Generic convergence of iterates for a class of nonlinear mappings

OF NONLINEAR MAPPINGS

SIMEON REICH AND ALEXANDER J. ZASLAVSKI

Received 4 March 2004

LetKbe a nonempty, bounded, closed, and convex subset of a Banach space. We show that the iterates of a typical element (in the sense of Baire’s categories) of a class of con-tinuous self-mappings ofK converge uniformly onK to the unique fixed point of this typical element.

1. Introduction

LetK be a nonempty, bounded, closed, and convex subset of a Banach space (X, · ). We consider the topological subspaceK⊂Xwith the relative topology induced by the norm · . Set

diam(K)=supx−y:x,y∈K. (1.1)

Denote byᏭthe set of all continuous mappingsA:K→K which have the following property:

(P1) for each>0, there existsx∈Ksuch that

_{Ax−x≤x−x+ ∀x∈K. (1.2)}

For eachA,B∈Ꮽ, set

d(A,B)=supAx−Bx:x∈K. (1.3)

Clearly, the metric space (Ꮽ,d) is complete.

In this paper, we use the concept of porosity [1,2,3,4,5,6] which we now recall. Let (Y,ρ) be a complete metric space. We denote byB(y,r) the closed ball of center y∈Y and radiusr >0. A subsetE⊂Y is called porous in (Y,ρ) if there existα∈(0, 1) andr0>0 such that for eachr∈(0,r0] and eachy∈Y, there existsz∈Yfor which

B(z,αr)⊂B(y,r)\E. (1.4)

A subset of the spaceY is calledσ-porous in (Y,ρ) if it is a countable union of porous subsets in (Y,ρ).

Since porous sets are nowhere dense, allσ-porous sets are of the first category. IfY is a finite-dimensional Euclidean spaceRn_{, thenσ-porous sets are of Lebesgue measure 0.}

To point out the difference between porous and nowhere dense sets, note that ifE⊂Y is nowhere dense,y∈Y, andr >0, then there are a pointz∈Yand a numbers >0 such thatB(z,s)⊂B(y,r)\E. If, however,Eis also porous, then for small enoughr, we can chooses=αr, whereα∈(0, 1) is a constant which depends only onE.

Our purpose in this paper is to establish the following result.

Theorem1.1. There exists a setᏲ⊂Ꮽsuch that the complementᏭ\Ᏺisσ-porous in (Ꮽ,d)and eachA∈Ᏺhas the following properties:

(i)there exists a unique fixed pointxA∈Ksuch that

Anx−→xA asn−→ ∞,uniformly∀x∈K; (1.5)

(ii)Ax−xA ≤ x−xAfor allx∈K;

(iii)for each>0, there exist a natural numbernandδ >0such that for each integer p≥n, eachx∈K, and eachB∈Ꮽsatisfyingd(B,A)≤δ,

_Bp_x−x

A≤. (1.6)

2. Auxiliary result

In this section, we present and prove an auxiliary result which will be used in the proof ofTheorem 1.1inSection 3.

Proposition2.1. LetA∈Ꮽand∈(0, 1). Then there existx¯∈KandB∈Ꮽsuch that d(A,B)≤,

x¯−Bx ≤ x¯−x ∀x∈K. (2.1)

Proof. Choose a positive number

0<8−12

diam(K) + 1−1. (2.2)

SinceA∈Ꮽ, there exists ¯x∈Ksuch that

Ax−x ≤ x¯ −x¯ +0 ∀x∈K. (2.3)

Letx∈K. There are three cases:

Ax−x¯ <; (2.4)

Ax−x ≥¯ , Ax−x¯ <x−x¯ ; (2.5) Ax−x ≥¯ , Ax−x ≥ x¯ −x.¯ (2.6)

First, we consider case (2.4). There exists an open neighborhoodVxofxinKsuch that

Defineψx:Vx→Kby

ψx(y)=x¯, y∈Vx. (2.8)

Clearly, for ally∈Vx,

0=ψx(y)−x¯≤ y−x¯ , Ay−ψx(y)= Ay−x¯ <. (2.9)

Consider now case (2.5). SinceAis continuous, there exists an open neighborhoodVxof

xinKsuch that

Ay−x¯ <y−x ∀y¯ ∈Vx. (2.10)

In this case, we defineψx:Vx→Kby

ψx(y)=Ay, y∈Vx. (2.11)

Finally, we consider case (2.6). Inequalities (2.6), (2.2), and (2.3) imply that

x−x ≥ Ax¯ −x −¯ 0>7

8. (2.12)

For eachγ∈[0, 1], set

z(γ)=γAx+ (1−γ) ¯x. (2.13)

By (2.13), (2.6), and (2.12), we have

_{z(0)−x¯=0, z(1)−x¯= Ax−x ≥ x¯ −x¯ >}7

8. (2.14)

By (2.2) and (2.14), there existsγ0∈(0, 1) such that _z

γ0−x¯= x−x −¯ 0. (2.15)

It now follows from (2.13), (2.15), and (2.3) that

γ0x−x¯ +0

≥γ0Ax−x =¯ γ0Ax+1−γ0x¯−x¯=zγ0−x¯= x−x −¯ 0, (2.16)

γ0≥x−x −¯ 0

x−x¯ +0 −1

=1−20

x−x¯ +0 −1

≥1−20x−x¯ −1_. (2.17)

Inequalities (2.17) and (2.12) imply that

γ0≥1−20

7 8

−1

By (2.13), (1.1), (2.18), and (2.2), _z

γ0−Ax=γ0Ax+1−γ0x¯−Ax=1−γ0Ax−x¯ ≤1−γ0diam(K)≤160(7)−1diam(K)

