SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Then there exists a subsequence (again denoted by)  $x_{k}(\cdot)$  converging to an absolutely continuous function  $x(\cdot)$  from  $I$  to  $X$  in the sense that
(i)  $x_{k}(\cdot)$  converges to  $x(\cdot)$  over compact subsets of  $I$
(ii)  $\dot{x}_k(\cdot)$  converges weakly to  $\dot{x} (\cdot)$  in  $L^1 (I,X)$
Proof. See [11, Chapter 0.3, Theorem 4].
From Lemma B.12 we can immediately check norm of all derivatives  $\dot{X}_{\lambda_m}$  are bounded by  $M_{dot}(T)$ . So condition (ii) holds with  $M_{dot}(T)$ . For condition (i), we prove  $\mathcal{F}_T$  is convergent in  $C([0,T],\mathbb{R}^n)$ .
Lemma B.14.  $\mathcal{F}_T$  is convergent sequence in  $C([0,T],\mathbb{R}^n)$ . In other words,  $\forall \epsilon > 0$ , there is  $N > 0$  such that  $n,m > N$  implies  $\sup_{t\in [0,T]}\| X_{\lambda_n}(t) - X_{\lambda_m}(t)\| < \epsilon$
Proof. We will show  $X_{\lambda_n}$  is Cauchy sequence in  $C([0,T],\mathbb{R}^n)$ . Let  $\nu, \lambda > 0$ . From (19), we see for  $t \in (0,T]$
$$
\begin{array}{l} \frac {d}{d t} \frac {1}{2} \| X _ {\nu} (t) - X _ {\lambda} (t) \| ^ {2} = \left\langle \dot {X} _ {\nu} (t) - \dot {X} _ {\lambda} (t), X _ {\nu} (t) - X _ {\lambda} (t) \right\rangle \\ = - \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), X _ {\nu} (t) - X _ {\lambda} (t) \right\rangle - \frac {\gamma}{t ^ {p}} \left\| X _ {\nu} (t) - X _ {\lambda} (t) \right\| ^ {2} \\ \leq - \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), X _ {\nu} (t) - X _ {\lambda} (t) \right\rangle . \\ \end{array}
$$
From definition of resolvent we know  $\mathbb{I} - \mathbb{J}_{\lambda \mathbb{A}} = \lambda \mathbb{A}_{\lambda}$ . And from Lemma B.10 (iv) we know  $\mathbb{A}_{\lambda}(X_{\lambda}(t)) \in \mathbb{A}(\mathbb{J}_{\lambda \mathbb{A}}(X_{\lambda}(t)))$ . Thus from monotone inequality we see
$$
\begin{array}{l} - \langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), X _ {\nu} (t) - X _ {\lambda} (t) \rangle \\ = - \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), \nu \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \lambda \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \right\rangle \\ - \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), \mathbb {J} _ {\nu \mathbb {A}} \left(X _ {\nu} (t)\right) - \mathbb {J} _ {\lambda \mathbb {A}} \left(X _ {\lambda} (t)\right) \right\rangle \\ \leq - \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), \nu \mathbb {A} _ {\nu} X _ {\nu} (t) - \lambda \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \right\rangle \\ = \left(\nu + \lambda\right) \left\langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right), \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \right\rangle - \left(\nu \left\| \mathbb {A} _ {\nu} X _ {\nu} (t) \right\| ^ {2} + \lambda \left\| \mathbb {A} _ {\lambda} X _ {\lambda} (t) \right\| ^ {2}\right). \\ \end{array}
$$
By Young's inequality
$$
\begin{array}{l} \left. (\nu + \lambda) \langle \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right), \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \rangle - \left(\nu \| \mathbb {A} _ {\nu} X _ {\nu} (t) \| ^ {2} + \lambda \| \mathbb {A} _ {\lambda} X _ {\lambda} (t) \| ^ {2}\right) \right. \\ \leq \nu \left(\| \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) \| ^ {2} + \frac {1}{4} \| \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \| ^ {2}\right) + \lambda \left(\| \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \| ^ {2} + \frac {1}{4} \| \mathbb {A} _ {\nu} \left(X _ {\nu} (t)\right) \| ^ {2}\right) \\ \left. - \left(\nu \| \mathbb {A} _ {\nu} X _ {\nu} (t) \| ^ {2} + \lambda \| \mathbb {A} _ {\lambda} X _ {\lambda} (t) \| ^ {2}\right) \right. \\ = \frac {1}{4} \left(\nu \| \mathbb {A} _ {\lambda} (X _ {\lambda} (t)) \| ^ {2} + \lambda \| \mathbb {A} _ {\nu} (X _ {\nu} (t)) \| ^ {2}\right). \\ \end{array}
$$
Now applying Lemma B.12 we have
$$
\frac {d}{d t} \frac {1}{2} \| X _ {\nu} (t) - X _ {\lambda} (t) \| ^ {2} \leq \frac {1}{4} (\nu + \lambda) M _ {\mathbb {A}} (T) ^ {2}.
