AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 683295 10.1155/2014/683295 683295 Research Article Implicit Approximation Scheme for the Solution of K-Positive Definite Operator Equation http://orcid.org/0000-0001-7155-5917 Shahzad Naseer 1 Rafiq Arif 2 http://orcid.org/0000-0002-9273-0830 Zegeye Habtu 3 Ćirić Ljubomir B. 1 Department of Mathematics King Abdulaziz University P.O. Box 80203, Jeddah 21859 Saudi Arabia kau.edu.sa 2 Department of Mathematics Lahore Leads University Lahore 54810 Pakistan leads.edu.pk 3 Departement of Mathematics University of Botswana Private Bag Box 00704, Gaborone Botswana ub.bw 2014 2332014 2014 05 12 2013 09 02 2014 23 3 2014 2014 Copyright © 2014 Naseer Shahzad et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We construct an implicit sequence suitable for the approximation of solutions of K-positive definite operator equations in real Banach spaces. Furthermore, implicit error estimate is obtained and the convergence is shown to be faster in comparsion to the explicit error estimate obtained by Osilike and Udomene (2001).

1. Introduction

Let E be a real Banach space and let J denote the normalized duality mapping from E to 2E* defined by (1)J(x)={f*E*:x,f*=x2,f*=x}, where E* denotes the dual space of E and ·,· denotes the generalized duality pairing. It is well known that if E* is strictly convex, then J is single valued. We will denote the single-valued duality mapping by j.

Let E be a Banach space. The modulus of smoothness of E is the function.

ρ E : [ 0 , ) [ 0 , ) defined by (2)ρE(t)=sup{12(x+y+x-y)-1:x1,yt}. The Banach space E is called uniformly smooth if (3)limt0ρE(t)t=0. A Banach space E is said to be strictly convex if for two elements x,yE which are linearly independent we have that x+y<x+y.

Let E1 be a dense subspace of a Banach space E. An operator T with domain D(T)E1 is called continuously E1-invertible if the range of T, R(T), with T in E considered as an operator restricted to E1, is dense in E and T has a bounded inverse on R(T).

Let E be a Banach space and let A be a linear unbounded operator defined on a dense domain, D(A), in E. An operator A will be called K positive definite (Kpd)  if there exist a continuously D(A)-invertible closed linear operator K with D(A)D(K) and a constant c>0 such that j(Kx)J(Kx), (4)Ax,j(Kx)cKx2,xD(A). Without loss of generality, we assume that c(0,1).

In , Chidume and Aneke established the extension of Kpd operators of Martynjuk  and Petryshyn [3, 4] from Hilbert spaces to arbitrary real Banach spaces. They proved the following result.

Theorem 1.

Let E be a real separable Banach space with a strictly convex dual E and let A be a Kpd operator with D(A)=D(K). Suppose (5)Ax,j(Ky)=Kx,j(Ay),x,yD(A). Then, there exists a constant α>0 such that for all xD(A)(6)AxαKx. Furthermore, the operator A is closed, R(A)=E, and the equation Ax=f has a unique solution for any given fE.

As the special case of Theorem 1 in which E=Lp (lp) spaces, 2p<, Chidume and Aneke  introduced an iteration process which converges strongly to the unique solution of the equation Ax=f, where A and K are commuting. Recently, Chidume and Osilike  extended the results of Chidume and Aneke  to the more general real separable q-uniformly smooth Banach spaces, 1<q<, by removing the commutativity assumption on A and K. Later on, Chuanzhi  proved convergence theorems for the iterative approximation of the solution of the Kpd operator equation Ax=f in more general separable uniformly smooth Banach spaces.

In , Osilike and Udomene proved the following result.

Theorem 2.

Let E be a real separable Banach space with a strictly convex dual and let A:D(A)EE be a Kpd operator with D(A)=D(K). Suppose Ax,j(Ky)=Kx,j(Ay) for all x,yD(A). Choose any ϵ1(0,c2/(1+α(1-c)+α2)) and define Tϵ:D(A)EE by (7)Tϵx=x+ϵK-1f-ϵK-1Ax. Then the Picard iteration scheme generated from an arbitrary x0D(A) by (8)xn+1=Tϵxn=Tϵnx0 converges strongly to the solution of the equation Ax=f. Moreover, if x* denotes the solution of the equation Ax=f, then (9)xn+1-x*(1-cϵ(1-c))nβ-1Kx0-Kx*.

