SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \left\| \tilde {f} _ {j} - f _ {j} \right\| _ {2} ^ {2} = \int_ {0} ^ {1} \left(\tilde {f} _ {j} (x) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (99) \\ = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \left(f _ {j} ^ {\prime} (u) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (100) \\ = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \left(n \int_ {u - 1 / n} ^ {u} f _ {j} (z) \mathrm {d} \lambda (z) - f _ {j} (x)\right) ^ {2} \mathrm {d} \lambda (x) (101) \\ \leq \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} n ^ {2} \left(\int_ {u - 1 / n} ^ {u} | f _ {j} (z) - f _ {j} (x) | \mathrm {d} \lambda (z)\right) ^ {2} \mathrm {d} \lambda (x) (102) \\ \leq \frac {C _ {v} ^ {2}}{n ^ {2}}. (103) \\ \end{array}
$$
Consider:
$$
\begin{array}{l} \lambda \left(\mathcal {E} _ {j, A} ^ {1} (\epsilon)\right) \cdot \frac {\epsilon}{2 \sqrt {2 k}} \leq \int_ {0} ^ {1} n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} (z) - A \tilde {f} _ {j} (z) \right| d \lambda (z) d \lambda (x) (104) \\ = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \left| A f _ {j} (z) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z) (105) \\ = \| A f _ {j} - A \tilde {f} _ {j} \| _ {1} \leq \frac {C _ {A} C _ {v}}{n}. (106) \\ \end{array}
$$
Thus selecting  $\epsilon >\sqrt{\frac{C_A C_v 4k\sqrt{2k}}{n}} = 2^{\frac{5}{4}}k^{\frac{3}{4}}(C_A C_v / n)^{\frac{1}{2}}$  gives  $\lambda (\overline{\mathcal{E}}_{j,A}(\epsilon))\leq \frac{\epsilon}{2k}$
Bounding  $\lambda (\mathcal{E}_{j,A}^2 (\epsilon))$  via Assumption 4.B. Notice that  $f_{j}$  is Lipschitz in [0,1] and thus by Assumption 4.B,  $Af_{j}$  is  $C_v$ -Lipschitz in [0,1] $E$ . Therefore we have, for  $x\in [0,1]\backslash E$  and  $z\in (y_x - 1 / n,y_x]\backslash E$  and:
$$
\left| A f _ {j} (x) - A f _ {j} (z) \right| \leq C _ {v} | x - z | \leq \frac {C _ {v}}{n}. \tag {107}
$$
When  $z \in E$ , we use the second condition in Assumption 4.B to get  $\| Af_j \mathbb{1}_E \|_1 \leq \frac{1}{n}$  and thus for any  $x \notin E$ :
$$
\begin{array}{l} n \int_ {y _ {x} - 1 / n} ^ {y _ {x}} | A f _ {j} (x) - A f _ {j} (z) | \mathrm {d} \lambda (z) (108) \\ = n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \cap E} | A f _ {j} (x) - A f _ {j} (z) | \mathrm {d} \lambda (z) + n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \backslash E} | A f _ {j} (x) - A f _ {j} (z) | \mathrm {d} \lambda (z) (109) \\ \leq \frac {C _ {v}}{n} + n \int_ {\left(y _ {x} - 1 / n, y _ {x} \right] \cap E} | A f _ {j} (x) | + | A f _ {j} (z) | d \lambda (z) (110) \\ \leq \frac {C _ {v}}{n} + | A f _ {j} (x) | n \lambda \left(E _ {x}\right) + \| A f _ {j} \mathbb {1} _ {E _ {x}} \| _ {1} (111) \\ \leq \frac {C _ {v} + 1}{n} + | A f _ {j} (x) | n \lambda \left(E _ {x}\right), (112) \\ \end{array}
$$
$$
\frac {C _ {v} + 1}{n} + \left| A f _ {j} (x) \right| n \lambda \left(E _ {x}\right) > \frac {\epsilon}{2 \sqrt {2 k}}, \tag {113}
$$
$$
