# Limit of matrix function

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.

• 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 $$AX + XA^T + Q = 0$$ – Omnomnomnom Dec 20 '18 at 1:46
• @Omnomnomnom: Right, thanks! However, I do not see how this can be helpful. – Ludwig Dec 20 '18 at 1:48
• Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$YA + A^TY + YQY = 0$$ – Omnomnomnom Dec 20 '18 at 1:49
• I'm not quite sure if that helps either – Omnomnomnom Dec 20 '18 at 1:55
• 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 $X_n=diag(\frac1n,0)$. – user1551 Dec 20 '18 at 4:14