SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof. Given that the cardinality of vertices of each color is the same, it is straightforward to map (via a Lebesgue-measure preserving bijection) vertices of  $V$  into equipartition of  $[0,1]$  into  $N$  pieces  $I_1,\ldots ,I_N$  such that each partition contains vertices of the same color and vertices from different partitions will have different colors. Let  $A$  be the normalized adjacency operator of  $G$  and  $f$  be a function with finite  $L^2$  norm such that  $f$  is constant on each  $I_{j}$  for each  $j$  from 1 to  $N$ . The goal is to show that  $Af$  is also constant on these pieces.
We show this by direct computation. Pick a vertex  $x$  and another vertex  $y$  from the same piece, say  $I_k$  for some particular  $k$ . By our construction,  $x$  and  $y$  have the same color since they come from the same piece. By our assumption, the multiset of their neighbors' colors is the same. However, since  $f$  is constant on each piece, we have  $f(z) = f(u)$  for  $u, v$  of the same color. Therefore, the multisets  $\{f(z) : (z, x) \in E\}$  and  $\{f(z) : (z, y) \in E\}$  is exactly the same. Taking the appropriate countable sum over each multiset thus result in the same number, i.e.  $Af(x) = Af(y)$ , which finishes the proof.
Note that we can drop the countable requirement of the previous proof by invoking an appropriate integral definition. Now we show that the hypercubes - an intermediate graph that is neither dense nor bounded degree, satisfies the requirements of Lemma 6.
Lemma 7 (Well-behaved operators - Polymer graphs). Let  $P$  be a polymer graphing on finite monomer  $n$ -vertex graph  $M = (V_M, E_M)$ . Let  $A$  be then normalized (by each vertex degree) adjacency operator of a graphing. For each  $N \in \mathbb{N}$ , there exists locally equivalent graphings:
1.  $P_N'$  such that  $A(P_n')$  satisfies Assumption 3.A with resolution set  $nd(N) = \{nx \in \mathbb{N} : x\alpha = N \text{ for some } \alpha \in \mathbb{N}\}$ .
2.  $P_N^{\prime \prime}$  such that  $A(P_N^{\prime \prime})$  satisfies Assumption 4.A with resolution set  $n[N]$ .
3.  $P_N^{\prime \prime \prime}$  such that  $A(P_N^{\prime \prime \prime})$  satisfies Assumption 4.B with resolution set  $n[N]$ .
Proof. The exact same technique as Lemma 5 (make copies of 'shrinked down' graphings to get the result with Assumption 3.A and find a small enough  $\delta(N)$  so that the wrap around has small measure to get the result with Assumption 4.A and Assumption 4.B) works. Intuitively, this is because we are replacing each node in the path graphing with a monomer graph.
To handle coloring at the monomers details, use Lemma 6 with the coloring that assigns the same color to the same node in each monomer (since each monomer is a copy of each other, there is an isomorphism between the nodes and by 'the same node' we meant under this isomorphism).
Lemma 8 (Well-behaved operators - Graphons). Let  $W:[0,1]^2\to [0,1]$  be a Lipschitz graphon, with Lipschitz constant  $C$ . In other words,  $W$  has finite  $L^2$  norm and is Lipschitz in both variables. Then the Hilbert-Schmidt integral operator  $H$  that defines the adjacency operator of  $W$  satisfies Assumption 3.B at any resolution when applied to graph(on) signal  $f$  with  $L^1$  norm 1.
Proof. The proof follows from definition. Take  $f \in L^{2}([0,1])$  such that  $\| f \|_{L^2} = 1$ , then:
$$
\begin{array}{l} \left| H f (x) - H f (y) \right| = \left| \int_ {0} ^ {1} W (x, z) f (z) \mathrm {d} z - \int_ {0} ^ {1} W (y, z) f (z) \mathrm {d} z \right| (38) \\ \leq \int_ {0} ^ {1} | W (x, z) - W (y, z)) | \cdot | f (z) | \mathrm {d} z (39) \\ \leq \int_ {0} ^ {1} C | x - y | \cdot | f (z) | \mathrm {d} z = C | x - y |, (40) \\ \end{array}
$$
where the first line is definition of the Hilbert-Schmidt operator  $H$ , the second line is triangle inequality and the last line is due to Lipschitzness of  $W$  and  $L^1$  norm of  $f$
Lemma 9 (Well-behaved operators - Hypercubes). Let  $C$  be a hypercube of dimension  $N$ . Then the normalized adjacency matrix of  $C$  satisfies assumption 3.A with resolutions  $2^{[n]}$  for each  $n < N$ .
