SlowGuess/NeurIPS_2023_OCR huggingface dataset:
We now wish to write  $\epsilon_{k + 1}$  in terms of  $\epsilon_{k}$ . As  $\epsilon_{k + 1}$  involves  $X_{\lambda ,\delta}^{k + 2}$ , we first write it explicitly.
$$
\begin{array}{l} X _ {\lambda , \delta} ^ {k + 2} = X _ {\lambda , \delta} ^ {k + 1} + 2 h F \left(X _ {\lambda , \delta} ^ {k + 1}, 2 h (k + 1)\right) \\ = \left\{ \begin{array}{l l} X _ {\lambda , \delta} ^ {k + 1} - \left(2 h \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k + 1}) + \frac {2 h}{\delta} (X _ {\lambda , \delta} ^ {k + 1} - X _ {0})\right) & 0 \leq k + 1 <   \frac {\delta}{2 h} \\ X _ {\lambda , \delta} ^ {k + 1} - \left(2 h \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k + 1}) + \frac {1}{k + 1} (X _ {\lambda , \delta} ^ {k + 1} - X _ {0})\right) & k + 1 \geq \frac {\delta}{2 h}. \end{array} \right. \\ \end{array}
$$
Now we find recursive inequality considering two cases.
(i)  $k + 1\geq \frac{\delta}{2h}$
Recall APPM (5) was defined as
$$
y _ {h, \lambda} ^ {k} = \frac {k}{k + 1} (2 x _ {h, \lambda} ^ {k} - y _ {h, \lambda} ^ {k - 1}) + \frac {1}{k + 1} X _ {0},
$$
and substituting  $y_{h,\lambda}^{k - 1} = x_{h,\lambda}^k + h\mathbb{A}_\lambda(x_{h,\lambda}^k)$ , we get a one line expression
$$
x _ {h, \lambda} ^ {k + 1} = \frac {k}{k + 1} x _ {h, \lambda} ^ {k} - h \left(\mathbb {A} _ {\lambda} \left(x _ {h, \lambda} ^ {k + 1}\right) + \frac {k}{k + 1} \mathbb {A} _ {\lambda} \left(x _ {h, \lambda} ^ {k}\right)\right) + \frac {1}{k + 1} X _ {0}.
$$
Rewriting  $X_{\lambda ,\delta}^{k + 2}$  to make easier to compare with above,
$$
\begin{array}{l} X _ {\lambda , \delta} ^ {k + 2} = \frac {k}{k + 1} X _ {\lambda , \delta} ^ {k + 1} - h \left(\mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 2}\right) + \frac {k}{k + 1} \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right)\right) + \frac {1}{k + 1} X _ {0} \\ + h \left(\mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 2}\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right)\right) - \frac {h}{k + 1} \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right). \\ \end{array}
$$
As  $\mathbb{A}_{\lambda}$  is  $\frac{1}{\lambda}$ -Lipschitz
$$
\epsilon_ {k + 1} \leq \frac {k}{k + 1} \epsilon_ {k} + \frac {h}{\lambda} \left(\epsilon_ {k + 1} + \frac {k}{k + 1} \epsilon_ {k}\right) + \frac {h}{\lambda} \left\| X _ {\lambda , \delta} ^ {k + 2} - X _ {\lambda , \delta} ^ {k + 1} \right\| + \frac {h}{k + 1} \left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right) \right\|.
$$
Organizing,
$$
\epsilon_ {k + 1} \leq \frac {1}{1 - \frac {h}{\lambda}} \left(\frac {k}{k + 1} \left(1 + \frac {h}{\lambda}\right) \epsilon_ {k} + \underbrace {h \left(\frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {k + 2} - X _ {\lambda , \delta} ^ {k + 1} \right\| + \frac {1}{k + 1} \left\| \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k + 1}) \right\|\right)} _ {:= e _ {k}}\right).
$$
(ii)  $k + 1 <   \frac{\delta}{2h}$
Rewriting  $X_{\lambda ,\delta}^{k + 2}$  we have
$$
\begin{array}{l} X _ {\lambda , \delta} ^ {k + 2} = \frac {k}{k + 1} X _ {\lambda , \delta} ^ {k + 1} - h \left(\mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 2}\right) + \frac {k}{k + 1} \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right)\right) + \frac {1}{k + 1} X _ {0} \\ + h \left(\mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 2}\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right)\right) + \frac {h}{k + 1} \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {k + 1}\right) + \left(\frac {1}{k + 1} - \frac {2 h}{\delta}\right) \left(X _ {\lambda , \delta} ^ {k + 1} - X _ {0}\right). \\ \end{array}
$$
The only difference between the case (i) is the last term. Therefore with same calculation, we get below inequality.
