# Approximation of Fixed Points and Best Proximity Points of Relatively Nonexpansive Mappings

Authors: Thabet Abdeljawad; Kifayat Ullah; Junaid Ahmad; Manuel De La Sen; Azhar Ulhaq
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this article, we study the Agarwal iterative process for finding fixed points and best proximity points of relatively nonexpansive mappings. Using the Von Neumann sequence, we establish the convergence result in a Hilbert space framework. We present a new example of relatively nonexpansive mapping and prove that its Agarwal iterative process is more efficient than the Mann and Ishikawa iterative processes.


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## Body

## 1. Introduction

LetE be a nonempty subset of a Banach space X. A self-map T of E is said to be nonexpansive mapping if(1)Tu−Tv≤u−v,for allu,v∈E.The class of nonexpansive mappings is important as an application point of view. One of the celebrated result of Kirk [1] states that any self nonexpansive mapping of closed bounded convex subset E of a reflexive Banach space has a fixed point provided that E has normal structure. This result was also independently proved in the same year by Browder [2] and Gohde [3] in uniformly convex Banach space (in short UCBS). After this celebrated result, many generalizations of nonexpansive mappings have been published [4–14]. Among the other things, one of the natural generalization of nonexpansive mappings was given by Eldred et al. [15] as follows. Let H and L be two nonempty subsets of a Banach space X. A self-map T of H∪L is said to be relatively nonexpansive if(2)Tu−Tv≤u−v,for allu∈Handv∈L.Iterative methods played a very important role in variational inequalities and many other areas of applied sciences (e.g., see [16–27] and others). One of the earlier iterative scheme is the Picard iteration process, un+1=Tun, which converges very well for Banach contraction mappings. However, this scheme is not suitable for finding fixed points of nonexpansive mappings and hence for the generalized nonexpansive mappings. Let E be a nonempty subset set of a Banach space X. In [30], Eldred and Praveen studied Mann [29] iterative process for finding fixed points and best proximity points of relatively nonexpansive mappings. In [30], Gopi and Pragadeeswara studied Ishikawa [31] iterative process for finding fixed points and best proximity points of relatively nonexpansive mappings.Motivated by the above work, we study the Agarwal [32] iterative process for finding fixed points and best proximity points of relatively nonexpansive mappings. We present a new example of relatively nonexansive mapping and prove that its Agarwal iterative process is more efficient than the Mann [29] and Ishikawa [31] iterative processes.Now, we present some notations which will be used in the sequel:(3)PHu=v∈H:u−v=du,H,dH,L=infu−v:u∈H,v∈L,H0=u∈H:u−v′=dH,Lfor somev′∈L,L0=v∈L:u′−v=dH,Lfor someu′∈H.Notice thatPHu is singleton, provided that H is closed convex in a reflexive and strictly convex space. Moreover, if H and L are a closed convex in a reflexive space, such that one of the H and L is bounded, then H0≠∅.A handful of definitions and theorems given below correspond to our results.Definition 1.
Suppose thatH and L be two nonempty subsets of a metric space. A point u∈H is said to be a best proximity point of the nonself-map T:H⟶L provided that(4)du,Tu=dH,L.Theorem 1 (see [15]).
SupposeH and L be two nonempty bounded closed convex subsets of a UCBS. Assume that T:H∪L⟶H∪L satisfies(i)
TH⊆LandTL⊆H(ii)
Tu−Tv≤u−vfor allu∈H,v∈L
Then, there existu,v∈H×Lsuch thatu−Tu=v−Tv=dH,L.Theorem 2 (see [15]).
SupposeH and L be two nonempty closed bounded convex subsets of a UCBS. If T:H∪L⟶H∪L satisfies the following:(i)
TH⊆HandTL⊆L(ii)
Tu−Tv≤u−vfor allu∈H,v∈L
Then, there existu0∈Handv0∈Lsuch thatTu0=u0,Tv0=v0,andu0−v0=dH,L.Theorem 3 (see [1–3]).
