Volume 2009, Article ID 586487,12pages doi:10.1155/2009/586487
Research Article
The Methods of Hilbert Spaces and Structure of
the Fixed-Point Set of Lipschitzian Mapping
Jarosław G ´ornicki
Department of Mathematics, Rzesz´ow University of Technology, P.O. Box 85, 35-595 Rzesz´ow, Poland
Correspondence should be addressed to Jarosław G ´ornicki,[email protected]
Received 16 May 2009; Accepted 25 August 2009
Recommended by William A. Kirk
The purpose of this paper is to prove, by asymptotic center techniques and the methods of Hilbert spaces, the following theorem. LetHbe a Hilbert space, letC be a nonempty bounded closed convex subset ofH, and letM an,kn,k≥1be a strongly ergodic matrix. IfT : C → C is a lipschitzian mapping such that lim infn→ ∞infm0,1,...∞k1an,k· Tkm
2
< 2, then the set of fixed points Fix T {x ∈ C : Tx x} is a retract of C. This result extends and improves the corresponding results of7, Corollary 9and8, Corollary 1.
Copyrightq2009 Jarosław G ´ornicki. 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.
1. Introduction
LetEbe a Banach space and letCbe a nonempty bounded closed convex subset ofE. We say that a mappingT:C → Cis nonexpansive if
Tx−Tyx−y for everyx, y∈C. 1.1
The result of Bruck 1 asserts that if a nonexpansive mapping T : C → C has a fixed point in every nonempty closed convex subset ofCwhich is invariant underT and ifCis convex and weakly compact, then Fix T {x ∈ C : Tx x}, the set of fixed points, is nonexpansive retract ofCi.e., there exists a nonexpansive mappingR:C → Fix Tsuch that
R|FixT I. A few years ago, the Bruck results were extended byT.Dom´ınguez Benavides and Lorenzo Ram´ırez2to the case of asymptotically nonexpansive mappings if the spaceE
was sufficiently regular.
Example 1.1Goebel3,4. LetFbe a nonempty closed subset ofC. Fixz∈F, 0< ε <1 and put
Txxε·distx, F·z−x, x∈C. 1.2
It is not difficult to see that Fix T Fand the Lipschitz constant ofT tends to 1 ifε↓0.
For more information on the structure of fixed point sets see4,5 and references therein.
In 1973, Goebel and Kirk 3 introduced the class of uniformly k-lipschitzian mappings, recall that a mappingT :C → Cisuniformlyk-lipschitzian,k1, if
Tnx−Tnykx−y for everyx, y∈C, n∈N, 1.3
and proved the following theorem.
Theorem 1.2. LetEbe a uniformly convex Banach space with modulus of convexityδEand letCbe
a nonempty bounded closed convex subset ofE. Suppose thatT :C → Cis uniformlyk-lipschitzian and
k
1−δE
1
k
<1. 1.4
ThenT has a fixed point inC.Note that in a Hilbert space,k <1/2√5.
Recently Se¸dłak and Wi´snicki 6 proved that under the assumptions of Theorem 1.2,
Fix T is not only connected but even a retract of C, and next the author proved the following theorem7, Corollary 9.
Theorem 1.3. Let H be a Hilbert space,C a nonempty bounded closed convex subset ofH, and
T : C → Ca uniformlyk-lipschitzian mapping withk < √2. Then T has a fixed point inCand
Fix Tis a retract ofC.
In this paper we shall continue this work. Precisely, by means of techniques of asymptotic centers and the methods of Hilbert spaces, we establish some result on the structure of fixed point sets for mappings with lipschitzian iterates in a Hilbert space. The class of mappings with lipschitzian iterates is importantly greater than the class of uniformly lipschitzian mappings; see8, Example 1.
2. Asymptotic Center
Denote byTthe Lipschitz norm ofT:
TsupTx −Ty
x−y :x, y∈C, x /y
Lifshitz9significantly extended Goebel and Kirk’s result and found an example of a fixed point free uniformly π/2−lipschitzian mapping which leaves invariant a bounded closed convex subset ofl2. The validity of Lifshitz’s Theorem in a Hilbert space for√2 k < π/2
remains open.
