SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \frac {d}{d t} \| X (t) - Y (t) \| ^ {2} = 2 \left\langle \dot {X} (t) - \dot {Y} (t), X (t) - Y (t) \right\rangle \\ = - 2 \left\langle \tilde {\mathbf {A}} (X (t)) - \tilde {\mathbf {A}} (Y (t)) + \beta (t) (X (t) - Y (t)) - \beta (t) \left(X _ {0} - Y _ {0}\right), X (t) - Y (t) \right\rangle \\ \leq - 2 \beta (t) \| X (t) - Y (t) \| ^ {2} + 2 \beta (t) \langle X _ {0} - Y _ {0}, X (t) - Y (t) \rangle \\ \leq - 2 \beta (t) \| X (t) - Y (t) \| ^ {2} + \beta (t) \left(\| X _ {0} - Y _ {0} \| ^ {2} + \| X (t) - Y (t) \| ^ {2}\right) \\ = - \beta (t) \left\| X (t) - Y (t) \right\| ^ {2} + \beta (t) \left\| X _ {0} - Y _ {0} \right\| ^ {2}. \\ \end{array}
$$
Now 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  $-\beta (t)\| X(t) - Y(t)\| ^2$  to the left hand side and multiplying both sides by  $C(t)$  , we have
$$
\begin{array}{l} \frac {d}{d t} \left(C (t) \| X (t) - Y (t) \| ^ {2}\right) = C (t) \frac {d}{d t} \| X (t) - Y (t) \| ^ {2} + C (t) \beta (t) \| X (t) - Y (t) \| ^ {2} \\ \leq C (t) \beta (t) \| X _ {0} - Y _ {0} \| ^ {2} = \frac {d}{d t} C (t) \| X _ {0} - Y _ {0} \| ^ {2} \\ \end{array}
$$
Integrating both sides from  $\epsilon > 0$  to  $t$  we have
$$
C (t) \left\| X (t) - Y (t) \right\| ^ {2} - C (\epsilon) \left\| X (\epsilon) - Y (\epsilon) \right\| ^ {2} \leq C (t) \left\| X _ {0} - Y _ {0} \right\| ^ {2} - C (\epsilon) \left\| X _ {0} - Y _ {0} \right\| ^ {2}.
$$
As  $C$  is nonnegative and nondecreasing,  $\lim_{\epsilon \to 0+} C(\epsilon)$  exists. Taking limit  $\epsilon \to 0+$  both sides we have
$$
C (t) \| X (t) - Y (t) \| ^ {2} \leq C (t) \| X _ {0} - Y _ {0} \| ^ {2}.
$$
Finally, dividing both sides by  $C(t)$  we conclude
$$
\left\| X (t) - Y (t) \right\| ^ {2} \leq \left\| X _ {0} - Y _ {0} \right\| ^ {2}.
$$
Proof for uniqueness can be done to generalized case (3).
Theorem B.1 (Uniqueness of solutions). If the solution for (3) exists, it is unique.
Proof. Suppose  $X_{1}, X_{2}$  are solutions of (3) with same initial value  $X_{0}$ . By Lemma 2.3, we have
$$
\left\| X _ {1} (t) - X _ {2} (t) \right\| \leq \left\| X _ {0} - X _ {0} \right\| = 0
$$
Therefore  $X_{1}(t) = X_{2}(t)$  for all  $t\in [0,\infty)$ , we get the desired result.
As (6) is special case of (3), uniqueness proof for Theorem 2.2 follows from Theorem B.1.
B.2 Proof of existence
The proof of existence needs tedious work due to the singularity of  $\frac{\gamma}{t^p}$  at  $t = 0$ , we provide our proof through subsections. Before we start, we provide a short outline of the proof. The proof is basically based on the proof provided in [65] and [11].
We first prove the case  $\mathbb{A}$  is Lipschitz. The differential inclusion becomes ODE when  $\mathbb{A}$  is Lipschitz, we can adopt similar argument done in [65]. We consider series of ODEs Lipschitz with respect to  $X$ , approximating  $\dot{X}(t) = -\mathbb{A}(X(t)) - \frac{\gamma}{t^p} (X_0 - X(t))$ . The approximated ODEs have solutions by classical theory of ODE, we obtain the true solution by considering proper subsequence of solutions.