Assume thatT be a self-map on a closed convex bounded subset of a UCBS. If T is nonexpansive, then T has a fixed point.Proposition 1 (see [33]).
Suppose thatX is a UCBS, α∈0,1 and ε>0, then for each d>0 and u,v∈X be such that u≤d,v≤d,u−v≥ε, then there exists some δ=δε/d>0 such that u+1−αv≤1−2δε/dα,1−αd.Lemma 1 (see [34]).
Let0<a<tn<b<1 for every n≥1. Assume that un and vn are sequence in a UCBS X such that un≤1,vn≤1. Define bn in X by bn=1−tnun+tnvn. If limn⟶∞bn=1, then limn⟶∞un−vn=0.Now, we are going to show that, under some appropriate assumptions, Agarwal’s [32] iteration converges to a fixed point of a given nonexpansive mapping. This result is useful for the upcoming main results.Theorem 4.
SupposeE be a nonempty bounded closed convex subset of a UCBS X, and assume that T be a self-map nonexpansive map of E. Choose u0∈E and set vn=1−βnun+βnTun,un+1=1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2 and ∑αnβn<∞. Then, limn⟶∞un−Tun=0. Moreover, if TE lies in a compact set, then un converges to a fixed point of T.Proof.
By Theorem3, there exists some v∈E such that Tv=v. Now,(5)vn−v=1−βnun+βnTun−v≤1−βnun−v+βnTun−v≤1−βnun−v+βnun−v=un−v,un+1−v=1−αnTun+αnTvn−v≤1−αnTun−v+αnTvn−v≤1−αnun−v+αnvn−v≤1−αnun−v+αnun−v=un−v.
It follows that the sequenceun−v is nonincreasing and bounded below by 0. Hence, we have un−v⟶d≥0.Case 1.
Ifun−v⟶0, then(6)un−Tun≤un−v+v−Tun=un−v+Tv−Tun≤un−v+un−v.
Ifn⟶∞, then un−Tun⟶0. Now,(7)un+1−un=1−αnTun+αnTvn−un=Tun−αnTun+αnTvn−un≤αnTun−Tvn+Tun−un≤αnvn−un+Tun−un=αn1−βnun+βnTun−un+Tun−un≤αnβnTun−un+Tun−un=αnβn+1Tun−un≤αnβn+1Tun−v+v−un=αnβn+1Tun−Tv+v−un≤αnβn+1un−v+v−un.
Ifn⟶∞, then un+1−un⟶0.Case 2.
Ifun−v⟶d>0, we need to show that un−Tun⟶0. Suppose not, then one have a subsequence unk of un and a positive real number ε such that unk−Tunk≥ε>0 for all k.
Since the modulus of convexityδ of X is continuous as well as increasing function, one can select some ξ>0 as small such that 1−cδε/d+ξd+ξ<d, where c>0.
Now, we selectk, such that unk−v≤d+ξ. By Proposition 1, we have(8)v−unk+1=v−1−αnkTunk+αnkTvnk=v−1−αnkTunk+αnkT1−βnkunk+βnkTunk=1−αnkv+αnkv−1−αnkTunk+αnkT1−βnkunk+βnkTunk≤1−αnkv−Tunk+αnkv−T1−βnkunk+βnkTunk=1−αnkTv−Tunk+αnkTv−T1−βnkunk+βnkTunk≤1−αnkv−unk+αnkv−1−βnkunk+βnkTunk=1−αnkd+ξ+αnk1−βnkv−unk+βnkv−Tunk≤1−αnkd+ξ+αnk1−2δεd+ξminβnk,1−βnkd+ξ=1−αnk+αnk−2αnkδεd+ξminβnk,1−βnkd+ξ=1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ.
Since there existh>0, such that 2minαnkβnk,αnk1−βnk≥h,(9)1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ≤1−hδεd+ξd+ξ.
Select very smallξ>0, we have 1−hδε/d+ξd+ξ<d, which is contradiction. This implies that the limn⟶∞un−Tun=0.
Now, we prove thatun+1−un⟶0. We have un+1−un≤αnβn+1Tun−un.
