=Paper= {{Paper |id=Vol-3179/Paper_14 |storemode=property |title=Theoretical Bound of the Complexity of Someextragradient-Type Algorithms for Variational Inequalities in Banach Spaces |pdfUrl=https://ceur-ws.org/Vol-3179/Paper_14.pdf |volume=Vol-3179 |authors=Serhii Denysov,Vladimir Semenov |dblpUrl=https://dblp.org/rec/conf/iti2/DenysovS21 }} ==Theoretical Bound of the Complexity of Someextragradient-Type Algorithms for Variational Inequalities in Banach Spaces== https://ceur-ws.org/Vol-3179/Paper_14.pdf
Theoretical Bound of the Complexity of Some Extragradient-
Type Algorithms for Variational Inequalities in Banach Spaces
Serhii Denysov and Vladimir Semenov
Taras Shevchenko National University of Kyiv, 64/13 Volodymyrska Street, Kyiv, 01161, Ukraine

                Abstract
                The work presents a study of three new extragradient-type algorithms for solving variational
                inequalities in a Banach space. Two algorithms are natural modifications of Tseng method and
                “Extrapolation from the Past” method for problems in Banach spaces, based on the generalized
                Alber projection. The third algorithm, called the operator extrapolation method, is a variant of
                forward-reflected-backward algorithm, where the generalized Alber projection is also used
                instead of the metric projection onto the admissible set. Advantage of the latter algorithm is
                only one calculation of the operator value and the generalized projection onto the feasible set
                on each iteration. For variational inequalities with monotone Lipschitz operators acting in a 2-
                uniformly convex and uniformly smooth Banach space O  1  estimates for the complexity in
                terms of the gap function are proved.

                Keywords 1
                variational inequality, monotone operator, extragradient algorithm, Extrapolation from the
                Past, gap function, complexity, 2-uniformly convex Banach space, uniformly smooth Banach
                space

1. Introduction
   Areas of operations research, data analysis, and mathematical physics produces many problems,
which can be written in the form of variational inequalities [1–5]. Prominent example is the saddle
problem that plays an important role in mathematical economics:
                                       min max L  p, q  ,
                                                               pP    qQ

where L : P  Q  R is a smooth convex-concave function, P  R n , Q  Rm – convex closed sets,
which can be formulated as
                           find x  C : Ax , x  x  0 x  C ,
where x   p, q   R n m , C  P  Q  Rnm , and mapping A : P  Q  Rnm has the next form:
                                           1 L  p, q  
                                     Ax  
                                            L  p, q  
                                                             .
                                           2              
   Let's recall another important example related to ranking and search on the Web. The problem of
finding the PageRank vector can be rewritten as follows
                                            minn Px  x  ,
                                                                        x


                                                                                       
where P is an n  n column-stochastic matrix,  n  x  R n :  i 1 xi  1, xi  0 , and p   max pi .
                                                                                                       n
                                                                                                              i 1,..., n

And the minimization problem can be transformed into a saddle point problem
                                       minn max  y, Px  x  ,
                                                                  x     y 1 1




Information Technology and Implementation (IT&I-2021), December 01–03, 2021, Kyiv, Ukraine
EMAIL: denisov.univ@gmail.com (S. Denysov); volodya.semenov@gmail.com (V. Semenov)
ORCID: 0000-0003-4397-4617 (S. Denysov); 0000-0002-3280-8245 (V. Semenov)
             ©️ 2022 Copyright for this paper by its authors.
             Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0).
             CEUR Workshop Proceedings (CEUR-WS.org)



                                                                                                                             144
where p 1   i 1 pi . The last saddle point problem can be rewritten as a bilinear minmax problem on
                 n


the product of standard simplexes
                                            minn max
                                             x   v
                                                   2n
                                                       Jv, Px  x  ,
where J is an n  2n -matrix of the form  I  I  , where I is the identity matrix. We got the following
variational inequality problem
                                                        
