SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Note the purpose was to obtain the first term, which is nonnegative due to monotonicity of  $\tilde{\mathbb{A}}$ . Now taking integration by parts to the second term we have
$$
\begin{array}{l} \int_ {t _ {0}} ^ {t} \left\langle \frac {d}{d s} \tilde {\mathbb {A}} (X (s)), C (s) \beta (s) W (s) \right\rangle d s \\ = \left[ \left\langle \tilde {\mathbf {A}} (X (W, s)), C (s) \beta (s) W (s) \right\rangle \right] _ {t _ {0}} ^ {t} - \int_ {t _ {0}} ^ {t} \left\langle \tilde {\mathbf {A}} (X (W, s)), \left(C (s) \beta (s) ^ {2} + C (s) \dot {\beta} (s)\right) W (s) + C (s) \beta (s) \dot {W} (s) \right\rangle d s \\ = \left[ \left\langle \tilde {\mathbf {A}} (X (W, s)), C (s) \beta (s) W (s) \right\rangle \right] _ {t _ {0}} ^ {t} + \int_ {t _ {0}} ^ {t} \beta (s) \left\| \dot {W} (s) \right\| ^ {2} d s + \int_ {t _ {0}} ^ {t} \left(\beta (s) ^ {2} + \dot {\beta} (s)\right) \left\langle \dot {W} (s), W (s) \right\rangle d s \\ = \left[ \left\langle \tilde {\mathbb {A}} (X (W, s)), C (s) \beta (s) W (s) \right\rangle \right] _ {t _ {0}} ^ {t} + \int_ {t _ {0}} ^ {t} \beta (s) \left\| \dot {W} (s) \right\| ^ {2} d s \\ + \left[ \left(\beta (s) ^ {2} + \dot {\beta} (s)\right) \frac {1}{2} \| W (s) \| ^ {2} \right] _ {t _ {0}} ^ {t} - \frac {1}{2} \int_ {t _ {0}} ^ {t} \left(2 \beta (s) \dot {\beta} (s) + \ddot {\beta} (s)\right) \| W (s) \| ^ {2} d s \\ \end{array}
$$
On the second equality, we used the fact  $\dot{W}(t) = -C(t)\tilde{\mathbf{A}}(X(W(t), t))$ . Note the fundamental theorem of calculus for  $\langle \tilde{\mathbf{A}}(X(W, s)), C(s)\beta(s)W(s) \rangle$  is valid since  $\tilde{\mathbf{A}}(X(W, s))$  is Lipschitz continuous and  $C(s)\beta(s)W(s)$  is continuously differentiable in  $[t_0, t]$ , so their inner product is absolutely continuous in  $[t_0, t]$ .
Observe the integrand in the last term can be rewritten as
$$
\left(2 \beta (s) \dot {\beta} (s) + \ddot {\beta} (s)\right) \| W (s) \| ^ {2} = C (s) ^ {2} \left(2 \beta (s) \dot {\beta} (s) + \ddot {\beta} (s)\right) \| X (s) - X _ {0} \| ^ {2} = \frac {d}{d s} \left(C (s) ^ {2} \dot {\beta} (s)\right) \| X (s) - X _ {0} \| ^ {2}.
$$
Now gathering the results, we conclude
$$
\begin{array}{l} E _ {1} \equiv \frac {1}{2} \left\| \dot {W} (t) \right\| ^ {2} - \int_ {t _ {0}} ^ {t} \beta (s) \left\| \dot {W} (s) \right\| ^ {2} d s + \int_ {t _ {0}} ^ {t} C (s) \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {W} (s) \right\rangle d s \\ = \frac {C (t) ^ {2}}{2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + \left[ \left\langle \tilde {\mathbf {A}} (X (W (s), s)), C (s) \beta (s) W (s) \right\rangle \right] _ {t _ {0}} ^ {t} + \left[ \left(\beta (s) ^ {2} + \dot {\beta} (s)\right) \frac {1}{2} \| W (s) \| ^ {2} \right] _ {t _ {0}} ^ {t} \\ + \int_ {t _ {0}} ^ {t} C (s) ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s - \frac {1}{2} \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(C (s) ^ {2} \dot {\beta} (s)\right) \| X (s) - X _ {0} \| ^ {2} d s \\ = \frac {C (t) ^ {2}}{2} \left(\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\beta (t) ^ {2} + \dot {\beta} (t)\right) \| X (t) - X _ {0} \| ^ {2}\right) \\ + \int_ {t _ {0}} ^ {t} C (s) ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s - \frac {1}{2} \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(C (s) ^ {2} \dot {\beta} (s)\right) \| X (s) - X _ {0} \| ^ {2} d s \\ - \underbrace {\frac {C (t _ {0}) ^ {2}}{2} \left(2 \beta (t _ {0}) \left\langle \tilde {\mathbf {A}} (X (t _ {0})), X (t _ {0}) - X _ {0} \right\rangle + \left(\beta (t _ {0}) ^ {2} + \dot {\beta} (t _ {0})\right) \| X (t _ {0}) - X _ {0} \| ^ {2}\right)} _ {= \text {c o n s t a n t}} \\ \end{array}
$$
Moving the constant terms to left hand side and naming  $E = E_{1} -$  constant, we get the desired result.
