SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \frac {2 V (t)}{1 - e ^ {- 2 \mu t}} = (e ^ {2 \mu t} - 1) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + 4 \mu \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle - 4 \mu^ {2} \| X (t) - X _ {0} \| ^ {2} \\ = (e ^ {\mu t} - 1) \underbrace {\left(\| \tilde {\mathbf {A}} (X (t)) \| ^ {2} - 4 \mu \left\langle \tilde {\mathbf {A}} (X (t)) , X (t) - X _ {0} \right\rangle + 4 \mu^ {2} \| X (t) - X _ {0} \| ^ {2}\right)} _ {= \left\| \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}) \right\| ^ {2}} \\ + e ^ {\mu t} (e ^ {\mu t} - 1) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + \underbrace {e ^ {\mu t} \left(4 \mu \left\langle \tilde {\mathbf {A}} (X (t)) , X (t) - X _ {0} \right\rangle - 4 \mu^ {2} \| X (t) - X _ {0} \| ^ {2}\right)} _ {= p (t)}. \\ \end{array}
$$
From the law of cosines, we have  $\| X(t) - X_0 \|^2 = \| X(t) - X_\star \|^2 - 2 \langle X(t) - X_0, X_0 - X_\star \rangle - \| X_\star - X_0 \|^2$ . Applying this to  $p(t)$  we have
$$
\begin{array}{l} p (t) = 4 \mu e ^ {\mu t} \big (\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {\star} \rangle - \langle \tilde {\mathbf {A}} (X (t)), X _ {0} - X _ {\star} \rangle - \mu \| X (t) - X _ {0} \| ^ {2} \big) \\ = 4 \mu e ^ {\mu t} \left(\langle \tilde {\mathbf {A}} (X (t)) + \mu (X (t) - X _ {\star}), X (t) - X _ {\star} \rangle - \langle \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}), X _ {0} - X _ {\star} \rangle - \mu \| X _ {0} - X _ {\star} \| ^ {2}.\right) \\ \end{array}
$$
$$
\begin{array}{l} \frac {2 V (t)}{1 - e ^ {- 2 \mu t}} = (e ^ {\mu t} - 1) \left\| \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}) \right\| ^ {2} + e ^ {\mu t} (e ^ {\mu t} - 1) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + p (t) \\ = \left(e ^ {\mu t} - 1\right) \| \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}) \| ^ {2} - 4 \mu e ^ {\mu t} \langle \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}), X _ {0} - X _ {\star} \rangle \\ + e ^ {\mu t} \left(e ^ {\mu t} - 1\right) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + 4 \mu e ^ {\mu t} \langle \tilde {\mathbf {A}} (X (t)) - \mu (X (t) - X _ {\star}), X (t) - X _ {\star} \rangle + 4 \mu^ {2} e ^ {\mu t} \| X _ {0} - X _ {\star} \| ^ {2} \\ = \left(e ^ {\mu t} - 1\right) \left\| \tilde {\mathbf {A}} (X (t)) - 2 \mu (X (t) - X _ {0}) - \frac {2 \mu e ^ {\mu t}}{e ^ {\mu t} - 1} (X _ {0} - X _ {\star}) \right\| ^ {2} - \frac {4 \mu^ {2} e ^ {2 \mu t}}{e ^ {\mu t} - 1} \| X _ {0} - X _ {\star} \| ^ {2} \\ + e ^ {\mu t} (e ^ {\mu t} - 1) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + 4 \mu e ^ {\mu t} \left(\left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {\star} \right\rangle - \mu \left\| X (t) - X _ {\star} \right\| ^ {2}\right) + 4 \mu^ {2} e ^ {\mu t} \| X _ {0} - X _ {\star} \| ^ {2}. \\ \end{array}
$$
Now since  $\tilde{\mathbb{A}}$  is  $\mu$ -strongly monotone,  $\langle \tilde{\mathbb{A}}(X(t)), X(t) - X_{\star} \rangle - \mu \|X(t) - X_{\star}\|^2 \geq 0$ . From previous section we know  $0 = V(0) \geq V(t)$  for all  $t > 0$ . Therefore for all  $t > 0$
$$
\begin{array}{l} 0 \geq \frac {2 V (t)}{1 - e ^ {- 2 \mu t}} \geq e ^ {\mu t} \left(e ^ {\mu t} - 1\right) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} + \left(4 \mu^ {2} e ^ {\mu t} - \frac {4 \mu^ {2} e ^ {2 \mu t}}{e ^ {\mu t} - 1}\right) \| X _ {0} - X _ {\star} \| ^ {2} \\ = e ^ {\mu t} \left(e ^ {\mu t} - 1\right) \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} - \frac {4 \mu^ {2} e ^ {\mu t}}{e ^ {\mu t} - 1} \| X _ {0} - X _ {\star} \| ^ {2}. \\ \end{array}
$$
Organizing, we conclude
$$
\| \tilde {\mathbf {A}} (X (t)) \| ^ {2} \leq 4 \left(\frac {\mu}{e ^ {\mu t} - 1}\right) ^ {2} \| X _ {0} - X _ {\star} \| ^ {2}.