       find x n , v   2n :  P*  I  J v , x  x   J *  I  P  x , v  v   0 x n v   2n .
    Note that often nonsmooth optimization problems can be effectively solved with algorithms for
variational inequalities, if former are reformulated as saddle point problems [6]. With the growing
popularity of generative adversarial networks (GANs) and other adversarial learning models, a steady
interest in algorithms for solving variational inequalities has arisen among specialists in the field of
machine learning [7, 8].
    The most widely known method for solving variational inequalities is so called Korpelevich
extragradient algorithm [9]. Many publications are devoted to the study of this algorithm and its
modifications [6, 10–16].
    An effective modern version of the extragradient method is Nemirovski mirror-prox method [6].
This method can be interpreted as a variant of the extragradient method with projection understood in
the sense of Bregman divergence. One more interesting method of dual extrapolation for solving
variational inequalities was proposed by Nesterov [10]. Adaptive variants of the Nemirovski proximal
mirror method were studied in [11–13].
    In the early 1980s, Popov proposed a modification of the classical Arrow-Hurwitz algorithm for
finding saddle points of convex-concave functions [17]. Recently Popov's algorithm for variational
inequalities has become well known among machine learning specialists under the name “Extrapolation
from the Past” [7, 8].
    A modification of Popov's method for solving variational inequalities with monotone operators was
studied in [18]. And in the article [19], a two-stage proximal algorithm for solving the equilibrium
programming problem is proposed, which is an adaptation of the method [17] to the general Ky Fan
inequalities. The algorithm from [19] uses Bregman divergence instead of Euclidean distance. Further
development of this circle of ideas led to the emergence of the so-called forward-reflected-backward
algorithm [20]:
                            xn1  PC  xn  n Axn  n1  Axn  Axn1   ,
and similar method [21]:
                           xn1  PC  xn  n Axn   n1  Axn  Axn1  .
    Recently, using theory of Banach spaces and constructions of their geometry [26–30], progress has
been achieved in the research of algorithms for problems above in Banach spaces [3, 23–25]. Extensive
material on this topic is contained in the book [3]. The next algorithm for solving variational inequalities
in a 2-uniformly convex and uniformly smooth Banach space was proposed in [22]:
                                    xn1  C J 1  Jxn   Axn  ,

where  С is Alber generalized projection operator [23],   0 , J is normalized dual mapping from
 E to E * . This method weakly converges for inversely strongly monotone (cocoercive) operators
 A : E  E * . Shehu [24] has recently extended Tseng’s result to 2-uniformly convex and uniformly
smooth Banach spaces. He proposed the next weakly converging process:
                                 yn   C J 1  xn  n Axn  ,
                                 
                                  xn 1  J  Jyn  n  Ayn  Axn   ,
                                             1



where n  0 is either chosen based on operator A Lipschitz constant value or calculated with a kind
of linear search procedure.



                                                                                                        145
    It should be noted that the early research on algorithms for solving variational inequalities was
usually concentrated on the study of convergence of algorithms and related questions of an asymptotic
nature [13–15, 17, 19–25].
    More recent studies are focused on estimating the number of iterations of algorithms required to
obtain an approximate solution of a given quality [6, 8, 10–12, 16, 18]. This direction of research was
initiated by the work of Nemirovski [6], where mentioned earlier mirror-prox algorithm was proposed
and O  1  complexity estimate in terms of the gap function was obtained for the class of problems with
monotone Lipschitz continuous operator.
  A fundamental question arose about the construction of an algorithm with O  1  complexity
estimation and single computation of the operator's value and projection onto the feasible set at the
iteration step. Algorithm [20] answers this question in the case of a Hilbert space.
    This work is devoted to the study of three new extragradient type algorithms for solving monotone
variational inequalities in a Banach space. The first two algorithms are natural modifications of Tseng's
method [14] and “Extrapolation from the Past” method [18] for problems in Banach spaces using the
generalized Alber projection. Iteration of each of these algorithms is more economical than iteration of
the extragradient method, because the first one uses single projection on iteration, and the second one
needs only one operator calculation. The third algorithm, called operator extrapolation method, is a
variant of the forward-reflected-backward algorithm, proposed by Malitsky and Tam [20]. Operator
extrapolation method also uses generalized Alber projection onto the feasible set. An attractive feature
of the algorithm is only one computation at the iterative step of the operator value and the generalized
projection. The O  1  complexity estimations are proved in terms of the gap function for variational
inequalities with monotone Lipschitz operators acting in a 2-uniformly convex and uniformly smooth
Banach space.
   The article has the following structure. Section 2 contains the necessary information on the geometry
of Banach spaces. Section 3 is devoted to variational inequalities. The algorithms are described in
Section 4. The formulations and proofs of complexity estimations are presented in Section 5.