Now, we definebn=un+1−v/un−v, vn=Tvn−v/un−v, and un=Tun−v/un−v. One can note that un≤1. Now,(10)Tvn−v=Tvn−Tv≤vn−v=1−βnun+βnTun−v=1−βnun−1−βnv+βnTun−v≤1−βnun−v+βnTun−Tv≤1−βnun−v+βnun−v=un−v.
Therefore,vn=Tvn−v/un−v≤un−v/un−v=1. From Agarwal’s iteration, we obtain un+1−v=1−αnTun−v+αnTvn−v. Dividing by un−v, we obtain(11)un+1−vun−v=1−αnTun−vun−v+αnTvn−vun−v.
Hence,bn=1−αnun+αnvn. Now, we show that bn⟶1. Now,(12)limn⟶∞bn=limn⟶∞un+1−vun−v=dd=1.
By Lemma1, un−vn⟶0. This implies that un−Tun⟶0. Therefore, un+1−un⟶0.
SinceTE is contained in a compact set, Tun has a subsequence Tunk that converges to point b∈H. Also, unk and unk+1 converge to b. This implies that un converge to b. Then, Tun⟶b. In particular, Tunk⟶b. Since T is continuous, implies that Tunk⟶Tb. Therefore, Tb=b.Theorem 5 (see [28]).
LetH and L be nonempty closed bounded convex subset of a UCBS. Let T:H∪L⟶H∪L satisfy(1)
TH⊆HandTL⊆L(2)
Tu−Tv≤u−vfor allu∈H,v∈LLetu0∈H and define un+1=Pn1−αnun+αnTvn, where αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2.Then,limn⟶∞un−Tun=0. Moreover, if TH lies in a compact set, un converges to a fixed point of T.Assume thatH be a convex closed subset of a Hilbert space X. Then, for u∈X, PHu is the nearest to u and element of H. Furthermore, PH is nonexpansive and distinguished by Kolmogorove’s criterion:(13)u−PHu,PHu−b≥0,for allu∈Xandb∈H.Assume thatH and L are two convex closed subsets of X. Set(14)Pu=PHPLu,for eachu∈X.Then, the sequencesPnu⊂H and PLPnu⊂L. When H and L are closed, the convergence of these sequences were established by Von Neumann in [35]. The sequences Pnu and PLPnu are called Von Neumann sequences (sometimes called alternating projection algorithm for two sets).Definition 2.
(see [36]). Suppose H and L are two nonempty convex closed subsets of a Hilbert space X. Then, H,L is called boundedly regular provided that, for every bounded subset S of X and for every ε>0 one can select a δ>0 such that(15)maxdu,H,du,L−z≤δ⇒du,L≤ε,∀u∈X,where z=PL−H¯0 is the displacement vector from the set H to set L (z is the unique vector such thatz=dH,L).Theorem 6 (see [36]).
SupposeH,L is boundedly regular; then, the Von Neumann sequence converges in norm.Theorem 7 (see [36]).
Assume that one of theH and L is boundedly compact; then, H,L is boundedly regular.Lemma 2 (see [37]).
Suppose thatH be a nonempty convex closed subset and L be a nonempty closed subset of a UCBS. Assume that un and bn be sequences in H and vn be a sequence in L such that(i)
un−vn⟶dH,L(ii)
bn−vn⟶dH,L
Then,un−bn converges to 0.Corollary 1 (see [37]).
SupposeH be a nonempty closed convex subset and L be a nonempty closed subset of a UCBS. Assume that un be a sequence in H and v0∈L such that un−v0⟶dH,L. Then, un converges to PHv0.Proposition 2 (see [15]).
SupposeH and L be two closed and convex subset of a Hilbert space X. Then, PLH⊆L, PHL⊆H, and PLu−PHv≤u−v for each u∈H and v∈L.Lemma 3.
SupposeH and L be two closed and convex subset of a Hilbert space X. Then, for each u∈X, we have(16)Pn+1u−b≤Pnu−b,forb∈H0∪L0.