Proof. We first display a mapping of the vertices in the hypercube over the interval  $[0,1]$  such that Assumption 3.A will be shown to hold. Recall that a hypercube vertices can be represented by a binary string of  $N$  numbers. Here, similarly, we will associate a vertex  $v$  of a hypercube with a number in  $[0,1]$  by adding a 0. in front of its binary representation. For example, 0.110101 (as a binary number) is a vertex correspond to the string 110101 when  $N = 6$ . We call this representation mapping  $f:V\to [0,1]$ . Two vertices  $u,v$  are connected by an edge iff  $f(v)$  and  $f(v)$  differs by exactly one digit in their binary representation.
To get a hypercube representation over the interval  $[0,1]$ , we simply take disconnected copies of (uncountably) infinitely many hypercubes described above. To be more precise, for a number  $x\in [0,1]$ , let  $A(x)$  be the  $N$ -letter binary string such that  $0.A$  (as a string) is the truncation of  $x$  to the  $N$ -th digit after the binary point. Then,  $x$  is connected to  $y\in [0,1]$  iff  $A(x)$  and  $A(y)$  differs in exactly one digit and  $x - A(x) = y - A(y)$ . For example, for  $N = 4$ , we connect 0.110101 with 0.010101, 0.100101, 0.111101 and 0.110001. Notice how the first four digits after the dots differ from the original number at exactly one letter, and the last few digits always stay the same. In this example,  $A(x) = 0.1101$  for  $x = 0.110101$ .
With this vertex representation in mind, we verify the conditions of Lemma 6. Let  $V$ . Fix  $n < N$  and divide [0,1] into  $2^{n}$  equipartitions and let vertices in the same partition enjoy the same color. We will also number the partition/color by the binary representation of their left hand endpoint. For example, the first interval is labelled 00..0 (  $n$  digits), the second interval is labelled 00..01 (  $n$  digits). Select a vertex  $x$  from a partition with label  $a_{1}a_{2}\ldots a_{n}$ . Then, by the definition of our hypercube representation, the neighbors of  $x$  have colors:
- Exactly 1 neighbor with color  $(a_{1} + 1 \mod 1)a_{2}\ldots a_{n}$ .
- Exactly 1 neighbor with color  $a_1(a_2 + 1 \mod 1)a_3 \ldots a_n$ .
.
- Exactly 1 neighbor with color  $a_1a_2 \ldots a_{n-1}(a_n + 1 \mod 1)$ .
- Remaining  $N - n$  neighbors all have color  $a_1a_2\ldots a_n$ .
Therefore, the multiset of neighbors' colors of  $x$  only depends on the fact that  $x$  comes from a partition with label  $a_1 \ldots a_n$  and thus, two vertices of the same color has the same multiset of neighbor's colors. Since the condition of Lemma 6 holds, we have the conclusion for this particular resolution  $n$ . Since  $n$  was chosen arbitrarily, the result holds for every  $n < N$ .
In future work, we formalize the way in which the above proof holds for limiting object of hypercube.
In this section, we prove a slightly more general version of Theorem 2.
Theorem 5 (General approximation theorem). Let  $A: \mathcal{F} \to \mathcal{F}$  be a  $P$ -operator satisfying Assumption 2 with constant  $C_A$ ; Assumption 5.  $A$  with constant  $C_c$  and resolutions in  $\mathcal{N}$ . Fix  $n \in \mathcal{N}$  and consider  $(k, C_v)$ -profiles. Let  $A_n: \mathcal{F}_n \to \mathcal{F}_n$  be a discretization of  $A$  as defined in Equation (9). Then:
$$
d _ {M} (A, A _ {n}) \leq 8 \left(\sqrt {\frac {C _ {A} C _ {v}}{n}} + \frac {C _ {v} + C _ {c}}{n}\right). \tag {41}
$$
If instead of Assumption 5.A,  $A$  satisfies Assumption 5.B with constant  $C_c$  and resolution set  $\mathcal{N}$ , then:
$$
d _ {M} (A, A _ {n}) \leq 8 \left(\sqrt {\frac {C _ {A} C _ {v} + 1}{n}} + \frac {C _ {v} + C _ {c} + 1}{n}\right). \tag {42}
$$
If instead of Assumption 5.A and 5.B,  $A$  satisfies Assumption 4.B with resolution set  $\mathcal{N}$  then:
$$
d _ {M} (A, A _ {n}) \leq 8 \left(\sqrt {\frac {C _ {A} C _ {v} + 1}{n}} + \frac {C _ {v} + 1}{n}\right). \tag {43}
$$
Proof of Theorem 2. Fix  $k \in \mathbb{N}$ . In order to derive an upper bound for  $d_{M}$ , we find an upper bound for the Hausdorff distance  $d_{H}$  between the two  $k$ -profiles of  $A$  and  $A_{n}$ .