$$
\epsilon_{k + 1}\leq \frac{1}{1 - \frac{h}{\lambda}}\left(\frac{k}{k + 1}\left(1 + \frac{h}{\lambda}\right)\epsilon_{k} + \underbrace{h\left(\frac{1}{\lambda}\left\|X_{\lambda,\delta}^{k + 2} - X_{\lambda,\delta}^{k + 1}\right\| + \frac{1}{k + 1}\left\| \mathbb{A}_{\lambda}(X_{\lambda,\delta}^{k + 1})\right\|\right) + \left|\frac{1}{k + 1} -\frac{2h}{\delta}\right|\left\|X_{\lambda,\delta}^{k + 1} - X_{0}\right\|}\right).
$$
From case (i) and (ii), by defining
$$
e _ {k} = \left\{ \begin{array}{l l} h \left(\frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {k + 2} - X _ {\lambda , \delta} ^ {k + 1} \right\| + \frac {1}{k + 1} \left\| \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k + 1}) \right\|\right) + \left| \frac {1}{k + 1} - \frac {2 h}{\delta} \right| \left\| X _ {\lambda , \delta} ^ {k + 1} - X _ {0} \right\| & 0 \leq k + 1 <   \frac {\delta}{2 h} \\ h \left(\frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {k + 2} - X _ {\lambda , \delta} ^ {k + 1} \right\| + \frac {1}{k + 1} \left\| \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k + 1}) \right\|\right) & k + 1 \geq \frac {\delta}{2 h}, \end{array} \right.
$$
we can write a recursive inequality for all  $k$  as below
$$
\epsilon_ {k + 1} \leq \frac {1}{1 - \frac {h}{\lambda}} \left(\frac {k}{k + 1} \left(1 + \frac {h}{\lambda}\right) \epsilon_ {k} + e _ {k}\right).
$$
D.3.2  $\sup_{0\leq i\leq \frac{T}{2h}}\epsilon_i = 3^{\frac{T}{\lambda}}\left(\mathcal{O}\left(\frac{h}{\lambda}\right) + \mathcal{O}\left(\frac{h}{\lambda} e^{2L_{\lambda,\delta}T}\right) + \mathcal{O}(h) + \mathcal{O}\left(\frac{h^2}{\lambda} e^{2L_{\lambda,\delta}T}\right) + \mathcal{O}\left(\frac{e^{2\delta L_{\lambda,\delta}}}{L_{\lambda,\delta}}\right) + \mathcal{O}(\delta)\right)$
We will now sum up above inequality. Multiplying  $(k + 1)\left(\frac{1 - \frac{h}{\lambda}}{1 + \frac{h}{\lambda}}\right)^{k + 1}$  both sides we have
$$
\left(\frac {1 - \frac {h}{\lambda}}{1 + \frac {h}{\lambda}}\right) ^ {k + 1} (k + 1) \epsilon_ {k + 1} \leq \left(\frac {1 - \frac {h}{\lambda}}{1 + \frac {h}{\lambda}}\right) ^ {k} k \epsilon_ {k} + \frac {\left(1 - \frac {h}{\lambda}\right) ^ {k + 1}}{\left(1 + \frac {h}{\lambda}\right) ^ {k}} (k + 1) e _ {k}.
$$
By summing up above inequality from 0 to  $k - 1$ , we have
$$
\left(\frac {1 - \frac {h}{\lambda}}{1 + \frac {h}{\lambda}}\right) ^ {k} k \epsilon_ {k} \leq \sum_ {i = 0} ^ {k - 1} \frac {\left(1 - \frac {h}{\lambda}\right) ^ {i + 1}}{\left(1 + \frac {h}{\lambda}\right) ^ {i}} (i + 1) e _ {i}.
