Suppose $X, B$ are both symmetric, positive definite real matrices. We also know all eigenvalues of $B$ are strictly smaller than $1$. I am wondering whether we can prove $X-B XB$ is positive semidefinite.
I tried to follow definition by considering $v^{\top}(X-BXB)v$ for all unit vectors. But I could not determine how this term $(Bv)^{\top} X (Bv)$ behaves as even if $Bv$ has norm smaller than $1$, it is also possible that $Bv$ is mapped to a vector corresponding to larger eigenvalue of $X$.
This question comes from that I am trying to determine whether ${\bf tr}(Y(X-B^{\top} X B))$ is always positive provided $X, Y, B$ are all positive definite and $B$ has spectral radius bounded by $1$. For this purpose, initially I used some crude bound \begin{align*} {\bf tr}(YX-Y B X B) \ge \lambda_{\min}(Y) {\bf tr}(X) - \lambda_{\max}(Y) {\bf tr}(BXB). \end{align*} It seems that $Y$ could determine the sign of the trace. But then it occurred to be me when the first equality holds the second must not hold. The bounds I am using seem to be too crude.
Your intuition that $Bv$ could be mapped to a vector with a larger $X$-norm (even though it has a smaller norm) is correct. Consider, for example, $$ X = \begin{bmatrix}1 & 0 \\ 0 & 1000\end{bmatrix}, \quad B = \begin{bmatrix}1/2 & 1/4 \\ 1/4 & 1/2\end{bmatrix}. $$ Here, I chose $X$ to treat the vectors $(1,0)$ and $(0,1)$ very unequally, and $B$ just as some random matrix with the properties you want.
Take $v = (1,0)$. Then $v^T X v$ is relatively small. But $Bv$ has a nonzero $y$-component, so $(Bv)^T X (Bv)$ is large. Therefore $v^T(X - BXB)v$ will be negative, and $X - BXB$ will not be positive semidefinite.
In this example, we can choose $Y$ so that $\operatorname{tr}(Y(X-BXB))$ will be negative. Since $(X-BXB)_{11}$ is negative but $(X-BXB)_{22}$ is positive, just ensure that $Y_{11}$ is much much larger than $Y_{22}$ to make the first of these features matter more than the second. For example, $$ Y = \begin{bmatrix}1000 & 0 \\ 0 & 1\end{bmatrix} $$ works.