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R E S E A R C H

Open Access

A regularized alternating direction method

of multipliers for a class of nonconvex

problems

Jin Bao Jian

1,2

, Ye Zhang

1

and Mian Tao Chao

1*

*Correspondence:

[email protected]

1College of Mathematics and Information Science, Guangxi University, Nanning, China Full list of author information is available at the end of the article

Abstract

In this paper, we propose a regularized alternating direction method of multipliers (RADMM) for a class of nonconvex optimization problems. The algorithm does not require the regular term to be strictly convex. Firstly, we prove the global

convergence of the algorithm. Secondly, under the condition that the augmented Lagrangian function satisfies the Kurdyka–Łojasiewicz property, the strong convergence of the algorithm is established. Finally, some preliminary numerical results are reported to support the efficiency of the proposed algorithm.

Keywords: Nonconvex optimization problems; Alternating direction method of multipliers; Kurdyka–Łojasiewicz property; Convergence

1 Introduction

In this paper, we consider the following nonconvex optimization problem

min f(x) +g(Ax), (1)

wheref :RnR∪ {+∞} is a proper lower semicontinuous function,g:RmR is a

continuous differentiable function with∇gLipschitz continuous and modulusL> 0, while ARm×nis a given matrix. When the functionsf andgare convex, the problem (1) can

be transformed into the split feasibility problem [1,2]. Problem (1) is equivalent to the following constraint optimization problem:

min f(x) +g(y),

s.t.Axy= 0. (2)

The augmented Lagrangian function of (2) is defined as follows:

(x,y,λ) =f(x) +g(y) –λ,Axy+

β

2 Axy

2, (3)

whereλRmis the Lagrangian parameter andβ> 0 is the penalty parameter.

(2)

The alternating direction method of multipliers (ADMM) was first proposed by Gabay and Mercier in 1970s, which is an effective algorithm for solving the two-block convex problems [3]. The iterative scheme of the classic ADMM for problem (2) is as follows:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

xk+1arg min{L

β(x,yk,λk)}, yk+1arg min{L

β(xk+1,y,λk)},

λk+1=λkβ(Axk+1yk+1).

(4)

Iff,gare convex functions, then the convergence of ADMM is well-understood and there are some recent convergence rate analysis results [4–8]. However, when the objective func-tion is nonconvex, ADMM does not necessarily converge. Recently, some scholars have proposed various improved ADMM for nonconvex problems, and analyzed their conver-gence [9–15]. In particular, Guo et al. [16,17] analyzed the strong convergence of classical ADMM for the nonconvex optimization problem of (2). Wang et al. [12,14] studied the convergence of the Bregman ADMM for the nonconvex optimization problems, where they need the augmented Lagrangian function with respect toxor the Bregman distance in thex-subproblem to be strongly convex.

The first formula of (4) has the following structure:

min

f(x) +gykλk,Axyk+β 2Axy

k2

. (5)

WhenAis not the identity matrix, the above problem may not be easy. Regularization is a popular technique to simplify the optimization problems [12,14,18]. For example, the regular term12 xxk 2

Gcould be added to the above problem (5), whereGis a symmetric

semidefinite matrix. Specifically, whenG=αIβAA, problem (5) is converted into the following form:

min

f(x) +α 2xb

k2

, (6)

with a certain knownbkRn. Since the first formula of (4) has the form of (6) withα= 1,

this paper considers the following regularized ADMM (in short, RADMM) for problem (2):

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

xk+1∈arg min{(x,yk,λk) +12 xxk 2G}, yk+1arg min{L

β(xk+1,y,λk)},

λk+1=λkβ(Axk+1yk+1),

(7)

whereGis a symmetric semidefinite matrix, x 2

G:=xGx.

The framework of this paper is as follows. In Sect.2, we present some preliminary ma-terials that will be used in this paper. In Sect.3, we prove the convergence of algorithm (7). In Sect.4, we report some numerical results. In Sect.5, we draw some conclusions.

