Stabilizability and detectability give the unique stabilizing positive-semidefinite CARE solution

By [F3], $R^{-1}$ exists and $G=G^{\ast }⪰0$, since $v^{\ast }Gv=(B^{\ast }v{)}^{\ast }{R}^{-1}(B^{\ast }v)\ge 0$. Regard the Hermitian matrices as a finite-dimensional real vector space and solve the initial-value problem $P^{\mathrm{\prime }}(s)=A^{\ast }P(s)+P(s)A-P(s)GP(s)+Q,P(0)=0.$ Its polynomial vector field is locally Lipschitz: on each bounded matrix ball, expanding the quadratic difference gives $PGP-HGH=(P-H)GP+HG(P-H)$, which is bounded linearly in $\mathrm{\parallel }P-H{\mathrm{\parallel }}_F$. Thus [F4] and [F5] give a unique maximal solution. The adjoint $P(s{)}^{\ast }$ satisfies the same IVP, so uniqueness makes $P(s)=P(s{)}^{\ast }$ on its interval.

Fix $T$ in this interval and let $x$ follow any admissible control on $[0,T]$. Put $S(t)=P(T-t)$. Differentiating $x(t{)}^{\ast }S(t)x(t)$ and substituting the ODE gives $x^{\ast }Qx+u^{\ast }Ru+\frac{d}{dt}(x^{\ast }Sx)=(u+R^{-1}{B}^{\ast }Sx{)}^{\ast }R(u+R^{-1}{B}^{\ast }Sx)\mathrm{.}$ Integrating, using $P(0)=0$, yields $J_T(u)=x_0^{\ast }P(T)x_0+{\int }_0^T(u+R^{-1}{B}^{\ast }Sx{)}^{\ast }R(u+R^{-1}{B}^{\ast }Sx)dt\mathrm{.}$ The candidate feedback makes $\dot{x}=(A-GS(t))x$, whose coefficient is continuous on $[0,T]$. If $F=\mathbb{R}$, [F7] gives its unique solution on the interval. If $F=\mathbb{C}$, write $x$ in real and imaginary parts and replace the coefficient $M$ by $\left[\begin{array}{cc}\mathrm{Re}M& -\mathrm{Im}M\\ \mathrm{Im}M& \mathrm{Re}M\end{array}\right]$; [F7] applies to this continuous real system, and its solution is exactly the complex feedback trajectory. Thus $u=-R^{-1}{B}^{\ast }Sx$ is admissible and makes the last integral zero, so $x_0^{\ast }P(T)x_0$ is the finite-horizon value and $P(T)⪰0$.

By stabilizability [F1], choose $K_0$ with $A_0=A-B{K}_0$ Hurwitz. Under $u=-K_{0}x$, [F6] bounds $x(t)=e^{t{A}_0}{x}_0$ by a square-integrable polynomial-exponential multiple of $\mathrm{\parallel }{x}_0\mathrm{\parallel }$. The control $K_{0}x(t)$ has the same property, so its cost on $[0,T]$ is at most $c\mathrm{\parallel }{x}_0{\mathrm{\parallel }}^2$ for one finite $c$ independent of $T$. Step 2.1 therefore gives $0⪯P(T)⪯cI$. If the maximal interval ended at finite $T_{\ast }$, the polynomial ODE field would be bounded on this bounded range, so $P^{\mathrm{\prime }}(s)$ would be bounded and every entry of $P(s)$ would have a finite limit as $s\uparrow T_{\ast }$. Applying [F4] at that limiting matrix extends the solution past $T_{\ast }$, contradicting maximality [F5]. Hence $P$ exists for every $s\ge 0$.

For $T_2>T_1$, every control on $[0,T_2]$ has nonnegative cost after $T_1$; its restriction to $[0,T_1]$ costs at least $x_0^{\ast }P(T_1)x_0$ by step 2.1. Taking infima gives $P(T_2)⪰P(T_1)$. Together with the bound in step 3.1, each quadratic form $v^{\ast }P(T)v$ increases to a finite limit. Applying this to the coordinate vectors and their pairwise sums shows every matrix entry converges to a Hermitian limit $X⪰0$.

