Endpoint maps and exact steering from a null-control operator

Put $A(r)=W(T,r)B(r)$. By F10, $\mathrm{\parallel }A(r)u\mathrm{\parallel }\le \mathrm{\parallel }W(T,r)\mathrm{\parallel }\mathrm{\parallel }B(r)u\mathrm{\parallel }\le MK\mathrm{\parallel }u\mathrm{\parallel }$, and taking the unit-ball supremum gives $\mathrm{\parallel }A(r)\mathrm{\parallel }\le MK$. For fixed $u\in U$, the increment of $A(r)u$ at $r_0$ is bounded by $M\mathrm{\parallel }B(r)u-B(r_0)u\mathrm{\parallel }+\mathrm{\parallel }(W(T,r)-W(T,r_0))B(r_0)u\mathrm{\parallel }$. Both terms tend to zero by the stated strong continuity hypotheses. Thus $r\mapsto A(r)u$ is continuous for every fixed $u$, without a separability hypothesis on $E$.

Let $Z_B$ denote the Borel-time space of F1 and let $Z$ use the completed time measure of F7. A Borel-strongly-measurable representative is also Lebesgue-strongly-measurable. Conversely, let $h$ be a representative in $Z$, and choose finite-range Lebesgue simple approximants $s_n$ converging to $h$ outside a Lebesgue null set, as in F6. Apply F8 to their countably many level sets and choose relative Borel replacements. Disjointifying each finite family, and putting zero on any remainder, gives finite-range Borel simple functions $t_n$ equal to $s_n$ off Borel null sets. Choose also a Borel null cover of the original exceptional set. Their countable union $N$ is Borel and null by F8. Define $\tilde{h}=h$ off $N$ and $\tilde{h}=0$ on $N$. The Borel simple functions $t_n{\mathbf{1}}_{N^c}$ converge pointwise in norm to $\tilde{h}$ everywhere, so $\tilde{h}$ is Borel-strongly-measurable and represents the original Lebesgue class. Two Borel-time representatives which agree Lebesgue-almost everywhere agree outside a Borel null set by F8, so this class map is also injective. It preserves sums and real scalar multiples. Scalar norm and inner-product integrals agree by simple approximation, since the completed measure agrees with the Borel measure on Borel sets; exceptional sets can be assigned zero before taking these integrals. Thus the class map is an onto real-linear isometry preserving the integral inner product. Transporting F1 through this map proves that the specified $Z$ is Hilbert and that the same finite-valued interval-step controls are dense in $Z$. No restriction of the admissible Lebesgue control classes is imposed.

For $n\ge 1$, partition $[0,T)$ into the $2^n$ intervals $I_{n,k}=[kT{2}^{-n},(k+1)T{2}^{-n})$ and put $q_n(r)=kT{2}^{-n}$ on $I_{n,k}$, with $q_n(T)=T$. Then $\mathrm{\mid }{q}_n(r)-r\mathrm{\mid }\le T{2}^{-n}$. If $f:[0,T]\to Y$ is a continuous curve in any Banach space $Y$, the function $f(q_n(r))$ is finite-valued and measurable: its level sets are unions of these measurable interval cells, possibly with the singleton $\{T\}$. Continuity gives $f(q_n(r))\to f(r)$ for every $r$. These are explicit simple approximants, so $f$, and its restriction to $(0,T)$, are strongly measurable by F6. In particular this applies to every curve $A(r)u$ from step 1.1.

For a strongly measurable representative $h$ of a class in $Z$, F6 supplies measurable finite-range simple functions $s_n$ with $s_n(r)\to h(r)$ off a measurable null set. Define $g_n(r)=A(q_n(r))s_n(r)$ on $(0,T)$. If $s_n$ takes the value $u_j$ on a measurable cell $H_{n,j}$, then $g_n$ takes the constant value $A(kT{2}^{-n})u_j$ on $H_{n,j}\cap I_{n,k}$; merging coincident values gives a measurable finite-range simple function. At each $r$ outside the exceptional null set, $\mathrm{\parallel }{g}_n(r)-A(r)h(r)\mathrm{\parallel }\le MK\mathrm{\parallel }{s}_n(r)-h(r)\mathrm{\parallel }+\mathrm{\parallel }(A(q_n(r))-A(r))h(r)\mathrm{\parallel }\to 0$. The second term tends to zero by step 1.1 applied to the fixed vector $h(r)$, while the first tends to zero by the chosen approximation. Therefore $r\mapsto A(r)h(r)$ is strongly measurable directly by F6, even when $E$ is not separable.

