SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Therefore we have  $\hat{f} (\xi) = 0$  for all  $\xi \in \mathbb{R}$
Now  $\hat{f} \equiv 0$  is clearly  $L^1(\mathbb{R},\mathbb{R})$ , we can apply Fourier inversion formula and conclude  $f \equiv 0$  almost everywhere.
However,  $f(s) = C(s)\beta (s)$  is not zero almost everywhere since  $\beta (s)$  is not constantly zero. Thus a contradiction, we get the desired result.
F.3 Proof of Theorem 4.4
F.3.1 Proof for the case  $p = 1, \gamma \geq 1$
Recall from (39), we know
$$
X (t) = \gamma e ^ {- t A} t ^ {- \gamma} \int_ {0} ^ {t} e ^ {s A} s ^ {\gamma - 1} d s X _ {0}.
$$
Without loss of generality, we consider the case  $X_0 = (1,0)^T$ .
(i)  $\gamma = 1$
Plugging  $\gamma = 1$  gives
$$
X (t) = \frac {e ^ {- t A}}{t} A ^ {- 1} \left[ e ^ {s A} \right] _ {0} ^ {t} X _ {0} = \frac {1}{t} A ^ {- 1} \left(I - e ^ {- t A}\right) (1, 0) ^ {T} = \frac {1}{t} A ^ {- 1} \left(1 - \cos t, \sin t\right) ^ {T}
$$
$$
\lim  _ {t \rightarrow \infty} \| t A (X (t)) \| = \lim  _ {t \rightarrow \infty} \left\| (1 - \cos t, \sin t) ^ {T} \right\| \neq 0.
$$
(ii)  $\gamma > 1$
We will show  $\lim_{t\to \infty}\| tA(X(t))\| = \gamma$ . With change of variable  $s = tv$  and integration by parts we have
$$
\begin{array}{l} \left(\frac {1}{\gamma}\right) t A (X (t)) = e ^ {- t A} A t ^ {- (\gamma - 1)} \int_ {0} ^ {t} e ^ {s A} s ^ {\gamma - 1} d s X _ {0} \\ = e ^ {- t A} t A \int_ {0} ^ {1} e ^ {t v A} v ^ {\gamma - 1} d v X _ {0} \\ = e ^ {- t A} \left[ e ^ {t v A} v ^ {\gamma - 1} \right] _ {0} ^ {1} X _ {0} - (\gamma - 1) e ^ {- t A} \int_ {0} ^ {1} e ^ {t v A} v ^ {\gamma - 2} d v X _ {0} \\ = X _ {0} - (\gamma - 1) e ^ {- t A} \left(\int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v, \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v\right) ^ {T} \\ \end{array}
$$
By the way, from  $\gamma > 1$  we have  $v^{\gamma - 2} \in L^{1}[0,1]$ , by Riemann-Lebesgue lemma we have
$$
\lim _ {t \to \infty} \int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v = \lim _ {t \to \infty} \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v = 0.
$$
Observe as  $e^{-tA}$  is a rotation, we know
$$
\left\| e ^ {- t A} \left(\int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v, \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v\right) \right\| = \left\| \left(\int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v, \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v\right) \right\|.
$$
Taking limit  $t\to \infty$  we know right hand side converges to zero, we conclude
$$
\lim _ {t \to \infty} e ^ {- t A} \left(\int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v, \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v\right) = 0.
$$
$$
\begin{array}{l} \lim  _ {t \rightarrow \infty} \| t A (X (t)) \| = \left\| \gamma X _ {0} - \gamma (\gamma - 1) \lim  _ {t \rightarrow \infty} e ^ {- t A} \left(\int_ {0} ^ {1} \cos (t v) v ^ {\gamma - 2} d v, \int_ {0} ^ {1} \sin (t v) v ^ {\gamma - 2} d v\right) ^ {T} \right\| \\ = \gamma \| X _ {0} \| = \gamma \\ \end{array}
$$
F.3.2 Proof for the case  $p = 1, \gamma < 1$
Since  $A$ ,  $e^{tA}$  are invertible, we see
$$
\lim  _ {t \to \infty} t ^ {2 \gamma} \| A (X (t)) \| ^ {2} \neq 0 \Longleftrightarrow \lim  _ {t \to \infty} t ^ {\gamma} \| A (X (t)) \| \neq 0 \Longleftrightarrow \lim  _ {t \to \infty} t ^ {\gamma} \left\| \frac {1}{\gamma} e ^ {t A} X (t) \right\| \neq 0.