Bounding  $\sup_{\eta \in \mathcal{S}_{k,C_v}(A)}\inf_{\eta_n\in \mathcal{S}_{k,C_v}(A_n)}d_{LP}(\eta ,\eta_n)$ . To bound the supinf quantity, we first select an arbitrary  $\overline{\eta}\in \mathcal{S}_{k,C_v}(A)$ . From this measure, we will construct a measure  $\overline{\eta}_n\in \mathcal{S}_{k,C_v}(A_n)$ . If we can upper bound  $d_{LP}(\overline{\eta},\overline{\eta}_n) < M$  then we have:
$$
\inf  _ {\eta_ {n} \in \mathcal {S} _ {k, C _ {v}} (A _ {n})} d _ {L P} (\bar {\eta}, \eta_ {n}) \leq d _ {L P} (\bar {\eta}, \bar {\eta} _ {n}) <   M, \text {f o r a l l} \bar {\eta} \in \mathcal {S} _ {k, C _ {v}} (A). \tag {44}
$$
If further  $M$  does not depend on the choice of  $\overline{\eta}$ , then we have
$$
\sup_{\eta \in \mathcal{S}_{k,C_{v}}(A)}\inf_{\eta_{n}\in \mathcal{S}_{k,C_{v}}(A_{n})}d_{LP}(\eta ,\eta_{n})\leq M.
$$
We now proceed with this plan. Fix an arbitrary  $\overline{\eta} \in \mathcal{S}_k(A)$ , by definition of  $k$ -profiles, there is a corresponding tuple  $F = (f_1, \ldots, f_k)$  with elements in  $\mathcal{F}_{\mathrm{reg}(C_v)}$  such that  $\mathcal{D}_A(F) = \overline{\eta}$ .
Form:
$$
F ^ {\prime} := \left\{\mathcal {F} _ {n, \operatorname {r e g} \left(C _ {v}\right)} \ni f _ {j} ^ {\prime}: u \mapsto n \int_ {u - 1 / n} ^ {u} f _ {j} \mathrm {d} \lambda \mid j \in [ k ] \right\}, \quad \bar {\eta} _ {n} := \mathcal {D} _ {A _ {n}} \left(F ^ {\prime}\right). \tag {45}
$$
That  $f_j' \in \mathcal{F}_{n,\mathrm{reg}(C_v)}$  is asserted in Lemma 10.
Bounding  $d_{LP}(\overline{\eta},\overline{\eta}_n)$  For some  $\epsilon >0$  to be we want to show that  $d_{LP}(\overline{\eta},\overline{\eta}_n)\leq \epsilon$ , which is equivalent to showing, for any  $U\in \mathcal{B}_{2k}$
$$
\bar {\eta} (U) \leq \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) + \epsilon \quad \text {a n d} \quad \bar {\eta} _ {n} (U) \leq \bar {\eta} \left(U ^ {\epsilon}\right) + \epsilon . \tag {46}
$$
Recall that  $U^{\epsilon}$  was defined in Eqn (18).