$$
Then integrating both sides from 0 to  $t$  we have
$$
\left\| X _ {\nu} (t) - X _ {\lambda} (t) \right\| ^ {2} \leq \frac {t}{2} (\nu + \lambda) M _ {\mathbb {A}} (T) ^ {2}. \tag {20}
$$
Now take  $\epsilon > 0$ . Then there is  $N > 0$  such that for  $n > N$ ,  $\lambda_n < \frac{\epsilon}{TM_{\mathrm{A}}(T)^2}$  holds. Then for  $t \in [0, T]$ ,  $n, m > N$ , we have
$$
\left\| X _ {\lambda_ {n}} (t) - X _ {\lambda_ {m}} (t) \right\| ^ {2} \leq \frac {t}{2} (\lambda_ {n} + \lambda_ {m}) M _ {\mathbb {A}} (T) ^ {2} <   \epsilon .
$$
Therefore we get the desired result.
Finally, we are ready to prove Proposition B.11, which implies the main theorem.
Proof of Proposition B.11. Take  $T > 0$ . We know  $X_{\lambda_n}$  uniformly converges on  $[0, T]$  by Lemma B.14. Name the limit as  $X$ , i.e. define  $X: [0, T] \to \mathbb{R}^n$  as  $X(t) = \lim_{n \to \infty} X_{\lambda_n}(t)$ . Then as  $X_{\lambda_n}(0) = X_0$  for all  $n \in \mathbb{N}$ , we see  $X$  satisfies the initial condition. It remains to show  $X$  satisfies (6) almost everywhere.
Recall  $\{\dot{X}_{\lambda_n}\}$  is bounded in  $L^{\infty}([0,T],\mathbb{R}^{n})$  by Lemma B.12. Thus we can apply Lemma B.13, there is a subsequence  $\{\dot{X}_{\lambda_{n_k}}\}$  converges weakly to  $\dot{X}$  in  $L^1 ([0,T],\mathbb{R}^n)$ . Furthermore we have  $\dot{X}\in L^{\infty}([0,T],\mathbb{R}^{n})$  and so  $\{\dot{X}_{\lambda_{n_k}}\}$  also converges weakly to  $\dot{X}$  in  $L^2 ([0,T],\mathbb{R}^n)$  as well.
For  $\lambda > 0$ , define  $f_{\lambda} \colon [0, T] \to \mathbb{R}^n$  as
$$
f _ {\lambda} (t) = \left\{ \begin{array}{l l} \frac {\gamma}{t ^ {p}} \left(X _ {\lambda} (t) - X _ {0}\right) & \text {i f} t > 0 \\ 0 & \text {i f} t = 0. \end{array} \right.
$$
Then for  $f\colon [0,T]\to \mathbb{R}^n$  defined as  $f(t) = \frac{\gamma}{t^p}\left(X(t) - X_0\right)$  for  $t > 0$  and  $f(0) = 0$ , we have  $\lim_{k\to \infty}f_{\lambda_{n_k}}(t) = f(t)$ . As  $\| f_{\lambda}(t)\| = \left\| \dot{X}_{\lambda}(t) + \mathbb{A}_{\lambda_n}(X_\lambda)(t)\right\| \leq M_{dot}(T) + M_{\mathbf{A}}(T)$  for  $t\in (0,T]$  by Lemma B.12, we have
$$
\left\| f _ {\lambda} (t) - f (t) \right\| ^ {2} \leq \left(\left\| f _ {\lambda} (t) \right\| + \left\| f (t) \right\|\right) ^ {2} \leq 4 \left(M _ {d o t} (T) + M _ {\mathbb {A}} (T)\right) ^ {2}.
