Smooth-boundary trace lifts and Sobolev Helmholtz layer extensions

Write $a(\xi )=(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. For a compactly supported smooth flat-space function $v$, Fourier inversion in the last coordinate and Cauchy–Schwarz give $\mathrm{\mid }\widehat{v(\cdot ,0)}(\xi ){\mathrm{\mid }}^2\le Ca(\xi {)}^{-1}\int (a(\xi {)}^2+4{\pi }^2{\tau }^2)\mathrm{\mid }\widehat{v}(\xi ,\tau ){\mathrm{\mid }}^{2}d\tau$. Integration in $\xi$ proves the flat $H^1\to H^{1\mathrm{/}2}$ trace estimate. The lift $\widehat{Lg}(\xi ,t)=\widehat{g}(\xi )e^{-a(\xi )\mathrm{\mid }t\mathrm{\mid }}\chi (t)$, with $\chi =1$ near zero and compactly supported, has trace $g$ and $H^1$ norm bounded by $C\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$, by integrating $e^{-2a\mathrm{\mid }t\mathrm{\mid }}$. In a $C^2$ graph chart $x=(z,t+\psi (z))$, the change of variables has determinant one and bounded derivative and inverse derivative; the chain rule bounds the two $H^1$ norms. Graph and chart distances and surface Jacobians are uniformly comparable, so the Slobodeckij double integral gives the corresponding $H^{1\mathrm{/}2}$ norm comparison. A finite partition of unity and smaller chart cutoffs therefore transport the flat trace and lifts to the surface. For each bounded physical chart piece and flat half-box use [F92]. A $C^2$ graph map has bounded derivative, inverse derivative and Jacobian, so the classical chain rule and change of variables bound composition on smooth approximants; passing their functions and derivatives in $L^2$ gives the weak composition formula and $H^1$ convergence for all inputs. This proves the one-sided trace/extension passage without assuming smoothness of an arbitrary Sobolev representative. For the normal-coordinate identity, smooth approximants from [F92] satisfy the classical one-variable FTC. Their traces converge by the flat trace estimate, and their normal-derivative integrals converge in tangential $L^2$, uniformly over a bounded normal interval, by Cauchy–Schwarz. Taking these limits supplies the $L^2$ slice representative and its integral identity for every normal parameter. A zero-trace $H^1$ function then has zero-extension in $H^1$ (the distributional boundary derivative is its trace times $n$, which vanishes), and the kernel of the trace is $H_0^1$: in a flat half-chart a zero-trace function satisfies $v(z,t)={\int }_0^t{\mathrm{\partial }}_{t}v(z,s)ds$, whence ${ϵ}^{-2}{\int }_{0<t<2ϵ}\mathrm{\mid }v{\mathrm{\mid }}^2\le C{\int }_{0<t<2ϵ}\mathrm{\mid }{\mathrm{\partial }}_{t}v{\mathrm{\mid }}^2\to 0$. Cutting it off for $t<ϵ$ therefore changes its $H^1$ norm by a quantity tending to zero; convolution of each resulting interior-supported function proves the claim. Transport this construction by the graph maps and sum the partition. For the trace interpretation, in a flattened half-collar, first for smooth functions and then by $H^1$ approximation, the fundamental theorem of calculus and Cauchy–Schwarz give $\mathrm{\parallel }v(\cdot ,t)-\gamma v{\mathrm{\parallel }}_{L^2}^2\le t{\int }_0^t\mathrm{\parallel }{\mathrm{\partial }}_{s}v(\cdot ,s){\mathrm{\parallel }}_{L^2}^{2}ds$. In a $C^2$ graph chart, the map $(z,t)\mapsto \mathrm{\Phi }(z)+t{n}_{\mathrm{\Phi }(z)}$ is $C^1$ and at $t=0$ its tangent columns and unit-normal column are linearly independent. The Euclidean inverse-function theorem gives a local normal collar with bounded derivative, inverse derivative and Jacobian after shrinking to a compact subchart; a finite cover provides the required fixed bounds; transporting this estimate proves $L^2$ convergence of the normal slices to the Sobolev trace from either side. These arguments apply to $C^2$ charts directly and do not import any polyhedral trace theorem. To verify density of the declared ambient smooth densities in $H^{1\mathrm{/}2}(\mathrm{\Gamma })$, lift any chart datum by the constructed compactly supported global $H^1$ lift, mollify that lift in $H^1$ as in [F92], and apply the bounded trace estimate: the traces of the smooth approximants converge in the chart norm. They are also dense in $L^2(\mathrm{\Gamma })$ by chart cutoffs and ordinary mollification. Their density in $H^{-1\mathrm{/}2}$ follows by Hilbert duality: an annihilator in $H^{1\mathrm{/}2}$ would have zero $L^2$ pairing with every such density and hence be zero; Riesz identifies the dual of the negative space with the positive space.

