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HAL Id: hal-00779255

https://hal.univ-antilles.fr/hal-00779255

Submitted on 21 Jan 2013

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enlargement of maximal monotone operators

Abdellatif Moudafi, E. Elizabeth

To cite this version:

Abdellatif Moudafi, E. Elizabeth. Approximate inertial proximal methods using the enlargement of maximal monotone operators. International Journal of Pure and Applied Mathematics, Academic Publishing Ltd, 2003, 5 (3), pp.283-299. �hal-00779255�

(2)

using the enlargement of a maximal monotone operator

A. Moudafi and E. Elisabeth

Univeristé Antilles Guyane, DSI-GRIMAAG 97200 Schoelcher, Martinique, France.

[email protected]

Abstract:

An approximate procedure for solving the problem of nding a zero of a maximal monotone operator is proposed and its convergence is established under var- ious conditions. More precisely, it is shown that this method weakly converges under natural assumptions and strongly converges provided that either the inverse of the involved operator is Lipschitz continuous around zero or the interior of the solution set is nonempty. A particular attention is given to the convex minimization case.

AMSSubjectClassication:

Primary, 90C25; Secondary, 49M45, 65C25.

Key words.

monotone operators, elargements, proximal point algorithm, lo- cal Lipschitz continuity, approximate subdierential, convergence, convex min- imization.

1 Introduction and preliminaries

In this paper we will focus our attention on the classical problem of nding a zero of a maximal monotone operator

T

on a real Hilbert space

H

:

nd

x2H

such that

Tx30:

(1.1)

This is a well-known problem which includes, as special cases, optimization and min-max problems, complementarity problems, and variational inequalities.

One of the fondamental approaches to solving (1.1) is the proximal method, which generates the next iterates

xk +1

by solving the subproblem

02

k

T(x)+(x?x

k

);

(1.2)

where

xk

is the current iterate and

k

is a regularization parameter. The liter- ature on this subject is vast (see

[8]

for a survey).

1

(3)

Recently, an inertial proximal algorithm was proposed by Alvarez in

[1]

in the context of convex minimization. Afterwards, Attouch and Alvarez considered its extension to maximal monotone operators in

[2]

. It works as follows. Given

x

k ?1

;x

k

2 H

and two parameters

k 2 [0;1[

and

k >0

, nd

xk +1 2H

such that

k T(x

k +1 )+x

k +1

?x

k

?

k (x

k

?x

k ?1

)30:

(1.3)

It is well known that the proximal iteration may be interpreted as an implicit one-step discretisation method for the evolution dierential inclusion

dx

dt

(t)+T(x(t))30

a.e.

t0;

(1.4) where the parameter

k

is a (variable) stepsize. While the inspiration for (1.3) comes from the implicit discretization of the dierential system of the second- order in time, namely

d 2

x

dt 2

(t)+ dx

dt

(t)+T(x(t))30

a.e.

t0;

(1.5) where

>0

is a damping or a friction parameter.

Under appropriate conditions on

k

and

k

Attouch

&

Alvarez proved that if the solution set

S =T?1(0)

is nonempty, then for every sequence

fxkg

gener- ated by (1.3), there exists an

x2S

such that

fxkg

converges to

x

weakly in

H

as

k!1

.

For developing implementable computational techniques, it is of particular im- portance to treat the case when (1.3) is solved approximately. Before introduc- ing our approximate method, let us recall the following concepts which are of common use in the context of convex and nonlinear analysis. Troughout,

H

is a real Hilbert space,

h;i

denotes the associated scalar product and

jj

stands for the corresponding norm. An operator is said to be monotone if

hu?v ;x?y i0

whenever

u2T(x);v2T(y ):

It is said to be maximalmonotone if, in addition, the graph,

f(x;y )2H H:y2

T(x)g

, is not properly contained in the graph of any other monotone operator.

It is well-known that for each

x2 H

and

>0

there is a unique

z 2H

such that

x2(I+T)z

. The single-valued operator

JT :=(I+T)?1

is called the resolvent of

T

of parameter

. It is a nonexpansive mapping which is everywhere dened and satises:

z=JTz

, if and only if,

02Tz

. Let us also recall a notion which is clearly inspired by the approximate subdierential. In [16], Iusem, Burachik and Svaiter dened

T"(x)

, an

"

-enlargement of a monotone operator

T

, as

T

"

(x):=fv2H;hu?v ;y?xi?" 8y ;u2T(y )g;

(1.6) where

"0

. Since

T

is assumed to be maximal monotone,

T0(x)=T(x)

, for any

x

. Furthermore, directly from the denition it follows that

0" " )T

"

1

(x)T

"

2

(x):

(4)

Thus

T"

is an enlargement of

T

. The use of elements in

T"

instead of

T

allows an extra degree of freedom, which is very useful in various applications. On the other hand, setting

" = 0

one retrieves the original operator

T

, so that the classical method can be also treated. For all these reasons, we consider the following scheme: nd

xk +12H

such that

k T

"k

(x

k +1 )+x

k +1

?y

k

30:

(1.7)

where

yk :=xk+k(xk?xk ?1);k;k;"k

are nonnegative real numbers.

