Let $\Omega_{n}$ denote the cone of $n\times n$ real symmetric positive definite matrices, and consider $F:\Omega_{n} \mapsto \Omega_{n}$. For $X,Y \in \Omega_{n}$, the matrix valued function $F(\cdot)$ is called operator/matrix monotone if $F(X) \geq F(Y)$ whenever $X \geq Y$.
Given an operator monotone function $F(.)$, and $B\in\Omega_{n}$ with $\rm{trace}(B)=1$, under what additional condition on $B$, can we say that $G(X) = F(X) B F(X)$ is operator monotone?