$$
Therefore it is enough to observe  $\frac{1}{\gamma} e^{tA} t^{\gamma} X(t)$ . Again for  $X_0 = (1, 0)^T$
$$
\begin{array}{l} \frac {1}{\gamma} e ^ {t A} t ^ {\gamma} X (t) = \int_ {0} ^ {t} e ^ {s A} s ^ {\gamma - 1} d s X _ {0} \\ = \left(\int_ {0} ^ {t} \cos (s) s ^ {\gamma - 1} d s, \int_ {0} ^ {t} \sin (s) s ^ {\gamma - 1} d s\right) ^ {T}. \\ \end{array}
$$
Since  $s^{\gamma - 1}$  decrease monotonically to zero, we may apply similar argument with alternative series test.
Define  $a_{n} = \int_{(n - 1)\pi}^{n\pi}\sin (s)s^{\gamma -1}ds$  and name  $S_{n} = \sum_{k = 1}^{n}a_{n}$ . Then for  $m\in \mathbb{N}, t > 2m\pi$  we see
$$
S _ {2 m} \leq \int_ {0} ^ {t} \sin (s) s ^ {\gamma - 1} d s \leq S _ {2 m - 1}.
$$
By the way  $S = \lim_{n\to \infty}S_n$  exists by alternative series test, thus we have
$$
\lim  _ {t \rightarrow \infty} \int_ {0} ^ {t} \sin (s) s ^ {\gamma - 1} d s = S
$$
by squeeze theorem. Note  $S \geq S_2 > 0$ . With similar argument, we can conclude  $\lim_{t \to \infty} \int_0^t \cos(s) s^{\gamma - 1} ds$  also exists.
Therefore we conclude  $\lim_{t\to \infty}\| t^{\gamma}X(t)\| \neq 0$  since
$$
\lim  _ {t \rightarrow \infty} \| t ^ {\gamma} X (t) \| = \lim  _ {t \rightarrow \infty} \gamma \left\|\left(\int_ {0} ^ {t} \cos (s) s ^ {\gamma - 1} d s, \int_ {0} ^ {t} \sin (s) s ^ {\gamma - 1} d s\right) ^ {T} \right\| \geq \gamma S > 0.
$$
F.3.3 Proof for the case  $0 < p < 1$
Recall from (9), we have
$$
X (t) = \frac {e ^ {- t A}}{C (t)} \left(\int_ {0} ^ {t} e ^ {s A} C (s) \beta (s) d s + C (0) I\right) X _ {0}.
$$
Our goal is to show
$$
\lim  _ {t \to \infty} t ^ {2 p} \| A (X (t)) \| ^ {2} \neq 0.
$$
We first observe, it is enough to show
$$
\lim  _ {t \rightarrow \infty} \frac {1}{\beta (t) C (t)} \int_ {0} ^ {t} C (s) \beta (s) \sin s d s \neq 0.
$$
This follows from below facts.
(1) Since  $\beta (t) = \frac{\gamma}{tp}$  and  $A$  is linear and invertible,
$$
\lim _ {t \to \infty} t ^ {2 p} \| A (X (t)) \| ^ {2} = \lim _ {t \to \infty} \frac {\gamma^ {2}}{\beta (t) ^ {2}} \| A (X (t)) \| ^ {2} \neq 0 \quad \Longleftrightarrow \quad \lim _ {t \to \infty} \left\| \frac {1}{\beta (t)} X (t) \right\| \neq 0.
$$
(2) Recall from (31), we have  $C(t) = e^{\frac{\gamma}{1 - p} t^{1 - p}}$  for  $\beta(t) = \frac{\gamma}{t^p}$ , so  $C(0) = \lim_{t \to 0+} C(t) = 1$  and  $\lim_{t \to \infty} \frac{1}{\beta(t)C(t)} = 0$ . Thus  $\lim_{t \to \infty} \frac{1}{\beta(t)C(t)} \|C(0)I\| = 0$ , second term is ignorable. Therefore the problem reduces to
$$
\lim  _ {t \rightarrow \infty} \left\| \frac {e ^ {- t A}}{\beta (t) C (t)} \left(\int_ {0} ^ {t} C (s) \beta (s) e ^ {- s A} d s\right) X _ {0} \right\| \neq 0.