Put $P_k=-\mathrm{\Delta }-k^2$ and $G_k(x)=e^{ik\mathrm{\mid }x\mathrm{\mid }}\mathrm{/}(4\pi \mathrm{\mid }x\mathrm{\mid })$. Radial differentiation gives $P_k{G}_k=0$ for $x\mathrm{\ne }0$. On a punctured test-function support, integration by parts gives the inner boundary contribution ${\int }_{\mathrm{\mid }x\mathrm{\mid }=ϵ}(G_k{\mathrm{\partial }}_{r}v-v{\mathrm{\partial }}_r{G}_k)dS\to v(0)$, since $G_k=O({ϵ}^{-1})$ and ${\mathrm{\partial }}_r{G}_k=-1\mathrm{/}(4\pi {ϵ}^2)+O(1)$. Thus $P_k{G}_k={\delta }_0$ distributionally. Choose $\zeta \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ equal to one on a ball larger than every difference $x-y$ needed for the fixed bounded target and source sets. The function $g=\zeta G_k$ is in $L^1$ and $P_{k}g={\delta }_0+h$ with smooth compactly supported $h$ supported away from zero. Hence $(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2-k^2)\widehat{g}(\xi )=1+\widehat{h}(\xi )$. For large $\mathrm{\mid }\xi \mathrm{\mid }$ this gives $\mathrm{\mid }\widehat{g}\mathrm{\mid }\le C(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-1}$; for bounded $\xi$ the same estimate follows from $g\in L^1$ with a larger constant. Plancherel now proves $g\ast :H^{-1}({\mathbb{R}}^3)\to H^1({\mathbb{R}}^3)$ and $g\ast :L^2\to H^2$. For distributions of bounded support these coincide on the chosen target set with $G_k\ast$, since the kernels agree on all differences there.

For $\lambda \in H^{-1\mathrm{/}2}(\mathrm{\Gamma })$, the distribution $T_{\lambda }$ defined by $T_{\lambda }(v)=\lambda (v{\mathrm{\mid }}_{\mathrm{\Gamma }})$ (using the complex-bilinear distributional convention) is supported on $\mathrm{\Gamma }$ and is in $H^{-1}({\mathbb{R}}^3)$ by step 1.1. Thus $S_k\lambda =G_k\ast T_{\lambda }$ is globally $H^1$ near the surface by step 2.1; it has a common trace and solves $P_k{S}_k\lambda =0$ off $\mathrm{\Gamma }$. For smooth $\lambda$, Fubini identifies the convolution with the classical single-layer integral, because $1\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }$ is locally integrable in three dimensions.

