SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Since  $\mathbb{A}$  is monotone we have  $\langle \tilde{\mathbb{A}} x^{k + 1} - \tilde{\mathbb{A}} x^k, x^{k + 1} - x^k\rangle \geq 0$ , we see that the right-hand side of (56) is greater or equal to  $0$  if  $1 - \beta_{k} \geq 0$ . Thus it remains to show  $1 - \beta_{k} \geq 0$ . As the index is quite confusing, we provide it as a lemma to avoid confusion while proceeding the proof.
Lemma I.3. If  $\tilde{\mathbb{A}} x^k\neq 0$  for all  $k\geq 1$  , then
$$
\left\langle \tilde {\mathbf {A}} x ^ {k}, x ^ {k} - x ^ {0} \right\rangle <   0, \quad \beta_ {k} \in (0, 1), \quad \Phi^ {k + 1} \leq 0
$$
for all  $k\geq 1$
Proof. Proof by induction. From  $x^{1} = \mathbb{J}_{\mathbb{A}}y^{0} = \mathbb{J}_{\mathbb{A}}x^{0}$ , we have  $x^{1} + \tilde{\mathbb{A}} x^{1} = x^{0}$  and so  $x^{1} - x^{0} = -\tilde{\mathbb{A}} x^{1}$ . Applying these facts we have
$$
\begin{array}{l} \langle \tilde {\mathbf {A}} x ^ {1}, x ^ {1} - x ^ {0} \rangle = - \| \tilde {\mathbf {A}} x ^ {1} \| ^ {2} <   0, \\ \beta_ {1} = \frac {\| \tilde {\mathbf {A}} x ^ {1} \| ^ {2}}{- \langle \tilde {\mathbf {A}} x ^ {1} , x ^ {1} - x ^ {0} \rangle + \| \tilde {\mathbf {A}} x ^ {1} \| ^ {2}} = \frac {1}{2} \in (0, 1). \\ \end{array}
$$
As  $1 - \beta_{1} > 0$ , from (56) we have
$$
(1 - \beta_ {1}) \tilde {\Phi} ^ {1} - \Phi^ {2} = (1 - \beta_ {1}) \langle \tilde {\mathbf {A}} x ^ {2} - \tilde {\mathbf {A}} x ^ {1}, x ^ {2} - x ^ {1} \rangle \geq 0.
$$
As  $\tilde{\Phi}^1 = 0$  by definition, we have  $\Phi^2 \leq 0$ . Therefore the statement is true for  $k = 1$ .
Now suppose the statements are true for  $k$ . By induction hypothesis, we know
$$
0 \geq \Phi^ {k + 1} = \| \tilde {\mathbb {A}} x ^ {k + 1} \| ^ {2} + \beta_ {k} \langle \tilde {\mathbb {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \rangle .
$$
As  $\beta_{k} > 0$  from induction hypothesis and  $\tilde{\mathbb{A}} x^{k + 1}\neq 0$  by assumption, reorganizing  $\Phi^k\leq 0$  we have
$$
\langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {0} \rangle \leq - \frac {\| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2}}{\beta_ {k}} <   0.
$$
And therefore
$$
\beta_ {k + 1} = \underbrace {\frac {\| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2}}{- \langle \tilde {\mathbf {A}} x ^ {k + 1} , x ^ {k + 1} - x ^ {0} \rangle} + \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2}} _ {> 0} \in (0, 1).
$$
Since  $1 - \beta_{k + 1} > 0$ , by (56) we have
$$
(1 - \beta_ {k + 1}) \tilde {\Phi} ^ {k + 1} - \Phi^ {k + 2} = (1 - \beta_ {k + 1}) \langle \tilde {\mathbf {A}} x ^ {k + 2} - \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 2} - x ^ {k + 1} \rangle \geq 0.
$$
Since  $\tilde{\Phi}^{k + 1} = 0$  by definition, we have  $\Phi^{k + 2}\leq 0$ . Therefore the statements are true for  $k + 1$ , by induction, we get the desired result.
