In this paper, we introduce a new iterative midpoint rule for …nd- ing a solution of …xed point problem for a nonexpansive semigroup in real Hilbert spaces

C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat.

Volum e 69, N umb er 1, Pages 613–628 (2020) D O I: 10.31801/cfsuasm as.484452

ISSN 1303–5991 E-ISSN 2618-6470

http://com munications.science.ankara.edu.tr/index.php?series= A 1

A GENERALIZED NONLINEAR ITERATIVE ALGORITHM FOR THE EXPLICIT MIDPOINT RULE OF NONEXPANSIVE

SEMIGROUP

M. CHERAGHI, M. AZHINI, AND H.R. SAHEBI

Abstract. In this paper, we introduce a new iterative midpoint rule for …nd- ing a solution of …xed point problem for a nonexpansive semigroup in real Hilbert spaces. We establish a strong convergence theorem for the sequences generated by our proposed iterative scheme. Furthermore, we provide appli- cation to Fredholm integral equations. A numerical example is presented to illustrate the convergence result. Our results improve and extend the corre- sponding results in the literature.

1. Introduction

Let R denote the set of all real numbers, H be a real Hilbert space with inner product h:; :i and norm k:k and C be a nonempty closed convex subset of H. A mapping T : C ! C is said to be contraction if there exists a constant 2 (0; 1) such that kT (x) T (y)k kx yk, for all x; y 2 C. If = 1, T is called nonexpansive on C.

The …xed point problem (F P P ) for a nonexpansive mapping T is: To …nd x 2 C such that x 2 F ix(T ), where F ix(T ) is the …xed point set of the nonexpansive mapping T .

The explicit midpoint rule is one of the powerful numerical methods for solving ordinary di¤erential equations and di¤erential algebraic equations. For related works, we refer to [2, 3, 4, 5, 9, 10, 11, 16, 19, 20, 21, 22, 23, 25, 27, 28] and the references cited therein. For instance, consider the initial value problem for the di¤erential equation y⁰(t) = f (y(t)) with the initial condition y(0) = y0, where f is a continuous function from R^d to R^d. The explicit midpoint rule in which a

Received by the editors: November 16, 2018, Accepted: December 08, 2019.

2010 Mathematics Subject Classi…cation. Primary: 47H09, 47H10; Secondary: 47J20.

Key words and phrases. Nonexpansive semigroup, equilibrium problem,midpoint method, strongly positive linear bounded operator, …xed point, Hilbert space.

c 2 0 2 0 A n ka ra U n ive rsity C o m m u n ic a tio n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a t ic s a n d S ta t is t ic s

613

sequence fyng is generated by the following the recurrence relation 1

h(y_n+1 y_n) = f (y_n+1 y_n 2 ):

In 2015, Xu et al. [30] extended and generalized the results of Alghamdi et al. [1]

and applied the viscosity method on the midpoint rule for nonexpansive mappings and they gave the generalized viscosity explicit method:

xn+1= nf (xn) + _nxn+ (1 n)T (xn+ xn+1

2 ):

In 2016, Rizvi [24] introduced the following iterative method for the explicit mid- point rule of nonexpansive mappings:

xn+1= n f (xn) + (1 nB)T (xn+ xn+1

2 ):

A family S := fT (s) : 0 s < 1g of mappings from C into itself is called a nonexpansive semigroup on C if it satis…es the following conditions:

(1) T (0)x = x for all x 2 C

(2) T (s + t) = T (s)T (t) for all s; t 0

(3) kT (s)x T (s)yk kx yk for all x; y 2 C and s 0 (4) For all x 2 C; s ! T (s)x is continuous.

Plubtieng and Punpaeng [18] introduced and studied the following iterative method to prove a strong convergence theorem for F P P in a real Hilbert space:

x_n+1= _nf (x_n) + _nx_n+ (1 _{n n})_s¹

n

Rsn

0 T (s)x_nds; 8n 2 N:

where f is a contraction mapping and f ng and f ng are the sequences in (0; 1) and fsng is a positive real divergent sequence.