A more general approach was proposed by the present author using the methods of Hilbert spaces, asymptotic techniques, and strongly ergodic matrix. We recall that a matrix
M an,kn,k1is calledstrongly ergodicif
ifor alln, k an,k0,
iifor allk limn→ ∞an,k0,
iiifor alln ∞k1an,k1,
ivlimn→ ∞∞k1|an,k1−an,k|0.
Then we have the following theorem.
Theorem 2.1see8. LetCbe a nonempty bounded closed convex subset of a Hilbert space and let
M an,kn,k1be a strongly ergodic matrix. IfT :C → Cis a mapping such that
glim inf n→ ∞ minf0,1,...
∞
k1
an,k·Tkm
2
<2, 2.2
thenT has a fixed point inC.
This result generalizes Lifshitz’s Theoremin case of a Hilbert spaceand shows that the theorem admits certain perturbations in the behavior of the norm of successive iterations in infinite sets; see8, Example 1.
LetE be a Banach space. Recall that the modulus of convexityδE is the functionδE : 0,2 → 0,1defined by
δEε inf 1−1
2xy:x1,y1,x−yε
2.3
anduniform convexitymeansδEε>0 forε >0. A Hilbert spaceHis uniformly convex. This fact is a direct consequence of parallelogram identity.
Now we prove some version of Se¸dłak and Wi´snicki’s result 6, Lemma 2.1. LetC
be a nonempty bounded closed convex subset of a real Hilbert spaceH, letM an,kn,k1 be a strongly ergodix matrix, and letT : C → Cbe a mapping such that Tk 1 for all
k1,2, . . ., and
lim sup n→ ∞
∞
k1
an,k·Tk
2
Letx, y∈Cwe use
ry,Tkxlim sup n→ ∞
∞
k1
an,k·y−Tkx
2 ,
rC,Tkxinf
y∈Cr
y,Tkx
2.5
to denote the asymptotic radius of {Tkx} at y and the asymptotic radius of {Tkx} in C, respectively. It is well known in a Hilbert space8that the asymptotic center of {Tkx} in
C:
AC,Tkxy∈C:ry,TkxrC,Tkx 2.6
is a singleton.
Let A : C → Cdenote a mapping which associates with a given x ∈ Ca unique
z∈AC,{Tkx}, that is,zAx. The following Lemma is a crucial tool to proveTheorem 4.1.
Lemma 2.2. LetHbe a Hilbert space and letCbe a nonempty bounded closed convex subset ofH.
Then the mappingA:C → Cis continuous.
Proof. On the contrary, suppose that there existsx0∈Candε0>0 such that for allη >0 there
existsx1 ∈Csuch thatx1−x0 < ηandz1−z0ε0, where{z0} AC,{Tkx0},{z1} AC,{Tkx1}.
Fixη >0 and takex1 ∈Csuch that
x1−x0< η, z1−z0ε0. 2.7
LetR0rC,{Tkx0},R1rC,{Tkx1}andRlim
n→ ∞∞k1an,k· z1−Tkx02. Notice that
R0< R. 2.8
Chooseε >0. Then
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
z1−Tkx0 <
√
Rε,
z0−Tkx0 <R0ε <√Rε,
z0−z1 ≥ε0
2.9
for all but finitely manyk.
If, for example,z1−Tkx0
√
Rεfor all everyonek, then
z1−Tkx0 2
Multiplying both sides of this inequalityfor fixedkby suitable element of the matrixM
and summing up such obtained inequalities fork1, we have forn1,2, . . . ,
∞
k1
an,k·z1−Tkx0
2
Rε. 2.11
Taking the limit superior asn → ∞on each side, we get
Rlim sup n→ ∞
∞
k1
an,k·z1−Tkx0
2
Rε > R, 2.12
which is contradiction.
It follows by2.9and the properties ofδHthat
Tkx0−z1z0
2
1−δH
ε0
√
Rε
√
Rε,
Tkx
0−
z1z0
2 2
1−δH
ε0
√
Rε
2
Rε.
2.13
Multiplying both sides of this inequality by suitable elements of the matrixMand summing up such obtained inequalities fork1, taking the limit superior asn → ∞on each side, we get
R0<lim sup n→ ∞
∞
k1
an,k·Tkx0−
z1z0
2 2
1−δH
ε0
√
Rε
2
Rε.
