Volume 2010, Article ID 716971,14pages doi:10.1155/2010/716971
Research Article
Convergence of the Sequence of
Successive Approximations to a Fixed Point
Tomonari Suzuki
Department of Basic Sciences, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan
Correspondence should be addressed to Tomonari Suzuki,[email protected]
Received 29 September 2009; Accepted 21 December 2009
Academic Editor: Mohamed A. Khamsi
Copyrightq2010 Tomonari Suzuki. 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.
IfX, dis a complete metric space andTis a contraction onX, then the conclusion of the Banach-Caccioppoli contraction principle is that the sequence of successive approximations{Tnx}ofT starting from any pointx∈Xconverges to a unique fixed point. In this paper, using the concept ofτ-distance, we obtain simple, sufficient, and necessary conditions of the above conclusion.
1. Introduction
The following famous theorem is referred to as the Banach-Caccioppoli contraction principle. This theorem is very forceful and simple, and it became a classical tool in nonlinear analysis.
Theorem 1.1see Banach1and Caccioppoli2. LetX, dbe a complete metric space and let
T be a self contraction onX, that is, there existsr ∈ 0,1such thatdTx, Ty ≤ rdx, yfor all
x, y∈X. Then the following holds.
AT has a unique fixed pointz, and{Tnx}converges tozfor anyx∈X.
We note that the conclusion of Kannan’s fixed point theorem3is alsoA. See Kirk’s survey4. Recently, we obtained thatAholds if and only ifTis a strong Leader mapping
5,6.
Theorem 1.2see6. LetT be a mapping on a complete metric spaceX, d. Then the following are equivalent.
i(A) holds.
aForx, y∈Xandε >0, there existδ >0andν∈Nsuch that
dTix, Tjy< εδ⇒dTiνx, Tjνy< ε, 1.1
for alli, j∈N∪ {0}, whereT0is the identity mapping onX.
bForx, y∈X, there existν∈Nand a sequence{αn}in0,∞such that
dTix, Tjy< αn⇒d
Tiνx, Tjνy< 1
n, 1.2
for alli, j∈N∪ {0}andn∈N.
The following theorem is proved in7,8.
Theorem 1.3see Rus7and Subrahmanyam8. LetX, dbe a complete metric space and letT be a continuous mapping onX. Assume that there existsr ∈ 0,1satisfyingdTx, T2x ≤
rdx, Txfor allx∈X. Then the following holds.
B{Tnx}converges to a fixed point for everyx∈X.
We obtained a condition equivalent toBin9.
Theorem 1.4see9. LetT be a mapping on a complete metric spaceX, d. Then the following are equivalent.
i(B) holds.
iiThe following hold.
aForx∈Xandε >0, there existδ >0andν∈Nsuch that
dTix, Tjx< εδ⇒dTiνx, Tjνx< ε, 1.3
for alli, j∈N∪ {0}.
bForx, y∈X, there existν∈Nand a sequence{αn}in0,∞such that
dTix, Tjy< αn⇒d
Tiνx, Tjνy< 1
n, 1.4
for alli, j∈N∪ {0}andn∈N.
We sometimes call a mapping satisfying A a Picard operator 10. We also call a mapping satisfyingBaweakly Picard operator11–13.
2. Preliminaries
Throughout this paper, we denote byN,Z, andRthe sets of positive integers, integers and real numbers, respectively.
In 2001, Suzuki introduced the concept ofτ-distance in order to improve results in Tataru14, Zhong15,16, and others. See also17.
Definition 2.1see18. LetX, dbe a metric space. Then a functionpfromX×Xinto0,∞ is called aτ-distanceon X if there exists a functionη from X ×0,∞ into 0,∞ and the following are satisfied:
τ1px, z≤px, y py, zfor allx, y, z∈X,
τ2ηx,0 0 andηx, t ≥ t for all x ∈ X and t ∈ 0,∞, andη is concave and continuous in its second variable,
τ3limnxn x and limnsup{ηzn, pzn, xm : m ≥ n} 0 imply pw, x ≤
lim infnpw, xnfor allw∈X,
τ4limnsup{pxn, ym:m≥n}0 and limnηxn, tn 0 imply that limnηyn, tn 0,
τ5limnηzn, pzn, xn 0 and limnηzn, pzn, yn 0 imply that limndxn, yn 0.
