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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 pointxXconverges 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, Tyrdx, yfor all

x, yX. Then the following holds.

AT has a unique fixed pointz, and{Tnx}converges tozfor anyxX.

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.

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aForx, yXandε >0, there existδ >0andν∈Nsuch that

dTix, Tjy< εδdTiνx, Tjνy< ε, 1.1

for alli, j∈N∪ {0}, whereT0is the identity mapping onX.

bForx, yX, there existν∈Nand a sequence{αn}in0,∞such that

dTix, Tjy< αnd

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 allxX. Then the following holds.

B{Tnx}converges to a fixed point for everyxX.

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.

aForxXandε >0, there existδ >0andν∈Nsuch that

dTix, Tjx< εδdTiνx, Tjνx< ε, 1.3

for alli, j∈N∪ {0}.

bForx, yX, there existν∈Nand a sequence{αn}in0,∞such that

dTix, Tjy< αnd

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.

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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, zpx, y py, zfor allx, y, zX,

τ2ηx,0 0 andηx, tt for all xX and t ∈ 0,∞, andη is concave and continuous in its second variable,

τ3limnxn x and limnsup{ηzn, pzn, xm : mn} 0 imply pw, x

lim infnpw, xnfor allwX,

τ4limnsup{pxn, ym:mn}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, T2xrpx, Txfor allxX. 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, Tyrpx, y, 2.1

for allx, yX. 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 somezX, thenlimndxn, yn

0. In particular forx, y, zX, 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.

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Lemma 2.6. LetX, dbe a metric space and letpbe aτ-distance onX. Then for everyzXand

ε >0, there existsδ >0such thatpz, xδandpz, yδimply thatdx, yε.

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,fttgtfor allt∈0,∞; andfis concave and continuous.

Proof. It is clear thatf0 0, fttgt, andf is concave. We shall prove thatf is continuous at 0. Fixε > 0. Then there existsδ > 0 such that ε. Chooseτ > 0 with ττ/δε. Fixt∈0, τ. Letα1, α2, . . . , αn>0 ands1, s2, . . . , sn≥0 such thattni1αisiand

n

i1αi1. Sinceδ

{αi:siδ} ≤t, we have

t

n

i1

αimin

gsi,1 ≤t

si<δ

αigsi

siδ

αit

si<δ

αiε

siδ

αi

t

δτε 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 ∈ NandxX,xAn if and only ifTjx /A0 forj 0,1, . . . , n−1and

TnxA

0.

iiAmAnform, n∈N∪ {0}withm /n.

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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, T2xrpx, Tx, pTx, TyMpx, y, 3.1

for allx, yX. 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, T2xrpx, Tx, pTx, Typx, y, 3.3

for allx, yX. 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 xXsatisfiesTnxxfor somenN, thenTxx. Define a mappingTfromXontoFT

byTxlimnTnxforxX, whereFTis the set of all fixed points ofT. Define a mapping

CfromXinto the set of subsets ofXby

CxTx, T2x, T3x, . . . , Tx. 3.4

SinceTxis a fixed point ofT, we have

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Next, we define a functionffromXintoZ∪ {∞}satisfying

fTxfx 1, fx ∞ ⇐⇒Txx, 3.6

for allxX. We putfx ∞forxFT. It is obvious thatfTx fxfx 1 forxFT. Define a sequence{An}of subsets ofXby

A1T−1FT\FT, An1T−1An. 3.7

Then byLemma 2.9,

FTAn, AmAn, 3.8

form, n∈Nwithm /n. We putfxnforxAn. We note that

fTx

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∞ if xA1,

fx 1 ifx

n2

An.

3.9

Put

Y X\

FT

n∈N

An

. 3.10

It is obvious thatTYY,T−1Y Y, andYFT . So,

TmxTnx⇐⇒mn, 3.11

forxY andm, n∈N∪ {0}. Define an equivalence relation∼onY as follows:xyif and only if there existm, n ∈ N∪ {0}such thatTmx Tny. By Axiom of Choice, there exists a

mappingBonY such that

Bxx, xy⇐⇒BxBy. 3.12

LetuY withBuu. Then we putfTnu nfornN∪ {0}. Define a sequence{D n}of

subsets ofY by

D0

u, Tu, T2u, T3u, . . ., D1T−1D0\D0, Dn1T−1Dn. 3.13

Then we haveDmDn∅form, n∈N∪ {0}withm /n; and

{xY : xu}

n∈N∪{0}

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We putfxnforxYwithn∈NandxDn. We have definedf. We note thatfx∈N

implies thatxY.

