SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Again, we will make use of a heuristic argument to argue that the middle term can be controlled by controlling  $\lambda(E)$  since  $A$  sends Lipschitz functions to Lipschitz functions outside of  $E$  and has bounded  $L^1$  norm inside of  $E$ . Therefore:
$$
\left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} \leq \frac {3 \left(C _ {v} + 1\right) ^ {2}}{n ^ {2}} + 3 \| A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] - A [ A ^ {k - 1} f _ {2} ] \| _ {2} ^ {2}. \tag {179}
$$
Solve the recurrent to get the result in the statement of the Lemma.
Lemma 14. In the same setting as Lemma 13, recall that a tilde over a function in  $\mathcal{F}_n$  denotes its extension to  $\mathcal{F}$ . If  $A$  satisfies Assumption 2 with constant  $C_A$  and Assumption 5.A with constant  $C_c$  and resolution set  $\mathcal{N}$ . Let  $\Phi(h, A, \cdot) = \rho\left(\sum_{g \in n_l} \sum_{k=0}^{K-1} A^k \Phi_g(h, A, \cdot)\right)$  for some  $\Phi_g$  graphop neural network such that  $\|\widetilde{\Phi}_g(h, A_n, f) - \Phi_g(h, A, \tilde{f})\|_2 < M$  for all  $g \in [n_l]$ . Then:
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} ^ {2} \leq K ^ {2} n _ {l} ^ {2} 3 ^ {K} \left(K n ^ {- 2} C _ {c} ^ {2} C _ {A} ^ {2 K} + C _ {A} ^ {2 K} M\right). \tag {180}
$$
If instead  $A$  satisfies Assumption 5.B with constant  $C_c$  then the bound becomes:
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} ^ {2} \leq K ^ {2} n _ {l} ^ {2} 3 ^ {K} \left(K n ^ {- 2} \left(C _ {c} + 1\right) ^ {2} C _ {A} ^ {2 K} + C _ {A} ^ {2 K} M\right). \tag {181}
$$
If instead  $A$  satisfies Assumption 4.B then the bound becomes:
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} ^ {2} \leq K ^ {2} n _ {l} ^ {2} 3 ^ {K} \left(K n ^ {- 2} \left(C _ {v} + 1\right) ^ {2} C _ {A} ^ {2 K} + C _ {A} ^ {2 K} M\right). \tag {182}
$$
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} ^ {2} \tag {183}
$$
$$
= \sum_ {u \in \frac {1}{n} [ n ]} \int_ {u - 1 / n} ^ {u} \left(\rho \left(\sum_ {g} \sum_ {k} \widetilde {A _ {n} ^ {k} \Phi_ {g}} (h, A _ {n}, f) (x)\right) - \rho \left(\sum_ {g} \sum_ {k} A ^ {k} \Phi_ {g} (h, A, \tilde {f}) (x)\right)\right) ^ {2} d \lambda (x) \tag {184}
$$
$$
\leq \sum_ {u \in \frac {1}{n} [ n ]} \int_ {u - 1 / n} ^ {u} \left(\sum_ {g} \sum_ {k} \widetilde {A _ {n} ^ {k} \Phi_ {g}} (h, A _ {n}, f) (x) - A ^ {k} \Phi_ {g} (h, A, \tilde {f}) (x)\right) ^ {2} d \lambda (x) \tag {185}
$$
$$
\leq \sum_ {u \in \frac {1}{n} [ n ]} \int_ {u - 1 / n} ^ {u} \left(\sum_ {g} \sum_ {k} \left| \widetilde {A _ {n} ^ {k} \Phi_ {g}} (h, A _ {n}, f) (x) - A ^ {k} \Phi_ {g} (h, A, \tilde {f}) (x) \right|\right) ^ {2} d \lambda (x) \tag {186}
$$
$$
\leq K n _ {l} \sum_ {u \in \frac {1}{n} [ n ]} \int_ {u - 1 / n} ^ {u} \sum_ {g} \sum_ {k} \left(\left| \widetilde {A _ {n} ^ {k} \Phi_ {g}} (h, A _ {n}, f) (x) - A ^ {k} \Phi_ {g} (h, A, \tilde {f}) (x) \right|\right) ^ {2} d \lambda (x) \tag {187}
$$
$$