The metricdis aτ-distance onX. Many useful examples and propositions are stated in9,18–23and references therein. The following fixed point theorems are proved in18.
Theorem 2.2see18. LetXbe a complete metric space and letTbe a mapping onX. Assume that there exist aτ-distancepandr∈0,1such thatpTx, T2x≤rpx, Txfor allx∈X. Assume the following.
iIflimnsup{pxn, xm : m > n} 0,limnpxn, Txn 0, andlimnpxn, y 0, then
Tyy.
Then (B) holds. Moreover, ifTzz, thenpz, z 0.
Theorem 2.3see18. LetXbe a complete metric space and letTbe a mapping onX. Assume that
T is a contraction with respect to someτ-distancep, that is, there exist aτ-distancepandr ∈0,1
such that
pTx, Ty≤rpx, y, 2.1
for allx, y∈X. Then (A) andpz, z 0hold.
The following lemmas are useful in our proofs.
Lemma 2.4see18. LetX, dbe a metric space and letpbe aτ-distance onX. If sequences{xn}
and{yn}inXsatisfylimnpz, xn 0andlimnpz, yn 0for somez∈X, thenlimndxn, yn
0. In particular forx, y, z∈X, pz, x 0andpz, y 0imply thatxy.
Lemma 2.5see18. Let X, dbe a metric space and let pbe aτ-distance onX. If a sequence
{xn}inXsatisfieslimnsup{pxn, xm:m > n}0, then{xn}is a Cauchy sequence. Moreover if a
sequence{yn}inXsatisfieslimnpxn, yn 0, thenlimndxn, yn 0.
Lemma 2.6. LetX, dbe a metric space and letpbe aτ-distance onX. Then for everyz ∈Xand
ε >0, there existsδ >0such thatpz, x≤δandpz, y≤δimply thatdx, y≤ε.
Lemma 2.7. LetXbe a metric space and letpbe aτ-distance onX. Assume that a sequence{xn}in
Xsatisfieslimnsup{pxn, xm:m > n}0,limnpxn, y 0, andlimnpxn, z 0. Thenyz.
The following is proved at Page 442 of18. However we give a proof because we use reductio ad absurdum in18.
Lemma 2.8see18. Letgbe a nondecreasing function from0,∞into itself satisfyinginf{gt: t >0}0. Define a functionffrom0,∞into itself by
ft tsup
n
i1
αimin
gsi,1 :t n
i1
αisi, si≥0, αi>0, n
i1
αi1
. 2.2
Thenf0 0,ft≥tgtfor allt∈0,∞; andfis concave and continuous.
Proof. It is clear thatf0 0, ft ≥ tgt, andf is concave. We shall prove thatf is continuous at 0. Fixε > 0. Then there existsδ > 0 such that gδ ≤ ε. Chooseτ > 0 with ττ/δ≤ε. Fixt∈0, τ. Letα1, α2, . . . , αn>0 ands1, s2, . . . , sn≥0 such thattni1αisiand
n
i1αi1. Sinceδ
{αi:si≥δ} ≤t, we have
t
n
i1
αimin
gsi,1 ≤t
si<δ
αigsi
si≥δ
αi≤t
si<δ
αiε
si≥δ
αi
≤tε t
δ ≤τε τ δ ≤2ε.
2.3
Since α1, α2, . . . , αn > 0 and s1, s2, . . . , sn ≥ 0 are arbitrary, we obtain ft ≤ 2ε. Thus,
limt→0ft 0f0.
The following is obvious.
Lemma 2.9. LetTbe a mapping on a setX. LetA0be a subset ofXsuch thatTA0⊂A0. Define a sequence{An}of subsets ofXby
A1T−1A0\A0, An1T−1An. 2.4
Then the following hold.
iFor everyn ∈ Nandx ∈ X,x ∈An if and only ifTjx /∈A0 forj 0,1, . . . , n−1and
Tnx∈A
0.
iiAm∩An∅form, n∈N∪ {0}withm /n.