As the solution for Lipschitz  $\mathbb{A}$  is obtained, we can adopt similar arguments done in [11]. We first consider solution  $X_{\lambda}$  with Yosida approximation  $\mathbb{A}_{\lambda} = \frac{1}{\lambda} (\mathbb{I} - (\mathbb{I} + \lambda \mathbb{A})^{-1})$ , which is an approximation of  $\mathbb{A}$  that is Lipschitz continuous. Then we can obtain a subsequence  $X_{\lambda_n}$  converging to original differential inclusion.
B.2.1 Existence proof for Lipschitz A
Since we will approximate  $\mathbb{A}$  with Lipschitz continuous functions, we first consider the ODE with Lipschitz continuous  $\mathbb{A}$ .
Theorem B.2. (Existence of solution for Lipschitz  $\mathbb{A}$ )
Suppose  $\tilde{A}:\mathbb{R}^n\to \mathbb{R}^n$  is  $L$ -Lipschitz continuous function. Consider the differential equation, with initial value condition  $\tilde{X}(0) = X_0\in \operatorname{dom}(\tilde{A})$ ,
$$
\dot {\tilde {X}} (t) = - \tilde {A} (\tilde {X} (t)) - \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}). \tag {12}
$$
where  $\gamma, p > 0$ . Then there exists a unique solution  $\tilde{X} \in \mathcal{C}^1([0,\infty),\mathbb{R}^n)$  that satisfies (12) for all  $t \in (0,\infty)$ . Moreover, for  $\dot{\tilde{X}}(0)$  defined by  $\dot{\tilde{X}}(0) = \lim_{t \to 0+} \frac{\tilde{X}(t) - X_0}{t}$ , following is true
$$
\dot {\tilde {X}} (0) = \left\{ \begin{array}{l l} - \tilde {A} (X _ {0}) & i f 0 <   p <   1 \\ - \frac {1}{\gamma + 1} \tilde {A} (X _ {0}) & i f p = 1 \\ 0 & i f p > 1. \end{array} \right.
$$
The proof need some preparation. We will think of approximated solutions  $\tilde{X}_{\delta}$ , obtain a sequence that converges to solution  $\tilde{X}$ . Thus we first define  $\tilde{X}_{\delta}$ . From now, we will denote  $\tilde{A}$  as a  $L$ -Lipschitz monotone function.
Definition B.3. Let  $0 < \delta < 1$ . Consider
$$
\dot {\tilde {X}} _ {\delta} (t) = \left\{ \begin{array}{l l} - \tilde {A} (\tilde {X} _ {\delta} (t)) - \frac {\gamma}{\delta^ {p}} (\tilde {X} _ {\delta} (t) - X _ {0}) & 0 \leq t <   \delta \\ - \tilde {A} (\tilde {X} _ {\delta} (t)) - \frac {\gamma}{t ^ {p}} (\tilde {X} _ {\delta} (t) - X _ {0}) & t > \delta \end{array} \right. \tag {13}
$$
Since right hand side above is  $\left(L + \frac{\gamma}{\delta^p}\right)$ -Lipschitz with respect to  $\tilde{X}_{\delta}$ , the solution uniquely exists by classical ODE theory. Define the solution as  $\tilde{X}_{\delta}$ . Then for positive sequence  $\{\delta_m\}_{m\in \mathbb{N}}$  such that  $\delta_{m} < 1$  and  $\lim_{m\to \infty}\delta_m = 0$ , consider sequence  $\left\{\tilde{X}_{\delta_m}\right\}_{m\in \mathbb{N}}$ .
Before we start, we prove a useful lemma we will widely use for the cases that operator is Lipschitz continuous.
Lemma B.4. Let  $\tilde{A}\colon \mathbb{R}^n\to \mathbb{R}^n$  a Lipschitz continuous function. Suppose  $\beta \colon D\to [0,\infty)$  be a continuous function with  $D\subset [0,\infty)$ . Consider differential equation
$$
\dot {\tilde {X}} = - \tilde {A} (\tilde {X}) - \beta (t) (\tilde {X} - X _ {0}),
$$
with  $X_0 \in \mathbb{R}^n$ . Let  $\tilde{X} \colon D \to \mathbb{R}^n$  be a differentiable curve that satisfies above equation for  $t \in D$ . Then for all  $0 \leq a < b$  such that  $[a, b] \subset D$ ,  $\tilde{X}$  and the composition  $\tilde{A} \circ \tilde{X} : [a, b] \to \mathbb{R}^n$  are Lipschitz continuous.