Kang et al. [12] considerd an iterative algorithm in a Hilbert space as follows:

x_n+1= _n f (x_n) + _nx_n+ ((1 _n)I _nA) 1 sn

Z sn

0

T (s)x_nds:

Under the certain conditions, the sequence fxⁿg strongly converges to a unique solution of the variational inequality h( f A)x ; x x i 0; 8x 2 F ix(T ).

Motivated and inspired by the results mentioned and related literature in [1, 12, 24, 30], we propose an iterative midpoint algorithm based on the viscosity method for …nding a common element of the set of solutions of nonexpansive semigroup in Hilbert spaces. Then we prove strong convergence theorems that extend and improve the corresponding results of Rizvi [24], Xu [30], and others. Finally, we give an example and numerical result to illustrate our main result.

The rest of paper is organized as follows. The next section presents some pre- liminary results. Section 3 is devoted to introduce midpoint algorithm for solving it. The last section presents a numerical example to demonstrate the proposed algorithms.

2. Preliminaries

For each point x 2 H, there exists a unique nearest point of C, denote by PCx, such that kx P_Cxk kx yk for all y 2 C. PC is called the metric projection of H onto C. It is well known that P_C is nonexpansive mapping and is characterized by the following property:

hx P_Cx; y P_Cyi 0 (1)

Further, it is well known that every nonexpansive operator T : H ! H satis…es, for all (x; y) 2 H H, inequality

h(x T (x)) (y T (y)); T (y) T (x)i (1

2)k(T (x) x) (T (y) y)k²; (2) and therefore, we get, for all (x; y) 2 H F ix(T ),

h(x T (x)); (y T (y))i (1

2)k(T (x) x)k²; (3)

see, e.g. [8].

It is also known that H satis…es Opial’s condition [17], i.e., for any sequence fxng with xn* x, the inequality

lim inf

n!1 kxⁿ xk < lim inf_n

!1 kxⁿ yk (4)

holds for every y 2 H with y 6= x.

Lemma 1. [6] The following inequality holds in real space H:

kx + yk² kxk²+ 2hy; x + yi; 8x; y 2 H:

De…nition 2. A mapping M : C ! H is said to be monotone, if hMx M y; x yi 0; 8x; y 2 C:

M is called -inverse-strongly-monotone if there exist a positive real number such that

hMx M y; x yi kMx M yk²; 8x; y 2 C:

De…nition 3. A mapping B : H ! H is said to be strongly positive linear bounded operator, if there exists a constant > 0 such that hBx; xi kxk², 8x 2 H.

Lemma 4. [15] Assume that B is a strong positive linear bounded self adjoint operator on a Hilbert space H with coe¢ cient > 0 and 0 < kBk ¹. Then

kI Bk 1 .

Lemma 5. [26] Let C be a nonempty bounded closed convex subset of a Hilbert space H and let S = fT (s) : 0 s < 1g be a nonexpansive semigroup on C. For each x 2 C and t > 0. Then, for any 0 h < 1,

tlim!1sup

x2Ck1 t

Z t 0

T (s)xds T (h)(1 t

Z t

0 T (s)xds)k = 0:

Lemma 6. [29] Let fang be a sequence of nonnegative real numbers such that a_n+1 (1 _n)a_n + _n; n 0 where _n is a sequence in (0; 1) and _n is a sequence in R such that

(i) ¹_{n=1 n}= 1; (ii) lim sup_n!1 ⁿ

n 0 or ¹_{n=1 n}< 1.

Then limn!1an= 0.

3. Nonlinear midpoint algorithm

In this section, we prove a strong convergence theorem based on the explicit it- erative for …xed point of nonexpansive semigroup. We …rstly present the following uni…ed algorithm.

Let S = fT (s) : s 2 [0; +1)g be a nonexpansive semigroup on C such that F ix(S) 6= ;. Also f : C ! H be a -contraction mapping and B; D be strongly positive bounded linear self adjoint operators on H with coe¢ cients ₁; ₂> 0 such that 0 < < ¹ < + ¹, ₁ kBk 1 and kDk = 2 1.