= K n _ {l} \sum_ {g} \sum_ {k} \| \widetilde {A _ {n} ^ {k} \Phi_ {g}} (h, A _ {n}, f) (x) - A ^ {k} \Phi_ {g} (h, A, \tilde {f}) (x) \| _ {2} ^ {2} \tag {188}
$$
$$
\leq K n _ {l} \sum_ {g} \sum_ {k} \frac {3 ^ {k + 1} k C _ {c} ^ {2} C _ {A} ^ {2 k}}{n ^ {2}} + 3 ^ {k} C _ {A} ^ {2 k} \| \widetilde {\Phi_ {g}} (h, A _ {n}, f) - \Phi_ {g} (h, A, \tilde {f}) \| _ {2} ^ {2} \tag {189}
$$
$$
\leq n ^ {- 2} n _ {L} ^ {2} \cdot K ^ {3} \cdot C _ {c} ^ {2} \cdot 3 ^ {K} C _ {A} ^ {2 K} + K ^ {2} \cdot n _ {l} ^ {2} \cdot 3 ^ {K} C _ {A} ^ {2 K} \cdot M \tag {190}
$$
$$
= K ^ {2} n _ {l} ^ {2} 3 ^ {K} \left(K n ^ {- 2} C _ {c} ^ {2} C _ {A} ^ {2 K} + C _ {A} ^ {2 K} M\right), \tag {191}
$$
where in the first line, we use the fact that continuous extension commutes with finite element-wise sum and element-wise application of  $\rho$ , while expanding  $L^2$  norm; the second line uses Lipschitz property of  $\rho$ ; the third line uses triangle inequality; the fourth line uses equivalence of  $p$ -norms in finite dimensional vectors; the fifth line is again  $L^2$  norm definition; the sixth line applies Lemma 13 and the rest is algebra.
To get the rest of the cases, applies different versions of Lemma 13.
Lemma 15. Let  $\Phi$  be an  $L$ -layer graphop neural network in the same setting as Lemma 13. If  $A$  satisfies Assumption 5.A with constant  $C_c$  then:
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} \leq n ^ {- 1} \left(3 ^ {K} K n _ {\max } C _ {A} ^ {K}\right) ^ {L} \max  \left(C _ {v}, 3 ^ {K} K ^ {2} C _ {c} n _ {\max } C _ {A} ^ {K}\right). \tag {192}
$$
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} \leq n ^ {- 1} \left(3 ^ {K} K n _ {\max } C _ {A} ^ {K}\right) ^ {L} \max  \left(C _ {v}, 3 ^ {K} K ^ {2} \left(C _ {c} + 1\right) n _ {\max } C _ {A} ^ {K}\right). \tag {193}
$$
If instead  $A$  satisfies Assumption 4.2 then the bound becomes:
$$
\left\| \widetilde {\Phi} (h, A _ {n}, f) - \Phi (h, A, \tilde {f}) \right\| _ {2} \leq n ^ {- 1} \left(3 ^ {K} K n _ {\max } C _ {A} ^ {K}\right) ^ {L} \max  \left(C _ {v}, 3 ^ {K} K ^ {2} \left(C _ {v} + 1\right) n _ {\max } C _ {A} ^ {K}\right). \tag {194}
$$
Proof. Solve the recurrent in Lemma 14.
We now prove a more general version of Theorem 3
Theorem 6. Let  $A \in \mathcal{F}$  satisfying Assumption 2 with constant  $C_A$  and Assumption 5.A with constant  $C_c$  and resolution set  $\mathcal{N} \subseteq \mathbb{N}$ . Let  $n \in \mathcal{N}$  and form the discretization  $A_n$  as per Theorem 2. Let  $h$  be normalized such that  $|h| \leq 1$  element-wise and form the graphop neural network  $\Phi(h, A, \cdot): \mathcal{F} \to \mathcal{F}$  and  $\Phi(h, A_n, \cdot): \mathcal{F}_n \to \mathcal{F}_n$ . We have the following approximation bound:
$$
d _ {M} \left(\Phi (h, A, \cdot), \Phi (h, A _ {n}, \cdot)\right) \leq n ^ {- 1 / 2} P _ {1} \sqrt {\overline {{C _ {A}}} C _ {v} + C _ {c} n _ {\max } C _ {A} ^ {K} \cdot P _ {2}} + C _ {v} n ^ {- 1} \tag {195}
$$
where  $P_{1} = 3^{KL}$  and  $P_{2} = 3^{K}K^{2}$ .