Next, we define aτ-distancepby

px, y

⎧ ⎨ ⎩

rfxrfy ifyCx,

rfxrfy1 ify /Cx, 3.15

wherer∞0. We note thatpx, y<1 implies either of the following.

iTxxy.

iiThere existuY,k ∈ N, and ∈N∪ {∞}such thatBu u,k < l,x Tku, and

yTu.In this case,xY,uBx,fx k, andfy hold.

We shall show thatpis aτ-distance. Letx, y, zX. IfyCxandzCy, thenzCx. So we have

px, z rfxrfzrfxrfyrfyrfzpx, ypy, z. 3.16

Ify /Cxorz /Cy, then

px, zrfxrfz1≤rfxrfyrfyrfz1≤px, ypy, z. 3.17

These implyτ1. We shall define a functionη fromX ×0,∞into0,∞. ForxX\Y, we putηx, t t. ForxY, we putu Bx. Since{Tnu}converges toTu, there exists a

strictly increasing sequence{hun}inNsuch thatjhunimplies thatdTju, Tu≤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, ttgutbyLemma 2.8. In order to showτ3, we

assume that limnxn xand limnsup{ηzn, pzn, xm : mn} 0. Then without loss of

generality, we may assume that sup{ηzn, pzn, xm:mn}<1. Thus sup{pzn, xm:m

n} < 1 forn ∈ N. It is obvious thatxmCzn form, n ∈ Nwithmn. We consider the

following two cases.

iThere existsν∈Nsuch thatxnFTfornν.

iiThere exists a subsequence{xnj}of{xn}such thatxnj/FT.

In the first case, sinceFTCzν exactly consists of one element andxnFTCzν for

nν,xn holds for allnν. Sox . Thus,pw, x limnpw, xnholds for every

wX. In the second case, we note thatzn/FTfor alln ∈N. HenceznY. PutuP z1.

SincexnCz1, there exists a sequence{n}inN∪ {∞}such thatxnTnu. Sincenj ∈Nfor

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have limnkn∞and limnn ∞. So we obtainxTu. We note thatxCxnfor alln∈N.

LetwX. In the case wherexCw, 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:

mn}<1. We consider the following two cases.

iThere existsν∈Nsuch thatynFTfornν.

iiThere exists a subsequence{ynj}of{yn}such thatynj/FT.

In the first case, we have

lim

n→ ∞η

yn, tn

lim

n→ ∞tnnlim→ ∞ηxn, tn 0. 3.21

In the second case, as in the proof of τ3, there existuY, 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. IfynFT, thenηyn, t tηu, 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 existuY, k∈N, and∈N∪ {∞}such thatBuu,k < l,zTku, andxTu, we have

ηz, pz, x< 1 n gu

rhunηz, rhun. 3.23

Hence

rkr pz, x< rhun. 3.24

Thus, we obtaink > hunand > hun. So we have

dz, x dTku, TudTku, TudTu, Tu≤ 1 n

1 n

2

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Therefore

ηz, pz, x< 1 n, η

z, pz, y< 1 nd

x, y≤ 4

n, 3.26

which implyτ5. Therefore we have shown thatpis aτ-distance onX.

We shall show3.3. Letx, yX. SinceTxCx,T2xCTx,fTxfx 1, and

fT2xfTx 1, we have

pTx, T2xrfTxrfT2xrfx1rfTx1 rpx, Tx. 3.27

IfyCx, thenTyCTxholds. So we have

pTx, TyrfTxrfTyrfx1rfy1rpx, ypx, y. 3.28

Ify /Cx, then we have

pTx, TyrfTxrfTy1≤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.

aForxXandε >0, there existδ >0andν∈Nsuch that

pTix, Tjx< εδpTiνx, Tjνx< ε, 3.30

for alli, j∈N∪ {0}withi < j.

bForx, yX, there existν∈Nand a sequence{αn}in0,∞such that

pTix, Tjy< αnp

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, T2xrpx, Tx and

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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, T2xrpx, Tx, pTx, Ty<

0 p

x, y, 4.1

for allx, yX. Then (A) holds.