The most general iterative formula for approximating solutions of nonlinear equation and fixed point of nonlinear mapping is the Mann iterative method  which produces a sequence {xn} via the recursive approach xn+1=αnxn+(1-αn)Txn, for nonlinear mapping T:C=D(T)C, where the initial guess x0C is chosen arbitrarily. For convergence results of this scheme and related iterative schemes, see, for example, .

In , Xu and Ori introduced the implicit iteration process {xn}, which is the modification of Mann, generated by x0C, xn=αnxn-1+(1-αn)Tnxn, for   Ti,i=1,2,,N, nonexpansive mappings, and Tn=Tn(modN) and {αn}(0,1). They proved the weak convergence of this process to a common fixed point of the finite family of nonexpansive mappings in Hilbert spaces. Since then fixed point problems and solving (or approximating) nonlinear equations based on implicit iterative processes have been considered by many authors (see, e.g., ).

It is our purpose in this paper to introduce implicit scheme which converges strongly to the solution of the Kpd operator equation Ax=f in a separable Banach space. Even though our scheme is implicit, the error estimate obtained indicates that the convergence of the implicit scheme is faster in comparison to the explicit scheme obtained by Osilike and Udomene .

2. Main Results

We need the following results.

Lemma 3 (see [<xref ref-type="bibr" rid="B11">10</xref>]).

If E* is uniformly convex then there exists a continuous nondecreasing function b:[0,)[0,) such that b(0)=0, b(δt)δb(t) for all δ1 and (R)x+y2x2+2y,j(x)+max{x,1}yb(y), for all x,yE.

Lemma 4 (see [<xref ref-type="bibr" rid="B14">22</xref>]).

If there exists a positive integer N such that for all nN, n (the set of all positive integers), (10)ρn+1(1-θn)ρn+bn, then (11)limnρn=0, where θn[0,1), n=1θn= and bn=o(θn).

Remark 5 (see [<xref ref-type="bibr" rid="B3">6</xref>]).

Since K is continuously D(A) invertible, there exists a constant β>0 such that (12)Kxβx,xD(K)=D(A).

In the continuation c(0,1), α and β are the constants appearing in (4), (6), and (12), respectively. Furthermore, ϵ>0 is defined by (13)ϵ=c-ηα(1-η),η(0,c).

With these notations, we now prove our main results.

Theorem 6.

Let E be a real separable Banach space with a strictly convex dual and let A:D(A)EE be a Kpd operator with D(A)=D(K). Suppose Ax,j(Ky)=Kx,j(Ay) for all x,yD(A). Let x* denote a solution of the equation Ax=f. For arbitrary x0E, define the sequence {xn}n=0 in E by (14)xn=xn-1+ϵK-1f-ϵK-1Axn,n0. Then, {xn}n=0 converges strongly to x* with (15)xn-x*ρnβ-1Kx0-Kx*, where ρ=1-((c-η)/(α(1-η)+c-η))η(0,1). Thus, the choice η=c/2 yields ρ=1-(c2/(4α(1-c/2)+2c)). Moreover, x* is unique.

Proof.

The existence of the unique solution to the equation Ax=f comes from Theorem 1. From (4) we have (16)Ax-cKx,j(Kx)0, and from Lemma 1.1 of Kato , we obtain that (17)KxKx+γ(Ax-cKx), for all xD(A) and γ>0. Now, from (14), linearity of K and the fact that Ax*=f we obtain that (18)Kxn=Kxn-1+ϵf-ϵAxn=Kxn-1+ϵAx*-ϵAxn, which implies that (19)Kxn-1=Kxn-ϵAx*+ϵAxn, so that (20)Kxn-1-Kx*=Kxn-Kx*-ϵAx*+ϵAxn. With the help of (14) and Theorem 1, we have the following estimate: (21)Axn-Ax*=A(xn-x*)αK(xn-x*)=αKxn-Kx*=αKxn-1-Kx*-ϵ(Axn-Ax*)αKxn-1-Kx*+αϵAxn-Ax*, which gives (22)Axn-Ax*α1-αϵKxn-1-Kx*. Furthermore, inequality (20) can be rewritten as (23)Kxn-1-Kx*=(1+ϵ)(Kxn-Kx*)+ϵ(Axn-Ax*-c(Kxn-Kx*))-ϵ(1-c)(Kxn-Kx*)=(1+ϵ)[ϵ1+ϵKxn-Kx*+ϵ1+ϵ(Axn-Ax*-c(Kxn-Kx*))]-ϵ(1-c)(Kxn-Kx*)=(1+ϵ)[ϵ1+ϵKxn-Kx*+ϵ1+ϵ(Axn-Ax*-c(Kxn-Kx*))]-ϵ(1-c)(Kxn-1-Kx*)+ϵ2(1-c)(Axn-Ax*). In addition, from (17) and (22), we get that (24)Kxn-1-Kx*(1+ϵ)ϵ1+ϵKxn-Kx*+ϵ1+ϵ(Axn-Ax*-c(Kxn-Kx*))-ϵ(1-c)Kxn-1-Kx*-ϵ2(1-c)Axn-Ax*(1+ϵ)Kxn-Kx*-ϵ(1-c)Kxn-1-Kx*-ϵ2(1-c)α1-αϵKxn-1-Kx*, which implies that (25)Kxn-Kx**1+ϵ(1-c)+ϵ2(1-c)(α/(1-αϵ))1+ϵKxn-1-Kx*=ρKxn-1-Kx*, where (26)ρ=1+ϵ(1-c)+ϵ2(1-c)(α/(1-αϵ))1+ϵ=1-ϵ1+ϵ(c-ϵ(1-c)α1-αϵ)=1-ϵ1+ϵη=1-c-ηα(1-η)+c-ηη=1-c24α(1-c/2)+2c. From (25) and (26), we have that (27)Kxn-Kx*ρKxn-1-Kx*ρnK(x0-x*). Hence by Remark 5, we get that (28)xn-x*β-1Kxn-Kx*ρnβ-1Kx0-Kx*0, as n. Thus, xnx* as n.