Let $\mathrm{\Phi }(P)=A^{\ast }P+PA-PGP+Q$. By step 4.1 and continuity, $P^{\mathrm{\prime }}(T)=\mathrm{\Phi }(P(T))\to \mathrm{\Phi }(X)$. If $\mathrm{\Phi }(X)\mathrm{\ne }0$, polarization gives a vector $v$ with $v^{\ast }\mathrm{\Phi }(X)v\mathrm{\ne }0$. Then the derivative of the convergent scalar function $v^{\ast }P(T)v$ tends to a nonzero real number, so it is eventually bounded away from zero with constant sign; integration contradicts its convergence. Thus $\mathrm{\Phi }(X)=0$, so $X$ solves the CARE.

Set $K=R^{-1}{B}^{\ast }X$ and $A_c=A-BK=A-GX$. The CARE gives $A_c^{\ast }X+X{A}_c=-Q-K^{\ast }RK$. Along $\dot{x}=A_{c}x$, integration over $[0,T]$ gives ${\int }_0^T(\mathrm{\parallel }Cx(t){\mathrm{\parallel }}_2^2+u(t{)}^{\ast }Ru(t))dt=x_0^{\ast }X{x}_0-x(T{)}^{\ast }Xx(T)\le x_0^{\ast }X{x}_0\mathrm{.}$ The Cholesky factor $R=L{L}^{\ast }$ is invertible by [F3], so $u^{\ast }Ru=\mathrm{\parallel }{L}^{\ast }u{\mathrm{\parallel }}_2^2$ controls $\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ by the finite operator norm of $L^{-\ast }$. Hence $Cx,u\in L^2(0,\mathrm{\infty })$.

By detectability [F1], choose $L$ with $A_o=A-LC$ Hurwitz. Then $\dot{x}=A_{o}x+w$, where $w=Bu+LCx\in L^2$. Since $x$ is locally absolutely continuous, $w$ is locally integrable and the product rule gives $\frac{d}{dt}(e^{-t{A}_o}x(t))=e^{-t{A}_o}w(t)$ almost everywhere. Integrating this identity from $0$ to $t$ and multiplying by $e^{t{A}_o}$ proves $x(t)=e^{t{A}_o}{x}_0+{\int }_0^t{e}^{(t-s)A_o}w(s)ds$. From [F6], there are $M,\alpha >0$ with $\mathrm{\parallel }{e}^{t{A}_o}{\mathrm{\parallel }}_2\le M{e}^{-\alpha t}$ after absorbing the finite polynomial factor into a smaller exponential rate. The initial term tends to zero. For the integral, split at a fixed $S$: the portion $0\le s\le S$ tends to zero as $t\to \mathrm{\infty }$; by Cauchy–Schwarz the tail is at most $M(2\alpha {)}^{-1\mathrm{/}2}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(S,\mathrm{\infty })}$, which tends to zero as $S\to \mathrm{\infty }$. Thus $x(t)\to 0$. Since this holds for every initial state, no eigenvalue of $A_c$ can have nonnegative real part: an eigenvector for such an eigenvalue would not decay. Therefore $A_c$ is Hurwitz.

For any admissible control of finite infinite-horizon cost, $u,Cx\in L^2$. Repeating the stable-convolution argument in step 7.1 gives $x(T)\to 0$. The same square-completion identity as step 2.1, now with constant $S=X$, yields $J_T(u)=x_0^{\ast }X{x}_0-x(T{)}^{\ast }Xx(T)+{\int }_0^T(u+Kx{)}^{\ast }R(u+Kx)dt\mathrm{.}$ Taking $T\to \mathrm{\infty }$ gives $J_{\mathrm{\infty }}(u)\ge x_0^{\ast }X{x}_0$. The feedback in step 6.1 has $u+Kx=0$ and $x(T)\to 0$, hence attains equality.

Let $Y⪰0$ be any other CARE solution. Repeating steps 6.1–7.1 with $Y$ proves $A_Y=A-GY$ is Hurwitz. Subtracting the CAREs and using $XGX-YGY=PG(X-Y)+(X-Y)GY\text{with }P=X$ gives $A_c^{\ast }(X-Y)+(X-Y)A_Y=0$. Hence the derivative of $e^{t{A}_c^{\ast }}(X-Y)e^{t{A}_Y}$ is zero, so this matrix equals $X-Y$ for every $t\ge 0$. Both exponential factors tend to zero by [F6], forcing $X-Y=0$. Thus $X$ is the unique positive-semidefinite solution, and in particular the unique stabilizing one.