The three-dimensional single layer is bounded and coercive

For fixed $\lambda$, define $\mathrm{\ell }(v)=⟨\lambda ,\gamma v{⟩}_{\mathrm{\Gamma }}$ on ${\dot{H}}^1$. The trace estimate in [F1] gives $\mathrm{\mid }\mathrm{\ell }(v)\mathrm{\mid }\le C\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{{\dot{H}}^1}$. The gradient inner product makes ${\dot{H}}^1$ a real Hilbert space by its completion definition, so [F2] applied to $a(u,v)={\int }_{{\mathbb{R}}^3}\mathrm{\nabla }u\cdot \mathrm{\nabla }v$ gives a unique $u$ with $a(u,v)=\mathrm{\ell }(v)$ for every $v$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{H}}^1}\le C\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}$.

Testing with smooth functions compactly supported off $\mathrm{\Gamma }$ makes the right side zero, so $u$ is weakly harmonic in each side. The same weak equation for a global smooth test $v$ is the distributional identity $-\mathrm{\Delta }u=\lambda {\delta }_{\mathrm{\Gamma }}$, where $⟨\lambda {\delta }_{\mathrm{\Gamma }},v⟩=⟨\lambda ,\gamma v{⟩}_{\mathrm{\Gamma }}$. Splitting the energy integral across ${\mathrm{\Omega }}^{-}$ and ${\mathrm{\Omega }}^{+}$ and using the weak Green definitions in Weak Laplace single-layer potentials and energy spaces gives $⟨{\mathrm{\partial }}_n^{-}u-{\mathrm{\partial }}_n^{+}u,\gamma v⟩=⟨\lambda ,\gamma v⟩$; the right inverse in [F1] makes every $H^{1\mathrm{/}2}$ trace available, hence the conormal jump is $\lambda$. The global ${\dot{H}}^1$ representative has one trace, so ${\gamma }^{-}u={\gamma }^{+}u=\gamma u$.

By the weak equation tested with $v=u$, $⟨\lambda ,V\lambda ⟩=\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{H}}^1}^2$. For any $g\in H^{1\mathrm{/}2}(\mathrm{\Gamma })$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le 1$, choose the bounded lift $Lg$ from [F1]; then $\mathrm{\mid }⟨\lambda ,g⟩\mathrm{\mid }=\mathrm{\mid }\int \mathrm{\nabla }u\cdot \mathrm{\nabla }Lg\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{H}}^1}$. Taking the supremum over such $g$ gives $\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{H}}^1}$. Combining this with the energy identity proves coercivity; applying the trace bound of [F1] and the estimate in step 1.1 proves boundedness of $V$.

Now let $\lambda \in L^{\mathrm{\infty }}(\mathrm{\Gamma })$, and put $U(x)={\int }_{\mathrm{\Gamma }}{G}_3(x,y)\lambda (y)d{S}_y$. Such a density belongs to $H^{-1\mathrm{/}2}(\mathrm{\Gamma })$ because $\mathrm{\mid }{\int }_{\mathrm{\Gamma }}\lambda g\mathrm{\mid }\le \mathrm{\parallel }\lambda {\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2\le C\mathrm{\parallel }\lambda {\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$. The finite Lipschitz graph atlas gives $\mathrm{\mid }\mathrm{\Gamma }\cap B(x,r)\mathrm{\mid }\le C{r}^2$ for small $r$: each graph area density is bounded and its planar projection inside a radius-$r$ ball lies in a radius-$r$ disk, as proved in Sobolev pullback and Green identity on Lipschitz graphs. Dyadic annuli about $x$ therefore bound $\mathrm{\mid }U(x)\mathrm{\mid }$ locally by $C$ and $\mathrm{\mid }\mathrm{\nabla }U(x)\mathrm{\mid }$ by $C(1+\mathrm{\mid }\log d(x,\mathrm{\Gamma })\mathrm{\mid })$: on each annulus of radius $r$, the kernels $\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}$ and $\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}$ contribute at most $Cr$ and $C$, respectively, and there are at most $C(1+\mathrm{\mid }\log d\mathrm{\mid })$ annuli. A width-$\delta$ neighborhood of this compact Lipschitz boundary has volume at most $C\delta$: in finitely many smaller graph charts, distance less than $\delta$ implies a vertical gap at most $(1+L)\delta$, and vertical Fubini bounds each bounded-base strip by $C\delta$, with support margins ensuring the smaller charts cover that neighborhood. This is also the strip estimate of Sobolev pullback and Green identity on Lipschitz graphs. Summing dyadic distance strips shows $\mathrm{\nabla }U\in L^2$ near $\mathrm{\Gamma }$; at infinity $U=O(\mathrm{\mid }x{\mathrm{\mid }}^{-1})$ and $\mathrm{\nabla }U=O(\mathrm{\mid }x{\mathrm{\mid }}^{-2})$. The displayed gradient is also the global weak gradient: the first derivative of the locally integrable Newton kernel is locally volume integrable, so the weak derivative of its surface convolution is the same integral by Fubini. Hence $U\in L^6$ and $\mathrm{\nabla }U\in L^2$. To see that $U$ belongs to the completion ${\dot{H}}^1$, take cutoffs ${\chi }_R=1$ on $B_R$, supported in $B_{2R}$, with $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$: the gradient tail tends to zero and $R^{-2}{\int }_{R<\mathrm{\mid }x\mathrm{\mid }<2R}\mathrm{\mid }U{\mathrm{\mid }}^2=O(R^{-1})$. Each ${\chi }_{R}U$ is compactly supported in $H^1$ and convolution approximates it in $H^1$ by smooth compactly supported functions. The fundamental-solution identity [F3] and Fubini [F4], whose local absolute-integrability follows from $\mathrm{\mid }{\mathrm{\nabla }}_x{G}_3\mathrm{\mid }=O(\mathrm{\mid }x-y{\mathrm{\mid }}^{-2})$ in three dimensions, give $\int \mathrm{\nabla }U\cdot \mathrm{\nabla }v={\int }_{\mathrm{\Gamma }}\lambda \gamma v$ for smooth compactly supported $v$. Density and [F1] extend this to all $v\in {\dot{H}}^1$. Uniqueness in step 1.1 gives $U=u$, proving the kernel representation.