SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \left\| \frac {\tilde {X} (t + h) - \tilde {X} (t)}{h} - \frac {\tilde {X} _ {\delta} (t + h) - \tilde {X} _ {\delta} (t)}{h} \right\| \\ \leq \frac {1}{| h |} \left(\left\| \tilde {X} (t + h) - \tilde {X} _ {\delta} (t + h) \right\| + \left\| \tilde {X} (t) - \tilde {X} _ {\delta} (t) \right\|\right) \leq 4 \frac {\delta}{| h |} \tilde {M} _ {d o t} (T). \\ \end{array}
$$
Also from (14) and since  $\tilde{A}$  is  $L$ -Lipschitz continuous,
$$
\begin{array}{l} \left\| \dot {\tilde {X}} _ {\delta} (t) + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right) \right\| = \left\| - \left(\tilde {A} (\tilde {X} _ {\delta} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} _ {\delta} (t) - X _ {0}\right)\right) + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right) \right\| \\ \leq L \left\| \tilde {X} _ {\delta} (t) - \tilde {X} (t) \right\| + \frac {\gamma}{t ^ {p}} \left\| \tilde {X} _ {\delta} (t) - \tilde {X} (t) \right\| \leq 2 \delta \left(L + \frac {\gamma}{t ^ {p}}\right) \tilde {M} _ {d o t} (T). \\ \end{array}
$$
Now from Lemma B.4 we have  $\dot{X}_{\delta}(t)$  is absolutely continuous, thus
$$
\begin{array}{l} \left\| \frac {\tilde {X} _ {\delta} (t + h) - \tilde {X} _ {\delta} (t)}{h} - \dot {\tilde {X}} _ {\delta} (t) \right\| = \left\| \frac {\int_ {t} ^ {t + h} \left(\dot {\tilde {X}} _ {\delta} (s) - \dot {\tilde {X}} _ {\delta} (t)\right) d s}{h} \right\| \\ = \frac {1}{| h |} \left\| \int_ {t} ^ {t + h} \int_ {t} ^ {s} \ddot {\tilde {X}} _ {\delta} (u) d u d s \right\| \leq \frac {1}{| h |} \left| \int_ {t} ^ {t + h} \int_ {t} ^ {s} \left\| \ddot {\tilde {X}} _ {\delta} (u) \right\| d u d s \right| \\ \end{array}
$$
Observe, for almost every  $u \in [t - |h|, t + |h|] \subset [0, T]$  by Lemma B.6 we have
$$
\begin{array}{l} \left\| \ddot {\tilde {X}} _ {\delta} (u) \right\| = \left\| - \frac {d}{d u} \tilde {A} (\tilde {X} _ {\delta} (u)) + \frac {p \gamma}{u ^ {p + 1}} \left(\tilde {X} _ {\delta} (u) - X _ {0}\right) - \frac {\gamma}{u ^ {p}} \dot {X} _ {\delta} (u) \right\| \\ = \left\| \frac {d}{d u} \tilde {A} (\tilde {X} _ {\delta} (u)) \right\| + \left\| \frac {p \gamma}{u ^ {p + 1}} \int_ {0} ^ {u} \dot {\tilde {X}} _ {\delta} (v) d v \right\| + \left\| \frac {\gamma}{u ^ {p}} \dot {X} _ {\delta} (u) \right\| \\ \leq L \tilde {M} _ {d o t} (T) + \frac {p \gamma}{(t - | h |) ^ {p + 1}} \int_ {0} ^ {T} \tilde {M} _ {d o t} (T) d v + \frac {\gamma}{(t - | h |) ^ {p}} \tilde {M} _ {d o t} (T) \\ = \left(L + \frac {p \gamma T}{(t - | h |) ^ {p + 1}} + \frac {\gamma}{(t - | h |) ^ {p}}\right) \tilde {M} _ {d o t} (T) =: M. \\ \end{array}
$$
Note  $M$  is independent of  $h$  or  $\delta$ . While obtaining the inequality, we used the fact that  $\tilde{A}(\tilde{X}_{\delta}(\cdot))$  is  $L\tilde{M}_{\text{dot}}(T)$ -Lipschitz continuous in  $[0,T]$ . Now
$$
\frac {1}{| h |} \left| \int_ {t} ^ {t + h} \int_ {t} ^ {s} \left\| \ddot {\bar {X}} _ {\delta} (u) \right\| d u d s \right| \leq \frac {1}{| h |} \left| \int_ {t} ^ {t + h} \int_ {t} ^ {s} M d u d s \right| = \frac {1}{| h |} M \left| \int_ {t} ^ {t + h} (s - t) d s \right| = \frac {| h |}{2} M.