\left| A f _ {j} (x) \right| > \frac {1}{n \lambda \left(E _ {x}\right)} \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {v} + 1}{n}\right) \tag {114}
$$
Since  $1 / n\geq \| Af_j\mathbb{1}_E\| _1$  , we have:
$$
\begin{array}{l} 1 \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {v} + 1}{n}\right) \int_ {\mathcal {E} _ {j, A} ^ {2}} \frac {1}{\lambda \left(E _ {x}\right)} \mathrm {d} \lambda (x) + 0 (115) \\ \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {v} + 1}{n}\right) \sum_ {u \in [ n ] / n} \frac {1}{\lambda \left(E _ {u}\right)} \int_ {\mathcal {E} _ {j, A} ^ {2} \cap (u - 1 / n, u ]} 1 \mathrm {d} \lambda (x) (116) \\ \geq \left(\frac {\epsilon}{2 \sqrt {2 k}} - \frac {C _ {v} + 1}{n}\right) n \lambda \left(\mathcal {E} _ {j, A} ^ {2}\right). (117) \\ \end{array}
$$
Thus choosing  $\epsilon > 4\sqrt{2k} (C_v + 1) / n$  means that:
$$
1 \geq \frac {\epsilon}{4 \sqrt {2 k}} n \lambda \left(\mathcal {E} _ {j, A} ^ {2}\right), \tag {118}
$$
$$
\lambda \left(\mathcal {E} _ {j, A} ^ {2}\right) \leq \frac {4 \sqrt {2 k}}{\epsilon n}. \tag {119}
$$
Putting everything together via Assumption 4.B. Thus, we can choose  $\bar{\epsilon} = 8k\left(\sqrt{\frac{C_A C_v + 1}{n}} + \frac{C_v + 1}{n}\right)$  to get:
$$
\lambda \left(\mathcal {E} ^ {\prime} (\epsilon)\right) \leq \sum_ {j = 1} ^ {k} \frac {\bar {\epsilon}}{k} = \bar {\epsilon}, \tag {120}
$$
$$
d _ {L P} \left(\bar {\eta}, \bar {\eta} _ {n}\right) \leq \bar {\epsilon}. \tag {121}
$$
$$
\inf  _ {\eta_ {n} \in \mathcal {S} _ {k, C _ {v}} (A _ {n})} d _ {L P} (\bar {\eta}, \eta_ {n}) \leq d _ {L P} (\bar {\eta}, \bar {\eta} _ {n}) \leq \bar {\epsilon}. \tag {122}
$$
$$
\sup  _ {\eta \in \mathcal {S} _ {k, \mathcal {C} _ {v}} (A)} \inf  _ {\eta_ {n} \in \mathcal {S} _ {k, \mathcal {C} _ {v}} (A _ {n})} d _ {L P} (\eta , \eta_ {n}) \leq \bar {\epsilon}. \tag {123}
$$
Bounding  $\sup_{\eta_n\in S_{k,C_v}(A_n)}\inf_{\eta \in S_{k,C_v}(A)}d_{LP}(\eta ,\eta_n)$ . In this direction, we proceed identically, but now choose  $\overline{\eta}_n$  arbitrarily in  $S_{k,C_v}(A_n)$ . By definition of  $(k,C_v)$ -profiles, there exists a tuple  $F = (f_{1},\ldots ,f_{k})$  each in  $\mathcal{F}_{n,\mathrm{reg}(C_v)}$  such that  $\overline{\eta}_n = \mathcal{D}_{A_n}(F)$ .
$$
F ^ {\prime} := \left\{\mathcal {F} _ {\operatorname {r e g} \left(C _ {v}\right)} \ni f _ {j} ^ {\prime}: x \mapsto \left(1 - y _ {x} n + x n\right) f _ {j} \left(y _ {x}\right) + \left(y _ {x} n - x n\right) f _ {j} \left(y _ {x} - 1 / n\right) \right\}, \tag {124}
$$
$$
\bar {\eta} = \mathcal {D} _ {A} \left(F ^ {\prime}\right), \tag {125}
$$
where  $y_{x} = \lceil xn \rceil / n$  for each  $j \in [k]$ . That  $f_{j}' \in \mathcal{F}_{\mathrm{reg}(C_{v})}$  is asserted in Lemma 11. Intuitively,  $f_{j}'$  is the continuous piecewise linear function that interpolates  $f_{j}$ , created by joining  $f_{j}(u - 1/n)$  and  $f_{j}(u)$  with a line segment for each  $u \in [n] / n$ . Note also that  $f_{j}'(u) = f_{j}(u)$  for all  $u \in [n] / n$ .