E.3 Proof of Corollary 3.3
E.3.1  $V$  is nonincreasing when  $\bar{\mathbb{A}}$  is Lipshitz continuous monotone
We first check it is true for the case  $\tilde{\mathbb{A}}$  is Lipschitz continuous. Then from Proposition 3.2 we can write  $V(t)$  as
$$
V (t) = E - 2 \int_ {t _ {0}} ^ {t} C (s) ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s
$$
with  $E$  in Proposition 3.2. As  $\tilde{\mathbb{A}}$  is monotone, from (1) we know  $C(s)^2\left\langle \frac{d}{ds}\tilde{\mathbb{A}} (X(s)),\dot{X} (s)\right\rangle \geq 0$  holds almost everywhere. Therefore for  $h > 0$
$$
V (t + h) - V (t) = - 2 \int_ {t} ^ {t + h} C (s) ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s \leq 0,
$$
we see  $V$  is a nonincreasing function. Therefore for all  $t > 0$ ,  $V(t) \leq \lim_{\epsilon \to 0+} V(\epsilon)$ . It remains to show the limit  $\lim_{\epsilon \to 0+} V(\epsilon)$  exists.
E.3.2 Calculation of  $V(0) = \lim_{t\to 0 + }V(t)$  for Lipschitz continuous monotone  $\mathring{\mathrm{A}}$
In this section, we calculate  $\lim_{t\to 0 + }V(t)$  when  $\tilde{\mathbf{A}}$  is Lipshitz continuous. Recall  $V$  was defined as
From now we denote  $V$  for the case  $t_0 = 0$  as  $V^0$ .
We first check
$$
C (t) = \left\{ \begin{array}{l l} t ^ {\gamma} & p = 1 \\ e ^ {\frac {\gamma}{1 - p}} t ^ {1 - p} & p > 0, p \neq 1. \end{array} \right. \tag {31}
$$
$$
\int_ {1} ^ {t} \frac {\gamma}{s} d s = \gamma \log t
$$
$$
\int_ {0} ^ {t} \frac {\gamma}{s ^ {p}} d s = \frac {\gamma}{1 - p} t ^ {1 - p} \quad \text {f o r} 0 <   p <   1
$$
$$
\int_ {\infty} ^ {t} \frac {\gamma}{s ^ {p}} d s = \frac {\gamma}{1 - p} t ^ {1 - p} \qquad \text {f o r} p > 1,
$$
we see  $C(t)$  defined above agrees with the definition of  $C(t)$  for each case. Note
$$
\lim  _ {t \to 0 +} C (t) = \left\{ \begin{array}{l l} 0 & p \geq 1 \\ 1 & 0 <   p <   1. \end{array} \right.
$$
Now we will show
$$
\lim  _ {t \rightarrow 0 +} V ^ {0} (t) = \lim  _ {t \rightarrow 0 +} \frac {C (t) ^ {2}}{2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} = \left\{\begin{array}{l l}0&\text {i f} p \geq 1\\\frac {\left\| \tilde {\mathbf {A}} (X _ {0}) \right\| ^ {2}}{2}&\text {i f} 0 <   p <   1.\end{array}\right. \tag {32}
$$
To do so, we first show for  $t > 0$
$$
\lim _ {\epsilon \to 0 +} \int_ {\epsilon} ^ {t} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} d s <   \infty .
$$
We first provide an elementary fact as a lemma.