$$
Therefore by dominated convergence theorem we have
$$
\lim _ {k \to \infty} \int_ {0} ^ {T} \left\| f _ {\lambda_ {n _ {k}}} (t) - f (t) \right\| ^ {2} d t = \int_ {0} ^ {T} \lim _ {k \to \infty} \left\| f _ {\lambda_ {n _ {k}}} (t) - f (t) \right\| ^ {2} d t = 0,
$$
we conclude  $f_{\lambda_{n_k}}$  strongly converges to  $f$  in  $L^2 ([0,T],\mathbb{R}^n)$
Now consider  $F_{\lambda}:[0,T]\to \mathbb{R}^{n}$  defined as
$$
F _ {\lambda} (t) = \left\{ \begin{array}{l l} \dot {X} _ {\lambda} (t) + \frac {\gamma}{t ^ {p}} (X _ {\lambda} (t) - X _ {0}) & \text {i f} t > 0 \\ - \mathbb {A} _ {\lambda} (X _ {0}) & \text {i f} t = 0 \end{array} \right.
$$
Note since  $X_{\lambda}$  are solution to ODE (19), we have
$$
F _ {\lambda} (t) = - \mathbb {A} _ {\lambda} (X _ {\lambda} (t)).
$$
Then for  $F:[0,T]\to \mathbb{R}^n$  defined as
$$
F (t) = \left\{ \begin{array}{l l} \dot {X} (t) + \frac {\gamma}{t ^ {p}} (X (t) - X _ {0}) & \text {i f} t > 0 \\ - m (\mathsf {A} (X _ {0})) & \text {i f} t = 0, \end{array} \right.
$$
we see  $\{F_{\lambda_{n_k}}\}_{k\in \mathbb{N}}$  converges weakly to  $F$  in  $L^2 ([0,T],\mathbb{R}^n)$
On the other hand, by Lemma B.10 (iv) and the fact  $-F_{\lambda_{n_k}}(t) = \mathbb{A}_{\lambda_{n_k}}(X_{\lambda_{n_k}}(t))$  , we see
$$
- F _ {\lambda_ {n _ {k}}} (t) \in \mathbb {A} (\mathbb {J} _ {\lambda_ {n _ {k}} \mathbb {A}} (X _ {\lambda_ {n _ {k}}} (t))).
$$
Observe, from the definition of  $\mathbb{A}_{\lambda_n}$  and Lemma B.12, we see
$$
\left\| X _ {\lambda_ {n}} (t) - \mathbb {J} _ {\lambda_ {n} \mathbf {A}} \left(X _ {\lambda_ {n}} (t)\right) \right\| = \lambda_ {n} \left\| \mathbf {A} _ {\lambda_ {n}} \left(X _ {\lambda_ {n}} (t)\right) \right\| \leq \lambda_ {n} M _ {\mathbf {A}} (T).
$$
Since  $X_{\lambda_n}$  converges to  $X$  in  $\mathcal{C}([0,T],\mathbb{R}^n)$ , by taking  $n\to \infty$  above inequality we see  $\mathbb{J}_{\lambda_n\mathbb{A}}(X_{\lambda_n})$  also converges to  $X$  in  $\mathcal{C}([0,T],\mathbb{R}^n)$ .
Now for  $\mathcal{A}:L^2 ([0,T],\mathbb{R}^n)\to L^2 ([0,T],\mathbb{R}^n)$  defined as  $(\mathcal{A}(x))(t) = \mathbb{A}(x(t))$  almost everywhere, by [11, Chapter 3.1, Proposition 4],  $\mathcal{A}$  is maximal monotone since  $\mathbb{A}$  is maximal monotone. Since  $-F_{\lambda_k}$  weakly converges to  $-F$  in  $L^2 ([0,T],\mathbb{R}^n)$
and  $\mathbb{J}_{\lambda_k}\mathbb{A}(X_{\lambda_k})$  strongly converges to  $X$  in  $L^2 ([0,T],\mathbb{R}^n)$ , by [11, Chapter 3.1, Proposition 2] we have  $-F\in \mathcal{A}(X)$  in  $L^2 ([0,T],\mathbb{R}^n)$ . Therefore for almost all  $t\in (0,T]$  we have
$$
- \left(\dot {X} (t) + \frac {\gamma}{t ^ {p}} \left(X (t) - X _ {0}\right)\right) \in \mathbb {A} (X (t)).