$$
(3) Since  $e^{tA}$  is linear and invertible, problem reduces to
$$
\lim  _ {t \rightarrow \infty} \left\| \frac {1}{\beta (t) C (t)} \left(\int_ {0} ^ {t} C (s) \beta (s) e ^ {- s A} d s\right) X _ {0} \right\| \neq 0.
$$
(4) Again without loss of generality, let  $X_0 = (1,0)^T$ . Recalling  $e^{-sA} = \begin{pmatrix} \cos s & \sin s \\ -\sin s & \cos s \end{pmatrix}$ , focusing on one component we see it is sufficient to show
$$
\lim _ {t \rightarrow \infty} \frac {1}{\beta (t) C (t)} \int_ {0} ^ {t} C (s) \beta (s) \sin s d s \neq 0.
$$
Now we prove the statement. For convenience, name  $D(t) = \beta (t)C(t)$ . We want to show
$$
\lim _ {t \to \infty} \underbrace {\int_ {0} ^ {t} \frac {D (s)}{D (t)} \sin s d s} _ {= S (t)} \neq 0.
$$
We first observe
$$
\lim_{t\to \infty}\sup_{\delta \in [0,\delta ]}\left|\frac{D(t + \delta)}{D(t + \pi)} -1\right| = 0.
$$
To do so, we first show  $\lim_{t\to \infty}\frac{D(t + \delta)}{D(t + \pi)} = 1$  for  $\delta \in [0,\pi ]$ . Take  $\delta \in [0,\pi ]$ . Observe
$$
\frac {D (t + \delta)}{D (t + \pi)} = \frac {\beta (t + \delta) C (t + \delta)}{\beta (t + \pi) C (t + \pi)} = \left(\frac {t + \pi}{t + \delta}\right) ^ {p} \frac {e ^ {\frac {\gamma}{1 - p} (t + \delta) ^ {1 - p}}}{e ^ {\frac {\gamma}{1 - p} (t + \pi) ^ {1 - p}}} = \left(\frac {t + \pi}{t + \delta}\right) ^ {p} e ^ {\frac {\gamma}{1 - p} (t + \delta) ^ {1 - p} \left(1 - \left(\frac {t + \pi}{t + \delta}\right) ^ {1 - p}\right)}.
$$
Considering L'ospital's rule for the exponent, we see
$$
\lim  _ {t \rightarrow \infty} (t + \delta) ^ {1 - p} \left(1 - \left(\frac {t + \pi}{t + \delta}\right) ^ {1 - p}\right) = \lim  _ {t \rightarrow \infty} \frac {1 - \left(\frac {t + \pi}{t + \delta}\right) ^ {1 - p}}{(t + \delta) ^ {p - 1}} = \lim  _ {t \rightarrow \infty} \frac {\frac {- \pi (p - 1) \left(\frac {t + \pi}{t + \delta}\right) ^ {- p}}{(t + \delta) ^ {2}}}{(p - 1) (t + \delta) ^ {p - 2}} = \lim  _ {t \rightarrow \infty} \frac {- \pi}{(t + \pi) ^ {p}} = 0.
$$
As the exponent reaches to zero, we conclude  $\lim_{t\to \infty}\frac{D(t + \delta)}{D(t + \pi)} = 1$
$$
\dot {D} (t) = C (t) \left(\beta (t) ^ {2} + \dot {\beta} (t)\right) = e ^ {\frac {\gamma}{1 - p} t ^ {1 - p}} \gamma t ^ {- 2 p - 1} (\gamma t - p t ^ {p}),
$$
we see  $D(t)$  is nondecreasing for  $t \geq \left(\frac{\gamma}{p}\right)^{\frac{1}{p - 1}}$  since  $0 < p < 1$ . Therefore for  $t \geq \left(\frac{\gamma}{p}\right)^{\frac{1}{p - 1}}$  we have  $\sup_{\delta \in [0, \delta]} \left|\frac{D(t + \delta)}{D(t + \pi)} - 1\right| = \max \left\{\left|\frac{D(t)}{D(t + \pi)} - 1\right|, \left|\frac{D(t + \delta)}{D(t + \pi)} - 1\right|\right\}$ , so it reaches zero as  $t \to \infty$ .