$$
Now consider  $\delta = h^2$  with  $h$  small enough that satisfies  $|h| < 1$  and  $|h| + h^2 < t$ . Then the conditions  $|h| < t$ ,  $0 < \delta < \min \{1, t - |h|\}$  hold, above arguments are valid. Gathering above results, we have
$$
\begin{array}{l} \left\| \frac {\tilde {X} (t + h) - \tilde {X} (t)}{h} + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right) \right\| \\ \leq 2 \sqrt {2} | h | \tilde {M} _ {d o t} (T) + \sqrt {2} | h | ^ {2} \left(L + \frac {\gamma}{t ^ {p}}\right) \tilde {M} _ {d o t} (T) + \frac {| h |}{2} M = \mathcal {O} (| h |). \\ \end{array}
$$
which implies the desired result.
(2) The value and continuity of  $\dot{\hat{X}}(t)$  at  $t = 0$ .
Define  $C(t)$  as
$$
C (t) = \left\{ \begin{array}{l l} t ^ {\gamma} & p = 1 \\ e ^ {\frac {\gamma}{1 - p}} t ^ {1 - p} & p > 0, p \neq 1. \end{array} \right.
$$
Then we see for  $t > 0$
$$
\frac {d}{d t} \left(C (t) (\tilde {X} (t) - X _ {0})\right) = C (t) \left(\dot {\tilde {X}} (t) + \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0})\right) = - C (t) \tilde {A} (\tilde {X} (t)).
$$
$$
C (t) (\tilde {X} (t) - X _ {0}) - C (\epsilon) (\tilde {X} (\epsilon) - X _ {0}) = - \int_ {\epsilon} ^ {t} C (s) \tilde {A} (\tilde {X} (s)) d s.
$$
As  $C$  is a nondecreasing function and bounded below by 0,  $\lim_{\epsilon \to 0 + }C(\epsilon)$  exists, taking limit  $\epsilon \rightarrow 0+$  we have
$$
C (t) (\tilde {X} (t) - X _ {0}) = - \int_ {0} ^ {t} C (s) \tilde {A} (\tilde {X} (s)) d s.
$$
Dividing both sides by  $tC(t)$ , with change of variable  $v = s / t$ , we have
$$
\frac {\tilde {X} (t) - X _ {0}}{t} = - \int_ {0} ^ {t} \frac {C (s)}{C (t)} \tilde {A} (\tilde {X} (s)) \frac {d s}{t} = - \int_ {0} ^ {1} \frac {C (t v)}{C (t)} \tilde {A} (\tilde {X} (t v)) d v.
$$
$$
\frac {C (t v)}{C (t)} = \left\{ \begin{array}{l l} v ^ {\gamma} & p = 1 \\ e ^ {\frac {\gamma}{1 - p} t ^ {1 - p} \big (v ^ {1 - p} - 1 \big)} & p \neq 1, p > 0. \end{array} \right.
$$
Note  $\frac{\gamma}{1 - p} t^{1 - p}\left(v^{1 - p} - 1\right) \leq 0$  for  $v \in [0,1]$ , since  $\frac{\gamma}{1 - p}$  and  $\left(v^{1 - p} - 1\right)$  has opposite sign either  $0 < p < 1$  or  $p > 1$ . Therefore  $e^{\frac{\gamma}{1 - p} t^{1 - p}\left(v^{1 - p} - 1\right)} \leq 1$ . Also,  $\tilde{A}(\tilde{X}(tv))$  is bounded for  $v \in [0,1]$  since  $A(\tilde{X}(\cdot))$  is continuous by Lemma B.4.