$$
H.2.3 Informal derivation of ODE from the method
Assume  $\mathbb{A}\colon \mathbb{R}^n\to \mathbb{R}^n$  be a continuous monotone operator. In [54], the method OS-PPM is presented as
$$
x ^ {k} = \mathbb {J} _ {h \mathbf {A}} y ^ {k - 1}
$$
$$
y ^ {k} = \left(1 - \frac {1}{s _ {k}}\right) \left\{x ^ {k} - \frac {1}{\nu} (y ^ {k - 1} - x ^ {k}) \right\} + \frac {1}{s _ {k}} y ^ {0}
$$
where  $y^0 = x^0$ ,  $\nu = 1 + 2h\mu$  and  $s_k = 1 + \nu^2 + \dots + \nu^{2k} = \frac{\nu^{2k+2} - 1}{\nu^2 - 1}$ . Using  $y^{k-1} = x^k + h\mathbb{A}x^k$ , substituting  $y^k$  and  $y^{k-1}$  this method can be expressed in a single line,
$$
x ^ {k + 1} + h \mathbb {A} x ^ {k + 1} = \left(1 - \frac {1}{s _ {k}}\right) \left(x ^ {k} - \frac {h}{\nu} \mathbb {A} x ^ {k}\right) + \frac {1}{s _ {k}} x ^ {0}.
$$
Reorganizing and dividing both sides by  $h$ , we have
$$
\frac {x ^ {k + 1} - x ^ {k}}{h} = - \mathbb {A} x ^ {k + 1} - \left(\frac {1}{\nu} - \frac {1}{\nu s _ {k}}\right) \mathbb {A} x ^ {k} - \frac {1}{h s _ {k}} (x ^ {k} - x ^ {0}).
$$
Identifying  $x^0 = X_0$ ,  $2hk = t$ ,  $x^k = X(t)$ ,
$$
\frac {X (t + 2 h) - X (t)}{h} = - \mathbb {A} (X (t + 2 h)) - \left(\frac {1}{\nu} - \frac {1}{\nu s _ {k}}\right) \mathbb {A} (X (t)) - \frac {1}{h s _ {k}} (X (t) - X _ {0}).
$$
Now observe
$$
\begin{array}{l} \lim _ {h \to 0 ^ {+}} h s _ {k} = \lim _ {h \to 0 ^ {+}} h \frac {\nu^ {2 k + 2} - 1}{\nu^ {2} - 1} = \lim _ {h \to 0 ^ {+}} h \frac {(1 + 2 h \mu) ^ {2 k + 2} - 1}{(1 + 2 h \mu) ^ {2} - 1} \\ = \lim _ {h \rightarrow 0 ^ {+}} \frac {(1 + 2 h \mu) ^ {2} (1 + 2 h \mu) ^ {t / h} - 1}{4 \mu (1 + h \mu)} = \frac {e ^ {2 \mu t} - 1}{4 \mu} \\ \lim  _ {h \rightarrow 0 ^ {+}} \frac {1}{\nu} = \lim  _ {h \rightarrow 0 ^ {+}} \frac {1}{1 + 2 h \mu} = 1 \\ \end{array}
$$
$$
\lim _ {h \to 0 ^ {+}} \frac {1}{\nu s _ {k}} = \lim _ {h \to 0 ^ {+}} \frac {1}{\nu} \frac {1}{h s _ {k}} h = 1 \times \frac {4 \mu}{e ^ {2 \mu t} - 1} \times 0 = 0.
