MZU09jacLd/MZU09jacLd.md

Inexact Alternating Direction Method Of Multipliers With Efficient Local Termination Criterion For Cross-Silo Federated Learning

Anonymous authors Paper under double-blind review

Abstract

Federated learning has attracted increasing attention in the machine learning community at the past five years. In this paper, we propose a new cross-silo federated learning algorithm with fast convergence guarantee to solve the machine learning models with nonsmooth regularizers. To solve this type of problems, we design an inexact federated alternating direction method of multipliers (ADMM). This method enables each agent to solve a strongly convex local problem. We introduce a new local termination criterion that can be quickly satisfied when using efficient solvers such as stochastic variance reduced gradient (SVRG). We prove that our method has faster convergence than existing methods. Moreover, we show that our proposed method has sequential convergence guarantees under the Kurdyka-Łojasiewicz (KL) assumption. We conduct experiments using both synthetic and real datasets to demonstrate the superiority of our new methods over existing algorithms.

1 Introduction

Federated learning (FL) is an emerging research paradigm in which multiple agents collaborate to solve a machine learning problem. Cross-silo FL is an important subclass where the participating agents are pre-defined silos, such as organizations or institutions (e.g., hospitals and banks) (Kairouz et al., 2021a).

Typically, there are around 2-100 agents in this setting. Cross-silo federated learning finds significant applications in many domains such as medical and healthcare, finance, and manufacturing (Nandury et al., 2021; Huang et al., 2022; Yang et al., 2019). In a cross-silo federated learning (FL) task, each agent possesses a specific portion of the data, which they use to train their machine learning model locally. Once the local training is completed, all agents send their outputs to a central server. The server then aggregates these outputs and sends an update back to the participating agents. Most FL works focus on the following federated composite optimization (Kairouz et al., 2021b; McMahan et al., 2017b; Pathak & Wainwright, 2020).

$$\underset{x\in {\mathbb{R}}^n}{\min }\sum _{i=1}^p{f}_i(x)+g(x),\backslash operatorname*\{min\}\_\{x\backslash in\backslash mathbb\{R\}^\{n\}\}\backslash sum\_\{i=1\}^\{p\}f\_\{i\}(x)+g(x),$$

$$(1)(1)$$ fi(x) + g(x), (1) where each fi: R n → R is smooth (probably nonconvex) and Li-smooth, and g : R n → R ∪ {+∞} is a proper closed convex regularizer. In machine learning applications, fiis the loss function of the agent i's local data sets and g can be ℓ1-regularizer, grouped ℓ1-regularizer, nuclear-norm regularizer (for matrix variable) (Candès & Recht, 2009; Bao et al., 2022), the indicator function of a convex constraint (Yuan et al., 2021; Bao et al., 2022), etc. Problem (1) is called the federated composite optimization in Yuan et al. (2021). In Yuan et al. (2021), the federated dual averaging (FedDualAvg) was proposed as an early attempt to deal with the nonsmooth g. Bao et al. (2022) proposed a fast federated dual averaging for problem (1) with a strongly convex f. Although FedAvg, FedProx, FedDualAvg, and their variants have intuitive approaches to distribute tasks and aggregate local outputs, they face limitations in both theory and practice. For instance, Braverman et

Table 1: Comparison in the inner updates of federated splitting methods. i Local Solver		Local Complexity
	fi	g
	Model	Local Termination Criterion	Assumptions on ϵ t
FedSplit	t+1 i − Proxfi (˜x t i)∥ ≤ ϵ t i	ϵ t i ≤ O(ϵ)	GD	log(ϵ−1 )
Pathak & Wainwright (2020)	SC	0	∥x
FedPD	t+1 i )∥ 2 ≤ ϵ t i	ϵ t i ≤ O(ϵ)	GD (SGD)	log(ϵ−1 ) (ϵ−1 )
Zhang et al. (2021)	NC	0	E∥∇Li(x t+1	t	t	t
FedDR	i	− Proxfi (˜x i)∥ ≤ ϵ i	1 p Pp i=1 PT t=0 ϵ i ≤ O(1)	-	-
Tran-Dinh et al. (2021)	NC NS	∥x
	t+1	t i)∥ ≤ r∥x t+1	t
	∥x i	− Proxfi (˜x	i	− x i∥	None	-	-
FedADMM1	t+1
Gong et al. (2022)	NC	0	∥∇Li(x i )∥ 2 ≤ ϵ t i	ϵ t i ≤ O(ϵ)	-	-
FedADMM2	t+1 i )∥ 2 ≤ ϵ t i	ϵ t+1 i ≤ νiϵ t ; νi ∈ [1/2, 1)	-	log[(ϵ t+1 i )−1 ]
Zhou & Li (2022)	NC	0	∥∇Li(x	i
FedADMM3	t+1 i − Proxfi (˜x t i)∥ ≤ ϵ t i	1 Pp PT t=0 ϵ t i ≤ O(1)	-	-
Wang et al. (2022)	NC NS	∥x	p	i=1
	i	t+1	t	t	t
FIAELT(Ours)	NC NS E t∥x i	− Proxfi (˜x i)∥ 2 ≤ ri∥x i − Proxfi (˜x i)∥ 2	None	SVRG	O(1)

Table 2: Comparison in the server updates of the federated splitting methods in Table 1. SC = Strongly Convex, NC = Nonconvex, NS = Nonsmooth. ϵ is the same as in Table 1.

Model Convergence

fi g Gradient Sequence

FedSplit SC 0 - Linear

FedPD NC 0 O(T −1) + ϵ -

FedDR NC NS O(T −1) -

FedADMM1 NC 0 O(T −1) + ϵ -

FedADMM2 NC 0 O(T −1) - FedADMM3 NC NS O(T −1) -

FIAELT(Ours) NC NS O(T −1) Linear when θ ∈ (0,

1

2

)

al. McMahan et al. (2017a) demonstrated that FedAvg can diverge in certain scenarios. Even when FedAvg converges, as shown in Pathak & Wainwright (2020), the resulting fixed points may not necessarily be stationary points of the original problem. Additionally, the analysis in Yuan et al. (2021); Li et al. (2020a); Reddi et al. (2021) often assumes that the dissimilarity between agents is bounded, which may not hold in real-world applications. These shortcomings of existing methods motivate the exploration of federated splitting methods for solving (1). In general, the idea behind splitting methods in federated learning is to establish a connection between (1) and a constrained problem of the form:

$$\underset{X}{\min }\sum _{i=1}^p{f}_i(x_i)+g(x_1)\mathrm{s}\mathrm{.}\mathrm{t}\mathrm{.}{x}_1=x_2=\cdots =x_p,\backslash operatorname*\{min\}\_\{X\}\backslash sum\_\{i=1\}^\{p\}f\_\{i\}(x\_\{i\})+g(x\_\{1\})\{\backslash mathrm\{~s.t.~\}\}x\_\{1\}=x\_\{2\}=\backslash cdots=x\_\{p\},$$ $$\left(2\right)\backslash left(2\backslash right)$$

fi(xi) + g(x1) s.t. x1 = x2 = · · · = xp, (2) where X = (x1, x2*, . . . , x*p).

Popular splitting methods in federated learning include FedSplit Pathak & Wainwright (2020), FedDR TranDinh et al. (2021), FedPD Zhang et al. (2021), and ADMM based federated learning methods, Gong et al.

(2022); Zhou & Li (2021); Zhang et al. (2021); Yue et al. (2021); Zhou & Li (2022). FedDR considers nonzero regularizer g while FedSplit, FedPD, and FedADMM deal with the unregularized case where g = 0, which can not apply to the applications where regularizers are needed to induce sparse parameters Zou & Hastie (2005); Yuan et al. (2021) or low rank matrices Candès & Recht (2009); Bao et al. (2022).

At each round t of federated splitting methods, each agent needs to find x t+1 ito approximate the proximal operator of each fi for the current point x˜ t i (denoted as Proxfi (˜x t i )), via a number of local updates with a certain termination criterion. However, the number of local updates (defined as local complexity) required by existing criteria is either unexplored or tends to infinity with an infinitesimal tolerance ϵ as the number of server updates T increases, as shown in Table 1. Therefore, a more advanced criterion that leads to a known constant number of more efficient local updates is much desired, which is an important goal of this work.

Moreover, existing federated splitting methods on nonconvex optimization with nonsmooth regularizer g also only focus on the convergence rate of the gradient but ignore the convergence of the generated sequences to a desired critical point. Zhou & Li (2022); Yue et al. (2021) proves that the accumulation point is critical point for regularized case (g = 0) but the convergence rate is still unknown. To obtain sequential convergence rate for nonsmooth regularizer g ̸= 0 is also an important goal of this work.

1.1 Our Contributions

To fulfill the above two goals, we propose a novel splitting method called Federated Inexact ADMM with Efficient Local Termination (FIAELT) for the nonconvex nonsmooth composed optimization problem (1) in the context of cross-silo federated learning, based on the equivalence between (1) and an np-dimensional constrained problem (4). Compared with existing works on federated splitting methods, we summarize our contributions as follows.

• For the local update of our algorithm, we propose a new criterion E i t∥x t+1 i − Proxfi (˜x t i )∥ 2 ≤ ri∥x t i − Proxfi (˜x t i )∥ 2(see Algorithm 1 for detail) where the tolerance ri ∈ (0, 1) does not need to be infinitesimal with large number T of communication rounds. Hence, our local complexity can be O(1), which outperforms existing splitting methods with an unexplored or large number of local updates (see Table 1 for comparison).

At the same time, we keep the state-of-the-art gradient convergence rate O(1/T) in the server updates (see Table 2).

• Furthermore, we demonstrate that FIAELT has sequential convergence properties in the deterministic case. Specifically, we prove that any accumulation point of the sequence generated at the server of FIAELT is a stationary point of (1). Moreover, we prove that FIAELT achieves global convergence under KurdykaŁojasiewicz (KL) geometry, which covers a wide range of functions in practice. Specifically, the server updates and the outputs of local servers converge in finitely many communications when the KL exponent α of the potential function is 0. These sequences converge linearly when α ∈ (0, 1 2 ). These sequences converge sublinearly when α ∈ ( 1 2 , 1). In the analysis, our proposed new criterion plays a key role. To the best of our knowledge, FIAELT is the first federated learning method that has sequential convergence rate in nonconvex nonsmooth settings.

• Finally, we conducted experiments involving the training of fully-connected neural networks. In these experiments, we compared our method against existing splitting methods as well as other state-of-the-art Federated methods. The experimental results revealed that our method is competitive and consistently outperforms the other approaches in terms of training loss, training accuracy, and testing accuracy. These findings indicate the superior performance and effectiveness of our proposed method in the task of training fully-connected neural networks.

1.2 Related Work

The literature of federated learning is rich. In this work, we only focus on the splitting methods in federated learning. A comparison between our method and existing splitting methods is summarized in Table 1.

In Pathak & Wainwright (2020), FedSplit was proposed. It implements the Peaceman-Rachford splitting P method for (2). Pathak & Wainwright (2020) analyzed the proposed method in the case where g = 0 and i fiis strongly convex. Pathak & Wainwright (2020) showed that when the error between the local output and the Proxfiis under a threshold ϵ, the sequence generated at the server by FedSplit linearly converges to an inexact solution of (1) up to an error determined by ϵ. They also applied the FedSplit to a strongly convex majorization of the original problem. In this setting, they showed a complexity of O˜( √ϵ) to obtain an ϵ-optimal function value. However, in general convex settings, it assumes FedSplit locally computes Proxfi exactly, which is unrealistic when the local server solves large-scale problems.

When g = 0, there are several work on federated ADMM, Zhang et al. (2021); Gong et al. (2022); Zhou & Li (2022); Elgabli et al. (2022). Gong et al. (2022) proposed FedADMM that randomly selects agents to attend each round. The ith agent terminates the local iterations when the norm of the local gradient of the current iterate is under a threshold ϵi. When there is an upper bound ϵ for {ϵi}, they showed FedADMM has a complexity of O(ϵ −1) + O(ϵ) to reach an ϵ-surrogate stationary point. When fi's are twice differentiable, ADMM is applied in designing a second-order FL method in Elgabli et al. (2022). Zhou & Li (2022) proposed an inexact ADMM for federated learning problems. At round t, the ith agent terminates the local updates when the norm of the local gradient is under a threshold ϵ t i . They assume {ϵ t i }t decreases exponentially, i.e., ϵ t+1 i ≤ νiϵ t i with νi ∈ [ 1 2 , 1). They showed that the generated sequence accumulates at the stationary point.

By further assuming the accumulation point of the generated sequence is isolated, they show the generated sequence converges globally. Compared with this work, we do not assume the accumulation point of the generated sequence to be isolated when we analyze the sequential convergence of our method.

When g ̸= 0, Tran-Dinh et al. (2021) proposed FedDR that applies the Douglas-Rachford (DR) splitting algorithms for (2). They combined the DR method with randomized block-coordinate strategies and asynchronous implementation. They estimated the complexity of FedDR under different termination criteria for local updates.The termination criteria in Tran-Dinh et al. (2021) test whether the distance between the prox of f and its approximation can be bounded by a certain value. However, this distance is unable to check in practice, especially when we use stochastic gradient methods for local updates. Yue et al. (2021) also considered the case where g ̸= 0. Specifically, they considered the case when g is the Bregman distance.

Assuming the Hessian of fi's in (1) being Lipschitz continuous, Yue et al. (2021) showed any accumulation point of the generated sequence is a stationary point. Yue et al. (2021) also showed the proposed method has a complexity of O(ϵ −1) to reach an ϵ-stationary point.

2 Preliminaries

In this paper, we denote R n the n-dimensional Euclidean space with inner product ⟨·, ·⟩ and Euclidean norm ∥ · ∥. We denote the set of all positive numbers as R++. We denote the distance from a point a to a set A as d(a, A). For a random variable ξ defined on a probability space (Ξ, Σ, P), we denote its expectation as Eξ.

Given an event A, the conditional expectation of ξ is denoted as E(ξ|A).

An extended-real-valued function f : R n → [−∞, ∞] is said to be proper if domf = {x ∈ R n : f(x) < ∞} is not empty and f never equals −∞. We say a proper function f is closed if it is lower semicontinuous. We define the indicator function of a closed set A as δA(x), which is zero when x ∈ A and ∞ otherwise.

We define the regular subdifferential of a proper function f : R n → [−∞, ∞] at x ∈ domf as ∂fˆ (x) := nξ∈R n:lim infz→x, z̸=x f(z)−f(x)−⟨ξ,z−x⟩ ∥z−x∥ ≥0 oThe (limiting) subdifferential of f at x ∈ domf is defined as ∂f(x):=nξ ∈ R n:∃x k f→x, ξk→ξ with ξ k ∈∂fˆ (x k),∀k o, where x k f→ x means both x k → x and f(x k) → f(x).

For x ̸∈ domf, we define ∂fˆ (x) = ∂f(x) = ∅. We denote dom∂f := {x : ∂f(x) ̸= ∅}. For a differential function h : R m × R n → R l, we denote ∇xL(x, y) and ∇yL(x, y) as the partial derivatives with respect to x and y correspondingly. We defined the normal cone of a set A at x as NA(x) := ∂δA(x). For a proper function f : R n → [−∞, ∞], we denote the proximal operator of f as Proxαf (x) = Arg minz∈Rnf(z) + 1 2α ∥z − x∥ 2 .

Consider a problem min f + g, where f is a smooth function and g is properly closed convex. We say x is a stationary point of this problem when 0 ∈ ∇f(x) + ∂g(x). We say x is an ε-stationary point if d 2(0, ∇f(x) + ∂g(x)) ≤ ε.

We next introduce the KL property used in analyzing the sequential convergence. Let Ψa be defined as the set of concave functions ψ : [0, a) → [0, ∞) satisfying ψ(0) = 0, being continuously differentiable on (0, a), and satisfying ψ ′ > 0 on (0, a).

Definition 1 (Kurdyka-Łojasiewicz property and exponent). A proper closed function f : R n → (−∞, ∞] is said to satisfy the Kurdyka-Łojasiewicz (KL) property at an xˆ ∈ dom∂f if there are a ∈ (0, ∞], a neighborhood V of xˆ and a φ ∈ Ψa such that for any x ∈ V with f(ˆx) < f(x) < f(ˆx) + a*, it holds that* ψ ′(f(x) − f(ˆx))dist(0, ∂f(x)) ≥ 1. If f *satisfies the KL property at* xˆ ∈ dom∂f and ψ *can be chosen as* ψ(ν) = a0ν 1−α for some a0 > 0 and α ∈ [0, 1), then we say that f satisfies the KL property at xˆ with exponent α. A proper closed function f satisfying the KL property with exponent α ∈ [0, 1) at every point in dom∂f is called a KL function with exponent α.

Functions satisfying KL property includes proper closed semi-algebraic functions, the quadratic loss function plus possibly nonconvex piecewise linear regularizers Attouch et al. (2010); Li & Pong (2018); Attouch et al. (2013); Zeng et al. (2021).

3 Federated Inexact Admm With Efficient Termination Criterion

We relate the problem (1) to (2). For (2), we view it as the following np-dimensional problem:

$$\underset{X\in {\mathbb{R}}^{n_p}}{\min }F(X)+G(X),\backslash operatorname*\{min\}\_\{X\backslash in\backslash mathbb\{R\}^\{n\_\{p\}\}\}\backslash \; F(X)+G(X),$$ $$\mathbf{\Sigma }\backslash mathbf\{\backslash Sigma\}$$

$\downarrow$ . F(X) + G(X), (3) where X = (x1, x2*, . . . , x*p) with each xi ∈ R n, F(X) := Pp i=1 fi(xi) with fi's in (1), G(X) := g(x1) +δC(X) with C := {X : x1 = · · · = xp} and g in (1).

The following proposition establishes the relation between (3) and (1).

Proposition 1. If X∗ = (x ∗ 1 , . . . , x∗p ) is a stationary point of (3), then x ∗ 1 is a stationary point of (1).

Furthermore, if X = (x1, . . . , xp) is an ε*-stationary point of* (1), then x1 is a pε*-stationary point of* (1).

Based on this relation, we consider ADMM to solve (3). Rewrite (3) as the following equivalent problem:

$$\underset{X,Y\in {\mathbb{R}}^{n_P}}{\min }F(X)+G(Y)\mathrm{s}\mathrm{.}\mathrm{t}\mathrm{.}X=Y\mathrm{.}\backslash operatorname*\{min\}\_\{X,Y\backslash in\backslash mathbb\{R\}^\{n\_\{P\}\}\}\backslash \; F(X)+G(Y)\{\backslash mathrm\{~s.~t.~\}\}X=Y.$$

$$\mathbf{\partial })\backslash mathbf\{\backslash partial\})$$ F(X) + G(Y ) s. t. X = Y. (4) The augmented lagrangian function of (4) is defined as:

$$L_{\beta }(X,Y,Z):=F(X)+G(Y)+⟨X-Y,Z⟩+\frac{\beta }{2}\mathrm{\parallel }X-Y{\mathrm{\parallel }}^2\mathrm{.}L\_\{\backslash beta\}(X,Y,Z):=F(X)+G(Y)+\backslash langle\; X-Y,Z\backslash rangle+\backslash frac\{\backslash beta\}\{2\}\backslash |X-Y\backslash |^\{2\}.$$ 2. (5) Given a starting point (X0, Y 0, Z0) ∈ R np × R np × R np and τ, β > 0, the ADMM for (3) is as follows:

$$\{\begin{array}{c}{X}^{t+1}=\arg {\min }_X{L}_{\beta }(X,Y^t,Z^t),\\ Z^{t+1}=Z^t+\tau \beta (X^{t+1}-Y^t),\\ Y^{t+1}=\arg {\min }_Y{L}_{\beta }(X^{t+1},Y,Z^{t+1})\mathrm{.}\end{array}\backslash begin\{cases\}X^\{t+1\}=\backslash arg\backslash min\_\{X\}L\_\{\backslash beta\}(X,Y^\{t\},Z^\{t\}),\backslash \backslash \; Z^\{t+1\}=Z^\{t\}+\backslash tau\backslash beta(X^\{t+1\}-Y^\{t\}),\backslash \backslash \; Y^\{t+1\}=\backslash arg\backslash min\_\{Y\}L\_\{\backslash beta\}(X^\{t+1\},Y,Z^\{t+1\}).\backslash end\{cases\}$$ $$\mathbf{\Sigma }\backslash mathbf\{\backslash Sigma\}$$ (6) Now we give an equivalent form of the third equation in (6) as follows.