Algorithm 7. For given x02 C arbitrary, let the sequence fxⁿg be generated by:

x_n+1= _n f (x_n) + _nDx_n+ ((1 _n)I _nD _nB) 1 sn

Z sn

0

T (s)(x_n+ x_n+1 2 )ds:

(5) where f ⁿg, f ng, f ⁿg are the sequence in (0; 1) such that ^{n n} and fsⁿg [s; 1) with s > 0 satisfying conditions:

(C1)

nlim!1 n = lim

n!1 n = lim

n!1 n = 0; ¹_n=1 n = ¹_{n=1 n}= 1;

(C2)

X1 n=1

j ^{n n 1}j < 1 or lim

n!1 n+1

n

= 1;

X1 n=1

j n n 1j < 1 or lim

n!1 n+1

n

= 1;

X1 n=1

j n n 1j < 1 or lim

n!1 n+1

n

= 1;

(C3)

nlim!1sn = 1; sup

n2Njsⁿ⁺¹ snj is bounded.

Lemma 8. For any 0 < < ¹ < + ¹, there exists a unique …xed point for sequence fxng.

Proof. As a matter of fact, for …xed x 2 C, we can de…ne the sequence fPⁿ: H ! Hg as follows:

P_n(x) := _n f (x) + _nDx + ((1 _n)I _nD _nB) 1 s_n

Z sn

0 T (s)x ds; 8x 2 H:

We may assume without loss of generality that _n (1 _{n n}kDk)kBk ¹. Since B and D are linear bounded self adjoint operators, we have

kBk = supfjhBx; xij : x 2 H; kxk = 1g;

kDk = supfjhDx; xij : x 2 H; kxk = 1g and observe that

h((1 n)I _nD _nB)x; xi = (1 n)hx; xi nhDx; xi nhBx; xi

1 n nkDk ⁿkBk 0:

Therefore, (1 n)I _nD nB is positive. Then, by strong positivity of B and D, we get

k(1 n)I _nD _nBk = supfh((1 n)I _nD _nB)x; xi : x 2 H; kxk = 1g

= supf(1 n)hx; xi nhDx; xi

nhBx; xi : x 2 H; kxk = 1g

1 n n 2 n 1

1 _{n 2 n 1}:

(6) For any x; y 2 C

kPnx P_nyk n kf(x) f (y)k + nkDkkx yk +k(1 ⁿ)I _nD nBk1

s_n Z sn

0 kT (s)x T (s)ykds

n kx yk + n 2kx yk + (1 n 2 n 1)kx yk

= (1 ( ₁ ) _n)kx yk:

Therefore, Banach contraction principle guarantees that P_nhas a unique …xed point in H, and so the iteration (5) is well de…ned.

Lemma 9. Let p 2 F ix(S). Then the sequence fxng generated by Algorithm 7 is bounded.

Proof. Let p 2 F ix(S), we obtain kxn+1 pk = k n f (x_n) + _nDx_n

+((1 _n)I _nD _nB) 1 s_n

Z sn

0

T (s)(xn+ xn+1

2 )ds pk

nk f(xⁿ) Bpk + nkDxⁿ Dpk + ⁿkpk +k((1 ⁿ)I _nD nB)kk1

sn

Z sn

0

T (s)(xn+ xn+1

2 ) T (s)pkds

n(k f(xⁿ) f (p)k + k f(p) Bpk) + nkDxⁿ Dpk + ⁿkpk +(1 _{n 2 n 1})kxn+ xn+1

2 pk

n kxⁿ pk + ⁿk f(p) Bpk + n 2kxⁿ pk + ⁿkpk +(1 _{n 2 n 1})

2 (kxⁿ pk + kxⁿ⁺¹ pk):

which implies that 1 + _{n 2}+ _{n 1}

2 kxⁿ⁺¹ pk ( n +1 + _{n 2 n 1}

2 )kxⁿ pk

+ _n(k f(p) Bpk + kpk):

Then

kxⁿ⁺¹ pk (1 2( ₁ ) n

1 + _{n 2}+ _{n 1})kxⁿ pk + 2 n( ₁ ) 1 + _{n 2}+ _{n 1}

k f(p) Bpk + kpk

1

maxfkxn pk;k f(p) Bpk + kpk

1 g (7)

...

maxfkx⁰ pk;k f(p) Bpk + kpk

1 g:

Hence fxⁿg is bounded.