2. Preliminaries
    Let us remind some basic terms and results from Banach spaces geometry, which are needed for us
to formulate and prove our results [23, 25–30].
    Everywhere later E is a real Banach space with norm  , E  – dual space for E , x , x – value
of functional x  E  on element x  E . Let’s also denote   norm in E  .
   Let SE  x  E : x  1 . Banach space E is called strictly convex, if for all x, y  S E and x  y
          x y
we have         1 . Convexity module of E is defined as
           2
                                       x y                          
                  E     inf 1         : x, y  S E , x  y    ,    0, 2 .
                                        2                            
Banach space E is called uniformly convex, if  E     0    0, 2 [26, 27]. Banach space E is
called 2-uniformly convex, if exists c  0 such that  E     c 2 for all    0, 2 [27]. Obviously 2-
uniformly convex space is uniformly convex. It’s known that uniformly convex Banach space is
reflexive [26-28].
    Banach space E is called smooth, if
                                                  x  ty  x
                                          lim                                                            (1)
                                           t 0       t
exists for all x, y  S E [26]. Banach space E is called uniformly smooth if limit (1) exists uniformly
for x, y  S E [26]. There is a duality between convexity and smoothness of E and it’s dual space E 


                                                                                                        146
[26,27]: E  – strictly convex  E – smooth; E  – smooth  E – strictly convex; E – uniformly
convex  E  – uniformly smooth; E – uniformly smooth  E  – uniformly convex. The first two
implications can be reversed for reflexive space E .
   Smoothness module of space E is defined as
                                         x  ty  x  ty                             
                      E  t   sup                                  1: x, y  S E  t  0 .
                                                          2                          
Uniform smoothness of Banach space E is equivalent to the relation limt 0  E  t  t 1  0 [27, 28].
Banach space E is called 2-uniformly smooth, if exists c  0 such that  E  t   ct 2 for all t  0 [27,
28]. Banach space E is 2-uniformly convex if and only if E  is 2-uniformly smooth [27–29].
   It is known, that Hilbert spaces and spaces L p ( 1  p  2 ) are 2-uniformly convex and uniformly
smooth ( L p are uniformly smooth for p  1,   and 2-uniformly smooth for p   2,   ) [27, 28].
                                            
   Multivalued mapping J : E  2E , which has the form

                                        
                                 Jx  x  E  : x , x  x  x
                                                                           2           2

                                                                                          ,
is called normalized dual mapping [27, 28].
    For Hilbert space J  I (identity). It is known [23, 27, 28] that: if space E is smooth, then mapping
 J is single-valued; if E is strictly convex, then J is injective and strictly monotone; if space E is
reflexive, then mapping J is surjective; if E is uniformly smooth, then J is uniformly continuous on
bounded subsets of E . The explicit form of the mapping J for spaces p , L p and W pm ( p  1,   ) is
given in [3, 23, 27, 28].
   Let E be a smooth Banach space. Let’s consider functional, introduced by Y. Alber in [23]:

                              d  x, y   x  2 Jy, x  y                         x, y  E.
                                                      2                    2



The next useful 3-point identity follows from the definition above:
                  d  x, y   d  x, z   d  z, y   2 Jz  Jy, x  z                  x, y, z  E.
If space E is strictly convex, then for x, y  E we have
                                                          d  x, y   0  x  y .

   Lemma 1 ([25]). Let E be a 2-uniformly convex smooth Banach space. Then for some   1
                                           1
                              d  x, y     x  y x, y  E .
                                                  2

                                           
                                                                 1
   For Banach spaces p , L p and W pm ( 1  p  2 ) we have        [29, 30]. For a Hilbert space,
                                                                p 1
inequality from Lemma 1 becomes identity with   1 .
   Lemma 2 ([29]). Let E be a 2-uniformly smooth Banach space. Then for some   0

                              x  y  x  2 Jx, y   y                             x, y  E .
                                    2             2                            2