Proposition 2. Consider (3). Let {(Xt+1, Y t+1, Zt+1)} be generated by (6). Suppose β > maxi Li*. Then* the solution of the problem in the third equation of (6) is (y1, . . . , y1) with y1 = Prox 1 βp g ( 1 p Pp i=1(x t+1 i + 1 β z t+1 i))).

P On the other hand, since F(X) in (3) is separable, we can write Lβ(X, Y, Z) in (5) as Lβ(X, Y, Z) = p i=1 Lβ,i(xi, yi, zi), where

$$L_{\beta ,i}(x_i,y_i,z_i):=f_i(x_i)+⟨x_i-y_i{z}_i⟩+\frac{\beta }{2}\mathrm{\parallel }{x}_i-y_i{\mathrm{\parallel }}^2\mathrm{.}L\_\{\backslash beta,i\}(x\_\{i\},y\_\{i\},z\_\{i\}):=f\_\{i\}(x\_\{i\})+\backslash langle\; x\_\{i\}-y\_\{i\}\backslash ,z\_\{i\}\backslash rangle+\{\backslash frac\{\backslash beta\}\{2\}\}\backslash |x\_\{i\}-y\_\{i\}\backslash |^\{2\}.$$

Therefore, the first equality in (6) can be rewritten as x t+1 i =x t+1 i,∗ where

$x_{i,*}^{t+1}:=\underset{x_{i}}{\operatorname{argmin}}L_{\beta,i}(x_{i},y^{t},z_{i}^{t});i=1,\ldots,p.$ In practice, (7) cannot be exactly solved as fiis usually a nonconvex loss function involving large training data. Hence, existing federated splitting methods inexactly solve (7) up to a certain local criterion. However, the computational complexities of the local updates required by these criteria are either unexplored or very large (see Table 1). To solve this limitation, we propose the following criterion.

$$\begin{array}{cccc}& {\mathbb{E}}_t^i\mathrm{\parallel }{x}_i^{t+1}-x_{i,\ast }^{t+1}{\mathrm{\parallel }}^2\le r_i\mathrm{\parallel }{x}_i^t-x_{i,\ast }^{t+1}{\mathrm{\parallel }}^2\mathrm{.}& & \text{(1)}\end{array}\backslash mathbb\{E\}\_\{t\}^\{i\}\backslash |x\_\{i\}^\{t+1\}-x\_\{i,*\}^\{t+1\}\backslash |^\{2\}\backslash leq\; r\_\{i\}\backslash |x\_\{i\}^\{t\}-x\_\{i,*\}^\{t+1\}\backslash |^\{2\}.\backslash tag\{1\}$$

where E i t denotes conditional expectation given the past trajectory {(x s i , ys, zs i ) : s = 0, 1*, . . . , t*}, and the tolerance ri ∈ (0, 1) does not need to be arbitrarily small to ensure O(1) local complexity even with stochastic gradient, as will be shown in the convergence analysis.

$$\left(7\right)\backslash left(7\backslash right)$$ $$(\mathfrak{g})(\{\backslash mathfrak\{g\}\})$$

Algorithm 1 Federated Inexact ADMM with Efficient Local Termination (FIAELT) for (1) 1: Input: β, τ > 0, ri > 0, mi ∈ N+, ηi > 0. (x 0 i , y0 i , z0 i ) and x¯ 0 = 1 p Pi x 0 i , z¯ 0 = 1 p Pi z 0 i for agents i = 1*, . . . , p*.

2: for iteration t = 0, 1*, . . . , T* − 1 do 3: for agent i = 1*, . . . , p* in parallel do 4: Find x t+1 ito approximately solve:

$$x_i^{t+1}\approx \underset{x_i}{\min }{L}_{\beta ,i}(x_i,y_i^t,z_i^t):=x_{i,\star }^{t+1}\mathrm{.}x\_\{i\}^\{t+1\}\backslash approx\backslash operatorname*\{min\}\_\{x\_\{i\}\}L\_\{\backslash beta,i\}(x\_\{i\},y\_\{i\}^\{t\},z\_\{i\}^\{t\}):=x\_\{i,\backslash star\}^\{t+1\}.$$ $$(\mathit{\delta })(\{\backslash boldsymbol\{\backslash delta\}\})$$ i,⋆ . (8) such that the criterion (9) is satisfied.

Upload ∆xi,t+1 = x t+1 i − x t i and ∆zi,t+1 = τ β(x t+1 i − y t i ) to the server.

5: end for 6: The server calculates x¯ t+1 = ¯x t+ 1 p Pi ∆xi,t+1, z¯ t+1 = ¯z t+ 1 p Pp i=1 ∆zi,t+1 and y t+1 = Prox 1 βp g (¯x t+1+ 1 β z¯ t+1), and broadcasts these variables to each agent.

7: end for We propose Algorithm 1 that implements the ADMM rule (6) in a federated way, where x t+1 iinexactly solves (7) with stochastic gradient methods.

When β > L := maxi Li, the local problem (8) is minimizing a strongly convex smooth function that has Lipscihtz continuous gradient. Hence, using the stochastic method called SVRG in Johnson & Zhang (2013), we obtain x t+1 that satisfies the following property.

Proposition 3. Consider (1). Set β > L := maxi Li*. Let* {(x t i , yt i , zt i )} be generated by Algorithm 1. Using SVRG in Johnson & Zhang (2013) with Option II with frequency mi, learning rate ηi*, and initialization* x t i for (8), such that

$$(10)(10)$$

Then criterion (9) is satisfied in at most k i t = log1/ρiβ+Li ri(β−Li) iterations of SVRG.

Remark 1. The above proposition shows that fixing any ri ∈ (0, 1), SVRG outputs an inexact solution of the local subproblem (8) within O(1) steps, independent of the number of communication rounds T. In contrast, the number of local updates required by other existing federated splitting methods is either unexplored or increases to infinity with T.

Remark 2. When (9) is deterministic, our subproblem degenerates to minimizing a strongly convex function.

According to the well know results, minimizing a strongly convex function with the simplest gradient descent method produce a linear convergent sequence of variables. Following the same analysis in the proofs of Proposition 3, we will have the local complexity of order O(1).

4 Convergence Analysis Of Algorithm 1

We analyze the convergence properties of the variables Xt:= [x t 1 ; . . . ; x t p ], Y t:= [y t 1 ; . . . ; y t p ], Z t:= [z t 1 ; . . . ; z t p ] generated by Algorithm 1. We also denote L := maxi Li, r := maxi ri, Xt+1 ∗:= [x t+1 1,∗ ; . . . ; x t+1 p,∗ ] and W = infX F(X) + infY G(Y ) > −∞ throughout the paper. First, the update rules of Algorithm 1 can be rewritten into the combined vectors Xt, Y t, Zt as follows.

We first show the following property.

Proposition 4. The update rules in Algorithm 1 satisfy

$$\begin{array}{c}\mathbb{E}\mathrm{\parallel }{X}^{t+1}-X_{\star }^{t+1}{\mathrm{\parallel }}^2\le r\mathrm{\parallel }{X}^t-X_{\star }^{t+1}{\mathrm{\parallel }}^2,\\ Z^{t+1}=Z^t+\tau \beta (X^{t+1}-Y^t),\\ Y^{t+1}={\min }_Y{L}_{\beta }(X^{t+1},Y,Z^{t+1}),\end{array}\backslash begin\{array\}\{l\}\{\{\backslash mathbb\{E\}\backslash |X^\{t+1\}-X\_\{\backslash star\}^\{t+1\}\backslash |^\{2\}\backslash leq\; r\backslash |X^\{t\}-X\_\{\backslash star\}^\{t+1\}\backslash |^\{2\},\}\}\backslash \backslash \; \{\{Z^\{t+1\}=Z^\{t\}+\backslash tau\backslash beta(X^\{t+1\}-Y^\{t\}),\}\}\backslash \backslash \; \{\{Y^\{t+1\}=\backslash operatorname*\{min\}\_\{Y\}L\_\{\backslash beta\}(X^\{t+1\},Y,Z^\{t+1\}),\}\}\backslash end\{array\}$$ 2, (11) t), (12) YLβ(Xt+1*, Y, Z*t+1), (13) $$\begin{array}{c}(11)\\ (12)\end{array}\backslash begin\{array\}\{l\}\{(11)\}\backslash \backslash \; \{(12)\}\backslash end\{array\}$$

$$(13)(13)$$

With Proposition 4, we can analyze {(Xt, Y t, Zt)} to analyze the convergence properties of Algorithm 1.

For {(Xt, Y t, Zt)}, we have the following theorem that is important in establishing our main convergence properties.

Proposition 5. Select hyperparameters β ≥ 5L, ri ∈ (0, 0.01], τ ∈ [1/2, 1). Denote Γ := 1−τ τ, Θ = 2β 2 + 4L 2, Λ := 4L 2. Υ := Θ τβ4r 1−2r and δ := 14 (β − L) − 2Υ*. Define*

$$H(X,Y,Z,X^{\mathrm{\prime }},Z^{\mathrm{\prime }}):=L_{\beta }(X,Y,Z)+\frac{\mathrm{\Gamma }}{\tau \beta }\mathrm{\parallel }Z-Z^{\mathrm{\prime }}{\mathrm{\parallel }}^2+\mathrm{{\rm Y}}\mathrm{\parallel }X-X^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{.}H(X,Y,Z,X^\{\backslash prime\},Z^\{\backslash prime\}):=L\_\{\backslash beta\}(X,Y,Z)+\backslash frac\{\backslash Gamma\}\{\backslash tau\backslash beta\}\backslash |Z-Z^\{\backslash prime\}\backslash |^\{2\}+\backslash Upsilon\backslash |X-X^\{\backslash prime\}\backslash |^\{2\}.$$ $${}^{+1},Y^{t+1},Z^{t+1},X^t,Z^t)\mathrm{.}The\{\}^\{+1\},Y^\{t+1\},Z^\{t+1\},X^\{t\},Z^\{t\}).\backslash ;\backslash ;\backslash ;T\; h\; e$$

and Ht+1 := EH(Xt+1, Y t+1, Zt+1, Xt, Zt). Then for t ≥ 1, it holds that δ ≥ 0.1L and

$$\begin{array}{cccc}& H_{t+1}\le H_t-\delta \mathbb{E}\mathrm{\parallel }{X}^{t+1}-X^t{\mathrm{\parallel }}^2-\frac{\beta }{2}\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\mathrm{.}& & \text{(14)}\end{array}H\_\{t+1\}\backslash leq\; H\_\{t\}-\backslash delta\backslash mathbb\{E\}\backslash |X^\{t+1\}-X^\{t\}\backslash |^\{2\}-\backslash frac\{\backslash beta\}\{2\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}.\backslash tag\{14\}$$

Hence, the sequence {Ht} converges to some H∗ ≥ W.

Thanks to Proposition 5, we have the following property with respect to the successive changes.

Corollary 1. Consider (1) and let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Then limt E∥Xt − Xt+1∥ 2 = limt E∥Y t+1 − Y t∥ 2 = limt E∥Z t+1 − Z t∥ 2 = lim E∥Y t − Xt∥ 2 = 0.

Remark 3. Corollary 1 together with Propositions 1 and 4 shows that the expectations of successive changes of {(x t 1 , . . . , xtp , yt, z1*, . . . , z*tp )} generated by Algorithm 1 also converge to 0.

Based on Proposition 5, {(Xt, Y t, Zt)} has the following convergence property.

Theorem 1. Select hyper-parameters per Proposition 5 hold and let H∗ be defined as in Proposition 5. Then

$$\begin{array}{cccc}& \sum _{t=0}^T\mathbb{E}\mathrm{\parallel }\mathrm{\nabla }F(Y^{t+1})+{\xi }^{t+1}{\mathrm{\parallel }}^2\le D(\mathrm{\parallel }\mathrm{\nabla }{L}_{\beta }(X^0,Y^0,Z^0){\mathrm{\parallel }}^2+\mathrm{\parallel }{X}^0-Y^0{\mathrm{\parallel }}^2)+D(L_{\beta }(X^0,Y^0,Z^0)-W),& & \text{(15)}\end{array}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |\backslash nabla\; F(Y^\{t+1\})+\backslash xi^\{t+1\}\backslash |^\{2\}\backslash leq\; D\backslash left(\backslash |\backslash nabla\; L\_\{\backslash beta\}(X^\{0\},Y^\{0\},Z^\{0\})\backslash |^\{2\}+\backslash |X^\{0\}-Y^\{0\}\backslash |^\{2\}\backslash right)+D\backslash left(L\_\{\backslash beta\}(X^\{0\},Y^\{0\},Z^\{0\})-W\backslash right),\backslash tag\{15\}$$

where

$$\begin{array}{cccc}& D:=\max \{3(L+\beta {)}^2\frac{2r}{1-2r},{(\frac{L}{\tau \beta }+1)}^2,(L+\beta {)}^2\}\cdot \max \{D_1,D_2,D_3\}& & \text{(16)}\end{array}D:=\backslash max\backslash \{3(L+\backslash beta)^\{2\}\backslash frac\{2r\}\{1-2r\},\backslash left(\backslash frac\{L\}\{\backslash tau\backslash beta\}+1\backslash right)^\{2\},(L+\backslash beta)^\{2\}\backslash \}\backslash cdot\backslash max\backslash \{D\_\{1\},D\_\{2\},D\_\{3\}\backslash \}\backslash tag\{16\}$$

with D1 := 2Γ+Θ 8r 1−2r +2 min{δ, 12 β}, D2 := (1 + Γ) 3(r+1) (L−β) 2 + D14 (L−β) 2 L+β+1 2 + 2τ β(Γ + 1) + Υ + (L−β) 2 8 , D3 := max{3, D12τ β(Γ + 1)}, Γ, Υ and Θ being defined in Proposition 5. Combining Theorem 1 with Proposition 1 and Proposition 3, we immediately obtain the following convergence rate of Algorithm 1.

Corollary 2. Select hyperparameters β = 5L, ri = 0.005, τ = 1/2 in Algorithm 1. Then the following convergence rate holds.

$$\frac{1}{1+T}\sum _{t=0}^T\mathbb{E}{d}^2(0,\sum _i\mathrm{\nabla }{f}_i(y^{t+1})+\mathrm{\partial }g(y^{t+1}))\backslash frac\{1\}\{1+T\}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}d^\{2\}(0,\backslash sum\_\{i\}\backslash nabla\; f\_\{i\}(y^\{t+1\})+\backslash partial\; g(y^\{t+1\}))$$ $$\leq p D\left(|\nabla L_{\beta}(X^{0},Y^{0},Z^{0})|^{2}+|X^{0}-Y^{0}|^{2}\right)+p D\left(L_{\beta}(X^{0},Y^{0},Z^{0})-W\right).$$

where D is the one defined in Theorem 1. Furthermore, the criterion (9) can be satisfied by implementing 10 iterations of SVRG Johnson & Zhang (2013) with Option II with frequency mi = 200*, learning rate* ηi =1 40L , and initialization x t i for (8).

Remark 4. Corollary 2 indicates that compared with existing federated methods, we keep the same state-ofthe-art convergence rate O(1/T) with T being the number of the communication round, while only O(1) local update steps for the local (8) is required.

Figure 1: Results on Synthetic-{(0,0), (0.5, 0.5), (1,1)} dataset.

4.1 Sequential Convergence In The Deterministic Case

In this section, we further investigate the convergence of the sequence {(Xt, Y t, Zt)} generated by Algorithm 1 when (9) holds deterministically, i.e., holds without the expectation. We first show the properties of the set of accumulation points of {(Xt, Y t, Zt, Xt−1, Zt−1)}.

Proposition 6. Consider (1) and let {(Xt, Y t, Zt)} be generated by Algorithm 1 with (9) holding deterministically. Suppose assumptions in Proposition 5 hold. Suppose {(Xt, Y t, Zt)} is bounded. Then any accumulation point of {Y t} is a stationary point of (3).

Combining Proposition 6 with Proposition 1 and Proposition 2, we immediately have the subsequential convergence of the sequence generated by FIAELT.

Corollary 3. Let {(x t 1 , . . . , xtp , yt, zt1 , . . . , ztp )} be generated by Algorithm 1 with (9) holding deterministically. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 6 hold. Then any accumulation point of {y t} is a stationary point of (1).

Next, we present the convergence rate of (Xt, Y t, Zt). Theorem 2. Consider (1) and Algorithm 1 with (9) holding deterministically. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Let H be defined as in Proposition 5 and suppose H is a KL function with exponent α ∈ [0, 1). Then {(Xt, Y t, Zt)} converges globally. Denoting (X∗, Y ∗, Z∗) := limt(Xt, Y t, Zt) and d t s := ∥(Xt, Y t, Zt)−(X∗, Y ∗, Z∗)∥*, then the followings hold. If* α = 0, then {d t s} converges finitely. If α ∈ (0, 1 2 ], then there exist b > 0, t1 ∈ N and ρ1 ∈ (0, 1) such that d t s ≤ bρt1 for t ≥ t1*. If* α ∈ ( 1 2 , 1), then there exist t2 and c > 0 such that d t s ≤ ct− 1 4α−2 for t ≥ t2.

Remark 5. Proposition 3 and Theorem 2 jointly illustrate that the local outputs {x t i }t and the server updates y t achieve global linear convergence towards a stationary point of (1) when the Kurdyka-Lojasiewicz (KL) exponent of function H is set to 12 . The precise determination of the KL exponent of H is interconnected

Figure 4: Results of our algorithm on FEMNIST dataset with different learning rates. (L1-norm regularize.) with another aspect involving the investigation of error bounds, which is beyond the boundaries of the present paper's scope. Interested readers are referred to sources such as Attouch et al. (2010); Li & Pong (2018); Attouch et al. (2013); Zeng et al. (2021) for more deeper insights.

5 Experimental Results

To evaluate the performance of our proposed FIAELT algorithm, we conduct experiments on both realistic and synthetic datasets. When g = 0 in (1), we compare our algorithm with FedDRTran-Dinh et al. (2021), FedPD Zhang et al. (2021), FedAvg McMahan et al. (2017b), FedAdmm Zhou & Li (2022). When g = λ| - ||1 for some \ E R++, we compare our algorithm with FedMid Yuan et al. (2021), FedDualAvg Yuan et al. (2021), and FedDR. Following FedDR Tran-Dinh et al. (2021), we choose the neural network as our model, and the details are deferred to the supplementary materials. For FedDR, FedPD, we refer to the code provided in Tran-Dinh et al. (2021), and we also re-implement the FedAdmm based on them. All experiments are running on the Linux-based server with the configuration: 8xA6000 GPU with 48GB memory each. To be in accordance with the theoretical analysis, we sample all the clients to perform updates for our algorithm in each communication round. We pick up hyper-parameters carefully and show the best results for each algorithm. For evaluation metrics, we use training loss, training accuracy, and test accuracy. Our code is available at https://anonymous.4open.science/r/FIAELT_TMLR-D6C7/. Results on synthetic datasets. Following the data generation process on Li et al. (2020a); Tran-Dinh et al. (2021), we generate three datasets: synthetic-{(0,0), (0.5, 0.5), (1,1)}. All agents perform updates at each communication round. Our algorithm is compared using synthetic datasets in both iid and non-iid settings. The performance of five algorithms on non-iid synthetic datasets is shown as Figure 1. Our algorithm can achieve better results than FedPD, FedAdmm, FedAvg, and FedDR on all three synthetic datasets. FEMNIST Cohen et al. (2017); Caldas et al. (2018) dataset is a more complex and federated extended MNIST. It has 62-class (26 upper-case and 26 lower-case letters, 10 digits) and the data is distributed to 200 devices. Figure 2 depicts the results of all 5 algorithms on FEMNIST. As it shows, FIAELT can achieve comparable training accuracy and loss value with FedDR. In comparison with FedAdmm, FedPD, and FedAvg, FIAELT has a significant improvement in both training accuracy and loss value. Our algorithm can also work much better with test accuracy than the other 4 algorithms.