Now, set tn:= _s¹

n

Rsn

0 T (s)(^xn+x₂ⁿ⁺¹)ds. Then ftⁿg and ff(xⁿ)g are bounded.

Lemma 10. The following properties are satisfying for the Algorithm 7 P1. limn!1kxⁿ⁺¹ xnk = 0:

P2. limn!1kxⁿ tnk = 0:

P3. limn!1kT (s)tⁿ tnk = 0:

Lemma 11. In order to prove P1, one can write ktn+1 t_nk = k 1

s_n+1 Z sn+1

0

T (s)(xn+1+ xn+2

2 )ds 1

s_n Z sn

0

T (s)(xn+ xn+1

2 )dsk

= k 1

s_n+1 Z sn+1

0

(T (s)(xn+1+ xn+2

2 ) T (s)(xn+ xn+1

2 ))ds +( 1

s_n+1 1 s_n)

Z sn

0

T (s)(xn+ xn+1

2 )ds

+ 1 sn+1

Z sn+1

sn

T (s)(xn+ xn+1

2 )dsk

= k 1

s_n+1 Z sn+1

0

(T (s)(xn+1+ xn+2

2 ) T (s)(xn+ xn+1

2 ))ds +( 1

s_n+1 1 s_n)

Z sn

0

(T (s)(xn+ xn+1

2 ) T (s)p)ds + 1

sn+1

Z sn+1

sn

(T (s)(xn+ xn+1

2 ) T (s)p)dsk kxn+1+ xn+2

2

xn+ xn+1

2 k + jsⁿ⁺¹ snjsⁿ

s_n+1s_n kxn+ xn+1

2 pk

+jsⁿ⁺¹ snj

s_n+1 kxn+ xn+1

2 pk1

2(kxⁿ⁺¹ xnk + kxⁿ⁺² xn+1k) +jsⁿ⁺¹ snj

s_n+1 (kxⁿ pk + kxⁿ⁺¹ pk): (8)

Next, we show that the sequence fxⁿg is asymptotically regular, i.e., limⁿ!1kxⁿ⁺² xn+1k = 0. By (8) we estimate that

kxⁿ⁺² xn+1k = k( ⁿ⁺¹ f (xn+1) + _n+1Dxn+1

+((1 n+1)I _n+1D n+1B) 1 s_n+1

Z sn+1

0

T (s)(xn+1+ xn+2

2 )ds)

( n f (xn) + _nDxn+ ((1 n)I _nD nB) 1 sn

Z sn

0

T (s)(xn+ xn+1

2 )ds)k

= k((1 ⁿ⁺¹)I _n+1D n+1B)( 1 sn+1

Z sn+1

0

T (s)(x_n+1+ x_n+2

2 )ds

1 sn

Z sn

0

T (s)(x_n+ x_n+1

2 )ds) + (( _n+ _nD + _nB) ( _n+1+ _n+1D + _n+1B)) 1

s_n Z sn

0

T (s)(xn+ xn+1

2 )ds + ( _{n+1 n}) f (x_n) + n+1( f (xn+1) f (xn)) + ( _{n+1 n})Dxn+ _n+1(Dxn+1 Dxn)k (1 _{n+1 2 n+1 1})ktn+1 t_nk + j n+1 njk1

s_n Z sn

0

T (s)(x_n+ x_n+1 2 )dsk +M j ^{n n+1}j + Nj n n+1j + ⁿ⁺¹ kf(xⁿ⁺¹) f (xn)k

(1 _{n+1 2 n+1 1})ktn+1 t_nk + j n+1 njk1 sn

Z sn

0

T (s)(x_n+ x_n+1 2 )dsk +M j n n+1j + Nj n n+1j + n+1 kxn+1 x_nk

1 _{n+1 2 n+1 1}

2 (kxn+1 x_nk + kxn+2 x_n+1k) +(1 _{n+1 2 n+1 1})jsⁿ⁺¹ snj

s_n+1 (kxⁿ pk + kxⁿ⁺¹ pk) + j ^{n+1 n}jktⁿk +M j ^{n n+1}j + Nj n n+1j + ⁿ⁺¹ kxⁿ⁺¹ xnk;