Lemma E.3. Let  $f \colon (0, \infty) \to \mathbb{R}$  is a continuous function. Suppose there is  $q < 1$ ,  $0 < l < \infty$  such that
$$
\limsup_{t\to 0 + }\left|f(t)t^{q}\right| <   l.
$$
Then for  $t > 0$ ,  $f \in L^{1}([0,t],\mathbb{R})$ .
Proof. Since  $\lim \sup_{s\to 0 + }f(s)s^q = l$ , there is  $\epsilon \in (0,t)$  such that
$$
0 <   s <   \epsilon \quad \Longrightarrow \quad | f (s) s ^ {q} | <   2 | l |.
$$
Then since  $q < 1$
$$
\int_ {0} ^ {\epsilon} | f (s) | d s = \int_ {0} ^ {\epsilon} | f (s) s ^ {q} | \frac {1}{s ^ {q}} d s \leq \int_ {0} ^ {\epsilon} \frac {2 | l |}{s ^ {q}} d s = \left[ \frac {2 | l |}{1 - q} s ^ {1 - q} \right] _ {0} ^ {\epsilon} = \frac {2 | l |}{1 - q} \epsilon^ {1 - q} <   \infty .
$$
By the way as  $f(t)$  is continuous on  $[\epsilon, t]$ ,  $M = \max_{s \in [\epsilon, t]} |f(s)|$  exists. Therefore,
$$
\int_ {0} ^ {t} | f (s) | d s = \int_ {0} ^ {\epsilon} | f (s) | d s + \int_ {\epsilon} ^ {t} | f (s) | d s \leq \frac {2 | l |}{1 - q} \epsilon^ {1 - q} + M (t - \epsilon) <   \infty .
$$
Applying Lemma E.3, we will show for  $t > 0$
$$
\left| s ^ {2} \frac {d}{d s} (C (s) ^ {2} \dot {\beta} (s)) \right| \in L ^ {1} ([ 0, t ], \mathbb {R}). \tag {33}
$$
$$
\frac {d}{d s} \left(C (s) ^ {2} \dot {\beta} (s)\right) = 2 C (s) ^ {2} \beta (s) \dot {\beta} (s) + C (s) ^ {2} \ddot {\beta} (s) = C (s) ^ {2} \left(- \frac {2 p \gamma^ {2}}{s ^ {2 p + 1}} + \frac {p (1 + p) \gamma}{s ^ {p + 2}}\right). \tag {34}
$$
(i)  $p > 1$
We first show  $\lim_{s\to 0 + }C(s)^2\frac{1}{s^n} = 0$  for all  $n > 0$ . Take  $n > 0$ . Then there is some  $k\in \mathbb{N}$  such that  $k(p - 1) > n$ .
With change of variable  $u = \frac{1}{s}$  and L'Hôpital's rule we see
$$
\begin{array}{l} \lim _ {s \to 0 +} C (s) ^ {2} \frac {1}{s ^ {n}} = \lim _ {u \to \infty} \frac {u ^ {n}}{e ^ {\frac {2 \gamma}{p - 1} u ^ {p - 1}}} \\ = \lim  _ {u \rightarrow \infty} \frac {n u ^ {n - 1}}{2 \gamma u ^ {p - 2} e ^ {\frac {2 \gamma}{p - 1} u ^ {p - 1}}} = \frac {n}{2 \gamma} \lim  _ {u \rightarrow \infty} \frac {u ^ {n + 1 - p}}{e ^ {\frac {2 \gamma}{p - 1} u ^ {p - 1}}} = \dots = \frac {\prod_ {m = 0} ^ {k - 1} (n - m (p - 1))}{(2 \gamma) ^ {k}} \lim  _ {u \rightarrow \infty} \frac {u ^ {n - k (p - 1)}}{e ^ {\frac {2 \gamma}{p - 1} u ^ {p - 1}}} = 0. \tag {35} \\ \end{array}
$$
$$
\lim  _ {s \rightarrow 0 +} \left| s ^ {2} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right)\right| = \lim  _ {s \rightarrow 0 +} \left| C (s) ^ {2} \left(- \frac {2 p \gamma^ {2}}{s ^ {2 p - 1}} + \frac {p (1 + p) \gamma}{s ^ {p}}\right)\right| = 0.