2 Preliminaries

For a vectorx= (x1,x2, . . . ,xn)∈Rn, we let x = (

n i=1x2i)

1

2, x 1=n

i=1|xi|, and x 1 2 = (ni=1|xi|

1

(3)

matrix. For a subsetSRnand a pointxRn, ifSis nonempty, letd(x,S) =inf

yS yx .

WhenS=∅, we setd(x,S) = +∞for allx. A functionf :Rn→(–∞, +∞] is said to be proper, if there exists at least onexRnsuch thatf(x) < +. The effective domain off is

defined throughdomf ={xRn|f(x) < +∞}.

Definition 2.1([19]) The functionf :RnR∪ {+∞}is lower semicontinuous atx, if¯

f(x)¯ ≤lim infx→¯xf(x). Iff is lower semicontinuous at every pointxRn, thenf is called a

lower semicontinuous function.

Definition 2.2([19]) Letf:RnR∪ {+∞}be a proper lower semicontinuous function.

(i) The Fréchet subdifferential, or regular subdifferential, off atx∈domf, denoted by ˆ

∂f(x), is the set of all elementsuRnwhich satisfy

ˆ

∂f(x) =

uRnlim y=xyinf→x

f(y) –f(x) –u,yx yx ≥0

,

whenx∈/domf, letˆf(x) =∅;

(ii) The limiting-subdifferential, or simply the subdifferential, off atx∈domf, denoted by∂f(x), is defined as

∂f(x) =uRn| ∃xkx,fxkf(x),uk∈ ˆ∂fxku,k→ ∞.

Proposition 2.1([20]) Let f :RnR∪ {+∞}be a proper lower semicontinuous function,

then

(i) ˆf(x)⊆∂f(x)for eachxRn,where the first set is closed and convex while the second one is only closed;

(ii) Letukf(xk)andlim

k→∞(xk,uk) = (x,u),thenu∂f(x); (iii) A necessary condition forxRnto be a minimizer off is

0∈∂f(x); (8)

(iv) Ifg:RnRis continuously differentiable,then(f+g)(x) =f(x) +g(x)for any

x∈domf.

A point that satisfies (8) is called a critical point or a stationary point. The set of critical points off is denoted by critf.

Lemma 2.1([21]) Suppose that H(x,y) =f(x) +g(y),where f :RnR∪ {+∞} and g:

RmR are proper lower semicontinuous functions,then

∂H(x,y) =∂xH(x,y)×∂yH(x,y) =∂f(x)×∂g(y),

for all(x,y)∈domH=domf×domg.

(4)

Lemma 2.2([22]) Let function h:RnR be continuously differentiable and its gradient

h be Lipschitz continuous with modulus L> 0,then

h(y) –h(x) –h(x),yxL 2 yx

2, for all x,yRn.

Definition 2.3 We say that (x∗,y∗,λ∗) is a critical point of the augmented Lagrangian

function(·) of (2) if it satisfies

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∗∈∂f(x∗), ∇g(y∗) =λ∗, Ax∗–y∗= 0.

(9)

Obviously, (2) is equivalent to 0∈∂Lβ(x∗,y∗,λ∗).

Definition 2.4([21] (Kurdyka–Łojasiewicz property)) Letf :RnR∪ {+∞}be a proper

lower semicontinuous function. If there existη∈(0, +∞], a neighborhoodU ofx, and aˆ concave functionϕ: [0,η)→R+satisfying the following conditions:

(i) ϕ(0) = 0;

(ii) ϕis continuously differentiable on(0,η)and continuous at 0; (iii) ϕ(s) > 0for alls∈(0,η);

(iv) ϕ(f(x) –f(x))d(0,ˆ ∂f(x))≥1, for allxU∩[f(x) <ˆ f(x) <f(x) +ˆ η], thenf is said to have the Kurdyka–Łojasiewicz (KL) property atx.ˆ

Lemma 2.3([23] (Uniform KL property)) LetΦη be the set of concave functions which satisfy(i), (ii)and(iii)in Definition2.4.Suppose that f:RnR∪ {+∞}is a proper lower

semicontinuous function andΩis a compact set.If f(x)≡a for all xΩand f satisfies the KL property at each point ofΩ,then there existε> 0,η> 0,andϕΦηsuch that

ϕf(x) –ad0,∂f(x)≥1,

for all x∈ {xRn|d(x,Ω) <ε} ∩[x:a<f(x) <a+η].