$$
Reorganizing the result with respect to  $\dot{X}$ , we have following is true for almost all  $t\in (0,T]$
$$
\dot {X} (t) \in - \mathbb {A} (X (t)) - \frac {\gamma}{t ^ {p}} (X (t) - X _ {0}).
$$
Since  $T > 0$  was arbitrary, we conclude  $X$  satisfies above inclusion for almost all  $t \in (0,\infty)$ .
C Proof of existence and uniqueness of the solution of (10)
As uniqueness comes from Theorem B.1, we only need to show the existence. What we need for the existence proof are
(i) Nonincreasing function  $\tilde{U}(t)$  which contains  $\left\| \dot{\tilde{X}} \right\|^2$  as in Lemma B.5.
(ii) Uniform boundedness of  $\dot{\hat{X}}_{\delta}(t)$  for  $t\in [0,T]$  as shown in Lemma B.6.
(iii)  $\dot{\tilde{X}}(0) = \lim_{t \to 0+} \frac{\tilde{X}(t) - X_0}{t} = \lim_{t \to 0+} \dot{\tilde{X}}(t)$  as shown in the existence proof of Lipschitz case.
(iv) Uniform boundedness of  $\left\| \dot{X}_{\lambda}(t)\right\|$  and  $\| \mathbb{A}_{\lambda}(X_{\lambda}(t))\|$  for  $t\in [0,T]$  as shown in Lemma B.12
We now show these steps can be also done to the  $\beta (t) = \frac{2\mu}{e^{2\mu t} - 1}$ .
(i) Nonincreasing function  $\tilde{U}(t)$  which contains  $\left\| \dot{\tilde{X}} \right\|^2$
$$
\dot {\tilde {X}} (t) = - \tilde {A} (\tilde {X} (t)) - \frac {2 \mu}{e ^ {2 \mu t} - 1} (\tilde {X} (t) - X _ {0})
$$
for almost all  $t$  we have
$$
\ddot {\tilde {X}} (t) = - \frac {d}{d t} \tilde {A} (\tilde {X} (t)) + \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} e ^ {2 \mu t} (\tilde {X} (t) - X _ {0}) - \frac {2 \mu}{e ^ {2 \mu t} - 1} \dot {\tilde {X}} (t)
$$
$$
\tilde {U} (t) = e ^ {- 2 \mu t} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}
$$
Therefore for almost all  $t > 0$
$$
\begin{array}{l} \dot {\bar {U}} (t) = 2 e ^ {- 2 \mu t} \left<   \dot {\bar {X}} (t), \ddot {\bar {X}} (t) \right> - 2 \mu e ^ {- 2 \mu t} \left\| \dot {\bar {X}} (t) \right\| ^ {2} \\ + 2 \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle - \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {3} 2 e ^ {2 \mu t} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - 2 e ^ {- 2 \mu t} \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle + 2 \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle - 2 e ^ {- 2 \mu t} \frac {2 \mu}{e ^ {2 \mu t} - 1} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} \\ - 2 \mu e ^ {- 2 \mu t} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + 2 \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle - \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {3} 2 e ^ {2 \mu t} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - 2 \mu e ^ {- 2 \mu t} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} - 2 e ^ {- 2 \mu t} \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle - \frac {2 \mu e ^ {- 2 \mu t}}{e ^ {2 \mu t} - 1} \left\| \dot {\tilde {X}} (t) - \frac {2 \mu e ^ {2 \mu t}}{e ^ {2 \mu t} - 1} \left(\tilde {X} (t) - X _ {0}\right) \right\| ^ {2} \\ \leq 0. \\ \end{array}
$$
(ii) Uniform boundedness of  $\dot{\hat{X}}_{\delta}(t)$
As (13), we define  $\dot{\tilde{X}}_{\delta}(t)$  as the solution of
$$
\dot {\tilde {X}} _ {\delta} (t) = \left\{ \begin{array}{l l} - \tilde {A} (\tilde {X} _ {\delta} (t)) - \frac {2 \mu}{e ^ {2 \mu \delta} - 1} (\tilde {X} _ {\delta} (t) - X _ {0}) & 0 \leq t \leq \delta \\ - \tilde {A} (\tilde {X} _ {\delta} (t)) - \frac {2 \mu}{e ^ {2 \mu t} - 1} (\tilde {X} _ {\delta} (t) - X _ {0}) & t \geq \delta \end{array} \right.
