The smooth Helmholtz combined field is a compact perturbation of half the identity

In surface coordinates put $r=\mathrm{\mid }x-y\mathrm{\mid }$. The $C^2$ graph bounds give $\mathrm{\mid }{n}_y\cdot (x-y)\mathrm{\mid }\le C{r}^2$ and $\mathrm{\mid }{n}_y-n_x\mathrm{\mid }\le Cr$. Therefore both kernels $L_V=G_k(x-y)$ and $L_K={\mathrm{\partial }}_{n_y}{G}_k(x-y)$ satisfy $\mathrm{\mid }L(x,y)\mathrm{\mid }\le C\mathrm{/}r$. Their first target-coordinate derivatives satisfy $\mathrm{\mid }{\mathrm{\partial }}_{x}L\mathrm{\mid }\le C\mathrm{/}{r}^2$: for $K$, differentiating the numerator tangentially gives $n_y\cdot t_x=(n_y-n_x)\cdot t_x=O(r)$, and differentiating the radial factors gives at worst $O(r^2)O(r^{-4})$. Fixed $k$ only changes the constants. Localize targets and sources by chart cutoffs; for near-diagonal pieces use a common graph chart and absorb its bounded surface Jacobian into the source kernel. Off-diagonal pieces have bounded first target derivatives. Each resulting chart kernel $l(z,w)$, extended with cutoffs by zero, thus has size $C\mathrm{/}\mathrm{\mid }z-w\mathrm{\mid }$ and target derivative size $C\mathrm{/}\mathrm{\mid }z-w{\mathrm{\mid }}^2$.

For a target translation $h$, $\delta =\mathrm{\mid }h\mathrm{\mid }<1\mathrm{/}4$, consider $l(z+h,w)-l(z,w)$. Where $\mathrm{\mid }z-w\mathrm{\mid }\le 2\delta$, integrate each size bound separately (including the ball around $z+h$), obtaining $C\delta$. On the complement the segment from $z$ to $z+h$ stays at distance at least $\mathrm{\mid }z-w\mathrm{\mid }\mathrm{/}2$, so the derivative bound and polar integration give $C\delta {\int }_{2\delta }^C{r}^{-1}dr\le C\delta (1+\mathrm{\mid }\log \delta \mathrm{\mid })$. This bounds both the supremum over $z$ of its absolute integral in $w$ and the supremum over $w$ of its absolute integral in $z$. Cauchy–Schwarz in the weighted kernel integral then proves the Schur estimate $\mathrm{\parallel }Tf{\mathrm{\parallel }}_2\le C\mathrm{\parallel }f{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }(Tf)(\cdot +h)-Tf{\mathrm{\parallel }}_2\le C\delta (1+\mathrm{\mid }\log \delta \mathrm{\mid })\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{.}$ Indeed $\mathrm{\mid }\int lf{\mathrm{\mid }}^2\le (\int \mathrm{\mid }l\mathrm{\mid })(\int \mathrm{\mid }l\mathrm{\mid }\mathrm{\mid }f{\mathrm{\mid }}^2)$ and integration uses the two bounds. The cutoff support is fixed and bounded.

For any $0<s<1$, integrate the square of the last translation bound against $\mathrm{\mid }h{\mathrm{\mid }}^{-2-2s}dh$ for $\mathrm{\mid }h\mathrm{\mid }<1$; polar coordinates give the finite integral $C{\int }_0^1{r}^{1-2s}(1+\mathrm{\mid }\log r\mathrm{\mid }{)}^{2}dr$. The $\mathrm{\mid }h\mathrm{\mid }\ge 1$ part is bounded by the $L^2$ estimate. Hence each localized $T$ maps $L^2$ boundedly to $H^s$ in the difference norm, in particular for $s=3\mathrm{/}4$. To check the equivalent Fourier norm directly, apply Plancherel to differences and Tonelli: the inner integral $\int \mathrm{\mid }{e}^{2\pi i\xi \cdot h}-1{\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^{-2-2s}dh$ equals $c_s\mathrm{\mid }\xi {\mathrm{\mid }}^{2s}$ by scaling and rotation, with $0<c_s<\mathrm{\infty }$ since near zero the numerator is $O(\mathrm{\mid }h{\mathrm{\mid }}^2)$ and at infinity it is bounded. This also identifies the half-order norm in [F3].

A bounded family of functions supported in a fixed compact chart and bounded in $H^{3\mathrm{/}4}$ is precompact in $H^{1\mathrm{/}2}$. Here is a direct verification. For a smooth compactly supported mollifier ${\rho }_{ϵ}$, $\mathrm{\mid }1-\widehat{\rho }(ϵ\xi )\mathrm{\mid }\le C\min (ϵ\mathrm{\mid }\xi \mathrm{\mid },1)$ and thus $\mathrm{\parallel }f-{\rho }_{ϵ}\ast f{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le C{ϵ}^{1\mathrm{/}4}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{3\mathrm{/}4}}$ by the Fourier weights. For fixed $ϵ$, the convolutions and their first two derivatives are uniformly bounded and have one fixed bounded support, by Cauchy–Schwarz with the fixed derivatives of ${\rho }_{ϵ}$. They are precompact in $C^1$ on a containing rectangle: choose increasingly fine finite grids, take convergent subsequences of the finitely many value and first-derivative samples, and use the uniform first/second derivative bounds to extend sample convergence uniformly. A diagonal subsequence gives convergence of both values and first derivatives. Uniform $C^1$ convergence on a fixed compact support implies $H^{1\mathrm{/}2}$ convergence, since the difference-norm integrand near the diagonal is bounded by $\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\mid }z-w{\mathrm{\mid }}^{-1}$ and away from the diagonal by a constant times $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\mid }z-w{\mathrm{\mid }}^{-3}$. The uniform mollification error now gives finite nets for the original family in $H^{1\mathrm{/}2}$, hence compact closure in its complete Hilbert norm.

Since $H^{1\mathrm{/}2}(\mathrm{\Gamma })$ embeds boundedly in $L^2(\mathrm{\Gamma })$, steps 1.1–4.1 and the finite chart partition make both boundary integrals compact on $H^{1\mathrm{/}2}(\mathrm{\Gamma })$. The single-layer integral agrees on smooth densities with $V_k$ of [F1]. For smooth densities, the normal-slice limits computed in [F2] are the Sobolev traces of [F1]: the planar half-Poisson terms have uniformly bounded mass, and the integrable $C\mathrm{/}r$ curvature remainders and smooth-density remainders have uniform integral bounds on the compact surface, so the pointwise slice limits also converge in $L^2$ by dominated convergence. The normal-slice estimate proved in [F1] identifies this $L^2$ limit with the weak trace. Consequently [F2] identifies the double-layer exterior trace with $\frac{1}{2}I+K_k$. Boundedness and density extend both identities to every $H^{1\mathrm{/}2}$ density. Therefore $A_{k,\eta }=\frac{1}{2}I+K_k-i\eta V_k$, and its displayed compact perturbation follows by addition and scalar multiplication of the compact maps just proved.