## 2. Main Results

Theorem 8.
SupposeH and L be nonempty bounded closed convex subsets of a UCBS X and assume that T:H∪L⟶H∪L such that(i)
TH⊆HandTL⊆L(ii)
Tu−Tv≤u−v,for allu∈H,v∈L
Selectu0∈H and set vn=1−βnun+βnTun,un+1=1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2 and ∑αnβn<∞. Suppose dun,H0⟶0, thenlimn⟶∞un−Tun=0. Moreover, if TH lies in a compact set, then un converges to a fixed point of T.Proof.
IfdH,L=0, then H0=L0=H∩L, and by Theorem 4, we can establish the theorem from the fact that T:H∩L⟶H∩L is nonexpansive. Let dH,L>0. By Theorem 2, there exists v∈L0 such that Tv=v. Now,(17)vn−z=1−βnun+βnTun−v≤1−βnun−v+βnTun−Tv≤1−βnun−v+βnun−v=un−v,which implies that(18)un+1−v=1−αnTun+αnTvn−v≤1−αnTun−v+αnTvn−v≤1−αnun−v+αnvn−v≤1−αnun−v+αnun−v=un−v.
Hence, the sequenceun−v is nonincreasing. So, one can select a d>0 such that limn⟶∞un−v=d. Assume that there is a subsequence unk of un and an ε>0 such that unk−Tunk≥ε>0 for every k.
However, the modulus of convexityδ of X is continuous and increasing function, and we choose ξ>0 as small that 1−cδε/d+ξd+ξ<d, where c>0.
Now, we choosek, such that unk−v≤d+ξ. By Proposition 1, we have(19)v−unk+1=v−1−αnkTunk+αnkTvnk=v−1−αnkTunk+αnkT1−βnkunk+βnkTunk=1−αnkv+αnkv−1−αnkTunk+αnkT1−βnkunk+βnkTunk≤1−αnkv−Tunk+αnkv−T1−βnkunk+βnkTunk=1−αnkTv−Tunk+αnkTv−T1−βnkunk+βnkTunk≤1−αnkd+ξ+αnkv−1−βnkunk+βnkTunk=1−αnkd+ξ+αnk1−βnkv−unk+βnkv−Tunk≤1−αnkd+ξ+αnk1−2δεd+ξminβnk,1−βnkd+ξ=1−αnk+αnk−2αnkδεd+ξminβnk,1−βnkd+ξ=1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ.
Since there existh>0 such that 2minαnkβnk,αnk1−βnk≥h,(20)1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ≤1−hδεd+ξd+ξ.
Suppose choosing very smallξ>0, we have 1−hδε/d+ξd+ξ<d, which is a contradiction. This implies that limn⟶∞un−Tun=0.
Now, we prove thatun+1−un⟶0. We have un+1−un≤αnβn+1Tun−un. Now, we define bn=un+1−z/un−v,vn=Tvn−v/un−v, and un=Tun−v/un−v. One can note that un≤1. Now,(21)Tvn−v=Tvn−Tv≤vn−v=1−βnun+βnTun−v=1−βnun−1−βnv+βnTun−v≤1−βnun−v+βnTun−Tv≤1−βnun−v+βnun−v=un−v.
Therefore,vn=Tvn−v/un−v≤un−v/un−v=1. From Agarwal’s iteration, we obtain un+1−v=1−αnTun−v+αnTvn−v. Dividing by un−v, we obtain(22)un+1−vun−v=1−αnTun−vun−v+αnTvn−vun−v.
Then,bn=1−αnun+αnvn. Now, we prove that bn⟶1. Now,(23)limn⟶∞bn=limn⟶∞un+1−vun−v=dd=1.
By Lemma1, un−vn⟶0. This shows that un−Tun⟶0. Therefore, un+1−un⟶0.
SinceTH is contained in a compact set, Tun has a subsequence Tunk that converges to point b∈H. Also, unk and unk+1 converge to b.