So we can apply dominated convergence theorem and take limit  $t \to 0+$ . Since
$$
\lim  _ {t \to 0 +} \frac {C (t v)}{C (t)} = \left\{ \begin{array}{l l} 1 & 0 <   p <   1 \\ v ^ {\gamma} & p = 1 \\ 0 & p > 1 \end{array} \right.
$$
for  $v\neq 0$  , we have
$$
\dot {\tilde {X}} (0) = \lim  _ {t \rightarrow 0 +} \frac {\tilde {X} (t) - X _ {0}}{t} = - \int_ {0} ^ {1} \lim  _ {t \rightarrow 0 +} \left(\frac {C (t v)}{C (t)} \tilde {A} (\tilde {X} (t v))\right) d v = \left\{\begin{array}{l l}- \tilde {A} (X _ {0})&0 <   p <   1\\- \frac {1}{\gamma + 1} \tilde {A} (X _ {0})&p = 1\\0&p > 1.\end{array}\right. \tag {15}
$$
We now check  $\dot{\tilde{X}}(t)$  is continuous at  $t = 0$ .
For  $t > 0$ , we know
$$
\dot {\tilde {X}} (t) = - \tilde {A} (\tilde {X} (t)) - \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}).
$$
$$
\lim  _ {t \rightarrow 0 +} \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}) = \lim  _ {t \rightarrow 0 +} \left(\gamma t ^ {1 - p} \cdot \frac {\tilde {X} (t) - X _ {0}}{t}\right) = \left\{\begin{array}{l l}0&0 <   p <   1\\- \frac {\gamma}{\gamma + 1} \tilde {A} (X _ {0})&p = 1.\end{array}\right. \tag {16}
$$
Now from ODE  $\dot{\tilde{X}}(t) = -\tilde{A}(\tilde{X}(t)) - \frac{\gamma}{t^p} (\tilde{X}(t) - X_0)$ , by taking  $\lim_{t \to 0+}$  we have
$$
\lim_{t\to 0 + }\dot{\tilde{X}} (t) = -\tilde{A} (X_{0}) - \lim_{t\to 0 + }\Bigl(\frac{\gamma}{t^{p}} (\tilde{X} (t) - X_{0})\Bigr) = \left\{ \begin{array}{ll} - \tilde{A} (X_{0}) & 0 <   p <   1\\ -\frac{1}{\gamma + 1}\tilde{A} (X_{0}) & p = 1. \end{array} \right.
$$
Therefore  $\dot{\tilde{X}}(t)$  is continuous at  $t = 0$ .
For  $p > 1$ , we don't know the value of  $\lim_{t\to 0}\frac{\gamma}{tp} (\tilde{X} (t) - X_0)$ . Thus we first find the limit.
Let's go back to
Recall  $\dot{C}(t) = \frac{\gamma}{t^p} C(t)$ . By taking integration by parts for the right hand side, we have
$$
\int_ {0} ^ {t} C (s) \tilde {A} (\tilde {X} (s)) d s = \frac {t ^ {p}}{\gamma} C (t) \tilde {A} (\tilde {X} (t)) - \int_ {0} ^ {t} p \frac {s ^ {p - 1}}{\gamma} C (s) \tilde {A} (\tilde {X} (s)) d s - \int_ {0} ^ {t} \frac {s ^ {p}}{\gamma} C (s) \frac {d}{d s} \tilde {A} (\tilde {X} (s)) d s.
$$
Where we know  $\tilde{A} (\tilde{X} (s))$  is differentiable almost everywhere from Lemma B.4. Now, divide both sides by  $\frac{t^pC(t)}{\gamma}$ . Then for  $s = tv$  we have
$$
\begin{array}{l} \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}) = - \tilde {A} (\tilde {X} (t)) + \int_ {0} ^ {t} \frac {p s ^ {p - 1}}{t ^ {p}} \frac {C (s)}{C (t)} \tilde {A} (\tilde {X} (s)) d s + \int_ {0} ^ {t} \frac {s ^ {p}}{t ^ {p}} \frac {C (s)}{C (t)} \left(\frac {d}{d s} \tilde {A} (\tilde {X} (s))\right) d s \\ = - \tilde {A} (\tilde {X} (t)) + \int_ {0} ^ {1} p v ^ {p - 1} \frac {C (t v)}{C (t)} \tilde {A} (\tilde {X} (s)) d v + \int_ {0} ^ {1} v ^ {p} \frac {C (t v)}{C (t)} \left(\frac {d}{d s} \tilde {A} (\tilde {X} (s))\right) t d v. \\ \end{array}
$$
From Proposition B.7 and since  $\tilde{X}$  satisfies (12) for  $t > 0$ , for  $s \in (0, t]$  we have
$$
\dot {\tilde {X}} (s) = \tilde {A} (\tilde {X} (s)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (s) - X _ {0}\right) = \lim  _ {m \to \infty} \left(\tilde {A} (\tilde {X} _ {\delta_ {m}} (s)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} _ {\delta_ {m}} (s) - X _ {0}\right)\right) = \lim  _ {m \to \infty} \dot {\tilde {X}} _ {\delta_ {m}} (s).
