A Sobolev function vanishing on scattered sites has an h^ν bound

Since $\mathrm{\Omega }$ is a nonempty open domain, it contains a ball $B(z,r)$. For $d\ge 1$ this ball contains the infinitely many distinct points $z+\frac{r}{2n}{e}_1$, $n\ge 1$, by [F6]. The finite set $X$ omits one such point $x$, so ${\min }_{y\in X}\mathrm{\mid }x-y\mathrm{\mid }>0$ and therefore $h_X\ge {\min }_{y\in X}\mathrm{\mid }x-y\mathrm{\mid }>0$. In particular $h^{-1}$ is defined. Fix any whole-space extension $U=U_g$ of $u$, and split $g=g_{\mathrm{l}\mathrm{o}}+g_{\mathrm{h}\mathrm{i}}$ with $g_{\mathrm{l}\mathrm{o}}=g{\mathbf{1}}_{\{\mathrm{\mid }\xi \mathrm{\mid }\le h^{-1}\}}$. Let $V=U_{g_{\mathrm{l}\mathrm{o}}}$ and $W=U_{g_{\mathrm{h}\mathrm{i}}}$. By Cauchy--Schwarz [F5], polar integration [F3], and $s=\nu +d\mathrm{/}2$, $\mathrm{\parallel }W{\mathrm{\parallel }}_{\mathrm{\infty }}\le {({\int }_{\mathrm{\mid }\xi \mathrm{\mid }>h^{-1}}(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-s}d\xi )}^{1\mathrm{/}2}\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\le C{h}^{\nu }\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\mathrm{.}$ Indeed, for $r\ge h^{-1}\ge 1$ the radial tail is at most $C{\int }_{h^{-1}}^{\mathrm{\infty }}{r}^{-2\nu -1}dr=C^{\mathrm{\prime }}{h}^{2\nu }$.

For each multi-index $\alpha$ with $\mathrm{\mid }\alpha \mathrm{\mid }=m$, the bounded frequency support makes $D^{\alpha }V$ well defined and continuous by differentiation under an absolutely integrable Fourier integral. Cauchy--Schwarz [F5] and polar coordinates give $\mathrm{\parallel }{D}^{\alpha }V{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{({\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le h^{-1}}\mathrm{\mid }\xi {\mathrm{\mid }}^{2m}(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-s}d\xi )}^{1\mathrm{/}2}\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\le C{h}^{\nu -m}\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\mathrm{.}$ For the last estimate, the radial integral on $r\le 1$ is finite and on $1\le r\le h^{-1}$ is bounded by $C{\int }_1^{h^{-1}}{r}^{2m-2\nu -1}dr\le C^{\mathrm{\prime }}{h}^{-2(m-\nu )}$, using $m>\nu$.

Fix $x\in \mathrm{\Omega }$ and choose the stencil from [F2]. Taylor-expand the globally defined smooth $V$ at $x$ to degree $m-1$. The segment from $x$ to each $y_j$ lies in ${\mathbb{R}}^d$, where $V$ is smooth, so [F3] bounds each remainder by $C\mathrm{\mid }{y}_j-x{\mathrm{\mid }}^m{\max }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}\mathrm{\parallel }{D}^{\alpha }V{\mathrm{\parallel }}_{\mathrm{\infty }}$. The polynomial part cancels under the stencil's reproduction identities. Therefore $\mid V(x)-{\sum }_j{\lambda }_{j}V(y_j)\mid \le CB(Rh{)}^m{h}^{\nu -m}\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\le C^{\mathrm{\prime }}{h}^{\nu }\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}\mathrm{.}$

Since $U(y_j)=u(y_j)=0$, we have $V(y_j)=-W(y_j)$. Thus $U(x)=(V(x)-{\sum }_j{\lambda }_{j}V(y_j))+W(x)-{\sum }_j{\lambda }_{j}W(y_j)\mathrm{.}$ Use step 3.1, step 1.1 and ${\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }\le B$ to obtain $\mathrm{\mid }u(x)\mathrm{\mid }=\mathrm{\mid }U(x)\mathrm{\mid }\le C^{\mathrm{\prime }\mathrm{\prime }}{h}^{\nu }\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s}$. The constants do not depend on $x$ or the chosen extension. Taking the supremum over $x\in \mathrm{\Omega }$, then the infimum over all extensions $U$, gives the claimed quotient-norm estimate.