$$
d _ {M} \left(\Phi (h, A, \cdot), \Phi (h, A _ {n}, \cdot)\right) \leq n ^ {- 1 / 2} P _ {1} \sqrt {\overline {{C _ {A}}} C _ {v} + \left(C _ {c} + 1\right) n _ {\max } C _ {A} ^ {K} \cdot P _ {2}} + \left(C _ {v} + 1\right) n ^ {- 1} \tag {196}
$$
$$
d _ {M} \left(\Phi (h, A, \cdot), \Phi (h, A _ {n}, \cdot)\right) \leq n ^ {- 1 / 2} P _ {1} \sqrt {\overline {{C _ {A}}} C _ {v}} + \left(C _ {v} + 1\right) n ^ {- 1} \tag {197}
$$
Proof. The proof structure is similar to the proof of Theorem 2. For brevity of exposition, we will write  $h$  as the (shared) set of parameters in the graphop neural networks and shorten  $\Phi \coloneqq \Phi(h, A, \cdot)$  and  $\Phi_n \coloneqq \Phi(h, A_n, \cdot)$ . Fix  $k \in \mathbb{N}$ , we will bound  $\sup_{\eta \in S_{k,C_v}(\Phi)} \inf_{\overline{\eta}_n \in S_{k,C_v}(\Phi_n)} d_{LP}(\overline{\eta}, \overline{\eta}_n)$ .
Fix arbitrary  $\overline{\eta} \in \mathcal{S}_{k,C_v}(\Phi)$ . By definition of a  $(k,C_v)$ -profile, there exists  $f_1,\ldots ,f_k \in L_{\mathrm{reg}(C_v)}^\infty ([0,1])$  such that  $\overline{\eta} = \mathcal{D}_{\Phi}(f_1,\dots ,f_k)$ . For all  $j \in [k]$ , let  $\mathcal{F}_n \ni f_j' : [n] / n \ni u \mapsto \int_{u - 1 / n}^u f_j(z)\mathrm{d}\lambda (z)$ . That  $f_j' \in L_{\mathrm{reg}(C_v)}^\infty ([n] / n)$  is shown in Lemma 10. Set  $\overline{\eta}_n = \mathcal{D}_{\Phi_n}(f_1',\ldots ,f_k')$ .
Bounding  $d_{LP}(\overline{\eta},\overline{\eta}_n)$ . Fix some  $\epsilon >0$  to be specified later, by definition of  $d_{LP}$ , we need to bound, for each  $U\in \mathcal{B}_k$ :
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) = \int_ {0} ^ {1} \mathbb {1} _ {\left(f _ {1} (x), \dots , \Phi f _ {k} (x)\right) \in U} \mathrm {d} \lambda (x) - \sum_ {u \in [ n ] / n} \frac {1}{n} \mathbb {1} _ {\left(f _ {1} ^ {\prime} (u), \dots , \Phi_ {n} f _ {k} ^ {\prime} (u)\right) \in U ^ {\epsilon}}. \tag {198}
$$
Notice that since  $\lambda$  is the Lebesgue measure, one can denote  $y_{x} = \lceil xn\rceil /n$  and write:
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) = \int_ {0} ^ {1} \mathbb {1} _ {\left(f _ {1} (x), \dots , \Phi f _ {k} (x)\right) \in U} - \mathbb {1} _ {\left(f _ {1} ^ {\prime} \left(y _ {x}\right), \dots , \Phi_ {n} f _ {k} ^ {\prime} \left(y _ {x}\right)\right) \in U ^ {\epsilon}} \mathrm {d} \lambda (x). \tag {199}
$$
Since the integrand is only positive when  $(f_{1}(x),\ldots ,\Phi f_{k}(x))\in U$  and  $(f_1'(y_x),\dots,\Phi_nf_k'(y_x))\notin U^\epsilon$  and this conjunction only happens when  $\| (f_1(x),\dots,\Phi f_k(x)) - (f_1'(y_x),\dots,\Phi_nf_k'(y_x))\| _2 > \epsilon$ , we have:
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) \leq \lambda (\mathcal {E} (\epsilon)) \leq \sum_ {j = 1} ^ {k} \lambda \left(\mathcal {E} _ {j} ^ {0} (\epsilon)\right) + \lambda \left(\mathcal {E} _ {j} ^ {1} (\epsilon)\right), \tag {200}
$$
$$
\mathcal {E} (\epsilon) := \left\{x: \| \left(f _ {1} (x), \dots , \Phi f _ {k} (x)\right) - \left(f _ {1} ^ {\prime} \left(y _ {x}\right), \dots , \Phi_ {n} f _ {k} ^ {\prime} \left(y _ {x}\right)\right) \| _ {2} > \epsilon \right\}, \tag {201}
$$
$$
\mathcal {E} _ {j} ^ {z} (\epsilon) = \left\{x: \left| \Phi^ {z} f _ {j} (x) - \Phi_ {n} ^ {z} f _ {j} ^ {\prime} (y _ {x}) \right| > \frac {\epsilon}{\sqrt {2 k}} \right\}, \text {f o r} z \in \{0, 1 \}, j \in [ k ], \tag {202}
$$
where  $B^0$  is the identity operator for any operator  $B$  in the appropriate space.
