Weak Laplace single-layer potentials and energy spaces

Write $G_3$ for the Newton kernel of The three-dimensional Laplace Newton kernel. For a smooth density $\lambda$ on $\mathrm{\Gamma }$, the classical single-layer potential is $S\lambda (x)={\int }_{\mathrm{\Gamma }}{G}_3(x,y)\lambda (y)d{S}_y,x\in {\mathbb{R}}^3\setminus \mathrm{\Gamma }\mathrm{.}$

Let ${\dot{H}}^1({\mathbb{R}}^3)$ be the completion of $C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ in the norm $\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{H}}^1}=\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2({\mathbb{R}}^3)}$, using its $L^6$ representative and the trace from The H1 trace and lifting theorem on a compact Lipschitz boundary. For $\lambda \in H^{-1\mathrm{/}2}(\mathrm{\Gamma })$, consider the equation for $u\in {\dot{H}}^1({\mathbb{R}}^3)$ ${\int }_{{\mathbb{R}}^3}\mathrm{\nabla }u\cdot \mathrm{\nabla }vdx=⟨\lambda ,\gamma v{⟩}_{\mathrm{\Gamma }}(v\in {\dot{H}}^1({\mathbb{R}}^3))\mathrm{.}$ Call such a function a weak single-layer potential and write $S\lambda =u$ when it exists. Define the exterior energy space $E^{+}({\mathrm{\Omega }}^{+})=\{u\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathrm{\Omega }}^{+}):u\in L^6({\mathrm{\Omega }}^{+}),\mathrm{\nabla }u\in L^2({\mathrm{\Omega }}^{+})\}$. For a weakly harmonic function $u^{-}\in H^1({\mathrm{\Omega }}^{-})$, define its weak outward conormal by $⟨{\mathrm{\partial }}_n^{-}{u}^{-},g⟩={\int }_{{\mathrm{\Omega }}^{-}}\mathrm{\nabla }{u}^{-}\cdot \mathrm{\nabla }{L}^{-}g$, where $L^{-}g$ is any $H^1$ trace lift. For a harmonic $u^{+}\in E^{+}({\mathrm{\Omega }}^{+})$, use the same fixed normal $n$ by $⟨{\mathrm{\partial }}_n^{+}{u}^{+},g⟩=-{\int }_{{\mathrm{\Omega }}^{+}}\mathrm{\nabla }{u}^{+}\cdot \mathrm{\nabla }{L}^{+}g$ for a compactly supported local lift. In either case the value is independent of the lift: the difference of two lifts has zero trace, and weak harmonicity makes its gradient pairing zero. The exterior formula also gives ${\int }_{{\mathrm{\Omega }}^{+}}\mathrm{\nabla }{u}^{+}\cdot \mathrm{\nabla }{v}^{+}=-⟨{\mathrm{\partial }}_n^{+}{u}^{+},{\gamma }^{+}{v}^{+}⟩$ for every $v^{+}\in E^{+}({\mathrm{\Omega }}^{+})$. Indeed, subtract a compactly supported lift of ${\gamma }^{+}{v}^{+}$; the remaining zero-trace finite-energy function is approximated in gradient norm by compactly supported zero-trace functions using radial cutoffs: $\mathrm{\parallel }{v}^{+}\mathrm{\nabla }{\chi }_R{\mathrm{\parallel }}_2\le C\mathrm{\parallel }{v}^{+}{\mathrm{\parallel }}_{L^6(\{R<\mathrm{\mid }x\mathrm{\mid }<2R\})}\to 0$. Boundary-chart flattening, zero extension, inward translation, and convolution approximate each zero-trace cutoff by interior smooth tests, so weak harmonicity makes its gradient pairing zero. This proves the energy identity without a flux assumption at infinity. Existence, uniqueness, agreement with the displayed kernel for smooth densities, and the conormal jump are proved in The three-dimensional single layer is bounded and coercive ↗. The weak definition fixes the normalization at infinity through $u\in L^6({\mathbb{R}}^3)$; arbitrary additive constants are excluded.