Results with the L1 norm. Following FedDR Tran-Dinh et al. (2021), we also consider the composite setting with g(x) := 0.01∥x∥1 to verify our algorithm by selecting different learning rates and the number of local SGD epochs. We conduct the experiment on the FEMNIST dataset and we show the results as Figure 3. As we can see from the training loss and training accuracy, FIAELT has competitive efficiency with FedDR and outperform FedDualAvg and FedMid. In addition, in testing accuracy, FIAELT outperforms all the other methods. Figure 5 shows how different learning rates affect the performance of our FIAME on the FEMNIST dataset.

6 Conclusion

In this paper, we propose a federated inexact ADMM with a new local termination criterion. This criterion is efficient and can be satisfied within iterations unrelated to the communication rounds, particularly when using stochastic gradient methods as the local solver. Our new method has the best-known complexity while having efficient local updates. Additionally, we provide proof that the proposed method has sequential convergence guarantees in the deterministic case. Under KL assumptions, we demonstrate that the whole generated sequence can converge sublinearly, linearly, or even finitely. Our experiments consistently demonstrate that the proposed method consistently outperforms state-of-the-art methods, especially in terms of testing accuracy.

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Dataset	Size(Input x FC layer x Output)
Synthetic	60 x 32 x 10
MNIST	784 x 128 x 10
FEMNIST	784 x 128 x 26

Table 3: The details of the neural networks in our numerical experiments.

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A Supplement For Experiment

The details of the training models. For all datasets, we apply neural networks with only Fullyconnected (FC) layers as training models. The size of the models is shown as Table 3. Our code is available at https://anonymous.4open.science/r/FIAELT-8CC5/. Hyperparameter choosing. The learning rates are 0.012 for synthetic datasets, and 0.009 for FEMNIST.

For FedPD, FedDR, and FedProx, we follow Tran-Dinh et al. (2021) to select the hyper-parameters, including µ for FedProx, η for FedPD, and η, α for FedDR. As for FedMid Yuan et al. (2021) and FedDualAvg Yuan et al. (2021), we also select the hyper-parameters working best for plotting the performance and comparison. Additional Results with Different Learning Rates Figure 5 shows how different learning rates affect the performance of our FIAME on the FEMNIST dataset.

Figure 6: Results on Synthetic-{(0,0), (0.5, 0.5), (1,1)} dataset.

Figure 7: Results on FEMNIST dataset.

Additional Results Comparing Fiame With Non-Admm Based Fl Algorithms A.1

We compare our method with FedAvg Li et al. (2020b), SCAFFOLD Karimireddy et al. (2020), FedSkip Fan et al. (2022). Results on synthetic datasets. Following the data generation process on Li et al. (2020a); Tran-Dinh et al. (2021), we generate three datasets: synthetic-{(0,0), (0.5, 0.5), (1,1)}. All agents perform updates at each communication round. Our algorithm is compared using synthetic datasets in both iid and non-iid settings. The performance of 4 algorithms on non-iid synthetic datasets is shown as Figure 6. Our algorithm can achieve better results than FedAvg, SCAFFOLD, FedSkip on all three synthetic datasets.

Results on FEMNIST dataset. FEMNIST Cohen et al. (2017); Caldas et al. (2018) dataset is a more complex and federated extended MNIST. It has 62-class (26 upper-case and 26 lower-case letters, 10 digits) and the data is distributed to 200 devices. Figure 7 depicts the results of all 4 algorithms on FEMNIST. As it shows, compared with the other 3 methods, FIAME has a significant improvement in both training accuracy and loss value. Our algorithm can also work much better with test accuracy than the other 3 algorithms.

B Convergence Analysis Of Algorithm 1

Proposition 1. If X∗ = (x ∗ 1 , . . . , x∗p ) is a stationary point of (3), then x ∗ 1 is a stationary point of (1).

Furthermore, if X = (x1, . . . , xp) is an ε*-stationary point of* (1), then x1 is a pε*-stationary point of* (1).

Proof. Note that

$\mathfrak{C}={(x_{1},\ldots,x_{p}):\ x_{1}-x_{2}=0,\ x_{2}-x_{3}=0,\ \ldots,\ x_{p-1}-x_{p}=0},.$ Using Theorem 6.14 of Rockafellar & Wets (1998), we have

$$N_{\mathfrak{C}}=\{\sum _{i=1}^{p-1}{\lambda }_i(0,\dots ,0,\underset{\text{the}{i}_{\text{th}}\text{coordinate}}{\underbrace{1}},-1,0,\dots ,0):({\lambda }_1,\dots ,{\lambda }_{p-1})\in {\mathbb{R}}^{p-1}\},N\_\{\backslash mathfrak\{C\}\}=\backslash left\backslash \{\backslash sum\_\{i=1\}^\{p-1\}\backslash lambda\_\{i\}(0,\backslash ldots,0,\backslash underbrace\{1\}\_\{\backslash text\{the\$i\_\{\backslash text\{th\}\}\$coordinate\}\},-1,0,\backslash ldots,0):\backslash \; (\backslash lambda\_\{1\},\backslash ldots,\backslash lambda\_\{p-1\})\backslash in\backslash mathbb\{R\}^\{p-1\}\backslash right\backslash \},$$

where 1 is the vector in R p whose coordinates are all one.

This together with Corollary 10.9, Proposition 10.5 shows that for any Y ∈ dom∂G, ∂G(Y ) can be represetned as

$$\{(\xi ,0,\dots ,0)+\sum _{i=1}^{p-1}{\lambda }_i(0,\dots ,0,\underset{i_0}{\underbrace{1}},-1,0,\dots ,0):\xi \in \mathrm{\partial }g(y_1),({\lambda }_1,\dots ,{\lambda }_{p-1})\in {\mathbb{R}}^{p-1}\}\mathrm{.}\backslash left\backslash \{(\backslash xi,0,\backslash ldots,0)+\backslash sum\_\{i=1\}^\{p-1\}\backslash lambda\_\{i\}(0,\backslash ldots,0,\backslash underbrace\{1\}\_\{i\_\{0\}\},-1,0,\backslash ldots,0):\backslash ;\backslash xi\backslash in\backslash partial\; g(y\_\{1\}),\backslash ;(\backslash lambda\_\{1\},\backslash ldots,\backslash lambda\_\{p-1\})\backslash in\backslash mathbb\{R\}^\{p-1\}\backslash right\backslash \}.$$ $$(17)(17)$$ $$(18)(18)$$ . (17) Suppose Y ∗ = (y ∗ 1 , . . . , y∗ p ) is a stationary point of (3). Then Y ∗ ∈ dom∂G ⊆ domG. Thus, y ∗ 1 = · · · = y ∗ p .

In addition, it holds that

$0\in\nabla F(Y^{})+\partial G(Y^{})$ $$=(\nabla f_{1}(y^{}),\ldots,\nabla f_{p}(y^{}))+(\partial g(y_{1}^{*}),0,\ldots,0)+\sum_{i=1}^{p}\lambda_{i}(0,\ldots,0,\underbrace{1}{t{i_{k}}},-1,0,\ldots,0),$$ $\lambda_{i}$ is the $i_{i_{1}}$-norm of $\lambda_{i}$.
where the second equality uses (17) together with Exercise 8.8 and Proposition 10.5 of Rockafellar & Wets (1998). The above relation is equivalent to

$$\begin{array}{c}0\in \mathrm{\nabla }{f}_1(y^{\ast })+\mathrm{\partial }g(y_1^{\ast })+{\lambda }_1\mathbf{1}\\ 0=\mathrm{\nabla }{f}_2-{\lambda }_1\mathbf{1}+{\lambda }_2\mathbf{1}\end{array}\backslash begin\{array\}\{l\}\{\{0\backslash in\backslash nabla\; f\_\{1\}(y^\{*\})+\backslash partial\; g(y\_\{1\}^\{*\})+\backslash lambda\_\{1\}\{\backslash bf1\}\}\}\backslash \backslash \; \{\{0=\backslash nabla\; f\_\{2\}-\backslash lambda\_\{1\}\{\backslash bf1\}+\backslash lambda\_\{2\}\{\backslash bf1\}\}\}\backslash end\{array\}$$ $$(19)(19)$$ ... $$\begin{array}{c}\\ 0=\mathrm{\nabla }{f}_{p-1}-{\lambda }_{p-2}\mathbf{1}+{\lambda }_{p-1}\mathbf{1}\\ 0=\mathrm{\nabla }{f}_p(y^{\ast })-{\lambda }_{p-1}\mathbf{1.}\end{array}\backslash begin\{array\}\{l\}\{\{\backslash vdots\}\}\backslash \backslash \; \{\{0=\backslash nabla\; f\_\{p-1\}-\backslash lambda\_\{p-2\}\{\backslash bf1\}+\backslash lambda\_\{p-1\}\{\backslash bf1\}\}\}\backslash \backslash \; \{\{0=\backslash nabla\; f\_\{p\}(y^\{*\})-\backslash lambda\_\{p-1\}\{\backslash bf1\}.\}\}\backslash end\{array\}$$

Substituting λ1 in (19) using the rest equality in the above relation, we have that

$$0\in \sum _i\mathrm{\nabla }{f}_i(y^{\ast })+\mathrm{\partial }g(y_1^{\ast })\mathrm{.}0\backslash in\backslash sum\_\{i\}\backslash nabla\; f\_\{i\}(y^\{*\})+\backslash partial\; g(y\_\{1\}^\{*\}).$$

Thus y ∗is a stationary point of (1).

Now, suppose Y = (y1*, . . . , y*p) is a ε-stationary point of (3). Then Y ∈ dom∂G ⊆ domG. Thus, y1 = · · · = yp and $$\epsilon \ge d^2(0,\mathrm{\nabla }F(Y)+\mathrm{\partial }G(Y))\mathrm{.}\backslash varepsilon\backslash geq\; d^\{2\}(0,\backslash nabla\; F(Y)+\backslash partial\; G(Y)).$$ 2(0, ∇F(Y ) + ∂G(Y )). (20) Using (17) and Proposition 10.5 of Rockafellar & Wets (1998), we have that d 2(0, ∇F(Y ) + ∂G(Y ))

= min ξ∈∂g(y1),λ∈Rp−1 ∥∇f1(y1) + ξ + λ11∥ 2 + Xp−2 i=2 ∥∇f1(y1) + λi1 − λi−11∥ 2 + ∥∇fp(y1) − λp−11∥ 2 (21) ≥ min ξ∈∂g(y1),λ∈Rp−1 1 p ∥ X i ∇fi(y1) + ξ∥ 2∥ 2 = min ξ∈∂g(y1) 1 p ∥ X i ∇fi(y1) + ξ∥ 2∥ 2 = 1 p d 2(0, X i ∇fi(y1) + ∂g(y1)). $\square$ . $$(23)(23)$$ $$\begin{array}{c}(24)\\ (25)\end{array}\backslash begin\{array\}\{l\}\{(24)\}\backslash \backslash \; \{(25)\}\backslash end\{array\}$$ This together with (20) shows that y1 is a pε-stationary point.

B.1 Proofs Of Proposition 2

The problem in updating Y t+1 in (6) is a constrained problem:

In (8) is a concatenated problem: $\min\limits_{Y},g(y_1)+\left<Z^t,X^{t+1}-Y\right>+\frac{\beta}{2}|X^{t+1}-Y|^2$ s.t. $y_2=y_3=\cdots=y_p=y_1.$ thanks you claim an issue.Than author used here. $$(22)(22)$$ Since β > L, the objective in the above problem is strongly convex. Thus, there exists a unique solution (y1, y2*, . . . , yp) to (22). Denote the Lagrange multiplier for the above problem as W = (w1, w2, . . . , w*p).

Then the Karush–Kuhn–Tucker condition for the above problem is

$\alpha$ condition for the above problem is $$0\in\partial g(y_{1})-z_{1}^{t+1}-\beta(x_{1}^{t+1}-y_{1})-\sum_{i=2}^{p}w_{i}$$ $$0=-z_{i}^{t+1}+w_{i}-\beta(x_{i}^{t+1}-y_{i}),,i=2,\ldots,p$$ $$y_{i}=y_{1},,,i=2,\ldots,p.$$ Combining (24) with (25) gives

With (20) gives $ \sum_{i=2}^p w_i=\beta\sum_{i=2}^p(x_i^{t+1}-y_i)+\sum_{i=2}^p z_i^{t+1}=\beta\sum_{i=2}^p x_i^{t+1}-(p-1)\beta y_1+\sum_{i=2}^p z_i^{t+1}.$ with (22) shows that... This together with (23) shows that

$$\beta \sum _{i=2}^p{x}_i^{t+1}-(p-1)\beta y_1+\sum _{i=2}^p{z}_i^{t+1}+z_1^{t+1}+\beta x_1^{t+1}\in \mathrm{\partial }g(y_1)+\beta y_1,\backslash beta\backslash sum\_\{i=2\}^\{p\}x\_\{i\}^\{t+1\}-(p-1)\backslash beta\; y\_\{1\}+\backslash sum\_\{i=2\}^\{p\}z\_\{i\}^\{t+1\}+z\_\{1\}^\{t+1\}+\backslash beta\; x\_\{1\}^\{t+1\}\backslash in\backslash partial\; g(y\_\{1\})+\backslash beta\; y\_\{1\},$$

which is equivalent to

$$\frac{1}{p}\sum _{i=1}^p(x_i^{t+1}+\frac{1}{\beta }{z}_i^{t+1})\in \frac{1}{\beta p}\mathrm{\partial }g(y_1)+y_1\mathrm{.}\{\backslash frac\{1\}\{p\}\}\backslash sum\_\{i=1\}^\{p\}(x\_\{i\}^\{t+1\}+\{\backslash frac\{1\}\{\backslash beta\}\}z\_\{i\}^\{t+1\})\backslash in\{\backslash frac\{1\}\{\backslash beta\; p\}\}\backslash partial\; g(y\_\{1\})+y\_\{1\}.$$

This implies that y1 ∈ Prox 1 βp g ( 1 p Pp i=1(x t+1 i + 1 β z t+1 i)). Recalling (25), we deduce that the solution of the problem in the third equation of (6) is (y1*, . . . , y*1) with y1 = Prox 1 βp g ( 1 p Pp i=1(x t+1 i + 1 β z t+1 i))).

Proposition 3. Consider (1). Set β > L := maxi Li*. Let* {(x t i , yt i , zt i )} be generated by Algorithm 1. Using

SVRG in Johnson & Zhang (2013) with Option II with frequency mi, learning rate ηi*, and initialization* x t i for (8), such that Then criterion (9) is satisfied in at most k i t = log1/ρiβ+Li ri(β−Li) iterations of SVRG.

Proof. Note that L(x, yt i , zt i ) is strongly convex with modulos β−Li and ∇L(x, yt i , zt i ) is Lipschitz continuous with modulos Li + β. Let ρi:= 1 (β−Li)η(1−2ηi(β+Li))mi +2ηi(β+Li) 1−2ηi(β+Li) , where mi and ηiis the frequency and learning rate in SVRG respectively. Using Theorem 1 of Johnson & Zhang (2013), there exists large m such that E

t iLβ,i(x t+1 i, yt, zt i ) − Lβ,i(x t+1 i,⋆ , yt, zt i ) ≤ ρ kt i Lβ,i(x t i , yt, zt i )−Lβ,i(x t+1 i,⋆ , yt, zt i )(26) Combing this with the strong convexity of L(x, yt i , zt i ) and the Lipschitz continuity of ∇L(x, yt i , zt i ), we have

that $$\mathbb{E}{i}^{t}|x{i}^{t+1}-x_{i,}^{t+1}|^{2}\leq\frac{\beta+L_{i}}{\beta-L_{i}}\rho_{i}^{t_{i}}|x_{i}^{t}-x_{i,}^{t+1}|^{2}\leq r_{i}|x_{i}^{t}-x_{i,*}^{t+1}|^{2},\tag{27}$$ where the second inequality is based on $\frac{\beta+L_{i}}{\beta-L_{i}}\rho_{i}^{t_{i}}\leq r_{i}$. This completes the proof.

C Proof For Convergence Analysis

To prove the results in Section Convergence Analysis of Algorithm 1, we first present the following well known facts for strongly convex functions, see Theorem 2 in Karimi et al. (2016) for example.

Proposition 7. Let f : R n → R be a strongly convex function with modulus µ*. Suppose in addition that* f is smooth and has Lipschitz continuous gradient with modulus L*. Then there exists unique minimizer* x ∗ that minimize f and it holds that

$$\mathrm{\parallel }\mathrm{\nabla }f(x){\mathrm{\parallel }}^2\ge 2\mu (f(x)-f(x^{\ast }))\ge {\mu }^2\mathrm{\parallel }x-x^{\ast }{\mathrm{\parallel }}^2\mathrm{.}\backslash |\backslash nabla\; f(x)\backslash |^\{2\}\backslash geq2\backslash mu\backslash left(f(x)-f(x^\{*\})\backslash right)\backslash geq\backslash mu^\{2\}\backslash |x-x^\{*\}\backslash |^\{2\}.$$

Proposition 2. Consider (3). Let {(Xt+1, Y t+1, Zt+1)} be generated by (6). Suppose β > maxi Li*. Then* the solution of the problem in the third equation of (6) is (y1, . . . , y1) with y1 = Prox 1 βp g ( 1 p Pp i=1(x t+1 i + 1 β z t+1 i))).

The second and third relation in this proposition are obvious. We only need show that Xtsatisfies (11).

Using (9) and the definition that r = maxi r, we have

$${\mathbb{E}}_i^t\mathrm{\parallel }{x}_i^{t+1}-x_{i,\star }^{t+1}{\mathrm{\parallel }}^2\le r_i\mathrm{\parallel }{x}_i^t-x_{i,\star }^{t+1}{\mathrm{\parallel }}^2\le r\mathrm{\parallel }{x}_i^t-x_{i,\star }^{t+1}{\mathrm{\parallel }}^2,\backslash mathbb\{E\}\_\{i\}^\{t\}\backslash |x\_\{i\}^\{t+1\}-x\_\{i,\backslash star\}^\{t+1\}\backslash |^\{2\}\backslash leq\; r\_\{i\}\backslash |x\_\{i\}^\{t\}-x\_\{i,\backslash star\}^\{t+1\}\backslash |^\{2\}\backslash leq\; r\backslash |x\_\{i\}^\{t\}-x\_\{i,\backslash star\}^\{t+1\}\backslash |^\{2\},$$

summing i = 1*, . . . , p*, we obtain (11).

C.1 Details And Proofs Of Proposition 5

Before proving Proposition 5, we first present several properties of the problem:

$$\underset{X}{\min }{L}_{\beta }(X,Y^t,Z^t),\backslash operatorname*\{min\}\_\{X\}L\_\{\backslash beta\}(X,Y^\{t\},Z^\{t\}),$$ $$(28)(28)$$ XLβ(X, Y t, Zt), (28) where Y t and z t are defined as in Proposition 4.