3 Convergence analysis

In this section, we prove the convergence of algorithm (7). Throughout this section, we assume that the sequence{zk:= (xk,yk,λk)}is generated by RADMM (7). Firstly, the global

convergence of the algorithm is established by the monotonically nonincreasing sequence {(zk)}. Secondly, the strong convergence of the algorithm is proved under the condition that(·) satisfies the KL property. From optimality conditions of each subproblem in (7), we have

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

0∈∂f(xk+1) –Aλk+βA(Axk+1–yk) +G(xk+1–xk),

0 =∇g(yk+1) +λkβ(Axk+1yk+1),

λk+1=λkβ(Axk+1–yk+1).

(5)

That is,

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Aλk+1–βA(yk+1yk) –G(xk+1xk)f(xk+1),

λk+1=g(yk+1),

λk+1=λkβ(Axk+1yk+1).

(11)

We need the following basic assumptions on problem (2).

Assumption 3.1

(i) f :RnR∪ {+∞}is proper lower semicontinuous;

(ii) g:RmRis continuously differentiable, withg(u) –g(v)L uv ,

u,vRn;

(iii) β> 2Landδ:=β–2LLβ2 > 0; (iv) G0andG+AA0.

The following lemma implies that sequence{(zk)}is monotonically nonincreasing.

Lemma 3.1

zk+1L β

zkδykyk+12 –1

2x

k+1xk2

G. (12)

Proof From the definition of the augmented Lagrangian function(·) and the third for-mula of (11), we have

xk+1,yk+1,λk+1=

xk+1,yk+1,λk+ λkλk+1,Axk+1–yk+1

=

xk+1,yk+1,λk+ 1

βλ

kλk+12

(13)

and

xk+1,yk+1,λk

xk+1,yk,λk

=gyk+1–gykλk,ykyk+1–β 2Ax

k+1yk2

+β 2Ax

k+1yk+12

. (14)

From Assumption3.1(ii), Lemma2.2and (11), we have

gyk+1–gyk≤ –λk+1,yk+1–yk+L 2y

kyk+12

. (15)

Inserting (15) into (14) yields

xk+1,yk+1,λk

xk+1,yk,λk

λkλk+1,yk+1–ykβ 2Ax

k+1yk2

+β 2Ax

k+1yk+12+L 2y

(6)

Sinceλk+1=λkβ(Axk+1yk+1), we have

Axk+1–yk+1=1

β

λkλk+1 (17)

and

Axk+1–yk=1

β

λkλk+1–ykyk+1.

It follows that

λkλk+1,yk+1–ykβ 2Ax

k+1yk2

= –β 2y

k+1yk2 1

2βλ

kλk+12.

(18)

Combining (16), (17) and (18), we have

xk+1,yk+1,λk

xk+1,yk,λk≤–βL 2 y

k+1yk2. (19)

From –λk+1=∇g(yk+1) and Assumption3.1(ii), we have

1

βλ

kλk+12 ≤L2

βy

k+1yk2

. (20)

Adding (13), (19) and (20), one has

xk+1,yk+1,λk+1≤

xk+1,yk,λk

βL

2 –

L2

β

yk+1–yk2.

Sincexx+1is the optimal solution of the first subproblem of (7), one has

xk+1,yk,λk

xk,yk,λk–1 2x

k+1xk2

G.

Thus

xk+1,yk+1,λk+1

xk,yk,λk

βL

2 –

L2

β

ykyk+12–1 2x

k+1xk2

G.

Lemma 3.2 If the sequence{zk}is bounded,then

+∞

k=0

zkzk+12< +∞.