Bounding  $d_{LP}(\overline{\eta},\overline{\eta}_n)$ . As with the previous direction, we start with an arbitrary  $U\in \mathcal{B}_k$  and write down the differences:
$$
\bar {\eta} _ {n} (U) - \bar {\eta} \left(U ^ {\epsilon}\right) = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \mathbb {1} _ {F _ {A _ {n}} (u) \in U} - \mathbb {1} _ {F _ {A} ^ {\prime} (x) \in U ^ {\epsilon}} \mathrm {d} \lambda (x), \tag {126}
$$
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) = \sum_ {u \in [ n ] / n} \int_ {u - 1 / n} ^ {u} \mathbb {1} _ {F _ {A} ^ {\prime} (x) \in U} - \mathbb {1} _ {F _ {A _ {n}} (u) \in U ^ {\epsilon}} \mathrm {d} \lambda (x). \tag {127}
$$
Define the events:
$$
\mathcal {E} _ {U} ^ {1} (\epsilon) = \left\{x: F _ {A _ {n}} \left(y _ {x}\right) \in U \wedge F _ {A} ^ {\prime} (x) \notin U ^ {\epsilon} \right\} \tag {128}
$$
$$
\mathcal {E} _ {U} ^ {2} (\epsilon) = \left\{x: F _ {A} ^ {\prime} (x) \in U \wedge F _ {A _ {n}} \left(y _ {x}\right) \notin U ^ {\epsilon} \right\} \tag {129}
$$
$$
\mathcal {E} ^ {\prime} (\epsilon) = \left\{x: \| F _ {A _ {n}} (y _ {x}) - F _ {A} ^ {\prime} (x) \| _ {2} > \epsilon \right\} \tag {130}
$$
$$
\mathcal {E} _ {j} (\epsilon) = \left\{x: \left| f _ {j} \left(y _ {x}\right) - f _ {j} ^ {\prime} (x) \right| > \epsilon / \sqrt {2 k} \right\}, \quad j = 1.. k \tag {131}
$$
$$
\mathcal {E} _ {j, A} (\epsilon) = \left\{x: \left| A _ {n} f _ {j} \left(y _ {x}\right) - A f _ {j} ^ {\prime} (x) \right| > \epsilon / \sqrt {2 k} \right\}, \quad j = 1.. k. \tag {132}
$$
$$
\bar {\eta} (U) - \bar {\eta} _ {n} \left(U ^ {\epsilon}\right) \leq \lambda \left(\mathcal {E} _ {U} ^ {1}\right) \quad \text {a n d} \quad \bar {\eta} _ {n} (U) - \bar {\eta} \left(U ^ {\epsilon}\right) \leq \lambda \left(\mathcal {E} _ {U} ^ {2}\right). \tag {133}
$$
$$
\mathcal {E} ^ {\prime} (\epsilon) \subseteq \bigcup_ {j = 1} ^ {k} \mathcal {E} _ {j} (\epsilon) \cup \mathcal {E} _ {j, A} (\epsilon), \tag {134}
$$
Bounding  $\lambda (\mathcal{E}_j(\epsilon))$ . Since  $f_j^\prime$  is  $C_v$ -Lipschitz for all  $j$  (Lemma 11), we have:
$$
\left| f _ {j} \left(y _ {x}\right) - f _ {j} ^ {\prime} (x) \right| = \left| f _ {j} ^ {\prime} \left(y _ {x}\right) - f _ {j} ^ {\prime} (x) \right| \leq \frac {C _ {v}}{n}. \tag {135}
$$
Bounding  $\lambda (\mathcal{E}_{j,A})$ . Let  $\mathcal{F}_{[-1,1]}\ni \tilde{f}$  be the extension of  $f$  defined as  $\tilde{f} (x) = f(\lceil xn\rceil /n)$  for all  $x\in [0,1]$ . Note that  $\tilde{f}$  is not continuous in general and hence not Lipschitz. We have for any  $x\in [0,1]$ :
$$
\begin{array}{l} \left| A f _ {j} ^ {\prime} (x) - A _ {n} f _ {j} \left(y _ {x}\right) \right| = \left| A f _ {j} ^ {\prime} (x) - n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} A \tilde {f} _ {j} (z) \mathrm {d} \lambda (z) \right| (136) \\ \leq n \int_ {y _ {x} - \frac {1}{n}} ^ {y _ {x}} \left| A f _ {j} ^ {\prime} (x) - A \tilde {f} _ {j} (z) \right| \mathrm {d} \lambda (z), (137) \\ \end{array}