Now we prove desired statement. Proof by contradiction. Suppose  $\lim_{t\to \infty}S(t) = 0$ . Then for  $0 < \epsilon < \frac{1}{2}$ , there is  $T_{1} > 0$  such that  $t > T_{1}$  implies  $|S(t)| < \epsilon$ . Now for  $t > T_{1}$ , observe
$$
\begin{array}{l} 2 \epsilon > | S (t + \pi) - S (t) | = \left| \int_ {0} ^ {t + \pi} \frac {D (s)}{D (t + \pi)} \sin s d s - \int_ {0} ^ {t} \frac {D (s)}{D (t)} \sin s d s \right| \\ = \left| \frac {1}{D (t + \pi)} \left(\int_ {0} ^ {t + \pi} D (s) \sin s d s - \int_ {0} ^ {t} D (s) \sin s d s\right) + \left(\frac {1}{D (t + \pi)} - \frac {1}{D (t)}\right) \int_ {0} ^ {t} D (s) \sin s d s \right| \\ \geq \left| \int_ {t} ^ {t + \pi} \frac {D (s)}{D (t + \pi)} \sin s d s \right| - \left| \left(\frac {D (t)}{D (t + \pi)} - 1\right) \int_ {0} ^ {t} \frac {D (s)}{D (t)} \sin s d s \right|. \\ \end{array}
$$
On the other hand, there is  $T_{2}$  such that  $t > T_{2}$  implies  $\sup_{\delta \in [0,\pi ]}\left|\frac{D(t + \delta)}{D(t + \pi)} -1\right| < \frac{\epsilon}{2}$ . Now for  $t = 2n\pi >\max \{T_1,T_2\}$ ,
$$
2 \epsilon > \left| \int_ {2 n \pi} ^ {(2 n + 1) \pi} \frac {D (s)}{D (t + \pi)} \sin s d s \right| - \left| \frac {D (t)}{D (t + \pi)} - 1 \right| \times | S (t) | > \int_ {2 n \pi} ^ {(2 n + 1) \pi} (1 - \epsilon) | \sin s | d s - \epsilon = 2 - 2 \epsilon .
$$
This contradicts the fact  $\epsilon < \frac{1}{2}$ , we prove the assumption  $\lim_{t\to \infty}\int_0^t S(t) = 0$  is not true. Therefore  $\lim_{t\to \infty}\int_0^t S(t)\neq 0$ .
G Proof of convergence analysis for discrete counterpart
G.1 Correspondence between discrete method in Theorem 5.1 and continuous model (6)
To check the correspondence of the method with (6), we provide an informal derivation. Assume operator  $\mathbb{A}$  is continuous. Then we have  $y^{k} = x^{k + 1} + \mathbb{A}x^{k + 1}$ , by substituting  $y^{k}$  and  $y^{k - 1}$ , the method can be equivalently written as
$$
x ^ {k + 1} + \mathbb {A} x ^ {k + 1} = \frac {k ^ {p}}{k ^ {p} + \gamma} \left(x ^ {k} - \mathbb {A} x ^ {k}\right) + \frac {\gamma}{k ^ {p} + \gamma} x ^ {0}. \tag {40}
$$
This can be considered as a special case of below method, with  $h = 1$
$$
x ^ {k + 1} + h \mathbb {A} x ^ {k + 1} = \frac {k ^ {p}}{k ^ {p} + h ^ {1 - p} \gamma} (x ^ {k} - h \mathbb {A} x ^ {k}) + \frac {h ^ {1 - p} \gamma}{k ^ {p} + h ^ {1 - p} \gamma} x ^ {0}.
$$
Dividing by  $h$  both sides and reorganizing, we have
$$
\frac {x ^ {k + 1} - x ^ {k}}{h} = - \mathbb {A} x ^ {k + 1} - \frac {h ^ {p} k ^ {p}}{h ^ {p} k ^ {p} + h \gamma} \mathbb {A} x ^ {k} - \frac {\gamma}{h ^ {p} k ^ {p} + h \gamma} \left(x ^ {k} - x ^ {0}\right).