Proof Since{zk}is bounded,{zk}has at least one cluster point. Letz= (x,y,λ) be a

(7)

hence

z∗≤lim inf kj→+∞

zkj. (21)

Thus{(zkj)}is bounded from below. Furthermore, by Lemma3.1, sequence{(zk)}is nonincreasing, and so{(zk)}is convergent. Moreover,(z∗)≤(zk), for allk.

On the other hand, summing up of (12) fork= 0, 1, 2, . . . ,p, it follows that

1 2

p

k=0

xk+1–xk2G+δ p

k=0

yk+1–yk2≤

z0–

zp+1

z0–

z

< +∞.

Sinceδ> 0,G0, andpis chosen arbitrarily,

+∞

k=0

yk+1–yk2< +∞, +∞

k=0

xk+1–xk2G< +∞. (22)

From (20) we have

+∞

k=0

λk+1–λk2< +∞.

Next we prove+k=0 xk+1xk 2< +. Fromλk+1=λkβ(Axk+1yk+1), we have

λk+1–λk=λkλk–1+βAxkAxk+1+βyk+1–yk.

Then

βAxkAxk+12

=λk+1–λkλkλk–1–βyk+1–yk2

≤3λk+1–λk2+λkλk–12+β2ykyk+12. (23)

Therefore, we have+k=0xk+1xk 2

AA< +∞. Taking into account (22), we have

+∞

k=0

xk+1–xk(2AA+G)< +∞.

SinceAA+G0 (see Assumption3.1(iv)), one hask+∞=0 xk+1xk 2< +.

Therefore,+k=0zk+1zk 2< +.

Lemma 3.3 Define

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

εxk+1=βA(ykyk+1) +A(λkλk+1) –G(xk+1–xk),

εyk+1=λk+1–λk,

(8)

Then(εk+1

x ,εky+1,ελk+1)∈∂Lβ(zk+1).Furthermore,if AA0,then there existsτ> 0such that

d0,∂Lβ

zk+1≤τyk+1–yk+ykyk–1, k≥1.

Proof By the definition of(·), one has

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∂xLβ(zk+1) =∂f(xk+1) –Aλk+1+βA(Axk+1–yk+1),

∂yLβ(zk+1) =∇g(yk+1) +λk+1–β(Axk+1–yk+1),

∂λLβ(zk+1) = –(Axk+1–yk+1).

(24)

Combining (24) and (11), we get

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

βA(ykyk+1) +A(λkλk+1) –G(xk+1xk)

xLβ(zk+1),

λk+1λk

yLβ(zk+1), 1

β(λ

k+1λk)

λLβ(zk+1).

(25)

From Lemma2.1, one has (εk+1

x ,εyk+1,ελk+1)∈∂Lβ(zk+1).

On the other hand, it is easy to see that there existsτ1> 0 such that

εkx+1,εyk+1,ελk+1≤τ1yk+1–yk+λk+1–λk+xk+1–xk. (26)

Due toAA0 and (23), there existsτ2> 0 such that

xk+1–xkτ2yk+1–yk+λk+1–λk+λkλk–1, k≥1. (27)

Since (εxk+1,εky+1,ελk+1)∈∂Lβ(zk+1), from (26), (20) and (27), there existsτ > 0 such that d0,∂Lβ

zk+1≤εxk+1,εky+1,εkλ+1≤τyk+1–yk+ykyk–1,

k≥1.

Theorem 3.1(Global convergence) LetΩdenote the cluster point set of the sequence{zk}.

If{zk}is bounded,then

(i) Ωis a nonempty compact set,andd(zk,Ω)→0,ask→+∞, (ii) Ω⊆crit,

(iii) (·)is constant onΩ,andlimk→+∞(zk) =(z∗)for allz∗∈Ω. Proof (i) From the definitions ofΩandd(zk,Ω), the claim follows trivially.