3. Condition (B)
In this section, we discuss ConditionB.
Theorem 3.1. LetXbe a complete metric space and letTbe a mapping onX. Assume that there exist aτ-distancep,r ∈0,1, andM∈0,∞such that
pTx, T2x≤rpx, Tx, pTx, Ty≤Mpx, y, 3.1
for allx, y∈X. Then (B) holds. Moreover, ifTzz, thenpz, z 0.
Proof. Assume that limnsup{pxn, xm:m > n}0, limnpxn, Txn 0, and limnpxn, y 0.
Then we have
pxn, Ty
≤pxn, Txn p
Txn, Ty
≤pxn, Txn Mp
xn, y
, 3.2
and hence, limnpxn, Ty 0. ByLemma 2.7, we obtainTy y. ByTheorem 2.2, we obtain
the desired result.
As a direct consequence ofTheorem 3.1, we obtain the following.
Corollary 3.2. LetXbe a complete metric space and letTbe a mapping onX. Assume that there exist aτ-distancepandr∈0,1such that
pTx, T2x≤rpx, Tx, pTx, Ty≤px, y, 3.3
for allx, y∈X. Then (B) holds.
Corollary 3.2characterizes ConditionB.
Theorem 3.3. LetT be a mapping on a metric spaceX, dsuch that (B) holds. Then there exist a
τ-distancepandr∈0,1satisfying3.3.
Proof. Letr ∈ 0,1be fixed. We note that every periodic point is a fixed point. That is, if x∈XsatisfiesTnxxfor somen∈N, thenTxx. Define a mappingT∞fromXontoFT
byT∞xlimnTnxforx∈X, whereFTis the set of all fixed points ofT. Define a mapping
CfromXinto the set of subsets ofXby
CxTx, T2x, T3x, . . . , T∞x. 3.4
SinceT∞xis a fixed point ofT, we have
Next, we define a functionffromXintoZ∪ {∞}satisfying
fTx≥fx 1, fx ∞ ⇐⇒Txx, 3.6
for allx∈X. We putfx ∞forx∈FT. It is obvious thatfTx fx ∞fx 1 forx∈FT. Define a sequence{An}of subsets ofXby
A1T−1FT\FT, An1T−1An. 3.7
Then byLemma 2.9,
FT∩An∅, Am∩An ∅, 3.8
form, n∈Nwithm /n. We putfx −nforx∈An. We note that
fTx
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
∞ if x∈A1,
fx 1 ifx∈
∞
n2
An.
3.9
Put
Y X\
FT
n∈N
An
. 3.10
It is obvious thatTY⊂Y,T−1Y Y, andY∩FT ∅. So,
TmxTnx⇐⇒mn, 3.11
forx∈Y andm, n∈N∪ {0}. Define an equivalence relation∼onY as follows:x∼yif and only if there existm, n ∈ N∪ {0}such thatTmx Tny. By Axiom of Choice, there exists a
mappingBonY such that
Bx∼x, x∼y⇐⇒BxBy. 3.12
Letu∈Y withBuu. Then we putfTnu nforn∈N∪ {0}. Define a sequence{D n}of
subsets ofY by
D0
u, Tu, T2u, T3u, . . ., D1T−1D0\D0, Dn1T−1Dn. 3.13
Then we haveDm∩Dn∅form, n∈N∪ {0}withm /n; and
{x∈Y : x∼u}
n∈N∪{0}
We putfx −nforx∈Ywithn∈Nandx∈Dn. We have definedf. We note thatfx∈N
implies thatx∈Y.
Next, we define aτ-distancepby
px, y
⎧ ⎨ ⎩
rfxrfy ify∈Cx,
rfxrfy1 ify /∈Cx, 3.15
wherer∞0. We note thatpx, y<1 implies either of the following.
iTxxy.
iiThere existu∈ Y,k ∈ N, and ∈N∪ {∞}such thatBu u,k < l,x Tku, and
yTu.In this case,x∈Y,uBx,fx k, andfy hold.