Moreover if  $\dot{\beta}(t)$  is well-defined and bounded for almost all  $t \in [a, b]$ , then  $\dot{\bar{X}}$  is Lipschitz continuous in  $[a, b]$ . Thus if  $\beta$  is twice differentiable,  $\dot{\bar{X}}$  is Lipschitz continuous.
As Lipschitz continuous functions,  $\tilde{X},\tilde{A}\circ \tilde{X}$  and  $\dot{\tilde{X}}$  are absolutely continuous functions.
Proof. We first prove  $\tilde{X}$  is Lipschitz continuous. As  $\tilde{A},\tilde{X},\beta$  are continuous in  $[a,b]$  we see  $\tilde{A} (\tilde{X} (t)) + \beta (t)(\tilde{X} (t) - X_0)$  is continuous in  $[a,b]$ , thus
$$
M _ {1} = \max _ {t \in [ a, b ]} \left\| \dot {X} (t) \right\| = \max _ {t \in [ a, b ]} \left\| \tilde {A} (\tilde {X} (t)) + \beta (t) (\tilde {X} (t) - X _ {0}) \right\|
$$
exists. Since its derivative is bounded by  $M_1 < \infty$ , we have  $\tilde{X}$  is  $M_1$ -Lipschitz continuous. As composition of two Lipschitz continuous functions,  $\tilde{A} \circ \tilde{X}$  is Lipschitz continuous.
First observe if  $\beta$  is twice differentiable,  $\dot{\beta}$  is bounded in  $[a,b]$  as it is continuous. Now if  $\dot{\beta} (t)$  is bounded for  $t\in [a,b]$ , i.e.  $\left|\dot{\beta} (t)\right|\leq M_2$  for some  $M_2 > 0$ , for  $M_3 = \max_{t\in [a,b]}|\beta (t)|$  we have
$$
\left\| \frac {d}{d t} (\beta (t) (\tilde {X} (t) - X _ {0})) \right\| = \left\| \dot {\beta} (t) (\tilde {X} (t) - X _ {0}) \right\| + \left\| \beta (t) \dot {\tilde {X}} (t) \right\| \leq M _ {2} M _ {1} | b - a | + M _ {3} M _ {1}
$$
Therefore  $\beta (t)(\tilde{X} (t) - X_0)$  is  $(M_2M_1|b - a| + M_3M_1)$ -Lipschitz continuous. Thus as a sum of Lipschitz continuous functions,  $\dot{\tilde{X}}$  is Lipschitz continuous.
We will show for every  $T > 0$ , the set of derivatives  $\left\{\dot{\tilde{X}}_{\delta} \mid \delta \in (0,1)\right\}$  is uniformly bounded on  $[0,T]$  and  $\left\{\tilde{X}_{\delta_m}\right\}_{m \in \mathbb{N}}$  converges uniformly on  $[0,T]$ . We first prove the boundedness of derivatives. To do so, we first prove a useful lemma.
Lemma B.5. For  $0 < a < b$ , suppose  $\tilde{X} : [a, b] \to \mathbb{R}^n$  satisfies ODE (12) for  $t \in [a, b]$ . Define  $\tilde{U} : [a, b] \to \mathbb{R}$  as
$$
\tilde {U} _ {1} (t) = \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma p}{t ^ {p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}
$$
$$
\tilde {U} _ {2} (t) = \frac {1}{t ^ {p - 1}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {\gamma p}{t ^ {2 p}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2}
$$
Then  $\tilde{U}_1$  is nonincreasing if for  $0 < p \leq 1$ , and  $\tilde{U}_2$  is nonincreasing for  $p > 1$ .
