PII. S0161171203110198 http://ijmms.hindawi.com © Hindawi Publishing Corp.
ON THE EQUIVALENCE OF MANN AND ISHIKAWA
ITERATION METHODS
B. E. RHOADES and STEFAN M. SOLTUZ
Received 19 October 2001
We show that certain Mann and Ishikawa iteration schemes are equivalent for various classes of functions.
2000 Mathematics Subject Classification: 47H10.
The Mann iterative scheme was invented in 1953, see [7], and was used to obtain convergence to a fixed point for many functions for which the Banach principle fails. For example, the first author in [8] showed that, for any contin-uous selfmap of a closed and bounded interval, the Mann iteration converges to a fixed point of the function.
In 1974, Ishikawa [5] devised a new iteration scheme to establish conver-gence for a Lipschitzian pseudocontractive map in a situation where the Mann iteration process failed to converge.
LetXbe a Banach space. The Mann iteration is defined by
x0∈X, xn+1=
1−αnxn+αnT xn, n≥0, (1)
where theαn∈(0,1), for alln≥0.
The Ishikawa iteration scheme is defined by
u0∈X, un+1=
1−αnun+αnT vn,
vn=1−βnun+βnT un, n≥0, (2)
where
0≤αn≤βn≤1, ∀n≥0. (3)
In specific situations, additional conditions are placed on(αn)nand(βn)n. In [9], the first author modified the definition of Ishikawa by replacing (3) by
0≤αn,βn≤1, ∀n≥0. (4)
Of course, having established the convergence of an Ishikawa method using (4), we obtain as a corollary the convergence of the corresponding Mann iteration method by setting eachβn=0.
A reasonable conjecture is that, wheneverT is a function for which Mann iteration converges, so does the Ishikawa iteration. Given the large variety of functions and spaces, such a global statement is, of course, not provable. How-ever, in this paper, we do show that, for several classes of functions, Mann and Ishikawa iteration procedures are equivalent.
Picard iteration is defined byp0∈Xandpn+1=T pn, whereT is a selfmap ofX.
Theorem1. LetXbe a normed space and letBbe a nonempty convex subset ofX,T:B→B, withT satisfying
T x−T y ≤kmaxx−y,x−T x,y−T y,x−T y,y−T x (5)
for allx,y∈B,0≤ k <1. Suppose thatT possesses a fixed pointp∈B. Then Picard iteration and certain Mann and Ishikawa iteration schemes converge strongly top.
Proof. Letp0∈Band definepn+1=T pn,n≥0. From [3], it follows that
(pn)nis Cauchy inB. Hence, it converges to a pointx∗∈B. From (5),
pn+1−p=T pn−T p
≤kmaxpn−p,pn−T pn,
pn−T p,T pn−p
=kmaxpn−p,pn−pn+1, pn−p,pn+1−p.
(6)
Taking the limit asn→ ∞yieldsp−x∗ ≤kp−x∗, which implies that p=x∗, that is,(pn)nconverges strongly top.
In [9], it was noted that [8, Theorem 6] could be extended to maps satisfying (5), that is, Mann iteration of aT satisfying (5) withαn∈(0,1)and bounded away from zero converges strongly to the unique fixed pointpofT.
In [12], it was shown that, forT satisfying (5) withαn∈(0,1), the Ishikawa method, with eachαn>0 andαn= ∞, converges strongly top.
As shown in [10], inequality (5) is one of the most general contractive con-ditions for a single map.
We need the following lemma.
Lemma2(see [11]). Let(an)n be a nonnegative sequence that satisfies the inequality
an+1≤
whereλn∈(0,1)for eachn∈N,∞n=1λn= ∞, andσn=nλn,limn=0. Then liman=0.
We are able now to prove the following result.
Theorem3. LetXbe a normed space,Ka nonempty closed convex subset ofX, andT a Lipschitzian selfmap ofKwith Lipschitz constantL≤1. Suppose thatT has a fixed point p∈B. Letx0=u0∈K, and definexnandun by (1) and (2), withαn,βnsatisfying (4), (i) limαn=limβn=0, and (ii) αn= ∞. Then the following are equivalent:
(a) the Mann iteration converges strongly top, (b) the Ishikawa iteration converges strongly top.
Proof. That (b) implies (a) is obvious settingβn=0 in (2). We prove that (a) implies (b). From (1) and (2),
xn+1−un+1=1−αn
xn−un+αnT xn−T vn
≤1−αnxn−un+αnLT xn−T vn
=1−αnxn−un+αnL1−βnun+βnT un−xn
=1−αnxn−un+αnL1−βnun−xn+βnT un−xn
≤1−αn1−L1−βnxn−un+αnβnM,
(8)
for some positiveMsince(T un−xn)n is bounded. This fact is obvious if we prove that(un)nis bounded. A simple induction lead us to
un+1≤
1−αnun+αnT vn
≤1−αnun+αnL1−βnun+βnT un ≤1−αnun+αnL1−βnun+αnβnLT un
=un≤ ··· ≤u0.