We shall show thatpis aτ-distance. Letx, y, z∈X. Ify∈Cxandz∈Cy, thenz∈Cx. So we have
px, z rfxrfz ≤rfxrfyrfyrfzpx, ypy, z. 3.16
Ify /∈Cxorz /∈Cy, then
px, z≤rfxrfz1≤rfxrfyrfyrfz1≤px, ypy, z. 3.17
These implyτ1. We shall define a functionη fromX ×0,∞into0,∞. Forx ∈ X\Y, we putηx, t t. Forx ∈ Y, we putu Bx. Since{Tnu}converges toT∞u, there exists a
strictly increasing sequence{hun}inNsuch thatj ≥hunimplies thatdTju, T∞u≤1/n
forj∈N∪ {∞}. Since limnhun ∞, we can define a nondecreasing functiongufrom0,∞
into0,1such thatgurhun 1/n. It is obvious thatgu0 limt→0gut 0. Put
ηx, t tsup
n
i1
αigusi:t n
i1
αisi, si ≥0, αi>0, n
i1
αi1
. 3.18
Thenηx, tsatisfiesτ2and ηx, t ≥ tgutbyLemma 2.8. In order to showτ3, we
assume that limnxn xand limnsup{ηzn, pzn, xm : m ≥ n} 0. Then without loss of
generality, we may assume that sup{ηzn, pzn, xm:m≥n}<1. Thus sup{pzn, xm:m≥
n} < 1 forn ∈ N. It is obvious thatxm ∈ Czn form, n ∈ Nwithm ≥ n. We consider the
following two cases.
iThere existsν∈Nsuch thatxn∈FTforn≥ν.
iiThere exists a subsequence{xnj}of{xn}such thatxnj/∈FT.
In the first case, sinceFT∩Czν exactly consists of one element andxn ∈ FT∩Czν for
n ≥ ν,xn xν holds for alln ≥ ν. Sox xν. Thus,pw, x limnpw, xnholds for every
w ∈X. In the second case, we note thatzn/∈FTfor alln ∈N. Hencezn ∈ Y. PutuP z1.
Sincexn ∈Cz1, there exists a sequence{n}inN∪ {∞}such thatxnTnu. Sincenj ∈Nfor
have limnkn∞and limnn ∞. So we obtainxT∞u. We note thatx∈Cxnfor alln∈N.
Letw∈X. In the case wherex∈Cw, we have
pw, x rfw lim
n→ ∞
rfwrfxn≤lim inf
n→ ∞ pw, xn. 3.19
In the other case, wherex /∈Cw, we havexn/∈Cw, and hence,
pw, x rfw1 lim
n→ ∞
rfwrfxn1 lim
n→ ∞pw, xn. 3.20
Therefore we have shownτ3. Let us proveτ4. We assume that limnsup{pxn, ym:m≥
n}0 and limnηxn, tn 0. Without loss of generality, we may assume that sup{pxn, ym:
m≥n}<1. We consider the following two cases.
iThere existsν∈Nsuch thatyn∈FTforn≥ν.
iiThere exists a subsequence{ynj}of{yn}such thatynj/∈FT.
In the first case, we have
lim
n→ ∞η
yn, tn
lim
n→ ∞tn≤nlim→ ∞ηxn, tn 0. 3.21
In the second case, as in the proof of τ3, there existu ∈ Y, a sequence {kn} in N, and
a sequence {n} in N∪ {∞} such thatBu u,xn Tknu, and yn Tnu. We note that
ηxn, t ηu, t. Ifyn ∈FT, thenηyn, t t≤ηu, t. Ifyn/∈FT, thenηyn, t ηu, t.
Therefore
lim
n→ ∞η
yn, tn
≤ lim
n→ ∞ηu, tn nlim→ ∞ηxn, tn 0. 3.22
Let us proveτ5. We assume thatηz, pz, x < 1/n. We note thatpz, x< 1. In the case whereTz z x, we havedz, x 0 < 1/n. In the other case, where there existu ∈ Y, k∈N, and∈N∪ {∞}such thatBuu,k < l,zTku, andxTu, we have
ηz, pz, x< 1 n gu
rhun≤ηz, rhun. 3.23
Hence
rkr pz, x< rhun. 3.24
Thus, we obtaink > hunand > hun. So we have
dz, x dTku, Tu≤dTku, T∞udTu, T∞u≤ 1 n
1 n
2
Therefore
ηz, pz, x< 1 n, η
z, pz, y< 1 n ⇒d
x, y≤ 4
n, 3.26
which implyτ5. Therefore we have shown thatpis aτ-distance onX.