Fix any  $U\in \mathcal{B}_{2k}$  , we have
$$
\bar {\eta} (U) = \int_ {\mathbb {R} ^ {2 k}} \mathbb {1} _ {U} \mathrm {d} \mathcal {D} _ {A} (F) = \int_ {0} ^ {1} \mathbb {1} _ {F _ {A} \in U} \mathrm {d} \lambda , \tag {47}
$$
$$
\bar {\eta} _ {n} \left(U ^ {\epsilon}\right) = \int_ {\mathbb {R} ^ {2 k}} \mathbb {1} _ {U ^ {\epsilon}} \mathrm {d} \mathcal {D} _ {A _ {n}} \left(F ^ {\prime}\right) = \sum_ {u \in [ n ] / n} \frac {1}{n} \mathbb {1} _ {F _ {A _ {n}} ^ {\prime} \in U ^ {\epsilon}}. \tag {48}
$$
Subtracting both sides yield:
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \mathbb {1} _ {F _ {A} (x) \in U} - \mathbb {1} _ {F _ {A n} ^ {\prime} (u) \in U ^ {\epsilon}} \mathrm {d} \lambda (x). \tag {49}
$$
$$
\bar {\eta} _ {n} (U) - \bar {\eta} \left(U ^ {\epsilon}\right) = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \mathbb {1} _ {F _ {A n} ^ {\prime} (u) \in U} - \mathbb {1} _ {F _ {A} (x) \in U ^ {\epsilon}} \mathrm {d} \lambda (x) \tag {50}
$$
Let  $y_{x} = \lceil x n \rceil / n$ . Recall that  $E \subset [0,1]$  defined in Assumption 5.B and 4.B is the set of  $x$  where Lipschitzness of the image under  $A$  may fail. Let  $E = \emptyset$  if we are using Assumption 5.A alternatively. Define the events:
$$
\mathcal {E} _ {U} ^ {1} (\epsilon) = \left\{x: F _ {A} (x) \in U \wedge F _ {A _ {n}} ^ {\prime} \left(y _ {x}\right) \notin U ^ {\epsilon} \right\} \tag {51}
$$
$$
\mathcal {E} _ {U} ^ {2} (\epsilon) = \left\{x: F _ {A _ {n}} ^ {\prime} \left(y _ {x}\right) \in U \wedge F _ {A} \notin U ^ {\epsilon} \right\} \tag {52}
$$
$$
\mathcal {E} ^ {\prime} (\epsilon) = \left\{x: \| F _ {A} (x) - F _ {A _ {n}} ^ {\prime} \left(y _ {x}\right) \| _ {2} > \epsilon \right\} \tag {53}
$$
$$
\mathcal {E} _ {j} (\epsilon) = \left\{x: \left| f _ {j} (x) - f _ {j} ^ {\prime} \left(y _ {x}\right) \right| > \epsilon / \sqrt {2 k} \right\}, \quad j = 1.. k \tag {54}
$$
$$
\mathcal {E} _ {j, A} (\epsilon) = \left\{x: \left| A f _ {j} (x) - A _ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) \right| > \epsilon / \sqrt {2 k} \right\}, \quad j = 1.. k. \tag {55}
$$
Using this notation, one also has
$$
\bar {\eta} (U) - \bar {\eta} _ {n} (U ^ {\epsilon}) \leq \lambda (\mathcal {E} _ {U} ^ {1}) \quad \text {a n d} \quad \bar {\eta} _ {n} (U) - \bar {\eta} (U ^ {\epsilon}) \leq \lambda (\mathcal {E} _ {U} ^ {2}). \tag {56}
$$
It is straightforward to see  $\mathcal{E}_U^l (\epsilon)\subseteq \mathcal{E}'(\epsilon),l = 1,2$  for any  $U$  by definition of  $U^{\epsilon}$ . Furthermore,
$$
\mathcal {E} ^ {\prime} (\epsilon) \subseteq \bigcup_ {j = 1} ^ {k} \mathcal {E} _ {j} (\epsilon) \cup \mathcal {E} _ {j, A} (\epsilon), \tag {57}
$$
since if all  $2k$  dimensions are bounded in absolute value by  $\epsilon / \sqrt{2k}$  then the Euclidean distance of the vector is bounded by  $\epsilon$ .
Therefore it suffices to bound  $\lambda (\mathcal{E}_j(\epsilon)) + \lambda (\mathcal{E}_{j,A}(\epsilon))$  for each  $j\in [k]$
Bounding  $\lambda (\mathcal{E}_j(\epsilon))$ . Since  $f_{j}$  is  $C_v$ -Lipschitz for all  $j$ , we have:
$$
\begin{array}{l} \left| f _ {j} (x) - f _ {j} ^ {\prime} \left(y _ {x}\right) \right| = \left| f _ {j} (x) - n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} f _ {j} (z) \mathrm {d} \lambda (z) \right| (58) \\ = n \left| \int_ {y _ {x} - 1 / n} ^ {y _ {x}} f _ {j} (x) - f _ {j} (z) \mathrm {d} \lambda (z) \right| (59) \\ = n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} | f _ {j} (x) - f _ {j} (z) | \mathrm {d} \lambda (z) (60) \\ \leq n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} \frac {C _ {v}}{n} \mathrm {d} \lambda (z) = \frac {C _ {v}}{n}, (61) \\ \end{array}
$$
where the first line is definition of  $f_{j}^{\prime}$ , the second line is because  $\lambda ((u - 1 / n,u]) = \frac{1}{n}$ , the third line is by triangle inequality and the last line is because of Lipschitzness of  $f_{j}$ .