Proposition 8. Consider (1). Let (Xt, Y t, Zt) be defined as in Proposition 4. Let β ≥Pi Li*. Denote* Xt+1 ⋆:= minX Lβ(X, Y t, Zt+1).

1 Then the following statements hold:

(i) Denote $e^{t+1}=X^{t+1}-X^{t+1}{\star}$. Then there exists $\xi^{t+1}\in\partial G(Y^{t+1})$ such that_ $$0=\mathrm{\nabla }F(X_{\star }^{t+1})+Z^t+\beta (X_{\star }^{t+1}-Y^t)\iff -Z^t-\beta (X^{t+1}-e^{t+1}-Y^t)=\mathrm{\nabla }F(X_{\star }^{t+1})0=\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\})+Z^\{t\}+\backslash beta(X\_\{\backslash star\}^\{t+1\}-Y^\{t\})\backslash Leftrightarrow-Z^\{t\}-\backslash beta(X^\{t+1\}-e^\{t+1\}-Y^\{t\})=\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\})$$ ⋆) (29) and $$0={\xi }^{t+1}-Z^{t+1}-\beta (X^{t+1}-Y^{t+1})0=\backslash xi^\{t+1\}-Z^\{t+1\}-\backslash beta(X^\{t+1\}-Y^\{t+1\})$$ t+1) (30) (ii) It holds that

$$Z^{t+1}=(1-\tau )Z^t+\beta \tau e^{t+1}+\tau \mathrm{\nabla }F(X_{\star }^{t+1})Z^\{t+1\}=(1-\backslash tau)Z^\{t\}+\backslash beta\backslash tau\; e^\{t+1\}+\backslash tau\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\})$$

⋆) (31) 1The existence and uniqueness of X t+1 ⋆ are thanks to β ≥ maxi Li and Proposition 7.

$$(29)(29)$$ $$(30)(30)$$ $$(31)(31)$$

(iii) Let r = maxi ri*. It holds that*

$$\mathbb{E}\mathrm{\parallel }{e}^t{\mathrm{\parallel }}^2\le \frac{2r}{1-2r}\mathbb{E}\mathrm{\parallel }{x}^t-x^{t-1}{\mathrm{\parallel }}^2\backslash mathbb\{E\}\backslash |e^\{t\}\backslash |^\{2\}\backslash leq\{\backslash frac\{2r\}\{1-2r\}\}\backslash mathbb\{E\}\backslash |x^\{t\}-x^\{t-1\}\backslash |^\{2\}$$ $$(32)(32)$$

Proof. (i) follows from the first optimality condition of (28) and (13). Combining (29) with (12), we have that

$$-Z^t-\frac{1}{\tau }(Z^{t+1}-Z^t)+\beta e^{t+1}=-Z^t-\beta (X^{t+1}-e^{t+1}-Y^t)=\mathrm{\nabla }F(X_{\star }^{t+1})\mathrm{.}-Z^\{t\}-\backslash frac\{1\}\{\backslash tau\}(Z^\{t+1\}-Z^\{t\})+\backslash beta\; e^\{t+1\}=-Z^\{t\}-\backslash beta(X^\{t+1\}-e^\{t+1\}-Y^\{t\})=\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\}).$$ $$\Leftrightarrow Z^{t+1}=(1-\tau)Z^{t}+\beta\tau e^{t+1}+\tau\nabla F(X_{\star}^{t+1}).$$

Now, we bound E∥e t∥ 2. Denote e t i := x t i − x t i∗ . Then using (27), we have that

$\mathbb{E}{t-1}|e{i}^{t}|^{2}\leq r_{i}|x_{i}^{t-1}-x_{i*}^{t}|^{2}\leq2r_{i}(|x_{i}^{t}-x_{i}^{t-1}|^{2}+|e_{i}^{t}|^{2})$.
where c ′ i := β+Li β−Li . Denote c ′ = maxi c ′ i , ρ := maxi ρi and kt := maxi k i t , r = maxi ri. Summing both sides of the above inequality from i = 1*, . . . , p*, we obtain that

$${\mathbb{E}}_{t-1}\mathrm{\parallel }{e}^t{\mathrm{\parallel }}^2\le 2r(\mathrm{\parallel }{x}^t-x^{t-1}{\mathrm{\parallel }}^2+{\mathbb{E}}_{t-1}\mathrm{\parallel }{e}^t{\mathrm{\parallel }}^2)\mathrm{.}\backslash mathbb\{E\}\_\{t-1\}\backslash |e^\{t\}\backslash |^\{2\}\backslash leq2r(\backslash |x^\{t\}-x^\{t-1\}\backslash |^\{2\}+\backslash mathbb\{E\}\_\{t-1\}\backslash |e^\{t\}\backslash |^\{2\}).$$

Taking expectation on both sides over all randomness and rearranging the above inequality we obtain (32).

Now, we are ready to prove Proposition 5.

Proposition 5. Select hyperparameters β ≥ 5L, ri ∈ (0, 0.01], τ ∈ [1/2, 1). Denote Γ := 1−τ τ, Θ = 2β 2 + 4L 2, Λ := 4L 2. Υ := Θ τβ4r 1−2r and δ := 14 (β − L) − 2Υ*. Define*

$$H(X,Y,Z,X^{\mathrm{\prime }},Z^{\mathrm{\prime }}):=L_{\beta }(X,Y,Z)+\frac{\mathrm{\Gamma }}{\tau \beta }\mathrm{\parallel }Z-Z^{\mathrm{\prime }}{\mathrm{\parallel }}^2+\mathrm{{\rm Y}}\mathrm{\parallel }X-X^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{.}H(X,Y,Z,X^\{\backslash prime\},Z^\{\backslash prime\}):=L\_\{\backslash beta\}(X,Y,Z)+\backslash frac\{\backslash Gamma\}\{\backslash tau\backslash beta\}\backslash |Z-Z^\{\backslash prime\}\backslash |^\{2\}+\backslash Upsilon\backslash |X-X^\{\backslash prime\}\backslash |^\{2\}.$$ $${}^{t+1},Y^{t+1},Z^{t+1},X^t,Z^t)\mathrm{.}T\{\}^\{t+1\},Y^\{t+1\},Z^\{t+1\},X^\{t\},Z^\{t\}).\backslash ;\backslash ;T$$

and Ht+1 := EH(Xt+1, Y t+1, Zt+1, Xt, Zt). Then for t ≥ 1, it holds that δ ≥ 0.1L and

$$\begin{array}{cccc}& H_{t+1}\le H_t-\delta \mathbb{E}\mathrm{\parallel }{X}^{t+1}-X^t{\mathrm{\parallel }}^2-\frac{\beta }{2}\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\mathrm{.}& & \text{(1)}\end{array}H\_\{t+1\}\backslash leq\; H\_\{t\}-\backslash delta\backslash mathbb\{E\}\backslash |X^\{t+1\}-X^\{t\}\backslash |^\{2\}-\backslash frac\{\backslash beta\}\{2\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}.\backslash tag\{1\}$$ $$(14)(14)$$

Hence, the sequence {Ht} converges to some H∗ ≥ W.

Proof. Note that

EtLβ(Xt+1, Y t, Zt) − Lβ(Xt, Y t, Zt) = Lβ(Xt+1, Y t, Zt) − Lβ(Xt+1 ⋆, Y t, Zt) + Lβ(Xt+1 ⋆, Y t, Zt) − Lβ(Xt, Y t, Zt) ≤ ρ ktLβ(Xt, Y t, Zt) − Lβ(Xt+1 ⋆, Y t, Zt)+ Lβ(Xt+1 ⋆, Y t, Zt) − Lβ(Xt, Y t, Zt) ≤ ρ ktLβ(Xt, Y t, Zt) − Lβ(Xt+1 ⋆, Y t, Zt)− β − L 2∥Xt − Xt+1 ⋆ ∥ 2 ≤ ρ ktLβ(Xt, Y t, Zt) − Lβ(Xt+1 ⋆, Y t, Zt)− β − L 4 Et∥Xt − Xt+1∥ 2 + β − L 2 Et∥Xt+1 − Xt+1 ⋆ ∥ 2 ≤ ρ kt β + L 2∥Xt − Xt+1 ⋆ ∥ 2 − β − L 4 Et∥Xt − Xt+1∥ 2 + β − L 2 Et∥e t+1∥ 2, (33) where the first inequality makes use of (26), the second inequality is because Lβ(X, Y t, Zt) is strongly convex with modulus β − maxi Li and Xt+1 ⋆is the minimizer of minX Lβ(X, Y t, Zt), the third inequality uses Young's inequality, the last inequality uses the Lipschitz continuity of ∇XLβ(X, Y t, Zt).

Using the fact that ∥Xt − Xt+1 ⋆ ∥ 2 ≤ 2Et∥Xt − Xt+1∥ 2 + 2Et∥e t+1∥ 2, (33) can be further passed to EtLβ(Xt+1, Y t, Zt) − Lβ(Xt, Y t, Zt) ≤ 2ρ kt β + L 2 Et∥Xt − Xt+1∥ 2 + 2ρ kt β + L 2 Et∥e t+1∥ 2 − β − L 4 Et∥Xt − Xt+1∥ 2 + β − L 2 Et∥e t+1∥ 2 = 2ρ kt β + L 2− β − L 4 Et∥Xt − Xt+1∥ 2 + 2ρ kt β + L 2+ β − L 2 Et∥e t+1∥ 2 (34) ≤ 2ρ kt β + L 2− β − L 4+ 2ρ kt β + L 2+ β − L 2 2r 1 − 2r Et∥Xt − Xt+1∥ 2 = ρ kt 1 − 2r (β + L) − 1 4 −r 1 − 2r (β − L) Et∥Xt − Xt+1∥ 2 where the second inequality uses (32). Next, using (12), we have

$$\begin{array}{cccc}& L_{\beta }(X^{t+1},Y^t,Z^{t+1})-L_{\beta }(X^{t+1},Y^t,Z^t)=\frac{1}{\tau \beta }\mathrm{\parallel }{Z}^{t+1}-Z^t{\mathrm{\parallel }}^2& & \text{(35)}\end{array}L\_\{\backslash beta\}(X^\{t+1\},Y^\{t\},Z^\{t+1\})-L\_\{\backslash beta\}(X^\{t+1\},Y^\{t\},Z^\{t\})=\backslash frac\{1\}\{\backslash tau\backslash beta\}\backslash |Z^\{t+1\}-Z^\{t\}\backslash |^\{2\}\backslash tag\{35\}$$

When τ ∈ (0, 1), combining (31) and the convexity of ∥ · ∥2, we have that

$$\begin{array}{c}\mathrm{\parallel }{Z}^{t+1}-Z^t{\mathrm{\parallel }}^2\le (1-\tau )\mathrm{\parallel }{Z}^t-Z^{t-1}{\mathrm{\parallel }}^2+\tau \mathrm{\parallel }\beta (e^{t+1}-e^t)+\mathrm{\nabla }(F(X_{\ast }^{t+1})-F(X_{\ast }^t)){\mathrm{\parallel }}^2\\ \le (1-\tau )\mathrm{\parallel }{Z}^t-Z^{t-1}{\mathrm{\parallel }}^2+2\tau {\beta }^2\mathrm{\parallel }{e}^{t+1}-e^t{\mathrm{\parallel }}^2+2\tau \mathrm{\parallel }\mathrm{\nabla }(F(X_{\ast }^{t+1})-F(X_{\ast }^t)){\mathrm{\parallel }}^2\\ \le (1-\tau )\mathrm{\parallel }{Z}^t-Z^{t-1}{\mathrm{\parallel }}^2+2\tau {\beta }^2\mathrm{\parallel }{e}^{t+1}-e^t{\mathrm{\parallel }}^2+2\tau L^2\mathrm{\parallel }{X}_{\ast }^{t+1}-X_{\ast }^t{\mathrm{\parallel }}^2,\end{array}\backslash begin\{array\}\{l\}\{\{\backslash |Z^\{t+1\}-Z^\{t\}\backslash |^\{2\}\backslash leq(1-\backslash tau)\backslash |Z^\{t\}-Z^\{t-1\}\backslash |^\{2\}+\backslash tau\backslash |\backslash beta(e^\{t+1\}-e^\{t\})+\backslash nabla(F(X\_\{*\}^\{t+1\})-F(X\_\{*\}^\{t\}))\backslash |^\{2\}\}\}\backslash \backslash \; \{\{\backslash leq(1-\backslash tau)\backslash |Z^\{t\}-Z^\{t-1\}\backslash |^\{2\}+2\backslash tau\backslash beta^\{2\}\backslash |e^\{t+1\}-e^\{t\}\backslash |^\{2\}+2\backslash tau\backslash |\backslash nabla(F(X\_\{*\}^\{t+1\})-F(X\_\{*\}^\{t\}))\backslash |^\{2\}\}\}\backslash \backslash \; \{\{\backslash leq(1-\backslash tau)\backslash |Z^\{t\}-Z^\{t-1\}\backslash |^\{2\}+2\backslash tau\backslash beta^\{2\}\backslash |e^\{t+1\}-e^\{t\}\backslash |^\{2\}+2\backslash tau\; L^\{2\}\backslash |X\_\{*\}^\{t+1\}-X\_\{*\}^\{t\}\backslash |^\{2\},\}\}\backslash end\{array\}$$

where the second inequality uses the Young's inequality for product, and the last inequality uses the Lipschitz continuity of ∇F. Rearranging the above inequality, we have that

∥Z t+1 − Z t∥ 2 ≤ 1 − τ τ ∥Z t − Z t−1∥ 2 − ∥Z t+1 − Z t∥ 2+ 2β 2∥e t+1 − e t∥ 2 + 2L 2∥Xt+1 ⋆ − Xt⋆∥ 2 ≤ 1 − τ τ ∥Z t − Z t−1∥ 2 − ∥Z t+1 − Z t∥ 2+ 2β 2∥e t+1 − e t∥ 2 (36) + 2L 2(1 + κ 2)∥Xt+1 − Xt∥ 2 + (1 + κ −2)∥e t+1 − e t∥ 2 = 1 − τ τ ∥Z t − Z t−1∥ 2 − ∥Z t+1 − Z t∥ 2+2β 2 + 4L 2∥e t+1 − e t∥ 2 + 4L 2∥Xt+1 − Xt∥ 2, $$(37)(37)$$ where κ > 0 and the last inequality uses the definition of e t+1 and Young's inequality for products.

Using the definition of Γ, Θ and Λ, (36) becomes

$${\parallel Z^{t+1}-Z^t\parallel }^2\le \mathrm{\Gamma }({\parallel Z^{t-1}-Z^t\parallel }^2-{\parallel Z^{t+1}-Z^t\parallel }^2)+\mathrm{\Theta }\mathrm{\parallel }{e}^t-e^{t+1}{\mathrm{\parallel }}^2+\mathrm{\Lambda }{\parallel X^t-X^{t+1}\parallel }^2\mathrm{.}\backslash left\backslash |Z^\{t+1\}-Z^\{t\}\backslash right\backslash |^\{2\}\backslash leq\backslash Gamma\backslash left(\backslash left\backslash |Z^\{t-1\}-Z^\{t\}\backslash right\backslash |^\{2\}-\backslash left\backslash |Z^\{t+1\}-Z^\{t\}\backslash right\backslash |^\{2\}\backslash right)+\backslash Theta\backslash |e^\{t\}-e^\{t+1\}\backslash |^\{2\}+\backslash Lambda\backslash left\backslash |X^\{t\}-X^\{t+1\}\backslash right\backslash |^\{2\}.$$ 2. (37) Now, combining (34), (35) and (37), we obtain that

$${\mathbb{E}}_t{L}_{\beta }(X^{t+1},Y^t,Z^{t+1})\backslash mathbb\{E\}\_\{t\}L\_\{\backslash beta\}(X^\{t+1\},Y^\{t\},Z^\{t+1\})$$ EtLβ(Xt+1, Y t, Zt+1) ≤ Lβ(Xt, Y t, Zt) + ρ kt 1 − 2r (β + L) − 1 4 −r 1 − 2r (β − L) Et∥Xt − Xt+1∥ 2 + Γ τ β ∥Z t−1 − Z t∥ 2 − Et∥Z t+1 − Z t∥ 2+ Θ τ β Et∥e t − e t+1∥ 2 + Λ τ β Et Xt − Xt+1 2 = Lβ(Xt, Y t, Zt) + ρ kt 1 − 2r (β + L) − 1 4 −r 1 − 2r (β − L) ∥Xt − Xt+1∥ 2 + Γ τ β ∥Z t−1 − Z t∥ 2 − Et∥Z t+1 − Z t∥ 2+ Θ τ β Et∥e t − e t+1∥ 2. Taking expectations with respect to X t, the above inequality implies

with respect to $\lambda^{t}$, the above inequality implies $$\begin{split}&\mathbb{E}L_{\beta}(X^{t+1},Y^{t},Z^{t+1})\leq\mathbb{E}L_{\beta}(X^{t},Y^{t},Z^{t})\ &+\left(\frac{\rho^{t_{k}}}{1-2r}(\beta+L)-\left(\frac{1}{4}-\frac{r}{1-2r}\right)(\beta-L)\right)\mathbb{E}|X^{t}-X^{t+1}|^{2}\ &+\frac{\Gamma}{r\beta}\left(\mathbb{E}|Z^{t-1}-Z^{t}|^{2}-\mathbb{E}|Z^{t+1}-Z^{t}|^{2}\right)+\frac{\Theta}{r\beta}\mathbb{E}|e^{t}-e^{t+1}|^{2}.\end{split}\tag{38}$$ (39) we obtain that Combining (32) with (38), we obtain that

ELβ(Xt+1, Y t, Zt+1) ≤ ELβ(Xt, Y t, Zt) + ρ kt 1 − 2r (β + L) − 1 4 −r 1 − 2r (β − L) E∥Xt − Xt+1∥ 2 (39) + Γ τ β E∥Z t−1 − Z t∥ 2 − E∥Z t+1 − Z t∥ 2 + Θ τ β 4r 1 − 2r E∥Xt − Xt−1∥ 2 + Θ τ β 4r 1 − 2r E∥Xt − Xt+1∥ 2. i , L = maxi Li, ρ = maxi ρi, r = maxi ri and k i satisfies β+L ρ kt i ≤ ri. This implies Recall that kt = mini k t t β−L

$${\rho }^{k_t}\le \frac{\beta -L}{\beta +L}r\mathrm{.}\backslash rho^\{k\_\{t\}\}\backslash leq\backslash frac\{\backslash beta-L\}\{\backslash beta+L\}r.$$

This together with (39) shows that

The general case ($\alpha$) shows that $$\mathbb{E}L_{\beta}(X^{t+1},Y^{t},Z^{t+1})\leq\mathbb{E}L_{\beta}(X^{t},Y^{t},Z^{t})-\frac{1}{4}(\beta-L)\mathbb{E}|X^{t}-X^{t+1}|^{2}$$ $$+\frac{\Gamma}{\tau\beta}\left(\mathbb{E}|Z^{t-1}-Z^{t}|^{2}-\mathbb{E}|Z^{t+1}-Z^{t}|^{2}\right)+\underbrace{\frac{\Theta}{\tau\beta}\frac{4r}{1-2r}}_{\Upsilon}\mathbb{E}|X^{t}-X^{t-1}|^{2}+\frac{\Theta}{\tau\beta}\frac{4r}{1-2r}\mathbb{E}|X^{t}-X^{t+1}|^{2}.$$ 2.(40) Finally, using the definition of δ and Υ, (40) further implies

$$\mathbb{E}{L}_{\beta }(X^{t+1},Y^t,Z^{t+1})\backslash mathbb\{E\}L\_\{\backslash beta\}(X^\{t+1\},Y^\{t\},Z^\{t+1\})$$ $$\leq\mathbb{E}L_{\beta}(X^{t},Y^{t},Z^{t})-\delta\mathbb{E}|X^{t}-X^{t+1}|^{2}$$ $$+\frac{\Gamma}{\tau\beta}\left(\mathbb{E}|Z^{t-1}-Z^{t}|^{2}-\mathbb{E}|Z^{t+1}-Z^{t}|^{2}\right)$$ $$+\Upsilon\left(\mathbb{E}|X^{t}-X^{t-1}|^{2}-\mathbb{E}|X^{t+1}-X^{t}|^{2}\right).$$ $$(40)(40)$$ $$(41)(41)$$ $$\left(42\right)\backslash left(42\backslash right)$$