$$
Identifying  $hk = t$ ,  $x^{k} = X(t)$  and taking  $h \to 0$ , we have
$$
\dot {X} (t) = - 2 \mathbb {A} (X (t)) - \frac {\gamma}{t ^ {p}} (X - X _ {0}).
$$
As monotonicity is preserved for scalar multiple, rescaling  $2\mathbb{A}$  to  $\mathbb{A}$  does not change the class of operators that the ODE covers. For notation simplicity, replacing  $2\mathbb{A}$  to  $\mathbb{A}$  we have
$$
\dot {X} (t) = - \mathbb {A} (X (t)) - \frac {\gamma}{t ^ {p}} (X - X _ {0}).
$$
G.2 Proof of boundedness of  $\| x^k\|$
While proving Theorem 5.1, we need a upper bound for  $\| x^k\|$ . Therefore we first prove following lemma.
Lemma G.1. Let  $\mathbb{A}$  be a maximal monotone operator. Consider a method
$$
x ^ {k} = \mathbb {J} _ {\mathbf {A}} y ^ {k - 1}
$$
for  $k = 1,2,\ldots$ , with sequence  $\{\beta_k\}_{k\in \mathbb{N}}, 0 \leq \beta_k \leq 1$  and initial condition  $y^0 = x^0 \in \mathbb{R}^n$ . Then following holds
$$
\left\| x ^ {k} - x ^ {\star} \right\| \leq \left\| x ^ {0} - x ^ {\star} \right\|.
$$
for  $x^{\star}\in \mathrm{Zer}\mathbb{A}$ . And so,  $\| x^{k} - x^{0}\| \leq 2\| x^{0} - x^{\star}\|$
Proof. Recall from Lemma D.2, we know for  $\mathbb{T} = \mathbb{R}_{\mathrm{A}} = 2\mathbb{J}_{\mathrm{A}} - \mathbb{I}$  given method is equivalent to below method
$$
y ^ {k + 1} = \beta_ {k} y ^ {0} + (1 - \beta_ {k}) \mathbb {T} y ^ {k}.
$$
Note that  $x^{\star} \in \operatorname{Zer} \mathbb{A} \Leftrightarrow 0 \in \mathbb{A}x^{\star} \Leftrightarrow x^{\star} = \mathbb{J}_{\mathbb{A}}x^{\star} \Leftrightarrow x^{\star} = \mathbb{T}x^{\star} \Leftrightarrow x^{\star} \in \operatorname{Fix} \mathbb{T}$ . We first prove  $\left\| y^{k} - x^{\star}\right\| \leq \left\| y^{0} - x^{\star}\right\|$  by induction. The statement is trivially true when  $k = 0$ . Now suppose the statement is true for  $k \in \mathbb{N}$ . Then
$$
\begin{array}{l} \left\| y ^ {k + 1} - x ^ {\star} \right\| = \left\| \beta_ {k} \left(y ^ {0} - x ^ {\star}\right) + \left(1 - \beta_ {k}\right) \left(\mathbb {T} y ^ {k} - x ^ {\star}\right) \right\| \\ \leq \beta_ {k} \left\| y ^ {0} - x ^ {\star} \right\| + (1 - \beta_ {k}) \left\| \mathbf {T} y ^ {k} - x ^ {\star} \right\| \\ \leq \beta_ {k} \| y ^ {0} - x ^ {\star} \| + (1 - \beta_ {k}) \| y ^ {k} - x ^ {\star} \| \\ \leq \beta_ {k} \| y ^ {0} - x ^ {\star} \| + (1 - \beta_ {k}) \| y ^ {0} - x ^ {\star} \| = \| y ^ {0} - x ^ {\star} \|. \\ \end{array}
$$
The first inequality is just triangular inequality, the second inequality comes from the fact  $x^{\star} \in \operatorname{Fix}\mathbb{T}$  and  $\mathbb{T}$  is nonexpansive, and the last inequality is from induction hypothesis. By induction, we have  $\| y^{k} - x^{\star} \| \leq \| y^{0} - x^{\star} \|$  for  $k = 0, 1, \ldots$ .