where M := supfk_s¹_nRsn

0 T (s)(^xn+x₂ⁿ⁺¹)dsk + kf(xn)kg and N := supfk_s¹_nRsn

0 T (s)(^xn+x₂ⁿ⁺¹)dsk + kxnkg. Then

(1 + _{n+1 1}+ _{n+1 2})kxⁿ⁺² xn+1k (1 _{n+1 2}+ (2 ₁) n+1)kxⁿ⁺¹ xnk +(1 _{n+1 2 n+1 1})2jsn+1 s_nj

sn+1 (kxn pk +kxn+1 pk) + 2j n+1 njktnk +2M j ^{n n+1}j + 2Nj n n+1j:

Therefore

kxn+2 x_n+1k (1 2( _{n+1 2}+ ( ₁ ) _n+1)

1 + _{n+1 1}+ _{n+1 2} )kxn+1 x_nk +(1 _{n+1 2 n+1 1}

1 + _{n+1 1}+ _{n+1 2})(2jsⁿ⁺¹ snj

sn+1 )(kxⁿ pk + kxⁿ⁺¹ pk)

+ 2

1 + _{n+1 1}+ _{n+1 2}j n+1 njktnk + 2M

1 + _{n+1 1}+ _{n+1 2}j n n+1j

+ 2N

1 + _{n+1 1}+ _{n+1 2}j n n+1j:

Lemma 6 and (C1)-(C2) implies

nlim!1kxn+1 x_nk = 0: (9)

And similarly, we have

nlim!1kxⁿ⁺² xn+1k = 0: (10)

Also by (8), (9),(10) and (C3) we have lim_n!1ktn+1 t_nk = 0.

In order to prove P2, one can write kxⁿ tnk kxⁿ⁺¹ xnk

+k ⁿ f (xn) + _nDxn+ ((1 n)I _nD nB)tn tnk kxn x_n+1k + nk f(xn) Bt_nk + n 2kxn t_nk + nktnk:

Then

(1 _{n 2})kxⁿ tnk kxⁿ xn+1k + ⁿk f(xⁿ) Btnk + ⁿktⁿk:

By (C1) and (9), we obtain

nlim!1kxⁿ tnk = 0: (11)

In order to prove P3, set K := fw 2 C : kw pk kx⁰ pk+ ₁¹ (k f(p) Bpk+

kpk)g. Then K is a nonempty bounded closed convex subset of C which is T (s)- invariant for each s 2 [0; +1) and contains fxng. So, without loss of generality, we may assume that S := fT (s) : s 2 [0; +1)g is a nonexpansive semigroup on K.

kT (s)xⁿ xnk = kT (s)xⁿ T (s) 1 s_n

Z sn

0

T (s)(xn+ xn+1

2 )ds

+T (s)1 sn

Z sn

0

T (s)(xn+ xn+1

2 )ds

1 sn

Z sn

0

T (s)(x_n+ x_n+1

2 )ds + 1 sn

Z sn

0

T (s)(x_n+ x_n+1

2 )ds x_nk kT (s)xn T (s) 1

s_n Z sn

0

T (s)(x_n+ x_n+1 2 )dsk

+kT (s)1 sn

Z sn

0

T (s)(xn+ xn+1

2 )ds 1

sn

Z sn

0

T (s)(xn+ xn+1

2 )dsk +k1

sn

Z sn

0

T (s)(x_n+ x_n+1

2 )ds x_nk kxn

1 s_n

Z sn

0

T (s)(xn+ xn+1

2 )dsk +kT (s)1

s_n Z sn

0

T (s)(xn+ xn+1

2 )ds 1

s_n Z sn

0

T (s)(xn+ xn+1

2 )dsk +k1

sn

Z sn

0

T (s)(xn+ xn+1

2 )ds xnk

= 2k1 sn

Z sn

0

T (s)(x_n+ x_n+1

2 )ds x_nk +kT (s)1

s_n Z sn

0

T (s)(xn+ xn+1

2 )ds 1

s_n Z sn

0

T (s)(xn+ xn+1

2 )dsk Since ^xn+x₂ⁿ⁺¹ 2 C, from (11) and Lemma 5, we obtain limn!1kT (s)xn x_nk = 0.