Sincedun,H0⟶0, there exists bn⊆H0, such that un−bn⟶0. Therefore, bnk⟶b, which gives that b∈H0.
LetD=dH,L and choose p∈L0 such that b−p=D.
We haveunk−p⟶b−p=D, and unk−p≥Tunk−Tp⟶b−Tp. So, b−Tp=D. By strict convexity of the norm, Tp=p. It follows that Tb=b.Corollary 2.
SupposeH and L are two nonempty bounded closed convex subsets of a UCBS X, and assume that T:H∪L⟶H∪L is such that(1)
TH⊆HandTL⊆L(2)
Tu−Tv≤u−vfor allu∈H,v∈L
Chooseu0∈H and set vn=1−βnun+βnTun,un+1=1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2 and ∑αnβn<∞. Then, limn⟶∞un−Tun=0. Moreover, if TH contained in a compact set, then un converges to a fixed point of T.Corollary 3.
SupposeH and L are two nonempty bounded closed convex subsets of a Hilbert space X, and assume that T:H∪L⟶H∪L be a relatively nonexpansive mapping such that(1)
TH⊆HandTL⊆L(2)
Tu−Tv≤u−vfor allu∈H,v∈L
Chooseu0∈H and set vn=1−βnun+βnTun,un+1=Pn1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2 and ∑αnβn<∞. Then, limn⟶∞un−Tun=0. Moreover, if TH is mapped into a compact subset of L, then un converges to a fixed point of T.Proof.
One can note thatPn1−αnTun+αnTvn=1−αnTun+αnTvn, and by Theorem 8, the result follows.
Now, we present a new example of relatively nonexpansive mappings and prove that its Agarwal [32] iterative process is better than the Mann [29] and Ishikawa [31] iterative processes.Example 1.
TakeX=ℝ2,(24)H=u,0:−4≤u≤−3,L=u′,0:3≤u′≤4.
Define(25)T:H⟶HbyTu,0=u−45,0,T:L⟶LbyTu′,0=u′+45,0.
Letu,0∈H and u′,0∈L. Then,(26)Tu,0−Tu′,0=u−45,0−u′+45,0=u−u′−85,0=u−u′−852+0≤u−u′2=u,0−u′,0.
From the above process, we getT is relatively nonexpansive mapping. The iterative values for u0=−1.5 are shown below in Table 1 and Figure 1.Table 1
Sequences generated by Agarwal, Ishikawa, and Mann iterative algorithms.
AgarwalIshikawaMannu0−1.5−1.5−1.5u1−1.028000072000000−1.068000072000000−1.100000400000000u2−1.001568008064010−1.009248019584011−1.020000160000320u3−1.000087808677378−1.001257731995140−1.004000048000192u4−1.000004917298577−1.000171051732452−1.000800012800077u5−1.000000275369428−1.000023263060245−1.000160003200025u6−1.000000015420727−1.000003163779543−1.000032000768007u7−1.000000000863563−1.000000430274473−1.000006400179202u8−1.000000000048359−1.000000058517390−1.000001280040961u9−1.000000000002708−1.000000007958373−1.000000256009216u10−1.000000000000151−1.000000001082346−1.000000051202048u11−1.000000000000008−1.000000000147198−1.000000010240450u12−1−1.000000000020019−1.000000002048098u13−1−1.000000000002722−1.000000000409621u14−1−1.000000000000370−1.000000000081924u15−−1.000000000000050−1.000000000016385u16−1−1.000000000000007−1.000000000003277u17−−1.000000000000001−1.000000000000655u18−1−1−1.000000000000131u19−1−1−1.000000000000026u20−1−1−1.000000000000005Figure 1
Convergence behavior of Agarwal (blue), Ishikawa (magenta), and Mann (cyan) iterates to the fixed pointv=−1 of the mapping T.Remark 1.
From Table1 and Figure 1, we see that Agarwal iterates converges faster to −1 than the Ishikawa and Mann iterates for the class of relatively nonexpansive mappings.
The stronger version for the approximation of fixed point by using Von Neumann sequences are follows.Theorem 9.