2.14
Moreover,
Tkx0−z12Tkx0−Tkx1Tkx1−z12
Tkx
0−Tkx122Tkx0−Tkx1·Tkx1−z1Tkx1−z12
Tk2· x
0−x122Tk· x0−x1 ·Tkx1−z1Tkx1−z12
Tk22Tk· x
0−x1 ·diamCTkx1−z1 2
3Tk2·diamC· x0−x1Tkx1−z12.
Multiplying both sides of this inequality by suitable elements of the matrixMand summing up such obtained inequalities fork1, taking the limit superior asn → ∞on each side, we get
Rlim sup n→ ∞
∞
k1
an,k·Tkx0−z1
2
3·diam C· x0−x1 ·lim sup
n→ ∞
∞
k1
an,k·Tk
2
lim sup n→ ∞
∞
k1
an,k·Tkx1−z1 2
3·B·diam C·ηR1ε.
2.16
Similarly,
R1<lim sup
n→ ∞ ∞
k1
an,k·Tkx1−z0 2
3·diam C· x1−x0 ·lim sup n→ ∞
∞
k1
an,k·Tk2lim sup n→ ∞
∞
k1
an,k·Tkx0−z02
3·B·diamC·ηR0ε.
2.17
From2.16and2.17, we have
R3·B·diamC·ηR1ε <6·B·diamC·η2·εR0. 2.18
IfR0 0, then from2.18it followsR 0. This is contradiction with2.8. IfR0 > 0, then combining2.18with2.14and applying the monotonicity ofδH, we obtain
R0<
1−δH
ε0
6·B·diamC·η3·εR0
2
6·B·diamC·η3·εR0. 2.19
Lettingη, ε↓0 and using the continuity ofδH, we conclude that
1
1−δH
ε0
R0
2
<1. 2.20
3. The Methods of Hilbert Spaces
LetM,T be as above. We define functionals
du lim sup n→ ∞
∞
k1
an,k·u−Tku
2 ,
rx lim sup n→ ∞
∞
k1
an,k·x−Tku2,
3.1
whereu, x ∈ C. LetzinCbe an asymptotic center of{Tku}k1 with respect tor·andC, which minimizes the functionalrxoverxinCfor fixu∈C.
Lemma 3.1. One hasrzdu.
Proof. It is consequence of the above definitions.
Lemma 3.2. One hasz−u2du.
Proof. For anyk∈N, we have
z−u22z−Tku2Tku−u2
−zu−2Tku2
2z−Tku22Tku−u2.
3.2
Multiplying both sides of this inequality by suitable elements of the matrixMand summing up such obtained inequalities fork1, taking the limit superior asn → ∞on each side, we get
z−u2 2 lim sup n→ ∞
∞
k1
an,k·z−Tku
2
2 lim sup n→ ∞
∞
k1
an,kTku−u
2
2rz du4du.
3.3
Lemma 3.3. One hasrTszTs2·rzfor alls∈N.
Proof. Fixs∈N, then we have
∞
k1
an,kTsz−Tku
2
s
k1
an,kTsz−Tku
2
Ts2· ∞
ks1
an,kz−Tk−su
2
s k1
an,kTsz−Tku2Ts2· ∞
k1
an,kz−Tk−su2− s k1
an,kz−Tk−su2
.
Since the matrixMis strongly ergodic,
s
k1
an,kTsz−Tku
2
−→0,
s k1
an,kz−Tk−su2−→0,
3.5
asn → ∞, we get thesis.
Lemma 3.4. One hasrz z−x2rxfor everyx∈C.
Proof. Forx∈Cand 0< t <1, we have
tx 1−tz−Tku2tx−Tku2 1−tz−Tku2−t1−tx−z2. 3.6
Multiplying both sides of this inequality by suitable elements of the matrixMand summing up such obtained inequalities fork1, taking the limit superior asn → ∞on each side, we get
lim sup n→ ∞
∞
k−1
an,ktx 1−tz−Tku
2
t·lim sup n→ ∞
∞
k1
an,kx−Tku
2
1−t·lim sup n→ ∞
∞
k1
an,kz−Tku
2
−t1−tx−z2.