$$
From Lemma B.6 we know  $\left\| \dot{\tilde{X}}_{\delta_m}(s)\right\| \leq \tilde{M}_{dot}(t)$  for  $s\in [0,t]$ , taking limit  $m\to \infty$  we have  $\left\| \dot{\tilde{X}} (s)\right\| \leq \tilde{M}_{dot}(t)$  for  $s\in [0,t]$ . Thus  $\tilde{X} (s)$  becomes  $\tilde{M}_{dot}(t)$ -Lipschitz continuous in  $s\in [0,t]$ . And so  $(\tilde{A}\circ \tilde{X})(s)$  becomes  $L\tilde{M}_{dot}(t)$ -Lipschitz continuous in  $s\in [0,t]$ , we have  $\left\| \frac{d}{ds}\tilde{A} (\tilde{X} (s))\right\| \leq L\tilde{M}_{dot}(t)$  for almost all  $s\in [0,t]$ . Moreover  $C(tv) / C(t)$  is bounded for  $v\in [0,1]$  since  $C$  is a nonnegative nondecreasing function.
Therefore we can again apply dominated convergence theorem. Reminding  $\lim_{t\to 0 + }\frac{C(tv)}{C(t)} = 0$  for  $p > 1$ , taking limit  $t\to 0+$  we have
$$
\lim  _ {t \rightarrow 0 +} \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}) = - \tilde {A} (X _ {0}) + 0 = - \tilde {A} (X _ {0}). \tag {17}
$$
Finally we have
$$
\lim  _ {t \to 0 +} \dot {\tilde {X}} (t) = - \tilde {A} (X _ {0}) - \lim  _ {t \to 0 +} \left(\frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0})\right) = - \tilde {A} (X _ {0}) - (- \tilde {A} (X _ {0})) = 0 = \dot {\tilde {X}} (0).
$$
Before we move on to original inclusion, we prove important corollaries that bound  $\left\| \dot{\tilde{X}}(t) \right\|$  and  $\left\| \tilde{A}(\tilde{X}(t)) \right\|$  uniformly on  $[0, T]$ . Note the main difference between following corollary and Lemma B.6 is that the dependency on Lipschitz constant  $L$  is dropped for the case  $p > 1$ , which will be crucial in the next section.
Corollary B.8. Denote  $\tilde{X}$  as the solution of (12). Then for  $T > 0$ , following inequality is true for  $t \in [0, T]$ .
$$
\left\| \dot {\tilde {X}} (t) \right\| \leq \left\{ \begin{array}{l l} \left\| \tilde {A} (X _ {0}) \right\| & 0 <   p <   1 \\ \frac {1}{\sqrt {\gamma + 1}} \left\| \tilde {A} (X _ {0}) \right\| & p = 1 \\ \sqrt {\frac {p}{\gamma} T ^ {p - 1}} \left\| \tilde {A} (X _ {0}) \right\| & p > 1. \end{array} \right.
$$
Proof. (i)  $0 < p \leq 1$
As  $\tilde{X}$  is the solution for (12), from Lemma B.5 we know
is a nonincreasing function. From (15), (16) and continuity of  $\dot{X} (t)$  at  $t = 0$ , we have
$$
\lim  _ {\epsilon \rightarrow 0 +} \tilde {U} _ {1} (\epsilon) = \left\{\begin{array}{l l}\left\| \tilde {A} \left(X _ {0}\right) \right\| ^ {2}&0 <   p <   1\\\frac {1}{\gamma + 1} \left\| \tilde {A} \left(X _ {0}\right) \right\| ^ {2}&p = 1.\end{array}\right. \tag {18}
$$
Therefore  $\left\| \dot{\tilde{X}}(t) \right\|^2 \leq \tilde{U}_1(t) \leq \lim_{\epsilon \to 0+} \tilde{U}_1(\epsilon)$  for  $t > 0$ , we get the desired result.