$$
Note  $\epsilon_0$  vanished as it is multiplied with 0. Reorganizing with respect to  $\epsilon_{k}$  we have
$$
\begin{array}{l} \epsilon_ {k} \leq \left(\frac {1 + \frac {h}{\lambda}}{1 - \frac {h}{\lambda}}\right) ^ {k} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} \frac {\left(1 - \frac {h}{\lambda}\right) ^ {i + 1}}{\left(1 + \frac {h}{\lambda}\right) ^ {i}} (i + 1) e _ {i} \\ \leq \left(\frac {1 + \frac {h}{\lambda}}{1 - \frac {h}{\lambda}}\right) ^ {\frac {T}{2 h}} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} = \left(\left(\frac {1 + \frac {h}{\lambda}}{1 - \frac {h}{\lambda}}\right) ^ {\frac {\lambda}{h}}\right) ^ {\frac {T}{2 \lambda}} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} \\ \end{array}
$$
For second inequality follows from the fact  $0 < \frac{h}{\lambda} < \frac{1}{2}$ , which implies  $0 < \left(\frac{1 - \frac{h}{\lambda}}{1 + \frac{h}{\lambda}}\right)^{i+1} \leq 1$  and  $1 < \frac{1 + \frac{h}{\lambda}}{1 - \frac{h}{\lambda}}$ . Observe  $f(x) = \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{x}}$  is nondecreasing in  $x \in (0,1)$  since
$$
f ^ {\prime} (x) = - \frac {\left(\frac {1 + x}{1 - x}\right) ^ {\frac {1}{x}} \left(\left(x ^ {2} - 1\right) \log \left(\frac {1 + x}{1 - x}\right) + 2 x\right)}{x ^ {2} \left(x ^ {2} - 1\right)}.
$$
Therefore, from  $f\left(\frac{1}{2}\right) = 9$  we have
$$
\epsilon_ {k} \leq f \left(\frac {1}{2}\right) ^ {\frac {T}{2 \lambda}} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} = 3 ^ {\frac {T}{\lambda}} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i}.
$$
Now we show
$$
\frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} = \mathcal {O} \left(\frac {h}{\lambda}\right) + \mathcal {O} \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} (h) + \mathcal {O} \left(\frac {h ^ {2}}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} (\delta) + \mathcal {O} \left(\frac {e ^ {2 \delta L _ {\lambda , \delta}}}{L _ {\lambda , \delta}}\right)
$$
for  $0 < T < \infty$ ,  $0 \leq k < \frac{T}{2h}$ . Name  $N_{k} = \min \left\{k, \left\lfloor \frac{\delta}{2h} \right\rfloor \right\}$ . From the definition of  $e_{i}$  we have
$$
\begin{array}{l} \frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} \\ = \frac {1}{k} \sum_ {i = 0} ^ {k - 1} h (i + 1) \left(\frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {i + 2} - X _ {\lambda , \delta} ^ {i + 1} \right\| + \frac {1}{i + 1} \left\| \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {i + 1}) \right\|\right) + \frac {1}{k} \sum_ {i = 0} ^ {N _ {k} - 1} (i + 1) \left| \frac {1}{i + 1} - \frac {2 h}{\delta} \right| \left\| X _ {\lambda , \delta} ^ {i + 1} - X _ {0} \right\| \\ \leq \sum_ {i = 0} ^ {k - 1} h \left(\frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {i + 2} - X _ {\lambda , \delta} ^ {i + 1} \right\| + \frac {1}{k} \left\| \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {i + 1}) \right\|\right) + \frac {1}{k} \sum_ {i = 0} ^ {N _ {k} - 1} \left\| X _ {\lambda , \delta} ^ {i + 1} - X _ {0} \right\|, \\ \end{array}
$$
where inequality follows from the fact  $\frac{i + 1}{k} \leq 1$  for  $0 \leq i \leq k - 1$  and  $\frac{1}{i + 1} \geq \frac{2h}{\delta}$  for  $0 \leq i \leq N_k - 1$ .
