For a fixed positive definite $A$ anb vector $x$, is $B\mapsto x^T B^{1/2} A B^{1/2} x$ always concave?

Question

Let $x\in R^d$ and $A\in R^{d\times d}$ positive definite. Is the map $$ B \mapsto x^T B^{1/2} A B^{1/2} x $$ always concave? One known result that gives a little hope is the Lieb inequality (cf. https://arxiv.org/abs/quant-ph/0404126 for instance) on the concavity of $trace[A^T \sqrt B A \sqrt B]$.

Answer 1

No.

For a counterexample, consider $x = (0, 1)$, $A = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$ (this is not positive definite, but can be approximated with positive definite matrices). Then for $B_{\pm} = \begin{bmatrix} 1/2 & \pm1/2 \\ \pm1/2 & 1/2\end{bmatrix}$ we have $x^{T} B_{\pm}^{1/2} A B_{\pm}^{1/2} x > 0$ but $x^{T} (B_{+} + B_{-})^{1/2} A (B_{+} + B_{-})^{1/2} x = 0$.

More generally, if the claim was true, then for any orthogonal basis $\{x_{1}, x_{2}, \ldots, x_{d}\}$ the functions $f_{i} : B \to x_{i}^{T} B^{1/2} A B^{1/2} x_{i}$ would all be concave, and hence would be their sum $f_{1} + f_{2} + \ldots + f_{d} = Tr(B^{1/2} A B^{1/2}) = Tr(A B) =$ linear. But sum of concave functions can only be linear if all the summands are affine so $B \mapsto x^{T} B^{1/2} A B^{1/2} x$ would need to be always linear.