Proof. ByTheorem 3.1,Bholds. Moreover, ifTxx, thenpx, x 0. Letz, wXbe 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, T2xrpx, Tx, pTx, Ty< px, y, 4.3

for allx, yXwithx /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

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ByLemma 2.4, we obtainTy y. ByTheorem 2.2,Bholds. Letz, wX 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, T2xrpx, Txholds. Fixx, yX. We consider the following two cases:

iTxxandTyy,

iieitherTx /xorTy /y.

In the first case,xyholds byA. Since

px, x pTx, T2xrpx, 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

IfyCx, thenTyCTxholds. So we have

pTx, TyrfTxrfTyrfx1rfy1< rfxrfypx, y. 4.9

Ify /Cx, then we have

pTx, TyrfTxrfTy1< 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, yXwithx /y.

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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, yXandε >0, there existδ >0andν∈Nsuch that

pTix, Tjy< εδpTiνx, Tjνy< ε, 4.11

for alli, j∈N∪ {0}withi < j.

bForx, yX, there existν∈Nand a sequence{αn}in0,∞such that

pTix, Tjy< αnp

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, yX.

ivThere exist a τ-distance p and r ∈ 0,1 such that pTx, T2xrpx, Tx and

pTx, Ty< px, yfor allx, yXwithx /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 intoRsatisfyingaA|xa|2 < with inner productx, y

aAxayafor allx, yH. Define a subsetXofHby

X{0} ∪

aA

{anea: n∈N}

, 5.1

whereeaHis defined byeaa 1 andeab 0 forbA\ {a}. Define a mappingTonX

by

T00, Tanea an1ea. 5.2

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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, Tyrpx, yfor allx, yX. 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≤rκnd0, x≤ 1

n. 5.4

We chooseαAsuch thatα2κn1 >1/n. Fixν∈Nwithrκνp0, α1≤1. Then we have

p0, α2κν1 p

T2κν0, T2κνα

1

r2κνp0, α

1rκν, 5.5

and hence,

1

ν < α2κν1d0, α2κν1≤ 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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14 D. Tataru, “Viscosity solutions of Hamilton-Jacobi equations with unbounded nonlinear terms,”

Journal of Mathematical Analysis and Applications, vol. 163, no. 2, pp. 345–392, 1992.

15 C.-K. Zhong, “On Ekeland’s variational principle and a minimax theorem,”Journal of Mathematical Analysis and Applications, vol. 205, no. 1, pp. 239–250, 1997.

16 C.-K. Zhong, “A generalization of Ekeland’s variational principle and application to the study of the relation between the weak P.S. condition and coercivity,”Nonlinear Analysis: Theory, Methods & Applications, vol. 29, no. 12, pp. 1421–1431, 1997.

17 L.-J. Lin and W.-S. Du, “Ekeland’s variational principle, minimax theorems and existence of nonconvex equilibria in complete metric spaces,”Journal of Mathematical Analysis and Applications, vol. 323, no. 1, pp. 360–370, 2006.

18 T. Suzuki, “Generalized distance and existence theorems in complete metric spaces,” Journal of Mathematical Analysis and Applications, vol. 253, no. 2, pp. 440–458, 2001.

19 O. Kada, T. Suzuki, and W. Takahashi, “Nonconvex minimization theorems and fixed point theorems in complete metric spaces,”Mathematica Japonica, vol. 44, no. 2, pp. 381–391, 1996.

20 T. Suzuki, “Several fixed point theorems concerningτ-distance,”Fixed Point Theory and Applications, no. 3, pp. 195–209, 2004.

21 T. Suzuki, “Contractive mappings are Kannan mappings, and Kannan mappings are contractive mappings in some sense,”Commentationes Mathematicae. Prace Matematyczne, vol. 45, no. 1, pp. 45– 58, 2005.

22 T. Suzuki, “The strong Ekeland variational principle,”Journal of Mathematical Analysis and Applications, vol. 320, no. 2, pp. 787–794, 2006.

23 T. Suzuki, “On the relation between the weak Palais-Smale condition and coercivity given by Zhong,”

References

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