Now let's observe each term. From Lemma B.12 and Lemma D.1 we know
$$
\left\| \dot {X} _ {\lambda , \delta} (t) \right\| \leq \sqrt {2} \| \mathbb {A} _ {\lambda} (X _ {0}) \| \leq \sqrt {2} \| m (\mathbb {A} (X _ {0})) \|
$$
$$
\| \mathbb {A} _ {\lambda} (X (t)) \| \leq \| \mathbb {A} _ {\lambda} (X _ {0}) \| \leq \| m (\mathbb {A} (X _ {0})) \|
$$
$$
\left\| X _ {\lambda , \delta} (t) - X _ {0} \right\| \leq 2 \left\| X _ {0} - X _ {\star} \right\|.
$$
Name  $M = \max \left\{\sqrt{2}\| m(\mathbb{A}(X_0))\| ,2\| X_0 - X_\star \| \right\} .$
(i)  $\frac{h}{\lambda}\left\| X_{\lambda ,\delta}^{i + 2} - X_{\lambda ,\delta}^{i + 1}\right\|$  First observe
$$
\begin{array}{l} h \sum_ {i = 0} ^ {k - 1} \frac {1}{\lambda} \left\| X _ {\lambda , \delta} (2 h (i + 2)) - X _ {\lambda , \delta} (2 h (i + 1)) \right\| \\ = \frac {h}{\lambda} \sum_ {i = 0} ^ {k - 1} \left\| \int_ {2 h (i + 1)} ^ {2 h (i + 2)} \dot {X} _ {\lambda , \delta} (t) d t \right\| \leq \frac {h}{\lambda} \sum_ {i = 0} ^ {k - 1} \int_ {2 h (i + 1)} ^ {2 h (i + 2)} \left\| \dot {X} _ {\lambda , \delta} (t) \right\| d t \leq \frac {h}{\lambda} k (2 h) M \leq \frac {h}{\lambda} T M. \\ \end{array}
$$
$$
\begin{array}{l} h \sum_ {i = 0} ^ {k - 1} \frac {1}{\lambda} \left\| X _ {\lambda , \delta} ^ {i + 2} - X _ {\lambda , \delta} ^ {i + 1} \right\| \\ \leq \frac {h}{\lambda} \sum_ {i = 0} ^ {k - 1} \left(\left\| X _ {\lambda , \delta} ^ {i + 2} - X _ {\lambda , \delta} (2 h (i + 2)) \right\| + \| X _ {\lambda , \delta} (2 h (i + 2)) - X _ {\lambda , \delta} (2 h (i + 1)) \| + \left\| X _ {\lambda , \delta} (2 h (i + 1)) - X _ {\lambda , \delta} ^ {i + 1} \right\|\right) \\ \leq \frac {h}{\lambda} \left(T M + 2 \sum_ {i = 0} ^ {k - 1} h e ^ {2 L _ {\lambda , \delta} T}\right) \\ \leq \frac {h}{\lambda} \left(T M + T e ^ {2 L _ {\lambda , \delta} T}\right). \\ \end{array}
$$
$$
\frac {h}{\lambda} \sum_ {i = 0} ^ {k - 1} \left\| X _ {\lambda , \delta} ^ {i + 2} - X _ {\lambda , \delta} ^ {i + 1} \right\| = \mathcal {O} \left(\frac {h}{\lambda}\right) + \mathcal {O} \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right).
$$
(ii)  $\frac{h}{k}\left\| \mathbb{A}_{\lambda}(X_{\lambda ,\delta}^{i + 1})\right\|$
Since  $\mathbb{A}_{\lambda}$  is  $\frac{1}{\lambda}$ -Lipschitz continuous and from (28)
$$
\begin{array}{l} \sum_ {i = 0} ^ {k - 1} \frac {h}{k} \left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {i + 1}\right) \right\| \leq \frac {h}{k} \sum_ {i = 0} ^ {k - 1} \left(\left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {i}\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} (2 h i)\right) \right\| + \| \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} (2 h i)\right) \|\right) \\ \leq \frac {h}{k} \sum_ {i = 0} ^ {k - 1} \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T} + M\right) = h \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T} + M\right). \\ \end{array}
$$
$$
\sum_ {i = 0} ^ {k - 1} \frac {h}{k} \left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta} ^ {i + 1}\right) \right\| = \mathcal {O} (h) + \mathcal {O} \left(\frac {h ^ {2}}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right).
$$
(iii)  $\frac{1}{k}\sum_{i = 0}^{N_k - 1}\left\| X_{\lambda ,\delta}^{i + 1} - X_0\right\|$  for  $N_{k} = \min \left\{k,\left\lfloor \frac{\delta}{2h}\right\rfloor \right\}$ .