3.7
Sincerzrtx 1−tz, we obtain
rzt·rx 1−t·rz−t1−tx−z2,
rzrx−1−tx−z2.
3.8
Takingt↓0, we get,rz z−x2rx.
4. Main Result
We are now in position to prove our main result.
Theorem 4.1. LetCbe a nonempty bounded closed convex subset of a Hilbert space and letM
an,kn,k1be a strongly ergodic matrix. IfT :C → Cis a mapping such that
glim inf
n→ ∞ minf0,1,...
∞
k1
an,k·Tkm
2
<2, 4.1
Proof. Let{ni}and{mi}be sequences of natural numbers such that
g lim
i→ ∞
∞
k1 ani,k·
Tkmi2<2. 4.2
ByTheorem 2.1, Fix T /∅. For any x ∈ Cwe can inductively define a sequence{zj}in the following manner:z1is the unique point inCthat minimizes the functional
lim sup i→ ∞
∞
k1 ani,k·
y−Tkmix2 4.3
overy∈C,andzj1is the unique point inCthat minimizes the functional
lim sup i→ ∞
∞
k1 ani,k·
y−Tkmiz
j
2
4.4
overy∈C, that is,zjAjx,j1,2, . . . .First we prove the following inequality:
dzg−1 du, 4.5
where
du lim sup i→ ∞
∞
k1
u−Tkmiu2, 4.6
andzis the asymptotic center inCwhich minimizes the functional
rx lim sup i→ ∞
∞
k1
x−Tkmiu2 4.7
overxinC.
In fact, we put inLemma 3.4xTpz. Then byLemma 3.3, we get
rz z−Tpz2rTpzTp2·rz,
z−Tpz2Tp2−1·rz. 4.8
Forpmkiwe have
z−Tkmiz2
Tkmi2−1
and hence
dz lim sup i→ ∞
∞
k1 ani,k·
z−Tkmiz2
lim i→ ∞
∞
k1 ani,k·
Tkmi2−1
·rz
g−1·rz by Lemma 3.1
g−1·du.
4.10
Next byLemma 3.2and inequality4.5, we have
zj1−zjAj1x−Ajx2g−1j
dx2·αj·diamC, 4.11
whereαg−1<1 forx∈C,j1,2, . . . .Thus
sup x∈C
Apx−Ajx α
j
1−α·2·
diam C−→0 ifp, j−→ ∞, 4.12
which implies that the sequence{Ajx}converges uniformly to a function
Rx lim
j→ ∞A
jx, x∈C.
4.13
It follows fromLemma 2.2thatR:C → Cis continuous. Moreover,
Rx−TkmiRx22Rx−Ajx2Ajx−TkmiRx2
−RxTkmiRx−2Ajx2
2Rx−Ajx22Ajx−TkmiRx2
2Rx−Ajx222Ajx−TkmiAjx22TkmiAjx−TkmiRx2
24Tkmi2
·Rx−Ajx24Ajx−TkmiAjx2.
Multiplying both sides of this inequalities by suitable elements of the matrixMand summing up such obtained inequalities fork1, taking the limit superior asi → ∞on each side, we get
dRx lim sup i→ ∞
∞
k1 ani,k·
Rx−TkmiRx2
24 lim i→ ∞
∞
k1 ani,k·
Tkmi2
Rx−Ajx2
4 lim sup i→ ∞
∞
k1
ani,k·A
jx−TkmiAjx2
24gRx−Ajx24dAjx by 4.5
24gRx−Ajx24g−1j·dx → 0 ifj → ∞.
4.15
Thus,dRx 0. This implies thatRx TRx; see8for details. ThusRx TRxfor every
x∈CandRis a retraction ofConto Fix T.
IfM an,kn,k1is the Cesaro matrix, that is, forn1,2, . . . ,
an,k ⎧ ⎨ ⎩ 1
n for k1,2, . . . , n,
0 for kn1,
4.16
then we have the following corollary.
Corollary 4.2. LetCbe a nonempty bounded closed convex subset of a Hilbert space. IfT :C → C
is a mapping such that
glim inf n→ ∞ minf0,1,...
1
n
n
k1
Tkm2<2, 4.17
thenFix T {x∈C:Txx}is a retract ofC.
References
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