Therefore

kT (s)tn t_nk kT (s)tn T (s)x_nk + kT (s)xn x_nk + kxn t_nk ktⁿ xnk + kT (s)xⁿ xnk + kxⁿ tnk:

Then we have limn!1kT (s)tⁿ tnk = 0.

4. Convergence algorithm

Theorem 12. The Algorithm (5) converges strongly z 2 F ix(S), which is a unique solution of the variational inequality h( f B)z; y zi 0, for all y 2 F ix(S).

Proof. Let q = P_{F ix(S)}. We get

kq(I B + f )(x) q(I B + f )(y)k k(I B + f )(x) (I B + f )(y)k kI Bkkx yk + kf(x) f (y)k (1 ₁)kx yk + kx yk

= (1 ( ₁ ))kx yk:

Then q(I B + f ) is a contraction mapping from H into itself. Therefore by Banach contraction principle, there exists z 2 H such that z = q(I B + f )z = P_{F ix(S)}(I B + f )z.

We show that h( f B)z; xn zi 0. To show this inequality, we choose a subsequence ftⁿig of ftⁿg such that

lim sup

n!1 h( f B)z; t_n zi = lim

i!1h( f B)z; t_ni zi: (12) Since ftnig is bounded, there exists a subsequence ftn_ijg of ftnig K which converges weakly to some w 2 C. Without loss of generality, we can assume that

t_ni * w. Now, we prove that w 2 F ix(S). Assume that w =2 F ix(S). Since t_ni* w and T (s)w 6= w, from Opial’s conditions (4) and Lemma 10 (P3), we have

lim inf

i!1 ktni wk < lim inf

i!1 ktni T (s)wk lim inf

i!1 (ktni T (s)t_nik + kT (s)tni T (s)wk) lim inf

i!1 ktⁿi wk;

which is a contradiction. Thus, we obtain w 2 F ix(S). Now from (1), we have lim sup

n!1 h( f B)z; x_n zi = lim sup

n!1 h( f B)z; t_n zi lim sup

i!1 h( f B)z; tni zi (13)

= h( f B)z; w zi 0:

Now we prove that xn is strongly convergence to z.

kxⁿ⁺¹ zk² = nh f(xⁿ) Bz; xn+1 zi + nhDxⁿ Dz; xn+1 zi

nhz; xⁿ⁺¹ zi + h((1 ⁿ)I _nD nB)(tn z); xn+1 zi

n( hf(xn) f (z); x_n+1 zi + h f(z) Bz; x_n+1 zi) + _nkDkkxⁿ zkkxⁿ⁺¹ zk ⁿkzkkxⁿ⁺¹ zk

+k(1 n)I _nD _nBkktn zkkxn+1 zk

n kxⁿ zkkxⁿ⁺¹ zk + ⁿh f(z) Bz; xn+1 zi + _{n 2}kxⁿ zkkxⁿ⁺¹ zk ⁿkzkkxⁿ⁺¹ zk +(1 _{n 2 n 1})kx_n+ x_n+1

2 zkkxⁿ⁺¹ zk

n kxⁿ zkkxⁿ⁺¹ zk + ⁿh f(z) Bz; xn+1 zi + _{n 2}kxn zkkxn+1 zk nkzkkxn+1 zk

+1 _{n 2 n 1}

2 (kxⁿ zk + kxⁿ⁺¹ zk)kxⁿ⁺¹ zk

= 1 + _{n 2} n( ₁ 2 )

2 kxn zkkxn+1 zk + nh f(z) Bz; x_n+1 zi

nkzkkxⁿ⁺¹ zk +1 _{n 2 n 1}

2 kxⁿ⁺¹ zk² 1 + _{n 2 n}( ₁ 2 )