Bounding  $\lambda (\mathcal{E}_j^1 (\epsilon))$ . Fix  $j\in [k]$ , we have:
$$
\int_ {0} ^ {1} | \Phi f _ {j} (x) - \Phi_ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) | \mathrm {d} x \leq \sqrt {\int_ {0} ^ {1} \left(\Phi f _ {j} (x) - \Phi_ {n} f _ {j} ^ {\prime} \left(y _ {x}\right)\right) ^ {2}} \mathrm {d} x \tag {203}
$$
$$
= \left\| \Phi f _ {j} - \widetilde {\Phi_ {n}} f _ {j} ^ {\prime} \right\| _ {L ^ {2}} \tag {204}
$$
$$
\leq n ^ {- 1} \left(3 ^ {K} K n _ {\max } C _ {A} ^ {K}\right) ^ {L} \max  \left(C _ {v}, 3 ^ {K} K ^ {2} C _ {c} n _ {\max } C _ {A} ^ {K}\right). \tag {205}
$$
where the last line uses Lemma 15 under Assumption 5.A with constant  $C_c$ . Similar results are obtained for the other assumptions.
We also have:
$$
\int_ {0} ^ {1} \left| \Phi f _ {j} (x) - \Phi_ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) \right| \mathrm {d} x \geq \lambda \left(\mathcal {E} _ {j} ^ {1} (\epsilon)\right) \cdot \frac {\epsilon}{\sqrt {2 k}} + 0. \tag {206}
$$
Thus selecting
$$
\epsilon > \sqrt {2 k \sqrt {2 k} \cdot n ^ {- 1} \left(3 ^ {K} K n _ {\max } C _ {A} ^ {K}\right) ^ {L} \max  \left(C _ {v}, 3 ^ {K} K ^ {2} C _ {c} n _ {\max } C _ {A} ^ {K}\right)} \tag {207}
$$
makes  $\lambda (\mathcal{E}_j^1 (\epsilon)) <   \frac{\epsilon}{2k}$
Bounding  $\lambda (\mathcal{E}_j^0 (\epsilon))$ . This step simply uses Lipschitzness of the graph signal to bound its discretization and is thus identical to that in Theorem 2. In short, choosing  $\epsilon >\sqrt{2k} C_v / n$  gives  $\lambda (\mathcal{E}_j^0 (\epsilon)) = 0$ .
Bounding  $\sup_{\eta_n\in S_k,C_v(\Phi_n)}\inf_{\eta \in S_k,C_v(\Phi)}d_{LP}(\eta ,\eta_n)$ . This direction uses the same technique as in Theorem 2.
Putting everything together Defining  $\overline{C_A} = (n_{\mathrm{max}}K C_A^K)^L$  gives a choice of:
$$
\bar {\epsilon} = n ^ {- 1 / 2} P _ {1} \sqrt {\overline {{C _ {A}}} C _ {v} + C _ {c} n _ {\max } C _ {A} ^ {K} \cdot P _ {2}} + C _ {v} n ^ {- 1} \tag {208}
$$
For other choices of assumptions, we obtain similar bounds.
E.1 Proof of Theorem 4
Because of completeness of  $d_H$  in the space of closed subsets of  $\mathcal{P}(\mathbb{R}^{2k})$  for every  $k\in \mathbb{N}$ , the statement in Theorem 4 is equivalent to showing that for each  $k\in \mathbb{N}$ , we have  $\mathcal{S}_{k,L(n)}(A_n)$  converges to  $\mathcal{S}_k(A)$  in  $d_H$ . We do this via a mollification argument.
