Noise-robust neural-operator approximation for EIT — Examples and Counterexamples

The square $[7\mathrm{/}16,9\mathrm{/}16{]}^2$ has side $1\mathrm{/}8$ and lies in ${\mathrm{\Omega }}^{\mathrm{\prime }}$ because its corner radius from $(1\mathrm{/}2,1\mathrm{/}2)$ is $\sqrt{2}\mathrm{/}16<1\mathrm{/}8$. The grid spacing is $s_N=1\mathrm{/}(8q)$ and $r_N=s_N\mathrm{/}4$. Distinct cell centers are separated by at least $s_N>2r_N$, and each disk lies inside its cell; hence the disks are disjoint and contained in ${\mathrm{\Omega }}^{\mathrm{\prime }}$. Thus $1\le {\gamma }_N\le 1+a=5\mathrm{/}4<3\mathrm{/}2$ and ${\gamma }_N=1$ on $\mathrm{\Omega }\setminus {\mathrm{\Omega }}^{\mathrm{\prime }}$.

For a disk $B(c,r)$ and a smooth compactly supported vector field $\phi$, polar coordinates give ${\int }_{B(c,r)}\mathrm{div}\phi dx=r{\int }_0^{2\pi }\phi (c+r(\cos \theta ,\sin \theta ))\cdot (\cos \theta ,\sin \theta )d\theta$. Indeed, in polar variables $\mathrm{div}\phi ={\rho }^{-1}{\mathrm{\partial }}_{\rho }(\rho {\phi }_r)+{\rho }^{-1}{\mathrm{\partial }}_{\theta }{\phi }_{\theta }$; integrating against $\rho d\rho d\theta$ makes the angular derivative integrate to zero and the inner radial term vanish. The distributional definition therefore gives $D{\mathbf{1}}_{B(c,r)}=-\nu {\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\partial }B(c,r)}$, whose variation has mass $2\pi r$. Summing these measures over the disjoint disks shows $D{\mathbf{1}}_{E_N}$ is finite with variation mass $2\pi r_{N}N$, so [F3] gives ${\mathbf{1}}_{E_N}\in BV(\mathrm{\Omega })$ and [F2] gives the stated upper bound.

The annuli of radii $(r_N\mathrm{/}2,3r_N\mathrm{/}2)$ about the chosen centers are pairwise disjoint: their outer radii sum to $3r_N=3s_N\mathrm{/}4<s_N$, while distinct centers are at least $s_N$ apart. They lie in $\mathrm{\Omega }$, since each center is within $\sqrt{2}\mathrm{/}16$ of $(1\mathrm{/}2,1\mathrm{/}2)$ and $\sqrt{2}\mathrm{/}16+3r_N\mathrm{/}2\le \sqrt{2}\mathrm{/}16+3\mathrm{/}64<1\mathrm{/}4$. On each annulus choose a smooth radial cutoff equal to one at radius $r_N$, vanishing outside the annulus and between zero and one. The sum of the corresponding radial unit-vector fields is smooth, compactly supported in $\mathrm{\Omega }$, has norm at most one, and equals the outward normal on every circle. Its flux is $2\pi r_{N}N$, so [F2] gives the matching lower bound. Since $D{\gamma }_N=aD{\mathbf{1}}_{E_N}$ by [F3] and $a>0$, step 1.2 gives $TV({\mathbf{1}}_{E_N};\mathrm{\Omega })=2\pi r_{N}N$ and $TV({\gamma }_N;\mathrm{\Omega })=2\pi a{r}_{N}N=\pi aN\mathrm{/}(16q)$.

Since $q=\lceil \sqrt{N}\rceil \le 2\sqrt{N}$, step 2.1 gives $TV({\gamma }_N;\mathrm{\Omega })\ge (\pi a\mathrm{/}32)\sqrt{N}$. The value and collar bounds were checked in step 1.1, and every finite union of disks has the finite perimeter computed in step 2.1. Thus each conductivity is individually finite-BV while the total variations diverge, so no fixed $R$ contains the whole family in $X_2(R)$.

The finite set $S$ is closed in the Hilbert metric and $d(x_{\ast },S)>0$. Define $\phi (x)=d(x,S)\mathrm{/}d(x_{\ast },S)$, $F_0(x)=0$ and $F_1(x)=\phi (x)y_{\ast }$. Distance to a closed set is 1-Lipschitz, so both maps are continuous. For every $x_i\in S$, $\phi (x_i)=0$, hence $F_0(x_i)=F_1(x_i)=0$; at $x_{\ast }$, $F_1(x_{\ast })-F_0(x_{\ast })=y_{\ast }$.

The construction shows that finite exact sample values do not determine an arbitrary continuous map on the ambient Hilbert input space. The distinction in the Example follows from the domains of the two claims: the continuum theorem quantifies over a compact subset of the full kernel $L^2$ space, whereas a finite-element experiment only evaluates finitely many discretized inputs and outputs. The finite-map construction is not asserted to belong to the generalized-FNO class or to contradict the continuum theorem.