Let $A\in\mathbb{R}^{n\times n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix, i.e., $X\succeq 0$. Consider the following matrix-valued function $$\tag{1}\label{1} F(X) = \left(\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t\right)^{-1/2} X \left(\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t\right)^{-1/2} $$ which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $\cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{\cdot}$ is the matrix exponential).

Note that \eqref{1} is a continuous function which is not defined for any $X\succeq 0$ such that $\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(\cdot)$ in the set of positive semidefinite matrices.

In more formal terms, let $\bar{X}\succeq 0$ be such that $\int_0^{\infty} e^{At} \bar{X} e^{A^\top t} \mathrm{d} t$ is singular, and let $\{X_n\}_{n\ge 0}$, $X_n\succ 0$, be any sequence such that $\lim_{n\to \infty} X_n = \bar{X}$. Does $ \lim_{n\to \infty} F(X_n)$ exist and is finite?

My question is motivated by the special case of scalar matrices $A=\alpha I$, $\alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.

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    $\begingroup$ Whenever it's defined, the function $g(Q) = \int_0^\infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$ $\endgroup$ – Omnomnomnom Dec 20 '18 at 1:46
  • $\begingroup$ @Omnomnomnom: Right, thanks! However, I do not see how this can be helpful. $\endgroup$ – Ludwig Dec 20 '18 at 1:48
  • $\begingroup$ Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$ $\endgroup$ – Omnomnomnom Dec 20 '18 at 1:49
  • $\begingroup$ I'm not quite sure if that helps either $\endgroup$ – Omnomnomnom Dec 20 '18 at 1:55
  • $\begingroup$ I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2,\ \bar{X}=0$ and $X_n=diag(\frac1n,0)$. $\endgroup$ – user1551 Dec 20 '18 at 4:14

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