LetH and L be nonempty bounded closed convex subsets of a Hilbert space X and suppose T:H∪L⟶H∪L is such that(1)
TH⊆HandTL⊆L(2)
Tu−Tv≤u−vfor allu∈H,v∈L
Letu0∈H and define vn=1−βnun+βnTun,un+1=Pn1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… with restriction ∑n⟶∞αnβn<∞ and ε∈0,1/2 and ∑αnβn<∞. Then, limn⟶∞un−Tun=0. Moreover, if TH is lies in a compact set andun−Tvn⟶0, then un converges to a fixed point of T.Proof.
IfdH,L=0, then H0=L0=H∩L and T:H∩L⟶H∩L is nonexpansive with un+1=Pn1−αnTun+αnT1−βnun+βnTun=1−αnTun+αnT1−βnun+βnTun, the usual Agarwal’s iteration. So, let us take that dH,L>0. By Theorem 2, there exist v∈L0 such that Tv=v. Now,(27)vn−v=1−βnun+βnTun−v≤1−βnun−v+βnTun−v≤1−βnun−v+βnun−v=un−v,which implies that(28)un+1−v=Pn1−αnTun+αnTvn−v≤1−αnTun+αnTvn−v≤1−αnTun−v+αnTvn−v≤1−αnTun−Tv+αnTvn−Tv≤1−αnun−v+αnvn−v≤1−αnun−v+αnun−v=un−v.
Hence, thee sequence un−v is nonincreasing. So, one can choose some d>0 with limn⟶∞un−v=d. Assume that one can find a subsequence, namely, unk of un and some positive ε such that unk−Tunk≥ε>0 for every k. Since the modulus of convexity of δ of X is continuous as well as increasing function, one may choose ξ>0 as small such that 1−cδε/d+ξd+ξ<d, where c>0.
Now, we selectk, such that unk−v≤d+ξ. By Proposition 1, we have(29)v−unk+1=v−Pnk1−αnkTunk+αnkTvnk≤v−1−αnkTunk+αnkTvnk=v−1−αnkTunk+αnkT1−βnkunk+βnkTunk=1−αnkv+αnkv−1−αnkTunk+αnkT1−βnkunk+βnkTunk≤1−αnkv−Tunk+αnkv−T1−βnkunk+βnkTunk=1−αnkTv−Tunk+αnkTv−T1−βnkunk+βnkTunk≤1−αnkv−unk+αnkv−1−βnkunk+βnkTunk=1−αnkd+ξ+αnk1−βnkv−unk+βnkv−Tunk≤1−αnkd+ξ+αnk1−2δεd+ξminβnk,1−βnkd+ξ=1−αnk+αnk−2αnkδεd+ξminβnk,1−βnkd+ξ=1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ.
Since there existh>0 such that 2minαnkβnk,αnk1−βnk≥h:(30)1−2δεd+ξminαnkβnk,αnk1−βnkd+ξ≤1−hδεd+ξd+ξ.
If we select a smallξ>0, then 1−hδε/d+ξd+ξ<d which is clearly a contradiction. This implies that limn⟶∞un−Tun=0
Since the setTH is contained in a compact set, so the sequence Tun has a subsequence Tunk such that it converges to a some point z0∈H. Also, unk converge to z0. From the given sequence, one has(31)unk+1−unk=Pnk1−αnkTunk+αnkTvnk−unk≤1−αnkTunk+αnkTvnk−unk=Tunk−αnkTunk+αnkTvnk−unk≤αnkTunk−Tvnk+Tunk−unk≤αnkvnk−unk+Tunk−unk=αnk1−βnunk+βnTunk−unk+Tunk−unk=αnkβnkTunk−unk+Tunk−unk=αnkβnk+1Tunk−unk∥.
SinceTunk−unk⟶0 which implies that unk+1−unk⟶0. Therefore, unk+1⟶z0, which implies that un⟶z0. Also, we have Tvnk⟶z0 as k⟶∞.
