Limiting Carleman estimate and right solvability on a product domain

Put $s=\phi$ and first set $q=0$. For a real smooth weight $\psi$, direct differentiation gives $P_{\psi ,h,0}=e^{\psi \mathrm{/}h}(-h^2{\mathrm{\Delta }}_g)e^{-\psi \mathrm{/}h}=-h^2{\mathrm{\Delta }}_g-\mathrm{\mid }{\mathrm{\nabla }}_g\psi {\mathrm{\mid }}^2+2h{\mathrm{\nabla }}_g\psi \cdot {\mathrm{\nabla }}_g+h{\mathrm{\Delta }}_g\psi \mathrm{.}$ For $\psi =s$, $\mathrm{\mid }\mathrm{\nabla }s\mathrm{\mid }=1$ and $\mathrm{\Delta }s=0$, so $P_{s,h,0}=A+C$ with $A=-h^2{\mathrm{\Delta }}_g-1$ self-adjoint and $C=2h{\mathrm{\partial }}_s$ skew-adjoint on compactly supported functions.

Convexify by $\psi =f(s)=s+\frac{h}{2\epsilon }{s}^2$, where $\epsilon >0$ is fixed small and then $h\mathrm{/}\epsilon$ is small. Since $M$ is compact, choose $h$ so $f^{\mathrm{\prime }}=1+hs\mathrm{/}\epsilon >1\mathrm{/}2$ on $M$. The conjugated operator is $P_{\psi ,h,0}=A_{\epsilon }+C_{\epsilon }$ with $A_{\epsilon }=-h^2{\mathrm{\Delta }}_g-(f^{\mathrm{\prime }}{)}^2,C_{\epsilon }=2h{f}^{\mathrm{\prime }}{\mathrm{\partial }}_s+h{f}^{\mathrm{\prime }\mathrm{\prime }},f^{\mathrm{\prime }\mathrm{\prime }}=h\mathrm{/}\epsilon \mathrm{.}$ Here $A_{\epsilon }$ is self-adjoint and $C_{\epsilon }$ is skew-adjoint. Product independence of $s$ makes $-{\mathrm{\Delta }}_g=-{\mathrm{\partial }}_s^2-{\mathrm{\Delta }}_{g_0}$, with ${\mathrm{\Delta }}_{g_0}$ commuting with $C_{\epsilon }$. Direct differentiation in the $s$ coordinate therefore gives the exact commutator $[A_{\epsilon },C_{\epsilon }]=4h(f^{\mathrm{\prime }}{)}^2{f}^{\mathrm{\prime }\mathrm{\prime }}-4h^3{f}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\partial }}_s^2\mathrm{.}$ For compactly supported $w$, integration by parts yields $\mathrm{\parallel }{P}_{\psi ,h,0}w{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{A}_{\epsilon }w{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{C}_{\epsilon }w{\mathrm{\parallel }}_2^2+4h{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }(f^{\mathrm{\prime }})w{\mathrm{\parallel }}_2^2+4h^3{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }{\mathrm{\partial }}_{s}w{\mathrm{\parallel }}_2^2\ge c{h}^2{\epsilon }^{-1}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2\mathrm{.}$ The same identity implies $\mathrm{\parallel }{A}_{\epsilon }w{\mathrm{\parallel }}_2\le \mathrm{\parallel }{P}_{\psi ,h,0}w{\mathrm{\parallel }}_2$. Since $\mathrm{\parallel }h\mathrm{\nabla }w{\mathrm{\parallel }}_2^2=⟨A_{\epsilon }w,w⟩+\mathrm{\parallel }{f}^{\mathrm{\prime }}w{\mathrm{\parallel }}_2^2$, the $L^2$ bound just obtained gives $\mathrm{\parallel }w{\mathrm{\parallel }}_2+\mathrm{\parallel }h\mathrm{\nabla }w{\mathrm{\parallel }}_2\le C_{\epsilon }{h}^{-1}\mathrm{\parallel }{P}_{\psi ,h,0}w{\mathrm{\parallel }}_2$.

The convexified and unconvexified weights are related by multiplication with $m=e^{s^2\mathrm{/}(2\epsilon )}$: $P_{\psi ,h,0}(mw)=m{P}_{s,h,0}w$. On compact $M$, $m$ and $m^{-1}$ are bounded, so step 2.1 yields $\mathrm{\parallel }w{\mathrm{\parallel }}_2\le C{h}^{-1}\mathrm{\parallel }{P}_{s,h,0}w{\mathrm{\parallel }}_2$. Adding the potential changes the conjugated operator by $h^{2}q$. Its norm on $w$ is at most $h^2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }w{\mathrm{\parallel }}_2$, which is absorbed in the preceding estimate provided $C_{0}h\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}2$, where $C_0$ is the bound for $q=0$. Choose $h_0>0$ below this threshold and the bound ensuring $h_0{\max }_M\mathrm{\mid }{x}_1\mathrm{\mid }\mathrm{/}\epsilon <1\mathrm{/}2$ in step 2.1. Taking the smaller threshold and larger constant for the two signs proves the estimate uniformly for every $0<h\le h_0$.

Let $P=P_{\phi ,h,q}$ and apply the a priori estimate also to $P^{\ast }=P_{-\phi ,h,q}$, which has the same real potential and the opposite allowed sign. On $E=P^{\ast }(C_c^{\mathrm{\infty }}(M^{\circ }))\subset L^2(M)$ define the complex-linear functional $\mathrm{\ell }(P^{\ast }w)={\int }_M\bar{F}wd{V}_g$. The estimate makes this well-defined: $P^{\ast }w=0$ implies $w=0$. It also gives $\mathrm{\mid }\mathrm{\ell }(P^{\ast }w)\mathrm{\mid }\le \mathrm{\parallel }F{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_2\le C{h}^{-1}\mathrm{\parallel }F{\mathrm{\parallel }}_2\mathrm{\parallel }{P}^{\ast }w{\mathrm{\parallel }}_2$. By [A1] extend $\mathrm{\ell }$ to $L^2(M)$ with the same bound; by [A2] there is $R\in L^2(M)$ with $\mathrm{\ell }(z)={\int }_{M}z\bar{R}d{V}_g$ and $\mathrm{\parallel }R{\mathrm{\parallel }}_2\le C{h}^{-1}\mathrm{\parallel }F{\mathrm{\parallel }}_2$. Taking complex conjugates gives ${\int }_{M}R\overline{P^{\ast }w}d{V}_g={\int }_{M}F\bar{w}d{V}_g$ for every test $w$, which is precisely $PR=F$ distributionally on $M^{\circ }$.