First, observe
$$
\left\| X _ {0} - X _ {\lambda , \delta} (t) \right\| = \left\| \int_ {0} ^ {t} \dot {X} _ {\lambda , \delta} (s) d s \right\| \leq \int_ {0} ^ {t} \left\| \dot {X} _ {\lambda , \delta} (s) \right\| d s \leq t M.
$$
From (27)
$$
\begin{array}{l} \left\| X _ {\lambda , \delta} ^ {i + 1} - X _ {0} \right\| \leq \left\| X _ {\lambda , \delta} ^ {i + 1} - X _ {\lambda , \delta} (2 h (i + 1)) \right\| + \| X _ {\lambda , \delta} (2 h (i + 1)) - X _ {0} \| \\ \leq h \left(1 + 4 h L _ {\lambda , \delta}\right) ^ {i + 1} + 2 h (i + 1) M. \\ \end{array}
$$
Now as  $i + 1\leq N_{k} = \min \left\{k,\left\lfloor \frac{\delta}{2h}\right\rfloor \right\}$
$$
\begin{array}{l} \frac {1}{k} \sum_ {i = 0} ^ {N _ {k} - 1} \left\| X _ {\lambda , \delta} ^ {i + 1} - X _ {0} \right\| \leq \frac {1}{k} \sum_ {i = 0} ^ {N _ {k} - 1} h \left((1 + 4 h L _ {\lambda , \delta}) ^ {i + 1} + 2 (i + 1) M\right) \\ \leq \frac {h}{k} \frac {\left(1 + 4 h L _ {\lambda , \delta}\right) ^ {N _ {k}} - 1}{4 h L _ {\lambda , \delta}} + 2 h M \sum_ {i = 0} ^ {N _ {k} - 1} \frac {i + 1}{k} \\ \leq \frac {e ^ {4 h L _ {\lambda , \delta} N _ {k}}}{k L _ {\lambda , \delta}} + 2 h N _ {k} M \\ \leq \frac {e ^ {2 \delta L _ {\lambda , \delta}}}{L _ {\lambda , \delta}} + 2 \delta M = \mathcal {O} \left(\frac {e ^ {2 \delta L _ {\lambda , \delta}}}{L _ {\lambda , \delta}}\right) + \mathcal {O} (\delta). \\ \end{array}
$$
From (i), (ii), (iii) we have
$$
\frac {1}{k} \sum_ {i = 0} ^ {k - 1} (i + 1) e _ {i} = \mathcal {O} \left(\frac {h}{\lambda}\right) + \mathcal {O} \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} (h) + \mathcal {O} \left(\frac {h ^ {2}}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} \left(\frac {e ^ {2 \delta L _ {\lambda , \delta}}}{L _ {\lambda , \delta}}\right) + \mathcal {O} (\delta).
$$
$$
\epsilon_ {k} = 3 ^ {\frac {T}{\lambda}} \left(\mathcal {O} \left(\frac {h}{\lambda}\right) + \mathcal {O} \left(\frac {h}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} (h) + \mathcal {O} \left(\frac {h ^ {2}}{\lambda} e ^ {2 L _ {\lambda , \delta} T}\right) + \mathcal {O} \left(\frac {e ^ {2 \delta L _ {\lambda , \delta}}}{L _ {\lambda , \delta}}\right) + \mathcal {O} (\delta)\right).
$$
D.4 Derivation of ODE (4) from EAG and FEG
D.4.1 Derivation from EAG
For  $L$ -Lipschitz continuous monotone operator  $\mathbb{A}$  and stepsize  $h > 0$ , EAG-C [75] is defined as
$$
z ^ {k + \frac {1}{2}} = z ^ {k} - \frac {1}{k + 2} (z ^ {k} - z ^ {0}) - h \mathbb {A} (z ^ {k})
$$
$$
z ^ {k + 1} = z ^ {k} - \frac {1}{k + 2} (z ^ {k} - z ^ {0}) - h \mathbb {A} (z ^ {k + \frac {1}{2}}).