(9)
Withan:= xn−un,λn:=αn(1−L(1−βn))∈(0,1), andσn:=αnβnM, for eachn∈N, the inequality ofLemma 2is satisfied. Therefore,
limxn−un=0. (10)
Since (a) is true, using (10),
un−p≤xn−p+xn−un, (11)
LetXbe an arbitrary Banach space and letJbe the normalized duality map fromXinto 2X∗. A mapT with domainD(T )and rangeR(T )is calledstrongly pseudocontractive (pseudocontractive) if, for eachx,y ∈D(T ), there exists
j(x−y)∈J(x−y)and at >1(t=1)such that
T x−T y,j(x−y)≤1 tx−y
2. (12)
Equivalently, there exists a constantt >1 such that
(I−T )x−(I−T )y,j(x−y)≥t−1 t x−y
2. (13)
If we setk=(t−1)/t, then the above inequality can be written in the form
(I−T−kI)x−(I−T−kI)y,j(x−y)≥0 (14)
and, from a result of Kato [6],
x−y ≤x−y+r(I−T−kI)x−(I−T−kI)y, (15)
for allx,y∈Xandr >0.
Theorem4. LetKbe a closed convex subset of an arbitrary Banach space
Xand letT be a Lipschitzian strongly pseudocontractive selfmap ofK. Letx0=
u0∈K, and xn andun be defined by (1) and (2), withαn,βn, satisfying (4) and conditions (i) and (ii) ofTheorem 3. Letpbe the fixed point ofT. Then the following are equivalent:
(a) the Mann iterative scheme converges top, (b) the Ishikawa iteration scheme converges top.
Proof. The existence of a fixed pointpcomes from [4, Corollary 1] which holds in an arbitrary Banach space. That (b) implies (a) is obvious settingβn=0. Without loss of generality, we may assume that the Lipschitz constant L ofT is greater than or equal to 1. IfL∈(0,1], then the result follows from Theorem 3.
To prove that (a) implies (b), it is necessary to expressun+1−xn+1in terms of (15). Using (1), (2), and the identity which appears as [2, formula (10), page 782], we obtain
un−xn=un+1+αnun−αnT vn−xn+1−αnxn+αnT xn
=1+αnun+1+αn(I−T−kI)un+1−(1−k)αnun
+(2−k)α2
n
un−T vn+αnT un+1−T vn
−(2−k)α2
n
xn−T xn−αnT xn+1−T xn
=1+αnun+1−xn+1
+αn(I−T−kI)un+1−(I−T−kI)xn+1
−(1−k)αnun−xn+(2−k)α2
nun−T vn−xn+T xn +αnT un+1−T vn−T xn+1+T xn.
(16)
Using the triangular inequality and (15), un−xn
≥1+αnun+1−xn+1
+1+αnαn(I−T−kI)un+1−(I−T−kI)xn+1
−(1−k)αnun−xn−(2−k)α2
nun−T vn−xn+T xn −αnT un+1−T vn−T xn+1+T xn
≥1+αnun+1−xn+1−(1−k)αnun−xn
−(2−k)α2
nun−T vn−xn+T xn−αnT un+1−T vn−T xn+1+T xn. (17)
Solving the above inequality forun+1−xn+1gives
un+1−xn+1≤1+(1−k)αn
1+αn un−xn
+(2−k)α2n
1+αn un−T vn−xn+T xn
+1+αnαnT un+1−T vn−T xn+1+T xn
≤
1+(1−k)αn
1+αn un−xn+(2−k)α
2
nun−T vn
+(2−k)α2
nT xn−xn+αnT un+1−T vn
+αnT xn+1−T xn,
(18)
un−T vn≤un−xn+xn−T xn+T xn−T vn. (19)
LetLdenote the Lipschitz constant forT. Then,
T xn−T vn≤Lxn−vn, (20)
vn−xn=1−βn
un+βnT un−xn ≤1−βnun−xn+βnT un−xn
≤1−βnun−xn+βnT un−T xn+T xn−xn ≤1−βnun−xn+βnLun−xn+βnT xn−xn =1−βn+βnLun−xn+βnT xn−xn.
