Noise-robust neural-operator approximation for EIT

The local graph charts of the smooth boundary form an open cover. Shrinking them gives graph-chart neighborhoods whose coordinate images have compact closure inside Euclidean balls; compactness selects a finite subcover. Rescale each coordinate ball affinely into $(1\mathrm{/}4,3\mathrm{/}4{)}^{d-1}$. On each compact chart closure, the map and inverse have bounded derivatives, so the restricted coordinate maps are bi-Lipschitz. Apply [F1] to the finite cover and combine partition functions sharing one chart to obtain nonnegative smooth ${\theta }_j$ compactly supported in $U_j$, with ${\sum }_j{\theta }_j=1$. Since ${\sum }_j{\theta }_j^2\ge J^{-1}$, ${\psi }_j={\theta }_j\mathrm{/}({\sum }_i{\theta }_i^2{)}^{1\mathrm{/}2}$ is smooth, has the same support, and satisfies ${\sum }_j{\psi }_j^2=1$.

For each pair set ${\mathrm{\Phi }}_{jk}(x,y)=({\varphi }_j(x),{\varphi }_k(y))$ and ${\chi }_{jk}(x,y)={\psi }_j(x){\psi }_k(y)$. Define $(Gh{)}_{jk}(z)=({\chi }_{jk}h)({\mathrm{\Phi }}_{jk}^{-1}z)$ on ${\mathrm{\Phi }}_{jk}(U_j\times U_k)$ and zero elsewhere, and define $(Lv)(x,y)={\sum }_{j,k}{\chi }_{jk}(x,y)v_{jk}({\mathrm{\Phi }}_{jk}(x,y))$ on the chart pairs, zero outside. These are finite sums. The chart Jacobians and surface-density factors are bounded above and below on compact supports by bi-Lipschitz regularity and [F2]–[F3], with Fubini [F5] for the product chart factors. Thus the pullbacks, zero extensions and smooth multiplications are bounded on $L^2$, so $G$ and $L$ are bounded linear maps.

Substitution in the finite sum gives $(LGh)(x,y)={\sum }_{j,k}{\chi }_{jk}(x,y{)}^{2}h(x,y)=({\sum }_j{\psi }_j(x{)}^2)({\sum }_k{\psi }_k(y{)}^2)h(x,y)=h(x,y)$ almost everywhere. Thus $LG=I$ as maps on $L^2(M)$.

For smooth $h$, the product and chain rules show that first derivatives of $(Gh{)}_{jk}$ are bounded chart coefficients times $h$ and its first tangential derivatives. The Jacobian bounds in step 2.1 give $\mathrm{\parallel }Gh{\mathrm{\parallel }}_{\oplus H^1}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1(M)}$ and $\mathrm{\parallel }Lv{\mathrm{\parallel }}_{H^1(M)}\le C\mathrm{\parallel }v{\mathrm{\parallel }}_{\oplus H^1}$. Since the weights vanish near patch edges, zero extension remains smooth. Density in the completion definition extends these maps boundedly to $H^1$.

By [F4], $P_N$ is a contraction on the finite direct sum and converges strongly to the identity. The image $G(C)$ is compact by step 2.1. For $\epsilon >0$, take a finite $\epsilon \mathrm{/}3$-net $v_1,\dots ,v_s$ for $G(C)$ and then take $N$ large enough that $\mathrm{\parallel }{P}_N{v}_i-v_i\mathrm{\parallel }<\epsilon \mathrm{/}3$ for all $i$. For any $v\in G(C)$ choose $v_i$ with $\mathrm{\parallel }v-v_i\mathrm{\parallel }<\epsilon \mathrm{/}3$; contraction gives $\mathrm{\parallel }{P}_{N}v-v\mathrm{\parallel }\le \mathrm{\parallel }{P}_N(v-v_i)\mathrm{\parallel }+\mathrm{\parallel }{P}_N{v}_i-v_i\mathrm{\parallel }+\mathrm{\parallel }{v}_i-v\mathrm{\parallel }<\epsilon$. This proves compact-uniform convergence.

