The backward LQR adjoint equation has a unique solution

The forcing $g(s):=Q(T-s)(x_u(T-s)-x_d(T-s))$ lies in $L^2(0,T;{\mathbb{R}}^n)$ by [F6]: $Q$ is essentially bounded, $x_u$ is continuous on a compact interval and hence bounded, and $x_d\in L^2$. Also $C(s):=A(T-s{)}^T\in L^{\mathrm{\infty }}$ by [F6]. Apply [F1] with $D(s)=I$ and initial value $q(0)=P(x_u(T)-x_T)$ to obtain an absolutely continuous $q$ satisfying the forward Volterra integral equation and $\dot{q}=Cq+g$ almost everywhere.

Set $p(t)=q(T-t)$. Each coordinate remains absolutely continuous because reflection preserves interval lengths and disjoint interiors in the defining finite-family condition of Absolute continuity on a compact interval. At points where the coordinate derivatives exist, the difference-quotient definition and the chain rule for $t\mapsto T-t$ give $\dot{p}(t)=-\dot{q}(T-t)$; [F6] carries the exceptional null set to a null set. The forward equation for $q$ therefore gives $-\dot{p}(t)=A(t{)}^{T}p(t)+Q(t)(x_u(t)-x_d(t))$ almost everywhere, and $p(T)=q(0)=P(x_u(T)-x_T)$. Thus $p$ satisfies [F2].

If $\tilde{p}$ is another adjoint solution, reflection gives a coordinatewise absolutely continuous $\tilde{q}(s)=\tilde{p}(T-s)$ as in step 2.1, with $\tilde{q}(0)=q(0)$ and $\dot{\tilde{q}}=C\tilde{q}+g$ almost everywhere by [F6]. By [F4], $\tilde{q}$ is continuous and bounded componentwise. Since $C\in L^{\mathrm{\infty }}$, $g\in L^2$ and [F5] gives $g\in L^1$, the function $h=C\tilde{q}+g$ belongs to $L^1$ componentwise. Apply [F3] to obtain $\tilde{q}(s)=q(0)+{\int }_0^s(C(r)\tilde{q}(r)+g(r))dr$ for every $s$. This is exactly the forward Volterra equation used to construct $q$ in step 1.1. Its actual integral-equation uniqueness in [F1] gives $\tilde{q}=q$, hence $\tilde{p}=p$.