$$
Again with same arguments of Lemma B.6 we have  $\left\| \tilde{X}_{\delta}(\delta) - X_0\right\| \leq \delta \left\| \tilde{A} (X_0)\right\|$  and  $\left\| \dot{\tilde{X}}_{\delta}(\delta)\right\| \leq \left\| \tilde{A} (X_0)\right\|$ . Now for  $t\in [0,T]$
$$
\begin{array}{l} \left\| \dot {\tilde {X}} _ {\delta} (t) \right\| \leq \sqrt {e ^ {2 \mu t} \tilde {U} (t)} \leq \sqrt {e ^ {2 \mu t} \tilde {U} (\delta)} = e ^ {\mu t} \sqrt {e ^ {- 2 \mu \delta} \left\| \dot {\tilde {X}} _ {\delta} (\delta) \right\| ^ {2} + \left(\frac {2 \mu}{e ^ {2 \mu \delta} - 1}\right) ^ {2} \left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| ^ {2}} \\ \leq e ^ {\mu t} \sqrt {e ^ {- 2 \mu \delta} \left\| \tilde {A} (X _ {0}) \right\| ^ {2} + \left(\frac {2 \mu \delta}{e ^ {2 \mu \delta} - 1}\right) ^ {2} \left\| \tilde {A} (X _ {0}) \right\| ^ {2}} \\ \leq e ^ {\mu T} \sqrt {2} \left\| \tilde {A} (X _ {0}) \right\|. \\ \end{array}
$$
(iii)  $\dot{\tilde{X}}(0) = \lim_{t \to 0+} \frac{\tilde{X}(t) - X_0}{t} = \lim_{t \to 0+} \dot{\tilde{X}}(t)$
Define  $C(t) \coloneqq 1 - e^{-2\mu t}$ . Then  $\dot{C}(t) = C(t)\beta(t)$  for  $\beta(t) = \frac{2\mu}{e^{2\mu t} - 1}$ . And since
$$
\lim _ {t \to 0 +} \frac {C (t v)}{C (t)} = \lim _ {t \to 0 +} \frac {1 - e ^ {- 2 \mu t v}}{1 - e ^ {- 2 \mu t}} = v,
$$
with same argument done to arrive (15), we have
$$
\dot {\tilde {X}} (0) = \lim  _ {t \rightarrow 0 +} \frac {\tilde {X} (t) - X _ {0}}{t} = - \int_ {0} ^ {1} \lim  _ {t \rightarrow 0 +} \left(\frac {C (t v)}{C (t)} \tilde {A} (\tilde {X} (t v))\right) d v = - \frac {1}{2} \tilde {A} (X _ {0})
$$
Now from ODE  $\dot{\tilde{X}}(t) = -\tilde{A}(\tilde{X}(t)) - \frac{2\mu}{e^{2\mu t} - 1} (\tilde{X}(t) - X_0)$ , by taking limit both sides by  $t \to 0+$  we have
$$
\lim _ {t \to 0 +} \dot {\tilde {X}} (t) = - \tilde {A} (\tilde {X} (0)) - \lim _ {t \to 0 +} \frac {2 \mu t}{e ^ {2 \mu t} - 1} \frac {\tilde {X} (t) - X _ {0}}{t} = - \frac {1}{2} \tilde {A} (X _ {0})
$$
Therefore,  $\lim_{t\to 0 + }\frac{\tilde{X}(t) - X_0}{t} = \lim_{t\to 0 + }\dot{\tilde{X}} (t).$
(iv) Uniform boundedness of  $\left\| \dot{X}_{\lambda}(t)\right\|$  and  $\| \mathbb{A}_\lambda (t)\|$  for  $t\in [0,T]$
Recall  $U(t) = e^{-2\mu t}\left\| \dot{X}_{\lambda}(t)\right\|^{2} + \left(\frac{2\mu}{e^{2\mu t} - 1}\right)^{2}\| X_{\lambda}(t) - X_{0}\|^{2}$  is nonincreasing. So from (iii), we have
$$