   For Banach spaces      p   , L p and W pm ( 2  p   ) we have   p  1 [29, 30]. For a Hilbert space,
inequality from Lemma 2 becomes identity with   1 .
    Let K be a non-empty closed and convex subset of a reflexive, strictly convex and smooth space
 E . It is known [23] that for each x  E there exists unique point z  K such that
                                                d  z , x   inf d  y , x  .
                                                                yK




                                                                                                           147
   This point z is denoted by  K x , and the corresponding operator  K : E  K is called generalized
projection of E onto K (generalized Alber projection) [23]. Note that for a Hilbert space  K
coincides with the metric projection onto the set K .
   Lemma 3 ([23]). Let K – closed and convex subset of a reflexive, strictly convex and smooth space
E , x  E , z  K . Then
                           z  K x            Jz  Jx, y  z  0 y  K .

   Inequality from Lemma 3 is equivalent to the next one [23]:
                           d  y,  K x   d   K x, x   d  y, x  y  K .

   Remark 1. The main element of the algorithms studied below is the calculation of a new point
                                              x    K J 1  Jx  x 
by known x  E and x  E  . From Lemma 3 and mentioned 3-point identity follows the inequality
fundamental for the analysis of algorithms.

                   d  y, x    d  y, x   d  x  , x  2 x , y  x    y  K .

   Basic information about monotone operators and variational inequalities in Banach spaces can be
found in [1, 3, 23, 28].

3. Variational inequalities
   Let E be a 2-uniformly convex and uniformly smooth Banach space, C is a non-empty subset of
space E , A is an operator, acting from E to E * . Let’s consider the variational inequality:
                                   find x  C :       Ax, y  x  0 y  C .                           (2)

Let’s denote set of solutions of (2) as S .
   We need the next assumptions:
       set C  E is convex and closed;
       operator A : E  E * is a monotone and Lipschitz continuous with constant L  0 on C ;
       set S is non-empty.
   Let’s consider dual variational inequality:
                                     find x  C :       Ay, x  y  0 y  C .                          (3)

     We will denote set of solutions of (3) as S d . Note that set S d is convex and closed. Inequality (3)
is sometimes called weak or dual formulation of (2) (or Minty inequality), and solutions of (3) are called
weak solutions for (2). For monotone operator A we always have S  S d . In our setting we have
 S d  S [1].
     Variational inequality (2) can be formulated as a fixed-point problem [23]:
                                              x  C J 1  Jx   Ax  ,                              (4)

with   0 . Formulation (4) is useful, as it leads to obvious algorithmic idea. Procedure
                                              xn1  C J 1  Jxn   Axn                            (5)

was studied in [22] for inversely strongly monotone operators A : E  E * . However, for Lipschitz
continuous monotone operators, algorithm (5) generally does not converge. Numerous modifications
of the extragradient algorithm [6, 9–16,24] can be used for such conditions.
    In this paper, we focus on three algorithms: the Tseng method [14], the extrapolation from the past
method [18] and more recent forward-reflected-backward algorithm [20]. We consider their natural

                                                                                                       148
modifications for problems in Banach spaces using Alber generalized projection instead of the metric
projection.
   The goal is to estimate the number of iterations of algorithms necessary to obtain an approximate
solution of a given quality. The quality of approximate solution x  C of variational inequality (2) will
be measured using the non-negative gap function [6]
                                       Gap  x   sup Ay, x  y .                                    (6)
                                                  yC

Obviously, for definition (6) to be correct, feasible set C should be bounded.
  The next lemma holds:
  Lemma 4. Let operator A be monotone. If x  C is a solution of (2), then Gap  x   0 . And vice
versa, if for some x  C we have Gap  x   0 , then x is a solution of (2).
   Everywhere below we assume, that set C  E is bounded.

4. Algorithms
   Let us study the next iterative extragradient-type algorithms for finding solutions of variational
inequality problem (2).
   Algorithm 1. Modified Tseng method ([24]).
   Select x1  E , n  0 . Set n  1 .
   1. Calculate
                                     yn  C J 1  Jxn  n Axn  .