We will impose the followingtolerance criteria on the term

"k

which are standard in the literature:

+1

X

k =1

k

"

k

<+1;

(1.8)

and

p

k

"

k

k kx

k +1

?y

k

k

with

+1X

k =1

k

<+1:

(1.9)

The rst condition is typically needed to establish global convergence, while the second is required for local linear rate of convergence result under additional natural assumptions.

The remainder of the paper is organized as follows: In section 2, we present a weak convergence result for the sequence generated by (1.7) under criterion (1.8). We also consider various conditions for which the convergence is strong.

In section 3, we present an application to convex minimization and study the convergence of a perturbed version of (1.7) as well as conditions ensuring the convergence for both the values and the iterates.

2 The main results

2.1 A weak convergence result

Theorem 2.1 Let

fxkgH

be a sequence such that

02x

k +1

?x

k

?

k (x

k

?x

k ?1 )+

k T

"k

(x

k +1

);k=1;2;

where

T : H!P(H)

is a maximal monotone operator with

S :=T?1(0) 6=;

, and the parameters

k;k

and

"k

satisfy:

1.

9>0

such that

8k2IN;k

.

2.

92[0;1[

such that

8k2IN;0k

.

3.

P+1k =1k"k <+1

.

(5)

If the followingconditionholds

+1

X

k =1

kj

x

k?

x

k ?1j2

<

+1

; (2.10)

then, there exists

x

2

S

such thatf

x

kgweaklyconverges to

x

as

k

!+1.

Pro of

. Fix x

2

S

=

T

?1(0)

, since

02

x

k +1?

x

k?

k(

x

k?

x

k ?1)+

k

T

"k(

x

k +1)

, from denition (1.6) it follows that

h

x

k +1?

x

k?

k(

x

k?

x

k ?1)

;x

k +1?

x

i

k

"

k

:

Dene the auxiliary real sequence '

k := 12j

x

k?

x

j2

. It is direct to check that

h

x

k +1?

x

k?

k(

x

k?

x

k ?1)

;x

k +1?

x

i =

'

k +1?

'

k+12j

x

k +1?

x

kj2

?

kh

x

k?

x

k ?1

;x

k +1?

x

i

; and since

h

x

k?

x

k ?1

;x

k +1?

x

i = h

x

k?

x

k ?1

;x

k?

x

i+h

x

k?

x

k ?1

;x

k +1?

x

ki

=

'

k?

'

k ?1+12j

x

k?

x

k ?1j2+h

x

k?

x

k ?1

;x

k +1?

x

ki

; it follows that

'

k +1?

'

k?

k(

'

k?

'

k ?1) ?12j

x

k +1?

x

kj2+

kh

x

k?

x

k ?1

;x

k +1?

x

ki

+ k

2

j

x

k?

x

k ?1j2+

k

"

k

= ?

1

2

j

x

k +1?

x

k?

k(

x

k?

x

k ?1)j2

+ k+

2

k

2

j

x

k?

x

k ?1j2+

k

"

k

: Hence

'

k +1?

'

k?

k(

'

k?

'

k ?1)?1

2

j

v

k +1j2+

kj

x

k?

x

k ?1j2+

k

"

k

: (2.11) Setting

k :=

'

k?

'

k ?1

and

k:=

kj

x

k?

x

k ?1j2+

k

"

k

, we obtain

k +1

k

k+

k

k[

k]++

k

; where

[

t

]+:=

max

(

t;

0)

, and consequently

[

k +1]+

[

k]++

k

; with

2[0

;

1[

given by (2).

The rest of the proof follows that given in

[2]

and is presented here for com- pleteness and to convey the idea in

[2]

. The latter inequality yields

[

k +1]+

k[

1]++k ?1X

i

k ?i

;

(6)

and therefore

1

X

k =1

[

k +1

]

1

1

?

([

1

]

+

+

X1

k =1

k

)

;

which is nite thanks to (3) and (2.10). Consider the sequence dened by

t

k

:=

'k?Pki=1

[

i

]

+

. Since

'k

0 and

Pki=1

[

i

]

+<

+

1

, it follows that

tk

is bounded from below. But

t

k +1

=

'k +1?

