Decaying Dirichlet kernel and the exact closed-component exception

On the component containing the end, truncate at a sufficiently large coordinate sphere $r=R$. No component of this truncation can be closed: a boundaryless component would be open and closed in the connected end component and separated from its nonempty end. Put $a_R={\sup }_{r=R}\mathrm{\mid }u\mathrm{\mid }$. On the truncation, both $u-a_R$ and $-u-a_R$ have nonpositive boundary values and satisfy $({\mathrm{\Delta }}_h-V)(\pm u-a_R)=V{a}_R\ge 0$. Comparison gives $\mathrm{\mid }u\mathrm{\mid }\le a_R$. Weighted decay makes $a_R\to 0$, so $u=0$ there. The same comparison on each compact component with nonempty boundary gives zero directly, using $a_R=0$.

Let $C$ be a closed compact connected component. On each coordinate chart of $C$, put $A^{ij}=m{h}^{ij}$. Applying only the pointwise operator formula in [F4], with its auxiliary potential zero, and expanding the derivative gives ${\mathrm{\Delta }}_{h}u=m^{-1}{\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)$. Hence the actual equation with the stated $V$ gives $-{\mathrm{\partial }}_i(A^{ij}{\mathrm{\partial }}_{j}u)+mVu=0$. Here $u$ is $C^{4,\alpha }$ locally and $V$ is continuous; only multiplication by $V$ is used, and no derivative of $V$ is taken. This application therefore does not require the actual potential to be smooth.

Choose a finite cover of $C$ by relatively compact coordinate neighborhoods $U_a$. By [F1] and [F6] choose a subordinate smooth partition ${\chi }_a$; since $C$ is compact, each support is compactly contained in its chart. In the $a$th chart multiply the equation of step 1.2 by ${\chi }_{a}u$. The coordinate products ${\chi }_{a}u{A}^{ij}{\mathrm{\partial }}_{j}u$ are $C^1$ with compact support in the open chart image. After extension by zero, integration of each $i$th derivative first in the $i$th coordinate gives zero by the fundamental theorem of calculus. Expanding those derivatives and using the density of [F5] therefore gives ${\int }_C({\chi }_a\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}_h^2+u{h}^{ij}{\mathrm{\partial }}_{j}u{\mathrm{\partial }}_i{\chi }_a+{\chi }_{a}V{u}^2)d{\mu }_h=0$, where $\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}_h^2=h^{ij}{\mathrm{\partial }}_{i}u{\mathrm{\partial }}_{j}u$ and the integrands are zero off the chart support. The middle term is the intrinsic inverse-metric pairing of $du$ with $d{\chi }_a$, multiplied by $u$. Sum these finitely many identities: ${\sum }_a{\chi }_a=1$ and ${\sum }_{a}d{\chi }_a=0$ cancel the middle terms and give ${\int }_C(\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}_h^2+V{u}^2)d{\mu }_h=0$. The positive density makes this argument valid even when $C$ is nonorientable.

The two integrands in the energy identity are continuous and nonnegative. If either were positive at a point, it would have positive integral on a smaller coordinate ball because the density is positive. Thus $du=0$ and $V{u}^2=0$ everywhere on $C$. The fundamental theorem of calculus along chart segments makes $u$ locally constant, so connectedness makes it constant on $C$. That constant is zero if $V$ is nonzero anywhere on $C$; if $V\equiv 0$ every constant solves the equation. Every component other than the end component lies in the compact core and is closed, hence compact. There are only finitely many such components: the open components of the manifold cover the compact core and have a finite subcover. Functions that are constant on any of the closed components with $V\equiv 0$ and zero on all other components are locally smooth, vanish on the boundary and on the end, and have finite norm on the fixed compact chart subsets. They therefore belong to $X_{\beta }$ and solve the equation. Combined with step 1.1, this proves the converse, the exact kernel classification and the connected-exterior conclusion.