   2.   If yn  xn , then STOP. Else calculate
                                     xn1  J 1  Jyn  n  Ayn  Axn   .
   3. Set n : n  1 and go to step 1.
   Algorithm 1 is a modification of forward-backward-forward method by P. Tseng [14] for problems
in Banach spaces, where generalized Alber projection is used instead of the metric one.
   The weak convergence of the Algorithm 1 in 2-uniformly convex and uniformly smooth Banach
space is proved in [24].
   Note that in the case of a Hilbert space and without constraints, the Algorithm 1 coincides with the
Korpelevich extragradient method.
   Algorithm 2. Extrapolation from the Past.
   Select x1  y0  E , n  0 . Set n  1 .
   1. Calculate
                                    yn  C J 1  Jxn  n Ayn 1  .
   2. Calculate
                                    xn1  C J 1  Jxn  n Ayn  ,

       if xn 1  yn  xn , then STOP. Else set n : n  1 and go to step 1.
   Algorithm 2 is a modification of L.D. Popov algorithm [17] for problems in Banach spaces using
generalized Alber projection operator instead of the metric one.
   The convergence of Algorithm 2 in a Hilbert space and in Euclidean space with Bregman divergence
instead of Euclidean distance is proved in [18, 19].
   Algorithm 3. Operator extrapolation.
   Select x0  x1  E , n  0 . Set n  1 .
   1. Calculate
                           xn1  C J 1  Jxn  n Axn  n1  Axn  Axn1   .

   2. If xn 1  xn  xn 1 , then STOP. Else set n : n  1 and go to step 1.
   Algorithm 3 is a modification of modern "forward-reflected-backward algorithm" [20] for
variational inequalities in Banach spaces.
   Remark 2. Some text Algorithm 3 can be represented in form, which is similar to Algorithm 2:


                                                                                                     149
                                      xn   C J 1  Jyn  n 1 Axn 1  ,
                                      
                                       yn 1  J  Jxn  n 1  Axn  Axn 1   .
                                                  1



   This formulation indicates conceptual similarity of Algorithms 1 and 3. More precisely, the operator
extrapolation algorithm is obtained from Algorithm 1 in the same way, as Algorithm 2 can be obtained
from the analogue of the extragradient algorithm
                                           yn   C J 1  Jxn  n Axn  ,
                                           
                                            xn 1   C J  Jxn  n Ayn  .
                                                           1



   Now let us turn to the analysis of the algorithms, namely, to the estimation of the number of iterations
required to obtain an approximate solution of the variational inequality (2) with a given value of the
gap function.

5. Analysis
   We will prove, that Algorithms 1 – 3, mentioned above, require O  LD
                                                                         iterations to obtain feasible

point x  C , for which Gap  x    , where   0 and D  supa ,bC d  a, b    .
   Theorem 1. Let  xn  ,  yn  are sequences, generated by Algorithm 1. Suppose that n  0, 
                                                                                                 1 
                                                                                                   L
                                                                                                     .           
                                                               y the next inequality holds
                                                                N

Then for the sequence of Cesaro means z N                      n 1       n       n

                                                              
                                                                  N
                                                                    n 1       n

                                                            1
                                     Gap  z N                    sup yC d  y, x1  .
                                                      2 n 1 n
                                                            N


   Proof. For arbitrary y  C we have
                       
      d  y, xn 1   d y , J 1  Jyn  n  Ayn  Axn     
                        y  2 Jyn  n  Ayn  Axn  , y  Jyn  n  Ayn  Axn  * 
                             2                                                              2



                            y  2 Jyn , y  2n Ayn  Axn , y  Jyn  n  Ayn  Axn  * 
                                 2                                                                2



                                  d  y, yn   yn        2n Ayn  Axn , y  Jyn  n  Ayn  Axn  * .
                                                      2                                                 2


Using Lemma 2, we get an estimation
                      d  y, xn 1   d  y, yn   2n Ayn  Axn , yn  y  n2 Ayn  Axn * .
                                                                                             2
                                                                                                                 (7)
Let’s use 3-point identity to transform d  y, yn 
                        d  y, yn   d  y, xn   d  xn , yn   2 Jxn  Jyn , y  xn .
and now use this in (7)
          d  y, xn1   d  y, xn   d  xn , yn  
                                     2 Jxn  Jyn , y  xn  2n Ayn  Axn , yn  y  n2 Ayn  Axn * .
                                                                                                             2



From Lipschitz continuity of A , equality d  xn , yn   2 Jyn  Jxn , yn  xn  d  yn , xn  and Lemma 1
we have
                           d  y, xn 1   d  y, xn   1  n2 L2  d  yn , xn  
                                                                2 Jxn  Jyn  n Axn  n Ayn , yn  y .
Identity
                                                 yn  C J 1  Jxn  n Axn 
is equivalent to inequality
                                               Jxn  n Axn  Jyn , yn  y  0 .