$$
By Lemma E.3, we conclude  $\int_0^t\left|s^2\frac{d}{ds}\left(\frac{C(s)^2\dot{\beta}(s)}{2}\right)\right|ds < \infty$
(ii)  $p = 1$
Since  $C(s) = s^{\gamma}$ , we see
$$
\lim  _ {s \rightarrow 0 +} \left| s ^ {2} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \cdot s ^ {1 - \gamma} \right| = \lim  _ {s \rightarrow 0 +} \left| s ^ {2} \cdot \gamma (\gamma - 1) s ^ {2 \gamma - 3} \cdot s ^ {1 - \gamma} \right| = \gamma | \gamma - 1 | \lim  _ {s \rightarrow 0 +} s ^ {\gamma} = 0.
$$
Since  $1 - \gamma < 1$ , by Lemma E.3 we conclude  $\int_0^t \left| s^2 \frac{d}{ds} \left( \frac{C(s)^2 \dot{\beta}(s)}{2} \right) \right| ds < \infty$ .
(iii)  $0 < p < 1$
Since  $\lim_{s\to 0 + }C(s) = \lim_{s\to 0 + }e^{\frac{\gamma}{1 - p}s^{1 - p}} = 1$  we see
$$
\limsup_{s\to 0 + }\left|s^{2}\frac{d}{ds}\left(\frac{C(s)^{2}\dot{\beta}(s)}{2}\right)\cdot s^{p}\right| = \limsup_{s\to 0 + }C(s)^{2}\left| - 2p\gamma^{2}s^{1 - p} + p(1 + p)\gamma \right| = p(1 + p)\gamma .
$$
Since  $p < 1$ , by Lemma E.3 we conclude  $\int_0^t \left| s^2 \frac{d}{ds} \left( \frac{C(s)^2 \dot{\beta}(s)}{2} \right) \right| ds < \infty$ .
Naming the bound in Corollary B.8 as  $M(T)$ , we know for  $0 < s < T$
$$
\left\| \frac {X (s) - X _ {0}}{s} \right\| \leq \frac {\int_ {0} ^ {s} \left\| \dot {X} (u) \right\| d u}{s} \leq \frac {\int_ {0} ^ {s} M (T) d u}{s} = M (T).
$$
Therefore applying (33), for  $0 < t < T$  we have
$$
\int_ {0} ^ {t} \left| \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} \right| d s \leq M (T) ^ {2} \int_ {0} ^ {t} \left| s ^ {2} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \right| d s <   \infty .
$$
Hence we know  $V^0 (t)$  is well defined and since  $\lim_{t\to 0 + }\int_0^t\frac{d}{ds}\left(\frac{C(s)^2\dot{\beta}(s)}{2}\right)\| X(s) - X_0\| ^2 ds = 0$ , we have
$$
\lim _ {t \to 0 +} V ^ {0} (t) = \lim _ {t \to 0 +} \frac {C (t) ^ {2}}{2} \left(\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \beta (t) \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\beta (t) ^ {2} + \dot {\beta} (t)\right) \left\| X (t) - X _ {0} \right\| ^ {2}\right).
$$
Now we are ready to show the desired result.
As we know  $X(t) - X_0$  and  $\tilde{\mathbb{A}}(X(t))$  are bounded from Lemma D.1 and Corollary B.9. Therefore from (35) we have
$$
0 = \lim  _ {t \to 0 +} C (t) ^ {2} = \lim  _ {t \to 0 +} C (t) ^ {2} \beta (t) = \lim  _ {t \to 0 +} C (t) ^ {2} \left(\beta (t) ^ {2} + \dot {\beta} (t)\right),
$$
therefore  $\lim_{t\to 0 + }V^0 (t) = 0$
(ii)  $0 < p \leq 1$
As  $\left\| \frac{X(t) - X_0}{t}\right\| \leq M(T) < \infty$  for  $0 < t < T$ , we see
$$
\begin{array}{l} \limsup_{t\to 0 + }C(t)^{2}\beta (t)\left\langle \tilde{\mathbf{A}} (X(t)),X(t) - X_{0}\right\rangle \leq \gamma \limsup_{t\to 0 + }C(t)^{2}t^{1 - p}\left\| \tilde{\mathbf{A}} (X(t))\right\| M(T) = 0 \\ \limsup_{t\to 0 + }C(t)^{2}\left(\beta (t)^{2} + \dot{\beta} (t)\right)\| X(t) - X_{0}\|^{2}\leq \gamma \limsup_{t\to 0 + }C(t)^{2}\left(\gamma t^{2 - 2p} - t^{1 - p}\right)M(T)^{2} = 0. \\ \end{array}
$$
$$
\lim _ {t \to 0 +} V ^ {0} (t) = \lim _ {t \to 0 +} \frac {C (t) ^ {2}}{2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} = \left\{ \begin{array}{l l} 0 & \mathrm {i f} p = 1 \\ \frac {\left\| \tilde {\mathbf {A}} (X _ {0}) \right\| ^ {2}}{2} & \mathrm {i f} 0 <   p <   1. \end{array} \right.