(ii) Let z∗ = (x∗,y∗,λ∗)∈ Ω, then there is a subsequence {zkj} of {zk}, such that

limkj+zkj=z. Sincexk+1is a minimizer of functionL

β(x,yk,λk) +12 xxk 2G for the variablex, one has

xk+1,yk,λk+1 2x

k+1xk2 G

x∗,yk,λk+1 2x

xk2 G,

that is,

xk+1,yk,λk

x∗,yk,λk+1 2x

xk2 G

1 2x

k+1xk2

(9)

Lemma3.1implies thatlimk→∞ xk+1–xk 2G= 0. Since(·) is continuous with respect toyandλ, we have

lim sup kj→+∞

zkj+1=lim sup

kj→+∞

xkj+1,ykj,λkj

≤lim sup kj→+∞

x∗,yk,λk

=

z∗. (29)

On the other hand, sinceL(·) is lower semicontinuous,

lim inf kj→+∞

zkj+1≥L β

z∗. (30)

Combining (29) and (30), we getlimkj→+∞(z

kj) =L

β(z∗). Thenlimkj→+∞f(x

kj) =f(x).

By taking the limitkj→+∞in (11), we have

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∗∈∂f(x∗), ∇g(y∗) = –λ∗, Ax∗–y∗= 0.

that is,z∗∈crit.

(iii) Let z∗ ∈ Ω. There exists {zkj} such that lim

kj→+∞z

kj = z. Combining

limkj→+∞(z

kj) =L

β(z∗) and the fact that{(zk)}is monotonically nonincreasing, for allz∗∈Ω, we have

lim k→+∞

zk=

z∗,

and so(·) is constant onΩ.

Theorem 3.2(Strong convergence) If{zk}is bounded,AA0,andLβ(z)satisfies the KL property at each point ofΩ,then

(i) +k=0zk+1zk < +,

(ii) The sequence{zk}converges to a stationary point ofLβ(·).

Proof (i) Letz∗∈Ω. From Theorem3.1, we havelimk+(zk) =(z∗). We consider two cases:

(a) There exists an integerk0, such that(zk0) =(z∗). From Lemma3.1, we have 1

2x

k+1xk2

G+δy

k+1yk2

zk

zk+1≤

zk0L

β

z∗= 0, kk0.

Then, one has xk+1xk 2

G = 0,yk+1=yk, kk0. From (20), one hasλk+1=λk,k>k0. Furthermore, from (23) andAA0, we havexk+1=xk,k>k

(10)

(b) Suppose that(zk) >(z∗),k≥1. From Theorem3.1(i), it follows that forε> 0, there existsk1> 0, such thatd(zk,Ω) <ε, for allk>k1. Sincelimk→+∞(zk) =(z∗), for givenη> 0, there existsk2> 0, such that(zk) <(z∗) +η, for allk>k2. Consequently, one has

dzk,Ω<ε,

z∗<

zk<

z∗+η, for allk>k˜=max{k1,k2}.

It follows from the KL property, that

ϕLβ

zk

zd0,∂Lβ

zk≥1, for allk>k.˜ (31)

By the concavity ofϕand since(zk) –(zk+1) = ((zk) –(z∗)) – ((zk+1) –(z∗)), we have

ϕLβ

zk

z∗–ϕLβ

zk+1–

z

ϕLβ

zk

z

zk

zk+1.

(32)

Letp,q=ϕ((zp) –(z∗)) –ϕ((zq) –(z∗)). Combiningϕ((zk) –(z∗)) > 0, (31) and (32), we have

zk

zk+1≤ k,k+1

ϕ((zk) –(z∗))

d0,∂Lβ

zkk,k+1.

From Lemma3.3, we obtain

zk

zk+1≤τykyk–1+yk–1–yk–2k,k+1.