Proof. From Lemma B.4 we can check  $\tilde{U}_1$  and  $\tilde{U}_2$  are absolutely continuous in  $[a,b]$ . Therefore it is enough to check the derivative is nonpositive for almost all  $t$ . From Lemma B.4, we can differentiate (12) both sides. Thus for almost all  $t$  we have
$$
\ddot {\tilde {X}} (t) = - \frac {d}{d t} \tilde {A} (\tilde {X} (t)) + \frac {\gamma p}{t ^ {p + 1}} (\tilde {X} (t) - X _ {0}) - \frac {\gamma}{t ^ {p}} \dot {\tilde {X}} (t).
$$
Recall from monotonicity of  $\tilde{A}$  and (1), we know  $\left\langle \dot{\tilde{X}}(t), \frac{d}{dt} \tilde{A}(\tilde{X}(t)) \right\rangle \geq 0$  for almost all  $t$ . Therefore for almost all  $t$ ,
(i)  $0 < p \leq 1$
$$
\begin{array}{l} \dot {\tilde {U}} _ {1} (t) = 2 \left\langle \dot {\tilde {X}} (t), \ddot {\tilde {X}} (t) \right\rangle + \frac {2 \gamma p}{t ^ {p + 1}} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle - \frac {\gamma p (p + 1)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - 2 \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle - \frac {2 \gamma}{t ^ {p}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {2 \gamma p}{t ^ {p + 1}} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle \\ + \frac {2 \gamma p}{t ^ {p + 1}} \left\langle \dot {\tilde {X}} (t), \tilde {X} (t) - X _ {0} \right\rangle - \frac {2 \gamma p ^ {2}}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} - \frac {\gamma p (1 - p)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - 2 \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle - \frac {2 \gamma}{t ^ {p}} \left\| \dot {\tilde {X}} (t) - \frac {p}{t} (\tilde {X} - X _ {0}) \right\| ^ {2} - \frac {\gamma p (1 - p)}{t ^ {p + 2}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ \leq 0. \\ \end{array}
$$
(ii)  $p > 1$
$$
\begin{array}{l} \dot {\tilde {U}} _ {2} (t) = - \frac {p - 1}{t ^ {p}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {2}{t ^ {p - 1}} 2 \left\langle \dot {\tilde {X}} (t), \ddot {\tilde {X}} (t) \right\rangle + \frac {2 \gamma p}{t ^ {2 p}} \left\langle \dot {\tilde {X}} (t), \tilde {X} - X _ {0} \right\rangle - \frac {2 \gamma p}{t ^ {2 p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - \frac {p - 1}{t ^ {p}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} - \frac {2}{t ^ {p - 1}} \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle \\ - \frac {2 \gamma}{t ^ {2 p - 1}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} + \frac {2 \gamma p}{t ^ {2 p}} \left\langle \dot {\tilde {X}} (t), \tilde {X} - X _ {0} \right\rangle + \frac {2 \gamma p}{t ^ {2 p}} \left\langle \dot {\tilde {X}} (t), \tilde {X} - X _ {0} \right\rangle - \frac {2 \gamma p}{t ^ {2 p + 1}} \left\| \tilde {X} (t) - X _ {0} \right\| ^ {2} \\ = - \frac {p - 1}{t ^ {p}} \left\| \dot {\tilde {X}} (t) \right\| ^ {2} - \frac {2}{t ^ {p - 1}} \left\langle \dot {\tilde {X}} (t), \frac {d}{d t} \tilde {A} (\tilde {X} (t)) \right\rangle - \frac {2 \gamma}{t ^ {2 p - 1}} \left\| \dot {\tilde {X}} (t) - \frac {p}{t} (\tilde {X} - X _ {0}) \right\| ^ {2} \\ \leq 0. \\ \end{array}
$$
We now are ready to prove uniform boundedness of derivatives  $\hat{X}_{\delta}$ .
Lemma B.6. Take  $T > 0$ . For all  $t \in [0, T]$  and  $\delta \in (0, 1)$ , below inequality holds.
$$
\left\| \dot {\tilde {X}} _ {\delta} (t) \right\| \leq \tilde {M} _ {d o t} (T) = \left\{ \begin{array}{l l} \sqrt {1 + \gamma p} \left\| \tilde {A} (X _ {0}) \right\| & 0 <   p \leq 1 \\ \sqrt {T ^ {p - 1} \left(\frac {1}{2 \gamma} + 2 p (2 L ^ {2} + 2 \gamma + 1)\right)} \left\| \tilde {A} (X _ {0}) \right\| & p \geq 1. \end{array} \right.
$$
Proof. We prove first statement by considering two cases.