Given $\varphi \in H^{1\mathrm{/}2}(\mathrm{\Gamma })$, take a bounded interior lift $F_0$ from step 1.1. On $H_0^1({\mathrm{\Omega }}^{-})$ apply [F2] to $b(v,w)=\int (\mathrm{\nabla }v\cdot \overline{\mathrm{\nabla }w}+v\bar{w})$ and the functional $-b(F_0,w)$. This produces $F=F_0+v$ with trace $\varphi$, $(-\mathrm{\Delta }+1)F=0$, and $\mathrm{\parallel }F{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$. Define $q\in H^{-1\mathrm{/}2}(\mathrm{\Gamma })$ by $⟨q,g⟩=b(F,Lg)$ for an interior lift $Lg$. Independence follows since differences of lifts are in $H_0^1$, and boundedness follows from step 1.1. For $f=F{1}_{{\mathrm{\Omega }}^{-}}$ and any smooth compactly supported test $v$, two weak integrations by parts give $P_{k}f(v)=-(1+k^2){\int }_{{\mathrm{\Omega }}^{-}}Fv+q(v{\mathrm{\mid }}_{\mathrm{\Gamma }})-{\int }_{\mathrm{\Gamma }}\varphi {\mathrm{\partial }}_{n}v$. In distributions, the last term is minus the source of the classical double layer. Convolving this equality with $G_k$ is legitimate because $f$ has compact support, and $G_k\ast P_{k}f=(P_k{G}_k)\ast f=f$. Consequently define $D_k\varphi =S_{k}q-(1+k^2)G_k\ast f-f\mathrm{.}$ For smooth $\varphi$, the differentiated fundamental-solution identity and distributional convolution identify this expression away from $\mathrm{\Gamma }$ with ${\int }_{\mathrm{\Gamma }}{\mathrm{\partial }}_{n_y}{G}_k(x-y)\varphi (y)d{S}_y$. The first term is $H^1$ near $\mathrm{\Gamma }$, the second is $H^2$ by step 2.1, and $f$ restricts to $F$ and zero on the two sides. This proves the asserted bounded piecewise $H^1$ extension and its uniqueness by density. It also shows its trace jump is $-\varphi$.

Each layer solves $P_{k}u=0$ on either side by its source identity. For any such interior restriction define $⟨q^{-},g⟩={\int }_{{\mathrm{\Omega }}^{-}}(\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }Lg}-k^{2}u\overline{Lg})$. Differences of lifts are in $H_0^1$, so the weak PDE proves independence; step 1.1 gives $\mathrm{\parallel }{q}^{-}{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$. For the exterior choose a compactly supported lift in a bounded collar and use minus this integral, because the exterior's outward normal on $\mathrm{\Gamma }$ is $-n$. This gives $q^{+}$ with the same bound. They give Green's identity for every interior $H^1$ test and every exterior $H^1$ test supported in a bounded collar and vanishing on its artificial outer boundary. More generally, on a bounded truncation $U=B_R\setminus \overline{{\mathrm{\Omega }}^{-}}$ with the sphere separated from $\mathrm{\Gamma }$, subtract such a compact inner trace lift from a test $v\in H^1(U)$. The remaining test has zero trace on $\mathrm{\Gamma }$, and integration by parts near the smooth outer sphere, together with the weak PDE for zero-trace tests, gives ${\int }_U(\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }v}-k^{2}u\bar{v})={\int }_{\mathrm{\partial }{B}_R}{\mathrm{\partial }}_{r}u\bar{v}dS-⟨q^{+},{\gamma }^{+}v⟩$ whenever $u$ is smooth near that sphere. Approximation gives the same identity with a weak outer conormal if necessary. No boundary term at infinity is discarded. All these identities use the declared sesquilinear convention.

Away from $\mathrm{\Gamma }$, the kernel $G_k(x-y)$ and every target derivative are smooth in the ambient source variable $y$; after restriction to the $C^2$ surface they have uniformly bounded $H^{1\mathrm{/}2}$ norms on compact target sets. The source normal in $D_k$ is $C^1$, which suffices for its integral against $\varphi \in L^2$. Thus all target derivatives can be passed through these pairings and integrals, proving smoothness in $x$ there. Uniformly for bounded $y$ and $r\to \mathrm{\infty }$, $G_k(r\omega -y)=O(r^{-1})$ and $({\mathrm{\partial }}_r-ik)G_k(r\omega -y)=O(r^{-2})$ together with any fixed finite number of $y$ derivatives; these follow by differentiating $\mathrm{\mid }r\omega -y\mathrm{\mid }=r-\omega \cdot y+O(r^{-1})$. The same estimates hold after one source normal derivative. The $H^{1\mathrm{/}2}(\mathrm{\Gamma })$ norm of the first remainder is $O(r^{-2})$ by its uniform value and first source derivative bounds, so pairing with $\lambda \in H^{-1\mathrm{/}2}$ proves the single-layer radiation condition. The double-layer radiation condition follows by integration against $\varphi \in L^2(\mathrm{\Gamma })$ and the source normal derivative remainder bound. This proves all assertions.