For $w\in H^1({\mathbb{T}}^n)$, weak integration by parts against a character gives $\widehat{{\mathrm{\partial }}_{\mathrm{\ell }}w}(k)=2\pi i{k}_{\mathrm{\ell }}\widehat{w}(k)$. Parseval [F4] yields $\mathrm{\parallel }(I-P_N)w{\mathrm{\parallel }}_2^2={\sum }_{\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}>N}\mathrm{\mid }\widehat{w}(k){\mathrm{\mid }}^2\le (2\pi N{)}^{-2}{\sum }_k\mathrm{\mid }2\pi k{\mathrm{\mid }}^2\mathrm{\mid }\widehat{w}(k){\mathrm{\mid }}^2\le (2\pi N{)}^{-2}\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2^2$ for $N\ge 1$. Apply this to each component of $Gh$ and use step 3.2. For $N=0$, $\mathrm{\parallel }Gh{\mathrm{\parallel }}_2\le \mathrm{\parallel }Gh{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1}$; enlarging $C_1$ proves the stated tail estimate.

Fix an integer $m\ge 1$ and $a\in {\mathbb{R}}^n$. Since $\sigma$ is not a polynomial, ${\sigma }^{(m)}$ is not identically zero; choose $b_m$ with ${\sigma }^{(m)}(b_m)\mathrm{\ne }0$. For $h\mathrm{\ne }0$ define $R_{m,h}(x):=\frac{1}{{\sigma }^{(m)}(b_m)h^m}{\sum }_{j=0}^m(-1{)}^{m-j}\left(\genfrac{}{}{0px}{}{m}{j}\right)\sigma (b_m+jh(a\cdot x))$. This is a finite linear combination of permitted units. The repeated-integral formula for the $m$th forward difference gives $R_{m,h}(x)=\frac{(a\cdot x{)}^m}{{\sigma }^{(m)}(b_m)}{\int }_{[0,1{]}^m}{\sigma }^{(m)}(b_m+h(a\cdot x)(t_1+\cdots +t_m))dt$. Because $a\cdot x$ is bounded on compact $K$ and ${\sigma }^{(m)}$ is uniformly continuous on a compact interval around $b_m$, $R_{m,h}\to (a\cdot x{)}^m$ uniformly on $K$ as $h\to 0$.

For $m\ge 1$ and arbitrary real $x_1,\dots ,x_m$, the polarization identity $x_1\cdots x_m=\frac{1}{2^{m}m!}{\sum }_{ϵ\in \{-1,1{\}}^m}({\prod }_{j=1}^m{ϵ}_j)({\sum }_{j=1}^m{ϵ}_j{x}_j{)}^m$ expresses each monomial of total degree $m$ in $n$ variables as a finite linear combination of powers of linear forms. Step 1.1 places every such power in the uniform closure of the network family, and that closure is a vector space because finite network sums are represented by concatenating their hidden units. Constants are present through $c_0$. Hence every real polynomial restricted to $K$ belongs to that closure.

The restrictions of real polynomials to $K$ form a unital real function algebra and separate distinct points, since a coordinate function distinguishes any two different points. Compact subsets of Euclidean space are compact Hausdorff, so [L1] makes these polynomial restrictions uniformly dense in $C(K,\mathbb{R})$. The network closure contains them by step 2.1 and is therefore all of $C(K,\mathbb{R})$.

For a nonnegative function $\delta$ with $\delta (0)=0$, consider the finite weighted inequality ${\sum }_{i,j}{a}_i{a}_j[\delta (\mathrm{\parallel }{v}_i\mathrm{\parallel }{)}^2+\delta (\mathrm{\parallel }{v}_j\mathrm{\parallel }{)}^2-\delta (\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }{)}^2]\ge 0$ for Hilbert vectors $v_i$ and nonnegative scalars $a_i$. The function $\delta (t)=t$ satisfies it, since the sum is $2\mathrm{\parallel }{\sum }_i{a}_i{v}_i{\mathrm{\parallel }}^2$. Each truncated function ${\delta }_A(t)=\min \{t,A\}$, $A>0$, also satisfies it. Indeed, for the subsum with BOTH $\mathrm{\parallel }{v}_i\mathrm{\parallel },\mathrm{\parallel }{v}_j\mathrm{\parallel }<A$, replacing the negative truncated-distance square by the negative full-distance square can only decrease the sum, and this subsum is then $2\mathrm{\parallel }{\sum }_{\mathrm{\parallel }{v}_i\mathrm{\parallel }<A}{a}_i{v}_i{\mathrm{\parallel }}^2\ge 0$. Every remaining summand is separately nonnegative, because at least one of its two positive squares is $A^2$ while its negative square is at most $A^2$. The property is closed under pointwise limits, and under taking $\delta =({\sum }_{k=1}^N{c}_k{\delta }_k^2{)}^{1\mathrm{/}2}$ with $c_k\ge 0$, since the weighted expression is linear in ${\delta }^2$.