                                                                                                                 150
So, we have
                  d  y, xn 1   d  y, xn   1  n2 L2  d  yn , xn   2n Ayn , yn  y .
From monotonicity of A operator it follows
                d  y, xn 1   d  y, xn   1  n2 L2  d  yn , xn   2n Ay, yn  y .                             (8)
Let’s transform (8) as
                  2n Ay, yn  y  d  y, xn   d  y, xn 1   1  n2 L2  d  yn , xn  .                           (9)
Summing (9) over n from 1 to N , we get
                                                   N
                                                 2 n Ay, yn  y  d  y, x1  ,
                                                  n 1
which leads to
                                                                                1
                                            Ay, z N  y                               d  y, x1  ,                       (10)
                                                                  2 n 1 n
                                                                                N



            y . Passing in (10) to the supremum over y  C , we obtain
                  N

where z          n 1       n       n

           
         N          N
                      n 1       n

                                                                        1
                                             Gap  z N                             sup yC d  y, x1  ,
                                                              2 n 1 n
                                                                            N


which was required to prove. ■
  Corollary 1. Let  xn  ,  yn  are the sequences, generated by Algorithm 1 with n  2 
                                                                                           1
                                                                                              L
                                                                                                . Then
for the sequence of means z N  N1  n 1 yn the next inequality holds:
                                                         N



                                                                  L  sup yC d  y, x1 
                                              Gap  z N                                              .
                                                                                       N

   For algorithm 2 we have the next result.
   Theorem 2. Let  xn  ,  yn  are the sequences, generated by Algorithm 2. Let n  0, 2L1  . Then         
                                                           y the next inequality holds:
                                                             N

for the Cesaro means sequence z N                           n 1       n       n

                                                          
                                                               N
                                                                 n 1       n

                                                         sup yC d  y, x1   1L d  x1 , y0 
                                         Gap  zN                                                          .
                                                                                    2 n1 n
                                                                                        N


   Proof. For arbitrary y  C we have
                             d  y, xn1   d  y, xn   d  xn1 , xn  2n Ayn , y  xn1 .
From monotonicity of A follows
           Ayn , y  xn1  Ayn , y  yn  Ayn , yn  xn1  Ay, y  yn  Ayn , yn  xn 1 .
So, we have
    d  y, xn 1   d  y, xn   d  xn1 , xn  2n Ayn , yn  xn 1  2n Ay, y  yn 
                                              d  y, xn   d  xn 1 , xn  2n Ayn1 , yn  xn 1 
                                                                        2n Ayn  Ayn1 , yn  xn 1  2n Ay, y  yn .   (11)
Let’s write 2n Ayn 1 , yn  xn 1 as
        2n Ayn1 , yn  xn1  2 Jxn  n Ayn1  Jyn , xn1  yn  2 Jyn  Jxn , xn 1  yn .
From the inclusion xn 1  C and Lemma 3 follows inequality
                                 Jxn  n Ayn1  Jyn , xn1  yn  0 .
As a result, we have


                                                                                                                           151
               2n Ayn1 , yn  xn1  2 Jyn  Jxn , xn1  yn  d  xn1 , xn   d  xn1 , yn   d  yn , xn  .                                            (12)
Estimating the right side of (11) with (12), we come to the next inequality:
                               d  y, xn 1   d  y, xn   d  xn 1 , yn   d  yn , xn  
                                                                         2n Ayn  Ayn1 , yn  xn 1  2n Ay, y  yn .                                       (13)
Now let’s estimate term 2n Ayn  Ayn1 , yn  xn1 . We get
              2n Ayn  Ayn 1 , yn  xn 1  2n Ayn 1  Ayn * xn 1  yn  2n L yn 1  yn xn 1  yn 
                           1                    1              2
                   2n L      yn 1  yn           xn 1  yn  
                                            2

