The bounded measurable linear ODE has a unique affine state map

For any Lebesgue-integrable vector function $h$ with $\int \mathrm{\parallel }h{\mathrm{\parallel }}_2<\mathrm{\infty }$, write $I=\int h$. If $I=0$, the norm inequality is immediate. Otherwise put $q=I\mathrm{/}\mathrm{\parallel }I{\mathrm{\parallel }}_2$. Then $\mathrm{\parallel }I{\mathrm{\parallel }}_2=\int q^{T}h\le \int \mathrm{\mid }{q}^{T}h\mathrm{\mid }\le \int \mathrm{\parallel }h{\mathrm{\parallel }}_2$, by [F1], linearity componentwise, and $\mathrm{\parallel }q{\mathrm{\parallel }}_2=1$. Thus the vector integral estimate needed below holds for these Lebesgue integrands. Now put $g=Bu\in L^1(0,T;{\mathbb{R}}^n)$; this follows from $B\in L^{\mathrm{\infty }}$, $u\in L^2$, and [F5]. Choose a finite partition $0=t_0<\cdots <t_N=T$ with $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}(t_{j+1}-t_j)<1$ on each subinterval when $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}>0$; if $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}=0$, use $N=1$. Given the already constructed value $x(t_j)$, define $\mathcal{T}z(t)=x(t_j)+{\int }_{t_j}^t(A(s)z(s)+g(s))ds$ on $C([t_j,t_{j+1}];{\mathbb{R}}^n)$. By [F2], $\mathcal{T}$ maps continuous functions to continuous functions, and the vector-integral estimate just proved gives $\mathrm{\parallel }\mathcal{T}z-\mathcal{T}w{\mathrm{\parallel }}_{\mathrm{\infty }}\le q_j\mathrm{\parallel }z-w{\mathrm{\parallel }}_{\mathrm{\infty }}$ with $q_j=\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}(t_{j+1}-t_j)<1$.

Starting from the constant function $z_0=x(t_j)$, let $z_{k+1}=\mathcal{T}{z}_k$. Then $\mathrm{\parallel }{z}_{k+1}-z_k{\mathrm{\parallel }}_{\mathrm{\infty }}\le q_j^k\mathrm{\parallel }{z}_1-z_0{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F7], the geometric majorant has finite sum, so the iterates are Cauchy in the supremum norm. Each coordinate converges uniformly by [F3], and the finite coordinate limit is a continuous vector function $z$. The integral estimate in step 1.1 makes $\mathcal{T}$ continuous in the supremum norm, hence $z=\mathcal{T}z$. If $z,w$ are two fixed points, $\mathrm{\parallel }z-w{\mathrm{\parallel }}_{\mathrm{\infty }}\le q_j\mathrm{\parallel }z-w{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $z=w$. Solving the finitely many subintervals in order gives one continuous solution on $[0,T]$, and the same uniqueness on each subinterval proves global uniqueness.

The integrand $Ax+Bu$ lies in $L^1$. Its indefinite integral is absolutely continuous by [F2], as required in the Statement. Under [A1], the $L^1$ fundamental theorem in [F2] differentiates the integral equation almost everywhere and yields $\dot{x}=Ax+Bu$. The same equation is independent of the chosen representatives of $A,B,u$ by [F6].

From the integral equation and step 1.1, $\mathrm{\parallel }x(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }{x}_0{\mathrm{\parallel }}_2+\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1}+\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_0^t\mathrm{\parallel }x(s){\mathrm{\parallel }}_{2}ds$. Apply [F4] with $c=\mathrm{\parallel }{x}_0{\mathrm{\parallel }}_2+\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1}$ and then [F5]; this gives the displayed supremum bound.

Linearity of the integral [F6] shows that a linear combination of two zero-initial-state solutions is the solution for the same linear combination of controls; uniqueness from step 2.1 makes the zero-initial-state map linear. Step 3.2 with $x_0=0$ bounds this linear map in the $L^2$ and supremum norms, while the fixed initial-state contribution is independent of $u$. Hence the full map is affine and bounded as claimed.