Successive Approximations to a Fixed Point

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.

iiT is a strong Leader mapping, that is, the following hold.

aForx, y∈Xandε >0, there existδ >0 andν∈Nsuch that

d

Tⁱx, T^jy

< εδ⇒d

T^iνx, T^jνy

< 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

d

Tⁱx, T^jy

< αn⇒d

T^iνx, T^jνy

< 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 >0 andν∈Nsuch that d

Tⁱx, T^jx

< εδ⇒d

T^iνx, T^jνx

< ε, 1.3

for alli, j∈N∪ {0}.

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

d

Tⁱx, T^jy

< α_n⇒d

T^iνx, T^jνy

< 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 satisfyingBa weakly Picard operator11–13.

We cannot tell that the conditionsiiof Theorems1.2and1.4are simple. Motivated by this, we obtain simpler conditions which are equivalent to ConditionsAandB.

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-distance onX 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 inf_npw, xnfor allw∈X,

τ4lim_nsup{pxn, y_m:m≥n}0 and lim_nηxn, t_n 0 imply that lim_nηyn, t_n 0, τ5lim_nηzn, pzn, x_n 0 and lim_nηzn, pzn, y_n 0 imply that lim_ndxn, y_n 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.

iIf lim_nsup{pxn, x_m : m > n} 0, lim_npxn, Tx_n 0, and lim_npxn, 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

p

Tx, Ty

≤rp x, y

, 2.1

for allx, y∈X. Then (A) andpz, z 0 hold.

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}inXsatisfy lim_npz, xn 0 and lim_npz, yn 0 for somez∈X, then lim_ndxn, y_n 0. In particular forx, y, z∈X, pz, x 0 andpz, y 0 imply thatxy.

Lemma 2.5see18. LetX, dbe a metric space and let pbe aτ-distance onX. If a sequence {xn}inXsatisfies lim_nsup{pxn, x_m:m > n}0, then{xn}is a Cauchy sequence. Moreover if a sequence{yn}inXsatisfies limnpxn, yn 0, then limndxn, yn 0.

The following lemmas are easily deduced from Lemmas2.4and2.5.

Lemma 2.6. LetX, dbe a metric space and letpbe aτ-distance onX. Then for everyz ∈Xand ε >0, there existsδ >0 such 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 Xsatisfies limnsup{pxn, xm:m > n}0, limnpxn, y 0, and limnpxn, 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 satisfying inf{gt: t >0}0. Define a functionffrom0,∞into itself by

ft tsup _n

i1

αimin

gsi,1 :tⁿ

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 thatt_n

i1αisiand _n

i1α_i1. Sinceδ

{αi:s_i≥δ} ≤t, we have

tⁿ

i1

α_imin

gsi,1 ≤t

si<δ

α_igsi

si≥δ

α_i≤t

si<δ

α_iε

si≥δ

α_i

≤tε t

δ ≤τε τ δ ≤2ε.

2.3

Since α₁, α₂, . . . , α_n > 0 and s₁, s₂, . . . , s_n ≥ 0 are arbitrary, we obtain ft ≤ 2ε. Thus, lim_t→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, A_n1T⁻¹An. 2.4

Then the following hold.

iFor everyn ∈ Nandx ∈ X,x ∈A_n if and only ifT^jx /∈A₀ forj 0,1, . . . , n−1 and Tⁿx∈A₀.

iiAm∩An∅form, n∈N∪ {0}withm /n.

iiiTA_n1 A_nfor everyn∈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

p

Tx, T²x

≤rpx, Tx, p

Tx, Ty

≤Mp x, y

, 3.1

for allx, y∈X. Then (B) holds. Moreover, ifTzz, thenpz, z 0.

Proof. Assume that lim_nsup{pxn, x_m:m > n}0, lim_npxn, Tx_n 0, and lim_npxn, y 0.