$$
From (i) and (ii) we get the desired conclusion
$$
V ^ {0} (0) = \lim  _ {t \to 0 +} V ^ {0} (t) = \left\{ \begin{array}{l l} 0 & \text {i f} p \geq 1 \\ \frac {\| \tilde {\mathbf {A}} (X _ {0}) \| ^ {2}}{2} & \text {i f} 0 <   p <   1, \end{array} \right.
$$
and therefore for general  $t_0\geq 0$
$$
V (0) = \lim  _ {t \rightarrow 0 +} V (t) = \lim  _ {t \rightarrow 0 +} V ^ {0} (t) - \int_ {0} ^ {t _ {0}} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} d s
$$
is well-defined.
E.3.3  $V(t) \leq \lim_{n \to \infty} V_{\lambda_n}(0)$  holds for  $t \in S$  and general maximal monotone  $\mathbb{A}$
Define  $S$  as defined in Lemma E.1. Take  $t \in S$ , let  $T > t$ . Let  $\{\lambda_n\}_{n \in \mathbb{N}}$  be a positive sequence  $\lambda_{n}$  that  $\lim_{n \to \infty} \lambda_{n} = 0$ ,  $X_{\lambda_n}$  converges to  $X$  uniformly on  $[0, T]$  and  $\dot{X}_{\lambda_n}$  converges weakly to  $\dot{X}$  in  $L^2([0, T], \mathbb{R}^n)$ . Recall existence of such sequence was guaranteed by Proposition B.11.
Recall we denoted  $X_{\lambda}$  as the solution of the ODE (19). Denote  $V_{\lambda}$  as  $V$  for the case  $\mathbb{A} = \mathbb{A}_{\lambda}$ , i.e.
$$
\begin{array}{l} V _ {\lambda} (t) = \frac {C (t) ^ {2}}{2} \left(\| \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right) \| ^ {2} + 2 \beta (t) \langle \mathbb {A} _ {\lambda} \left(X _ {\lambda} (t)\right), X _ {\lambda} (t) - X _ {0} \rangle + \left(\beta (t) ^ {2} + \dot {\beta} (t)\right) \| X _ {\lambda} (t) - X _ {0} \| ^ {2}\right) \\ - \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X _ {\lambda} (t) - X _ {0} \| ^ {2} d s. \\ \end{array}
$$
Note equality  $\dot{X} (t) = -\bar{\mathbf{A}} (X(t)) - \beta (t)(X(t) - X_0)$  holds since  $t\in S$  , therefore we have
$$
V (t) = \frac {C (t) ^ {2}}{2} \left(\left\| \dot {X} (t) \right\| ^ {2} + \dot {\beta} (t) \left\| X (t) - X _ {0} \right\| ^ {2}\right) - \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} d s.
$$
The goal of this section is to show
$$
V (t) \leq \lim  _ {n \to \infty} \sup  _ {n \to \infty} V _ {\lambda_ {n}} (t) \leq \lim  _ {n \to \infty} \sup  _ {n \to \infty} V _ {\lambda_ {n}} (0) = \lim  _ {n \to \infty} V _ {\lambda_ {n}} (0).