(1)  $t\in [0,\delta ]$
From Lemma B.4, we know  $\dot{\hat{X}}_{\delta}$  is absolutely continuous and so  $\left\| \dot{\hat{X}}_{\delta} \right\|^2$  is absolutely continuous as well. Differentiating (13), for almost all  $t \in (0, \delta)$  we have
$$
\ddot {\tilde {X}} _ {\delta} = - \frac {d}{d t} \tilde {A} (\tilde {X} _ {\delta} (t)) - \frac {\gamma}{\delta^ {p}} \dot {\tilde {X}} _ {\delta}.
$$
Now for almost all  $t\in (0,\delta)$
$$
\frac {d}{d t} \left\| \dot {\tilde {X}} _ {\delta} \right\| ^ {2} = 2 \left\langle \dot {\tilde {X}} _ {\delta}, \ddot {\tilde {X}} _ {\delta} \right\rangle = - 2 \left\langle \dot {\tilde {X}} _ {\delta}, \frac {d}{d t} \tilde {A} (\tilde {X} _ {\delta} (t)) \right\rangle - \frac {2 \gamma}{\delta^ {p}} \left\| \dot {\tilde {X}} _ {\delta} \right\| ^ {2} \leq - \frac {2 \gamma}{\delta^ {p}} \left\| \dot {\tilde {X}} _ {\delta} \right\| ^ {2} \leq 0.
$$
(i)  $0 <   p\leq 1$
From above we know  $\left\| \dot{\hat{X}}_{\delta}\right\|$  is nonincreasing in  $t\in [0,\delta ]$  , we have  $\left\| \dot{\hat{X}}_{\delta}(t)\right\| \leq \left\| \dot{\hat{X}}_{\delta}(0)\right\| = \left\| \tilde{A} (X_0)\right\|$
Integrating  $\frac{d}{dt}\left\| \dot{\tilde{X}}_{\delta}\right\|^{2} \leq -\frac{2\gamma}{\delta^{p}}\left\| \dot{\tilde{X}}_{\delta}\right\|^{2}$  we have
$$
\begin{array}{l} \left\| \dot {\hat {X}} _ {\delta} (t) \right\| ^ {2} - \left\| \dot {\hat {X}} _ {\delta} (0) \right\| ^ {2} \leq - \frac {2 \gamma}{\delta^ {p}} \int_ {0} ^ {t} \left\| \dot {\hat {X}} _ {\delta} (s) \right\| ^ {2} d s \\ \leq - \frac {2 \gamma}{\delta^ {p}} \int_ {0} ^ {t} \left\| \dot {\bar {X}} _ {\delta} (t) \right\| ^ {2} d s = - \frac {2 \gamma t}{\delta^ {p}} \left\| \dot {\bar {X}} _ {\delta} (t) \right\| ^ {2}. \\ \end{array}
$$
Organizing with respect to  $\left\| \dot{\bar{X}}_{\delta}(t)\right\|$  we have
$$
\left\| \dot {\tilde {X}} _ {\delta} (t) \right\| \leq \frac {\left\| \dot {\tilde {X}} _ {\delta} (0) \right\|}{\sqrt {1 + \frac {2 \gamma t}{\delta^ {p}}}} = \sqrt {\frac {\delta^ {p}}{\delta^ {p} + 2 \gamma t}} \left\| \tilde {A} (X _ {0}) \right\| \leq \sqrt {\frac {\delta^ {p}}{2 \gamma t}} \left\| \tilde {A} (X _ {0}) \right\|.
$$
(2)  $t > \delta$
Note for  $t > \delta$  (12) holds, we can apply Lemma B.5.
From (i) we know  $\left\| \dot{\tilde{X}}_{\delta}(t)\right\| \leq \left\| \tilde{A} (X_0)\right\|$  for  $t\in [0,\delta ]$  . Therefore  $\left\| \dot{\tilde{X}}_{\delta}(\delta)\right\| \leq \left\| \tilde{A} (X_0)\right\|$  and we see
$$
\left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| = \left\| \int_ {0} ^ {\delta} \dot {\tilde {X}} _ {\delta} (s) d s \right\| \leq \int_ {0} ^ {\delta} \left\| \dot {\tilde {X}} _ {\delta} (s) \right\| d s \leq \delta \left\| \tilde {A} (X _ {0}) \right\|.