Next, noting that Y t+1 is the minimizer of (13) which is β-strongly convex, it holds that

$$\mathbb{E}{L}_{\beta }(X^{t+1},Y^{t+1},Z^{t+1})\le \mathbb{E}{L}_{\beta }(X^{t+1},Y^t,Z^{t+1})-\frac{\beta }{2}\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\mathrm{.}\backslash mathbb\{E\}L\_\{\backslash beta\}(X^\{t+1\},Y^\{t+1\},Z^\{t+1\})\backslash leq\backslash mathbb\{E\}L\_\{\backslash beta\}(X^\{t+1\},Y^\{t\},Z^\{t+1\})-\backslash frac\{\backslash beta\}\{2\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}.$$ Summing (42) and (41), we have that $$\mathbb{E}L_{\beta}(X^{t+1},Y^{t+1},Z^{t+1})$$ $$\leq\mathbb{E}L_{\beta}(X^{t},Y^{t},Z^{t})-\delta\mathbb{E}|X^{t}-X^{t+1}|^{2}+\frac{\Gamma}{r\beta}\left(\mathbb{E}|Z^{t-1}-Z^{t}|^{2}-\mathbb{E}|Z^{t+1}-Z^{t}|^{2}\right)$$ $$\quad+\Upsilon\left(\mathbb{E}|X^{t}-X^{t-1}|^{2}-\mathbb{E}|X^{t+1}-X^{t}|^{2}\right)-\frac{\beta}{2}\mathbb{E}|Y^{t+1}-Y^{t}|^{2}.$$ Rearranging the above inequality and recalling the definition of $H(X,Y,Z,X^{\prime},Z^{\prime})$, we have that above inequality and recalling the definition of $H(X,Y,Z,X',Z')$, we have $;;;\mathbb{E}H(X^{t+1},Y^{t+1},Z^{t+1},X^t,Z^t)$ $;;;\leq\mathbb{E}H(X^t,Y^t,Z^t,X^{t-1},Z^{t-1})-\delta\mathbb{E}|X^t-X^{t+1}|^2-\frac{\beta}{2}\mathbb{E}|Y^{t+1}-Y^t|^2$. Now we prove {Ht} is convergent. Inequality (14) implies that {Ht} is nonincreasing. Since F and G are bounded from below, we denote W = inf F + inf G . Now we show that Ht ≥ W for all t. Suppose to the contrary that there exists t0 such that Ht0 < W. Since (14) implies Ht is nonincreasing, it hold that

$$\sum _{t=1}^T(H_t-W)\le \sum _{t=1}^{t_0-1}(H_t-W)+(T-t_0+1)(H_{t_0}-W)\mathrm{.}\backslash sum\_\{t=1\}^\{T\}(H\_\{t\}-W)\backslash leq\backslash sum\_\{t=1\}^\{t\_\{0\}-1\}(H\_\{t\}-W)+(T-t\_\{0\}+1)(H\_\{t\_\{0\}\}-W).$$

Thus

$$\begin{array}{cccc}& \underset{T\to \mathrm{\infty }}{\lim }\sum _{t=1}^T(H_t-W)=-\mathrm{\infty }\mathrm{.}& & \text{(1)}\end{array}\backslash lim\_\{T\backslash to\backslash infty\}\backslash sum\_\{t=1\}^\{T\}(H\_\{t\}-W)=-\backslash infty.\backslash tag\{1\}$$ $$(43)(43)$$

On the other hand, using (41), for t ≥ 1, it holds that

Ht − W ≥ EH(Xt+1, Y t+1, Zt+1, Xt, Zt) − W (a) ≥ ELβ(Xt+1, Y t, Zt+1) − W ≥ EF(Xt+1) + G(Y t) + Xt+1 − Y t, Zt+1− W ≥ EXt+1 − Y t, Zt+1 (b) =1 τ β EZ t+1 − Z t, Zt+1=1 τ β E∥Z t+1∥ 2 − E∥Z t∥ 2 + E∥Z t+1 − Z t∥ 2 ≥1 τ β (E∥Z t+1∥ 2 − E∥Z t∥ 2). where (a) makes use of the definition of Ht and Lβ, (b) uses (12). Summing the above inequality from t = 0 to T and take T to the infinity, we have that

$$\begin{array}{c}{\lim }_{T\to \mathrm{\infty }}{\sum }_{t=1}^T(H_t-W)\ge {\lim }_{T\to \mathrm{\infty }}{\sum }_{t=1}^T\frac{1}{\tau \beta }(\mathrm{\parallel }{Z}^{t+1}{\mathrm{\parallel }}^2-\mathrm{\parallel }{Z}^t{\mathrm{\parallel }}^2)\\ =\frac{1}{\tau \beta }{\lim }_{T\to \mathrm{\infty }}(\mathbb{E}\mathrm{\parallel }{Z}^{T+1}{\mathrm{\parallel }}^2-\mathbb{E}\mathrm{\parallel }{Z}^0{\mathrm{\parallel }}^2)\ge -\frac{1}{\tau \beta }\mathrm{\parallel }{Z}^0{\mathrm{\parallel }}^2>-\mathrm{\infty },\end{array}\backslash begin\{array\}\{l\}\{\{\backslash operatorname*\{lim\}\_\{T\backslash to\backslash infty\}\backslash sum\_\{t=1\}^\{T\}(H\_\{t\}-W)\backslash geq\backslash operatorname*\{lim\}\_\{T\backslash to\backslash infty\}\backslash sum\_\{t=1\}^\{T\}\backslash frac\{1\}\{\backslash tau\backslash beta\}(\backslash |Z^\{t+1\}\backslash |^\{2\}-\backslash |Z^\{t\}\backslash |^\{2\})\}\}\backslash \backslash \; \{\{\backslash \; =\backslash frac\{1\}\{\backslash tau\backslash beta\}\backslash operatorname*\{lim\}\_\{T\backslash to\backslash infty\}(\backslash mathbb\{E\}\backslash |Z^\{T+1\}\backslash |^\{2\}-\backslash mathbb\{E\}\backslash |Z^\{0\}\backslash |^\{2\})\backslash geq-\backslash frac\{1\}\{\backslash tau\backslash beta\}\backslash |Z^\{0\}\backslash |^\{2\}>-\backslash infty,\}\}\backslash end\{array\}$$

which contradicts with (43). Therefore, Ht is bounded from below. This together with (14) gives that {Ht} is convergent.

C.2 Details And Proofs Of Corollary 1

Thanks to Proposition 5, we have the following properties with respect to the successive changes.

Corollary 4. Consider (1) and let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Then the following statements hold.

(i) It holds that

$$\begin{array}{cccc}& \sum _{t=0}^T\mathbb{E}\mathrm{\parallel }{X}^t-X^{t+1}{\mathrm{\parallel }}^2+\sum _{t=0}^T\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\le \frac{L_{\beta }(X^0,Y^0,Z^0)+C-H_{\ast }}{\min \{\delta ,\frac{\beta }{2}\}}\mathrm{.}& & \text{(44)}\end{array}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |X^\{t\}-X^\{t+1\}\backslash |^\{2\}+\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}\backslash leq\backslash frac\{L\_\{\backslash beta\}(X^\{0\},Y^\{0\},Z^\{0\})+C-H\_\{*\}\}\{\backslash min\backslash \{\backslash delta,\backslash frac\{\backslash beta\}\{2\}\backslash \}\}.\backslash tag\{44\}$$

and

$$\begin{array}{cccc}& \sum _{t=0}^T\mathbb{E}\mathrm{\parallel }{Z}^t-Z^{t+1}{\mathrm{\parallel }}^2\le (1+\mathrm{\Gamma })\frac{3(r+1)}{(L-\beta {)}^2}\mathrm{\parallel }\mathrm{\nabla }{L}_{\beta }(X^0,Y^0,Z^0){\mathrm{\parallel }}^2+3\mathrm{\parallel }{X}^0-Y^0{\mathrm{\parallel }}^2& & \text{(45)}\end{array}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |Z^\{t\}-Z^\{t+1\}\backslash |^\{2\}\backslash leq(1+\backslash Gamma)\backslash frac\{3(r+1)\}\{(L-\backslash beta)^\{2\}\}\backslash |\backslash nabla\; L\_\{\backslash beta\}(X^\{0\},Y^\{0\},Z^\{0\})\backslash |^\{2\}+3\backslash |X^\{0\}-Y^\{0\}\backslash |^\{2\}\backslash tag\{45\}$$ $$+2\left(\Gamma+2\Theta\frac{2r}{1-2r}\right)\frac{L_{\beta}(X^{0},Y^{0},Z^{0})+C-W}{\min{\delta,\frac{\beta}{2}}},$$

where C := 2τ β(Γ + 1)∥X0 −Y 0∥ 2 +4 (L−β) 2 L+β+1 2 + 2τ β(Γ + 1) + Υ + (L−β) 2 8 ∥∇XLβ(X0, Y 0, Z0)∥ 2.

with Θ and Γ being defined as in Proposition 5.

$$\delta \sum _{t=1}^{T-1}\mathbb{E}\mathrm{\parallel }{X}^t-X^{t+1}{\mathrm{\parallel }}^2+\frac{\beta }{2}\sum _{t=1}^{T-1}\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\backslash delta\backslash sum\_\{t=1\}^\{T-1\}\backslash mathbb\{E\}\backslash |X^\{t\}-X^\{t+1\}\backslash |^\{2\}+\backslash frac\{\backslash beta\}\{2\}\backslash sum\_\{t=1\}^\{T-1\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}$$ $$\leq L_{\beta}(X^{0},Y^{0},Z^{0})+C-H_{T}\leq L_{\beta}(X^{0},Y^{0},Z^{0})+C-H_{*},$$ $$(49)(49)$$

Rearranging the above inequality, we have that

(40), we have $H_{T}\leq L_{\beta}(X^{0},Y^{0},Z^{0})+C$ $-\delta\sum_{t=1}^{T}\mathbb{E}|X^{t}-X^{t+1}|^{2}-\frac{\beta}{2}\sum_{t=1}^{T}\mathbb{E}|Y^{t+1}-Y^{t}|^{2}$ $\leq H_{1}-\delta\sum_{t=1}^{T-1}\mathbb{E}|X^{t}-X^{t+1}|^{2}-\frac{\beta}{2}\sum_{t=1}^{T-1}\mathbb{E}|Y^{t+1}-Y^{t}|^{2}$.
Now we bound H1. Note that H1 = ELβ(X1, Y 1, Z1) + Γ τ β E∥Z 1 − Z 0∥ 2 + ΥE∥X1 − X0∥ 2 (i) ≤ ELβ(X1, Y 0, Z1) + Γ τ β E∥Z 1 − Z 0∥ 2 + ΥE∥X1 − X0∥ 2 (ii) ≤ ELβ(X1, Y 0, Z0) + Γ + 1 τ β E∥Z 1 − Z 0∥ 2 + ΥE∥X1 − X0∥ 2 (iii) ≤ E Lβ(X0, Y 0, Z0) + ∇XLβ(X0, Y 0, Z0) ⊤(X1 − X0) + L + β 2∥X1 − X0∥ 2+ τ β(Γ + 1)E∥X1 − Y 0∥ 2 + ΥE∥X1 − X0∥ 2 ≤ Lβ(X0, Y 0, Z0) + 12 ∥∇XLβ(X0, Y 0, Z0)∥ 2 + 2τ β(Γ + 1)∥X0 − Y 0∥ 2 + L + β + 1 2+ 2τ β(Γ + 1) + ΥE∥X1 − X0∥ 2 (iv) ≤ Lβ(X0, Y 0, Z0) + 2τ β(Γ + 1)∥X0 − Y 0∥ 2 +4 (L − β) 2 L + β + 1 2+ 2τ β(Γ + 1) + Υ + (L − β) 2 8 ∥∇XLβ(X0, Y 0, Z0)∥ 2, (48) where (i) uses (42), (ii) uses (35), (iii) uses the property that Lβ(X, Y, ·) is (L + β)-smooth, and (iv) uses the following inequality. E∥X1 − X0∥ 2 ≤ 2E∥X1 − X1 ∗ ∥ 2 + 2E∥X0 − X1 ∗ ∥ 2 ≤ 4E∥X0 − X1 ∗ ∥ 2 ≤4 (L − β) 2 ∥∇XLβ(X0, Y 0, Z0)∥ 2 Thus, summing (47) and (48), we have $$\begin{array}{cccc}& H_T\le H_1-\delta \sum _{t=1}^T\mathbb{E}\mathrm{\parallel }{X}^t-X^{t+1}{\mathrm{\parallel }}^2-\frac{\beta }{2}\sum _{t=1}^T\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2& & \text{(1)}\end{array}H\_\{T\}\backslash leq\; H\_\{1\}-\backslash delta\backslash sum\_\{t=1\}^\{T\}\backslash mathbb\{E\}\backslash |X^\{t\}-X^\{t+1\}\backslash |^\{2\}-\backslash frac\{\backslash beta\}\{2\}\backslash sum\_\{t=1\}^\{T\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}\backslash tag\{1\}$$ $$\leq H_{1}-\delta\sum_{t=1}^{T-1}\mathbb{E}|X^{t}-X^{t+1}|^{2}-\frac{\beta}{2}\sum_{t=1}^{T-1}\mathbb{E}|Y^{t+1}-Y^{t}|^{2}$$ $$(46)(46)$$ $$(47)(47)$$ Proof. Summing (14) from t = 1 to T, it holds that

holds that $$\lim_{t}\mathbb{E}|X^{t}-X^{t+1}|^{2}=\lim_{t}\mathbb{E}|Y^{t+1}-Y^{t}|^{2}=\lim_{t}\mathbb{E}|Z^{t+1}-Z^{t}|^{2}=\lim_{t}\mathbb{E}|Y^{t}-X^{t}|^{2}=0.$$ 2 = 0. (46) (ii) It holds that where the second inequality is because {Ht} is nonincreasing and convergent. This implies (44).

Taking T in the above inequality to infinity, we deduce that

where the last inequality is because {Ht} is convergent. Therefore, we have {E∥Xt − Xt+1∥ 2}, and limt E∥Y t+1 − Y t∥ 2 are summable and

$$\begin{array}{cccc}& \underset{t}{\lim }\mathbb{E}\mathrm{\parallel }{X}^t-X^{t+1}{\mathrm{\parallel }}^2=\underset{t}{\lim }\mathbb{E}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2=0.& & \text{(50)}\end{array}\backslash lim\_\{t\}\backslash mathbb\{E\}\backslash |X^\{t\}-X^\{t+1\}\backslash |^\{2\}=\backslash lim\_\{t\}\backslash mathbb\{E\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}=0.\backslash tag\{50\}$$

In addition, summing (37) from t = 1 to T, we have that

X T t=0 E∥Z t − Z t+1∥ 2 ≤ (1 + Γ)∥Z 0 − Z 1∥ 2 + ΘX T t=1 E∥e t − e t+1∥ 2 + ΓX T t=1 E∥Xt − Xt+1∥ 2 ≤ (1 + Γ)∥Z 0 − Z 1∥ 2 + 2Θ 2r 1 − 2r X T t=1 E∥Xt − Xt−1∥ 2+(Γ + 2Θ 2r 1 − 2r ) X T (51) t=0 E∥Xt−Xt+1∥ 2 ≤ (1 + Γ)∥Z 0 − Z 1∥ 2 + 2 Γ + 2Θ 2r 1 − 2r X T t=0 E∥Xt − Xt+1∥ 2 ≤ (1 + Γ)∥Z 0 − Z 1∥ 2 + 2 Γ + 2Θ 2r 1 − 2r Lβ(X0, Y 0, Z0) + C − H∗ min{δ, β 2 } , where the second inequality uses (32). Recall the definition of Z 1, we have that $$\mathbb{E}\mathrm{\parallel }{Z}^1-Z^0{\mathrm{\parallel }}^2=\mathbb{E}\mathrm{\parallel }{X}^1-Y^0{\mathrm{\parallel }}^2\le 3\mathbb{E}\mathrm{\parallel }{X}^1-X_{\star }^1{\mathrm{\parallel }}^2+3\mathrm{\parallel }{X}_{\star }^1-X^0{\mathrm{\parallel }}^2+3\mathrm{\parallel }{X}^0-Y^0{\mathrm{\parallel }}^2\backslash mathbb\{E\}\backslash |Z^\{1\}-Z^\{0\}\backslash |^\{2\}=\backslash mathbb\{E\}\backslash |X^\{1\}-Y^\{0\}\backslash |^\{2\}\backslash leq3\backslash mathbb\{E\}\backslash |X^\{1\}-X^\{1\}\_\{\backslash star\}\backslash |^\{2\}+3\backslash |X^\{1\}\_\{\backslash star\}-X^\{0\}\backslash |^\{2\}+3\backslash |X^\{0\}-Y^\{0\}\backslash |^\{2\}$$ $$\leq3r|X^{0}-X^{1}{\star}|^{2}+3|X^{1}{\star}-X^{0}|^{2}+3|X^{1}{\star}-Y^{0}|^{2}$$ $$\leq\frac{3(r+1)}{(L-\beta)^{2}}|\nabla L{\beta}(X^{0},Y^{0},Z^{0})|^{2}+3|X^{0}-Y^{0}|^{2}.$$ This together with (51) gives

$$\begin{array}{cccc}& \sum _{t=0}^T\mathbb{E}\mathrm{\parallel }{Z}^t-Z^{t+1}{\mathrm{\parallel }}^2\le (1+\mathrm{\Gamma })\frac{3(r+1)}{(L-\beta {)}^2}\mathrm{\parallel }\mathrm{\nabla }{L}_{\beta }(X^0,Y^0,Z^0){\mathrm{\parallel }}^2+3\mathrm{\parallel }{X}^0-Y^0{\mathrm{\parallel }}^2& & \text{(52)}\end{array}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |Z^\{t\}-Z^\{t+1\}\backslash |^\{2\}\backslash leq(1+\backslash Gamma)\backslash frac\{3(r+1)\}\{(L-\backslash beta)^\{2\}\}\backslash |\backslash nabla\; L\_\{\backslash beta\}(X^\{0\},Y^\{0\},Z^\{0\})\backslash |^\{2\}+3\backslash |X^\{0\}-Y^\{0\}\backslash |^\{2\}\backslash tag\{52\}$$ $$+2\left(\Gamma+2\Theta\frac{2r}{1-2r}\right)\frac{L_{\beta}(X^{0},Y^{0},Z^{0})+C-H_{\star}}{\min{\delta,\frac{\beta}{2}}}.$$

Taking T in the above inequality to infinity we deduce that {E∥Z t − Z t+1∥ 2} is summable and using (12), we have that

$$\lim \mathbb{E}\mathrm{\parallel }{Y}^t-X^{t+1}{\mathrm{\parallel }}^2=\underset{t}{\lim }\mathbb{E}\mathrm{\parallel }{Z}^t-Z^{t+1}{\mathrm{\parallel }}^2=0.\backslash operatorname*\{lim\}\backslash mathbb\{E\}\backslash |Y^\{t\}-X^\{t+1\}\backslash |^\{2\}=\backslash operatorname*\{lim\}\_\{t\}\backslash mathbb\{E\}\backslash |Z^\{t\}-Z^\{t+1\}\backslash |^\{2\}=0.$$

This together with (50) gives that

$$\lim \mathbb{E}\mathrm{\parallel }{Y}^t-X^t{\mathrm{\parallel }}^2=0.\backslash operatorname*\{lim\}\backslash mathbb\{E\}\backslash |Y^\{t\}-X^\{t\}\backslash |^\{2\}=0.$$

C.3 Details And Proofs Of Theorem 1

Here, we prove Theorem 1.