$$
Dividing the second line by  $h$  and reorganizing we have
$$
\begin{array}{l} \frac {z ^ {k + 1} - z ^ {k}}{h} = - \mathbb {A} \left(z ^ {k + \frac {1}{2}}\right) - \frac {1}{h (k + 2)} \left(z ^ {k} - z ^ {0}\right) \\ = - \mathbb {A} (z ^ {k}) - \frac {1}{h (k + 2)} \left(z ^ {k} - z ^ {0}\right) - \left(\mathbb {A} (z ^ {k + \frac {1}{2}}) - \mathbb {A} (z ^ {k})\right). \\ \end{array}
$$
Identify  $hk = t$ ,  $z^k = X(t)$ ,  $z^0 = X_0$ . As  $z^k$  is a converging sequence [76], it is bounded. Thus we see
$$
\begin{array}{l} \left\| \mathbb {A} \left(z ^ {k + \frac {1}{2}}\right) - \mathbb {A} \left(z ^ {k}\right) \right\| \leq L \left\| z ^ {k + \frac {1}{2}} - z ^ {k} \right\| \\ = L \left\| \frac {h}{h (k + 2)} \left(z ^ {k} - z ^ {0}\right) + h \mathbb {A} \left(z ^ {k}\right) \right\| \\ = L h \left\| \frac {1}{t + 2 h} \left(X (t) - X _ {0}\right) + \mathbb {A} (X (t)) \right\| = \mathcal {O} (h). \\ \end{array}
$$
Therefore taking limit  $h\to 0+$  we have
$$
\dot {X} (t) = - \mathbb {A} (X (t)) - \frac {1}{t} (X (t) - X _ {0}).
$$
D.4.2 Derivation from FEG
For  $L$ -Lipschitz continuous monotone operator  $\mathbb{A}$  and stepsize  $h > 0$ , FEG [42] is defined as
$$
z ^ {k + \frac {1}{2}} = z ^ {k} - \frac {1}{k + 1} \left(z ^ {k} - z ^ {0}\right) - \frac {k}{k + 1} h \mathbb {A} \left(z ^ {k}\right)
$$
$$
z ^ {k + 1} = z ^ {k} - \frac {1}{k + 1} (z ^ {k} - z ^ {0}) - h \mathbb {A} (z ^ {k + \frac {1}{2}}).
$$
$$
\begin{array}{l} \frac {z ^ {k + 1} - z ^ {k}}{h} = - \mathbb {A} \left(z ^ {k + \frac {1}{2}}\right) - \frac {1}{h (k + 1)} \left(z ^ {k} - z ^ {0}\right) \\ = - \mathbb {A} (z ^ {k}) - \frac {1}{h (k + 1)} \left(z ^ {k} - z ^ {0}\right) - \left(\mathbb {A} (z ^ {k + \frac {1}{2}}) - \mathbb {A} (z ^ {k})\right). \\ \end{array}
$$
$$
\begin{array}{l} \left\| \mathbb {A} \left(z ^ {k + \frac {1}{2}}\right) - \mathbb {A} \left(z ^ {k}\right) \right\| \leq L \left\| z ^ {k + \frac {1}{2}} - z ^ {k} \right\| \\ = L \left\| \frac {h}{h (k + 1)} \left(z ^ {k} - z ^ {0}\right) + h \frac {h k}{h (k + 1)} \mathbb {A} \left(z ^ {k}\right) \right\| \\ = L h \left\| \frac {1}{t + h} (X (t) - X _ {0}) + \frac {t}{t + h} \mathbb {A} (X (t)) \right\| = \mathcal {O} (h). \\ \end{array}
$$
Therefore taking limit  $h \to 0+$  we have
Proof of convergence analysis for monotone  $\mathbb{A}$
E.1 Extending  $\tilde{\mathbf{A}}$  to  $[0,\infty)$
Lemma E.1. Suppose  $\mathbb{A}$  is a maximal monotone operator. Let  $X$  be the solution for (3). Define  $S$  as
$$
S = \left\{t \in [ 0, \infty) \mid \dot {X} (t) \in - \mathbb {A} (X (t)) - \beta (t) (X (t) - X _ {0}) i s t r u e \right\}.
$$
Then as  $X$  is the solution for (3), we know  $[0, \infty) \backslash S$  is of measure zero.