$$
Taking limit  $h\to 0^{+}$  and organizing,
$$
2 \dot {X} (t) = - \mathbb {A} (X (t)) - (1 + 0) \mathbb {A} (X (t)) - \frac {4 \mu}{e ^ {2 \mu t} - 1} (X (t) - X _ {0}).
$$
By diving both sides by 2, we get the desired anchor ODE
$$
\dot {X} (t) = - \mathbb {A} (X (t)) - \frac {2 \mu}{e ^ {2 \mu t} - 1} (X (t) - X _ {0}).
$$
I Proof omitted in Section 7
I.1 Proof for Theorem 7.1
$$
\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} \leq 4 \beta (t) ^ {2} \| X _ {0} - X _ {\star} \| ^ {2} = \mathcal {O} (\beta (t) ^ {2}),
$$
$$
\dot {X} (t) = - \tilde {\mathbf {A}} (X (t)) - \beta (t) (X (t) - X _ {0})
$$
$$
\beta (t) = - \frac {\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}}{2 \left\langle \tilde {\mathbf {A}} (X (t)) , X (t) - X _ {0} \right\rangle}.
$$
By definition of  $\beta (t)$ ,  $\Phi (t)$  defined in (8) becomes zero, i.e.
$$
0 \equiv \Phi (t) = \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), X - X _ {0} \right\rangle .
$$
Plugging  $\Phi (t) = 0$  to inequality (8) we have
$$
0 \geq \frac {1}{2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} - 2 \beta (t) ^ {2} \| X _ {0} - X _ {\star} \| ^ {2}.
$$
Reorganizing, we get the desired result. Other results need some works, we provide the proof with steps into subsections.
I.1.1  $\beta (t) > 0$  for  $t > 0$
Taking inner product with  $\tilde{\mathbf{A}}(X)$  to the ODE and applying  $\frac{1}{2}\Phi(t) = 0$  we have
$$
\left\langle \dot {X} (t), \tilde {\mathbf {A}} (X (t)) \right\rangle = - \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} - \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle = - \frac {1}{2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}.
$$
As  $\mathbb{A}$  is assumed to be continuous, taking limit  $t\to 0+$  we have
$$
\lim _ {t \to 0 +} \Big \langle \dot {X} (t), \tilde {\mathbb {A}} (X (t)) \Big \rangle = \Big \langle \lim _ {t \to 0 +} \dot {X} (t), \tilde {\mathbb {A}} (X _ {0}) \Big \rangle = - \frac {1}{2} \lim _ {t \to 0 +} \big \| \tilde {\mathbb {A}} (X (t)) \big \| ^ {2} = - \frac {1}{2} \big \| \tilde {\mathbb {A}} (X _ {0}) \big \| ^ {2}.
$$
On the other hand, by assumption we have  $\lim_{t\to 0 + }\dot{X} (t) = \lim_{t\to 0 + }\frac{X(t) - X_0}{t}$  and  $\| \tilde{\mathbf{A}} (X_0)\| \neq 0$  , thus
$$
\lim  _ {t \rightarrow 0 +} \frac {1}{t} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle = \left\langle \tilde {\mathbf {A}} (X _ {0}), \lim  _ {t \rightarrow 0 +} \dot {X} (t) \right\rangle = - \frac {1}{2} \left\| \tilde {\mathbf {A}} (X _ {0}) \right\| ^ {2} <   0. \tag {50}
$$
Therefore there is  $\epsilon > 0$  such that  $\langle \tilde{\mathbf{A}}(X(t)), X(t) - X_0 \rangle < 0$  for  $t \in (0, \epsilon)$ , thus for  $t \in (0, \epsilon)$  we have  $\langle \tilde{\mathbf{A}}(X(t)), X(t) - X_0 \rangle \neq 0$  so  $\beta(t)$  is well-defined and satisfies
$$
\beta (t) = - \frac {\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}}{2 \left\langle \tilde {\mathbf {A}} (X (t)) , X (t) - X _ {0} \right\rangle} > 0.