Suppose  $\| \tilde{\mathbb{A}} x^k\| ^2\neq 0$  for all  $k\geq 1$ . From the lemma we know  $\Phi^{k + 1}\leq 0$  for  $k\geq 1$ , so with the same argument of (8) we have for  $x^{\star}\in \mathrm{Zer}\mathbb{A}$
$$
\begin{array}{l} 0 \geq \Phi^ {k + 1} = \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} + \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {k + 1} - x ^ {\star} \rangle - \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {0} - x ^ {\star} \rangle \\ \geq \| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} - \beta_ {k} \langle \tilde {\mathbf {A}} x ^ {k + 1}, x ^ {0} - x ^ {\star} \rangle \\ \geq | \tilde {\mathbb {A}} x ^ {k + 1} | ^ {2} - \left(\frac {1}{2} \| \tilde {\mathbb {A}} x ^ {k + 1} \| ^ {2} + \frac {\beta_ {k} ^ {2}}{2} \| x ^ {0} - x ^ {\star} \| ^ {2}\right) = \frac {1}{2} \| \tilde {\mathbb {A}} x ^ {k + 1} \| ^ {2} - \frac {\beta_ {k} ^ {2}}{2} \| x ^ {0} - x ^ {\star} \| ^ {2}. \\ \end{array}
$$
Organizing, we get
$$
\| \tilde {\mathbf {A}} x ^ {k + 1} \| ^ {2} \leq \beta_ {k} ^ {2} \| x ^ {0} - x ^ {\star} \| ^ {2}.
$$
Now we show the upper bound of  $\beta_{k}$ . Observe from (56) and the fact  $\bar{\Phi} = 0$ , we have
$$
\begin{array}{l} (1 - \beta_ {k}) \left\langle \tilde {\mathbb {A}} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle = (1 - \beta_ {k}) \tilde {\Phi} ^ {k} - \Phi^ {k + 1} \\ = \frac {\beta_ {k}}{\beta_ {k + 1}} \tilde {\Phi} ^ {k + 1} - \Phi^ {k + 1} = \left(\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1\right) \left\| \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}. \\ \end{array}
$$
As  $1 - \beta_{k} \neq 0$  for  $k \geq 1$  by Lemma I.3, dividing both sides by  $1 - \beta_{k}$ , we get the discrete counterpart of (52),
$$
\left\langle \tilde {\mathbf {A}} x ^ {k + 1} - \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle = \frac {\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1}{1 - \beta_ {k}} \left\| \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}. \tag {57}
$$
(i) When  $\mathbb{A}$  is monotone.
From (57) and monotonicity of  $\mathbb{A}$  we have
$$
0 \leq \left\langle \tilde {\mathbf {A}} x ^ {k + 1} - \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle = \frac {\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1}{1 - \beta_ {k}} \left\| \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}.
$$
From Lemma I.3 we have  $1 - \beta_{k} > 0$ , therefore
$$
\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1 \geq 0 \quad \Longrightarrow \quad \frac {1}{\beta_ {k}} + 1 \leq \frac {1}{\beta_ {k + 1}}.
$$
Summing up, as  $\beta_{1} = \frac{1}{2}$  and  $\frac{1}{\beta_k} > 0$  we have
$$
\frac {1}{\beta_ {1}} + (k - 1) = k + 1 \leq \frac {1}{\beta_ {k}} \quad \Longrightarrow \quad \beta_ {k} \leq \frac {1}{k + 1}.
$$
(ii) When  $\mathbb{A}$  is  $\mu$ -strongly monotone.
Since  $\mathbb{A}$  is  $\mu$ -strongly monotone, from (57) we have
$$
\mu \left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2} \leq \left\langle \tilde {\mathbf {A}} x ^ {k + 1} - \tilde {\mathbf {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle = \frac {\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1}{1 - \beta_ {k}} \left\| \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}.