Define  $\tilde{\mathbb{A}} x^{k} = y^{k - 1} - x^{k}$ , then  $\tilde{\mathbb{A}} x^{k}\in \mathbb{A}x^{k}$  since  $\mathbb{J}_{\mathbb{A}}y^{k - 1} = x^k$ . Observe for  $k = 1,2,\ldots$ , from monotone inequality we have
$$
\left\| y ^ {k - 1} - x ^ {\star} \right\| ^ {2} = \left\| \tilde {\mathbb {A}} x ^ {k} + (x ^ {k} - x ^ {\star}) \right\| ^ {2} = \left\| \tilde {\mathbb {A}} x ^ {k} \right\| ^ {2} + 2 \left\langle \tilde {\mathbb {A}} x ^ {k}, x ^ {k} - x ^ {\star} \right\rangle + \left\| x ^ {k} - x ^ {\star} \right\| ^ {2} \geq \left\| x ^ {k} - x ^ {\star} \right\| ^ {2}.
$$
Therefore we conclude
$$
\left\| x ^ {k} - x ^ {\star} \right\| \leq \left\| y ^ {k - 1} - x ^ {\star} \right\| \leq \left\| y ^ {0} - x ^ {\star} \right\| = \left\| x ^ {0} - x ^ {\star} \right\|.
$$
$$
\left\| x ^ {k} - x ^ {0} \right\| \leq \left\| x ^ {k} - x ^ {\star} \right\| + \left\| x ^ {\star} - x ^ {0} \right\| \leq 2 \left\| x ^ {0} - x ^ {\star} \right\|.
$$
G.3 Proof of Theorem 5.1
The outline of the proofs originate from continuous proofs. To simplify calculations, instead of directly deriving discrete counterpart of Proposition 3.2, we consider rescaled conservation law for each cases. By considering dilated coordinate  $W_{1} = X - X_{0}$  for  $p > 1$ ,  $W_{2} = t^{p}\left(X - X_{0}\right)$  for  $0 < p < 1$ ,  $W_{3} = t\left(X - X_{0}\right)$  for  $p = 1$ ,  $\gamma \geq 1$  and  $W_{4} = t^{\gamma}\left(X - X_{0}\right)$  for  $p = 1$ ,  $0 < \gamma < 1$ , we obtain below conservation laws.
$$
\begin{array}{l} E _ {1} \equiv \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + \frac {2 \gamma}{t ^ {p}} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\frac {\gamma^ {2}}{t ^ {2 p}} + \frac {\gamma p}{t ^ {p + 1}}\right) \| X (t) - X _ {0} \| ^ {2} \\ + \int_ {0} ^ {t} \left\langle \frac {d}{d s} \tilde {\mathbf {A}} (X (s)), \dot {X} (s) \right\rangle d s + \int_ {0} ^ {t} \frac {\gamma}{s ^ {p}} \left\| \dot {X} (s) + \frac {p}{s} (X (s) - X _ {0}) \right\| ^ {2} d s + \frac {1}{2} \int_ {0} ^ {t} \frac {\gamma p (p - 1)}{s ^ {p + 2}} \| X (s) - X _ {0} \| ^ {2} d s \\ \end{array}
$$
$$
\begin{array}{l} E _ {2} \equiv t ^ {2 p} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \gamma t ^ {p} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\gamma^ {2} - \gamma p t ^ {p - 1}\right) \left\| X (t) - X _ {0} \right\| ^ {2} \\ + \int_ {0} ^ {t} 2 s ^ {2 p} \left\langle \frac {d}{d s} \tilde {\mathbb {A}} (X (s)), \dot {X} (s) \right\rangle d s + \int_ {0} ^ {t} 2 s ^ {p} (\gamma - p s ^ {p - 1}) \left\| \dot {X} (s) \right\| ^ {2} d s + \int_ {0} ^ {t} \gamma p (p - 1) s ^ {p - 2} \| X (s) - X _ {0} \| ^ {2} d s \\ \end{array}
$$
$$
\begin{array}{l} E _ {3} \equiv t ^ {2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \gamma t \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \gamma (\gamma - 1) \left\| X (t) - X _ {0} \right\| ^ {2} \\ + \int_ {t _ {0}} ^ {t} 2 s ^ {2} \left\langle \frac {d}{d s} \tilde {\mathbb {A}} (X (s)), \dot {X} (s) \right\rangle d s + \int_ {t _ {0}} ^ {t} 2 s (\gamma - 1) \| \dot {X} (s) \| ^ {2} d s \\ \end{array}
$$
$$
\begin{array}{l} E _ {4} \equiv t ^ {2 \gamma} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \gamma t ^ {2 \gamma - 1} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \gamma (\gamma - 1) t ^ {2 \gamma - 2} \left\| X (t) - X _ {0} \right\| ^ {2} \\ + \int_ {0} ^ {t} 2 s ^ {2 \gamma} \left\langle \frac {d}{d s} \tilde {\mathsf {A}} (X (s)), \dot {X} (s) \right\rangle d s + \int_ {0} ^ {t} 2 \gamma (\gamma - 1) s ^ {2 \gamma - 3} \left\| X (s) - X _ {0} \right\| ^ {2} d s. \\ \end{array}
