Constrained optimization over a trace functional

Question

Let $A\in\mathbb{R}^{n\times n}$ be a stable matrix (i.e., the eigenvalues of $A$ have negative real parts). Consider the following optimization problem in $X \in \mathbb{R}^{n \times n}$

$$\begin{array}{ll} \text{maximize} & \mbox{tr} \left(P X \right)\\ \text{subject to} & \mbox{tr}(X) = 1\\ & X \succeq 0\end{array}$$

where $X \succeq 0$ denotes that $X$ is positive semidefinite, and $P$ is the solution of the following Lyapunov equation $$ AP+PA^\top =-X. $$

Question: Does the above-formulated problem admit a closed-form solution?

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An equivalent formulation. We can rewrite the above optimization problem in the following equivalent way. Consider the optimization problem in $P \in \mathbb{R}^{n \times n}$, $P\succeq 0$,

$$\begin{array}{ll} \text{minimize} & \mbox{tr} \left(PAP \right)\\ \text{subject to} & \mbox{tr}(PA) = -1/2\\ & AP+PA^\top \preceq 0\\ & P \succeq 0\end{array}$$

Answer 1

This is not really an answer, but an observation too long for a comment. The OP may already know this ...

From the solution of Lyapunov equation $P = \int_{0}^{\infty} e^{At}Xe^{A^{\top}t} {\mathrm{d}}t$, we have ${\rm{tr}}(PX) = \int_{0}^{\infty} {\rm{tr}}(YX){\mathrm{d}}t$, where $Y := e^{At}Xe^{A^{\top}t} \succeq 0$.

One can then get a bound by using the fact that for positive semi-definite $Y,X$, we have $\lambda_{\min}(Y){\rm{tr}}(X) \leq {\rm{tr}}(YX) \leq \lambda_{\max}(Y){\rm{tr}}(X)$. Using this inequality with ${\rm{tr}}(X)=1$ gives an upper bound:

$$ {\rm{tr}}(PX) \leq \int_{0}^{\infty} \parallel e^{At} X^{1/2}\parallel_{2}^{2} {\rm{d}}t \leq \lambda_{\max}(X)\int_{0}^{\infty} \parallel e^{At}\parallel_{2}^{2} {\rm{d}}t$$

where the last step used sub-multiplicative property of 2-norm. Not sure how tight is this.