Thus, choosing  $\epsilon >\sqrt{2k} C_v / n$  means that  $\lambda (\mathcal{E}_j(\epsilon)) = 0$ . (We can tighten this bound by only assuming that  $f_{j}$  is  $C_v$ -Lipschitz outside a set of small measure.)
Bounding  $\lambda (\mathcal{E}_{j,A})$ . Let  $\mathcal{F}_{[-1,1]}\ni \tilde{f}$  be the extension of  $f^{\prime}$  defined as  $\tilde{f} (x) = f^{\prime}(\lceil xn\rceil /n)$  for all  $x\in [0,1]$ . Note that  $\tilde{f}$  is not continuous in general and hence not Lipschitz. We have
for any  $x\in [0,1]$
$$
\begin{array}{l} \left| A f _ {j} (x) - A _ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) \right| = \left| A f _ {j} (x) - n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} A \tilde {f} _ {j} (z) \mathrm {d} \lambda (z) \right| (62) \\ \leq n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} | A f _ {j} (x) - A \tilde {f} _ {j} (z) | \mathrm {d} \lambda (z), (63) \\ \end{array}
$$
where we used uniformity of  $\lambda$  and triangle inequality. From here, we proceed slightly differently depending on the specific assumptions.
Proof via Assumption 5.A or 5.B. If  $A$  satisfies Assumption 5.A or Assumption 5.B, then we have for each  $x$ :
$$
\begin{array}{l} \left| A f _ {j} (x) - A _ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) \right| \leq n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} (x) - A \tilde {f} _ {j} (x) \right| + \left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) (64) \\ \leq \left| A f _ {j} (x) - A \tilde {f} _ {j} (x) \right| + n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z). (65) \\ \end{array}
$$
Define the following events:
$$
\mathcal {E} _ {j, A} ^ {1} (\epsilon) := \left\{x: \left| A f _ {j} (x) - A \tilde {f} _ {j} (x) \right| > \frac {\epsilon}{2 \sqrt {2 k}} \right\} \tag {66}
$$
$$
\mathcal {E} _ {j, A} ^ {2} (\epsilon) := \left\{x \notin E: n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} \left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) > \frac {\epsilon}{2 \sqrt {2 k}} \right\} \tag {67}
$$
Then it is clear that  $\mathcal{E}_{j,A} \subseteq \mathcal{E}_{j,A}^{1} \cup \mathcal{E}_{j,A}^{2} \cup E$  and thus  $\lambda(\mathcal{E}_{j,A}) \leq \lambda(\mathcal{E}_{j,A}^{1}) + \lambda(\mathcal{E}_{j,A}^{2}) + \lambda(E)$ .