is a nonincreasing function. However, as we don't know the value of  $\lim_{t\to 0 + }\frac{1}{t^{p - 1}}\left\| \dot{\bar{X}} (t)\right\| ^2$  , we first calculate it. To do so, we consider
$$
\tilde {U} _ {1} (t) = \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma p}{t ^ {p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}.
$$
In the proof of Lemma B.5, we have observed its derivative becomes
$$
\begin{array}{l} \dot {\tilde {U}} _ {1} (t) = - 2 \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle - \frac {2 \gamma}{t ^ {p}} \left\| \dot {\tilde {X}} (t) - \frac {p}{t} (\tilde {X} (t) - X _ {0}) \right\| ^ {2} - \frac {\gamma p (1 - p)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ \leq \frac {\gamma p (p - 1)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}. \\ \end{array}
$$
For  $\epsilon > 0$ , integrating from  $\epsilon$  to  $t$  we have
$$
\tilde {U} _ {1} (t) \leq \tilde {U} _ {1} (\epsilon) + \int_ {\epsilon} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s.
$$
We consider taking limit  $\epsilon \to 0+$ . Observe from (17) and (15) we know  $\lim_{t\to 0+}\frac{\tilde{X} - X_0}{t^p} = -\frac{\tilde{A}(X_0)}{\gamma}$  and  $\left\| \dot{\tilde{X}} (0)\right\| = 0$  and therefore we have
$$
\lim _ {\epsilon \rightarrow 0 +} \tilde {U} _ {1} (\epsilon) = 0 + \lim _ {\epsilon \rightarrow 0 +} \frac {\gamma p}{\epsilon^ {2 p}} \left\| \tilde {X} (\epsilon) - X _ {0} \right\| ^ {2} \cdot \epsilon^ {p - 1} = \frac {p}{\gamma} \left\| \tilde {A} (X _ {0}) \right\| ^ {2} \lim _ {\epsilon \rightarrow 0 +} \epsilon^ {p - 1} = 0.
$$
Moreover as  $\lim_{t\to 0 + }\frac{\|\tilde{X}(t) - X_0\|^2}{t^{2p}} = \frac{\|\tilde{A}(X_0)\|^2}{\gamma^2}$ , there is  $\delta > 0$  such that  $0 < s < \delta$  implies  $\frac{\|\tilde{X}(t) - X_0\|^2}{s^{2p}} \leq \frac{2\|\tilde{A}(X_0)\|^2}{\gamma^2}$ . Recalling  $p > 1$ , for  $0 < \epsilon < \delta$  we have
$$
\begin{array}{l} \int_ {0} ^ {\epsilon} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s \leq \int_ {0} ^ {\epsilon} \frac {\gamma p (p - 1)}{s ^ {2 - p}} \frac {2 \left\| \tilde {A} (X _ {0}) \right\| ^ {2}}{\gamma^ {2}} d s \\ = \frac {2 p (p - 1) \left\| \tilde {A} \left(X _ {0}\right) \right\| ^ {2}}{\gamma} \left[ \frac {1}{p - 1} s ^ {p - 1} \right] _ {0} ^ {\epsilon} = \frac {2 p \left\| \tilde {A} \left(X _ {0}\right) \right\| ^ {2}}{\gamma} \epsilon^ {p - 1}, \\ \end{array}
$$
$$
\int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s \leq \frac {2 p \left\| \tilde {A} (X _ {0}) \right\| ^ {2}}{\gamma} \epsilon^ {p - 1} + \int_ {\epsilon} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s <   \infty .
$$
Thus the integral is well-defined when  $\epsilon \to 0+$ . Taking limit  $\epsilon \to 0+$  we have
$$
\tilde {U} _ {1} (t) \leq \int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s.