\begin{array}{l} e ^ {- 2 \mu t} \left\| \dot {X} _ {\lambda} (t) \right\| ^ {2} + \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \| X _ {\lambda} (t) - X _ {0} \| ^ {2} \leq \lim  _ {t \rightarrow 0 +} \left(e ^ {- 2 \mu t} \left\| \dot {X} _ {\lambda} (t) \right\| ^ {2} + \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \| X _ {\lambda} (t) - X _ {0} \| ^ {2}\right) \\ = \frac {1}{4} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} + \frac {1}{4} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} = \frac {1}{2} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} \\ \end{array}
$$
Therefore we have from Lemma B.10 (iii)
$$
e ^ {- 2 \mu t} \left\| \dot {X} _ {\lambda} (t) \right\| ^ {2} \leq \frac {1}{2} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} \leq \frac {1}{2} \| m (\mathbb {A} (X _ {0})) \| ^ {2} \quad \Longrightarrow \quad \left\| \dot {X} _ {\lambda} (t) \right\| \leq \frac {e ^ {\mu T}}{\sqrt {2}} \| m (\mathbb {A} (X _ {0})) \|, \tag {21}
$$
and by Young's inequality
$$
\begin{array}{l} \left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \right\| ^ {2} \leq \left\| \dot {X} _ {\lambda} (t) + \frac {2 \mu}{e ^ {2 \mu t} - 1} \left(X _ {\lambda} (t) - X _ {0}\right) \right\| ^ {2} \\ = 2 \left(\left\| \dot {X} _ {\lambda} (t) \right\| ^ {2} + \left(\frac {2 \mu}{e ^ {2 \mu t} - 1}\right) ^ {2} \| X _ {\lambda} (t) - X _ {0} \| ^ {2}\right) \\ \leq 2 \left(\frac {e ^ {2 \mu T}}{2} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} + \frac {1}{2} \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2}\right) = (e ^ {2 \mu T} + 1) \| \mathbb {A} _ {\lambda} (X _ {0}) \| ^ {2} \leq (e ^ {2 \mu T} + 1) \| m (\mathbb {A} (X _ {0})) \| ^ {2}. \\ \end{array}
$$
$$
\left\| \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \right\| \leq \sqrt {e ^ {2 \mu T} + 1} \| m (\mathbb {A} \left(X _ {0}\right)) \|. \tag {22}
$$
D Omitted proofs for derivation of anchor ODE (4)
D.1 Preparation for the proof of Theorem 2.1
We first provide the boundedness of trajectories as a lemma. As mentioned in the discussion after Lemma 2.3, boundedness of trajectories is an immediate corollary of Lemma 2.3. However, to address cases that are slightly more generalized, we present a proof using an argument similar to the one used in the proof of Lemma 2.3. Note the proof argument of following lemma is valid for the solution of differential equation (11) with satisfying the assumptions in Theorem 7.1 as well.
Lemma D.1. (Boundedness of solutions) Suppose  $X(\cdot)$  is the solution of the differential inclusion (3). Then for all  $X_{\star} \in \mathrm{Zer}\mathbb{A}$ ,  $t \in [0, \infty)$ , following holds.
And so,  $\| X(t) - X_0 \| \leq 2 \| X_0 - X_\star \|$ .
Proof. It is trivial for  $t = 0$ , so we may assume  $t > 0$ .