Define  $\tilde{\mathbb{A}}(X) \colon S \to \mathbb{R}^n$  as
$$
\tilde {\mathbf {A}} (X) (t) = - \dot {X} (t) - \beta (t) (X (t) - X _ {0}) \tag {29}
$$
and denote  $\tilde{\mathbf{A}}(X(t)) = \tilde{\mathbf{A}}(X)(t)$ . Assume for every  $T > 0$ , there is  $M > 0$  such that for all  $t \in S \cap [0, T]$
$$
\left\| \tilde {\mathbf {A}} (X (t)) \right\| \leq M.
$$
Then  $\tilde{\mathbb{A}}(X)$  can be extended to  $t \in [0, \infty)$  with satisfying following properties.
(i)  $\tilde{\mathbb{A}} (X(t))\in \mathbb{A}(X(t))$  for all  $t\in [0,\infty)$
(ii) For  $t \in [0, \infty) \backslash S$ , there is a sequence  $\{t_k\}_{k \in \mathbb{N}}$  such that  $t_k \in S$ ,  $\lim_{k \to \infty} t_k = t$  and  $\tilde{\mathbb{A}}(X(t_k))$  converges to  $\tilde{\mathbb{A}}(X(t))$ .
Proof. Take  $t \in [0, \infty)$ , and take a sequence  $\{t_n\}_{n \in \mathbb{N}}$  such that  $t_n \in S$  and  $\lim_{n \to \infty} t_n = t$ . As a converging sequence  $t_n$  is bounded, there is  $T > 0$  such that  $t_n \in [0, T]$ . For that  $T$ , we have  $\| \tilde{\mathbf{A}}(X(t_n)) \| \leq M$  by the assumption. As  $n \mapsto \tilde{\mathbf{A}}(X(t_n))$  is a bounded sequence, there is a subsequence  $\{t_{n_k}\}_{k \in \mathbb{N}}$  such that  $k \mapsto \tilde{\mathbf{A}}(X(t_{n_k}))$  converges. Name the limit as  $u = \lim_{k \to \infty} \tilde{\mathbf{A}}(X(t_{n_k}))$ . On the other hand, as  $X$  is a continuous curve we have  $\lim_{k \to \infty} X(t_{n_k}) = X(t)$ . Then since  $\mathbb{A}$  is maximally monotone, from [15, Proposition 20.38] we have  $(X(t), u) \in \operatorname{Gra} \mathbb{A}$ . Defining  $\tilde{\mathbf{A}}(X(t))$  as  $u$ , we get the desired result.
Corollary E.2. Let  $X$  be the solution for (6). Then  $\tilde{\mathbb{A}}(X)$  defined as (29) has extension with properties stated in Lemma E.1.
Proof. First consider the case  $\beta(t) = \frac{\gamma}{t^p}$ ,  $p > 0$ ,  $\gamma > 0$ . Take  $T > 0$ . It is enough to show there is  $M > 0$  such that for  $t \in S \cap [0, T]$
Recall from Proposition B.11 for Yosida approximation  $\mathbb{A}_{\lambda}$ , we denoted  $X_{\lambda}$  as the solution of
$$
\dot {X} _ {\lambda} = - \mathbb {A} _ {\lambda} (X _ {\lambda}) - \frac {\gamma}{t ^ {p}} (X _ {\lambda} - X _ {0}),
$$
and we have shown there is a sequence  $\{\lambda_n\}_{n\in \mathbb{N}}$  such that  $X_{\lambda_n}$  uniformly converges to  $X$  on  $[0,T]$ .
From Lemma B.12, we see for  $h \neq 0$
$$
\left\| \frac {X _ {\lambda} (t + h) - X _ {\lambda} (t)}{h} \right\| \leq \frac {\int_ {t} ^ {t + h} \left\| \dot {X} _ {\lambda} (s) \right\| d s}{h} \leq \frac {\int_ {t} ^ {t + h} M _ {d o t} (T) d s}{h} = M _ {d o t} (T),
$$
$$
\left\| \frac {X (t + h) - X (t)}{h} \right\| = \lim  _ {\lambda \to 0 +} \left\| \frac {X _ {\lambda} (t + h) - X _ {\lambda} (t)}{h} \right\| \leq M _ {d o t} (T).
$$
Therefore for  $t \in S \cap [0, T]$ ,  $\dot{X}(t) = \lim_{h \to 0} \frac{X(t + h) - X(t)}{h}$  holds, we conclude
$$
\left\| \dot {X} (t) \right\| = \lim  _ {h \rightarrow 0} \left\| \frac {X (t + h) - X (t)}{h} \right\| \leq M _ {d o t} (T).
