Proof of Lyapunov Representation and Positive Semidefiniteness for Linear Matrix Equations

Preliminaries. We use: entrywise sums and products of continuous functions are continuous (Sum and Product Rules for One-Dimensional Derivatives and Continuity); matrix products are rearranged with Associativity of the Matrix Product; the transpose reversal $(UV)^{\top}=V^{\top}U^{\top}$ and the transpose-dot identity $y\cdot(Mz)=(M^{\top}y)\cdot z$ of claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals; and the identity $x\cdot\bigl(\bigl(\int_a^tU(r)\,dr\bigr)x\bigr)=\int_a^t x\cdot(U(r)x)\,dr$ of claim 6 there.

Claim 1. Uniqueness and existence. Identify $k\times k$ matrices with $\mathbb{R}^{k^{2}}$ as in the proof conventions of Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations. The map $F(t,X)=A(t)X+XA(t)^{\top}+C(t)$ is composition continuous, and Lipschitz in $X$ with constant $2k^{3}\alpha$ where $\alpha$ bounds the entries of $A$ (entry estimates from claims 1 and 2 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals; Extreme Value Theorem on a Compact Interval for the bound). By Global Existence and Uniqueness for Lipschitz Ordinary Differential Equations in Integral Form there is exactly one continuous solution $P$.

Representation. Let $\Phi,\Psi$ be as in Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations and put $Y(t)=\int_a^t\Psi(r)C(r)\Psi(r)^{\top}\,dr$ (entrywise; the integrand has continuous entries). By claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals the entries of $Y$ are continuous on all of $[a,b]$, and by Fundamental Theorem of Calculus, Part I in One Dimension they are differentiable at interior points with $Y'=\Psi C\Psi^{\top}$. Define $\widehat P(t)=\Phi(t)\bigl(P_0+Y(t)\bigr)\Phi(t)^{\top}$; its entries are continuous on $[a,b]$. On $(a,b)$: $\Phi'=A\Phi$, and $(\Phi^{\top})'=(\Phi')^{\top}=(A\Phi)^{\top}=\Phi^{\top}A^{\top}$ (transposition commutes with entrywise differentiation, being a relabelling of entries); the product and sum rules (Sum and Product Rules for One-Dimensional Derivatives and Continuity, applied entrywise to the finite sums defining the products) give

using $\Phi\Psi=I_k$ (identity matrix) and $\Psi^{\top}\Phi^{\top}=(\Phi\Psi)^{\top}=I_k$. Since the entries of the right side are continuous on $[a,b]$ and $\widehat P(a)=P_0$, each entry of $\widehat P$ is an antiderivative of the corresponding entry, and Fundamental Theorem of Calculus, Part II in One Dimension, applied on $[a,t]$ for each $t$ (degenerate $t=a$ by the convention of Mean-Square Riemann Integral of a Family of Random Variables), yields the integral equation for $\widehat P$; by the uniqueness just proved, $P=\widehat P$.

Claim 2. Suppose $P_0^{\top}=P_0$ and $C(t)^{\top}=C(t)$ for all $t$. Transposing the integral equation entrywise (the transpose of an entrywise integral is the entrywise integral of the transpose, a relabelling) and using $(AP)^{\top}=P^{\top}A^{\top}$ and $(PA^{\top})^{\top}=AP^{\top}$ (claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals) shows that $t\mapsto P(t)^{\top}$ satisfies the same integral equation; by uniqueness, $P(t)^{\top}=P(t)$.

Claim 3. Suppose additionally that $P_0$ and every $C(t)$ are positive semidefinite. Fix $t$ and $x\in\mathbb{R}^{k}$, and put $w=\Phi(t)^{\top}x$. By the transpose-dot identity (twice) and the representation,

and by claim 6 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals and the transpose-dot identity again,

the integrand being continuous and nonnegative (monotonicity of the integral via Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval and Linearity and Monotonicity of the Lebesgue Integral; degenerate $t=a$ by the convention of Mean-Square Riemann Integral of a Family of Random Variables). Also $w\cdot(P_0w)\ge0$. Hence $x\cdot(P(t)x)\ge0$; with the symmetry from claim 2, $P(t)$ is positive semidefinite. $\blacksquare$