$$
Define  $r_k = \frac{\left\|x^{k + 1} - x^k\right\|^2}{\left\|\tilde{\mathbf{A}}x^{k + 1}\right\|^2}$  for  $k = 0,1,\ldots$  Dividing both sides by  $\left\| \tilde{\mathbf{A}} x^{k + 1}\right\|^2$  and organizing, we have
$$
r _ {k} \mu \leq \frac {\frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1}{1 - \beta_ {k}} \quad \Longrightarrow \quad r _ {k} \mu (1 - \beta_ {k}) \leq \frac {\beta_ {k}}{\beta_ {k + 1}} - \beta_ {k} - 1 = \beta_ {k} \left(\frac {1}{\beta_ {k + 1}} - 1\right) - \beta_ {k} - (1 - \beta_ {k}).
$$
Dividing both sides by  $\beta_{k}$  and reorganizing, we get a recursive inequality for  $\frac{1}{\beta_k} - 1$
$$
\left(1 + r _ {k} \mu\right) \left(\frac {1}{\beta_ {k}} - 1\right) + 1 \leq \frac {1}{\beta_ {k + 1}} - 1. \tag {58}
$$
We now prove an upper bound of  $\beta_{k}$  from above inequality.
Lemma I.4. Suppose  $\mathbb{A}$  be a  $\mu$ -strongly monotone operator. Let  $\beta_{k}$  be a sequence defined as Theorem 7.2 and let  $r_k = \frac{\left\|x^{k + 1} - x^k\right\|^2}{\left\|\tilde{\mathbf{A}}x^{k + 1}\right\|^2}$  for  $k = 0,1,\ldots$ . Then following holds for  $k = 1,2,\ldots$
$$
\beta_ {k} \leq \frac {1}{\sum_ {j = 1} ^ {k - 1} \prod_ {i = j} ^ {k - 1} (1 + r _ {i} \mu) + 2}.
$$
Proof. First observe, the statement is equivalent to
$$
\frac {1}{\beta_ {k}} - 1 \geq \sum_ {j = 1} ^ {k - 1} \prod_ {i = j} ^ {k - 1} (1 + r _ {i} \mu) + 1.
$$
The proof can be done by induction with (58).
When  $k = 1$ , recalling  $\beta_{1} = \frac{1}{2}$  from the proof of Lemma I.3, we can check the inequality is true.
Now suppose the inequality is true for  $k = m$ . Then from (58) we have
$$
\begin{array}{l} \frac {1}{\beta_ {m + 1}} - 1 \geq (1 + r _ {m} \mu) \left(\frac {1}{\beta_ {m}} - 1\right) + 1 \\ \geq \left(1 + r _ {m} \mu\right) \left(\sum_ {j = 1} ^ {m - 1} \prod_ {i = j} ^ {m - 1} \left(1 + r _ {i} \mu\right) + 1\right) + 1 \\ = \left(\sum_ {j = 1} ^ {m - 1} \prod_ {i = j} ^ {m} (1 + r _ {i} \mu) + (1 + r _ {m} \mu)\right) + 1 = \sum_ {j = 1} ^ {m} \prod_ {i = j} ^ {m} (1 + r _ {i} \mu) + 1. \\ \end{array}
$$
Therefore, we get the desired result.
Under the identification considered in Appendix I.2, the continuous counterpart of  $r_k$  is
$$
r _ {k} = \frac {\left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2}}{\left\| h \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}} = 4 \frac {\left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2}}{(2 h) ^ {2}} \frac {1}{\left\| \tilde {\mathbf {A}} x ^ {k + 1} \right\| ^ {2}} \quad \xrightarrow {h \to 0 +} \quad \frac {4 \left\| \dot {X} (t) \right\| ^ {2}}{\left\| \tilde {\mathbf {A}} (X (t)) \right\| ^ {2}},
$$
and is greater or equal to 1 by (54). We obtained an exponential convergence rate in continuous setup from this fact. In the same spirit, we can get an exponential convergence rate for discrete setup if there is a positive lower bound for  $r_k$ , we provide it as a corollary of Lemma I.4.