$$
From Lemma B.5 we know  $\tilde{U}_1(t) = \left\| \dot{\tilde{X}}_\delta (t)\right\|^2 +\frac{\gamma p}{t^{p + 1}}\left\| \tilde{X}_\delta (t) - X_0\right\|^2$  is nonincreasing. Therefore  $\left\| \dot{\tilde{X}}_{\delta}(t)\right\| ^2\leq$ $\tilde{U}_1(\delta)$  for  $t\geq \delta$ . Since  $\delta < 1$  we have
$$
\begin{array}{l} \left\| \dot {\tilde {X}} _ {\delta} (t) \right\| \leq \sqrt {\tilde {U} _ {1} (\delta)} = \sqrt {\left\| \dot {\tilde {X}} _ {\delta} (\delta) \right\| ^ {2} + \frac {\gamma p}{\delta^ {p + 1}} \left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| ^ {2}} \\ \leq \sqrt {\left\| \tilde {A} (X _ {0}) \right\| ^ {2} + \gamma p \delta^ {1 - p} \left\| \tilde {A} (X _ {0}) \right\| ^ {2}} \leq \sqrt {1 + \gamma p} \left\| \tilde {A} (X _ {0}) \right\|. \\ \end{array}
$$
From (i) we know  $\left\| \dot{\tilde{X}}_{\delta}(t)\right\| \leq \sqrt{\frac{\delta^p}{2\gamma t}}\left\| \tilde{A} (X_0)\right\|$  for  $t\in [0,\delta ]$  . Therefore  $\left\| \dot{\tilde{X}}_{\delta}(\delta)\right\| \leq \sqrt{\frac{\delta^{p - 1}}{2\gamma}}\left\| \tilde{A} (X_0)\right\|$  and we see
$$
\left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| \leq \int_ {0} ^ {\delta} \left\| \dot {\tilde {X}} _ {\delta} (s) \right\| d s \leq \int_ {0} ^ {\delta} \sqrt {\frac {\delta^ {p}}{2 \gamma s}} \left\| \tilde {A} (X _ {0}) \right\| d s = \sqrt {\frac {2 \delta^ {p + 1}}{\gamma}} \left\| \tilde {A} (X _ {0}) \right\|.
$$
Applying (13) and recalling the fact  $\tilde{A}$  is  $L$ -Lipschitz, we have
$$
\begin{array}{l} \frac {\gamma^ {2}}{\delta^ {2 p}} \| X _ {\delta} (\delta) - X _ {0} \| ^ {2} = \left\| \tilde {A} (X _ {\delta} (\delta)) + \dot {X} _ {\delta} (\delta) \right\| ^ {2} = \left\| \tilde {A} (X _ {\delta} (\delta)) - \tilde {A} (X _ {0}) + \tilde {A} (X _ {0}) + \dot {X} _ {\delta} (\delta) \right\| ^ {2} \\ \leq 2 \left\| \tilde {A} \left(X _ {\delta} (\delta)\right) - \tilde {A} \left(X _ {0}\right) \right\| ^ {2} + 2 \left\| \tilde {A} \left(X _ {0}\right) + \dot {X} _ {\delta} (\delta) \right\| ^ {2} \\ \leq 2 L ^ {2} \left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| ^ {2} + 4 \left\| \tilde {A} (X _ {0}) \right\| ^ {2} + 4 \left\| \dot {X} _ {\delta} (\delta) \right\| ^ {2} \\ \leq \left(\frac {4}{\gamma} L ^ {2} \delta^ {p + 1} + 4 + \frac {2 \delta^ {p - 1}}{\gamma}\right) \left\| \tilde {A} (X _ {0}) \right\| ^ {2}. \\ \end{array}
$$
From Lemma B.5 we know  $\tilde{U}_2(t) = \frac{1}{t^{p - 1}}\left\| \dot{\tilde{X}} (t)\right\| ^2 +\frac{\gamma p}{t^{2p}}\left\| \tilde{X} (t) - X_0\right\| ^2$  is nonincreasing. Therefore  $\frac{\|\dot{\tilde{X}}_{\delta}(t)\|^{2}}{t^{p - 1}}\leq$ $\tilde{U}_2(\delta)$  for  $t\geq \delta$ . Since  $\delta < 1$  we have