Bounding  $\lambda (\mathcal{E}_{j,A}^{1}(\epsilon))$  via Assumption 5.A or 5.B. Because of Assumption 2, we have:  $\| A\tilde{f}_j - Af_j\| _2\leq C_A\| \tilde{f}_j - f_j\| _2$  . By  $L^p$  norms inequality, we have
$$
\left\| A \tilde {f} _ {j} (x) - A f _ {j} (x) \right\| _ {1} \leq C _ {A} \| \tilde {f} _ {j} - f _ {j} \| _ {2} \leq \frac {C _ {A} C _ {v}}{n}. \tag {68}
$$
where the last inequality is due to
$$
\begin{array}{l} \left\| \tilde {f} _ {j} - f _ {j} \right\| _ {2} ^ {2} = \int_ {0} ^ {1} \left(\tilde {f} _ {j} (x) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (69) \\ = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \left(f _ {j} ^ {\prime} (u / n) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (70) \\ = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \left(n \int_ {(u - 1) / n} ^ {u / n} f _ {j} (z) \mathrm {d} \lambda (z) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (71) \\ \leq \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} n ^ {2} \left(\int_ {u - 1 / n} ^ {u} | f _ {j} (z) - f _ {j} (x) | \mathrm {d} \lambda (z)\right) ^ {2} \mathrm {d} \lambda (x) (72) \\ \leq \frac {C _ {v} ^ {2}}{n ^ {2}}. (73) \\ \end{array}
$$
$$
\begin{array}{l} \frac {C _ {A} C _ {v}}{n} \geq \int_ {0} ^ {1} | A \tilde {f} _ {j} (x) - A f _ {j} (x) | d x (74) \\ = \int_ {\mathcal {E} _ {j, A} ^ {1} (\epsilon)} | A \tilde {f} _ {j} (x) - A f _ {j} (x) | d x + \int_ {[ 0, 1 ] \backslash \mathcal {E} _ {j, A} ^ {1} (\epsilon)} | A \tilde {f} _ {j} (x) - A f _ {j} (x) | d x (75) \\ \geq \frac {\epsilon}{2 \sqrt {2 k}} \lambda \left(\mathcal {E} _ {j, A} ^ {1} (\epsilon)\right) + 0. (76) \\ \end{array}
$$
Thus selecting  $\epsilon > 2\sqrt{(\sqrt{2}k\sqrt{k}C_A C_v) / n} = 2^{\frac{5}{4}}k^{\frac{3}{4}}(C_A C_v / n)^{\frac{1}{2}}$  gives  $\lambda(\mathcal{E}_{j,A}^{1}(\epsilon)) \leq \frac{\epsilon}{2k}$ .
Bounding  $\lambda (\mathcal{E}_{j,A}^2 (\epsilon))$  via Assumption 5.A or 5.B. Notice that  $\tilde{f}_j$  is constant in each  $(u - 1 / n,u]$  and thus by Assumption 5.A or 5.B,  $A\tilde{f}_j$  is  $C_c$ -Lipschitz in each  $(u - 1 / n,u]\backslash E$ . Therefore we have, for  $x\in [0,1]\backslash E$  and  $z\in (y_{x} - 1 / n,y_{x}]\backslash E$ :
$$
\left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \leq C _ {c} | x - z | \leq \frac {C _ {c}}{n}. \tag {77}
$$
When  $z \in E$ , we use the second condition in Assumption 5.B to get  $\| A\tilde{f}_j\mathbb{1}_E\|_1 \leq \frac{1}{n}$  and thus for any  $x \notin E$ :
$$
n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} | A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) | \mathrm {d} \lambda (z) \tag {78}
$$
$$
= n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \cap E} \left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) + n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \backslash E} \left| A \tilde {f} _ {j} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) \tag {79}
$$
$$
\leq \frac {C _ {c}}{n} + n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \cap E} | A \tilde {f} _ {j} (x) | + | A \tilde {f} _ {j} (z) | \mathrm {d} \lambda (z) \tag {80}
$$
$$
\leq \frac {C _ {c}}{n} + | A \tilde {f} _ {j} (x) | n \lambda \left(E _ {x}\right) + \| A \tilde {f} _ {j} \mathbb {1} _ {E _ {x}} \| _ {1} \tag {81}
$$
$$
\leq \frac {C _ {c} + 1}{n} + | A \tilde {f} _ {j} (x) | n \lambda \left(E _ {x}\right), \tag {82}
$$
where  $E_{x} = (y_{x} - 1 / n, y_{x}] \cap E$ .