$$
where we used uniformity of  $\lambda$  and triangle inequality. The last thing that we need to show is:
$$
\begin{array}{l} \left\| f _ {j} ^ {\prime} - \tilde {f} _ {j} \right\| _ {2} ^ {2} = \int_ {0} ^ {1} \left(\left(1 - y _ {x} n + x n\right) f \left(y _ {x}\right) + \left(x n - y _ {x} n\right) f \left(y _ {x} - 1 / n\right) - f \left(y _ {x}\right)\right) ^ {2} \mathrm {d} \lambda (x) (138) \\ = \int_ {0} ^ {1} n ^ {2} \left(y _ {x} - x\right) ^ {2} \left(f \left(y _ {x}\right) - f \left(y _ {x} - 1 / n\right)\right) ^ {2} \mathrm {d} \lambda (x) (139) \\ \leq \frac {C _ {v} ^ {2}}{n ^ {2}}. (140) \\ \end{array}
$$
From here, by a word-for-word argument, we can show that the same choice of  $\overline{\epsilon}$  does the trick to make  $\lambda (\mathcal{E}'(\epsilon))\leq \overline{\epsilon}$ . This works since we only use the fact that  $f_{j}^{\prime}\in \mathcal{F}_{\mathrm{reg}(C_{v})}$  as well as assumption conditions in the previous proof.
Bounding  $d_{M}$ . We have:
$$
d _ {H} \left(\mathcal {S} _ {k} (A), \mathcal {S} _ {k} \left(A _ {n}\right)\right) = \max  \left(\sup  _ {\overline {{\eta}} _ {n} \in \mathcal {S} _ {k} \left(A _ {n}\right)} \inf  _ {\eta \in \mathcal {S} _ {k} (A)} d _ {L P} \left(\overline {{\eta}} _ {n}, \eta\right), \sup  _ {\overline {{\eta}} \in \mathcal {S} _ {k} (A)} \inf  _ {\eta_ {n} \in \mathcal {S} _ {k} \left(A _ {n}\right)} d _ {L P} \left(\overline {{\eta}}, \eta_ {n}\right)\right) \leq \bar {\epsilon}. \tag {141}
$$
$$
d _ {M} (A, A _ {n}) \leq \sum_ {k = 1} ^ {\infty} \frac {8 k \left(\sqrt {\frac {C _ {A} C _ {v} + 1}{n}} + \frac {C _ {v} + 1}{n}\right)}{2 ^ {k}} \leq 8 \left(\sqrt {\frac {C _ {A} C _ {v} + 1}{n}} + \frac {2 C _ {c} + C _ {v} + 1}{n}\right). \tag {142}
$$
Lemma 10. Fix  $n \in \mathbb{N}$ . Fix  $f \in \mathcal{F}_{reg(C_v)}$ . Define the restriction  $f' : \frac{1}{n}[n] \to [-1,1] : u \mapsto n\int_{u - 1/n}^{u} f(z)\mathrm{d}\lambda(z)$ . Then  $f' \in \mathcal{F}_{n,reg(C_v)}$ .
Proof. Firstly, we have for any  $u \in [n] / n$ :
$$
\left| f ^ {\prime} (u) \right| \leq n \int_ {u - 1 / n} ^ {u} | f (z) | \mathrm {d} \lambda (z) \leq n \int_ {u - 1 / n} ^ {u} 1 \mathrm {d} \lambda (z) = 1. \tag {143}
$$
Therefore  $\| f' \|_{L^2([n]/n)} \leq n / n = 1$  and thus  $f'$  is measurable. Finally, for any  $u < u' \in [n] / n$ :
$$
\left| f ^ {\prime} (u) - f ^ {\prime} \left(u ^ {\prime}\right) \right| \leq n \int_ {u - 1 / n} ^ {u} | f (z) - f (z + \left(u ^ {\prime} - u\right)) | d \lambda (z) \leq C _ {v} \left(u ^ {\prime} - u\right), \tag {144}
$$
where we use Lipschitz property of  $f$  in the last inequality. This shows that  $f'$  is also  $C_v$ -Lipschitz.