                          2 2                    2              
                              L
                             n
                               2
                                                                      2 L
                                                                             
                                      2 yn 1  xn  2  2 xn  yn  n xn 1  yn 
                                                    2                2

                                                                         2
                                                                                 2
                                                                                                     
                                                                     
                                                              n L yn 1  xn
                                                                                     2
                                                                                           1  2  x  y   n           n
                                                                                                                              2
                                                                                                                                   2 xn 1  yn
                                                                                                                                                   2
                                                                                                                                                       .       (14)
Here we have used auxiliary inequalities

                                           ab  2 a2  21 2 b2 ,  a  b   2a2  2  2 b2 .
                                                    2                            2
                                                                                                                
Estimating norms in (14) using inequality from Lemma 1, we get
                           2n Ayn  Ayn1 , yn  xn 1  n L d  xn , yn 1  

                                                                                                    
                                                                             n L 1  2 d  yn , xn   n L 2 d  xn1 , yn  .                             (15)
Using (15) in (13), we get
                                                                                    
      y, xn 1   d  y, xn   1  n L 2 d  xn 1 , yn   1  n L 1  2 d  yn , xn                         
                                                                                              n L d  xn , yn1   2n Ay, y  yn .                         (16)
Let’s rewrite (16) as
           2n Ay, yn  y   d  y, xn   n L d  xn , yn1     d  y, xn1   n1L d  xn1 , yn   

                                                                
                                                         1   L n 1  2n         d  x , y   1   L 1  2  d  y , x  . (17)
                                                                                              n 1       n                        n                    n    n

Summing (17) over n from 1 to N , we get
                                              N
                                            2 n Ay, yn  y  d  y, x1   1 L d  x1 , y0  ,
                                             n 1
and so
                                                                                  d  y, x1   1 L d  x1 , y0 
                                                             Ay, z N  y                                                             ,                         (18)
                                                                                             2 n1 n
                                                                                                     N



            y . Passing in (18) to the supremum over y  C , we obtain
                    N

where z            n 1       n       n

           
          N           N
                        n 1       n

                                                              sup yC d  y, x1   1L d  x1 , y0 
                                             Gap  zN                                                              ,
                                                                                 2 n1 n
                                                                                      N


which was required to prove. ■
  Corollary 2. Let  xn  ,  yn  are sequences, generated by Algorithm 2 with n  3 1 L . Then for the
means sequence z N  N1  n 1 yn the next inequality holds:
                                               N


                                                               L 32 sup yC d  y, x1   16 d  x1 , y0 
                                            Gap  z N                                                              .
                                                                                     N

    Let’s study Algorithm 3.



                                                                                                                                                                152
       Theorem 3. Let  xn  be a sequence, generated by Algorithm 3. Let n  0, 2 1 L  . Then for the                     
                                                               x
                                                                     N
                                                                     n 1 n n 1
sequence of Cesaro means z N 1                                                      the next inequality holds
                                                                
                                                                        N
                                                                         n 1     n

                                                                                      1
                                                       Gap  z N 1                        sup d  y, x1  .
                                                                                  2 n 1 n yC
                                                                                     N


       Proof. For sequence  xn  , generated by algorithm 3, the next inequality holds
           2 n Axn  n1  Axn  Axn1  , y  xn1  d  y, xn   d  xn1 , xn   d  y, xn1                                     y  С . (19)
Let’s rewrite (19) the next way:
       d  y, xn   d  y, xn1   2n Axn1 , xn1  y 2n Axn 1  Axn , xn 1  y 
                          2n 1 Axn  Axn 1 , xn  y  2n1 Axn  Axn 1 , xn 1  xn  d  xn 1 , xn  .                                        (20)
Summing (20) over n from 1 to N , we get
     d  y, x1   d  y, xN 1  
                                                        N
                                                 2 n Axn 1 , xn 1  y 2N AxN 1  AxN , xN 1  y 
                                                       n 1
                                                                                            N
                                                                                             2n 1 Axn  Axn 1 , xn 1  xn  d  xn 1 , xn   . (21)
                                                                                            n 1
Using Lipschitz continuity of A and Lemma 1 we get
 N

