Let $P\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix, i.e., $P=P^\top$ and $\langle Px,x\rangle\geq0$ for all $x\in\mathbb{R}^n$.
Assume that, for a certain $A\in\mathbb{R}^{n\times n}$ there exists a $\mu\in\mathbb{R}$ such that: $$PA+A^\top P<2\mu P.$$
Let $F(\varepsilon)\in\mathbb{R}^{n\times n}$ be such that $\lim_{\varepsilon\to0}\|F(\varepsilon)\|_2=0$. I would like to prove/disprove the following.
Lemma. There exists an $\varepsilon^*>0$ and a $\delta(\epsilon)\in\mathbb{R}$, with $\lim_{\varepsilon\to0}\delta(\varepsilon)=0$, such that for all $\varepsilon\in(0,\varepsilon^*)$ the following holds: $$P(A+F(\varepsilon))+(A+F(\varepsilon))^\top P< 2(\mu+\delta(\varepsilon))P.$$
I have tried to prove this result by extending robustness results for $P>0$. However, they all use (at some point in the proof) $P^{-1}$, which is not defined in our case (since $P\geq0$).