Definition 5 (Lipschitz mollifier). A Lipschitz mollifier in  $\mathbb{R}$  is a smooth (infinitely differentiable) function  $\phi : \mathbb{R} \to \mathbb{R}$  satisfying:
1.  $\int_{\mathbb{R}}\phi (x)\mathrm{d}\lambda (x) = 1$
2.  $\lim_{\epsilon \to 0}\phi_{\epsilon}(x)\coloneqq \lim_{\epsilon}\epsilon^{-1}\phi (x / \epsilon) = \delta (x)$  - the Dirac function.
3. Although not standard, we require  $\phi$  to be 1-Lipschitz and symmetric around 0  $(\phi (x) = \phi (-x))$
Given a measurable function  $f \in L_{[-1,1]}^{\infty}(\mathbb{R})$ , defines the convolution operation:
$$
f * \phi : x \rightarrow \int_ {\mathbb {R}} f (y) \phi (y - x) \mathrm {d} \lambda (y). \tag {209}
$$
The next result shows the existence of such a function:
Lemma 16. Let  $\phi : \mathbb{R} \to \mathbb{R}$  be:
$$
\phi (x) = \left\{ \begin{array}{l l} e ^ {- \left(1 - x ^ {2}\right) ^ {2}} / Z & i f | x | \leq 1 \\ 0 & o t h e r w i s e, \end{array} \right. \tag {210}
$$
where  $Z$  is a normalization constant to make sure that  $\int_{\mathbb{R}} \phi \, \mathrm{d}\lambda = 1$ . Then  $\phi$  is a Lipschitz mollifier.
Proof. The first property is built into the definition. The second property is obvious since the support of  $\phi(x / \epsilon)$  is  $[- \epsilon, \epsilon]$  and thus it converges to the Dirac function as  $\epsilon$  goes to 0. Lipschitz-ness can be seen by computing the first derivative (since the function is smooth) over  $(-1, 1)$  and see that it is bounded in  $[-1, 1]$ . Symmetry is also obvious since the function depends only on the absolute value of its argument.
Consequences of a Lipschitz mollifier include:
Lemma 17. Let  $\phi$  be a Lipschitz mollifier and  $\epsilon >0$ , then for any measurable  $f\in \mathcal{F}_{[-1,1]}^{\infty}(\mathbb{R})$ ,  $f*\phi_{\epsilon}$  is  $\max (1,\epsilon^{-2})$ -Lipschitz and  $\lim_{\epsilon \to 0}\| f - f*\phi_{\epsilon}\| _2 = 0$ .
Proof. For the first part of the statement, consider:
$$
\begin{array}{l} \left| f * \phi_ {\epsilon} (x) - f * \phi_ {\epsilon} (y) \right| = \left| \int_ {\mathbb {R}} f (z) \phi_ {\epsilon} (z - x) - f (z) \phi (z - y) \mathrm {d} \lambda (z) \right| (211) \\ \leq \int_ {\mathbb {R}} \left| \frac {f (z)}{\epsilon} \left(\phi \left(\frac {z - x}{\epsilon}\right) - \phi \left(\frac {z - y}{\epsilon}\right)\right) \right| d \lambda (z) (212) \\ \leq \frac {1}{\epsilon} \int_ {\mathbb {R}} \left| \frac {z - x}{\epsilon} - \frac {z - y}{\epsilon} \right| \mathrm {d} \lambda (z) (213) \\ \leq \frac {| x - y |}{\epsilon^ {2}} (214) \\ \end{array}
$$
For the second part, we have:
$$
\begin{array}{l} \left| f * \phi_ {\epsilon} (x) - f (x) \right| \leq \int_ {\mathbb {R}} \phi_ {\epsilon} (z - x) | f (z) - f (x) | \mathrm {d} \lambda (z) (215) \\ \leq \int_ {\mathbb {R}} \phi_ {\epsilon} (z) | f (z + x) - f (x) | \mathrm {d} \lambda (z), (216) \\ \end{array}
$$
where we use two changes of variables.