$$
(1)  $V(t)\leq \lim \sup_{n\to \infty}V_{\lambda_n}(t)$
First observe, from Lemma B.12 we know
$$
\frac {\| X _ {\lambda} (s) - X _ {0} \|}{s} \leq \frac {\int_ {0} ^ {s} \left\| \dot {X} _ {\lambda} (s) \right\| d s}{s} \leq \frac {\int_ {0} ^ {s} M _ {d o t} (T) d s}{s} = M _ {d o t} (T)
$$
holds for  $s \leq T$ . Thus from (33) we have
$$
\left| \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X _ {\lambda} (s) - X _ {0} \| ^ {2} \right| \leq \left| s ^ {2} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \right| M _ {d o t} (T) ^ {2} \in L ^ {1} ([ 0, T ], \mathbb {R}).
$$
Therefore applying dominated convergence theorem, we have for  $t_0 \in [0, T]$
$$
\lim  _ {n \rightarrow \infty} \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X _ {\lambda_ {n}} (s) - X _ {0} \| ^ {2} d s = \int_ {t _ {0}} ^ {t} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} d s. \tag {36}
$$
From elementary analysis, we can easily check  $\lim \sup_{n\to \infty}(a_n + b_n) = \lim \sup_{n\to \infty}a_n + \lim_{n\to \infty}b_n$  holds when  $\lim_{n\to \infty}b_n$  exists. Thus we have
$$
\limsup_{n\to \infty}V_{\lambda_{n}}(t) - V(t) = \frac{C(t)^{2}}{2}\left(\limsup_{n\to \infty}\left\| \dot{X}_{\lambda_{n}}(t)\right\|^{2} - \left\| \dot{X} (t)\right\|^{2}\right).
$$
Therefore it is suffices to show following lemma.
Lemma E.4. Suppose for  $T > 0$  and sequence  $\{\lambda_n\}_{n \in \mathbb{N}}$ ,  $X_{\lambda_n}$  converges to  $X$  uniformly on  $[0, T]$  and  $\dot{X}_{\lambda_n}$  converges weakly to  $\dot{X}$  in  $L^2([0, T], \mathbb{R}^n)$ . Let  $t \in S \cap [0, T]$ . Then following inequality is true
$$
\left\| \dot {X} (t) \right\| \leq \lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (t) \right\|.
$$
Proof. (i)  $\left\| \dot{X}(s) \right\| \leq \lim \sup_{n \to \infty} \left\| \dot{X}_{\lambda_n}(s) \right\|$  holds for almost every  $s \in [0, T]$ .
Let  $D$  be a measurable subset  $D\subset [0,T]$ . Since  $\dot{X}_{\lambda_n}\rightharpoonup \dot{X}$  in  $L^2 ([0,T],\mathbb{R}^n)$  and  $\chi_D\dot{X}\in L^2 ([0,T],\mathbb{R}^n)$ , we have
$$
\begin{array}{l} \int_ {D} \left\| \dot {X} (s) \right\| ^ {2} d s = \int_ {0} ^ {T} \left\langle \dot {X} (s), \chi_ {D} (s) \dot {X} (s) \right\rangle d s = \lim _ {n \rightarrow \infty} \int_ {0} ^ {T} \left\langle \dot {X} _ {\lambda_ {n}} (s), \chi_ {D} (s) \dot {X} (s) \right\rangle d s \\ = \limsup _ {n \to \infty} \int_ {D} \left\langle \dot {X} _ {\lambda_ {n}} (s), \dot {X} (s) \right\rangle d s \leq \limsup _ {n \to \infty} \int_ {D} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| \left\| \dot {X} (s) \right\| d s. \\ \end{array}
$$
The inequality comes from the Cauchy-Schwarz inequality. Now from Reverse Fatou's Lemma we have
$$
\lim  _ {n \rightarrow \infty} \sup  _ {D} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\|\left\| \dot {X} (s) \right\| d s \leq \int_ {D} \lim  _ {n \rightarrow \infty} \sup  _ {D} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\|\left\| \dot {X} (s) \right\| d s.
$$
Therefore combining two inequalities we have
$$
\int_ {D} \left(\lim  _ {n \rightarrow \infty} \sup  _ {s \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| - \left\| \dot {X} (s) \right\|\right) \| \dot {X} (s) \| d s \geq 0.