$$
\begin{array}{l} \frac {\left\| \dot {\tilde {X}} _ {\delta} (t) \right\|}{\sqrt {t ^ {p - 1}}} \leq \sqrt {\tilde {U} _ {2} (\delta)} = \sqrt {\frac {\left\| \dot {\tilde {X}} _ {\delta} (\delta) \right\| ^ {2}}{\delta^ {p - 1}} + \frac {\gamma p}{\delta^ {2 p}} \left\| \tilde {X} _ {\delta} (\delta) - X _ {0} \right\| ^ {2}} \\ \leq \sqrt {\frac {\left\| \tilde {A} (X _ {0}) \right\| ^ {2}}{2 \gamma} + p \left(4 L ^ {2} \delta^ {p + 1} + 4 \gamma + 2 \delta^ {p - 1}\right) \left\| \tilde {A} (X _ {0}) \right\| ^ {2}} \leq \sqrt {\frac {1}{2 \gamma} + 2 p \left(2 L ^ {2} + 2 \gamma + 1\right)} \left\| \tilde {A} (X _ {0}) \right\|. \\ \end{array}
$$
Therefore for  $t\in [0,T]$  we have
$$
\left\| \dot {\tilde {X}} _ {\delta} (t) \right\| \leq \sqrt {T ^ {p - 1} \left(\frac {1}{2 \gamma} + 2 p (2 L ^ {2} + 2 \gamma + 1)\right)} \left\| \tilde {A} (X _ {0}) \right\|
$$
From (1) and (2), we get the desired result.
We now show the sequence  $\left\{\tilde{X}_{\delta_m}\right\}_{m\in \mathbb{N}}$  converges uniformly on  $[0,T]$  for every  $T > 0$ . It is suffices to prove following proposition.
Proposition B.7. For  $T > 0$ , the sequence  $\left\{\tilde{X}_{\delta_m}\right\}_{m\in \mathbb{N}}$  is a Cauchy sequence with respect to supremum norm on  $[0,T]$ .
Proof. Take  $\epsilon > 0$ . We want to show, there is  $N \in \mathbb{N}$  such that if  $n, m > N$  then  $\left\| \tilde{X}_{\delta_n}(t) - \tilde{X}_{\delta_m}(t) \right\| \leq \epsilon$  for all  $t \in [0, \infty)$ .
Define  $d_{\delta, \nu}(t) = \frac{1}{2} \left\| \tilde{X}_{\delta}(t) - \tilde{X}_{\nu}(t) \right\|^2$ . Without loss of generality, we may assume  $\delta \geq \nu$ . With  $M_{dot}(T)$  defined in Lemma B.6, we will show  $d_{\delta, \nu}(t) \leq 2\delta^2 \tilde{M}_{dot}(T)^2$ .
First consider the case  $0 \leq t \leq \delta$ . By Lemma B.6, we have
$$
\begin{array}{l} \left\| \tilde {X} _ {\delta} (t) - \tilde {X} _ {\nu} (t) \right\| \leq \int_ {0} ^ {t} \left\| \dot {\tilde {X}} _ {\delta} (s) - \dot {\tilde {X}} _ {\nu} (s) \right\| d s \leq \int_ {0} ^ {t} \left(\left\| \dot {\tilde {X}} _ {\delta} (s) \right\| + \left\| \dot {\tilde {X}} _ {\nu} (s) \right\|\right) d s \\ \leq \int_ {0} ^ {t} 2 \tilde {M} _ {d o t} (T) d s = 2 t \tilde {M} _ {d o t} (T) \\ \end{array}
$$
Thus for  $t\in [0,\delta ]$  we have
$$
d _ {\delta , \nu} (t) \leq 2 t ^ {2} \tilde {M} _ {d o t} (T) ^ {2} \leq 2 \delta^ {2} \tilde {M} _ {d o t} (T) ^ {2}.