Theorem 3. Consider (1). Let {(x t 1 , . . . , xtp , yt, zt1 , . . . , ztp} be generated by Algorithm 1. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Then the following statements hold.

(i) There exists E > 0 such that

$|\nabla F(Y^{t+1})+\xi^{t+1}|\leq E\left(|X^{t+1}-X^{t}|+|Z^{t+1}-Z^{t}|+|Y^{t}-Y^{t+1}|\right).$ (53) In section (53) we have $$where{\xi }^{t+1}\in \mathrm{\partial }F(Y^{t+1})\mathrm{.}w\; h\; e\; r\; e\backslash ;\backslash xi^\{t+1\}\backslash in\backslash partial\; F(Y^\{t+1\}).$$ (ii) It holds that

$$\frac{1}{1+T}\sum _{t=0}^T\mathbb{E}{d}^2(0,\mathrm{\nabla }F(Y^{t+1})+\mathrm{\partial }G(Y^{t+1}))\backslash frac\{1\}\{1+T\}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}d^\{2\}(0,\backslash nabla\; F(Y^\{t+1\})+\backslash partial\; G(Y^\{t+1\}))$$ $$\leq\frac{1}{T+1}R\left((1+\Gamma)\frac{3(r+1)}{(L-\beta)^{2}}|\nabla L_{\beta}(X^{0},Y^{0},Z^{0})|^{2}+3|X^{0}-Y^{0}|^{2}\right)$$ $$+\frac{1}{T+1}R\left(2\Gamma+\Theta\frac{8r}{1-2r}+2\right)\frac{L_{\beta}(X^{0},Y^{0},Z^{0})+C-H_{}}{\min{\delta,\frac{\beta}{2}}},$$ are defined in Proposition 5. $H{}$ and $C$ is defined in Proposition 5. $\alpha$._ where Γ and Θ are defined in Proposition 5, H∗ and C is defined in Proposition 5 and Corollary 4 respectively, R := max{3(L + β) 2 2r 1−2r , L τβ + 12,(L + β) 2}.

Proof. Using (29), it hold that

$$0=\mathrm{\nabla }F(Y^{t+1})+\mathrm{\nabla }F(X_{\star }^{t+1})-\mathrm{\nabla }F(Y^{t+1})+Z^t+\beta (X_{\star }^{t+1}-Y^t)\mathrm{.}0=\backslash nabla\; F(Y^\{t+1\})+\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\})-\backslash nabla\; F(Y^\{t+1\})+Z^\{t\}+\backslash beta(X\_\{\backslash star\}^\{t+1\}-Y^\{t\}).$$

Summing this with (30), we have that

$0=\nabla F(Y^{t+1})+\xi^{t+1}+\nabla F(X^{t+1}{*})-\nabla F(Y^{t+1})+Z^{t}-Z^{t+1}+\beta(X^{t+1}{*}-X^{t+1})-\beta(Y^{t+1}-Y^{t})$.
This implies that

∥∇F(Y t+1) + ξ t+1∥ ≤ ∥∇F(Xt+1 ⋆) − ∇F(Y t+1)∥ + ∥Z t − Z t+1∥ + β∥Xt+1 ⋆ − Xt+1∥ + β∥Y t+1 − Y t∥ ≤ L∥Xt+1 ⋆ − Y t+1∥ + ∥Z t − Z t+1∥ + β∥Xt+1 ⋆ − Xt+1∥ + β∥Y t+1 − Y t∥ ≤ L∥Xt+1 ⋆ − Xt+1∥ + L∥Xt+1 − Y t∥ + (L + β)∥Y t − Y t+1∥ + ∥Z t − Z t+1∥ + β∥Xt+1 ⋆ − Xt+1∥ = (L + β)∥Xt+1 ⋆ − Xt+1∥ + L τ β + 1∥Z t+1 − Z t∥ + (L + β)∥Y t − Y t+1∥, $$(54)(54)$$ where the last equality uses (12). Using (32), we have that E∥Xt+1 ⋆ − Xt+1∥ 2 ≤ q 2r 1−2r E∥Xt+1 − Xt∥ 2.

Using this, (54) can be further passed to

In, (54) can be further passed to $$\mathbb{E}|\nabla F(Y^{t+1})+\xi^{t+1}|^2\leq(L+\beta)\sqrt{\frac{2r}{1-2r}}3\mathbb{E}|X^{t+1}-X^t|^2+\left(\frac{L}{\tau\beta}+1\right)3\mathbb{E}|Z^{t+1}-Z^t|^2$$ $$+(L+\beta)3\mathbb{E}|Y^t-Y^{t+1}|^2.$$ This together with Cauchy-Schwarz inequality, we have that

After what could easily be imaginary, we have that $$\mathbb{E}|\nabla F(Y^{t+1})+\mathbb{E}^{t+1}|^2\leq3(L+\beta)^2\frac{2r}{1-2r}\mathbb{E}|X^{t+1}-X^t|^2+\left(\frac{L}{\tau\beta}+1\right)^2\mathbb{E}|Z^{t+1}-Z^t|^2$$ $$+(L+\beta)^2\mathbb{E}|Y^t-Y^{t+1}|^2.$$ $$(55)(55)$$ This proves (53). Summing the above inequality from t = 0 to T, it holds that

X T t=0 E∥∇F(Y t+1) + ξ t+1∥ 2 ≤ 3(L + β) 22r 1 − 2r X T t=0 E∥Xt+1 − Xt∥ 2 + L τ β + 12X T t=0 E∥Z t+1 − Z t∥ 2 + (L + β) 2X T t=0 E∥Y t − Y t+1∥ 2 ≤ max{3(L + β) 22r 1 − 2r , L τ β + 12,(L + β) 2} · X T t=0 E∥Xt+1 − Xt∥ 2 + ∥Y t − Y t+1∥ 2 + ∥Z t+1 − Z t∥ 2 ! ≤ max{3(L + β) 22r 1 − 2r , L τ β + 12,(L + β) 2} · (1 + Γ) 3(r + 1) (L − β) 2 ∥∇Lβ(X0, Y 0, Z0)∥ 2 + 3∥X0 − Y 0∥ 2 + 2Γ + Θ 8r 1 − 2r + 2Lβ(X0, Y 0, Z0) + C − H∗ min{δ, β 2 } ! , 2 where $C:=2\tau\beta(\Gamma+1)|X^{0}-Y^{0}|^{2}+\frac{4}{(L-\beta)^{2}}\Big{(}\frac{L+\beta+1}{2}+2\tau\beta(\Gamma+1)+\Upsilon+\frac{(L-\beta)^{2}}{8}\Big{)}$, the last term will be $\tau\beta(\Gamma)=\frac{1}{2}\left(\frac{L+\beta+1}{2}+2\tau\beta(\Gamma+1)+\Upsilon+\frac{(L-\beta)^{2}}{8}\right)$.
· ∥∇XLβ(X0, Y 0, Z0)∥ 2, the last inequality uses (44) and (45). Dividing both sides with T + 1 and recalling ξ t+1 ∈ ∂G(Y t+1), we have the conclusion. Grouping the constants of ∥X0 − Y 0∥ 2, ∥∇XLβ(X0, Y 0, Z0)∥ 2, Lβ(X0, Y 0, Z0), we have that

$$\begin{array}{cccc}& \sum _{t=0}^T\mathbb{E}\mathrm{\parallel }\mathrm{\nabla }F(Y^{t+1})+{\xi }^{t+1}{\mathrm{\parallel }}^2& & \text{(56)}\end{array}\backslash sum\_\{t=0\}^\{T\}\backslash mathbb\{E\}\backslash |\backslash nabla\; F(Y^\{t+1\})+\backslash xi^\{t+1\}\backslash |^\{2\}\backslash tag\{56\}$$ $$\leq D\left(|\nabla L_{\beta}(X^{0},Y^{0},Z^{0})|^{2}+|X^{0}-Y^{0}|^{2}+L_{\beta}(X^{0},Y^{0},Z^{0})-W\right),$$ $$\mathrm{\square }\backslash square$$

where

$$\begin{array}{cccc}& D:=\max \{3(L+\beta {)}^2\frac{2r}{1-2r},{(\frac{L}{\tau \beta }+1)}^2,(L+\beta {)}^2\}\cdot \max \{D_1,D_2,D_3\}& & \text{(57)}\end{array}D:=\backslash max\backslash \{3(L+\backslash beta)^\{2\}\backslash frac\{2r\}\{1-2r\},\backslash left(\backslash frac\{L\}\{\backslash tau\backslash beta\}+1\backslash right)^\{2\},(L+\backslash beta)^\{2\}\backslash \}\backslash cdot\backslash max\backslash \{D\_\{1\},D\_\{2\},D\_\{3\}\backslash \}\backslash tag\{57\}$$

max{3, D12τ β(Γ + 1)}.

with D1 := 2Γ+Θ 8r 1−2r +2 min{δ, 12 β}, D2 := (1 + Γ) 3(r+1) $$\frac{(r+1)}{L-\beta {)}^2}+D_1\frac{4}{(L-\beta {)}^2}(\frac{L+\beta +1}{2}+2\tau \beta (\mathrm{\Gamma }+1)+\mathrm{{\rm Y}}+\frac{(L-\beta {)}^2}{8}),D_3:=0.\backslash frac\{(r+1)\}\{L-\backslash beta)^\{2\}\}+D\_\{1\}\backslash frac\{4\}\{(L-\backslash beta)^\{2\}\}\backslash Big(\backslash frac\{L+\backslash beta+1\}\{2\}+2\backslash tau\backslash beta(\backslash Gamma+1)+\backslash Upsilon+\backslash frac\{(L-\backslash beta)^\{2\}\}\{8\}\backslash Big),\backslash ;D\_\{3\}:=0.$$

C.3.1 Proofs Of Proposition 6 And Corollary 3

We provide the detailed version of Proposition 6 as follows.

Proposition 9. Consider (1). Let {(x t 1 , . . . , xtp , yt, zt1 , . . . , ztp} be generated by Algorithm 1. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Suppose {(Xt, Y t, Zt)} is bounded and denote the set of accumulation points of {(Xt, Y t, Zt, Xt−1, Zt−1)} as Ω*. The following statements hold:* (i) limt d((Xt, Y t, Zt, Xt−1, Zt−1), Ω) = 0.

(ii) Any accumulation point of {Y t} is a stationary point of (1).

(iii) H ≡ H∗ on Ω.

Proof. For (i), let Y ∗ be an accumulation point of {Y t} with Y ti → Y ∗. Using (29) and (30), there exists ξ ti ∈ G(Y ti ) such that

$$0=\mathrm{\nabla }F(X_{\ast }^{t_i})+Z^{t_i-1}+\beta (X_{\ast }^{t_i}-Y^{t_i-1})0=\backslash nabla\; F(X\_\{*\}^\{t\_\{i\}\})+Z^\{t\_\{i\}-1\}+\backslash beta(X\_\{*\}^\{t\_\{i\}\}-Y^\{t\_\{i\}-1\})$$ $$=\nabla F(Y^{t})+\nabla F(X_{}^{t_{i}})-\nabla F(Y^{t})+Z^{t_{i}-1}+\beta(X_{}^{t_{i}}-Y^{t_{i}-1}).$$

and $$0={\xi }^{t_i}-Z^{t_i}-\beta (X^{t_i}-Y^{t_i})\mathrm{.}0=\backslash xi^\{t\_\{i\}\}-Z^\{t\_\{i\}\}-\backslash beta(X^\{t\_\{i\}\}-Y^\{t\_\{i\}\}).$$ The above relations shows that ti)(58) above relations shows that $0=\nabla F(Y^t)+\xi^{ti}+\nabla F(X^t_i)-\nabla F(Y^t)+Z^{ti-1}-Z^{ti}+\beta(X^{t_i}-Y^{t_{i-1}})-\beta(X^{t_i}-Y^{t_i})$ $=\nabla F(Y^t)+\xi^{ti}+\nabla F(X^t_i)-\nabla F(Y^t)+\tau\beta(X^{t_i}-Y^{t_{i-1}})+\beta(X^{t_i}-Y^{t_{i-1}})-\beta(X^{t_i}-Y^{t_i})$ then we list as above as of (12). Now we have that $|X^t-Y^{t_i}-\beta_i Y^{t_i}-\beta_i Y^{t_i}|$ is a B-positive where the equality makes uses of (12). Now we show that limi ∥Xt⋆ − Xt∥ = 0. Using Proposition 7 and (11), we have that

$$\mathrm{\parallel }{e}^t{\mathrm{\parallel }}^2=\mathrm{\parallel }{X}_{\star }^t-X^t{\mathrm{\parallel }}^2\le \frac{2r}{1-2r}\mathrm{\parallel }{X}^t-X^{t-1}{\mathrm{\parallel }}^2\mathrm{.}\backslash |e^\{t\}\backslash |^\{2\}=\backslash |X\_\{\backslash star\}^\{t\}-X^\{t\}\backslash |^\{2\}\backslash leq\{\backslash frac\{2r\}\{1-2r\}\}\backslash |X^\{t\}-X^\{t-1\}\backslash |^\{2\}.$$

Since limt ∥Xt − Xt−1∥ = 0, we have that

$$(59)(59)$$ $$\underset{i}{\lim }\mathrm{\parallel }{X}_{\star }^t-X^t\mathrm{\parallel }=0.\backslash operatorname*\{lim\}\_\{i\}\backslash |X\_\{\backslash star\}^\{t\}-X^\{t\}\backslash |=0.$$ i∥Xt⋆ − Xt∥ = 0. (59) Next, we show that limi ∥Xt − Y t−1∥ = 0 . Using (12), it holds that

$${\parallel Z^t-Z^{t-1}\parallel }^2\backslash left\backslash |Z^\{t\}-Z^\{t-1\}\backslash right\backslash |^\{2\}$$ $$\leq\Gamma\left(\left|Z^{t-2}-Z^{t-1}\right|^{2}-\left|Z^{t}-Z^{t-1}\right|^{2}\right)+\Theta|e^{t-1}-e^{t}|^{2}+\Lambda\left|X^{t-1}-X^{t}\right|^{2}$$ $$\leq\Gamma\left(\left|Z^{t-2}-Z^{t-1}\right|^{2}-\left|Z^{t}-Z^{t-1}\right|^{2}\right)+\Theta\frac{4r}{1-2r}|X^{t-1}-X^{t-2}|^{2}+(\Lambda+\frac{4r}{1-2r})\left|X^{t-1}-X^{t}\right|^{2}$$ where the first inequality uses (37) and the second inequality is due to (32). Summing the above inequality, from t = 1 to T, we have that X T 1=1 Z t − Z t−1 2≤ Γ∥Z t1−2 − Z t1−1∥ 2 − ∥Z tK − Z tK−1∥ 2 +1 τ β Θ 4r 1 − 2r X T 1=1 ∥Xt−1 − Xt−2∥ 2 + (Λ + 4r 1 − 2r ) X T 1=1 Xt−1 − Xt 2 ≤ Γ∥Z t1−2 − Z t1−1∥ 2 − ∥Z tK − Z tK−1∥ 2+ Θ 4r 1 − 2r X K i=1 ∥Xt−1 − Xt−2∥ 2 + (Λ + 4r 1 − 2r ) X K i=1 Xt−1 − Xt 2 ≤ Γ∥Z t1−2 − Z t1−1∥ 2 + Θ 4r 1 − 2r X T 1=1 ∥Xt−1 − Xt−2∥ 2 + (Λ + 4r 1 − 2r ) X T 1=1 Xt−1 − Xt 2.

$$(60)(60)$$

Taking K in the above inequality to infinity and recalling thatXt−1 − Xt 2is summable, we deduce that PT 1=1 ∥Z t − Z t−1∥ 2 < ∞. This together with (12) show that

$$\underset{t}{\lim }\mathrm{\parallel }{X}^t-Y^{t-1}\mathrm{\parallel }=\frac{1}{\tau \beta }\underset{t}{\lim }\mathrm{\parallel }{Z}^t-Z^{t-1}\mathrm{\parallel }=0.\backslash operatorname*\{lim\}\_\{t\}\backslash |X^\{t\}-Y^\{t-1\}\backslash |=\{\backslash frac\{1\}\{\backslash tau\backslash beta\}\}\backslash operatorname*\{lim\}\_\{t\}\backslash |Z^\{t\}-Z^\{t-1\}\backslash |=0.$$ t−1∥ = 0. (60) Next, we show that limt ∥Y t − Y t−1∥ = 0. Using (12) again, we have that

$$Y^t-Y^{t-1}=X^{t+1}-X^t-\frac{1}{\tau \beta }(Z^{t+1}-Z^t)-\frac{1}{\tau \beta }(Z^t-Z^{t-1})\mathrm{.}Y^\{t\}-Y^\{t-1\}=X^\{t+1\}-X^\{t\}-\backslash frac\{1\}\{\backslash tau\backslash beta\}(Z^\{t+1\}-Z^\{t\})-\backslash frac\{1\}\{\backslash tau\backslash beta\}(Z^\{t\}-Z^\{t-1\}).$$

This together with the fact that limt ∥Xt −Xt−1∥ = limt ∥Z t −Z t−1∥ = 0 implies that limt ∥Y t −Y t−1∥ = 0.

Since Y ti → Y ∗, combining (59), (60) and (46), we have that

$\lim_i Y^{t_i-1}=\lim_i X^{t_i}=\lim_i X^{t_i}=\lim_i Y^{t_i}=Y^{*}$. $$\mathrm{n}{Y}^{t_i}=Y^{\ast }\mathrm{.}\backslash mathrm\{n\}\backslash ,Y^\{t\_\{i\}\}=Y^\{*\}.$$ This together with the continuity of ∇F, the closedness of ∂G and (58) shows that

$$0\in \mathrm{\nabla }F(Y^{\ast })+\mathrm{\partial }G(Y^{\ast })\mathrm{.}0\backslash in\backslash nabla\; F(Y^\{*\})+\backslash partial\; G(Y^\{*\}).$$ $$(61)(61)$$

This completes the proof.