[

k +1

]

+?Xk

i=1

[

i

]

+'k +1?'k +1

+

'k?Xk

i=1

[

i

]

+

=

tk;

so that

ftkg

is nonincreasing. We thus deduce that

ftkg

is convergent and so is

f'

k

g

. On the other hand, from (2.11) we obtain the estimate 1 2

jvk +1j2'k?'k +1

+

[

k

]

+

+

k:

Passing to the limit in the latter inequality and taking into account that

f'kg

converges, [

k

]

+

and

k

go to zero as k tends to +

1

, we obtain

lim

k !+1 v

k +1

= 0

:

Now let

x

be a weak cluster point of

fxkg

. There exists a subsequence

fxg

which converges weakly to

x

and satises

v+1

+

T"

(

x+1

)

3

0. By denition (1.6), we have

h?

v

+1

?z;x

+1

?y i?"

8z2T

(

y

)

:

Passing to the limit, as

!

+

1

, we obtain

h?z;x

?y i

0

;

this being true for any

z 2 T

(

y

), from maximal monotonicity of

T

, it follows that 0

2 T

(

x

), that is

x2 S

. We end the proof by applying the well-known Opial's lemma [12].

Remark 2.1 1. Although condition (2.10) involves the iterates that are a priori unknown. In practice, as it was stressed by Alvarez & Attouch, it is easy to enforce it by applying an appropriate on-line rule. Further- more, condition (2.10) is automatically satised in some special cases (see proposition 2.1, [2]).

2. Under assumptions of theorem 2.1 and in view of its proof, it is clear that

fx

k

g

is bounded if, and only if, there exists at least one solution to (1.1).

(7)

2.2 Strong convergence results

First, we give a result showing that the criterion (1.9) ensures the strong con- vergence of the sequence generated by:

x

k +1

?y

k

+

kT"k

(

xk +1

)

3

0

;

(2.12) where

yk

=

xk

+

k

(

xk?xk ?1

),

"k

and

k

are positive reals.

Theorem 2.2 Let

fxkg

be any sequence generated by (1.7) using criterion (1.9) with

fkg

nondecreasing (

k " 1

+

1

). Assume that

fxkg

is bounded,

9 2

[0

;

1[ such that

8k 2 IN;

0

k

, and

T?1

is Lipschitz continuous around zero, i.e., problem (1.1) has the unique solution, say

x

, and there exist some constants

a>

0 and

>

0 such that

jv j;v2T

(

y

)

)jy?xj

ajv j:

Then, there is a real

2

]0

;

1[ and a range

K2IN

such that

jx

k +1

?xj

jxk?xj

+ (2

k

+

a

k

)

kjxk?xk ?1j 8kK :

Furthermore, if we assume that lim

k !+1kkjxk?xk ?1j

= 0

;

then, the sequence

fx

k

g

strongly converges to

x

.

Proof . The rst part of the proof follows that given in [15] and is presented here for completeness. The sequence

fxkg

being bounded, also satises condition 3) of theorem 2.1 for

"k

=

kjxk +1?ykj

, so the conclusions of theorem 2.1 are in force. Now, let ~

xk +1

be the exact solution of the

k

-th inertial proximal method, that is

k

Tx

~

k +1

+ ~

xk +1?yk 3

0

:

(2.13) Denition of

T"k

combined with relations (2.12) and (2.13) leads to

hy

k

?x

k +1

?

(

yk?x

~

k +1

)

;xk +1?x

~

k +1i?k"k;

which implies that

jx

~

k +1?xk +1j2k"k

. The latter together with (1.9) yield

jx

~

k +1?xk +1jkjxk +1?ykj:

(2.14) Therefore

jx

~

k +1?ykjjx

~

k +1?xk +1j

+

jxk +1?ykj

(1 +

k

)

jxk +1?ykj;

which, using the convergence of

vk +1

:=

xk +1?yk !

0 (theorem 2.1), implies

that

jx

~

k +1?ykj !

0. Because

fkg

is bounded away from zero, we further

(8)

conclude that

wk

=

1k

(

yk?x

~

k +1

)

!