4 (kxn zk²+ kxn+1 zk²) + nh f(z) Bz; xn+1 zi ⁿkzkkxⁿ⁺¹ zk

+1 _{n 2 n 1}

2 kxⁿ⁺¹ zk² 1 + _{n 2 n}( ₁ 2 )

4 kxⁿ zk²

+3 _{n 2} n(3 ₁ 2 )

4 kxⁿ⁺¹ zk²

+ _nh f(z) Bz; x_n+1 zi nkzkkxn+1 zk 1 + _{n 2 n}( ₁ 2 )

4 kxn zk²+3

4kxn+1 zk² + nh f(z) Bz; xn+1 zi ⁿkzkkxⁿ⁺¹ zk:

This implies that

4kxn+1 zk² (1 + _{n 2 n}( ₁ 2 ))kxn zk²+ 3kxn+1 zk² +4 nh f(z) Bz; xn+1 zi + 4 ⁿkzkkxⁿ⁺¹ zk:

Then

kxⁿ⁺¹ zk² (1 ( n( ₁ 2 ) _{n 2}))kxⁿ zk²

+4 nh f(z) Bz; xn+1 zi + 4 ⁿkzkkxⁿ⁺¹ zk

= (1 kn)kxⁿ zk²+ 4 nln; (14)

where kn = n( ₁ 2 ) + _{n 2}and ln= h f(z) Bz; xⁿ⁺¹ zi kzkkxⁿ⁺¹ zk.

Since limn!1 n = limn!1 n = 0 and ¹_n=0 n= ¹_{n=0 n}= 1, it is easy to see that limn!1kn = 0, ¹_n=0kn = 1 and lim supn!1ln 0: Hence, from (13) and (14) and Lemma 6, we deduce that xn ! z, where z = PF ix(S)(I B + f )z.

5. Numerical examples

In this section, we give some examples and numerical results for supporting our main theorem. All the numerical results have been produced in Matlab 2017 on a Linux workstation with a 3.8 GHZ Intel annex processor and 8 Gb of memory Example 13. Consider a Fredholm integral equation of the following form

x(t) = g(t) + Z t

0

F (t; k; x(k)) dk; t 2 [0; 1]; (15) where g is a continuous function on [0; 1] and F : [0; 1] [0; 1] R ! R is continuous.

Note that if F satis…es the Lipschitz continuity condition, i.e.,

jF (t; k; x) F (t; k; y)j jx yj; 8t; k 2 [0; 1]; x; y 2 R;

then equation (15) has at least one solution in L²[0; 1] (see [13]).

De…ne a mapping T (s) : L²[0; 1] ! L²[0; 1] by (T (s)x)(t) = e ^3s(g(t) +

Z t 0

F (t; k; x(k)) dk); t 2 [0; 1]:

It is easy to observe that S = fT (s) : s 2 [0; +1)g is a nonexpansive semigroup.

In fact, we have, for x; y 2 L²[0; 1], kT (s)x T (s)yk² =

Z 1

0 j(T (s)x)(t) (T (s)y)(t)j² dt

= Z 1

0 je ^3s Z 1

0

(F (t; k; x(k)) F (t; k; y(k))) dkj² dt Z 1

0

( Z 1

0 jx(k) y(k)j² dk) dt

= Z 1

0 jx(k) y(k)j² dk

= kx yk²:

This means that to …nd the solution of integral equation (15) is reduced to …nd a

…xed point of the nonexpansive semigroup S in L²[0; 1]. For any given function x₀2 L²[0; 1], de…ne a sequence of functions x_n in L²[0; 1] by

xn+1= n f (xn) + _nDxn+ ((1 n)I _nD nB)1 s_n

Z sn

0

T (s)(xn+ xn+1

2 )ds

satisfying the conditions of Algorithm 7. Then the sequence fxⁿg converges strongly in L²[0; 1] to the solution of integral equation (15) which is also a solution of the following variational inequality

h( f B)z; y zi 0; 8y 2 F ix(S):

Example 14. Let H = R, the set of all real numbers, with the inner product de…ned by hx; yi = xy; 8x; y 2 R, and induced usual norm j : j. Let C = [ 1; 3];

Let f (x) = ¹₉x; B(x) = ¹₄x; D(x) = x and let, for each x 2 C; T (s)x = _1+2s¹ x.