$$
And also from Lemma B.12, for  $t \in [0,T]$  we have
$$
\left\| \frac {\gamma}{t ^ {p}} (X _ {\lambda} (t) - X _ {0}) \right\| = \left\| \dot {X} _ {\lambda} (t) + \mathbb {A} _ {\lambda} (t) \right\| \leq \left\| \dot {X} _ {\lambda} (t) \right\| + \| \mathbb {A} _ {\lambda} (t) \| \leq M _ {d o t} (T) + M _ {\mathbb {A}} (T),
$$
$$
\left\| \frac {\gamma}{t ^ {p}} (X (t) - X _ {0}) \right\| = \lim  _ {\lambda \rightarrow 0 +} \left\| \frac {\gamma}{t ^ {p}} (X _ {\lambda} (t) - X _ {0}) \right\| \leq M _ {d o t} (T) + M _ {\mathbf {A}} (T).
$$
Gathering the result, for  $t \in S \cap [0, T]$  we have
$$
\left\| \tilde {\mathbf {A}} (X (t)) \right\| = \left\| \dot {X} (t) + \frac {\gamma}{t ^ {p}} (X (t) - X _ {0}) \right\| \leq \left\| \dot {X} (t) \right\| + \left\| \frac {\gamma}{t ^ {p}} (X (t) - X _ {0}) \right\| \leq 2 M _ {d o t} (T) + M _ {\mathbf {A}} (T).
$$
E.2 Proof of Proposition 3.2
We prove this theorem by deriving the energy with dilated coordinate  $W(t) = C(t)(X(t) - X_0)$  and conservation law from [67]. Think of dilated coordinate  $W(t) = C(t)(X(t) - X_0)$ . As we're considering the case  $\tilde{\mathbb{A}}$  is Lipschitz continuous, the differential inclusion (3) becomes ODE
$$
\dot {X} (t) = - \tilde {\mathbb {A}} (X (t)) - \beta (t) (X (t) - X _ {0}).
$$
Rewriting the ODE in terms of  $W$ , we have
$$
\frac {1}{C (t)} \left(\dot {W} (t) - \beta (t) W (t)\right) = - \tilde {\mathbf {A}} (X (W (t), t)) - \frac {\beta (t)}{C (t)} W (t)
$$
where  $X(W(t), t) = X(t) = \frac{W(t)}{C(t)} + X_0$ . Organizing, we have
$$
0 = \dot {W} (t) + C (t) \tilde {\mathbf {A}} (X (W (t), t)).
$$
From Lemma B.4 we know  $\tilde{\mathbb{A}}(X(W, t))$  is differentiable almost everywhere, by differentiating we obtain second order ODE which holds almost everywhere
$$
\begin{array}{l} 0 = \ddot {W} + C (t) \beta (t) \tilde {\mathbf {A}} (X (W (t), t)) + C (t) \frac {d}{d t} \tilde {\mathbf {A}} (X (W (t), t)) \\ = \ddot {W} - \beta (t) \dot {W} + C (t) \frac {d}{d t} \tilde {\mathbf {A}} (X (W (t), t)). \tag {30} \\ \end{array}
$$
Now by taking inner product with  $\dot{W}$  and integrating, we get equality which was referred as conservation law in [67]
$$
E _ {1} \equiv \frac {1}{2} \left\| \dot {W} (t) \right\| ^ {2} - \int_ {t _ {0}} ^ {t} \beta (s) \left\| \dot {W} (s) \right\| ^ {2} d s + \int_ {t _ {0}} ^ {t} C (s) \left\langle \frac {d}{d s} \tilde {\mathbb {A}} (X (s)), \dot {W} (s) \right\rangle d s.
$$
Since  $\dot{W}(t) = C(t)\left(\dot{X}(t) + \beta(t)(X(t) - X_0)\right)$ , we can rewrite the integrand in the last term as
$$
\int_ {t _ {0}} ^ {t} C (s) \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {W} (s) \right\rangle d s = \int_ {t _ {0}} ^ {t} C (s) ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s + \int_ {t _ {0}} ^ {t} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), C (s) \beta (s) W (s) \right\rangle d s.
$$