Bounded invertibility of the smooth-boundary Helmholtz combined field

Suppose $A_{k,\eta }\varphi =0$ and set $u=D_k\varphi -i\eta S_k\varphi$. Here $\varphi \in H^{1\mathrm{/}2}$ is a valid single-layer density by the continuous inclusions $H^{1\mathrm{/}2}\subset L^2\subset H^{-1\mathrm{/}2}$, the last inclusion following from Cauchy–Schwarz and the continuous $H^{1\mathrm{/}2}\to L^2$ embedding. By [F2], $u^{+}$ is outgoing, weak Helmholtz and has zero exterior trace. [F3] gives $u^{+}=0$, so also ${\mathrm{\partial }}_n^{+}u=0$.

By [F2], ${\gamma }^{-}u-{\gamma }^{+}u=-\varphi$, so ${\gamma }^{-}u=-\varphi$. The conormal jump is ${\mathrm{\partial }}_n^{-}u-{\mathrm{\partial }}_n^{+}u=-i\eta \varphi$, whence ${\mathrm{\partial }}_n^{-}u=-i\eta \varphi =i\eta {\gamma }^{-}u$. The products here use the $L^2$ realization of $H^{1\mathrm{/}2}$ as an element of its dual $H^{-1\mathrm{/}2}$.

The interior weak Green identity [F2] tested with $u^{-}$ gives ${\int }_{{\mathrm{\Omega }}^{-}}(\mathrm{\mid }\mathrm{\nabla }{u}^{-}{\mathrm{\mid }}^2-k^2\mathrm{\mid }{u}^{-}{\mathrm{\mid }}^2)dx=⟨{\mathrm{\partial }}_n^{-}u,{\gamma }^{-}u⟩=i\eta \mathrm{\parallel }{\gamma }^{-}u{\mathrm{\parallel }}_{L^2(\mathrm{\Gamma })}^2\mathrm{.}$ The left side is real. Taking imaginary parts and using real $\eta \mathrm{\ne }0$ gives ${\gamma }^{-}u=0$, and step 2.1 gives $\varphi =0$. Thus $A_{k,\eta }$ is injective.

By [F1], $2A_{k,\eta }=I-{\mathcal{K}}_{k,\eta }$ with compact ${\mathcal{K}}_{k,\eta }$. Injectivity and [F4] make $I-{\mathcal{K}}_{k,\eta }$ boundedly invertible, hence so is $A_{k,\eta }$. Its finite inverse norm gives the estimate. For a prescribed $f$, use $\varphi =A_{k,\eta }^{-1}f$; [F2] gives the outgoing solution and the operator identity gives its trace $f$. The difference of any two such solutions satisfies [F3], proving uniqueness.