Note that, forL≥1, 1−βn+βnL≤L. Substituting (21) into (20) and then (20) into (19) gives
un−T vn≤un−xn+xn−T xn+L
Lun−xn+βnT xn−xn =1+L2un−xn+1+LβnT xn−xn,
(22) T un+1−T vn≤Lun+1−vn=L1−αnun+αnT vn−vn
≤L1−αnun−vn+αnT vn−vn. (23)
Using (21),
T vn−vn≤T vn−T xn+T xn−xn+xn−vn
≤(1+L)xn−vn+T xn−xn
≤(1+L)Lxn−un+βnT xn−xn+T xn−xn
=(1+L)Lxn−un+(1+L)βn+1T xn−xn,
(24)
un−vn=un−
1−βnun−βnT un=βnun−T un
≤βnun−xn+xn−T xn+T xn−T un
≤βn(1+L)xn−un+xn−T xn.
(25)
Substituting (25) and (24) into (23), we obtain T un+1−T vn≤L
1−αnβn(1+L)un−xn+βnT xn−xn
+αnL(1+L)Lxn−un+(1+L)βn+1T xn−xn
≤L1−αnβn(1+L)+αnL2(1+L)xn−un
+βnL1−αn+αnL(1+L)βn+1xn−T xn.
(26)
Substituting (22) and (26) into (18) and using(1+αn)−1≤1−αn+α2
n, yields
xn+1−un+1≤
1+(1−k)αn1−αn+α2
nxn−un
+(2−k)α2
n
1+L2xn−un+1+LβnT xn−xn
+(2−k)α2
nxn−T xn
+αnL1−αnβn(1+L)+αnL2(1+L)xn−un
+αnβnL1−αn+αnL(1+L)βn+1xn−T xn
+αnLxn+1−xn
≤γnxn−un+δnxn−T xn+αnxn+1−xn,
where
δn=αn(2−k)2+Lβnαn+βnL1−αn+αnL(1+L)βn+1,
γn=1+(1−k)αn1−αn+α2
n
+(2−k)1+L2α2
n +αnL(1+L)βn1−αn+Lαn.
(28)
Note that
1+(1−k)αn1−αn+α2
n
=1−kαn+kα2
n+(1−k)α3n ≤1−kαn+kα2n+(1−k)α2n
=1−kαn+α2
n.
(29)
Therefore,
γn≤1−kαn+αn2αn+(2−k)1+L2αn
+L(1+L)βn1−αn+Lαn =1−kαn+αn2+(2−k)1+L2+L2(1+L)αn
+L(1+L)1−αnβn
≤1−kαn+αnMαn+βn,
(30)
whereM=2+(2−k)(1+L2)+L2(1+L).
Sinceαnandβnsatisfy (i), there exists an integerNsuch thatM(αn+βn)≤ k(1−k)for alln≥N.
Thus,
xn+1−un+1≤
1−k2αnxn−un
+αn(2−k)2+Lβnαn
+βnL1−αn+αnL(1+L)βn+1xn−T xn +Lxn+1−xn.
(31)
Withλn:=k2αn,an= xn−un, andn=the quantity in braces, we have
an+1≤1−λnan+nλn. (32)
Consequently,
un−p≤un−xn+xn−p →0 asn→ ∞, (33)
and the Ishikawa method converges.
Using the argument in [2], it follows that we also have corresponding theo-rem for Lipschitz strictly hemicontractive operators.
Let S :X →X be a Lipschitz operator with L >1. It is well known that the operatorS :X→X is strongly accretive if and only if(I−S) is strongly pseudocontractive operator, and conversely. Consider the equation Sx=f, wheref∈Xis given andS is a strongly accretive operator. A fixed point for
T x=f+(I−S)xwill be the solution ofSx=f, and conversely. If we consider in (1) and (2) the operatorT x=f+(I−S)x, then T will be strongly pseu-docontractive.Theorem 4assures the equivalence between the convergencies of Mann and Ishikawa iteration. We consider equationx+Sx=f, withS an accretive operator, that is,(I−S)is a pseudocontractive operator. A solution forx+Sx=fis a fixed point forT x=f−Sx, which is a strongly pseudocon-tractive operator. Replacing (1) and (2), we obtain the equivalence between the convergencies of Mann and Ishikawa iteration for an accretive operator. The solutions existences in the above two equations hold as in [2].
All our results hold for multivalued operators provided that this admit ap-propriate single-valued selections.
It has been shown in [1] that there exists a Lipschitzian pseudocontractive map defined on a compact subset ofR2for which an Ishikawa method, with
αn≤βn, converges to a fixed point, but for which no Mann iterative method converges. Therefore, it is not possible to extendTheorem 4to pseudocontrac-tive maps.
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B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
E-mail address:[email protected]
Stefan M. Soltuz: “T. Popoviciu” Institute of Numerical Analysis, P.O. Box 68-1, Cluj-Napoca 3400, Romania