Then we have p

xn, Ty

≤pxn, Txn p

Txn, Ty

≤pxn, Txn Mp xn, y

, 3.2

and hence, lim_npxn, 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

p

Tx, T²x

≤rpx, Tx, p

Tx, Ty

≤p x, 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

Cx

Tx, T²x, T³x, . . . , T^∞x

. 3.4

SinceT^∞xis a fixed point ofT, we have

y∈Cx⇒Cy⊂Cx. 3.5

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, A_n1T⁻¹An. 3.7

Then byLemma 2.9,

FT∩A_n∅, A_m∩A_n ∅, 3.8

form, n∈Nwithm /n. We putfx −nforx∈A_n. We note that

fTx

⎧⎪

⎪⎨

⎪⎪

⎩

∞ if x∈A1, fx 1 ifx∈^∞

n2

An. 3.9

Put

Y X\

FT

n∈N

A_n

. 3.10

It is obvious thatTY⊂Y,T⁻¹Y Y, andY∩FT ∅. So,

T^mxTⁿx⇐⇒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 thatT^mx 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{Dn}of subsets ofY by

D0

u, Tu, T²u, T³u, . . .

, D1T⁻¹D0\D0, D_n1T⁻¹Dn. 3.13

Then we haveDm∩Dn∅form, n∈N∪ {0}withm /n; and {x∈Y : x∼u}

n∈N∪{0}

D_n. 3.14

We putfx −nforx∈Ywithn∈Nandx∈Dn. We have definedf. We note thatfx∈N implies thatx∈Y.

Next, we define aτ-distancepby

p x, y

⎧⎨

⎩

r^fxr^fy ify∈Cx,

r^fxr^fy1 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 T^ku, 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 r^fxr^fz ≤r^fxr^fyr^fyr^fzp x, y

p y, z

. 3.16

Ify /∈Cxorz /∈Cy, then

px, z≤r^fxr^fz1≤r^fxr^fyr^fyr^fz1≤p x, y

p y, 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 thatdT^ju, T^∞u≤1/n forj∈N∪ {∞}. Since limnh_un ∞, we can define a nondecreasing functiong_ufrom0,∞ into0,1such thatg_ur^hun 1/n. It is obvious thatg_u0 lim_t→0g_ut 0. Put

ηx, t tsup _n

i1

αigusi:tⁿ

i1

αisi, si ≥0, αi>0, n

i1

αi1

. 3.18

Thenηx, tsatisfiesτ2and ηx, t ≥ tg_utbyLemma 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, x_m:m≥n}<1. Thus sup{pzn, x_m:m≥ n} < 1 forn ∈ N. It is obvious thatx_m ∈ Cz_n form, n ∈ Nwithm ≥ n. We consider the following two cases.

iThere existsν∈Nsuch thatxn∈FTforn≥ν.

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

In the first case, sinceFT∩Cz_ν exactly consists of one element andx_n ∈ FT∩Cz_ν 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 thatz_n/∈FTfor alln ∈N. Hencez_n ∈ Y. PutuP z₁. Sincex_n ∈Cz₁, there exists a sequence{n}inN∪ {∞}such thatx_nTⁿu. Since_nj ∈Nfor allj ∈N, there exists a sequence{kn}inNsuch thatz_n T^knu. Since lim_npzn, x_n 0, we

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 r^fw lim

n→ ∞

r^fwr^fxn

≤lim inf

n→ ∞ pw, xn. 3.19

In the other case, wherex /∈Cw, we havex_n/∈Cw, and hence, pw, x r^fw1 lim

n→ ∞

r^fwr^fxn1 lim

n→ ∞pw, xn. 3.20 Therefore we have shownτ3. Let us proveτ4. We assume that limnsup{pxn, ym:m≥ n}0 and lim_nηxn, t_n 0. Without loss of generality, we may assume that sup{pxn, y_m: m≥n}<1. We consider the following two cases.

iThere existsν∈Nsuch thaty_n∈FTforn≥ν.

iiThere exists a subsequence{ynj}of{yn}such thaty_nj/∈FT. In the first case, we have

n→ ∞limη yn, tn

lim

n→ ∞tn≤ lim

n→ ∞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,x_n T^knu, and y_n Tⁿu. We note that ηxn, t ηu, t. Ifyn ∈FT, thenηyn, t t≤ηu, t. Ifyn/∈FT, thenηyn, t ηu, t.