Corollary I.5. Consider the setup of Lemma I.4. Suppose there is  $r \geq 0$  such that  $r_k \geq r$  for  $k = 0, 1, \ldots$ . Then following is true for  $k = 1, 2, \ldots$ .
$$
\beta_ {k} \leq \frac {r \mu}{(1 + r \mu) ^ {k} - 1 + r \mu}.
$$
Proof. From  $r_k \geq r$  we have
$$
\begin{array}{l} \sum_ {j = 1} ^ {k - 1} \prod_ {i = j} ^ {k - 1} (1 + r _ {i} \mu) + 2 \geq \sum_ {j = 1} ^ {k - 1} \prod_ {i = j} ^ {k - 1} (1 + r \mu) + 2 \\ = \sum_ {j = 1} ^ {k - 1} (1 + r \mu) ^ {k - j} + 2 = \sum_ {l = 0} ^ {k - 1} (1 + r \mu) ^ {l} + 1 = \frac {(1 + r \mu) ^ {k} - 1 + r \mu}{r \mu}. \\ \end{array}
$$
Applying Lemma I.4, we get the desired result.
We now show  $r_k \geq \frac{1}{1 + L^2}$  holds when  $\mathbb{A}$  is furthermore  $L$ -Lipschitz continuous.
(iii) When  $\mathbb{A}$  is  $\mu$ -strongly monotone and  $L$ -Lipschitz continuous.
Recall from the proof of Lemma I.2, we know
$$
x ^ {k + 1} - x ^ {k} = - \mathbb {A} x ^ {k + 1} - \left(1 - \beta_ {k}\right) \mathbb {A} x ^ {k} - \beta_ {k} \left(x ^ {k} - x ^ {0}\right).
$$
Taking inner product with  $\tilde{\mathbb{A}} x^k$  both sides we have
$$
\left\langle \tilde {\mathbb {A}} x ^ {k}, x ^ {k + 1} - x ^ {k} \right\rangle = - \left\langle \mathbb {A} x ^ {k + 1}, \tilde {\mathbb {A}} x ^ {k} \right\rangle - \tilde {\Phi} ^ {k} = - \left\langle \mathbb {A} x ^ {k + 1}, \tilde {\mathbb {A}} x ^ {k} \right\rangle .
$$
From above equality we can check
$$
\left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2} + \left\| \mathbb {A} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k} \right\| ^ {2} = \left\| \mathbb {A} x ^ {k + 1} \right\| ^ {2} + \left\| x ^ {k + 1} - x ^ {k} + \tilde {\mathbb {A}} x ^ {k} \right\| ^ {2} \geq \left\| \mathbb {A} x ^ {k + 1} \right\| ^ {2}.
$$
As  $\mathbb{A}$  is  $L$ -Lipschitz continuous, we have
$$
\left(1 + L ^ {2}\right) \left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2} \geq \left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2} + \left\| \mathbb {A} x ^ {k + 1} - \tilde {\mathbb {A}} x ^ {k} \right\| ^ {2} \geq \left\| \mathbb {A} x ^ {k + 1} \right\| ^ {2}.
$$
Dividing both sides by  $\left(1 + L^2\right)\left\| \mathbb{A}x^{k + 1}\right\|^2$  we get a lowerbound for  $r_k$
$$
r _ {k} = \frac {\left\| x ^ {k + 1} - x ^ {k} \right\| ^ {2}}{\left\| \mathbb {A} x ^ {k + 1} \right\| ^ {2}} \geq \frac {1}{1 + L ^ {2}}.
$$
Applying Corollary I.5 we get the desired result
$$
\beta_ {k} \leq \frac {\mu / (1 + L ^ {2})}{\left(1 + \mu / (1 + L ^ {2})\right) ^ {k} - 1 + \mu / (1 + L ^ {2})}.
$$
Note if we take limit  $\mu \to 0^{+}$ , with substitution  $\alpha = \frac{\mu}{1 + L^2}$  we have
$$
\lim  _ {\mu \rightarrow 0 ^ {+}} \frac {\mu / (1 + L ^ {2})}{(1 + \mu / (1 + L ^ {2})) ^ {k} - 1 + \mu / (1 + L ^ {2})} = \frac {1}{\lim  _ {\alpha \rightarrow 0 ^ {+}} \frac {1}{\alpha} ((1 + \alpha) ^ {k} - 1) + 1} = \frac {1}{k + 1},
$$
which is the bound for the monotone case.