$$
As  $D$  was arbitrary measurable subset of  $[0,T]$ , we conclude for almost every  $s\in [0,T]$
$$
\left(\lim  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| - \left\| \dot {X} (s) \right\|\right)\left\| \dot {X} (s) \right\| \geq 0 \quad \Longrightarrow \quad \left\| \dot {X} (s) \right\| \leq \lim  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\|.
$$
(ii)  $\left\| \dot{X} (t)\right\| \leq \lim \sup_{n\to \infty}\left\| \dot{X}_{\lambda_n}(t)\right\|$  for  $t\in S\cap [0,T]$
Let  $t \in S \cap [0, T]$ . Then for  $h > 0$  such that  $t + h < T$ , since  $S$  is measure zero, from (i) we have
$$
\int_ {t} ^ {t + h} \left\| \dot {X} (s) \right\| d s \leq \int_ {t} ^ {t + h} \lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| d s.
$$
Now, for some  $a > 0$  consider
$$
U _ {\lambda_ {n}} (s) = \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| ^ {2} + \underbrace {\frac {\gamma p}{s ^ {p + 1}} \| X _ {\lambda_ {n}} (s) - X _ {0} \| ^ {2} + \int_ {a} ^ {s} \frac {\gamma p (1 - p)}{u ^ {p + 2}} \| X _ {\lambda_ {n}} (u) - X _ {0} \| ^ {2} d u} _ {= f _ {n} (s)}.
$$
Then from the proof Lemma B.5, we know
$$
\dot {U} _ {\lambda_ {n}} (s) = - 2 \left\langle \dot {X} _ {\lambda_ {n}} (s), \frac {d}{d s} \mathbb {A} _ {\lambda_ {n}} (X (s)) \right\rangle - \frac {2 \gamma}{s ^ {p}} \left\| \dot {X} _ {\lambda_ {n}} (s) - \frac {p}{s} (X _ {\lambda_ {n}} (s) - X _ {0}) \right\| ^ {2} \leq 0
$$
holds for almost every  $s$ , thus  $U_{\lambda_n}$  is nonincreasing.
By the way, as  $X_{\lambda_n}$  converges to  $X$  uniformly on  $[0,T]$ , using dominated convergence theorem we have
$$
\lim  _ {n \rightarrow \infty} f _ {n} (s) = \frac {\gamma p}{s ^ {p + 1}} \| X (s) - X _ {0} \| ^ {2} + \int_ {a} ^ {s} \frac {\gamma p (1 - p)}{u ^ {p + 2}} \| X (u) - X _ {0} \| ^ {2} d u.
$$
Denote  $f(s) = \lim_{n\to \infty}f_n(s)$ .
Using above facts, for  $s \in [t, t + h]$  we have
$$
\begin{array}{l} \limsup_{n\to \infty}\left\| \dot {X} _{\lambda_{n}}(s)\right\| = \limsup_{n\to \infty}\sqrt{U_{\lambda_{n}}(s) - f_{n}(s)} \\ \leq \lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} \sqrt {U _ {\lambda_ {n}} (t) - f _ {n} (s)} = \sqrt {\lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} U _ {\lambda_ {n}} (t) - f (s)}. \tag {37} \\ \end{array}
$$
$$
\| X (t + h) - X (t) \| \leq \int_ {t} ^ {t + h} \left\| \dot {X} (s) \right\| d s \leq \int_ {t} ^ {t + h} \lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (s) \right\| d s \leq \int_ {t} ^ {t + h} \sqrt {\lim  _ {n \rightarrow \infty} U _ {\lambda_ {n}} (t) - f (s)} d s.