The function $\eta (t)=\omega (\sqrt{t}{)}^2$ is nondecreasing, continuous, zero at zero and concave. Here is a proof that covers nonsmooth moduli. For integers $n\ge 1$, interpolate $\omega$ on grids $0,1\mathrm{/}n,2\mathrm{/}n,\dots ,n$ and extend constantly after $n$; these concave nondecreasing polygonal functions converge pointwise to $\omega$. On each affine piece ${\omega }_n(r)=a+br$, its slope $b$ is nonnegative and its intercept $a$ is nonnegative because concavity and ${\omega }_n(0)=0$ imply ${\omega }_n(r)\ge r{\omega }_n^{\mathrm{\prime }}(r)$. Thus ${\omega }_n(\sqrt{t}{)}^2=a^2+2ab\sqrt{t}+b^{2}t$ is concave on that piece. At a junction its one-sided derivatives are ${\omega }_n(r)b\mathrm{/}r$, so the slope drops when $b$ drops. Continuity at zero then proves global concavity for ${\eta }_n$, and passing to the pointwise limit proves it for $\eta$. Interpolate $\eta$ on the same type of grids and flatten the final slope to zero. Every such polygonal ${\eta }_n$ is a finite sum ${\sum }_k{c}_k\min \{t,t_k\}$ with $c_k\ge 0$: take $c_k$ to be the decrease of slope at the breakpoint $t_k$. Hence $\sqrt{{\eta }_n(t^2)}=({\sum }_k{c}_k\min \{t,\sqrt{t_k}{\}}^2{)}^{1\mathrm{/}2}$ has step 1.1's property. Passing to the limit gives that property for $\omega (t)=\sqrt{\eta (t^2)}$.

First suppose the index set is finite and nonempty. If $\omega$ is identically zero, the pairwise hypothesis makes all the $y_i$ equal and their common value works. Otherwise $\omega (t)>0$ for every $t>0$: if $\omega (t_{\ast })=0$ with $t_{\ast }>0$, monotonicity makes it zero on $[0,t_{\ast }]$, and concavity bounds every subsequent slope by zero, forcing it to be identically zero. If some $x_i=x$, then $y=y_i$ works directly by the pairwise hypothesis. In the remaining case all $r_i=\mathrm{\parallel }{x}_i-x\mathrm{\parallel }$ and $b_i=\omega (r_i)$ are positive. In the finite-dimensional affine span of the $y_i$, minimize $h(y)={\max }_i\mathrm{\parallel }y-y_i{\mathrm{\parallel }}^2\mathrm{/}{b}_i^2$. A minimum exists because this continuous function tends to infinity with $\mathrm{\parallel }y\mathrm{\parallel }$ in that span; its perpendicular projection from $H_2$ never increases any distance. Let $y$ be a minimizer and let $J$ be its active set. The origin belongs to the convex hull of $(y-y_i)\mathrm{/}{b}_i^2$, $i\in J$: otherwise that finite convex hull has a nonzero nearest point $z$, whose variational inequality gives $⟨z,(y-y_i)\mathrm{/}{b}_i^2⟩\ge \mathrm{\parallel }z{\mathrm{\parallel }}^2>0$ for every active $i$. Moving a sufficiently small distance in direction $-z$ strictly decreases all active functions and keeps the finitely many inactive ones below the old maximum, a contradiction.

Choose nonnegative coefficients ${\beta }_i$, summing to one over $J$, with ${\sum }_{i\in J}{\beta }_i(y-y_i)\mathrm{/}{b}_i^2=0$, and normalize $a_i=({\beta }_i\mathrm{/}{b}_i^2)\mathrm{/}({\sum }_{j\in J}{\beta }_j\mathrm{/}{b}_j^2)$. Then $\sum a_i=1$ and $y=\sum a_i{y}_i$. Write $h(y)={\lambda }^2$; every active index has $\mathrm{\parallel }y-y_i{\mathrm{\parallel }}^2={\lambda }^2{b}_i^2$. The Hilbert variance identity gives ${\sum }_{i,j\in J}{a}_i{a}_j\mathrm{\parallel }{y}_i-y_j{\mathrm{\parallel }}^2=2{\sum }_{i\in J}{a}_i\mathrm{\parallel }{y}_i-y{\mathrm{\parallel }}^2=2{\lambda }^2{\sum }_i{a}_i{b}_i^2$. Apply the pairwise hypothesis and step 2.1's weighted inequality to $v_i=x_i-x$. They bound that same sum above by ${\sum }_{i,j}{a}_i{a}_j\omega (\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }{)}^2\le 2{\sum }_i{a}_i\omega (\mathrm{\parallel }{v}_i\mathrm{\parallel }{)}^2=2{\sum }_i{a}_i{b}_i^2$. The latter scalar is positive, so $\lambda \le 1$. The minimizer therefore belongs to EVERY target ball, proving finite intersection.