We shall show3.3. Letx, y∈X. SinceTx∈Cx,T2x∈CTx,fTx≥fx 1, and
fT2x≥fTx 1, we have
pTx, T2xrfTxrfT2x ≤rfx1rfTx1 rpx, Tx. 3.27
Ify∈Cx, thenTy∈CTxholds. So we have
pTx, TyrfTxrfTy≤rfx1rfy1rpx, y≤px, y. 3.28
Ify /∈Cx, then we have
pTx, Ty≤rfTxrfTy1≤rfxrfy1px, y. 3.29
Therefore3.3holds.
Remark 3.4. We have proved that, for everyr ∈ 0,1, there exists aτ-distancep satisfying
3.3.
Combining Theorem 6 in9, we obtain the following.
Corollary 3.5. Let T be a mapping on a complete metric space X, d. Then the following are equivalent.
i(B) holds.
iiThere exists aτ-distanceponXsatisfying the following.
aForx∈Xandε >0, there existδ >0andν∈Nsuch that
pTix, Tjx< εδ⇒pTiνx, Tjνx< ε, 3.30
for alli, j∈N∪ {0}withi < j.
bForx, y∈X, there existν∈Nand a sequence{αn}in0,∞such that
pTix, Tjy< αn⇒p
Tiνx, Tjνy< 1
n, 3.31
for alln∈Nandi, j∈N∪ {0}withi > j.
iiiThere exist a τ-distance p and r ∈ 0,1 such that pTx, T2x ≤ rpx, Tx and
4. Condition (A)
In this section, we discuss ConditionA. Define a relation<
0 on0,∞as follows:s <0 tif and only if eithers t 0 ors < t
holds.
Theorem 4.1. LetXbe a complete metric space and letTbe a mapping onX. Assume that there exist aτ-distancepandr∈0,1such that
pTx, T2x≤rpx, Tx, pTx, Ty<
0 p
x, y, 4.1
for allx, y∈X. Then (A) holds.
Proof. ByTheorem 3.1,Bholds. Moreover, ifTxx, thenpx, x 0. Letz, w∈Xbe fixed points ofT. Then
pz, w pTz, Tw<
0 pz, w, 4.2
which implies thatpz, w 0. Sincepz, z 0, we obtainz wbyLemma 2.4. Thus the fixed point is unique.
Theorem 4.2. LetXbe a complete metric space and letTbe a mapping onX. Assume that there exist aτ-distancepandr∈0,1such that
pTx, T2x≤rpx, Tx, pTx, Ty< px, y, 4.3
for allx, y∈Xwithx /y. Then (A) holds.
Proof. In the case whereX consists of one element, the conclusion obviously holds. So we consider the other case. Assume that limnsup{pxn, xm:m > n}0, limnpxn, Txn 0, and
limnpxn, y 0. We consider the following two cases:
ixn/yfor sufficiently largen∈N,
iithere exists a sequence{xnj}of{xn}such thatxnjy.
In the first case, we have
pxn, Ty
≤pxn, Txn p
Txn, Ty
< pxn, Txn p
xn, y
4.4
for sufficiently largen, and hence, limnpxn, Ty 0. ByLemma 2.7, we obtainTyy. In the
second case, we have
py, Ty lim
j→ ∞p
xnj, Txnj
0, py, y lim
j→ ∞p
xnj, y
ByLemma 2.4, we obtainTy y. ByTheorem 2.2,Bholds. Letz, w ∈X be distinct fixed points ofT. Then
pz, w pTz, Tw< pz, w, 4.6
which implies a contradiction. Thus the fixed point is unique.
We shall show that Theorems4.1and4.2characterize ConditionA.