From Lemma3.1and the above inequality, we have

1 2x

k+1xk2

G+δy

k+1yk2

τykyk–1+yk–1–yk–2k,k+1, for allk>k.˜

Thus

yk+1yk2τ δy

kyk–1+yk–1yk–2

k,k+1, for allk>k.˜

Furthermore,

3yk+1–yk

≤2ykyk–1+yk–1–yk–2 1 2

3 2

τ δ

1 2

k,k+1

, for allk>k.˜

Using the fact that 2ab≤a2+b2, we obtain

3yk+1–yk

ykyk–1+yk–1–yk–2+9τ

(11)

Summing up the above inequality fork=k˜+ 1, . . . ,s, yields

3

s

k=k˜+1

yk+1–yk

s

k=k˜+1

ykyk–1+yk–1yk–2+9τ

4δk˜+1,s+1.

Thus

s

kk+1

yk+1–yk≤2yk˜+1–yk˜+yk˜–yk˜–1+9τ

4δk˜+1,s+1.

Notice thatϕ((zs+1) –(z∗)) > 0, so taking the limits→+∞, we have +∞

k=k˜+1

yk+1–yk

≤2yk˜+1–yk˜+yk˜–yk˜–1+9τ 4δϕ

z˜k+1–Lz∗. (34) Thus

+∞

kk+1

yk+1–yk< +∞.

It follows from (20) that

+∞

kk+1

λk+1–λk< +∞.

FromAA0, (23) and the above two formulas, we obtain

+∞

kk+1

xk+1xk< +.

Since

zk+1–zk=xk+1–xk2+yk+1–yk2+λk+1–λk2

1 2

xk+1–xk+yk+1–yk+λk+1–λk,

we know

m

kk+1

zk+1–zk< +∞.

(ii) From (i), we known that {zk} is a Cauchy sequence and so is convergent.

Theo-rem3.2(ii) follows immediately from Theorem3.1(ii).

(12)

Lemma 3.4 Suppose that AA0and

Γ := inf yRm

g(y) – 1

2L∇g(y) 2

> –∞.

If one of the following statements is true:

(i) f is coercive,i.e.,limx →+∞f(x) = +∞,

(ii) f is bounded from below andgis coercive,i.e.,infxRnf(x) > –∞and

limx →+∞g(x) = +∞,

then{zk}is bounded.

Proof (i) Suppose thatf is coercive. From Lemma3.1, we know that(zk)≤(z1) < +∞, for allk≥1. Combining with∇g(xk) = –λk, one has

z1≥fxk+gykλk,Axkyk+β 2Ax

kyk2

=fxk+gyk– 1 2βλ

k2+β

2

Axkyk–1

βλ k

2

=fxk+

gyk 1

2L∇g

yk2

+ 1 2L– 1 2β

λk2

+β 2

Axkyk– 1

βλ k

2

fxk+Γ +

1 2L– 1 2β

λk2+β 2

Axkyk– 1

βλ k

2

. (35)

Sinceβ> 2Landf is coercive, it is easy to see that{xk},{λk}, and{β

2 Axkyk– 1

βλk 2}are bounded. Furthermore,{yk}is bounded. Thus{zk}is bounded.

(ii) Similar as with (i), we have

z1≥fxk+gyk– 1 2βλ

k2 +β

2

Axkyk–1

βλ k

2

fxk+1 2g

yk+1 2Γ +

1 4L∇g

yk2– 1 2βλ

k2

+β 2

Axkyk– 1

βλ k

2

fxk+1 2g

yk+1 2Γ +

1 4L– 1 2β

λk2+β 2

Axkyk–1

βλ k

2 .

Notice thatβ> 2L, functionf is bounded from below,gis coercive and Assumption3.1(ii) holds, thus {yk}, {λk}, and {β

2 Axkyk– 1

βλk 2} are bounded. Since AA0, {xk} is

(13)

4 Numerical examples

In compressed sensing, one needs to find the sparsest solution of a linear system, which can be modeled as

min x 0,

s.t.Dx=b, (36)

whereDRm×nis the measuring matrix,bRmis observed data, x 0denotes the

num-ber of nonzero elements ofx, which is called thel0norm.