Remark 1. The restriction to  $\mathcal{F}_{[-1,1]}$  is necessary for this lemma to work because  $L^2([0,1])$  functions can blow up near 0.
Lemma 11. Fix  $n \in \mathbb{N}$ . Fix  $f \in \mathcal{F}_{n,reg(C_v)}$ . Let  $y_x \coloneqq \lceil xn \rceil / n$  for any  $x \in [0,1]$ . Define the extension  $f' : x \mapsto (1 - y_x n + xn)f(y_x) + (xn - y_x n)f(y_x - 1/n)$ . Then  $f' \in \mathcal{F}_{reg(C_v)}$ .
Proof. Firstly, since  $f'$  linearly interpolates between points of  $f$ , its range cannot exceed that of  $f$ . The restricted range immediately implies that the  $L^2$  norm is bounded by 1 since the support is also in [0, 1]. Finally, for any  $x < x' \in [0, 1]$ , if there is a  $u$  such that  $x, x' \in [u - 1/n, u)$  then the fact that the interpolation is linear means that the line segment from  $f'(x)$  to  $f'(x')$  shares the same slope as that from  $f(u)$  to  $f(u - 1/n)$ , which is at most  $C_v$  since  $f \in \mathcal{F}_{n,\mathrm{reg}(C_v)}$ . Otherwise, there exists a  $u < u' \in [n]/n$  such that  $x \in [u - 1/n, u)$  and  $x' \in [u' - 1/n, u')$ . We have:
$$
\left| f ^ {\prime} (x) - f ^ {\prime} \left(x ^ {\prime}\right) \right| \leq \left| f ^ {\prime} (x) - f ^ {\prime} (u) \right| + \left| f ^ {\prime} (u) - f ^ {\prime} \left(u ^ {\prime} - 1 / n\right) \right| + \left| f ^ {\prime} \left(u ^ {\prime} - 1 / n\right) - f ^ {\prime} \left(x ^ {\prime}\right) \right| \leq C _ {v} \left(x ^ {\prime} - x\right), \tag {145}
$$
which proves Lipschitzness of  $f'$ .
D.3 Proofs for Section 4.2
The following result characterizes the behaviors of Assumptions 2, 5.A, 5.B and 4.B under addition, multiplication by scalars, power, and element-wise composition with a 1-Lipschitz map:
Lemma 12. Fix  $k \in \mathbb{N}$  and  $\alpha \in \mathbb{R}$ . Recall that  $\rho : \mathbb{R} \to \mathbb{R}$  is a 1-Lipschitz map. Let  $A_1, A_2 : \mathcal{F} \to \mathcal{F}$  satisfy Assumption 2 with constant  $C_A^1$  and  $C_A^2$  respectively; and Assumption 5.A or 5.B or 4.B with constant  $C_c^1$  and  $C_c^2$  respectively with common resolution set  $\mathcal{N}$ . Then:
1.  $A_{1} + A_{2}$  satisfy Assumptions 2 and 5.A or 5.B or 4.B with constant  $(C_A^1 + C_A^2)$  and  $(C_c^1 + C_c^2)$  respectively.
2.  $\alpha A_{1}$  satisfies Assumption 2 and 5.A or 5.B or 4.B with constant  $|\alpha|C_{A}^{1}$  and  $|\alpha|C_{c}^{1}$  respectively.
3.  $\rho A_{1}$  (where the composition is done element-wise) satisfies Assumption 2 and 5.A or 5.B or 4.B with constant  $C_A^1$  and  $C_c^1$  respectively.
4. Furthermore, if  $C_c^1 = 0$  then  $A_2 \circ A_1$  satisfies Assumption 2 and 5.A 5.B or 4.B with constant  $C_A^1 C_A^2$  and  $C_c^2$  respectively.
Proof of Lemma 12. To recall,  $A_{1}, A_{2}: \mathcal{F} \to \mathcal{F}$  are  $P$ -operators that satisfies Assumption 2 with constant  $C_{A}^{1}$  and  $C_{A}^{2}$  respectively and Assumption 5.A or 5.B or 4.B with constant  $C_{c}^{1}$  and  $C_{c}^{2}$  respectively with common resolution set  $\mathcal{N}$ . We now show each part of the Lemma.