$$
Observe the denominator of  $\beta(t)$  is zero when  $\|\tilde{\mathbf{A}}(X(t))\| = 0$ . As  $\beta$  is assumed to be well-defined, we have  $\|\tilde{\mathbf{A}}(X(t))\| \neq 0$  for all  $t > 0$ . Since  $\beta$  is continuous as  $\mathbb{A}$  and  $X$  are continuous, by intermediate value theorem we have  $\|\tilde{\mathbf{A}}(X(t))\| > 0$  for all  $t > 0$ .
I.1.2 Proof for the main statements
We first show following lemma.
Lemma I.1. Following equality holds for almost every  $t > 0$ .
$$
\left(\dot {\beta} (t) + \beta (t) ^ {2}\right) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle = \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \dot {X} (t) \right\rangle . \tag {51}
$$
Proof. Since  $\tilde{\mathbb{A}}$  is Lipschitz continuous by assumption, by Lemma B.4 we know  $\tilde{\mathbb{A}}(X(t))$  is differentiable almost everywhere. Differentiating  $\Phi(t)$  we have
$$
\begin{array}{l} 0 = \dot {\Phi} (t) \\ = 2 \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \tilde {\mathbf {A}} (X (t)) \right\rangle + 2 \dot {\beta} (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + 2 \beta (t) \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), \dot {X} (t) \right\rangle \\ = 2 \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \tilde {\mathbf {A}} (X (t)) + \beta (t) (X (t) - X _ {0}) \right\rangle + 2 \dot {\beta} (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle \\ + 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), - \tilde {\mathbf {A}} (X (t)) - \beta (t) (X (t) - X _ {0}) \right\rangle \\ = - 2 \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \dot {X} (t) \right\rangle - 2 \beta (t) \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 (\dot {\beta} (t) - \beta (t) ^ {2}) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle \\ = - 2 \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \dot {X} (t) \right\rangle + 4 \beta (t) ^ {2} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + 2 \left(\dot {\beta} (t) - \beta (t) ^ {2}\right) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle \\ = - 2 \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X (t)), \dot {X} (t) \right\rangle + 2 \left(\dot {\beta} (t) + \beta (t) ^ {2}\right) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle . \\ \end{array}
$$
for  $t\in (0,\infty)$  almost everywhere. The equalities come from the ODE  $\dot{X} (t) = -\tilde{\mathbf{A}} (X(t)) - \beta (t)(X(t) - X_0)$  and the fact  $\Phi (t) = 0$  . Reorganizing, we have the desired equation (51).
Now we show the upper bounds of  $\beta(t)$  for each monotone and strongly monotone case. Observe from (51) and the definition of  $\beta(t)$ , we have for almost all  $t \in (0, \infty)$
$$
- \frac {\dot {\beta} (t) + \beta (t) ^ {2}}{2 \beta (t)} \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} = \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X), \dot {X} \right\rangle . \tag {52}
$$
(i) When  $\tilde{\mathbb{A}}$  is monotone.
From (52) and (1) we have for almost all  $t\in (0,\infty)$
$$
- \frac {\dot {\beta} (t) + \beta (t) ^ {2}}{2 \beta (t)} \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} = \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X), \dot {X} \right\rangle \geq 0.
$$
Since  $\beta (t) > 0$  we have
$$
\dot {\beta} (t) + \beta (t) ^ {2} \leq 0.
$$
almost everywhere. Now dividing both sides by  $\beta (t)^2$  we have
$$
1 \leq - \frac {\dot {\beta} (t)}{\beta (t) ^ {2}}
$$
holds almost everywhere. Since  $-\frac{\dot{\beta}(t)}{\beta(t)^2} = \frac{d}{dt}\left(\frac{1}{\beta(t)}\right)$ , integrating above inequality both side from  $\delta$  to  $t$  we have
$$
t - \delta \leq \frac {1}{\beta (t)} - \frac {1}{\beta (\delta)} \Longrightarrow \beta (t) \leq \frac {1}{t - \delta + \frac {1}{\beta (\delta)}}.