Problem (36) is NP-hard. In practical applications, one may relax thel0 norm to the l1norm orl1

2 norm, and consider their regularized versions, which lead to the following convex problem (37) and nonconvex problem (38):

min γ x 1+ y 2,

s.t.Dxy=b, (37)

and

min γ x 1 2 1 2

+ y 2,

s.t.Dxy=b. (38)

In this section, we will apply RADMM (7) to solve the above two problems. For sim-plicity, we setb= 0 throughout this section. Applying RADMM (7) to problem (37) with G=αIβDDyields

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

xk+1S(xkβ

αDDxk+ 1

αD(βyk+λk)); γ 2α),

yk+1= 1 2+β(βDx

k+1λk),

λk+1=λkβ(Dxk+1–yk+1),

(39)

where S(·;μ) ={(x1),(x2), . . . ,(xn)} is the soft shrinkage operator [24] defined as

follows:

(xi) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

xi+μ2, ifxi≤–μ2,

0, if|x|<μ2,

xiμ2, ifxiμ2.

Applying RADMM (7) to problem (38) withG=αIβDDyields

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

xk+1∈H(xkβαDDxk+α1D(βyk+λk));γα),

yk+1= 1

2+β(βDxk+1–λk),

λk+1=λkβ(Dxk+1–yk+1),

(14)
[image:14.595.116.479.81.225.2]

Figure 1Comparison the performance of HRADMM and SRADMM

whereH(·;μ) ={(x1),(x2), . . . ,(xn)}is the half-shrinkage operator [25] defined as

follows:

(xi) =

⎧ ⎨ ⎩

2xi 3 (1 +cos

2

3(πϕ(|xi|))), |xi|>

3

54 4 μ

2 3,

0, otherwise,

withϕ(x) =arccos(μ8(xi 3)

–32).

For simplicity, we denote algorithms (39) and (40) by SRADMM and HRADMM, respec-tively. The selection of relevant parameters in numerical experiments is given below. We now conduct an experiment to verify convergence of the nonconvex RADMM, and reveal its advantages in sparsity-inducing and efficiency through comparing the performance of HRADMM and SRADMM. In the experiment,m= 511,n= 512, the matrixDR511×512 is obtained by unitizing the matrix with randomly generated entries obeying the normal distributionN(0, 1), the noise vectorεN(0, 1), the recovery vectorr=Dx0+ε, and the regularization parameters areγ = 0.0015,β= 0.8,α= 2.5.

The experimental results are shown in Fig.1, where the restoration accuracy is measured by means of the mean squared error:

MSEx∗–xk=1 nx

xk,

MSEy∗–yk=1 ny

yk,

where (x∗,y∗) = (0, 0) is the optimal solution for the problems (37) and (38), respectively. Programming is performed on Matlab R2014a, a computer running the program is con-figured as follows: Windows 7 system, Inter(R) Core(TM) i7-4790 CPU 3.60 GHz, 4 GB memory. Numerical results show that algorithm (7) is efficient and stable. As shown in Fig.1, both sequencesxkandykwere fairly near the true solution. i.e., the convergence is

justified. It is readily seen that HRADMM converges faster than SRADMM.

5 Conclusion and outlook

(15)

Kurdyka–Łojasiewicz property, the strong convergence of the algorithm is analyzed. Fi-nally, the effectiveness of the algorithm is verified by numerical experiments.

Funding

This research was supported by the National Natural Science Foundation of China (Nos. 11601095, 11771383) and by the Natural Science Foundation of Guangxi Province (Nos. 2016GXNSFBA380185, 2016GXNSFDA380019 and

2014GXNSFFA118001).

Availability of data and materials

The authors declare that all data and material in the paper are available and veritable.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors wrote, read, and approved the final manuscript.

Authors’ information

JinBao Jian, Ph.D., professor, E-mail: [email protected]; Ye Zhang, E-mail: [email protected]; MianTao Chao, Ph.D., associate professor, E-mail: [email protected].

Author details

1College of Mathematics and Information Science, Guangxi University, Nanning, China.2College of Science, Guangxi University for Nationalities, Nanning, China.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 11 December 2018 Accepted: 1 July 2019 References

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Figure

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