NtD smoothing and Hilbert–Schmidt kernel bounds

Write $r=r_d\ge 0$, $n=d-1$ and $w_j=1+{\lambda }_j$ on the total-mean-zero subspace. The counting estimate [F1] implies $w_j^{-1\mathrm{/}2}\le C(1+j{)}^{-1\mathrm{/}n}$, after enlarging the constant for finitely many low modes. By [F5], $R={\mathcal{R}}_{\gamma }$ maps the $L^2$ unit ball into a uniformly bounded $H_{\mathrm{\Delta }}^1$ ball. If $P_J$ is the first $J$-mode projection, then $R_J=P_{J}R$ has rank at most $J$ and $\mathrm{\parallel }R-R_J\mathrm{\parallel }\le C(1+J{)}^{-1\mathrm{/}n}$. Likewise $S=C_M^{r\mathrm{/}2}$ has the rank-$J$ approximation $S_J=P_{J}S$ with $\mathrm{\parallel }S-S_J\mathrm{\parallel }\le C(1+J{)}^{-r\mathrm{/}n}$; for $r=0$ this means the bound one and does not assert that $S$ is compact.

The rank-at-most-$2J$ map $R_{J}S+(R-R_J)S_J$ approximates $RS$ with error $(R-R_J)(S-S_J)$, bounded by $C(1+J{)}^{-(1+r)\mathrm{/}n}$. The compact operator $RS$ therefore has singular values $s_j\le C(1+j{)}^{-(1+r)\mathrm{/}n}$. To justify this implication, apply [F7] to $(RS{)}^{\ast }(RS)$: truncating its orthogonal singular expansion gives a rank-$k$ approximation with error $s_{k+1}$, and any other rank-$k$ approximation has a kernel vector in the span of the first $k+1$ right singular vectors, so its error is at least $s_{k+1}$. Since $2(1+r)>n$ by the actual dimension-dependent choice of $r$, ${\sum }_j{s}_j^2<\mathrm{\infty }$ uniformly. Thus $A_{\gamma }=RS$ and its zero extension are Hilbert–Schmidt. Apply [F3] to obtain the unique $L^2$ kernel and norm equality. Input and output annihilation of the constant mode mean that its integral in either variable is zero, by testing the integral operator and its adjoint on the constant function.

Put $B={\mathcal{R}}_{\gamma }-{\mathcal{R}}_{{\gamma }_0}$ and $s=\mathrm{\parallel }{\kappa }_{\gamma }-{\kappa }_{{\gamma }_0}{\mathrm{\parallel }}_2$. By [F4], $s^2={\sum }_{i,j}\mathrm{\mid }{b}_{ij}{\mathrm{\mid }}^2{w}_j^{-r}$. Choose $L$ so large that both ${\sum }_i{w}_i^{1-2L}$ and ${\sum }_j{w}_j^{r+1-2L}$ converge by [F1]. Then [F5] gives $T:={\sum }_{i,j}\mathrm{\mid }{b}_{ij}{\mathrm{\mid }}^2{w}_i{w}_j^{r+1}\le C$, uniformly over every conductivity pair. Cauchy–Schwarz now yields ${\sum }_{i,j}{w}_i^{1\mathrm{/}2}{w}_j^{1\mathrm{/}2}\mathrm{\mid }{b}_{ij}{\mathrm{\mid }}^2\le (\sum \mathrm{\mid }{b}_{ij}{\mathrm{\mid }}^2{w}_j^{-r}{)}^{1\mathrm{/}2}(\sum \mathrm{\mid }{b}_{ij}{\mathrm{\mid }}^2{w}_i{w}_j^{r+1}{)}^{1\mathrm{/}2}\le Cs$. The left side is the squared Hilbert–Schmidt norm from $H_{\mathrm{\Delta }}^{-1\mathrm{/}2}$ to $H_{\mathrm{\Delta }}^{1\mathrm{/}2}$: the normalized input basis is $w_j^{1\mathrm{/}4}{e}_j$. Taking square roots and [F5]'s quotient norm equivalence gives $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathcal{L}({\dot{\mathcal{D}}}^{\ast },\dot{\mathcal{D}})}\le C{s}^{1\mathrm{/}2}$, also when $s=0$.

For clarity, the inverse assertion in [F6] uses no boundary pointwise differentiability: the Dirichlet weak solution with boundary class $g$ has Neumann functional ${\dot{\mathrm{\Lambda }}}_{\gamma }g$, and uniqueness of the weak Neumann problem gives ${\mathcal{R}}_{\gamma }{\dot{\mathrm{\Lambda }}}_{\gamma }g=g$; conversely the weak Neumann solution satisfies its own Dirichlet problem, giving ${\dot{\mathrm{\Lambda }}}_{\gamma }{\mathcal{R}}_{\gamma }F=F$. Therefore ${\dot{\mathrm{\Lambda }}}_{\gamma }-{\dot{\mathrm{\Lambda }}}_{{\gamma }_0}={\dot{\mathrm{\Lambda }}}_{\gamma }({\mathcal{R}}_{{\gamma }_0}-{\mathcal{R}}_{\gamma }){\dot{\mathrm{\Lambda }}}_{{\gamma }_0}$. Uniform DN operator bounds in [F6] and step 3.1 prove the displayed square-root comparison. Together with step 2.1 this establishes all assertions.