Smooth-manifold elliptic nonresonant solvability and regularity

Choose a real constant $Q>1+\mathrm{\parallel }q{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(M)}$ and let $A=-{\mathrm{\Delta }}_g+Q$ on $H_0^1(M)$, with form $a(w,z)={\int }_M(g(\mathrm{\nabla }w,\mathrm{\nabla }\bar{z})+Qw\bar{z})d{V}_g$. The form is bounded and coercive in the chartwise $H^1$ norm, so [F2] gives a bounded inverse $A_D^{-1}:(H_0^1(M){)}^{\ast }\to H_0^1(M)$.

Multiplication by $Q-q$ maps $H_0^1(M)$ boundedly to $L^2(M)$, and the inclusion $H_0^1(M)\hookrightarrow L^2(M)$ is compact: localize with the finite atlas and partition, apply [F3] on each bounded chart piece (using zero extension at a boundary face), then extract a finite diagonal subsequence and sum the localized limits. Hence $K:=A_D^{-1}(Q-q)$ is compact on $H_0^1(M)$.

For $w\in H_0^1(M)$, $(I-K)w=0$ is equivalent, after applying $A$, to $(-{\mathrm{\Delta }}_g+q)w=0$. The latter has only the zero solution by hypothesis, so $I-K$ is injective. By [F4], $I-K$ is boundedly invertible. This gives a unique weak zero-trace solution for every bounded functional on $H_0^1(M)$, in particular for every $F\in L^2(M)$ after subtracting any fixed boundary lift.

For $n=1$, the bounded endpoint trace, smooth finite-dimensional lift and zero-trace kernel follow from the fundamental theorem of calculus and Cauchy--Schwarz; specifically, $\mathrm{\mid }u(t){\mathrm{\mid }}^2\le t{\int }_0^t\mathrm{\mid }{u}^{\mathrm{\prime }}(r){\mathrm{\mid }}^{2}dr$ permits cutoff approximation at a zero endpoint, followed by interior mollification. For $n=0$ the boundary is empty, the trace space is zero and $H_0^1=H^1$. For $n\ge 2$, a finite collection of flattened boundary charts, the collar neighborhood, and [F5] give a bounded extension $E:H^{1\mathrm{/}2}(\mathrm{\partial }M)\to H^1(M)$ with trace $f$. Set $u=Ef+w$ and solve for $w\in H_0^1(M)$ with weak right-hand-side functional $z\mapsto {\int }_{M}F\bar{z}d{V}_g-a_q(Ef,z)$ by step 3.1. This proves existence for the stated data. The difference of any two solutions has zero global trace, so [F7] puts it in $H_0^1(M)$; the assumed trivial kernel then proves uniqueness.

First take zero boundary data and $F\in L^2$. For $n\ge 2$, in each boundary chart the collar flattens the boundary and the pulled-back coefficients are smooth and uniformly elliptic; localize $u$ by a cutoff, whose trace remains zero and hence lies in $H_0^1$ locally by [F7], and apply the $m=0$ case of [F6]. In interior charts apply [F8]. A finite cover gives $u\in H^2(M)$ and the estimate with the additional $L^2$ term. If $f$ is smooth, extend it smoothly in a boundary collar by keeping its chart representative constant in the normal variable near the boundary and multiplying by a cutoff; a finite partition patches these extensions to a smooth $Ef$. Then $w=u-Ef$ has zero trace and, by [F7], lies in $H_0^1(M)$; the same local estimates apply to $w$. For $n=1$, in a coordinate the weak equation has form $-{\mathrm{\partial }}_x(a{\mathrm{\partial }}_{x}u)+b{\mathrm{\partial }}_{x}u+cu=F$ with smooth $a>0$; it gives ${\mathrm{\partial }}_x(a{\mathrm{\partial }}_{x}u)\in L^2$, hence $u\in H^2$ with the corresponding estimate. For $n=0$, the equation is pointwise and the trivial-kernel hypothesis gives the estimate.

Fix $k\ge 1$ and suppose inductively that $u\in H^{k+1}(M)$ and $F\in H^k(M)$. If $n\ge 2$, in a boundary chart a cutoff localization has zero trace on the physical boundary and hence belongs locally to $H_0^1$ by [F7]; its equation has right-hand side $\chi F+[L,\chi ]u\in H^k$, since the commutator is first order with smooth coefficients. Apply [F6] at order $k$ to these boundary-localized equations; in interior charts apply [F8]. This gives $u\in H^{k+2}(M)$ with a bound by $C_k(\mathrm{\parallel }F{\mathrm{\parallel }}_{H^k}+\mathrm{\parallel }u{\mathrm{\parallel }}_{H^{k+1}}+\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2})$. If $n=1$, the coordinate equation in step 5.1 gives ${\mathrm{\partial }}_x(a{\mathrm{\partial }}_{x}u)=b{\mathrm{\partial }}_{x}u+cu-F$; under the induction hypothesis the right-hand side belongs to $H^k$, so $a{\mathrm{\partial }}_{x}u\in H^{k+1}$ and $u\in H^{k+2}$ with the analogous estimate. If $n=0$, all Sobolev orders coincide and the pointwise inverse gives it. For zero trace, the bounded inverse in step 3.1 controls $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$ by $\mathrm{\parallel }F{\mathrm{\parallel }}_{L^2}$; the induction hypothesis then absorbs the lower-order term and yields $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^{k+2}}\le C_k\mathrm{\parallel }F{\mathrm{\parallel }}_{H^k}$. For smooth nonzero $f$, apply the same induction to $u-Ef$. No differentiation of normal boundary values is used.