The Helmholtz double layer has a common weak conormal

For any piecewise $H^1$ Helmholtz solution $u$ with jumps $[\gamma u]={\gamma }^{-}u-{\gamma }^{+}u$ and $[q]=q^{-}-q^{+}$, apply the two weak Green identities of [F1] to a compactly supported smooth test $v$. The outer artificial boundary contributes zero, and the opposite orientations of the inner boundary give $⟨P_{k}u,v⟩=⟨[q],v{\mathrm{\mid }}_{\mathrm{\Gamma }}{⟩}_{\mathrm{b}\mathrm{i}\mathrm{l}}-{\int }_{\mathrm{\Gamma }}[\gamma u]{\mathrm{\partial }}_{n}vdS$. The subscript indicates the complex-bilinear version of weak duality; conjugating the test converts this to the sesquilinear convention. For $u=D_k\varphi$, the differentiated kernel source identity is $⟨P_k{D}_k\varphi ,v⟩={\int }_{\mathrm{\Gamma }}\varphi {\mathrm{\partial }}_{n}vdS$. Substituting $[\gamma u]=-\varphi$ from [F1] cancels these terms, leaving $⟨[q],v{\mathrm{\mid }}_{\mathrm{\Gamma }}{⟩}_{\mathrm{b}\mathrm{i}\mathrm{l}}=0$. Traces of smooth ambient tests are dense in $H^{1\mathrm{/}2}(\mathrm{\Gamma })$ by the chart lift/approximation construction in [F1], so $[q]=0$ in $H^{-1\mathrm{/}2}$.

For $u=S_k\lambda$, [F1] gives zero trace jump and $⟨P_k{S}_k\lambda ,v⟩=\lambda (v{\mathrm{\mid }}_{\mathrm{\Gamma }})$. The identity of step 1.1 reduces to $⟨[q]-\lambda ,v{\mathrm{\mid }}_{\mathrm{\Gamma }}{⟩}_{\mathrm{b}\mathrm{i}\mathrm{l}}=0$. The same density and boundedness give $[q]=\lambda$. Together with the trace assertions of [F1], this proves all four identities.