$$
By the way, from (50) we have
$$
\lim _ {t \to 0 +} t \beta (t) = - \lim _ {t \to 0 +} \frac {\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}}{2 \left\langle \tilde {\mathbf {A}} (X (t)) , \frac {X (t) - X _ {0}}{t} \right\rangle} = 1,
$$
thus  $\lim_{\delta \to 0 + }\beta (\delta) = \infty$  , so  $\lim_{\delta \to 0 + }\frac{1}{\beta(\delta)} = 0$  . Therefore taking limit  $\delta \rightarrow 0+$  we have
$$
\beta (t) \leq \lim _ {\delta \to 0 +} \frac {1}{t - \delta + \frac {1}{\beta (\delta)}} = \frac {1}{t}.
$$
(ii) When  $\tilde{\mathbb{A}}$  is  $\mu$ -strongly monotone.
From (52) and (2) for almost all  $t\in (0,\infty)$  we have
$$
- \frac {\dot {\beta} (t) + \beta (t) ^ {2}}{2 \beta (t)} \| \tilde {\mathbf {A}} (X (t)) \| ^ {2} = \left\langle \frac {d}{d t} \tilde {\mathbf {A}} (X), \dot {X} \right\rangle \geq \mu \| \dot {X} (t) \| ^ {2}. \tag {53}
$$
On the other hand, observe
$$
\begin{array}{l} \left\| \dot {X} (t) \right\| ^ {2} = \left\| \tilde {\mathbf {A}} (X (t)) + \beta (t) (X (t) - X _ {0}) \right\| ^ {2} \\ = \underbrace{\left\| \tilde{\mathbf{A}}(X(t))\right\|^{2} + 2\beta(t)\left\langle\tilde{\mathbf{A}}(X(t)),X(t) - X_{0}\right\rangle}_{= \Phi (t) = 0} + \beta (t)^{2}\left\| X(t) - X_{0}\right\|^{2} = \beta (t)^{2}\left\| X(t) - X_{0}\right\|^{2}. \\ \end{array}
$$
From Cauchy-Schwarz inequality we see
$$
\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} = - 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle \leq 2 \beta (t) \left\| \tilde {\mathbf {A}} (X (t)) \right\| \| X (t) - X _ {0} \|,
$$
therefore  $\| \tilde{\mathbf{A}} (X(t))\| \leq 2\beta (t)\| X(t) - X_0\|$ . Combining above observations, we have for almost all  $t > 0$
$$
\frac {4 \left\| \dot {X} (t) \right\| ^ {2}}{\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}} = \frac {4 \beta (t) ^ {2} \| X (t) - X _ {0} \| ^ {2}}{\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}} \geq 1. \tag {54}
$$
From (53) and (54), we get an inequality for  $\beta(t)$  and  $\dot{\beta}(t)$
$$
- \frac {\dot {\beta} (t) + \beta (t) ^ {2}}{2 \beta (t)} = \frac {\dot {\beta} (t)}{2 \beta (t)} + \frac {\beta (t)}{2} \geq \frac {\mu}{4}.
$$
Moving  $\frac{\beta(t)}{2}$  to the right-hand side and reorganizing, we have for almost all  $t > 0$
$$
1 \leq - \frac {\dot {\beta} (t)}{\beta (t) (\frac {\mu}{2} + \beta (t))} = - \frac {\dot {\beta} (t)}{\frac {\mu}{2}} \left(\frac {1}{\beta (t)} - \frac {1}{\frac {\mu}{2} + \beta (t)}\right) \quad \Longrightarrow \quad \frac {\mu}{2} \leq - \frac {\dot {\beta} (t)}{\beta (t)} + \frac {\dot {\beta} (t)}{\frac {\mu}{2} + \beta (t)}.