$$
Note  $\sqrt{\lim\sup_{n\to\infty}U_{\lambda_n}(t) - f(s)}$  is a continuous function with respect to  $s$ . Thus we have
$$
\begin{array}{l} \lim  _ {h \rightarrow 0} \frac {1}{h} \int_ {t} ^ {t + h} \sqrt {\lim  _ {n \rightarrow \infty} U _ {\lambda_ {n}} (t) - f (s)} d s = \sqrt {\lim  _ {n \rightarrow \infty} U _ {\lambda_ {n}} (t) - f (t)} \\ = \sqrt {\lim _ {n \to \infty} \left\| \dot {X} _ {\lambda_ {n}} (t) \right\| ^ {2} + \lim _ {n \to \infty} f _ {n} (t) - f (t)} = \lim _ {n \to \infty} \left\| \dot {X} _ {\lambda_ {n}} (t) \right\|. \\ \end{array}
$$
Finally as  $t \in S$ ,  $\dot{X}(t) = \lim_{h \to 0} \frac{X(t + h) - X(t)}{h}$  exists. Therefore
$$
\begin{array}{l} \left\| \dot {X} (t) \right\| = \lim  _ {h \rightarrow 0 +} \frac {\| X (t + h) - X (t) \|}{h} \\ \leq \lim  _ {h \rightarrow 0 +} \frac {1}{h} \int_ {t} ^ {t + h} \sqrt {\lim  _ {n \rightarrow \infty} U _ {\lambda_ {n}} (t) - f (s)} d s = \lim  _ {n \rightarrow \infty} \sup  _ {n \rightarrow \infty} \left\| \dot {X} _ {\lambda_ {n}} (t) \right\|, \\ \end{array}
$$
we conclude the desired result.
(2)  $\lim \sup_{n\to \infty}V_{\lambda_n}(t)\leq \lim \sup_{n\to \infty}V_{\lambda_n}(0)$
As  $\mathbb{A}_{\lambda_n}$  is Lipschitz continuous, from Appendix E.3.1 we know  $V_{\lambda_n}$  is nonincreasing, and thus
$$
V _ {\lambda_ {n}} (t) \leq \lim  _ {\epsilon \rightarrow 0 +} V _ {\lambda_ {n}} (\epsilon) = V _ {\lambda_ {n}} (0).
$$
Taking limsup both sides we get the desired result.
(3)  $\lim \sup_{n\to \infty}V_{\lambda_n}(0) = \lim_{n\to \infty}V_{\lambda_n}(0)$
From (32) and (ii) of Lemma B.10, we know
$$
\lim  _ {n \to \infty} V _ {\lambda_ {n}} ^ {0} (0) = \left\{ \begin{array}{l l} 0 & \text {i f} p \geq 1 \\ \frac {\| m (\mathbb {A} (X _ {0})) \| ^ {2}}{2} & \text {i f} 0 <   p <   1. \end{array} \right.
$$
And applying (36) we have
$$
\lim  _ {n \rightarrow \infty} V _ {\lambda_ {n}} (0) = \lim  _ {n \rightarrow \infty} V _ {\lambda_ {n}} ^ {0} (0) - \int_ {0} ^ {t _ {0}} \frac {d}{d s} \left(\frac {C (s) ^ {2} \dot {\beta} (s)}{2}\right) \| X (s) - X _ {0} \| ^ {2} d s.
$$
As the limit  $\lim_{n\to \infty}V_{\lambda_n}(0)$  exists, the limsup concides with the limit.
E.3.4 Proof for general maximal monotone  $\mathbb{A}$
First, we show  $V(t) \leq \lim_{n \to \infty} V_{\lambda_n}(0)$  holds for  $t \in [0, \infty)$ . Next, we show  $\lim_{n \to \infty} V_{\lambda_n}(0) = \lim_{t \to 0+} V(t)$ . Then we have
$$
V (t) \leq \lim  _ {n \rightarrow \infty} V _ {\lambda_ {n}} (0) = \lim  _ {t \rightarrow 0 +} V (t) = V (0),
$$
which is our desired result.
(i)  $V(t)\leq \lim_{n\to \infty}V_{\lambda_n}(0)$  holds for  $t\in [0,\infty)$
Take  $t \in [0, \infty)$ . We know the inequality is true when  $t \in S$ , thus assume  $t \notin S$ . Then from Lemma E.1, we know there is a sequence  $\{t_k\}_{k \in \mathbb{N}}$  such that  $t_k \in S$ ,  $\lim_{k \to \infty} t_k = t$  and  $\tilde{\mathbb{A}}(X(t_k))$  converges to  $\tilde{\mathbb{A}}(X(t))$ . As  $t_k \in S$ ,  $V(t_k) \leq \lim_{n \to \infty} V_{\lambda_n}(0)$  holds. Since  $\lim_{k \to \infty} V(t_k) = V(t)$ , taking limit  $k \to \infty$  to the inequality we get the desired result.
(ii)  $\lim_{n\to \infty}V_{\lambda_n}(0) = \lim_{t\to 0 + }V(t)$