Proof of Global Existence for the Backward Riccati Equation under Convexity Conditions

All products are matrix products, $(\cdot)^{\top}$ is the transpose, and we use freely the transpose rules of claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, the involutivity $(S^{\top})^{\top}=S$ (immediate from the definition of the transpose), and associativity (Associativity of the Matrix Product).

Step 1: reversed data. For $s\in[0,T]$ define

The map $s\mapsto T-s$ is continuous from $[0,T]$ to $[0,T]$ (Sums and Products of Continuous Real-Valued Functions), $t\mapsto R(t)^{-1}$ has continuous entries by Invertibility of Symmetric Positive Definite Matrices and claim 1 of Continuity of the Inverse of a Continuous Matrix Function, and precomposition with a continuous map preserves continuity (Composition of Continuous Euclidean Maps); hence all entries of $\check A,\check D,\check C$ are continuous, by Sums and Products of Continuous Real-Valued Functions.

Each $\check D(s)$ is symmetric positive semidefinite: symmetry follows from the transpose rules and the symmetry of $R^{-1}$ (Invertibility of Symmetric Positive Definite Matrices), and for $x\in\mathbb{R}^{l}$, with $y:=B(T-s)^{\top}x$, the transpose rule gives $x\cdot(\check D(s)x)=y\cdot\bigl(R(T-s)^{-1}y\bigr)\ge0$, since $R(T-s)^{-1}$ is positive definite (Invertibility of Symmetric Positive Definite Matrices), positive semidefiniteness including the value $0$ at $y=0$. Each $\check C(s)$ is symmetric positive semidefinite by hypothesis (evaluated at $T-s$; symmetry again by the transpose rules and symmetry of $Q$ and $R^{-1}$), and $F$ is symmetric positive semidefinite.

Step 2: an algebraic identity. For every symmetric real $l\times l$ matrix $S$ and every $r\in[0,T]$:

Indeed, $(SB+V)^{\top}=B^{\top}S^{\top}+V^{\top}=B^{\top}S+V^{\top}$ by symmetry of $S$, so

and the right-hand side of ($\dagger$) equals $A^{\top}S+SA-SBR^{-1}B^{\top}S-SBR^{-1}V^{\top}-VR^{-1}B^{\top}S-VR^{-1}V^{\top}+Q$ (all at $r$), which regroups exactly as $\check A(T-r)S+S\check A(T-r)^{\top}-S\check D(T-r)S+\check C(T-r)$, since $\check A(T-r)=A(r)^{\top}-V(r)R(r)^{-1}B(r)^{\top}$, $\check A(T-r)^{\top}=A(r)-B(r)R(r)^{-1}V(r)^{\top}$ (symmetry of $R^{-1}$), $\check D(T-r)=B(r)R(r)^{-1}B(r)^{\top}$, and $\check C(T-r)=Q(r)-V(r)R(r)^{-1}V(r)^{\top}$.

Step 3: existence. By Global Existence and Uniqueness for the Kalman Covariance Riccati Equation applied on $[0,T]$ with data $\check A$, $\check C$, $\check D$, and initial matrix $F$, there is exactly one assignment $P$ with continuous entries such that

and every $P(s)$ is symmetric positive semidefinite. Define $Z(t):=P(T-t)$; its entries are continuous (composition with the continuous map $t\mapsto T-t$) and every $Z(t)$ is symmetric positive semidefinite. Let $g(r)$ denote the integrand of the forward equation, with continuous entries, so that $Z(t)=P(T-t)=F+\int_0^{T-t}g(\rho)\,d\rho$. By claim 2 of Product Rule and Reflection for Indefinite Riemann Integrals on $[0,T]$ applied entrywise to the continuous function $r\mapsto g(T-r)$,

so $Z(t)=F+\int_t^T g(T-r)\,dr$. By ($\dagger$) with the symmetric matrix $S=P(T-r)=Z(r)$,

so $Z$ satisfies the backward Riccati equation.

Step 4: every continuous solution is symmetric. Let $Z'$ be any assignment with continuous entries satisfying the backward equation. Its integrand $g'(r):=A(r)^{\top}Z'(r)+Z'(r)A(r)-(Z'(r)B(r)+V(r))R(r)^{-1}(Z'(r)B(r)+V(r))^{\top}+Q(r)$ has continuous entries. Transposing the backward equation entrywise (transposition commutes with entrywise integration, being a relabeling of entries) and using $F^{\top}=F$, $Q^{\top}=Q$, the transpose rules, and the fact that $MR^{-1}M^{\top}$ is symmetric for any $l\times k$ matrix $M$ (symmetry of $R^{-1}$), we find that $Z'^{\top}$ satisfies

with the same quadratic term. Subtracting, the difference $\Delta:=Z'-Z'^{\top}$ satisfies the homogeneous linear equation $\Delta(t)=\int_t^T\bigl(A(r)^{\top}\Delta(r)+\Delta(r)A(r)\bigr)\,dr$. Reflect: by claim 2 of Product Rule and Reflection for Indefinite Riemann Integrals, $\check\Delta(s):=\Delta(T-s)$ satisfies $\check\Delta(s)=\int_0^s\bigl(A(T-\rho)^{\top}\check\Delta(\rho)+\check\Delta(\rho)A(T-\rho)\bigr)\,d\rho$. Set $\delta(s):=\sum_{i,j}|\check\Delta_{ij}(s)|$, a continuous function. The entries of $A$ are bounded on $[0,T]$ by Extreme Value Theorem on a Compact Interval; by the product entry bound (claim 2 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals) and the triangle inequality for Riemann integrals of continuous functions (claims 4-5 of the same lemma give the needed splitting and bounding of entrywise integrals), there is a real $C\ge0$ with $\delta(s)\le C\int_0^s\delta(\rho)\,d\rho$ for all $s$. By Gronwall's Lemma (Integral Form) with zero constant term, $\delta\equiv0$, i.e. $Z'(t)$ is symmetric for every $t$.

Step 5: uniqueness. Let $Z'$ be as in Step 4, so every $Z'(t)$ is symmetric, and define $P'(s):=Z'(T-s)$, with continuous entries. Evaluating the backward equation at $T-s$ and applying claim 2 of Product Rule and Reflection for Indefinite Riemann Integrals to the continuous $g'$ itself, $\int_{T-s}^{T}g'(r)\,dr=\int_0^{s}g'(T-\rho)\,d\rho$, whence

By ($\dagger$) with the symmetric matrix $S=Z'(T-\rho)=P'(\rho)$, read from right to left, $g'(T-\rho)=\check A(\rho)P'(\rho)+P'(\rho)\check A(\rho)^{\top}-P'(\rho)\check D(\rho)P'(\rho)+\check C(\rho)$, so $P'$ satisfies the forward equation of Step 3. By the uniqueness assertion of Global Existence and Uniqueness for the Kalman Covariance Riccati Equation, $P'=P$, hence $Z'=Z$. $\square$