$$
Moving  $\frac{\gamma p}{t^{p + 1}}\left\| \tilde{X} (t) - X_0\right\|^2$  to the right hand side, we get a inequality for  $\left\| \dot{\tilde{X}} (t)\right\|^2$
$$
\left\| \dot {\tilde {X}} (t) \right\| ^ {2} = \tilde {U} _ {1} (t) - \frac {\gamma p}{t ^ {p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \leq \int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s - \frac {\gamma p}{t ^ {p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}.
$$
Observe, by L'Hôpital's rule we have
$$
\lim _ {t \to 0 +} \frac {\int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} (s) - X _ {0} \right\| ^ {2} d s}{t ^ {p - 1}} = \lim _ {t \to 0 +} \frac {\frac {\gamma p (p - 1)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}}{(p - 1) t ^ {p - 2}} = \lim _ {t \to 0 +} \gamma p \frac {\left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}}{t ^ {2 p}} = \frac {p}{\gamma} \left\| \tilde {A} (X _ {0}) \right\| ^ {2}.
$$
Now dividing  $t^{p - 1}$  to previous inequality and taking limit, we conclude
$$
\begin{array}{l} \lim  _ {t \rightarrow 0 +} \frac {1}{t ^ {p - 1}} \left\| \dot {\tilde {X}} \right\| ^ {2} \leq \lim  _ {t \rightarrow 0 +} \frac {1}{t ^ {p - 1}} \int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \left\| \tilde {X} - X _ {0} \right\| ^ {2} d s - \lim  _ {t \rightarrow 0 +} \frac {\gamma p}{t ^ {2 p}} \left\| \tilde {X} - X _ {0} \right\| ^ {2} \\ = \frac {p}{\gamma} \left\| \tilde {A} (X _ {0}) \right\| ^ {2} - \frac {p}{\gamma} \left\| \tilde {A} (X _ {0}) \right\| ^ {2} \\ = 0. \\ \end{array}
$$
Thus we have  $\lim_{t\to 0 + }\frac{1}{t^{p - 1}}\left\| \dot{\bar{X}}\right\| ^2 = 0$  . Therefore,
$$
\lim  _ {\epsilon \rightarrow 0 +} \tilde {U} _ {2} (\epsilon) = \lim  _ {\epsilon \rightarrow 0 +} \left(\frac {1}{\epsilon^ {p - 1}} \left\| \dot {\tilde {X}} (\epsilon) \right\| ^ {2} + \frac {\gamma p}{\epsilon^ {2 p}} \left\| \tilde {X} (\epsilon) - X _ {0} \right\| ^ {2}\right) = \frac {p}{\gamma} \left\| \tilde {A} (X _ {0}) \right\| ^ {2}.
$$
From  $\frac{\left\|\dot{\hat{X}}\right\|^2}{t^{p-1}} \leq \lim_{\epsilon \to 0+} \tilde{U}_2(\epsilon) = \frac{p}{\gamma} \left\|\tilde{A}(X_0)\right\|^2$ , we conclude the desired result.
Corollary B.9. Denote  $\tilde{X}$  as the solution of (12). Then for  $T > 0$ , following inequality is true for  $t \in [0, T]$ .
$$
\left\| \tilde {A} (\tilde {X} (t)) \right\| \leq \left\{ \begin{array}{l l} \sqrt {(\gamma + 1) \left(1 + \frac {T ^ {1 - p}}{p}\right)} \left\| \tilde {A} (X _ {0}) \right\| & 0 <   p <   1 \\ \left\| \tilde {A} (X _ {0}) \right\| & p = 1 \\ \sqrt {\frac {p (\gamma + 1)}{\gamma} \left(T ^ {p - 1} + \frac {1}{p}\right)} \left\| \tilde {A} (X _ {0}) \right\| & p > 1. \end{array} \right.
$$
Proof. First observe,
$$
\begin{array}{l} \left\| \tilde {A} (\tilde {X} (t)) \right\| ^ {2} = \left\| \dot {\tilde {X}} (t) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right) \right\| ^ {2} \\ = \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + 2 \gamma \left\langle \dot {\tilde {X}} (t), \frac {\tilde {X} (t) - X _ {0}}{t ^ {p}} \right\rangle + \frac {\gamma^ {2}}{t ^ {2 p}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ \leq \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \gamma \left(\left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}}{t ^ {2 p}}\right) + \frac {\gamma^ {2}}{t ^ {2 p}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = (\gamma + 1) \left(\left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma}{t ^ {2 p}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}\right) \\ \end{array}
$$
The inequality comes from Young's inequality. Now we consider each case.