$$
Integrating above inequality both sides from  $\delta$  to  $t$  we have
$$
\frac {\mu}{2} (t - \delta) \leq \left[ - \log \beta (t) + \log \left(\frac {\mu}{2} + \beta (t)\right) \right] _ {\delta} ^ {t} = \log \frac {\frac {\mu}{2} + \beta (t)}{\beta (t)} - \log \frac {\frac {\mu}{2} + \beta (\delta)}{\beta (\delta)}.
$$
As observed in case (i) we know  $\lim_{\delta \to 0 + }\beta (\delta) = \infty$  , taking limit  $\delta \rightarrow 0+$  both sides we have
$$
\frac {\mu t}{2} \leq \log \frac {\frac {\mu}{2} + \beta (t)}{\beta (t)} = \log \left(1 + \frac {\mu / 2}{\beta (t)}\right) \Longrightarrow e ^ {\mu t / 2} \leq 1 + \frac {\mu / 2}{\beta (t)}.
$$
Organizing, we get the desired result.
$$
\beta (t) \leq \frac {\mu / 2}{e ^ {\mu t / 2} - 1}.
$$
I.2 Correspondence between the ODE (11) and the discrete method in Theorem 7.2
Observe, the case  $h = 1$  for below method corresponds to the method provided in Theorem 7.2.
$$
\beta_ {k} = \left\{ \begin{array}{l l} \frac {\| h \tilde {\mathbf {A}} x ^ {k} \| ^ {2}}{- \langle h \tilde {\mathbf {A}} x ^ {k} ,   x ^ {k} - x ^ {0} \rangle + \| h \tilde {\mathbf {A}} x ^ {k} \| ^ {2}} & \text {i f} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} \neq 0 \\ 0 & \text {i f} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} = 0 \end{array} \right.
$$
$$
y ^ {k} = \left(1 - \beta_ {k}\right) \left(2 x ^ {k} - y ^ {k - 1}\right) + \beta_ {k} x ^ {0}.
$$
We now show when  $\mathbb{A}$  is continuous and  $\| \tilde{\mathbb{A}} x^k\|^2 \neq 0$  for all  $k \geq 0$ , we obtain the ODE (11) when we take limit  $h \to 0+$ . Note when  $\mathbb{A}$  is continuous  $\mathbb{A}$  equals to  $\tilde{\mathbb{A}}$ .
From  $h\tilde{\mathbf{A}} x^{k + 1} + x^{k + 1} = y^k$ , substituting  $y^{k}$  and  $y^{k - 1}$  we get a single line expression
$$
h \tilde {\mathbf {A}} x ^ {k + 1} + x ^ {k + 1} = (1 - \beta_ {k}) \left(x ^ {k} - h \tilde {\mathbf {A}} x ^ {k}\right) + \beta_ {k} x ^ {0}.
$$
$$
\frac {x ^ {k + 1} - x ^ {k}}{h} = - \tilde {\mathbf {A}} x ^ {k + 1} - (1 - \beta_ {k}) \tilde {\mathbf {A}} x ^ {k} - \frac {\beta_ {k}}{h} \left(x ^ {k} - x ^ {0}\right).
$$
Identify  $x^0 = X_0$ ,  $2hk = t$ , and  $x^{k} = X(t)$ . Then we see
$$
\frac {\beta_ {k}}{h} = \frac {h ^ {2} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2}}{- h ^ {2} \langle \tilde {\mathbf {A}} x ^ {k} , x ^ {k} - x ^ {0} \rangle + h ^ {3} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2}} = \frac {\| \tilde {\mathbf {A}} (X (t)) \| ^ {2}}{- \langle \tilde {\mathbf {A}} (X (t)) , X (t) - X _ {0} \rangle + h \| \tilde {\mathbf {A}} (X (t)) \| ^ {2}}.
$$
Thus  $\beta_{k} = \mathcal{O}(h)$ , we have  $\lim_{h\to 0 + }\beta_{k} = 0$ . Now taking limit  $h\to 0+$  we have
$$
2 \dot {X} (t) = - 2 \tilde {\mathbb {A}} (X (t)) - \frac {\left\| \tilde {\mathbb {A}} (X (t)) \right\| ^ {2}}{\langle \tilde {\mathbb {A}} (X (t)) , X (t) - X _ {0} \rangle} (X (t) - X _ {0}).
$$
Dividing both sides by 2, we obtain (11).