Take  $X_{\star}\in \mathsf{ZerA}$ . By monotonicity of  $\mathbb{A}$  and Young's inequality, we get the following inequality.
$$
\begin{array}{l} \frac {d}{d t} \left\| X (t) - X _ {\star} \right\| ^ {2} = 2 \left\langle \dot {X} (t), X (t) - X _ {\star} \right\rangle = - 2 \left\langle \tilde {\mathbf {A}} (X (t)) + \beta (t) (X (t) - X _ {0}), X (t) - X _ {\star} \right\rangle \\ = - 2 \left\langle \tilde {\mathbf {A}} (X (t)) + \beta (t) (X (t) - X _ {\star}) - \beta (t) \left(X _ {0} - X _ {\star}\right), X (t) - X _ {\star} \right\rangle \\ \leq - 2 \beta (t) \| X (t) - X _ {\star} \| ^ {2} + 2 \beta (t) \langle X _ {0} - X _ {\star}, X (t) - X _ {\star} \rangle \\ \leq - 2 \beta (t) \| X (t) - X _ {\star} \| ^ {2} + \beta (t) \left(\| X _ {0} - X _ {\star} \| ^ {2} + \| X (t) - X _ {\star} \| ^ {2}\right) \\ = - \beta (t) \| X (t) - X _ {\star} \| ^ {2} + \beta (t) \| X _ {0} - X _ {\star} \| ^ {2}. \\ \end{array}
$$
Now again define  $C(t) = e^{\int_v^t\beta (s)ds}$  for some  $v > 0$  , then we see  $\dot{C} (t) = C(t)\beta (t)$  . Moving  $-\beta (t)\| X(t) - X_{\star}\|^{2}$  to the left hand side and multiplying both sides by  $C(t)$  , we have
$$
\begin{array}{l} \frac {d}{d t} \left(C (t) \| X (t) - X _ {\star} \| ^ {2}\right) = C (t) \frac {d}{d t} \| X (t) - X _ {\star} \| ^ {2} + C (t) \beta (t) \| X (t) - X _ {\star} \| ^ {2} \\ \leq C (t) \beta (t) \| X _ {0} - X _ {\star} \| ^ {2} = \frac {d}{d t} C (t) \| X _ {0} - X _ {\star} \| ^ {2}. \\ \end{array}
$$
$$
C (t) \| X (t) - X _ {\star} \| ^ {2} - C (\epsilon) \| X (\epsilon) - X _ {\star} \| ^ {2} \leq C (t) \| X _ {0} - X _ {\star} \| ^ {2} - C (\epsilon) \| X _ {0} - X _ {\star} \| ^ {2}.
$$
As  $\beta > 0$ , we have  $C$  is nonnegative and nondecreasing,  $\lim_{\epsilon \to 0+} C(\epsilon)$  exists. Taking limit  $\epsilon \to 0+$  both sides we have and dividing both sides by  $C(t)$  we conclude
$$
\left\| X (t) - X _ {\star} \right\| ^ {2} \leq \left\| X _ {0} - X _ {\star} \right\| ^ {2}.
$$
The latter statement holds directly from triangular inequality,
$$
\left\| X (t) - X _ {0} \right\| \leq \left\| X (t) - X _ {\star} \right\| + \left\| X _ {\star} - X _ {0} \right\| \leq 2 \left\| X _ {0} - X _ {\star} \right\|.
$$
Following lemma shows APPM is an instance of Halpern method. It is immediate from induction, but we state it as a lemma due as its importance.
Lemma D.2. Consider a method defined as
$$
x ^ {k + 1} = \mathbb {J} _ {\mathbb {A}} y ^ {k}
$$
$$
y ^ {k + 1} = \left(1 - \beta_ {k}\right) \left(2 x ^ {k + 1} - y ^ {k}\right) + \beta_ {k} x ^ {0},
$$
for  $k = 0,1,\ldots$ , with initial condition  $y^0 = x^0$ . Then for reflected resolvent  $\mathbb{R}_{\mathrm{A}}$  defined as  $\mathbb{R}_{\mathrm{A}} = 2\mathbb{J}_{\mathrm{A}} - \mathbb{I}$ , above method is equivalent to
$$
\tilde {y} ^ {k + 1} = \beta_ {k} \tilde {y} ^ {0} + \left(1 - \beta_ {k}\right) \mathbb {T} \tilde {y} ^ {k}
$$
when  $\mathbb{T} = \mathbb{R}_{\mathbf{A}}$ ,  $\tilde{y}^0 = y^0$ . Here equivalence means  $\tilde{y}^k = y^k$  holds for  $k = 0,1,\ldots$ .