(i)  $p = 1$
For this case, the terms on the right hand side exactly become  $\tilde{U}_1(t)$ . Therefore from (18)
$$
\left\| \tilde {A} (\tilde {X} (t)) \right\| ^ {2} \leq (\gamma + 1) \tilde {U} _ {1} (t) \leq (\gamma + 1) \lim  _ {\epsilon \rightarrow 0 +} \tilde {U} _ {1} (\epsilon) = \left\| \tilde {A} (X _ {0}) \right\| ^ {2}.
$$
(ii)  $0 < p < 1$
Recall from the proof of Corollary B.8 we know  $\tilde{U}_1(t) \leq \lim_{\epsilon \to 0+} \tilde{U}_1(\epsilon) = \left\| \tilde{A}(X_0) \right\|^2$ . Therefore we have  $\frac{\gamma p}{t^{p+1}} \left\| \tilde{X}(t) - X_0 \right\|^2 \leq \tilde{U}_1(t) \leq \left\| \tilde{A}(X_0) \right\|^2$ , applying Corollary B.8 we get
$$
\begin{array}{l} \left\| \tilde {A} (\tilde {X} (t)) \right\| ^ {2} \leq (\gamma + 1) \left(\left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma p}{t ^ {p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \cdot \frac {t ^ {1 - p}}{p}\right) \\ \leq (\gamma + 1) \left\| \tilde {A} (X _ {0}) \right\| ^ {2} \left(1 + \frac {T ^ {1 - p}}{p}\right). \\ \end{array}
$$
(iii)  $p > 1$
Recall from the proof of Corollary B.8 we know  $\tilde{U}_2(t) \leq \lim_{\epsilon \to 0+} \tilde{U}_2(\epsilon) = \frac{p}{\gamma} \left\| \tilde{A}(X_0) \right\|^2$ . Therefore we have
$\frac{\gamma p}{t^{2p}} \left\| \tilde{X}(t) - X_0 \right\|^2 \leq \tilde{U}_2(t) \leq \frac{p}{\gamma} \left\| \tilde{A}(X_0) \right\|^2$ , applying Corollary B.8 we get
$$
\begin{array}{l} \left\| \tilde {A} (\tilde {X} (t)) \right\| ^ {2} \leq (\gamma + 1) \left(\left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma p}{t ^ {2 p}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \cdot \frac {1}{p}\right) \\ \leq \frac {p (\gamma + 1)}{\gamma} \left(T ^ {p - 1} + \frac {1}{p}\right) \left\| \tilde {A} (X _ {0}) \right\| ^ {2}. \\ \end{array}
$$
B.2.2 Existence proof for general  $\mathbb{A}$
Now we move on to the original inclusion. As noticed before, we will approximate  $\mathbb{A}$  with a Lipschitz function  $\mathbb{A}_{\lambda}$  called Yosida approximation. We define and state some facts about  $\mathbb{A}_{\lambda}$  as a lemma, and use it without proof. For the ones who are interested in proofs, see [11, Chpater 3.1, Theorem 2].
Lemma B.10. Define  $\mathbb{A}_{\lambda}:\mathbb{R}^n\to \mathbb{R}^n$  as
$$
\mathbb {A} _ {\lambda} = \frac {1}{\lambda} \left(\mathbb {I} - \mathbb {J} _ {\lambda \mathbb {A}}\right) = \frac {1}{\lambda} \left(\mathbb {I} - (\mathbb {I} + \lambda \mathbb {A}) ^ {- 1}\right)
$$
This is so called Yosida approximation of  $\mathbb{A}$ . Followings are true.
(i)  $\mathbb{A}_{\lambda}$  is  $\frac{1}{\lambda}$ -Lipschitz continuous and maximal monotone.
(ii)  $\lim_{\lambda \to 0 + }\mathbb{A}_{\lambda}x = m(\mathbb{A}(x)).$
Here  $m(\mathbb{A}(x))$  is defined as  $m(\mathbb{A}(x)) = \Pi_{\mathbb{A}(x)}(0)$ , the element of  $\mathbb{A}(x)$  with minimal norm.