I.2.1 Correspondence between convergence rates in Theorem 7.1 and Theorem 7.2
From identification above, we see  $\beta(t)$  corresponds to  $\frac{\beta_k}{2h}$ . Therefore we see with identification  $x^0 = X_0$ ,  $x^\star = X_\star$ ,  $2hk = t$ , and  $x^k = X(t)$  we have
$$
\left\| h \tilde {\mathbf {A}} (x ^ {k + 1}) \right\| ^ {2} \leq \beta_ {k} ^ {2} \left\| x ^ {0} - x ^ {\star} \right\| ^ {2} = 4 h ^ {2} \frac {\beta_ {k} ^ {2}}{(2 h) ^ {2}} \left\| x ^ {0} - x ^ {\star} \right\| ^ {2} \quad \xrightarrow {\text {d i v i d e b y} h ^ {2} , h \to 0 +} \quad \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} \leq 4 \beta (t) ^ {2} \| X _ {0} - X _ {\star} \| ^ {2}.
$$
And we can also check that the bound  $\beta_{k} \leq \frac{1}{k + 1}$  for the monotone case, is equivalent to  $\frac{\beta_{k}}{2h} \leq \frac{1}{2hk + \mathcal{O}(h)}$  and corresponds to  $\beta(t) \leq \frac{1}{t}$  as well.
Now suppose  $\mathbb{A}$  is  $\mu$ -strongly monotone and  $L$ -Lipschitz continuous. Then  $h\mathbb{A}$  is  $h\mu$ -strongly monotone and  $hL$ -Lipschitz continuous, we have the following inequality from Theorem 7.2
$$
\frac {\beta_ {k}}{2 h} \leq \frac {\frac {\mu}{2 \left(1 + (h L) ^ {2}\right)}}{\left(1 + \frac {h \mu}{1 + (h L) ^ {2}}\right) ^ {k} - 1 + \frac {h \mu}{1 + (h L) ^ {2}}}. \tag {55}
$$
Recalling the identification  $2hk = t$ , we see
$$
\left(1 + \frac {h \mu}{1 + (h L) ^ {2}}\right) ^ {k} = \left(1 + \frac {h \mu}{1 + (h L) ^ {2}}\right) ^ {\frac {t}{2 h}} = \left(\left(1 + \frac {h \mu}{1 + (h L) ^ {2}}\right) ^ {\frac {1 + (h L) ^ {2}}{h \mu}}\right) ^ {\frac {\mu t}{2 (1 + (h L) ^ {2})}} \quad \xrightarrow {h \to 0 +} \quad e ^ {\mu t / 2}.
$$
Therefore identifying  $\beta(t) = \frac{\beta_k}{2h}$  and taking limit  $h \to 0+$  to the inequality (55), we have
I.3 Proof of Theorem 7.2
$$
\beta_ {k} = \left\{ \begin{array}{l l} \frac {\| \tilde {\mathbf {A}} x ^ {k} \| ^ {2}}{- \langle \tilde {\mathbf {A}} x ^ {k} ,   x ^ {k} - x ^ {0} \rangle + \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2}} & \text {i f} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} \neq 0 \\ 0 & \text {i f} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} = 0 \end{array} \right.
$$
$$
y ^ {k} = (1 - \beta_ {k}) (2 x ^ {k} - y ^ {k - 1}) + \beta_ {k} x ^ {0}.
$$
First, we assume  $\tilde{\mathbb{A}} x^k\neq 0$  for all  $k\geq 0$ . Define
$$
\tilde {\Phi} ^ {k} = \left(1 - \beta_ {k}\right) \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} + \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \rangle
$$
for  $k = 1,2,\ldots$  , and
$$
\Phi^ {k} = \left\| \tilde {\mathbf {A}} x ^ {k} \right\| ^ {2} + \beta_ {k - 1} \langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \rangle
$$
for  $k = 2,3,\ldots$  Note by definition of  $\beta_{k}$  , we have  $\tilde{\Phi}^k = 0$  . Our goal is to prove
$$
\Phi^ {k + 1} \leq 0.
$$
Then with the same argument with (8) we can conclude  $\| \tilde{\mathbb{A}} x^{k + 1}\| ^2\leq \beta_k^2\| x^0 -x^\star \| ^2$ . To do so, we first show following lemma.