$$
Lyapunov style proof can be obtained by considering below functions
$$
\begin{array}{l} U _ {1} (t) = \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + \frac {2 \gamma}{t ^ {p}} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\frac {\gamma^ {2}}{t ^ {2 p}} + \frac {\gamma p}{t ^ {p + 1}}\right) \| X (t) - X _ {0} \| ^ {2} \\ U _ {2} (t) = t ^ {2 p} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \gamma t ^ {p} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \left(\gamma^ {2} - \gamma p t ^ {p - 1}\right) \left\| X (t) - X _ {0} \right\| ^ {2} \\ U _ {3} (t) = t ^ {2} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + 2 \gamma t \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \gamma (\gamma - 1) \| X (t) - X _ {0} \| ^ {2} \\ U _ {4} (t) = t ^ {2 \gamma} \left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2} + \gamma t ^ {2 \gamma - 1} \left\langle \tilde {\mathbf {A}} (X (t)), X (t) - X _ {0} \right\rangle + \gamma (\gamma - 1) t ^ {2 \gamma - 2} \left\| X (t) - X _ {0} \right\| ^ {2}. \\ \end{array}
$$
The main blocks of the calculations correspond to continuous cases, but there are some 'errors' in terms of  $\tilde{\mathbf{A}} x^k$  and  $x^{k} - x^{0}$  occur due to discretization. The proofs are done by showing these 'errors' don't effect to the conclusions.
(0) Preparation.
For all cases, we will consider functions of the form
$$
U ^ {k} = a _ {k} \left\| \tilde {\mathbf {A}} x ^ {k} \right\| ^ {2} + b _ {k} \left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \right\rangle + c _ {k} \left\| x ^ {k} - x ^ {0} \right\| ^ {2}, \tag {41}
$$
and consider  $U^{k + 1} - U^k$ . As similar calculations will be repeated, we first organize the repeating calculations. That is, we prove following equality is true.
$$
\begin{array}{l} U ^ {k + 1} - U ^ {k} + \lambda_ {k} \left\langle \tilde {\mathbb {A}} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle - \tau_ {k} \left\langle \tilde {\mathbb {A}} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle \tag {42} \\ = \left(a _ {k + 1} - \lambda_ {k} \frac {k ^ {p} + \gamma}{k ^ {p}}\right) \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} + \left(\lambda_ {k} \frac {k ^ {p}}{k ^ {p} + \gamma} - a _ {k}\right) \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} \\ + \left(b _ {k + 1} - \lambda_ {k} \frac {\gamma}{k ^ {p}} - \tau_ {k} - c _ {k + 1} \frac {k ^ {p} + \gamma}{k ^ {p}}\right) \langle \tilde {\mathbb {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \rangle + (\tau_ {k} - c _ {k + 1}) \left\langle \tilde {\mathbb {A}} x ^ {k + 1}, x ^ {k} - x ^ {0} \right\rangle \\ + \left(\tau_ {k} - c _ {k + 1}\right) \left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {0} \right\rangle + \left(\lambda_ {k} \frac {\gamma}{k ^ {p} + \gamma} - b _ {k} - \tau_ {k} - c _ {k + 1} \frac {k ^ {p}}{k ^ {p} + \gamma}\right) \left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \right\rangle \\ - c _ {k + 1} \frac {\gamma}{k ^ {p}} \left\| x ^ {k + 1} - x ^ {0} \right\| ^ {2} - c _ {k + 1} \frac {\gamma}{k ^ {p} + \gamma} \left\| x ^ {k} - x ^ {0} \right\| ^ {2} + (c _ {k + 1} - c _ {k}) \left\| x ^ {k} - x ^ {0} \right\| ^ {2}. \\ \end{array}
$$
Observe, this can be shown by checking below equalities are true.