(iii)  $\| \mathbf{A}_{\lambda}(x)\| \leq \| m(\mathbf{A}(x))\|$
(iv)  $\forall x\in \mathbb{R}^n$ $\mathbb{A}_{\lambda}(x)\in \mathbb{A}(\mathbb{J}_{\lambda \mathbf{A}}x)$
Now we can state the proposition that proves existence of Theorem 2.2
Proposition B.11. For Yosida approximation  $\mathbb{A}_{\lambda}$ , consider the ODE
$$
\dot {X} _ {\lambda} (t) = - \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) - \frac {\gamma}{t ^ {p}} \left(X _ {\lambda} (t) - X _ {0}\right) \tag {19}
$$
with initial value condition  $X_{\lambda}(0) = X_0 \in \operatorname{dom}(\mathbb{A})$ . The solution uniquely exists by Theorem B.2, denote the solution as  $X_{\lambda}$ . Now for  $T > 0$  and a positive sequence  $\{\lambda_n\}_{n \in \mathbb{N}}$  such that  $\lim_{n \to \infty} \lambda_n = 0$ , define a sequence of solutions as  $\mathcal{F}_T = \{X_{\lambda_n} : [0, T] \to \mathbb{R}^n \mid m \in \mathbb{N}\}$ . Then there is a subsequence  $\{\lambda_{n_k}\}_{k \in \mathbb{N}}$  such that  $X_{\lambda_{n_k}}$  converges to the solution of (6) uniformly on  $[0, T]$ .
From Corollary B.8, Corollary B.9 and Lemma B.10 (iii), following lemma is immediate.
Lemma B.12. Let  $X_{\lambda}$  be the solution of (19). Then following is true for  $t \in [0, T]$  for all  $T > 0$ .
$$
\left\| \dot {X} _ {\lambda} (t) \right\| \leq M _ {d o t} (T) = \left\{ \begin{array}{l l} \| m (\mathbb {A} (X _ {0})) \| & 0 <   p <   1 \\ \frac {1}{\sqrt {\gamma + 1}}   \| m (\mathbb {A} (X _ {0})) \| & p = 1 \\ \sqrt {\frac {p}{\gamma} T ^ {p - 1}}   \| m (\mathbb {A} (X _ {0})) \| & p > 1 \end{array} \right.
$$
$$
\| \mathbb {A} _ {\lambda} (X _ {\lambda} (t)) \| \leq M _ {\mathbb {A}} (T) = \left\{ \begin{array}{l l} \sqrt {(\gamma + 1) \left(1 + \frac {T ^ {1 - p}}{p}\right)} \| m (\mathbb {A} (X _ {0})) \| & 0 <   p <   1 \\ \| m (\mathbb {A} (X _ {0})) \| & p = 1 \\ \sqrt {\frac {p (\gamma + 1)}{\gamma} \left(T ^ {p - 1} + \frac {1}{p}\right)} \| m (\mathbb {A} (X _ {0})) \| & p > 1. \end{array} \right.
$$
Proof. Replace  $\tilde{A}$  with  $\mathbb{A}_{\lambda}$  and  $\tilde{X}$  with  $X_{\lambda}$  in Corollary B.8 and Corollary B.9. Applying  $\| \mathbb{A}_\lambda (X_0)\| \leq \| m(\mathbb{A}(X_0))\|$  from Lemma B.10 (iii), we're done.
While we're concluding the existence of converging sequence, we will exploit following lemma from [11, Chapter 0.3, Theorem 4]. For convenience, we restate the lemma here.
Lemma B.13. Let us consider a sequence of absolutely continuous functions  $x_{k}(\cdot)$  from an interval  $I$  of  $\mathbb{R}$  to a Banach space  $X$  satisfying
(i)  $\forall t\in I,\{x_k(t)\}_{k\in \mathbb{N}}$  is a relatively compact subset of  $X$
(ii) there exists a positive function  $c(\cdot)\in L^1 (I,[0,\infty))$  such that, for almost all  $t\in I$ $\| \dot{x}_k(t)\| \leq c(t)$