1. We have, for any  $f,g\in \mathcal{F}$
$$
\begin{array}{l} \left\| \left(A _ {1} + A _ {2}\right) f - \left(A _ {1} + A _ {2}\right) g \right\| _ {2} \leq \left\| A _ {1} f - A _ {1} g \right\| _ {2} + \left\| A _ {2} f - A _ {2} g \right\| _ {2} (146) \\ \leq C _ {A} ^ {1} \| f - g \| _ {2} + C _ {A} ^ {2} \| f - g \| _ {2}, (147) \\ \end{array}
$$
where the first line is triangle inequality and the second line is Assumption 2.
Let  $f \in \mathcal{F}_{\mathrm{reg}(C_v)}$  piecewise constant on intervals  $[n] / n$  and  $x, y \in (u - 1 / n, u]$  for some  $n \in \mathcal{N}$  and  $u \in [n]$ , we have:
$$
\begin{array}{l} \left. \left| \left(A _ {1} + A _ {2}\right) f (x) - \left(A _ {1} + A _ {2}\right) f (y) \right| \leq \left| \left[ A _ {1} f \right] (x) - \left[ A _ {1} f \right] (y) \right| + \left| \left[ A _ {2} f \right] (x) - \left[ A _ {2} f \right] (y) \right| \right. (148) \\ \leq C _ {c} ^ {1} | x - y | + C _ {c} ^ {2} | x - y |, (149) \\ \end{array}
$$
where the first line is triangle inequality and the second line is Assumption 5.A.
2. We have, for any  $f,g\in \mathcal{F}$
$$
\begin{array}{l} \left\| \left(\alpha A _ {1}\right) f - \left(\alpha A _ {1}\right) g \right\| _ {2} \leq | \alpha | \cdot \left\| A _ {1} f - A _ {1} g \right\| _ {2} (150) \\ \leq | \alpha | C _ {A} ^ {1} \| f - g \| _ {2}, (151) \\ \end{array}
$$
where the first line is property of norm and the second line is Assumption 2.
Let  $f \in \mathcal{F}_{\mathrm{reg}(C_v)}$  piecewise constant on intervals  $[n] / n$ , and  $x, y \in (u - 1 / n, u]$  for some  $n \in \mathcal{N}$  and  $u \in [n]$ , we have:
$$
\begin{array}{l} \left| \left(\alpha A _ {1}\right) f (x) - \left(\alpha A _ {1}\right) f (y) \right| \leq \left| \alpha \left[ A _ {1} f \right] (x) - \alpha \left[ A _ {1} f \right] (y) \right| (152) \\ \leq | \alpha | C _ {c} ^ {1} | x - y |, (153) \\ \end{array}
$$
where the second line is Assumption 5.A.
3. We have, for any  $f,g\in \mathcal{F}$
$$
\begin{array}{l} \| (\rho A _ {1}) f - (\rho A _ {1}) g \| _ {2} \leq \| A _ {1} f - A _ {1} g \| _ {2} (154) \\ \leq C _ {A} ^ {1} \| f - g \| _ {2}, (155) \\ \end{array}
$$
where the first line is Lipschitz property of  $\rho$  and the second line is Assumption 2.
$$
\begin{array}{l} \left| \left(\rho A _ {1}\right) f (x) - \left(\rho A _ {1}\right) f (y) \right| \leq \left| \rho \left(\left[ A _ {1} f \right] (x)\right) - \rho \left(\left[ A _ {1} f \right] (y)\right) \right| (156) \\ \leq \left| A _ {1} f (x) - A _ {1} f (y) \right| \leq C _ {c} ^ {1} | x - y |, (157) \\ \end{array}
$$
where the second line is Lipschitz property of  $\rho$  and Assumption 5.A.
4. We have, for any  $f,g\in \mathcal{F}$
$$
\begin{array}{l} \left\| \left(A _ {2} A _ {1}\right) f - \left(A _ {2} A _ {1}\right) g \right\| _ {2} \leq C _ {A} ^ {2} \| A _ {1} f - A _ {1} g \| _ {2} (158) \\ \leq C _ {A} ^ {1} C _ {A} ^ {2} \| f - g \| _ {2}, (159) \\ \end{array}
$$
where the first line is Lipschitz property of  $A_{2}$  and the second line is that of  $A_{1}$ .