Now we prove (ii). Fix any (X∗, Y ∗, Z∗, X¯ ∗,Z¯∗) ∈ Ω. Then there exists {ti}i such that (Xti, Y ti, Zti, Xti−1, Y ti−1) converges to (X∗, Y ∗, Z∗, X¯ ∗,Z¯∗). Thanks to Proposition 5 (ii), we know that

$$H_{\ast }=\underset{i}{\lim }H(X^{t_i},Y^{t_i},Z^{t_i},X^{t_i-1},Y^{t_i-1})H\_\{*\}=\backslash operatorname*\{lim\}\_\{i\}H(X^\{t\_\{i\}\},Y^\{t\_\{i\}\},Z^\{t\_\{i\}\},X^\{t\_\{i\}-1\},Y^\{t\_\{i\}-1\})$$ iH(Xti, Y ti, Zti, Xti−1, Y ti−1) (61) and

$$\begin{array}{cccc}& H(X^{\ast },Y^{\ast },Z^{\ast },{\bar{X}}^{\ast },{\bar{Z}}^{\ast })=L_{\beta }(X^{\ast },Y^{\ast },Z^{\ast })=F(X^{\ast })+G(Y^{\ast })+⟨X^{\ast }-Y^{\ast },Z^{\ast }⟩+\frac{\beta }{2}\mathrm{\parallel }{X}^{\ast }-Y^{\ast }{\mathrm{\parallel }}^2\mathrm{.}& & \text{(62)}\end{array}H(X^\{*\},Y^\{*\},Z^\{*\},\backslash bar\{X\}^\{*\},\backslash bar\{Z\}^\{*\})=L\_\{\backslash beta\}(X^\{*\},Y^\{*\},Z^\{*\})=F(X^\{*\})+G(Y^\{*\})+\backslash langle\; X^\{*\}-Y^\{*\},Z^\{*\}\backslash rangle+\backslash frac\{\backslash beta\}\{2\}\backslash |X^\{*\}-Y^\{*\}\backslash |^\{2\}.\backslash tag\{62\}$$

Since Y tis the minimizer of (13), it holds that

$$G(Y^{t_i})+⟨X^{t_i}-Y^{t_i},Z^{t_i}⟩+\frac{\beta }{2}\mathrm{\parallel }{X}^{t_i}-Y^{t_i}{\mathrm{\parallel }}^2\le G(Y^{\ast })+⟨X^{t_i}-Y^{\ast },Z^{t_i}⟩+\frac{\beta }{2}\mathrm{\parallel }{X}^{t_i}-Y^{\ast }{\mathrm{\parallel }}^2\mathrm{.}G(Y^\{t\_\{i\}\})+\backslash left\backslash langle\; X^\{t\_\{i\}\}-Y^\{t\_\{i\}\},Z^\{t\_\{i\}\}\backslash right\backslash rangle+\{\backslash frac\{\backslash beta\}\{2\}\}\backslash |X^\{t\_\{i\}\}-Y^\{t\_\{i\}\}\backslash |^\{2\}\backslash leq\; G(Y^\{*\})+\backslash left\backslash langle\; X^\{t\_\{i\}\}-Y^\{*\},Z^\{t\_\{i\}\}\backslash right\backslash rangle+\{\backslash frac\{\backslash beta\}\{2\}\}\backslash |X^\{t\_\{i\}\}-Y^\{*\}\backslash |^\{2\}.$$

Taking the above inequality to infinity, we have that

$$\underset{i}{\mathrm{lim\hspace{0.17em}sup}}G(Y^{t_i})+⟨X^{\ast }-Y^{\ast },Z^{\ast }⟩+\frac{\beta }{2}\mathrm{\parallel }{X}^{\ast }-Y^{\ast }{\mathrm{\parallel }}^2\backslash limsup\_\{i\}G(Y^\{t\_\{i\}\})+\backslash langle\; X^\{*\}-Y^\{*\},Z^\{*\}\backslash rangle+\backslash frac\{\backslash beta\}\{2\}\backslash |X^\{*\}-Y^\{*\}\backslash |^\{2\}$$ $$=\limsup_{i}G(Y^{t_{i}})+\langle X^{t_{i}}-Y^{t_{i}},Z^{t_{i}}\rangle+\frac{\beta}{2}|X^{t_{i}}-Y^{t_{i}}|^{2}$$ $$\leq G(Y^{})+\langle X^{}-Y^{},Z^{}\rangle+\frac{\beta}{2}|X^{}-Y^{}|^{2}.$$

This together with the closedness of G shows that limi G(Y ti ) = G(Y ∗). This together with the continuity of F, Corollary 4 (ii) and (61) gives that

$$H_{\ast }=\underset{i}{\lim }H(X^{t_i},Y^{t_i},Z^{t_i},X^{t_i-1},Y^{t_i-1})H\_\{*\}=\backslash operatorname*\{lim\}\_\{i\}H(X^\{t\_\{i\}\},Y^\{t\_\{i\}\},Z^\{t\_\{i\}\},X^\{t\_\{i\}-1\},Y^\{t\_\{i\}-1\})$$ $H_{0}=\underset{i}{\max}(X^{},Y^{},Z^{},Z^{})$, $=F(X^{})+G(Y^{})+\langle X^{}-Y^{},Z^{}\rangle+\frac{\beta}{2}|X^{}-Y^{}|^{2}=H(X^{},Y^{},Z^{},\bar{X}^{},\bar{Z}^{})$, where the second equality uses (62).

Corollary 3. Let {(x t 1 , . . . , xtp , yt, zt1 , . . . , ztp )} be generated by Algorithm 1 with (9) holding deterministically. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 6 hold. Then any accumulation point of {y t} is a stationary point of (1).

Proof. From Proposition 2, we understand that Y t = (y t*, . . . , yt) for any t. Let y ∗ be any accumulation point of y t. Then Y ∗ = (y ∗, . . . , y*∗) is an accumulation point of {Y t}. Proposition 6 demonstrates that the Y ∗is a stationary point of (3). By applying Proposition 1, we deduce that y ∗is a stationary point of (1).

C.3.2 Details And Proofs For Theorem 2

To show the global convergence of the generated sequence, we first need to bound the subdifferential of ∂H(Xt+1, Y t+1, Zt+1, Xt, Zt).

$$\mathrm{\square }\backslash square$$

Lemma 1. Consider (1). Let {(x t 1 , . . . , xtp , yt, zt1 , . . . , ztp} be generated by Algorithm 1. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose (9) is satisfied deterministically (satisfied without expectation). Suppose assumptions in Proposition 5 hold. There exists D > 0 such that

$$d(0,\mathrm{\partial }H(X^{t+1},Y^{t+1},Z^{t+1},X^t,Z^t))\le D(\mathrm{\parallel }{X}^{t+1}-X^t\mathrm{\parallel }+\mathrm{\parallel }{Y}^{t+1}-Y^t\mathrm{\parallel }+\mathrm{\parallel }{Z}^{t+1}-Z^t\mathrm{\parallel })\mathrm{.}d(0,\backslash partial\; H(X^\{t+1\},Y^\{t+1\},Z^\{t+1\},X^\{t\},Z^\{t\}))\backslash leq\; D\backslash left(\backslash |X^\{t+1\}-X^\{t\}\backslash |+\backslash |Y^\{t+1\}-Y^\{t\}\backslash |+\backslash |Z^\{t+1\}-Z^\{t\}\backslash |\backslash right).$$

Proof. Using Exercise 8.8, Proposition 10.5 and Corollary 10.9 of RockWets98, it holds that

$$\mathrm{\partial }H(X,Y,Z,X^{\mathrm{\prime }},Z^{\mathrm{\prime }})\supseteq \left(\begin{array}{c}\mathrm{\nabla }F(X)\\ \mathrm{\partial }G(Y)\\ 0\\ 0\end{array}\right)+\left(\begin{array}{c}Z+\beta (X-Y)+\frac{\mathrm{\Theta }}{{\pi }^3}\frac{16\pi }{1-2\tau }(X-X^{\mathrm{\prime }})\\ -Z-\beta (X-Y)\\ X-Y+\frac{2\mathrm{\Gamma }}{{\pi }^3}(Z-Z^{\mathrm{\prime }})\\ -\frac{\mathrm{\Theta }}{{\pi }^3}\frac{16}{1-2\tau }(X-X^{\mathrm{\prime }})\\ -\frac{3\mathrm{\Gamma }}{{\pi }^3}(Z-Z^{\mathrm{\prime }})\mathrm{.}\end{array}\right)\mathrm{.}\backslash partial\; H(X,Y,Z,X^\{\backslash prime\},Z^\{\backslash prime\})\backslash supseteq\backslash begin\{pmatrix\}\backslash nabla\; F(X)\backslash \backslash \; \backslash partial\; G(Y)\backslash \backslash \; 0\backslash \backslash \; 0\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}Z+\backslash beta(X-Y)+\backslash frac\{\backslash Theta\}\{\backslash pi^\{3\}\}\backslash frac\{16\backslash pi\}\{1-2\backslash tau\}(X-X^\{\backslash prime\})\backslash \backslash \; -Z-\backslash beta(X-Y)\backslash \backslash \; X-Y+\backslash frac\{2\backslash Gamma\}\{\backslash pi^\{3\}\}(Z-Z^\{\backslash prime\})\backslash \backslash \; -\backslash frac\{\backslash Theta\}\{\backslash pi^\{3\}\}\backslash frac\{16\}\{1-2\backslash tau\}(X-X^\{\backslash prime\})\backslash \backslash \; -\backslash frac\{3\backslash Gamma\}\{\backslash pi^\{3\}\}(Z-Z^\{\backslash prime\}).\backslash end\{pmatrix\}.$$

Thus,

∂H(Xt+1, Y t+1, Zt+1, Xt, Zt) ⊇ ∇F(Xt+1) + Z t+1 + β(Xt+1 − Y t+1) + Θ τβ 16r 1−2r (Xt+1 − Xt) ∂G(Y t+1) − Z t+1 − β(Xt+1 − Y t+1) Xt+1 − Y t+1 + 2Γ τβ (Z t+1 − Z t) − Θ τβ 16r 1−2r (Xt+1 − Xt) − 2Γ τβ (Z t+1 − Z t) (63) ⊇ ∇F(Xt+1) + Z t+1 + β(Xt+1 − Y t+1) + Θ τβ 16r 1−2r (Xt+1 − Xt) 0 Xt+1 − Y t+1 + 2Γ τβ (Z t+1 − Z t) − Θ τβ 16r 1−2r (Xt+1 − Xt) − 2Γ τβ (Z t+1 − Z t)
where the seconde inclusion follows from (30).

Now, we bound each coordinate in the right hand side of the relation. For the first one, we denote At+1 := ∇F(Xt+1) + Z t+1 + β(Xt+1 − Y t+1) + Θ τβ 16r 1−2r (Xt+1 − Xt). Using (29), we have that

$$\begin{array}{c}{\mathcal{A}}^{t+1}\ni \mathrm{\nabla }F(X^{t+1})-\mathrm{\nabla }F(X_{\star }^{t+1})+(Z^{t+1}-Z^t)\\ \\ +\beta (X^{t+1}-Y^{t+1}-X_{\star }^{t+1}+Y^t)+\frac{\mathrm{\Theta }}{\tau \beta }\frac{16r}{1-2r}(X^{t+1}-X^t)\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\{\{\{\backslash mathcal\{A\}\}^\{t+1\}\backslash ni\backslash nabla\; F(X^\{t+1\})-\backslash nabla\; F(X\_\{\backslash star\}^\{t+1\})+(Z^\{t+1\}-Z^\{t\})\}\}\backslash \backslash \; \{\{\{\}\}\}\backslash \backslash \; \{\{\{\}+\backslash beta(X^\{t+1\}-Y^\{t+1\}-X\_\{\backslash star\}^\{t+1\}+Y^\{t\})+\backslash frac\{\backslash Theta\}\{\backslash tau\backslash beta\}\backslash frac\{16r\}\{1-2r\}(X^\{t+1\}-X^\{t\}).\}\}\backslash end\{array\}$$

Thus, we deduce that d 2(0, At+1) is bounded above by

$$4(L+\beta {)}^2\mathrm{\parallel }{X}^{t+1}-X^{t+1}{\mathrm{\parallel }}^2+4\mathrm{\parallel }{Z}^{t+1}-Z^t{\mathrm{\parallel }}^2+4{\beta }^2\mathrm{\parallel }{Y}^t-Y^{t+1}{\mathrm{\parallel }}^{2}4(L+\backslash beta)^\{2\}\backslash |X^\{t+1\}-X^\{t+1\}\backslash |^\{2\}+4\backslash |Z^\{t+1\}-Z^\{t\}\backslash |^\{2\}+4\backslash beta^\{2\}\backslash |Y^\{t\}-Y^\{t+1\}\backslash |^\{2\}$$ $$+\frac{4\Theta^{2}}{\tau^{2}\beta^{2}}\frac{64r^{2}}{(1-2r)^{2}}|X^{t+1}-X^{t}|^{2}\tag{64}$$

where we also make use of the Lipscitz continuity of ∇F.

For the third coordinate in (63), using (12), it holds that

$${\parallel X^{t+1}-Y^{t+1}+\frac{2\mathrm{\Gamma }}{\tau \beta }(Z^{t+1}-Z^t)\parallel }^2={\parallel \frac{1}{\tau \beta }(Z^{t+1}-Z^t)+Y^t-Y^{t+1}+\frac{2\mathrm{\Gamma }}{\tau \beta }(Z^{t+1}-Z^t)\parallel }^2\backslash left\backslash |X^\{t+1\}-Y^\{t+1\}+\backslash frac\{2\backslash Gamma\}\{\backslash tau\backslash beta\}(Z^\{t+1\}-Z^\{t\})\backslash right\backslash |^\{2\}=\backslash left\backslash |\backslash frac\{1\}\{\backslash tau\backslash beta\}(Z^\{t+1\}-Z^\{t\})+Y^\{t\}-Y^\{t+1\}+\backslash frac\{2\backslash Gamma\}\{\backslash tau\backslash beta\}(Z^\{t+1\}-Z^\{t\})\backslash right\backslash |^\{2\}$$ $$\leq2|Y^{t}-Y^{t+1}|^{2}+\frac{(1+2\Gamma)^{2}}{\tau^{2}\beta^{2}}|Z^{t+1}-Z^{t}|^{2}$$

This together with (63) and (64), we deduce that

d 2(0, ∂H(Xt+1, Y t+1, Zt+1, Xt, Zt)) ≤ 4(L + β) 2∥Xt+1 − Xt+1 ⋆ ∥ 2 + 4∥Z t+1 − Z t∥ 2 + 4β 2∥Y t − Y t+1∥ 2 + 4Θ2 τ 2β 2 64 ∗ 4r 2 (1 − 2r) 2 ∥Xt+1 − Xt∥ 2 + 2∥Y t − Y t+1∥ 2 + (1 + 2Γ)2 τ 2β 2∥Z t+1 − Z t)∥ 2 (65) +Θ2 τ 2β 2 64 ∗ 4r 2 (1 − 2r) 2 ∥Xt+1 − Xt∥ 2 + 4Γ2 τ 2β 2 ∥Z t+1 − Z t∥ 2. $$\mathrm{\square }\backslash square$$ Note that using 32, we have that

$$\begin{array}{cccc}& \mathrm{\parallel }{X}^{t+1}-X_{\ast }^{t+1}{\mathrm{\parallel }}^2\le \frac{2r}{1-2r}\mathrm{\parallel }{X}^{t+1}-X^t{\mathrm{\parallel }}^2\mathrm{.}& & \text{(66)}\end{array}\backslash |X^\{t+1\}-X^\{t+1\}\_\{*\}\backslash |^\{2\}\backslash leq\backslash frac\{2r\}\{1-2r\}\backslash |X^\{t+1\}-X^\{t\}\backslash |^\{2\}.\backslash tag\{66\}$$

Combining (65) with (66), we have that

d 2(0, ∂H(Xt+1, Y t+1, Zt+1, Xt, Zt)) ≤ 4(L + β) 22r 1 − 2r ∥Xt+1 − Xt∥ 2 + 4∥Z t+1 − Z t∥ 2 + 4β 2∥Y t − Y t+1∥ 2 + 4Θ2 τ 2β 2 64 ∗ 4r 2 (1 − 2r) 2 ∥Xt+1 − Xt∥ 2 + 2∥Y t − Y t+1∥ 2 + (1 + 2Γ)2 τ 2β 2∥Z t+1 − Z t)∥ 2 +Θ2 τ 2β 2 64 ∗ 4r 2 (1 − 2r) 2 ∥Xt+1 − Xt∥ 2 + 4Γ2 τ 2β 2 ∥Z t+1 − Z t∥ 2 = D′(∥Xt+1 − Xt∥ 2 + ∥Y t − Y t+1∥ 2 + ∥Z t+1 − Z t∥ 2), where D is the maximum of the coordinates of ∥Xt+1 − Xt∥ 2, ∥Y t − Y t+1∥ and ∥Z t+1 − Z t∥ 2 on the right hand side of the above inequality. Finally, using the fact that P3 i s 2 i ≤ (P3 i ai) 2for any a1, a2, a3 ≥ 0, the above inequality can be further passed to

$$d^2(0,\mathrm{\partial }H(X^{t+1},Y^{t+1},Z^{t+1},X^t,Z^t))\le D^{\mathrm{\prime }}(\mathrm{\parallel }{X}^{t+1}-X^t\mathrm{\parallel }+\mathrm{\parallel }{Y}^t-Y^{t+1}\mathrm{\parallel }+\mathrm{\parallel }{Z}^{t+1}-Z^t\mathrm{\parallel })\mathrm{.}d^\{2\}(0,\backslash partial\; H(X^\{t+1\},Y^\{t+1\},Z^\{t+1\},X^\{t\},Z^\{t\}))\backslash leq\; D^\{\backslash prime\}(\backslash |X^\{t+1\}-X^\{t\}\backslash |+\backslash |Y^\{t\}-Y^\{t+1\}\backslash |+\backslash |Z^\{t+1\}-Z^\{t\}\backslash |).$$

Taking square root on both sides of the above inequality we have the conclusion.

Now we are ready to prove Theorem 2. In fact, we already show the key properties that will be needed. They are Proposition 5, Corollary 4, Proposition C.3.1 and Lemma 1. The rest steps are routine. We follow the proofs in Borwein et al. (2017); Bolte et al. (2014); Li & Pong (2016) and include it only for completeness.

Theorem 2. Consider (1) and Algorithm 1 with (9) holding deterministically. Let (Xt, Y t, Zt) be defined as in Proposition 4. Suppose assumptions in Proposition 5 hold. Let H be defined as in Proposition 5 and suppose H is a KL function with exponent α ∈ [0, 1). Then {(Xt, Y t, Zt)} converges globally. Denoting (X∗, Y ∗, Z∗) := limt(Xt, Y t, Zt) and d t s := ∥(Xt, Y t, Zt)−(X∗, Y ∗, Z∗)∥*, then the followings hold. If* α = 0, then {d t s} converges finitely. If α ∈ (0, 1 2 ], then there exist b > 0, t1 ∈ N and ρ1 ∈ (0, 1) such that d t s ≤ bρt1 for t ≥ t1*. If* α ∈ ( 1 2 , 1), then there exist t2 and c > 0 such that d t s ≤ ct− 1 4α−2 for t ≥ t2.

Proof. We first show that {(Xt, Y t, Zt)} is convergent. If there exists t0 such that Ht0 = H∗. Since {Ht} is nonincreasing thanks to (14), we deduce that Ht = H∗ for all t ≥ t0. Using (14) again we have that for all t ≥ t0, it holds that Xt = Xt−1 = · · · = Xt0−1 and Y t = Y t−1 = · · · = Y t0. Recalling in (46) we have that limt(Xt − Y t) = 0, we have that Y t0 = Xt0−1. Thus, Xt+1 − Y t = Y t0 − Xt0−1 = 0 for all t ≥ t0. This together with (12), we deduce that Z t+1 = Z t = · · · = Z t0for all t ≥ t0. Therefore, when there exists t0 such that Ht0 = H∗, {(Xt, Y t, Zt)} converge finitely.