I.3.1 If there is  $k \geq 0$  such that  $\tilde{\mathbb{A}} x^k = 0$
Suppose  $\tilde{\mathbb{A}} x^{k} = 0$  for some  $k$ . Let  $x^{N}$  be the very first iterate such that  $\tilde{\mathbb{A}} x^{N} = 0$ . Then from previous arguments we know Theorem 7.2 is true for  $k < N$ . Thus it remains to show the statements are true for  $k \geq N$ .
From  $\tilde{\mathbf{A}} x^{N} = 0$  we know  $y^{N - 1} = x^{N} + \tilde{\mathbf{A}} x^{N} = x^{N}$ . And since  $\tilde{\mathbf{A}} x^{N} = 0$  implies  $\beta_{N} = 0$ , from the definition of the method we have  $y^{N} = 2x^{N} - y^{N - 1} = x^{N}$ . Therefore
$$
x ^ {N + 1} = \mathbb {J} _ {\mathrm {A}} y ^ {N} = \mathbb {J} _ {\mathrm {A}} x ^ {N} = x ^ {N},
$$
we conclude  $x^{k} = x^{N} \in \mathrm{Zer}\mathbb{A}$  for all  $k \geq N$ . Thus  $\beta_{k} = 0$ ,  $\left\| \tilde{\mathbf{A}} x^{k} \right\| = 0$  for all  $k \geq N$ , the Theorem 7.2 is trivially true for  $k \geq N$ .
J Details of experiment in Section 7.1
We solve a compressed sensing problem of Shi et al. [63] which is formulated as an  $\ell_1$ -regularized least-squared problem
$$
\underset {x \in \mathbb {R} ^ {d}} {\text {m i n i m i z e}} \quad \frac {1}{n} \sum_ {i = 1} ^ {n} \left\{\frac {1}{2} \| A _ {(i)} x - b _ {i} \| ^ {2} + \rho \| x \| _ {1} \right\}.
$$
We solve this problem in decentralized manner due to the problem setup where the network of local agents are as Figure 2 and each agents communicate only with their neighbors, the nodes connected to each agents by edge. We use Metropolis-Hastings matrix as our mixing matrix  $W \in \mathbb{R}^{n \times n}$  and apply PG-EXTRA. Let  $W_{i,j}$  denote  $(i,j)$ -th entry of  $W$  and  $N_i \subseteq \{1,2,\dots,n\}$  denote the index of the agents in the neighborhood of agent  $i$ . Consider
$$
\mathbf {x} _ {i} ^ {k + 1} = \operatorname {P r o x} _ {\alpha \rho \| \cdot \| _ {1}} \left(\sum_ {j \in N _ {i}} W _ {i, j} \mathbf {x} _ {j} ^ {k} - \alpha A _ {(i)} ^ {\intercal} \left(A _ {(i)} \mathbf {x} _ {i} ^ {k} - b _ {(i)}\right) - \mathbf {w} _ {i} ^ {k}\right)
$$
$$
\mathbf {w} _ {i} ^ {k + 1} = \mathbf {w} _ {i} ^ {k} + \frac {1}{2} \left(x _ {i} ^ {k} - \sum_ {j \in N _ {i}} W _ {i, j} \mathbf {x} _ {j} ^ {k}\right), \quad k = 0, 1, \dots \tag {PG-EXTRA}
$$
to the problem above. Under suitable choice of parameters, PG-EXTRA can be seen as a fixed-point iteration of an averaged operator with respect to  $\| \cdot \| _M$  [74, Theorem 2], where the metric matrix  $M$  is defined as
$$
M = \left[ \begin{array}{c c} (1 / \alpha) I & U ^ {\intercal} \\ U & \alpha I \end{array} \right],
$$
and  $U$  is a symmetric definite matrix with  $U^2 = \frac{1}{2} (I - W)$ . That is, denoting  $\mathbf{x}^k$ ,  $\mathbf{w}^k \in \mathbb{R}^{d \times n}$  as the vertical stack of  $\mathbf{x}_i^{k'}$ s and  $\mathbf{w}_i^{k'}$ s respectively [58, Chapter 11.3], PG-EXTRA can be rewritten as  $(\mathbf{x}^{k+1}, \mathbf{w}^{k+1}) = \mathbb{T}(\mathbf{x}^k, \mathbf{w}^k)$ . Using this  $\mathbb{T}$ , we proceed the experiment with the Halpern method
$$
\left(\mathbf {x} ^ {k + 1}, \mathbf {w} ^ {k + 1}\right) = \beta_ {k} \left(\mathbf {x} ^ {0}, \mathbf {w} ^ {0}\right) + \left(1 - \beta_ {k}\right) \mathbb {T} \left(\mathbf {x} ^ {k}, \mathbf {w} ^ {k}\right).