0. Using the Lipschitz continuity of

T?1

around 0, we have, for indices

k

suciently large, that:

jx

~

k +1?xj

ajwkj

=

a

k

jx

~

k +1?ykj:

(2.15) We further obtain

jy

k

?xj

2

=

jyk?x

~

k +1j2

+

jx

~

k +1?xj

2

+ 2

hyk?x

~

k +1;x

~

k +1?xi

=

jyk?x

~

k +1j2

+

jx

~

k +1?xj

2

+ 2

khwk;x

~

k +1?xi

jy

k

?x

~

k +1j2

+

jx

~

k +1?xj

2

(1 + (

ak

)

2

)

jx

~

k +1?xj

2:

(2.16)

Hence, by setting

k

:=

paa 2

+ 2

k

, we obtain

jx

~

k +1?xj

kjyk?xj:

Using the latter relation and (2.14), we further obtain

jx

k +1

?xj

jxk +1?x

~

k +1j

+

jx

~

k +1?xj

k jx

k +1

?y

k

j

+

kjyk?xj:

(2.17) Similarly,

jx

k +1

?y

k

j jx

k +1

?x

~

k +1j

+

jx

~

k +1?ykj

k jx

k +1

?y

k

j

+

jyk?xj;

where also (2.16) was used in the last inequality.

Therefore

jx

k +1

?y

k j

1

?

1

kjyk?xj:

Now, combining the latter relation with (2.17) and using the triangular inequal- ity, we obtain

jx

k +1

?xj

kjyk?

xjkjxk?xj

+

kkjxk?xk ?1j;

where

k

:=

1?kk

+

k

, from which we deduce easily, by taking into account conditions on

k

and

k

, the existence of a range

K

such that

jx

k +1

?xj

jxk?xj

+

k;

where

2

]0

;

1[ and

k

:= (2

k

+

ka

)

kjxk?xk ?1j

. Hence,

jx

k

?xj

kjx0?xj

+

Xk

j=1

j

k ?j :

The result follows from Ortega and Rheinboldt [14

;p:

338], since by hypothesis

lim

k !+1k

= 0

:

(9)

Remark 2.2 1. When

k

= 0 , we obtain the linear convergence estimate obtained by Solodov & Svaiter [15] which is strictly better than the one for the classical proximal algorithm, namely,

~

k

=

1?k+kk

( [14] , theorem 2).

2. The condition of Lipschitz continuity above holds true if

T?1

is globally Lipschitz continuous which is satised, for instance, when

T

is strongly monotone.

3. The condition of local Lipschitz continuity above is also satised if

T?1

is dierentiable at 0, that is,

T?1

(0) =

fxg

and

9A

:

H!H

a continuous linear transformation such that, for

>

0

T

?1

(0)

?

x

+

Awo

(

jw j

)

B

if

jw j;

where

B

stands for the closed unit ball.

We close this section with a special result showing that the Inertial Proximal Method can converge in nitely many iterations:

Theorem 2.3 Let

fxkg

be any sequence generated by (1.8) under the criterion (1.9) with

k

bounded away from zero. Suppose that

fxkg

is bounded and that

9x

2H

such that 0

2

i

ntTx:

(2.18) Then, for all

k

suciently large, it holds that

J T

k

(

xk

+

k

(

xk?xk ?1

)) =

x:

Moreover, the sequence

fxkg

strongly converges to

x

.

Proof . Let ~

xk +1

be the exact solution of the

k

-th inertial proximal method, that is

k

Tx

~

k +1

+ ~

xk +1?yk 3

0

:

Hypothesis (2.18) and ([14], theorem 3) imply the existence of a positive real

"

such that

jxj")T

?1

x

=

fxg:

But the hypothesis of the present theorem recovers those of theorem 2.1. there- fore, we know that

1k

(

yk ?x

~

k +1

)

!

0. Since ~

xk +1 2 T?1

(

1k

(

yk ?x

~

k +1

)), assumption (2.18) implies that, for

k

suciently large, ~

xk +1

=

x

. Thus the inertial proximal method in its exact form converges to

x

in a nite number of iterations from any starting points

x0

and

x1

. From which we deduce, for all

k

sucientely large, that

jx

k +1

?xj

2

=

jxk +1?x

~

k +1j2k"k:

The strong convergence of the sequence

fxkg

to

x

follows by passing to the limit

in the last inequality, since condition (1.8) implies that lim

k !+1k"k

= 0.

(10)

3 Convex Minimization

3.1 Approximate methods

An interesting case is obtained by taking

T

=

@f

,

@f

stands for the subdier- ential of a proper convex lower-semicontinuous function

f

:

X ! IR[f

+

1g

. Indeed,

@f

is well-known to be a maximalmonotone operator and problem (1.1) reduces to the one of nding a minimizer of the function

f

.