Now,Tunk−TPLz0≤unk−PLz0, which gives thatz0−TPLz0≤z0−PLz0. Therefore, TPLz0=PLz0.
Also,TPz0−PLz0=TPz0−TPLz0≤Pz0−PLz0. So, TPz0=Pz0
Now,TPLPz0−Pz0=TPLPz0−TPz0≤PLPz0−Pz0. Thus, TPLPz0=PLPz0.
For anyn, TPnz0=Pnz0 and TPLPnz0=PLPnz0. By Theorem 6, for each u∈H, the sequence Pnu converges to some wu∈H0. Now,(32)Twz0−PLwz0≤limn⟶∞Twz0−PLPnz0=limn⟶∞Twz0−TPLPnz0≤limn⟶∞wz0−PLPnz0=wz0−PLwz0.
So,Twz0−PLwz0≤wz0−PLwz0.
Therefore,Twz0=wz0, and similarly TPLwz0=PLwz0.
Now, we definegn:H⟶ℝ by gnu=Pnu−wu.
Sincewu−wv=limn⟶∞Pnu−Pnv∥≤∥u−v, then we conclude that w is continuous. Therefore, gnu is continuous and converges pointwise to zero. Since wu∈H0, by Lemma 3, we obtain gn+1≤gn. Therefore, gn converges uniformly on the compact set:(33)S=1−αnkTunk+αnkTvnk∪z0.
Therefore,(34)limn⟶∞Pnk1−αnkTunk+αnkTvnk−w1−αnkTunk+αnkTvnk=0.
Sincew1−αnkTunk+αnkTvnk⟶wz0, we get unk+1⟶wz0, which gives that wz0=z0. Therefore, Tz0=Twz0=wz0=z0, which completes the proof.
SupposeX is a Hilbert space and assume that T be as in Theorem 1. Consider PHT:H⟶H and PLT:L⟶L.
From Proposition2, PHTu−PLTv≤u−v for u∈H and v∈L, by Theorems 8 and 9, we give the following results on convergence of best proximity points.Corollary 4.
SupposeH and L are two nonempty bounded closed convex subsets of a Hilbert space X. Assume that T be as in Theorem 1. If TH is mapped into a compact subset of L, then for every u0∈H0 the sequence generated by by un+1=1−αnTun+αnPHT1−βnun+βnPHTun converges to u in H0 such that u−Tu=dH,LCorollary 5.
SupposeH and L are two nonempty bounded closed convex subsets of a Hilbert space X. Assume that T be as in Theorem 1. If TH is mapped into a compact subset of L, then for any u0∈H0 the sequence defined by un+1=1−αnTun+αnPHT1−βnun+βnPHTun converges to u in H0 such that u−Tu=dH,L provided dun,H0⟶0.Corollary 6.
SupposeH and L are two nonempty bounded closed convex subsets of a Hilbert space X. Assume that T be as in Theorem 1. If TH is mapped into a compact subset of L, then for every u0∈H0 the sequence generated by un+1=Pn1−αnTun+αnPHT1−βnun+βnPHTun converges to u in H0 such that u−Tu=dH,LProof.
The result follows by Corollary4.Corollary 7.
SupposeH and L are two nonempty bounded closed convex subsets of a Hilbert space X. Assume that T be as in Theorem 1. Choose u0∈H and set vn=1−βnun+βnTun,un+1=Pn1−αnTun+αnTvn, where βn,αn∈ε,1−ε,n=0,1,2,… and ε∈0,1/2 and ∑αnβn<∞. If TH is mapped into a compact subset of L and un−PHTvn⟶0, then un converges to u in H0 such that u−Tu=dH,L.Proof.
The result follows by Theorem9.


## 3. Conclusions

In this article, we have used the Agarwal iterative process for finding fixed points and best proximity points of relatively nonexpansive mappings. Using the Von Neumann sequence, we have established the convergence result in a Hilbert space framework. We have offered a new example of relatively nonexpansive mapping and proved that its Agarwal iterative process is more efficient than the Mann and Ishikawa iterative processes.



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