$$
When  $\mathbb{T}$  is an averaged operator, it is nonexpansive, we know  $\mathbb{T} = \mathbb{R}_{\mathbb{A}} = 2\mathbb{J}_{\mathbb{A}} - \mathbb{I}$  for some maximal monotone operator  $\mathbb{A}$  [54, Lemma 2.1]. Considering the equivalence discussed in Lemma D.2, we see above Halpern method is equivalent to our presented algorithms of the form
$$
x ^ {k + 1} = \mathbb {J} _ {\mathrm {A}} y ^ {k}
$$
by corresponding  $y^{k} = (\mathbf{x}^{k + 1},\mathbf{w}^{k + 1})$ . Note the operator norm  $\left\| \tilde{\mathbf{A}} x^k\right\| _M^2$  can be calculated by considering below equation
$$
\frac {1}{2} \left(\mathbb {T} y ^ {k - 1} - y ^ {k - 1}\right) = \mathbb {J} _ {\mathbb {A}} y ^ {k - 1} - y ^ {k - 1} = x ^ {k} - \left(\tilde {\mathbb {A}} x ^ {k} - x ^ {k}\right) = - \tilde {\mathbb {A}} x ^ {k}.
$$
We use the anchor coefficients  $\beta_{k} = \frac{1}{k + 1}$ ,  $\beta_{k} = \frac{\gamma}{kp + \gamma}$  with  $p = 1.5$ ,  $\gamma = 2.0$  and the adaptive choice of  $\beta_{k}$  in Theorem 7.2 with  $M$ -norm. Note, in the experiment the adaptive coefficient is calculated by considering below equation
$$
\begin{array}{l} \frac {1}{2} \frac {\left\| \mathbb {T} y ^ {k - 1} - y ^ {k - 1} \right\| _ {M} ^ {2}}{\left\| \mathbb {T} y ^ {k - 1} - y ^ {k - 1} \right\| _ {M} ^ {2} + \left\langle \mathbb {T} y ^ {k - 1} - y ^ {k - 1} , y ^ {k - 1} - x ^ {0} \right\rangle_ {M}} = \frac {1}{2} \frac {4 \left\| \tilde {\mathbb {A}} x ^ {k} \right\| _ {M} ^ {2}}{4 \left\| \tilde {\mathbb {A}} x ^ {k} \right\| _ {M} ^ {2} + \left\langle - 2 \tilde {\mathbb {A}} x ^ {k} , \tilde {\mathbb {A}} x ^ {k} + x ^ {k} - x ^ {0} \right\rangle_ {M}} \\ = \frac {\| \tilde {\mathbf {A}} x ^ {k} \| _ {M} ^ {2}}{- \left\langle \tilde {\mathbf {A}} x ^ {k} , x ^ {k} - x ^ {0} \right\rangle_ {M} + \| \tilde {\mathbf {A}} x ^ {k} \| _ {M} ^ {2}}. \\ \end{array}
$$
We choose the dimension of signal  $d = 100$ , the number of agents  $n = 20$ , the number of measurement for each agent  $m_{i} = 4$ ,  $\ell_1$ -regularization parameter  $\rho = 0.01$ , and algorithm parameter  $\alpha = 0.01$ .
K Broader Impacts
Our work focuses on the theoretical aspects of convex optimization algorithms. There are no negative social impacts that we anticipate from our theoretical results.
Our analysis concerns convex optimization. Although this assumption is standard in optimization theory, many functions that arise in machine learning practice are not convex.