In [1], Alvarez proposed the following approximate inertial proximal method:

k

@

"

k

f

(

xk +1

) +

xk +1?xk?k

(

xk?xk ?1

)

3

0

;

(3.19) where

@"kf

is the approximate subdierential of

f

. Since in the case

T

=

@f

the enlargement given in (1.6) is larger than the the approxiamte subdierential, i.e.

@

"

f

(

@f

)

"

, we can write

@"kf

(

xk +1

)

(

@f

)

"k

(

xk +1

)

;

which leads to

k

(

@f

)

"k

(

xk +1

) +

xk +1?xk?k

(

xk?xk ?1

)

3

0

;

(3.20) which is a particular case of the method proposed in this paper with

T

=

@f

. As a consequence of theorem 2.1 and theorem 2.2, we obtain the following convergence result which recover and completes a result by Alvarez [1].

Corollary 3.1 Let

fxkgH

be a sequence such that

0

2xk +1?xk?k

(

xk?xk ?1

) +

k@"kf

(

xk +1

)

;k

= 1

;

2

;

(3.21) where

f

is a proper closed convex function with Argmin

f 6

=

;

, and the param- eters

k;k

and

"k

satisfy:

1.

9>

0 such that

8k2IN;k

.

2.

92

[0

;

1[ such that

8k2IN;

0

k

. 3.

P+1k =1k"k <

+

1

.

If the following condition holds

+1

X

k =1

k jx

k

?x

k ?1 j

2

<

+

1;

(3.22)

then

fxkg

weakly converges to a minimizer of

f

and lim

k !+1

f

(

xk

) = inf

x2H f

(

x

) . Moreover, if we replace condition 3) of theorem 2.1 by criterion (1.9) and we assume in addition that

@f

is Lipschitz continuous around zero, the convergene is strong. The function

f

stands for the Fenchel conjuguate of

f

, namely,

f

(

x

) =

supx2H

(

hx;xi?f

(

x

)) .

Remark 3.1 The formula lim

k !+1

f

(

xk

) = inf

x2H

f

(

x

) is obtained from relation (3.19), denition of the approximate subdierential, lower semicontinuity of the function

f

and the fact that lim

k !+1 v

k +1

= 0 (see the proof of theorem 2.1).

(11)

3.2 A perturbed Inertial Proximal Method

When

f

is nonsmooth, subproblems (3.19) may be very hard to solve and several authors proposed to approximate

f

by a sequence

fk

of more tractable convex functions (see for example [6], [7] and [11]).

In this section we consider

fk;k

= 2

;

3

, a monotone decreasing sequence of proper closed convex function on

H

,

f

be such that

f

=

cl

(inf

kfk

). Now,

x

0

;x

1

be given in

H

, let us consider a sequence

fxkg

generated by the following perturbed method:

0

2xk?xk ?1?k

(

xk ?1?xk ?2

) +

k@"kfk

(

xk

)

;k

= 2

;

3

;

(3.23) where the parameters

k

,

"k

and

k

are nonnegative real numbers.

Theorem 3.1 Suppose that the sequence

fxkg

is bounded and that the following condition holds true

lim

k !+1

"

k

= 0 and lim

k !+1

k

p

k jx

k ?1

?x

k ?2

j

= 0

;

(3.24) then lim

k !+1fk

(

xk

) = inf

f

and every weak cluster point of

fxkg

is a mini- mizer of

f

.

Proof . For all

x2H

, by denition of the approximate subdierential, we have

f

k

(

x

)

fk

(

xk

)

?

1

k hx

k

?x

k ?1

?

k

(

xk ?1?xk ?2

)

;x?xki?"k:

(3.25) Setting

x

=

xk ?1

and taking into account the following inequalities

?hx

k

?x

k ?1

?

k

(

xk ?1?xk ?2

)

;xk ?1?xki 12jxk?xk ?1j2

?

k hx

k ?1

?x

k ?2

;x

k

?x

k ?1 i

=

?22kjxk ?1?xk ?2j2

+

12jxk?xk ?1?k

(

xk ?1?xk ?2

)

j2;

we infer that

f

k

(

xk

) + 1 2

kjvkj2fk

(

xk ?1

) + 2

2kkjxk ?1?xk ?2j2

+

k;

where

vk

:=

12jxk?xk ?1?k

(

xk ?1?xk ?2

)

j

.

This combined with the fact that

fk

decreases yields

f

k

(

xk

) + 1 2

kjvkj2fk ?1

(

xk ?1

) + 2

2kkjxk ?1?xk ?2j2

+

k:

(3.26)

Now let

I

=

fk2IN;fk

(

xk

)

>fk ?1

(

xk ?1

)

g

.

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