Therefore

nlim→ ∞η yn, tn

≤ lim

n→ ∞ηu, tn lim

n→ ∞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,zT^ku, andxTu, we have

η

z, pz, x

< 1 n gu

r^hun

≤η

z, r^hun

. 3.23

Hence

r^kr pz, x< r^hun. 3.24

Thus, we obtaink > hunand > hun. So we have dz, x d

T^ku, Tu

≤d

T^ku, T^∞u d

Tu, T^∞u

≤ 1 n 1

n 2

n. 3.25

Therefore

η

z, pz, x

< 1 n, η

z, p z, 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

p

Tx, T²x

r^fTxr^fT2x ≤r^fx1r^fTx1 rpx, Tx. 3.27

Ify∈Cx, thenTy∈CTxholds. So we have p

Tx, Ty

r^fTxr^fTy≤r^fx1r^fy1rp x, y

≤p x, y

. 3.28

Ify /∈Cx, then we have p

Tx, Ty

≤r^fTxr^fTy1≤r^fxr^fy1p x, 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 >0 andν∈Nsuch that p

Tⁱx, T^jx

< εδ⇒p

T^iνx, T^jνx

< ε, 3.30

for alli, j∈N∪ {0}withi < j.

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

p

Tⁱx, T^jy

< αn⇒p

T^iνx, T^jνy

< 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 pTx, Ty≤px, yfor allx, y∈X.

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

p

Tx, T²x

≤rpx, Tx, p

Tx, 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

p

Tx, T²x

≤rpx, Tx, p

Tx, Ty

< p x, 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 lim_nsup{pxn, x_m:m > n}0, lim_npxn, Tx_n 0, and limnpxn, y 0. We consider the following two cases:

ixn/yfor sufficiently largen∈N,

iithere exists a sequence{xnj}of{xn}such thatx_njy.

In the first case, we have p

x_n, Ty

≤pxn, Tx_n p

Tx_n, Ty

< pxn, Tx_n p x_n, y

4.4 for sufficiently largen, and hence, lim_npxn, Ty 0. ByLemma 2.7, we obtainTyy. In the second case, we have

p y, Ty

lim

j→ ∞p

x_nj, Tx_nj

0, p y, y

lim

j→ ∞p x_nj, y

0. 4.5

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 p

Tx, 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

r^fx1r^fy1< r^fxr^fy. 4.8

Ify∈Cx, thenTy∈CTxholds. So we have p

Tx, Ty

r^fTxr^fTy≤r^fx1r^fy1< r^fxr^fyp x, y

. 4.9

Ify /∈Cx, then we have p

Tx, Ty

≤r^fTxr^fTy1< r^fxr^fy1p x, 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.

Proof. The proof ofTheorem 4.3works.

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 >0 andν∈Nsuch that p

Tⁱx, T^jy

< εδ⇒p

T^iνx, T^jνy

< ε, 4.11

for alli, j∈N∪ {0}withi < j.

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

p

Tⁱx, T^jy

< αn⇒p

T^iνx, T^jνy

< 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 intoRsatisfying

a∈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, Tane_a a_n1e_a. 5.2

ThenAholds. However,Tis not a contraction with respect to anyτ-distancep.

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≤r^κn ⇒d0, x≤ 1

n. 5.4

We chooseα∈Asuch thatα_2κn1 >1/n. Fixν∈Nwithr^κνp0, α1e_α≤1. Then we have p0, α2κν1e_α p

T^2κν0, T^2κνα1e_α

≤r^2κνp0, α1e_α≤r^κν, 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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