Now let $I$ be arbitrary and nonempty, and fix $i_0\in I$. For every finite $J\subset I$ containing $i_0$, the set $C_J={\bigcap }_{i\in J}\bar{B}(y_i,b_i)$ is nonempty by step 4.1, closed and convex. Let $p_J$ be its nearest point to zero by [F1]. Since $C_J$ lies in the fixed ball with center $y_{i_0}$ and radius $b_{i_0}$, the numbers $\mathrm{\parallel }{p}_J{\mathrm{\parallel }}^2$ have a finite supremum $D$. By [A1], for each integer $n\ge 1$, choose finite $J_n$ with $\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2>D-1\mathrm{/}n$ and replace them by their cumulative unions, preserving this inequality. They are increasing. For $m\ge n$, [F1] gives $\mathrm{\parallel }{p}_{J_m}-p_{J_n}{\mathrm{\parallel }}^2\le \mathrm{\parallel }{p}_{J_m}{\mathrm{\parallel }}^2-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\le D-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2$, so they form a Cauchy sequence and converge to a point $p\in H_2$.

For any fixed $i\in I$, let $q_n=p_{J_n\cup \{i\}}$. Again [F1] gives $\mathrm{\parallel }{q}_n-p_{J_n}{\mathrm{\parallel }}^2\le \mathrm{\parallel }{q}_n{\mathrm{\parallel }}^2-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\le D-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\to 0$. Hence $q_n\to p$; since every $q_n$ lies in the closed ball $\bar{B}(y_i,b_i)$, so does $p$. This works for each $i$ without enumerating $I$, and proves the required full intersection. If $I$ is empty, zero belongs to its intersection by convention. This proves all cases.

Let $(U_j,{\varphi }_j{)}_{j=1}^J$ and $({\psi }_j{)}_{j=1}^J$ be the boundary charts and square partition supplied by [F1], with ${\varphi }_j(U_j)⋐(1\mathrm{/}4,3\mathrm{/}4{)}^n$. Define $\mathcal{G}f=(({\psi }_{j}f)\circ {\varphi }_j^{-1}{)}_{j=1}^J,\mathcal{L}(v_1,\dots ,v_J)={\sum }_{j=1}^J{\psi }_j(v_j\circ {\varphi }_j),$ where each chart component is extended by zero outside its chart image and each summand by zero off $U_j$. Since every ${\psi }_j$ is compactly supported in $U_j$, the zero extensions preserve smoothness for smooth inputs. Change of variables, the bounded chart Jacobians, and the product and chain rules show that $\mathcal{G}$ and $\mathcal{L}$ are bounded on $L^2$ and $H^1$. Direct substitution gives $\mathcal{L}\mathcal{G}f={\sum }_j{\psi }_j^{2}f=f$ almost everywhere. For $N\ge 1$, Fourier Parseval [F4] gives componentwise $\mathrm{\parallel }(I-P_N)\mathcal{G}f{\mathrm{\parallel }}_{\oplus L^2({\mathbb{T}}^n)}^2\le (2\pi N{)}^{-2}\mathrm{\parallel }\mathrm{\nabla }\mathcal{G}f{\mathrm{\parallel }}_{\oplus L^2({\mathbb{T}}^n)}^2\le C{N}^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1(M)}^2\mathrm{.}$ For $N=0$, boundedness of $\mathcal{G}$ on $H^1$ gives the same estimate after increasing $C$. Therefore $\mathrm{\parallel }f-\mathcal{L}{P}_N\mathcal{G}f{\mathrm{\parallel }}_2\le C(1+N{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1(M)}\mathrm{.}$ The finite Fourier sums $\mathcal{L}{P}_N\mathcal{G}f$ are smooth functions on $M$ and converge to $f$ in $L^2$, so smooth functions are dense in $L^2(M)$.

The estimate in step 1.1 proves $H^1(M)\hookrightarrow L^2(M)$ is compact: the finite-rank maps $\mathcal{L}{P}_N\mathcal{G}$ approximate the identity uniformly on the $H^1$ unit ball, and their images are bounded subsets of finite-dimensional spaces, hence have finite nets of every positive radius. Thus the unit ball is totally bounded in the complete space $L^2(M)$.