Next, we consider the case where Ht > H∗ for all t. Thanks to Proposition C.3.1 (iii), using Lemma 6 of Bolte et al. (2014), there exists r > 0, a > 0 and ψ ∈ Ψa such that

$${\psi }^{\mathrm{\prime }}(H(X,Y,Z,X^{\mathrm{\prime }}{Z}^{\mathrm{\prime }})-H_{\ast })d(0,\mathrm{\partial }H(X,Y,Z,X^{\mathrm{\prime }},Z^{\mathrm{\prime }}))\ge 1\backslash psi^\{\backslash prime\}(H(X,Y,Z,X^\{\backslash prime\}Z^\{\backslash prime\})-H\_\{*\})d(0,\backslash partial\; H(X,Y,Z,X^\{\backslash prime\},Z^\{\backslash prime\}))\backslash geq1$$

when d((X, Y, Z, X′, Z′), Ω) ≤ r and H∗ < H(X, Y, Z, X′, Z′) < H∗ +a. Thanks to Corollary 4 and Theorem 5, we know that there exists t1 such that when t > t1, d((Xt, Y t, Zt, Xt−1, Zt−1), Ω) ≤ r and H∗ < H(Xt, Y t, Zt, Xt−1, Zt−1) < H∗ + a. Thus, it holds that ψ ′(H((Xt, Y t, Zt, Xt−1, Zt−1) − H∗)d(0, ∂H((Xt, Y t, Zt, Xt−1, Zt−1)) ≥ 1. (67) Recaling (14), we have that Since ψ is concave, using the above inequality we have that

$$(X^t,Y^t,Z^t,X^{t-1},Z^{t-1}))\ge 1.(X^\{t\},Y^\{t\},Z^\{t\},X^\{t-1\},Z^\{t-1\}))\backslash geq1.$$ $$\delta \mathrm{\parallel }{X}^{t+1}-X^t{\mathrm{\parallel }}^2+\frac{\beta }{2}\mathrm{\parallel }{Y}^{t+1}-Y^t{\mathrm{\parallel }}^2\le H_t-H_{t+1}\backslash delta\backslash |X^\{t+1\}-X^\{t\}\backslash |^\{2\}+\backslash frac\{\backslash beta\}\{2\}\backslash |Y^\{t+1\}-Y^\{t\}\backslash |^\{2\}\backslash leq\; H\_\{t\}-H\_\{t+1\}$$ $$\leq\psi^{\prime}(H_{t}-H_{s})d(0,\beta H(X^{t},Y^{t},Z^{t},X^{t-1},Z^{t-1}))\left(H_{t}-H_{t+1}\right)\tag{68}$$ $$\leq\Delta_{0}^{t+1}d(0,\beta H(X^{t},Y^{t},Z^{t},X^{t-1},Z^{t-1}))$$ where the second inequality uses (67) and the last inequality uses the concavity of $\psi$. Using Lemma 1, we have $$(67)(67)$$ have from (68) that on (85) that $ \frac{1}{2}\min{\delta,\frac{\beta}{2}}\left(|X^{t+1}-X^t|+|Y^{t+1}-Y^t|\right)^2\leq\min{\delta,\frac{\beta}{2}}\left(|X^{t+1}-X^t|^2+|Y^{t+1}-Y^t|^2\right)$ $ \leq\delta|X^{t+1}-X^t|^2+\frac{\beta}{2}|Y^{t+1}-Y^t|^2$ $ \leq\Delta_{\theta}^{t+1}D\left(|X^t-X^{t-1}|+|Y^t-Y^{t-1}|+|Z^t-Z^{t-1}|\right)$ In fact, we will use the fact that $ |X^t|^2+\frac{\beta}{2}|Z^t|^2$ for every $ t\in\mathbb{R}$. $$(69)(69)$$ where the first inequality uses the fact that 12 (a + b) 2 ≤ a 2 + b 2for any a, b ∈ R. Now we bound ∥Z t − Z t−1∥. Using (31), we have that ∥Z t+1 − Z t∥ = |1 − τ |∥Z t − Z t−1∥ + βτ∥e t+1 − e t∥ + τ∥∇F(Xt+1 ⋆) − ∇F(Xt⋆ )∥ ≤ |1 − τ |∥Z t − Z t−1∥ + βτ∥e t+1 − e t∥ + τL∥Xt+1 ⋆ − Xt⋆∥ ≤ |1 − τ |∥Z t − Z t−1∥ + (β + L)τ∥e t+1 − e t∥ + τL∥Xt+1 − Xt∥ ≤ |1 − τ |∥Z t − Z t−1∥ + (β + L)τ4 (β − L) 2 ∥Xt − Xt−1∥ + τL∥Xt+1 − Xt∥ where the second inequality uses the definition of e t and last inequality uses (32). Rearranging the above inequality, it holds that

$$\mathrm{\parallel }{Z}^t-Z^{t-1}\mathrm{\parallel }\le \frac{1+\mathrm{\mid }1-\tau \mathrm{\mid }}{1-\mathrm{\mid }1-\tau \mathrm{\mid }}(\mathrm{\parallel }{Z}^t-Z^{t-1}\mathrm{\parallel }-\mathrm{\parallel }{Z}^t-Z^{t+1}\mathrm{\parallel })-\mathrm{\parallel }{Z}^t-Z^{t+1}\mathrm{\parallel }\backslash |Z^\{t\}-Z^\{t-1\}\backslash |\backslash leq\backslash frac\{1+|1-\backslash tau|\}\{1-|1-\backslash tau|\}\backslash left(\backslash |Z^\{t\}-Z^\{t-1\}\backslash |-\backslash |Z^\{t\}-Z^\{t+1\}\backslash |\backslash right)-\backslash |Z^\{t\}-Z^\{t+1\}\backslash |$$ $$+\frac{2}{1-|1-\tau|}(\beta+L)\tau\frac{4}{(\beta-L)^{2}}|X^{t}-X^{t-1}|+\frac{2}{1-|1-\tau|}\tau L|X^{t+1}-X^{t}|.$$

Plugging this bound into (69), we have that

1 2 min{δ, β 2 }∥Xt+1 − Xt∥ + ∥Y t+1 − Y t∥2 ≤ ∆ t+1 ψ D∥Xt − Xt−1∥ + ∥Y t − Y t−1∥ + ∆t+1 ψ D 1 + |1 − τ | 1 − |1 − τ | ∥Z t − Z t−1∥ − ∥Z t − Z t+1∥− ∥Z t − Z t+1∥ + ∆t+1 ψ D 2(β + L)τ 1 − |1 − τ | 4 (β − L) 2 ∥Xt − Xt−1∥ +2τL 1 − |1 − τ | ∥Xt+1 − Xt∥ ≤ ∆ t+1 ψ DD1 ∆1 t + ∆2 t , where

$$\begin{array}{c}{\mathrm{\Delta }}_{\psi }^{t+1}:=\psi (H_t-H_{\ast })-\psi (H_{t+1}-H_{\ast }),\\ D_1:=\mathrm{m}\mathrm{a}\mathrm{x}\{1+\frac{2(\beta +L)\tau }{1-\mathrm{\mid }1-\tau \mathrm{\mid }}\frac{4}{(\beta -L{)}^2},\frac{2\tau L}{1-\mathrm{\mid }1-\tau \mathrm{\mid }},1,\frac{1+\mathrm{\mid }1-\tau \mathrm{\mid }}{1-\mathrm{\mid }1-\tau \mathrm{\mid }}\},\\ {\mathrm{\Delta }}_t:=\mathrm{\parallel }{X}^t-X^{t-1}\mathrm{\parallel }+\mathrm{\parallel }{X}^{t+1}-X^t\mathrm{\parallel }+\mathrm{\parallel }{Y}^t-Y^{t-1}\mathrm{\parallel },\\ {\mathrm{\Delta }}_t^2:=(\mathrm{\parallel }{Z}^t-Z^{t-1}\mathrm{\parallel }-\mathrm{\parallel }{Z}^t-Z^{t+1}\mathrm{\parallel })-\mathrm{\parallel }{Z}^t-Z^{t+1}\mathrm{\parallel }\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\{\{\backslash Delta\_\{\backslash psi\}^\{t+1\}:=\backslash psi(H\_\{t\}-H\_\{*\})-\backslash psi(H\_\{t+1\}-H\_\{*\}),\}\}\backslash \backslash \; \{\{D\_\{1\}:=\backslash mathrm\{max\}\backslash \{1+\backslash frac\{2(\backslash beta+L)\backslash tau\}\{1-|1-\backslash tau|\}\backslash frac\{4\}\{(\backslash beta-L)^\{2\}\},\backslash frac\{2\backslash tau\; L\}\{1-|1-\backslash tau|\},1,\backslash frac\{1+|1-\backslash tau|\}\{1-|1-\backslash tau|\}\backslash \},\}\}\backslash \backslash \; \{\{\backslash Delta\_\{t\}:=\backslash |X^\{t\}-X^\{t-1\}\backslash |+\backslash |X^\{t+1\}-X^\{t\}\backslash |+\backslash |Y^\{t\}-Y^\{t-1\}\backslash |,\}\}\backslash \backslash \; \{\{\backslash Delta\_\{t\}^\{2\}:=\backslash left(\backslash |Z^\{t\}-Z^\{t-1\}\backslash |-\backslash |Z^\{t\}-Z^\{t+1\}\backslash |\backslash right)-\backslash |Z^\{t\}-Z^\{t+1\}\backslash |.\}\}\backslash end\{array\}$$

Rearranging the above inequality and taking square toot on both sides, we obtain that

$$\begin{array}{c}\mathrm{\parallel }{X}^{t+1}-X^t\mathrm{\parallel }+\mathrm{\parallel }{Y}^{t+1}-Y^t\mathrm{\parallel }\le \sqrt{\frac{2}{\min \{\delta ,\frac{\beta }{2}\}}}{\mathrm{\Delta }}_{\psi }^{t+1}D{D}_1({\mathrm{\Delta }}_t^1+{\mathrm{\Delta }}_t^2)\\ \le \frac{2}{\min \{\delta ,\frac{\beta }{2}\}}{\mathrm{\Delta }}_{\psi }^{t+1}D{D}_1+\frac{1}{4}({\mathrm{\Delta }}_t^1+{\mathrm{\Delta }}_t^2)\end{array}\backslash begin\{array\}\{l\}\{\{\backslash |X^\{t+1\}-X^\{t\}\backslash |+\backslash |Y^\{t+1\}-Y^\{t\}\backslash |\backslash leq\backslash sqrt\{\backslash frac\{2\}\{\backslash operatorname*\{min\}\backslash \{\backslash delta,\backslash frac\{\backslash beta\}\{2\}\backslash \}\}\}\backslash Delta\_\{\backslash psi\}^\{t+1\}D\; D\_\{1\}\backslash left(\backslash Delta\_\{t\}^\{1\}+\backslash Delta\_\{t\}^\{2\}\backslash right)\}\}\backslash \backslash \; \{\{\backslash leq\backslash frac\{2\}\{\backslash operatorname*\{min\}\backslash \{\backslash delta,\backslash frac\{\backslash beta\}\{2\}\backslash \}\}\backslash Delta\_\{\backslash psi\}^\{t+1\}D\; D\_\{1\}+\backslash frac\{1\}\{4\}\backslash left(\backslash Delta\_\{t\}^\{1\}+\backslash Delta\_\{t\}^\{2\}\backslash right)\}\}\backslash end\{array\}$$

where the second inequality uses the fact that √ab ≤ 1 2 (a + b) for any a, b > 0. Recalling the definitions of ∆1 t and ∆2 t , and rearranging the above inequality, we have that

∥Xt+1 − Xt∥ + ∥Y t+1 − Y t∥ ≤ s2 min{δ, β 2 } ∆ t+1 ψ DD1∆ ≤2 min{δ, β 2 } ∆ t+1 ψ DD1 + 1 4 ∥Xt − Xt−1∥ + ∥Xt+1 − Xt∥ + ∥Y t − Y t−1∥ + 1 4 ∥Z t − Z t−1∥ − ∥Z t − Z t+1∥ − ∥Z t − Z t+1∥ Further rearranging the above inequality, we have

$$(70)(70)$$ $$(71)(71)$$ Then, denoting ∆t+1 := ∥Xt+1 − Xt∥ + ∥Y $X^{t+1}-X^{t}|+|Y^{t+1}-Y^{t}|+D_{2}|Z^{t+1}-Z^{t}|$ (70) can be further passed to $$\frac{1}{4}\Delta_{t+1}\leq\frac{2}{\min{\delta,\frac{\beta}{2}}}\Delta_{\psi}^{t+1}DD_{1}+\frac{1}{4}\left(\Delta_{t}-\Delta_{t+1}\right)$$

Summing the above inequality from t = t1 + 1 to T, we have that

$$\begin{array}{c}\frac{1}{4}{\sum }_{t=t_1+1}^T{\mathrm{\Delta }}_{t+1}\le \frac{2}{\min \{\delta ,\frac{\beta }{2}\}}{\mathrm{\Delta }}_{\psi }^{t+1}D{D}_1+\frac{1}{4}({\mathrm{\Delta }}_{t_1+1}-{\mathrm{\Delta }}_{T+1})\\ \le \frac{2}{\min \{\delta ,\frac{\beta }{2}\}}\psi (H_t-H_{\ast })D{D}_1+\frac{1}{4}{\mathrm{\Delta }}_{t_1+1}\end{array}\backslash begin\{array\}\{l\}\{\{\backslash frac\{1\}\{4\}\backslash sum\_\{t=t\_\{1\}+1\}^\{T\}\backslash Delta\_\{t+1\}\backslash leq\backslash frac\{2\}\{\backslash operatorname*\{min\}\backslash \{\backslash delta,\backslash frac\{\backslash beta\}\{2\}\backslash \}\}\backslash Delta\_\{\backslash psi\}^\{t+1\}D\; D\_\{1\}+\backslash frac\{1\}\{4\}\backslash left(\backslash Delta\_\{t\_\{1\}+1\}-\backslash Delta\_\{T+1\}\backslash right)\}\}\backslash \backslash \; \{\{\backslash leq\backslash frac\{2\}\{\backslash operatorname*\{min\}\backslash \{\backslash delta,\backslash frac\{\backslash beta\}\{2\}\backslash \}\}\backslash psi(H\_\{t\}-H\_\{*\})D\; D\_\{1\}+\backslash frac\{1\}\{4\}\backslash Delta\_\{t\_\{1\}+1\}\}\}\backslash end\{array\}$$

where the last inequality uses the fact that P ψ > 0. Taking T in the above inequality to infinity, we see that ∞ t=t1+1 ∆t+1 < ∞. Thus {(Xt, Y t, Zt)} is convergent.

Next, we show the convergence rate of the generated sequence. Denote the limit of (Xt, Y t, Zt) as (X∗, Y ∗, Z∗). Define St =P∞ i=t+1 ∆i. Noting that ∥X∗ − Xt∥ + ∥Y ∗ − Y t∥ + ∥Z t − Z ∗∥ ≤ P∞ i=t ∆i = St, it suffices to show the convergence rate of St. Using (71), there exists D2 > 0 such that

$$\begin{array}{cccc}& S_t=\sum _{i=t}^{\mathrm{\infty }}{\mathrm{\Delta }}_i\le D_2(\psi (H_t-H_{\ast })-\psi (H_{t+1}-H_{\ast }))+({\mathrm{\Delta }}_t-{\mathrm{\Delta }}_{t+1})& & \text{(72)}\end{array}S\_\{t\}=\backslash sum\_\{i=t\}^\{\backslash infty\}\backslash Delta\_\{i\}\backslash leq\; D\_\{2\}\backslash left(\backslash psi(H\_\{t\}-H\_\{*\})-\backslash psi(H\_\{t+1\}-H\_\{*\})\backslash right)+\backslash left(\backslash Delta\_\{t\}-\backslash Delta\_\{t+1\}\backslash right)\backslash tag\{72\}$$ $$\leq D_{2}\psi(H_{t}-H_{})+\Delta_{t}=D_{2}\psi(H_{t}-H_{})+\left(S_{t-1}-S_{t}\right).$$

Now we bound ψ(Ht − H∗). From the KL assumption, ψ(w) = cw1−θ with some c >. Thanks to Theorem 5 (ii) and (14), we have from the KL inequality, it holds that

$$c(1-\theta )d(0,\mathrm{\partial }H(X^t,Y^t,Z^t,X^{t-1},Z^{t-1}))\ge (H_t-H_{\ast }{)}^{\theta }\mathrm{.}c(1-\backslash theta)d(0,\backslash partial\; H(X^\{t\},Y^\{t\},Z^\{t\},X^\{t-1\},Z^\{t-1\}))\backslash geq(H\_\{t\}-H\_\{*\})^\{\backslash theta\}.$$ θ. (73) Combining this with (1), we have that

$$(73)(73)$$ $$c(1-\theta )D(S_{t-1}-S_t)\ge (H_t-H_{\ast }{)}^{\theta }\mathrm{.}c(1-\backslash theta)D(S\_\{t-1\}-S\_\{t\})\backslash geq(H\_\{t\}-H\_\{*\})^\{\backslash theta\}.$$

This is equivalent to

$$c{(c(1-\theta )D(S_{t-1}-S_t))}^{\frac{1-\theta }{\theta }}\ge c(H_t-H_{\ast }{)}^{1-\theta }=\psi (H_t-H_{\ast })\mathrm{.}c\backslash left(c(1-\backslash theta)D(S\_\{t-1\}-S\_\{t\})\backslash right)^\{\backslash frac\{1-\backslash theta\}\{\backslash theta\}\}\backslash geq\; c(H\_\{t\}-H\_\{*\})^\{1-\backslash theta\}=\backslash psi(H\_\{t\}-H\_\{*\}).$$

Using this (72) can be further passed to

$$S_t\le D_3(S_{t-1}-S_t{)}^{\frac{1-\theta }{\theta }}+(S_{t-1}-S_t),S\_\{t\}\backslash leq\; D\_\{3\}(S\_\{t-1\}-S\_\{t\})^\{\backslash frac\{1-\backslash theta\}\{\backslash theta\}\}+(S\_\{t-1\}-S\_\{t\})\backslash ,,$$ $$\left(74\right)\backslash left(74\backslash right)$$

θ + (St−1 − St), (74) where D3 := D2c (c(1 − θ)D) 1−θ θ. Now we claim 1. When θ = 0, {(Xt, Y t, Zt)} converges finitely.

2. When θ ∈ (0, 1 2 ], there exist a > 0 and ρ1 ∈ (0, 1) such that St ≤ aρt1 .

3. When θ ∈ ( 1 2 , 1), there exists D4 such that St ≤ ct− 1−θ 2θ−1 for large t.

When θ = 0, we claim that there exists t such that Ht = H∗. Suppose to the contrary that Ht > H∗ for all t. Then, for large t, (73) holds, i.e., d(0, ∂H(Xt, Y t, Zt, Xt−1, Zt−1)) ≥1 c(1−θ) > 0. However, thanks to 1 and Corollary 4, we know that limt d(0, ∂H(Xt, Y t, Zt, Xt−1, Zt−1)) = 0, a contradiction. Therefore, there exists t such that Ht = H∗. From the argument in the beginning of this proof, we see that {(Xt, Y t, Zt)} converges finitely.

When θ ∈ (0, 1 2 ], we have 1−θ θ ≥ 1. Thanks to Corollary 4, we know that there exists t2 such that St−St−1 <

1. Thus, (74) can be further passed to St ≤ D3(St−1 − St) + (St−1 − St). This implies that

$$S_t\le \frac{D_3+1}{D_3+2}{S}_{t-1}\mathrm{.}S\_\{t\}\backslash leq\{\backslash frac\{D\_\{3\}+1\}\{D\_\{3\}+2\}\}S\_\{t-1\}.$$

Thus there exist a > 0 and ρ1 ∈ (0, 1) such that St ≤ aρt1 .

When θ ∈ ( 1 2 , 1), it holds that 1−θ θ < 1. From the last case, we know that St − St−1 < 1 when t > t2. Using (74), we have that St ≤ D3(St−1 − St) 1−θ θ + (St−1 − St) 1−θ θ = (D3 + 1) (St−1 − St) 1−θ θ. This implies that

$$S_t^{\frac{\theta }{1-\theta }}\le D_3^{\frac{\theta }{1+\theta }}(S_{t-1}-S_t)\mathrm{.}S\_\{t\}^\{\backslash frac\{\backslash theta\}\{1-\backslash theta\}\}\backslash leq\; D\_\{3\}^\{\backslash frac\{\backslash theta\}\{1+\backslash theta\}\}(S\_\{t-1\}-S\_\{t\}).$$

With this inequality, following the arguments in Theorem 2 of Attouch & Bolte (2009) starting from Equation (13) in Attouch & Bolte (2009), there exists c > 0 such that St ≤ ct− 1−θ 2θ−1 for large t. Thus, {S t} converges sublinearly.