  2 L Ax  Ax , x
n 1
            n 1        n               n 1    n 1    xn  d  xn 1 , xn   
          N
                                               1                1            2    1
          2n 1 L xn  xn 1 xn 1  xn      xn 1  xn      xn  xn 1       xN 1  xN 
                                                             2                                    2

         n 1                                 2               2                2 
                                                                                              1
                                                                                                xN 1  xN .
                                                                                                            2

                                                                                             2
Using this estimation in (21), we get
                                                              N
             d  y, x1   d  y, xN 1   2 n Axn 1 , xn 1  y 2N AxN 1  AxN , xN 1  y 
                                                              n 1

                                                                            1
                                                                             xN 1  xN 
                                                                                        2

                                                                           2
                       N
                                                                                                                  1
                    2 n Axn 1 , xn 1  y 2N L xN 1  xN xN 1  y                                          xN 1  xN 
                                                                                                                              2

                      n 1                                                                                       2
                                                        N
                                                 2 n Axn 1 , xn 1  y N L xN 1  y .
                                                                                                                 2

                                                       n 1
So, we can come to inequality
                                            N
                                       2 n Axn 1 , xn 1  y N L xN 1  y  d  y, xN 1   d  y, x1                           y  С .
                                                                                                       2
                                                                                                                                                      (22)
                                         n 1
Using monotonicity of A , we get
                                                N                                        N
                                                                                                              N    
                                                  Ax , x
                                                n 1
                                                       n        n 1       n 1    y   n Ay, xn 1  y    n  Ay, zN 1  y ,
                                                                                        n 1                  n 1 
                                                                                                                                                      (23)

                      x
                            N
                           n 1 n n 1
where z N 1                                   . Using estimation (22) in (23), we come to inequality
                       
                              N
                                n 1    n

                         N
                                        1       
            2   n  Ay, z N 1  y    N L  xN 1  y  d  y, x1                                               y  С ,
                                                            2

               n 1                          
from which follows


                                                                                                                                                       153
                                                                         1
                                                         N 
                  Gap  z N 1   sup Ay, z N 1  y   2 n  sup d  y, x1  ,
                                    yC                  n 1  yC
which was required to prove. ■
                                                                                       1
   Corollary 3. Let  xn  be a sequence, generated by Algorithm 3 with n                  . Then for the means
                                                                                      2 L
sequence z N 1  N1  n 1 xn 1 the next inequality holds
                       N


                                                       L
                                    Gap  z N 1        sup d  y, x1  .
                                                       N yC

6. Conclusions
   Three new extragradient-type algorithms for solving monotone variational inequalities in a Banach
space are studied in the paper.
   The first two algorithms are natural modifications of Tseng's method [14] and “Extrapolation from
the Past” method [18] for problems in Banach spaces using generalized Alber projection. Iteration of
each of these algorithms is more economical than iteration of the extragradient method. The first
algorithm has less projections, the second has less operator calculations.
   The third algorithm, called the method of operator extrapolation, is a variant of the Malitsky–Tam
forward-reflected-backward algorithm [20]. Generalized Alber projection is used instead of the metric
projection onto feasible set. An attractive feature of the algorithm is only one computation of the
operator value and of the generalized projection onto the feasible set at the iterative step. For variational
inequalities with monotone Lipschitz operators acting in a 2-uniformly convex and uniformly smooth
Banach space, O  1  complexity bounds are proved in terms of the gap function.
   Let us point out two actual questions, related to the current research. First, fast and stable algorithms
for computing generalized Alber projection for a wide range of sets are needed to efficiently apply
algorithms to nonlinear problems in Banach spaces. Second, all results were obtained for the class of
2-uniformly convex and uniformly smooth Banach spaces, which does not contain spaces L p and W pm
( 2  p   ), which are important for applications. It is highly desirable to get rid of this limitation.

Acknowledgements
    This work was supported by the Ministry of Education and Science of Ukraine (project
“Mathematical modeling and optimization of dynamic systems for defense, medicine and ecology”,
state registration number 0119U100337) and the National Academy of Sciences of Ukraine (project
“New methods of research of correctness and solution search for discrete optimization problems,
variational inequalities and their applications”, state registration number 0119U101608).

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