Put $q(u,v)={\int }_M(u\bar{v}+{\mathrm{\nabla }}_{M}u\cdot \overline{{\mathrm{\nabla }}_{M}v})dS$ on $H^1(M)$. For $f\in L^2(M)$, [F2] gives a unique $u=\tilde{K}f\in H^1(M)$ with $q(u,v)=⟨f,v⟩$ for every $v\in H^1(M)$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$. Let $J:H^1(M)\hookrightarrow L^2(M)$ and $K=J\tilde{K}$. By step 2.1, $K$ is compact. Hermitian symmetry of $q$ gives $⟨Kf,g⟩=⟨f,Kg⟩$ and $⟨Kf,f⟩=q(\tilde{K}f,\tilde{K}f)\ge 0$. If $Kf=0$, the weak equation and smooth density from step 1.1 imply $⟨f,v⟩=0$ for every $v\in L^2(M)$, so $f=0$. Thus $K$ is compact, positive, self-adjoint and injective.

By [F3], $K$ has an orthonormal eigenbasis $e_j$ with positive eigenvalues ${\mu }_j\to 0$; order these so that ${\mu }_j$ is nonincreasing. Put ${\lambda }_j={\mu }_j^{-1}-1$. From $K{e}_j={\mu }_j{e}_j$ and the weak equation, ${\int }_M{\mathrm{\nabla }}_M{e}_j\cdot \overline{{\mathrm{\nabla }}_{M}v}dS={\lambda }_j⟨e_j,v⟩$ for every $v\in H^1(M)$. Taking $v=e_j$ gives ${\lambda }_j={\int }_M\mathrm{\mid }{\mathrm{\nabla }}_M{e}_j{\mathrm{\mid }}^{2}dS\ge 0$, and these are the eigenvectors of the stated weak Laplace–Beltrami operator. In the form inner product $q$, the vectors $(1+{\lambda }_j{)}^{-1\mathrm{/}2}{e}_j$ are orthonormal and complete: a vector $v$ orthogonal to every $e_j$ in $q$ satisfies $(1+{\lambda }_j)⟨v,e_j⟩=0$ for every $j$, hence vanishes in $L^2$. Form Parseval gives $q(v,v)={\sum }_j(1+{\lambda }_j)\mathrm{\mid }⟨v,e_j⟩{\mathrm{\mid }}^2$, proving the $s=1$ norm identity.

Let $E_{\mathrm{\Lambda }}=\mathrm{span}\{e_j:{\lambda }_j\le \mathrm{\Lambda }\}$. For $u\in E_{\mathrm{\Lambda }}$, step 4.1 gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}^2\le (1+\mathrm{\Lambda })\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. If $\dim E_{\mathrm{\Lambda }}$ exceeded the rank of $P_N\mathcal{G}{\mathrm{\mid }}_{E_{\mathrm{\Lambda }}}$, rank-nullity would give nonzero $u\in E_{\mathrm{\Lambda }}$ with $P_N\mathcal{G}u=0$. Since $\mathcal{L}\mathcal{G}=I$ and $\mathcal{L}$ is bounded, step 1.1 then gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le C(1+N{)}^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C\sqrt{1+\mathrm{\Lambda }}(1+N{)}^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Choose $N\ge 2C\sqrt{1+\mathrm{\Lambda }}$ to contradict $u\mathrm{\ne }0$. Thus $N_M(\mathrm{\Lambda })=\dim E_{\mathrm{\Lambda }}\le \mathrm{rank}(P_N\mathcal{G})\le J(2N+1{)}^n\le C^{\mathrm{\prime }}(1+\mathrm{\Lambda }{)}^{n\mathrm{/}2}$.

For $p=n\mathrm{/}2+\tau$, group eigenvalues into $[2^k,2^{k+1})$. The bound from step 5.1 gives ${\sum }_j(1+{\lambda }_j{)}^{-p}\le C{\sum }_{k\ge 0}{2}^{-kp}{N}_M(2^{k+1})\le C^{\mathrm{\prime }}{\sum }_{k\ge 0}{2}^{-k(p-n\mathrm{/}2)}<\mathrm{\infty }$; the finitely many eigenvalues below $1$ contribute finitely. This proves the stated summability for every $\tau >0$. Defining $H_{\mathrm{\Delta }}^s(M)$ by the weighted coefficient norm in the Statement completes the spectral-scale claim.

