Remark 5: controllability above the Sobolev algebra threshold

Set $X_T=C([0,T];H)$ with the supremum norm, which is finite and satisfies the norm axioms by the time-space definition in [F1]. The interval is nonempty and $H$ is complete. Thus [F6] makes $X_T$ complete in $\rho$. Since $\rho (a,b)\le \mathrm{\parallel }a-b{\mathrm{\parallel }}_{X_T}$, any supremum-norm Cauchy sequence has a $\rho$ limit in $X_T$. If $\rho (a,b)<1$, then $\min \{\mathrm{\parallel }a(t)-b(t){\mathrm{\parallel }}_H,1\}\le \rho (a,b)<1$ for each $t$, so no pointwise distance is truncated and $\mathrm{\parallel }a-b{\mathrm{\parallel }}_{X_T}=\rho (a,b)$. Consequently convergence in $\rho$ implies convergence in the supremum norm, proving that $X_T$ is a complex Banach space. Its inclusion into itself has norm one. Unitarity gives $\mathrm{\parallel }W(\cdot )v_0{\mathrm{\parallel }}_{X_T}=\mathrm{\parallel }{v}_0{\mathrm{\parallel }}_H$, while the retarded estimate in [F3] gives the second abstract hypothesis.

Write $g(z)=\lambda z^{(m+1)\mathrm{/}2}{\bar{z}}^{(m-1)\mathrm{/}2}$. The scale identification, algebra estimate and conjugation invariance imply $\mathrm{\parallel }g(v){\mathrm{\parallel }}_H\le C\mathrm{\parallel }v{\mathrm{\parallel }}_H^m$ and, by telescoping the $m$ factors, $\mathrm{\parallel }g(v)-g(u){\mathrm{\parallel }}_H\le C(\mathrm{\parallel }v{\mathrm{\parallel }}_H^{m-1}+\mathrm{\parallel }u{\mathrm{\parallel }}_H^{m-1})\mathrm{\parallel }v-u{\mathrm{\parallel }}_H$. The energy rule shows $g(v)\in H_0^1$, so the scale identification makes these estimates estimates in $H$, not merely in $H^s$. The same difference estimate makes $t\mapsto g(v(t))$ norm-continuous in $H$; hence it is Bochner integrable on $[0,T]$. Integrating the pointwise bounds gives the third hypothesis with a factor $T$, and one sufficiently large constant works in all three hypotheses.

Apply the abstract control theorem to $X_T$ for the chosen $m$. It supplies ${\delta }_1$, the continuous control $k$ and an integral solution joining the prescribed endpoints. The cutoff gives the stated support. Put $f=ig(v)-i\phi k$. By the integral-solution formulation, $f$ is Bochner integrable in $H$. Moreover $f$ is continuous in $L^2$: step 2.1 makes $g(v)$ continuous in $H$, the spectral norm satisfies $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le \mathrm{\parallel }x{\mathrm{\parallel }}_H$ because $A\ge I$, and multiplication by $\phi$ has $L^2$ norm at most one. In particular the nonlinear and control terms are locally integrable.

Let $B$ be the generator on $H$. For every $x\in D(B)$ the right difference quotients converge to $Bx$ in $H$, hence in $L^2$ by the preceding norm inequality. The scale group is consistent with the $L^2$ group by [F3]. Its negative difference quotient, for $h>0$, is $(W(-h)x-x)\mathrm{/}(-h)=W(-h)(W(h)x-x)\mathrm{/}h$, which also converges in $L^2$ to $Bx$ by unitarity and strong continuity. Apply the mathematical flow-calculus lemma in [F7] to the self-adjoint ${\mathrm{\Delta }}_D$ on $L^2(\mathrm{\Omega })$: the two-sided limit forces $x\in D({\mathrm{\Delta }}_D)$ and $Bx=i{\mathrm{\Delta }}_{D}x$ in $L^2$. Thus this identification holds on the full $H$-generator domain, and its action there is $i\mathrm{\Delta }x$ in spatial distributions.

Write $U(t)={\int }_0^{t}v(r)dr$ and $P(t)={\int }_0^{t}f(r)dr$. By [F4], $U(t)\in D(B)$ and $v(t)=v_0+BU(t)+P(t)$. The continuous inclusion $H\to L^2$ commutes with these integrals by [F8]. For either continuous $L^2$ path $q=v$ or $q=f$, the norm of $({\int }_t^{t+h}q(r)dr)\mathrm{/}h-q(t)$ is at most ${\sup }_{r\text{ between }t\text{ and }t+h}\mathrm{\parallel }q(r)-q(t){\mathrm{\parallel }}_2$, which tends to zero. Hence $U^{\mathrm{\prime }}=v$ and $P^{\mathrm{\prime }}=f$ in $L^2$ on $(0,T)$. Step 4.1 turns the integrated identity into $v=v_0+i{\mathrm{\Delta }}_{x}U+P$ in space-time distributions. Differentiating it in time gives ${\mathrm{\partial }}_{t}v=i{\mathrm{\Delta }}_{x}v+f$: time and spatial distributional derivatives commute, as follows by applying their test-function definitions in either order. Substitution of $f$ yields $i{\mathrm{\partial }}_{t}v+{\mathrm{\Delta }}_{x}v+g(v)=\phi k$. Explicitly, for every $\eta \in C_c^{\mathrm{\infty }}((0,T)\times \mathrm{\Omega };\mathbb{C})$, the bilinear test identity is ${\int }_0^T{\int }_{\mathrm{\Omega }}[-iv{\mathrm{\partial }}_t\eta +v{\mathrm{\Delta }}_x\eta +(g(v)-\phi k)\eta ]dxdt=0$. Local integrability from step 3.1 makes each term meaningful.

Extend each boundary endpoint by $E$. Boundedness of $E$ makes both exterior endpoints smaller than ${\delta }_1$ when their $X^s$ norms are sufficiently small. Apply steps 3.1 and 5.1 and then the boundary-transfer proposition: restriction gives the equation and endpoint values, the obstacle trace vanishes, and the outer trace is a continuous $H^{s-1\mathrm{/}2}({\mathrm{\Gamma }}_1)$ control. This applies separately to $m=3$ and $m=5$, completing both assertions.