For  $x\in \mathcal{E}_{j,A}^{2}(\epsilon)$  , we have:
$$
\frac {C _ {c} + 1}{n} + \left| A \tilde {f} _ {j} (x) \right| n \lambda \left(E _ {x}\right) > \frac {\epsilon}{2 \sqrt {2 k}}, \tag {83}
$$
$$
\left| A \tilde {f} _ {j} (x) \right| > \frac {1}{n \lambda \left(E _ {x}\right)} \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {c} + 1}{n}\right) \tag {84}
$$
Since  $1 / n\geq \| A\tilde{f}_j\mathbb{1}_E\| _1$  , we have:
$$
\begin{array}{l} 1 \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {c} + 1}{n}\right) \int_ {\mathcal {E} _ {j, A} ^ {2}} \frac {1}{\lambda \left(E _ {x}\right)} \mathrm {d} \lambda (x) + 0 (85) \\ \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {c} + 1}{n}\right) \sum_ {u \in [ n ] / n} \frac {1}{\lambda \left(E _ {u}\right)} \int_ {\mathcal {E} _ {j, A} ^ {2} \cap (u - 1 / n, u ]} 1 \mathrm {d} \lambda (x) (86) \\ \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {c} + 1}{n}\right) n \lambda \left(\mathcal {E} _ {j, A} ^ {2}\right). (87) \\ \end{array}
$$
Thus choosing  $\epsilon > 4\sqrt{2k} (C_c + 1) / n$  means that:
$$
1 \geq \frac {\epsilon}{4 \sqrt {2 k}} n \lambda \left(\mathcal {E} _ {j, A} ^ {2}\right), \tag {88}
$$
$$
\lambda \left(\mathcal {E} _ {j, A} ^ {2}\right) \leq \frac {4 \sqrt {2 k}}{\epsilon n}. \tag {89}
$$
Finally, choosing  $\epsilon >\sqrt{\frac{16k\sqrt{2k}}{n}} = \frac{2^{\frac{9}{4}}k^{\frac{3}{4}}}{n^{\frac{1}{2}}}$  makes  $\lambda (\mathcal{E}_{j,A}^{2})\leq \frac{\epsilon}{4k}$ ; and choosing  $\epsilon >\frac{4k}{n}$  makes  $\lambda (E) < \frac{1}{n} < \frac{\epsilon}{4k}$ .
Putting everything together via Assumption 5.A or 5.B. Thus, we can choose  $\overline{\epsilon} = 8k\left(\sqrt{\frac{C_A C_v + 1}{n}} + \frac{C_v + C_c + 1}{n}\right)$  to get:
$$
\lambda \left(\mathcal {E} ^ {\prime} (\epsilon)\right) \leq \sum_ {j = 1} ^ {k} \frac {\bar {\epsilon}}{k} = \bar {\epsilon}, \tag {90}
$$
which allows us to conclude:
$$
d _ {L P} \left(\bar {\eta}, \bar {\eta} _ {n}\right) \leq \bar {\epsilon}. \tag {91}
$$
Since  $\overline{\eta}$  was chosen arbitrarily, we have for all  $\overline{\eta} \in S_k(A)$ ,
$$
\inf  _ {\eta_ {n} \in \mathcal {S} _ {k, C _ {v}} (A _ {n})} d _ {L P} (\bar {\eta}, \eta_ {n}) \leq d _ {L P} (\bar {\eta}, \bar {\eta} _ {n}) \leq \bar {\epsilon}. \tag {92}
$$
Thus we also have:
$$
\sup  _ {\eta \in \mathcal {S} _ {k, C _ {v}} (A)} \inf  _ {\eta_ {n} \in \mathcal {S} _ {k, C _ {v}} (A _ {n})} d _ {L P} (\eta , \eta_ {n}) \leq \bar {\epsilon}. \tag {93}
$$
Proof via Assumption 4.B. Here, we use the other triangle inequality to get for each  $x$ :
$$
\begin{array}{l} \left| A f _ {j} (x) - A _ {n} f _ {j} ^ {\prime} \left(y _ {x}\right) \right| \leq n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} (x) - A f _ {j} (z) \right| + \left| A f _ {j} (z) - A \tilde {f} _ {j} (z) \right| d \lambda (z) (94) \\ \leq n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} | A f _ {j} (x) - A f _ {j} (z) | \mathrm {d} \lambda (z) + n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} (z) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z). (95) \\ \end{array}
$$
$$
\mathcal {E} _ {j, A} ^ {1} (\epsilon) := \left\{x: n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} (z) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) > \frac {\epsilon}{2 \sqrt {2 k}} \right\} \tag {96}
$$
$$
\mathcal {E} _ {j, A} ^ {2} (\epsilon) := \left\{x \notin E: n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} | A f _ {j} (x) - A f _ {j} (z) | \mathrm {d} \lambda (z) > \frac {\epsilon}{2 \sqrt {2 k}} \right\} \tag {97}
$$
Bounding  $\lambda (\mathcal{E}_{j,A}^{1}(\epsilon))$  via Assumption 4.B. Because of Assumption 2, we have:  $\| A\tilde{f}_j - A f_j\| _2\leq C_A\| \tilde{f}_j - f_j\| _2$  . By  $L^p$  norms inequality, we have
$$
\left\| A \tilde {f} _ {j} - A f _ {j} \right\| _ {1} \leq C _ {A} \left\| \tilde {f} _ {j} - f _ {j} \right\| _ {2} \leq \frac {C _ {A} C _ {v}}{n}. \tag {98}
$$