Let  $f \in \mathcal{F}_{\mathrm{reg}(C_v)}$  piecewise constant on intervals  $[n] / n$ . Since  $A_{1}$  send constant pieces to constant pieces, the final implication follows direction from Assumption 5.A of  $A_{2}$ .
Lemma 13. Fix  $n \in \mathcal{N}$ ,  $k \in [K] \cup \{0\}$ ,  $A: \mathcal{F} \to \mathcal{F}$  satisfies Assumption 2 with constant  $C_A$  and Assumption 5.A with constant  $C_c$  and resolution set  $\mathcal{N}$ . Let  $n \in \mathcal{N}$ ,  $f_1 \in \mathcal{F}_n$ ,  $f_2 \in \mathcal{F}$ . Recall that  $\tilde{f} \in \mathcal{F}$  denotes the extension of  $f \in \mathcal{F}_n$  to  $[0,1]$  as  $\tilde{f}(x) = f(\lceil xn \rceil) / n$ . If  $\| \widetilde{f}_1 - f_2 \|_2 \leq M$  for some positive constant  $M$  then
$$
\left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} \leq \frac {3 ^ {k + 1} k C _ {c} ^ {2} C _ {A} ^ {2 k}}{n ^ {2}} + 3 ^ {k} C _ {A} ^ {2 k} M. \tag {160}
$$
If  $A$  satisfies Assumption 5.B with constant  $C_c$  instead then the bound becomes:
$$
\left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} \leq \frac {3 ^ {k + 1} k \left(C _ {c} + 1\right) ^ {2} C _ {A} ^ {2 k}}{n ^ {2}} + 3 ^ {k} C _ {A} ^ {2 k} M. \tag {161}
$$
If  $A$  satisfies Assumption 4.B instead then the bound becomes:
$$
\left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} \leq \frac {3 ^ {k + 1} k \left(C _ {v} + 1\right) ^ {2} C _ {A} ^ {2 k}}{n ^ {2}} + 3 ^ {k} C _ {A} ^ {2 k} M. \tag {162}
$$
Proof. When  $k = 0$ , the bound is vacuously true. We will hide the measure in integrals when it is clear from context in this proof.
Assume that the bound is correct up to some  $k - 1$ . We have:
$$
\begin{array}{l} \left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} (163) \\ = \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \left(A _ {n} ^ {k} f _ {1} (u) - A ^ {k} f _ {2} (z)\right) ^ {2} \mathrm {d} z (164) \\ = \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \left(n \int_ {u - \frac {1}{n}} ^ {u} A \widetilde {\left[ A _ {n} ^ {k - 1} f _ {1} \right]} (y) \mathrm {d} y - A ^ {k} f _ {2} (z)\right) ^ {2} \mathrm {d} z (165) \\ \end{array}
$$
Here we can use Jensen's inequality to get:
$$
\begin{array}{l} \left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} (166) \\ = n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \int_ {u - \frac {1}{n}} ^ {u} \left(A \widetilde {\left[ A _ {n} ^ {k - 1} f _ {1} \right]} (y) - A \left[ A ^ {k - 1} f _ {2} \right] (z)\right) ^ {2} \mathrm {d} y \mathrm {d} z (167) \\ \end{array}
$$
We proceed slightly differently based on the exact assumptions that we have. If  $A$  satisfies Assumption 5.A or 5.B then:
$$
\begin{array}{l} \left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} (169) \\ \leq n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \\ \int_ {u - \frac {1}{n}} ^ {u} \left(\left| A \left[ \widetilde {A _ {n} ^ {k - 1} f _ {1}} \right] (y) - A \left[ \widetilde {A _ {n} ^ {k - 1} f _ {1}} \right] (z) \right| + \left| A \left[ \widetilde {A _ {n} ^ {k - 1} f _ {1}} \right] (z) - A \left[ A ^ {k - 1} f _ {2} \right] (z) \right|\right) ^ {2} d y d z (170) \\ \leq n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \\ \int_ {u - \frac {1}{n}} ^ {u} \left(\frac {C _ {c} + 1}{n} + | A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] (z) | n \lambda \left(E _ {z}\right) + | A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] (z) - A [ A ^ {k - 1} f _ {2} ] (z) |\right) ^ {2} d y d z (171) \\ \leq \int_ {0} ^ {1} \left(\frac {C _ {c} + 1}{n} + | A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] (z) | n \lambda (E _ {z}) + | A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] (z) - A [ A ^ {k - 1} f _ {2} ] (z) |\right) ^ {2} d z (172) \\ \leq \frac {3 \left(C _ {c} + 1\right) ^ {2}}{n ^ {2}} + 3 \int_ {0} ^ {1} \left(A \left[ \widetilde {A _ {n} ^ {k - 1} f _ {1}} \right] (z) n \lambda \left(E _ {z}\right)\right) ^ {2} \mathrm {d} z + 3 \| A \left[ \widetilde {A _ {n} ^ {k - 1} f _ {1}} \right] - A \left[ A ^ {k - 1} f _ {2} \right] \| _ {2} ^ {2} \mathrm {d} z. (173) \\ \end{array}
$$
Here, we give a heuristic argument while the formal argument is exactly similar to that in Theorem 2. The term  $3\int_0^1 (A[\widetilde{A_n^{k - 1}f_1](z)n\lambda (E_z))^2\mathrm{d}z$  can be made as close to 0 as possible simply by changing the requirement on  $\lambda (E)$  to be smaller and smaller (see the discussion
after Assumption 4.A). At the same time, our assumptions ensure that since  $\widetilde{A_n^{k - 1}f_1}$  is a piecewise constant function, the action of  $A$  on it cannot become too wild (Lipschitz outside of  $E$  and bounded  $L^1$  norm inside  $E$ ). This heuristic argument, when made rigorous can help us conclude that:
$$
\left\| \widetilde {A _ {n} ^ {k}} f _ {1} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} \leq \frac {3 \left(C _ {c} + 1\right) ^ {2}}{n ^ {2}} + 3 C _ {A} ^ {2} \left\| \widetilde {A _ {n} ^ {k - 1}} f _ {1} - A ^ {k - 1} f _ {2} \right\| _ {2} ^ {2} \tag {174}
$$
Plug in the inductive hypothesis and solve the recurrent to get the bound.
Similarly, if we are instead using Assumption 4.B then we use the other triangle inequality to conclude:
$$
\begin{array}{l} \left\| \widetilde {A _ {n} ^ {k} f _ {1}} - A ^ {k} f _ {2} \right\| _ {2} ^ {2} (175) \\ \leq n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \\ \int_ {u - \frac {1}{n}} ^ {u} \left(\left| A \widetilde {\left[ A _ {n} ^ {k - 1} f _ {1} \right]} (y) - A \left[ A ^ {k - 1} f _ {1} \right] (y) \right| + \left| A \left[ A ^ {k - 1} f _ {1} \right] (y) - A \left[ A ^ {k - 1} f _ {2} \right] (z) \right|\right) ^ {2} d y d z (176) \\ \leq 2 n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \int_ {u - \frac {1}{n}} ^ {u} \left(A [ \widetilde {A _ {n} ^ {k - 1} f _ {1}} ] (y) - A [ A ^ {k - 1} f _ {1} ] (y)\right) ^ {2} \mathrm {d} y \mathrm {d} z \\ + 2 n \sum_ {u \in 1 / n [ n ]} \int_ {u - \frac {1}{n}} ^ {u} \int_ {u - \frac {1}{n}} ^ {u} \left(\frac {C _ {v} + 1}{n} + | A [ A ^ {k - 1} f _ {2} ] (z) | n \lambda (E _ {z})\right) ^ {2} d y d z (177) \\ \leq \frac {3 \left(C _ {v} + 1\right) ^ {2}}{n ^ {2}} + 3 \int_ {0} ^ {1} \left(A \left[ A ^ {k - 1} f _ {1} \right] (z) n \lambda \left(E _ {z}\right)\right) ^ {2} \mathrm {d} z + 3 \| A \widetilde {\left[ A _ {n} ^ {k - 1} f _ {1} \right]} - A \left[ A ^ {k - 1} f _ {2} \right] \| _ {2} ^ {2}. (178) \\ \end{array}
$$