Proof. Proof by induction. As  $\tilde{y}^0 = y^0$  by assumption, the statement is true for  $k = 0$ . Now suppose  $y^{k} = \tilde{y}^{k}$  holds for  $k\in \mathbb{N}$ , then
$$
\begin{array}{l} \tilde {y} ^ {k + 1} = \left(1 - \beta_ {k}\right) \mathbb {R} _ {\mathbf {A}} \tilde {y} ^ {k} + \beta_ {0} \tilde {y} ^ {0} = \left(1 - \beta_ {k}\right) \left(2 \mathbb {J} _ {\mathbf {A}} - \mathbb {I}\right) \tilde {y} ^ {k} + \beta_ {0} \tilde {y} ^ {0} \\ = \left(1 - \beta_ {k}\right) \left(2 \mathbb {J} _ {\mathbf {A}} - \mathbb {I}\right) y ^ {k} + \beta_ {0} y ^ {0} = \left(1 - \beta_ {k}\right) \left(2 x ^ {k + 1} - y ^ {k}\right) + \beta_ {0} y ^ {0} = y ^ {k + 1}. \\ \end{array}
$$
Thus  $\tilde{y}^{k + 1} = y^{k + 1}$ , the statement is true for  $k + 1$ . By induction, we get the desired result.
D.2 Proof of Theorem 2.1
Let  $\mathbb{A}\colon \mathbb{R}^n\to \mathbb{R}^n$  be a maximal monotone operator, and  $h,\lambda ,\delta >0$  . Again, denote  $\mathbb{A}_{\lambda} = \frac{1}{\lambda} (\mathbb{I} - \mathbb{A}_{\lambda \mathbf{A}}) = \frac{1}{\lambda} (\mathbb{I} - (\mathbb{I} + \lambda \mathbb{A})^{-1})$  Since various kind of terms appear in the proof, we first organize the terms and notations.
-  $X$ : Solution of differential inclusion,
$$
\dot {X} \in - \mathbb {A} (X) - \frac {1}{t} (X - X _ {0}).
$$
-  $X_{\lambda}$ : Solution of differential equation,
$$
\dot {X} _ {\lambda} = - \mathbb {A} _ {\lambda} (X _ {\lambda}) - \frac {1}{t} (X _ {\lambda} - X _ {0}).
$$
-  $X_{\lambda, \delta}$ : Solution of approximated differential equation,
$$
\dot {X} _ {\lambda , \delta} (t) = \left\{ \begin{array}{l l} - \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta}\right) (t) - \frac {1}{\delta} \left(X _ {\lambda , \delta} (t) - X _ {0}\right) & 0 \leq t <   \delta \\ - \mathbb {A} _ {\lambda} \left(X _ {\lambda , \delta}\right) (t) - \frac {1}{t} \left(X _ {\lambda , \delta} (t) - X _ {0}\right) & t \geq \delta . \end{array} \right. \tag {23}
$$
-  $X_{\lambda, \delta}^{k}$ : Sequence obtained by taking Euler discretization of ODE (23),
$$
X _ {\lambda , \delta} ^ {k + 1} = \left\{ \begin{array}{l l} X _ {\delta} ^ {k} - \left(2 h \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k}) + \frac {2 h}{\delta} (X _ {\lambda , \delta} ^ {k} - X _ {0})\right) & 0 \leq k <   \frac {\delta}{2 h} \\ X _ {\delta} ^ {k} - \left(2 h \mathbb {A} _ {\lambda} (X _ {\lambda , \delta} ^ {k}) + \frac {1}{k} (X _ {\lambda , \delta} ^ {k} - X _ {0})\right) & k \geq \frac {\delta}{2 h}. \end{array} \right.
$$
-  $x_{h,\lambda}^{k}$ : Sequence obtained from APPM with operator  $h\mathbb{A}_{\lambda}$ , i.e.
$$
x _ {h, \lambda} ^ {k} = \mathbb {J} _ {h \mathbb {A} _ {\lambda}} y _ {h, \lambda} ^ {k - 1}
$$
$$
y _ {h, \lambda} ^ {k} = \frac {k}{k + 1} (2 x _ {h, \lambda} ^ {k} - y _ {h, \lambda} ^ {k - 1}) + \frac {1}{k + 1} X _ {0}.
$$
-  $x_h^k$ : Sequence obtained from APPM with operator  $h\mathbb{A}$ , i.e.