$$
\begin{array}{l} \left\langle \tilde {\mathbb {A}} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle \tag {43} \\ = - \frac {k ^ {p} + \gamma}{k ^ {p}} \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} - \frac {\gamma}{k ^ {p}} \langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \rangle + \frac {k ^ {p}}{k ^ {p} + \gamma} \| \tilde {\mathbf {A}} x ^ {k} \| ^ {2} + \frac {\gamma}{k ^ {p} + \gamma} \langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \rangle \\ \end{array}
$$
$$
\begin{array}{l} \left\langle \tilde {\mathbb {A}} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle \tag {44} \\ = \left\langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \right\rangle + \left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \right\rangle - \left(\left\langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k} - x ^ {0} \right\rangle + \left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {0} \right\rangle\right) \\ \end{array}
$$
$$
\begin{array}{l} c _ {k + 1} \left\| x ^ {k + 1} - x ^ {0} \right\| ^ {2} - c _ {k} \left\| x ^ {k} - x ^ {0} \right\| ^ {2} \tag {45} \\ = - c _ {k + 1} \frac {k ^ {p} + \gamma}{k ^ {p}} \left\langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \right\rangle - c _ {k + 1} \left\langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k} - x ^ {0} \right\rangle \\ - c _ {k + 1} \left\langle \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {0} \right\rangle - c _ {k + 1} \frac {k ^ {p}}{k ^ {p} + \gamma} \left\langle \tilde {\mathbb {A}} x ^ {k}, x ^ {k} - x ^ {0} \right\rangle \\ - c _ {k + 1} \frac {\gamma}{k ^ {p}} \left\| x ^ {k + 1} - x ^ {0} \right\| ^ {2} - c _ {k + 1} \frac {\gamma}{k ^ {p} + \gamma} \left\| x ^ {k} - x ^ {0} \right\| ^ {2} + (c _ {k + 1} - c _ {k}) \left\| x ^ {k} - x ^ {0} \right\| ^ {2}. \\ \end{array}
$$
- Proof for (43)
Recall, the method was defined as
$$
x ^ {k + 1} = \mathbb {J} _ {\mathbb {A}} y ^ {k}.
$$
By considering  $x^{k + 1} + \tilde{\mathbf{A}} x^{k + 1} = y^k$ , substituting  $y^{k}$  and  $y^{k - 1}$  we can rewrite the method as
$$
x ^ {k + 1} + \tilde {\mathbf {A}} x ^ {k + 1} = \left(1 - \frac {\gamma}{k ^ {p} + \gamma}\right) \left(x ^ {k} - \tilde {\mathbf {A}} x ^ {k}\right) + \frac {\gamma}{k ^ {p} + \gamma} x ^ {0}. \tag {46}
$$
By multiplying  $k^p + \gamma$  to both sides and reorganizing we have,
$$
(k ^ {p} + \gamma) \left\{\left(x ^ {k + 1} - x ^ {0}\right) + \tilde {\mathbf {A}} x ^ {k + 1} \right\} = k ^ {p} \left\{\left(x ^ {k} - x ^ {0}\right) - \tilde {\mathbf {A}} x ^ {k} \right\}.
$$
By subtracting both sides by  $k^p (x^{k + 1} - x^0)$  and  $(k^p + \gamma)(x^k - x^0)$  respectively, above equation can be rewritten as
$$
\begin{array}{l} - k ^ {p} \left(\tilde {\mathbf {A}} x ^ {k} + \left(x ^ {k + 1} - x ^ {k}\right)\right) = \left(k ^ {p} + \gamma\right) \tilde {\mathbf {A}} x ^ {k + 1} + \gamma \left(x ^ {k + 1} - x ^ {0}\right) \\ - \left(k ^ {p} + \gamma\right) \left(\tilde {\mathbf {A}} x ^ {k + 1} + \left(x ^ {k + 1} - x ^ {k}\right)\right) = k ^ {p} \tilde {\mathbf {A}} x ^ {k} + \gamma \left(x ^ {k} - x ^ {0}\right). \\ \end{array}
$$
From above, we get the following