Square both sides and take integral over  $x$  and use Jensen inequality to get:
$$
\left\| f * \phi_ {\epsilon} - f \right\| _ {2} ^ {2} \leq \int_ {\mathbb {R}} \int_ {\mathbb {R}} \left(\phi_ {\epsilon} (z)\right) ^ {2} \left(f (z + x) - f (x)\right) ^ {2} \mathrm {d} \lambda (z) \mathrm {d} \lambda (x). \tag {217}
$$
Apply Fubini-Tonelli theorem to factorize the mollifier, we get:
$$
\begin{array}{l} \left\| f * \phi_ {\epsilon} - f \right\| _ {2} ^ {2} \leq \int_ {\mathbb {R}} \int_ {\mathbb {R}} (f (z + x) - f (x)) ^ {2} d \lambda (x) \left(\phi_ {\epsilon} (z)\right) ^ {2} d \lambda (z) (218) \\ = \int_ {\mathbb {R}} \left(\epsilon^ {- 1} \int_ {\mathbb {R}} (f (z + x) - f (x)) ^ {2} \mathrm {d} \lambda (x)\right) (\phi (z \epsilon^ {- 1})) ^ {2} \mathrm {d} \lambda (z). (219) \\ \end{array}
$$
Apply a final change of variable:
$$
\left\| f * \phi_ {\epsilon} - f \right\| _ {2} ^ {2} \leq \int_ {\mathbb {R}} \left(\int_ {\mathbb {R}} (f (z \epsilon + x) - f (x)) ^ {2} \mathrm {d} \lambda (x)\right) (\phi (z)) ^ {2} \mathrm {d} \lambda (z). \tag {220}
$$
Therefore, as  $\epsilon$  goes to 0, the inner integrand goes to 0. Since  $f$  and  $f*\phi_{\epsilon}$  are all bounded (as  $f\in [-1,1]$ ) for small enough  $\epsilon$ , we can apply dominated convergence to conclude that the integral itself goes to 0 and thus the 2-norm on the left hand side also goes to 0.
We are now ready to proceed with the proof of Theorem 4. Given an element  $\eta \in S_k(A)$ , there exists a set of functions  $F = \{f_1, \ldots, f_k\} \subseteq L_{[-1,1]}^{\infty}([0,1])$ . Each of these functions can be extended to  $\mathbb{R}$  by setting  $f(x) = f(0)$  if  $x \leq 0$  and  $f(x) = f(1)$  if  $x > 1$ . Call the extended function  $f'$ . Now we can apply mollification convolution to each of them to get a family of functions  $\{f_{j,\epsilon} := f_j' * \phi_\epsilon\}_{j \in [k], \epsilon > 0}$ . Recall that we have shown  $f_{j,n} := f_{j,1/\sqrt{L(n)}}$  to be  $L(n)$ -Lipschitz for each  $n \in \mathbb{N}$ . Let  $f_{j,n}''$  be the restriction of  $f_{j,n}$  to  $\mathcal{F}_n$ . Then  $f_{j,n}''$  is still  $L(n)$ -Lipschitz (in the metric on  $1/n[n]$  induced by metric on  $[0,1]$ ) and thus we can find a profile for  $F_n'' := \{f_{1,n}', \ldots, f_{k,n}'\}$  in  $\mathcal{S}_{k,L(n)}(A_n)$ .
Furthermore, by property of mollifier and the fact that  $L(n) \to \infty$  as  $n$  goes to infinity, we have  $\| f_{j,n} - f_j' \|_2 \to 0$  with  $n \to \infty$ . Since  $f_j'$  is constant outside [0,1], we also have  $\| f_{j,n,|[0,1]} - f_j \|_{\mathcal{F}} \to 0$  with  $n$ . Using the same proof technique as Theorem 2, we can conclude that:
$$
d _ {L P} \left(\mathcal {D} _ {A _ {n}} \left(F _ {n} ^ {\prime \prime}\right), \mathcal {D} _ {A} (F)\right) \xrightarrow {n \rightarrow \infty} 0, \tag {221}
$$
and thus  $\mathcal{S}_k(A) \subseteq \lim_n \overline{\mathcal{S}_{k,L(n)}(A_n)}$ . For the other direction, recall that  $\overline{\mathcal{S}_k(A)} = \lim_n \overline{\mathcal{S}_k(A_n)}$  and that  $\mathcal{S}_{k,L(n)}(A_n) \subseteq \mathcal{S}_k(A_n)$ . Together with completeness of  $d_H$ , we conclude that
$$
\lim  _ {n} d _ {H} \left(\mathcal {S} _ {k} (A), \mathcal {S} _ {k, L (n)} (A _ {n})\right) = 0.
$$
Conjecture 1 (Action convergence of graphop neural networks). Let  $(A_{n})_{n\in \mathbb{N}}$  be an action convergent sequence of graphops. Then  $(\Phi (h,A_n,\cdot))_{n\in \mathbb{N}}$  is an action convergent sequence of  $P$ -operators.