Theorem 4.3. LetT be a mapping on a metric spaceX, dsuch that (A) holds. Then there exist a
τ-distancepandr∈0,1satisfying4.1.
Proof. Letp,r,f, andCbe as in the proof ofTheorem 3.3. ThenpTx, T2x≤rpx, Txholds. Fixx, y∈X. We consider the following two cases:
iTxxandTyy,
iieitherTx /xorTy /y.
In the first case,xyholds byA. Since
px, x pTx, T2x≤rpx, Tx px, x, 4.7
we obtainpx, x 0. Thus,pTx, Ty px, y 0. In the second case, we note that either fx∈Zorfy∈Zholds. Thus
rfx1rfy1< rfxrfy. 4.8
Ify∈Cx, thenTy∈CTxholds. So we have
pTx, TyrfTxrfTy≤rfx1rfy1< rfxrfypx, y. 4.9
Ify /∈Cx, then we have
pTx, Ty≤rfTxrfTy1< rfxrfy1px, y. 4.10
Therefore4.1holds.
Theorem 4.4. LetT be a mapping on a metric spaceX, dsuch that (A) holds. Then there exist a
τ-distancepandr∈0,1satisfying4.3for allx, y∈Xwithx /y.
Combining Theorem 7 in9, we obtain the following.
Corollary 4.5. Let T be a mapping on a complete metric space X, d. Then the following are equivalent.
i(A) holds.
iiThere exists aτ-distanceponXsatisfying the following.
aForx, y∈Xandε >0, there existδ >0andν∈Nsuch that
pTix, Tjy< εδ⇒pTiνx, Tjνy< ε, 4.11
for alli, j∈N∪ {0}withi < j.
bForx, y∈X, there existν∈Nand a sequence{αn}in0,∞such that
pTix, Tjy< αn⇒p
Tiνx, Tjνy< 1
n, 4.12
for alln∈Nandi, j∈N∪ {0}withi > j.
iiiThere exist a τ-distance p and r ∈ 0,1 such that pTx, T2x ≤ rpx, Tx and
pTx, Ty<
0 px, yfor allx, y∈X.
ivThere exist a τ-distance p and r ∈ 0,1 such that pTx, T2x ≤ rpx, Tx and
pTx, Ty< px, yfor allx, y∈Xwithx /y.
5. Additional Result
Since Theorem 2.2 deduces Corollary 3.2, we can tell that Theorem 2.2 characterizes ConditionB. However, the following example tells thatTheorem 2.3does not characterize ConditionA.
Example 5.1. Let A be the set of all real sequences {an} such that an ∈ 0,∞ for n ∈ N, {an}is strictly decreasing, and{an}converges to 0. LetH be a Hilbert space consisting of
all the functionsxfromA intoRsatisfyinga∈A|xa|2 < ∞with inner productx, y
a∈Axayafor allx, y∈H. Define a subsetXofHby
X{0} ∪
a∈A
{anea: n∈N}
, 5.1
whereea∈His defined byeaa 1 andeab 0 forb∈A\ {a}. Define a mappingTonX
by
T00, Tanea an1ea. 5.2
Proof. It is obvious thatAholds. Arguing by contradiction, we assume thatTis a contraction with respect to someτ-distancep. That is, there exist aτ-distancepandr ∈0,1such that pTx, Ty≤rpx, yfor allx, y∈X. Since
p0,0 pT0, T0≤rp0,0, 5.3
we havep0,0 0. ByLemma 2.6, there exists a strictly increasing sequence{κn}inNsuch
that
p0, x≤rκn ⇒d0, x≤ 1
n. 5.4
We chooseα∈Asuch thatα2κn1 >1/n. Fixν∈Nwithrκνp0, α1eα≤1. Then we have
p0, α2κν1eα p
T2κν0, T2κνα
1eα
≤r2κνp0, α
1eα≤rκν, 5.5
and hence,
1
ν < α2κν1d0, α2κν1eα≤ 1
ν. 5.6
This is a contradiction.
Acknowledgments
The author is supported in part by Grant-in-Aid for Scientific Research from Japan Society for the Promotion of Science. The author wishes to express his gratitude to the referees for careful reading and giving a historical comment.
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