In , Chuanzhi provided the following result.

Theorem 7.

Let E be a real uniformly smooth separable Banach space, and let A:D(A)EE be a Kpd operator with D(A)=D(K). Suppose Ax,j(Ky)=Kx,j(Ay) for all x,yD(A). For arbitrary fE and x0D(A), define the sequence {xn}n=0 by (29)xn+1=xn+tnγn,γn=K-1f-K-1Axn,0tn12c,tn=0,limntn=0,b(αtn)2cBα,n0, where b(t) is as in (R), α is the constant appearing in inequality (6), c is the constant appearing in inequality (4), and (30)B=max{Kγ0,1}. Then, {xn}n=0 converges strongly to the unique solution of Ax=f.

However, its implicit version is as follows.

Theorem 8.

Let E be a real uniformly smooth separable Banach space, and let A:D(A)EE be a Kpd operator with D(A)=D(K). Suppose Ax,j(Ky)=Kx,j(Ay) for all x,yD(A). For arbitrary fE and x0D(A), define the sequence {xn}n=0 by (31)xn=xn-1+tnγn,(32)γn=K-1f-K-1Axn,(33)tn=,limntn=0,n0. Then, {xn}n=0 converges strongly to the unique solution of Ax=f.

Proof.

The existence of the unique solution to the equation Ax=f comes from Theorem 1. Using (31) and (32) we obtain (34)Kγn=Kγn-1-tnAγn. Consider (35)Kγn2=Kγn,j(Kγn)=Kγn-1-tnAγn,j(Kγn)=Kγn-1,j(Kγn)-tnAγn,j(Kγn)Kγn-1Kγn-ctnKγn2, which implies that (36)KγnKγn-1-ctnKγn. Hence, {Kγn}n=0 is bounded. Let (37)M1=supn0Kγn. Also from (6) it can be easily seen that {Aγn}n=0 is also bounded. Let (38)M2=supn0Aγn. Denote M=M1+M2; then M<.

By using (34) and Lemma 3, we have (39)Kγn2=Kγn-1-tnAγn2Kγn-12-2tnAγn,j(Kγn-1)+max{Kγn-1,1}tnAγnb(tnAγn)=Kγn-12-2tnAγn-1,j(Kγn-1)+2tnAγn-1-Aγn,j(Kγn-1)+max{Kγn-1,1}tnAγnb(tnAγn)(1-2ctn)Kγn-12+2tnAγn-1-AγnKγn-1+max{Kγn-1,1}αtnKγnb(αtnKγn)(1-2ctn)Kγn-12+2Mtnηn+max{M,1}α2M2tnb(tn), where (40)ηn=Aγn-1-Aγn. By using (6) and (34) we obtain that (41)Aγn-1-Aγn=A(γn-1-γn)αK(γn-1-γn)=αtnAγnMαtn0,asn. Thus, (42)ηn0asn. Denote (43)ρn=xn-p,θn=2ctn,σn=2Mtnηn+max{M,1}α2M2tnb(tn). Condition (33) assures the existence of a rank n0 such that θn=2ctn1, for all nn0. Since b(t) is continuous, so limnb(tn)=0 (by condition (33)). Now with the help of (33), (42), and Lemma 4, we obtain from (39) that (44)limnKγn=0. At last by Remark 5, γn0 as n; that is Axnf as n. Because A has bounded inverse, this implies that xnA-1f, the unique solution of Axn=f. This completes the proof.