Lemma I.2. For  $k \geq 1$ , following is true.
$$
(1 - \beta_ {k}) \tilde {\Phi} ^ {k} - \Phi^ {k + 1} = (1 - \beta_ {k}) \langle \tilde {\mathbf {A}} x ^ {k + 1} - \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \rangle . \tag {56}
$$
Proof. Recall  $y^{k} = x^{k + 1} + \tilde{\mathbb{A}} x^{k + 1}$ . Substituting  $y^{k}$  and  $y^{k - 1}$ , the method is equivalent to
$$
x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k + 1} = (1 - \beta_ {k}) (x ^ {k} - \tilde {\mathbf {A}} x ^ {k}) + \beta_ {k} x ^ {0}.
$$
From above we can get two different expression of  $(1 - \beta_{k})(x^{k + 1} - x^{k})$
$$
\begin{array}{l} (1 - \beta_ {k}) (x ^ {k + 1} - x ^ {k}) = - (1 - \beta_ {k}) (\tilde {\mathbf {A}} x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k}) - \beta_ {k} (\tilde {\mathbf {A}} x ^ {k + 1} + (x ^ {k + 1} - x ^ {0})) \\ = - \left(1 - \beta_ {k}\right) \left[ \left(\tilde {\mathbf {A}} x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k}\right) - \beta_ {k} \left(\tilde {\mathbf {A}} x ^ {k} - \left(x ^ {k} - x ^ {0}\right)\right) \right]. \\ \end{array}
$$
With reorganizing, first equality can be obtained by subtracting both sides by  $\beta_{k}(x^{k + 1} - x^{0})$  and the second equality can be obtained by multiplying both sides by  $(1 - \beta_{k})$ .
From this we have
$$
\begin{array}{l} (1 - \beta_ {k}) \langle \tilde {\mathbf {A}} x ^ {k + 1} - \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \rangle \\ = \langle \tilde {\mathbf {A}} x ^ {k + 1}, (1 - \beta_ {k}) \left(x ^ {k + 1} - x ^ {k}\right) \rangle - \langle \tilde {\mathbf {A}} x ^ {k}, (1 - \beta_ {k}) \left(x ^ {k + 1} - x ^ {k}\right) \rangle \\ = \left\langle \tilde {\mathbf {A}} x ^ {k + 1}, - (1 - \beta_ {k}) \left(\tilde {\mathbf {A}} x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k}\right) - \beta_ {k} \left\{\tilde {\mathbf {A}} x ^ {k + 1} + \left(x ^ {k + 1} - x ^ {0}\right) \right\} \right\rangle \\ - \left(1 - \beta_ {k}\right) \left\langle \tilde {\mathbf {A}} x ^ {k}, - \left(\tilde {\mathbf {A}} x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k}\right) + \beta_ {k} \left\{\tilde {\mathbf {A}} x ^ {k} - \left(x ^ {k} - x ^ {0}\right) \right\} \right\rangle \\ = - (1 - \beta_ {k}) \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} - \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k + 1}, \tilde {\mathbf {A}} x ^ {k + 1} + \left(x ^ {k + 1} - x ^ {0}\right) \rangle \\ + \left(1 - \beta_ {k}\right) \left(\| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} - \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k}, \tilde {\mathbf {A}} x ^ {k} - \left(x ^ {k} - x ^ {0}\right) \rangle\right) \\ = \left(1 - \beta_ {k}\right) \left(\left(1 - \beta_ {k}\right) \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} + \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \rangle\right) - \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} - \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \rangle \\ = (1 - \beta_ {k}) \tilde {\Phi} ^ {k} - \Phi^ {k + 1}. \\ \end{array}
$$