If $C=\mathrm{\varnothing }$, the zero operator from [F3] suffices. Otherwise let $P_N$ be the orthogonal projection onto real Fourier modes of frequency at most $N$. Strong convergence in [F1] is uniform on compact $C$: for $\eta >0$, choose a finite $\eta \mathrm{/}3$-net $f_1,\dots ,f_s$ for $C$, then take $N$ so that $\mathrm{\parallel }{P}_N{f}_i-f_i{\mathrm{\parallel }}_2<\eta \mathrm{/}3$ for every $i$. For $f$ within $\eta \mathrm{/}3$ of $f_i$, contraction of $I-P_N$ gives $\mathrm{\parallel }(I-P_N)f{\mathrm{\parallel }}_2<2\eta \mathrm{/}3$. Hence ${\sup }_{f\in C}\mathrm{\parallel }{P}_{N}f-f{\mathrm{\parallel }}_2\to 0$.

For each $f\in C$, let $m(f)$ be the least nonnegative integer such that $T$ has oscillation less than $\epsilon \mathrm{/}8$ on $B(f,2^{-m(f)})$, and set $r_f=2^{-m(f)}$. Continuity ensures that such an integer exists, so this defines all radii canonically without a simultaneous choice. Choose a finite subcover $B(f_i,r_i\mathrm{/}2)$ of $C$ by the HALF-radius balls, and take a positive $\delta <{\min }_i{r}_i\mathrm{/}4$. By step 1.1, increase $N$ until ${\sup }_{f\in C}\mathrm{\parallel }{P}_{N}f-f{\mathrm{\parallel }}_2<\delta$. For every $f\in C$ choose $i$ with $\mathrm{\parallel }f-f_i{\mathrm{\parallel }}_2<r_i\mathrm{/}2$; then $\mathrm{\parallel }{P}_{N}f-f_i{\mathrm{\parallel }}_2<3r_i\mathrm{/}4$, so both points lie in the full continuity ball $B(f_i,r_i)$ and $\mathrm{\parallel }T(P_{N}f)-T(f){\mathrm{\parallel }}_2<\epsilon \mathrm{/}8<\epsilon \mathrm{/}4$. The set $P_{N}C$ and its continuous image $T(P_{N}C)$ are compact. Fourier projections converge uniformly on the latter by the finite-net argument of step 1.1, so choose $M$ with ${\sup }_{z\in P_{N}C}\mathrm{\parallel }{P}_{M}T(z)-T(z){\mathrm{\parallel }}_2<\epsilon \mathrm{/}4$.

The range of $P_N$ is finite dimensional. In a real orthonormal trigonometric basis, define $g:P_{N}C\to {\mathbb{R}}^q$ by listing the coefficients of $P_{M}T(z)$. By [F2], approximate $g$ uniformly on compact $P_{N}C$ by a finite $\sigma$-network with Euclidean coefficient error less than $\epsilon \mathrm{/}8$; orthonormality makes the corresponding output-polynomial error less than $\epsilon \mathrm{/}8$ in $L^2$.

First apply the finite Fourier multiplier $P_N$ from step 2.1, retaining the WHOLE real Fourier polynomial, including both sine and cosine components of each nonzero frequency. Its coefficient vectors lie in a compact set, so all its pointwise values are uniformly bounded. Inject the finitely many retained real basis functions ${\phi }_j(x)$ as periodic bias channels. The multiplication map $(z,t)\mapsto zt$ is continuous on the compact ranges of $(P_{N}f)(x)$ and ${\phi }_j(x)$; approximate it by the pointwise networks of [F2], then use a zero-frequency multiplier to average each product. Since $\int (P_{N}f)(x){\phi }_j(x)dx$ is exactly the real coefficient of ${\phi }_j$, these constant channels approximate ALL desired coefficients without using an impossible translation-invariant cosine-only projection. Mandatory activations can be passed with arbitrarily small error on these bounded channels: choose $c$ with ${\sigma }^{\mathrm{\prime }}(c)\mathrm{\ne }0$, encode $z$ as $c+\tau z$, and decode affinely by $(\sigma (c+\tau z)-\sigma (c))\mathrm{/}(\tau {\sigma }^{\mathrm{\prime }}(c))$, which converges uniformly to $z$ on every bounded interval as $\tau \to 0$. Approximate the finite products and implement the coefficient network from step 3.1 using finitely many such layers. At each stage uniform continuity on the compact finite coefficient ranges propagates arbitrarily small errors through the finite composition; zero-mode averaging has norm one.