Remark 9.

( 1 ) According to the estimates (6–8) of Martynjuk , we have (45)Kxn+1-Kx*1+ϵ1(1-c)+αϵ12(1-c+α)1+ϵ1Kxn-Kx*=θKxn-Kx*, where (46)θ=1+ϵ1(1-c)+αϵ12(1-c+α)1+ϵ1=1-ϵ11+ϵ1(c-α(1-c+α)ϵ1)=1-ϵ11+ϵ1η, for η=c-α(1-c+α)ϵ1 or ϵ1=(c-η)/α(1-c+α), η(0,c). Thus, (47)θ=1-c-ηα(1-c+α)+c-ηη=1-c24α(1-c+α)+2c.(2) For α>c/2, we observe that (48)ρ=1-c24α(1-c/2)+2c=θ-4αc2(4α(1-c/2)+2c)(4α(1-c+α)+2c)(α-c2). Thus, the relation between Martynjuk  and our parameter of convergence, that is, between θ and ρ, respectively, is the following: (49)ρ<θ.

Despite the fact that our scheme is implicit, inequality (49) shows that the results of Osilike and Udomene  are improved in the sense that our scheme converges faster.

Example 10.

Suppose E=, D(A)=+, Ax=x, Kx=2I (x*=0 is the solution of Ax=f); then for the explicit iterative scheme due to Osilike and Udomene  we have (50)Kxn+1=Kxn-ϵ1Axn, which implies that (51)2xn+1=2xn-ϵ1xn, and hence (52)xn+1=(1-ϵ12)xn. Also for the implicit iterative scheme we have that (53)Kxn=Kxn-1-ϵAxn, which implies that (54)xn=11+ϵ/2xn-1. It can be easily seen that for c1/2 and α1/2, (4) and (6) are satisfied. Suppose c=1/4 and α=3/5; then η=0.125,   ϵ=(c-η)/α(1-η)=0.23810, ϵ1=(c-η)/α(1-c+α)=0.15432, ρ=0.97596, and θ=0.983288 and so ρ<θ. Take x0=0.1; then from (52) we have Table 1 and for (54) we get Table 2.

n    1 2 3 4 5 6 7 8
x n 0.0922 0.0851 0.07859 0.07528 0.06947 0.06411 0.05916 0.05459
n    1 2 3 4 5 6 7 8
x n 0.0893 0.0798 0.07136 0.06376 0.05698 0.05092 0.04550 0.04066
Example 11.

Let us take E=, D(A)=+, Ax=(1/4)x, Kx=2x (x*=0 is the solution of Ax=f); then for the explicit iterative scheme due to Osilike and Udomene  we have (55)Kxn+1=Kxn-ϵ1Axn, which implies that (56)2xn+1=2xn-ϵ14xn, and hence (57)xn+1=(1-ϵ18)xn. Also for the implicit iterative scheme we have that (58)Kxn=Kxn-1-ϵAxn, which implies that (59)xn=11+ϵ/8xn-1. It can be easily seen that for c1/8 and α1/8, (4) and (6) are satisfied. Suppose c=0.0625 and α=0.2; then η=0.03125, ϵ=(c-η)/α(1-η)=0.16129, ϵ1=(c-η)/α(1-c+α)=0.13736, ρ=0.99566, and θ=0.99623  and so ρ<θ. Take x0=0.01; then from (57) we have Table 3 and for (59) we get Table 4.

Even though our scheme is implicit we observe that it converges strongly to the solution of the Kpd operator equation Ax=f with the error estimate which is faster in comparison to the explicit error estimate obtained by Osilike and Udomene .

n    1 2 3 4 5 6 7 8
x n 0.0098 0.0096 0.00949 0.00933 0.00917 0.00901 0.00885 0.00870
n 1 2 3 4 5 6 7 8
x n 0.0098 0.0096 0.00941 0.00923 0.00905 0.00887 0.00869 0.00852
Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The first author, therefore, acknowledges his thanks to DSR for the financial support. This paper is dedicated to Professor Miodrag Mateljevi’c on the occasion of his 65th birthday.

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