 \\ 0 & 3\end{bmatrix},\quad
 C = \begin{bmatrix} 5 & 0 \\ 0 & 2\end{bmatrix}.
\end{equation\*}


With this choice, and for $f(X)=\text{tr} (B^\*X^2B)^{1/3}$ we see that $f((A+C)/2)- (f(A)+f(C))/2 \approx -0.0616$. 




---


You may find the following necessary condition interesting:



> 
> **(Prop. 5.1, [Hiai, 2013](http://www.sciencedirect.com/science/article/pii/S0024379513002826)).** Assume that $p,s\neq 0$. If $A \in \mathbb{P}\_2 \mapsto \text{tr}(B^\*A^pB)^s$ is concave for any invertible matrix $B \in \mathbb{C}^{2\times 2}$, then either $0<p\le 1$ and $0<s\le 1/p$, or $-1\le p \le 0$ and $1/p\le s < 0$.
> 
> 
> 