$$
Now we consider the case  $t \geq \delta$ . By monotonicity of  $\tilde{A}$ , we have
$$
\begin{array}{l} \frac {d}{d t} \frac {1}{2} \left\| \tilde {X} _ {\delta} (t) - \tilde {X} _ {\nu} (t) \right\| ^ {2} = \left\langle \dot {X} _ {\delta} (t) - \dot {X} _ {\nu} (t), \tilde {X} _ {\delta} (t) - \tilde {X} _ {\nu} (t) \right\rangle \\ = \left\langle - \left(\tilde {A} \left(\tilde {X} _ {\delta} (t)\right) - \tilde {A} \left(\tilde {X} _ {\nu} (t)\right)\right) - \frac {\gamma}{t ^ {p}} \left(X _ {\delta} (t) - X _ {0}\right) + \frac {\gamma}{t ^ {p}} \left(X _ {\nu} (t) - X _ {0}\right), X _ {\delta} (t) - X _ {\nu} (t) \right\rangle \\ \leq \left\langle - \frac {\gamma}{t ^ {p}} \left(X _ {\delta} (t) - X _ {0}\right) + \frac {\gamma}{t ^ {p}} \left(X _ {\nu} (t) - X _ {0}\right), X _ {\delta} (t) - X _ {\nu} (t) \right\rangle \\ = - \frac {\gamma}{t ^ {p}} \left\| X _ {\delta} (t) - X _ {\nu} (t) \right\| ^ {2} \\ \leq 0. \\ \end{array}
$$
Thus we have  $d_{\delta, \nu}(t) \leq d_{\delta, \nu}(\delta)$  for  $t \geq \delta$ .
Now combining two cases, we have
$$
d _ {\delta , \nu} (t) \leq 2 \delta^ {2} \tilde {M} _ {d o t} (T) ^ {2} \tag {14}
$$
Since  $\lim_{k\to \infty}\delta_n = 0$ , there is  $N\in \mathbb{N}$  such that  $m > n > N$  implies  $d_{\delta_n,\delta_m}\leq \frac{\epsilon^2}{2}$ , we're done.
We are now ready to prove Theorem B.2.
Proof. (1) Existence of solution.
From Proposition B.7, we know  $\left\{\tilde{X}_{\delta_m}\right\}_{m\in \mathbb{N}}$  converging uniformly on  $[0,T]$  for every  $T >$  . Denote the limit as  $\tilde{X}$  i.e. define  $\tilde{X}:[0,\infty)\to \mathbb{R}^n$  as
$$
\lim _ {m \to \infty} \tilde {X} _ {\delta_ {m}} (t) = \tilde {X} (t).
$$
We can check  $\tilde{X}(0) = X_0$  easily since  $\tilde{X}_{\delta_m}(0) = X_0$  for all  $m \in \mathbb{N}$ . It remains to show  $\tilde{X}$  satisfies (12).
Take  $t > 0$ . We wish to show
$$
\lim  _ {h \rightarrow 0} \left\| \frac {\tilde {X} (t + h) - \tilde {X} (t)}{h} + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right)\right\| = 0.
$$
Consider  $h, \delta, T$  such that  $0 < |h| < t$ ,  $0 < \delta < \min \{1, t - |h|\}$  and  $T > t + |h|$ . Then  $(t - |h|, t + |h|) \in [0, T]$  and  $\dot{\tilde{X}}_{\delta}(t) = -\tilde{A}(\tilde{X}_{\delta}(t)) - \frac{\gamma}{t^p} \left( \tilde{X}_{\delta}(t) - X_0 \right)$ . Consider inequality
$$
\begin{array}{l} \left\| \frac {\tilde {X} (t + h) - \tilde {X} (t)}{h} + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} (\tilde {X} (t) - X _ {0}) \right\| \\ \leq \left\| \frac {\tilde {X} (t + h) - \tilde {X} (t)}{h} - \frac {\tilde {X} _ {\delta} (t + h) - \tilde {X} _ {\delta} (t)}{h} \right\| + \left\| \frac {\tilde {X} _ {\delta} (t + h) - \tilde {X} _ {\delta} (t)}{h} - \dot {\tilde {X}} _ {\delta} (t) \right\| + \left\| \dot {\tilde {X}} _ {\delta} (t) + \tilde {A} (\tilde {X} (t)) + \frac {\gamma}{t ^ {p}} \left(\tilde {X} (t) - X _ {0}\right) \right\|. \\ \end{array}
$$
We now show right hand side goes to zero as  $h \to 0$ . The point of the proof is,  $\tilde{X}_{\delta}$  converges uniformly to  $\tilde{X}$  and  $\ddot{\tilde{X}}_{\delta}$  is uniformly bounded on  $[t - |h|, T]$ .
From (14) we have