For $f\in C([0,T];U)$, step 2.1 gives strong measurability of its restriction to $(0,T)$, while F7 gives ${\int }_0^T\mathrm{\parallel }f(r){\mathrm{\parallel }}_U^{2}dr\le T\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}^2<\mathrm{\infty }$. Hence $if=[f{\mathrm{\mid }}_{(0,T)}]$ defines a map $i:C([0,T];U)\to Z$ with $\mathrm{\parallel }if{\mathrm{\parallel }}_Z\le \sqrt{T}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Restriction and passage to equivalence classes preserve pointwise sums and real scalar multiples, so $i$ is bounded and real-linear. Every occurrence of $Lf$ for a continuous control consequently means $L(if)$.

Choose the representative $h$ to be zero on the exceptional null set of its strong simple approximation, as allowed by F6. The scalar function $f(r)=\mathrm{\parallel }h(r){\mathrm{\parallel }}_U$ is measurable: the norms of the simple approximants, also set to zero on that set, are measurable scalar simple functions converging pointwise to $f$. By the definition of $Z$, ${\int }_0^{T}f(r{)}^{2}dr=\mathrm{\parallel }h{\mathrm{\parallel }}_Z^2<\mathrm{\infty }$. The constant function $g(r)=1$ is measurable with ${\int }_0^{T}g(r{)}^{2}dr=T<\mathrm{\infty }$ for the time measure in F7. Apply F9 to $f,g$ on this restricted Lebesgue measure space, with conjugate exponents $p=q=2$, to obtain ${\int }_0^T\mathrm{\parallel }h(r){\mathrm{\parallel }}_{U}dr\le \sqrt{T}\mathrm{\parallel }h{\mathrm{\parallel }}_Z$. By step 1.1 and F10, $\mathrm{\parallel }A(r)h(r){\mathrm{\parallel }}_E\le MK\mathrm{\parallel }h(r){\mathrm{\parallel }}_U$; together with step 3.1, this gives the strong measurability and integrable norm needed in F2. Hence the endpoint integrand is Bochner integrable and $\mathrm{\parallel }Lh\mathrm{\parallel }\le MK\sqrt{T}\mathrm{\parallel }h{\mathrm{\parallel }}_Z$. Linearity follows from the integral's linearity. If $h$ and $\tilde{h}$ agree almost everywhere, their endpoint integrands also agree there; the integral norm bound applied to their difference gives $Lh=L\tilde{h}$. Thus $L$ is a well-defined bounded real-linear map on $Z$.

Put $Ov(r)=B(r{)}^{\ast }W(T,r{)}^{\ast }v$. By F3, both adjoints are bounded and real-linear, with $\mathrm{\parallel }B(r{)}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }B(r)\mathrm{\parallel }\le K$ and $\mathrm{\parallel }W(T,r{)}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }W(T,r)\mathrm{\parallel }\le M$. Thus $\mathrm{\parallel }Ov(r)\mathrm{\parallel }\le MK\mathrm{\parallel }v\mathrm{\parallel }$. The assumed strong measurability gives $Ov\in Z$, and pointwise real linearity and this estimate make $O:E\to Z$ bounded and real-linear. For a finite-valued step control $h$, commute the bounded functional $x\mapsto (x,v{)}_E$ with the Bochner integral and use the defining adjoint identities pointwise to get $(Lh,v{)}_E={\int }_0^T(h(r),Ov(r){)}_{U}dr=(h,Ov{)}_Z$. Density of such controls and the bounds for $L$ and $O$ extend this equality to every $h\in Z$. Uniqueness of the Hilbert adjoint therefore gives $O=L^{\ast }$.

The evolution identities give $W(T,0)W(0,T)=I_E$. With $y=x_0-W(0,T)x_T$, the null-control hypothesis gives $LNy=-W(T,0)y$. Consequently $W(T,0)x_0+LN(y)=W(T,0)W(0,T)x_T=x_T$. The composition defining $G_N$ is bounded and real-linear.

Apply [F4] with observation $O=L^{\ast }$ and observation space $Z$. Its adjoint is $L$, because the identity in step 5.1 also characterizes $O^{\ast }$ by symmetry of the real inner products. Thus $Q=O^{\ast }O=L{L}^{\ast }$ is invertible, $R=O{Q}^{-1}$ obeys $LR=I_E$, and $\mathrm{\parallel }R\mathrm{\parallel }\le c^{-1}$. Substitution into the endpoint formula gives $W(T,0)x_0+LG(x_0,x_T)=x_T$; boundedness follows from the bound for $R$ and boundedness of $W(T,0)$.

If $J$ is supplied, [F4] gives $R_C=J{Q}^{-1}$, a bounded map into $C([0,T];U)$ with $i{R}_C=R$. Step 3.2 supplies the bounded linear class map $i$ required in F4, so $L{R}_C=L(i{R}_C)=LR=I_E$. The same substitution proves continuous-control steering and boundedness of $G_C$.

If $B=-i{M}_{\phi }$, the endpoint formula in step 4.1 has integrand $W(T,r)(-i\phi h(r))$ by substitution. The proof used only real linearity, so it does not move $i$ through $W(T,r)$. This proves the stated conditional Schrödinger formulation.