The Hurwitz Lyapunov integral is the unique solution and has the stated positivity

Put $M(t)=e^{tA}$. For finite matrices $P,R$, $\mathrm{\parallel }PR{\mathrm{\parallel }}_F^2={\sum }_j\mathrm{\parallel }PR{e}_j{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }P{\mathrm{\parallel }}_2^2{\sum }_j\mathrm{\parallel }R{e}_j{\mathrm{\parallel }}_2^2=\mathrm{\parallel }P{\mathrm{\parallel }}_2^2\mathrm{\parallel }R{\mathrm{\parallel }}_F^2$ by the operator-norm definition and [F4]. The entry formula in [F4] gives $\mathrm{\parallel }{R}^{\ast }{\mathrm{\parallel }}_F=\mathrm{\parallel }R{\mathrm{\parallel }}_F$. The SVD in [F5] gives the same positive singular values for $P$ and $P^{\ast }$; [F6] therefore gives $\mathrm{\parallel }{P}^{\ast }{\mathrm{\parallel }}_2=\mathrm{\parallel }P{\mathrm{\parallel }}_2$. Applying the left-multiplication estimate to adjoints yields $\mathrm{\parallel }RP{\mathrm{\parallel }}_F\le \mathrm{\parallel }R{\mathrm{\parallel }}_F\mathrm{\parallel }P{\mathrm{\parallel }}_2$. Applying these inequalities successively gives $\mathrm{\parallel }M(t)QM(t{)}^{\ast }{\mathrm{\parallel }}_F\le \mathrm{\parallel }M(t){\mathrm{\parallel }}_2^2\mathrm{\parallel }Q{\mathrm{\parallel }}_F$.

For $X_T={\int }_0^{T}M(t)QM(t{)}^{\ast }dt$, [F2] and the product rule give $\frac{d}{dt}(M(t)QM(t{)}^{\ast })=AM(t)QM(t{)}^{\ast }+M(t)QM(t{)}^{\ast }{A}^{\ast }$. Integrating this identity yields $A{X}_T+X_T{A}^{\ast }=M(T)QM(T{)}^{\ast }-Q$.

By [F1], the right side in step 1.1 is bounded by a finite sum of polynomial multiples of $e^{-2\alpha t}$. Each such term is integrable on $[0,\mathrm{\infty })$ by repeated integration by parts. Every entry of the integrand is therefore absolutely integrable; define $X$ by the finitely many scalar improper entry integrals. The entrywise tails tend to zero, so $X_T\to X$ in Frobenius norm and the matrix integral converges absolutely in that norm. The integrand is self-adjoint because $Q=Q^{\ast }$.

By [F1], $M(T)QM(T{)}^{\ast }\to 0$ as $T\to \mathrm{\infty }$, and by step 2.1, $X_T\to X$. Taking limits in step 1.2 gives $AX+X{A}^{\ast }=-Q$.

Suppose $Q⪰0$. By [F3], $Q^{1\mathrm{/}2}$ exists and is self-adjoint, so $Q=(Q^{1\mathrm{/}2}{)}^{\ast }{Q}^{1\mathrm{/}2}$. Hence for every $v$, $v^{\ast }Xv={\int }_0^{\mathrm{\infty }}\mathrm{\parallel }{Q}^{1\mathrm{/}2}{e}^{t{A}^{\ast }}v{\mathrm{\parallel }}_2^{2}dt\ge 0$. Thus $X⪰0$. If $Q\succ 0$ and $v\mathrm{\ne }0$, the integrand at $t=0$ is $v^{\ast }Qv>0$ and is continuous, so it stays positive on an interval of positive length; therefore $v^{\ast }Xv>0$ and $X\succ 0$.

If $Y$ is another solution, $H=Y-X$ satisfies $AH+H{A}^{\ast }=0$. The derivative of $M(t)HM(t{)}^{\ast }$ is therefore zero by [F2], so it is constant and equals $H$ at $t=0$. By [F1] it tends to zero as $t\to \mathrm{\infty }$, hence $H=0$ and the solution is unique.

For $Q⪰0$ and nonzero $v$, the continuous integrand in step 3.2 has integral zero exactly when it vanishes for all $t\ge 0$: if it were positive at one time, continuity would make its integral over a neighboring interval positive. It vanishes at time $t$ exactly when $e^{t{A}^{\ast }}v\in \ker Q^{1\mathrm{/}2}$. Hence $v^{\ast }Xv>0$ for every nonzero $v$ exactly when no nonzero trajectory remains in that kernel for all $t\ge 0$.