Then there exists a unique sequence fxⁿg R generated by the iterative scheme xn+1 = ( 1

9p n+ 1

2n)xn (16)

+((1 1

(n + 1)²)I 1

2nD 1 pnB) 1

sn

Z sn

0

1

1 + 2s(xn+ xn+1

2 )ds

where _n = p¹

n; _n = _2n¹ ; _n = _(n+1)¹ 2 and s_n = n. Then fxng converges to f0g 2 F ix(S). f is contraction mapping with constant =¹₆ and B; D are strongly positive bounded linear operators with constant ₁ = ¹₅ on C. Therefore, we can choose = 1 which satis…es 0 < < < +¹. Furthermore, it is easy to observe that F ix(S) = f0g 6= ;. After simpli…cation, scheme (16) reduce to

xn+1=

1 9p

n+_2n¹ +_4n¹(1 _(n+1)¹ 2 1 2n

1 4p

n) ln(1 + 2n) 1 _4n¹(1 _(n+1)¹ 2 1

2n 1 4p

n) ln(1 + 2n) xn:

Following the proof of Theorem 12, we obtain that fxⁿg converges strongly to w = f0g 2 F ix(S).

Let H = R, the set of all real numbers, with the inner product de…ned by hx; yi = xy; 8x; y 2 R, and induced usual norm j : j. Let C = [0; 4]; Let f(x) = 10¹(x 3); B(x) = ¹₃x; D(x) = ¹₂x and let, for each x 2 C; T (s)x = e ^2sx. Then there exists a unique sequence fxng R generated by the iterative scheme

x_n+1 = 3

20n + 5(x_n 3) + 1 2p

n + 2x_n (17)

+((1 1

n²)I 1

pn + 2D 3 4n + 1B)1

s_n Z sn

0

e ^2s(x_n+ x_n+1

2 )ds

where _n = _4n+1³ ; _n = p¹

n+2; _n = _n¹2 and s_n = 2n. Then fxng converges to f0g 2 F ix(S). f is contraction mapping with constant =¹₉ and B; D are strongly positive bounded linear operators with constant ₁ = ¹₄ on C. Therefore, we can choose = 2 which satis…es 0 < < < +¹. Furthermore, it is easy to observe that F ix(S) = f0g 6= ;. After simpli…cation, scheme (17) reduce to

xn+1= (_20n+5³ +₂p¹ n+2

1

8n(e ⁴ⁿ 1)(1 _n¹2

1 2p

n+2 1

4n+1))xn 9 20n+5

1 + _8n¹ (e ⁴ⁿ 1)(1 _n¹2 1 2p

n+2 1

4n+1) :

Following the proof of Theorem 12, we obtain that fxⁿg converges strongly to w = f0g 2 F ix(S).

6. Conculsion

In this paper, we present a viscosity nonlinear midpoint algorithm for solving equilibrium problems in real Hilbert spaces. The methods propose a theoretical generalization of some existing results in the literature and primary numerical ex- periments also demonstrate the potential applicability of these methods. We es- tablish the algorithm’s strong convergence under mild and standard assumptions.

This work open the doors for many promising research directions such as obtaining error bound and convergence rate of our algorithms as well as extensions to Banach spaces.

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Current address : M. Cheraghi: Department of Mathematics, Science and Research Branch, Islamic Azad university, Tehran, Iran

E-mail address : mcheraghi98@gmail.com

ORCID Address: http://orcid.org/0000-0003-4878-9250

Current address : M. Azhini: Department of Mathematics, Science and Research Branch, Islamic Azad university, Tehran, Iran

E-mail address : mahdi.azhini@gmail.com

ORCID Address: http://orcid.org/0000-0001-6390-0266

Current address : H.R. Sahebi (Corresponding author): Department of Mathematics, Ashtian Branch, Islamic Azad University, Ashtian, Iran.

E-mail address : sahebi@aiau.ac.ir

ORCID Address: http://orcid.org/0000-0002-1944-5670