≤30diam(K)−1≤3 8,

(2.19)

_z

γ0−Ax≤3

8. (2.20)

Relations (2.15) and (2.20) imply that there exists an open neighborhoodVx ofxinK

such that for eachy∈Vx,

_z

γ0−Ay<, zγ0−x¯<y−x.¯ (2.21)

Defineψx:Vx→Kby

ψx(y)=z

γ0, y∈Vx. (2.22)

It is not difficult to see that in all three cases, we have defined an open neighborhoodVx

ofxinKand a continuous mappingψx:Vx→Ksuch that for eachy∈Vx,

_{Ay−ψx(y)<, x¯−ψx(y)≤ y−x.¯ (2.23)}

Since the metric spaceK with the metric induced by the norm is paracompact, there exists a continuous locally finite partition of unity{φi}i∈IonKsubordinated to{Vx}x∈K,

where eachφi:K→[0, 1],i∈I, is a continuous function such that for eachy∈K, there

is a neighborhoodUofyinKsuch that

U∩suppφi= ∅ (2.24)

only for a finite number ofi∈I;

i∈I

φi(x)=1, x∈K; (2.25)

and for eachi∈I, there isxi∈Ksuch that

suppφi

⊂Vxi. (2.26)

Here, supp(φ) is the closure of the set{x∈K:φ(x)=0}. Define

Bz=

i∈I

φi(z)ψxi(z), z∈K. (2.27)

Clearly,B:K→Kis well defined and continuous.

Letz∈K. There are a neighborhoodUofzinKandi1,...,in∈Isuch that

U∩suppφi

= ∅ for anyi∈I\i1,...,in

We may assume, without loss of generality, that

z∈suppφip

, p=1,...,n. (2.29)

Then

n

p=1

φip(z)=1, Bz=

n

p=1

φip(z)ψxip(z). (2.30)

Relations (2.26), (2.29), and (2.23) imply that forp=1,...,n, the following relations also hold:z∈Vxip,

_Az−ψx

ip(z)<, x¯−ψxip(z)≤ x¯−z. (2.31)

By (2.31) and (2.30),

Bz−Az =

n

p=1

φip(z)ψxip(z)−Az ≤

n

p=1

φip(z)ψxip(z)−Az<,

_x¯−Bz= x¯−

n

p=1

φip(z)ψxip(z) ≤

n

p=1

φip(z)x¯−ψxip(z)≤ x¯−z,

Bz−Az<, x¯−Bz ≤ x¯−z.

(2.32)

Proposition 2.1is proved.

3. Proof of Theorem1.1

For eachC∈Ꮽandx∈K, setC0_{x=x. For each natural numbern, denote byᏲ}

nthe set

of allA∈Ꮽwhich have the following property:

(P2) there exist ¯x∈K, a natural numberq, and a positive numberδ >0 such that

x¯−Ax ≤ x¯−x+n−1 _{∀x∈K, (3.1)}

and such that, for eachB∈Ꮽsatisfyingd(B,A)≤δ, and eachx∈K,

_Bq_x−x¯≤n−1_{. (3.2)}

Define

Ᏺ= ∩∞

n=1Ᏺn. (3.3)

Lemma3.1. LetA∈Ᏺ. Then there exists a unique fixed pointxA∈Ksuch that

(i)An_x→x

Aasn→ ∞, uniformly onK;

(ii)

(iii)for each>0, there exist a natural numberqandδ >0such that, for eachB∈Ꮽ

satisfyingd(B,A)≤δ, eachx∈K, and each integeri≥q, _Bi_x−x

A≤. (3.5)

Proof. Letnbe a natural number. SinceA∈Ᏺ⊂Ᏺn, it follows from (P2) that there exist

xn∈K, an integerqn≥1, and a numberδn≥0 such that

_{xn−Ax≤xn−x+n}−1 _{∀x∈K, (3.6)}

and we have the following property:

(P3) for eachB∈Ꮽsatisfyingd(B,A)≤δn, and eachx∈K,

_Bqn_x−_x

n≤ 1

n. (3.7)

Property (P3) implies that for eachx∈K,Aqn_x−_x

n ≤1/n. This fact implies, in turn,

that for eachx∈K,

_Ai_x−x

n≤1

n for any integeri≥qn. (3.8)

Sincen is any natural number, we conclude that for eachx∈K,{Ai_x}∞

i=1 is a Cauchy sequence and there exists limi→∞Aix. Inequality (3.8) implies that for eachx∈K,

lim

i→∞A

i_x−x n

≤1

n. (3.9)

Sincenis again an arbitrary natural number, we conclude further that limi→∞Aixdoes not depend onx. Hence, there isxA∈Ksuch that

xA=lim i→∞A

i_{x ∀x∈K. (3.10)}

By (3.9) and (3.10),

_xA−xn≤1

n. (3.11)

Inequalities (3.11) and (3.6) imply that for eachx∈K,

_{Ax−xA≤Ax−xn+xn−xA≤}1

n+Ax−xn≤ 1

n+x−xn+ 1 n ≤2

n+x−xA+xA−xn≤x−xA+ 3 n,

(3.12)

so that

_{Ax−xA≤x−xA+}3

Sincenis an arbitrary natural number, we conclude that

_{Ax−xA≤x−xA for eachx∈K. (3.14)}

Let>0. Choose a natural number

n >8

. (3.15)

Property (P3) implies that

_Bi_x−x

n≤1_n (3.16)

for eachx∈K, each integeri≥qn, and eachB∈Ꮽsatisfyingd(B,A)≤δn. Inequalities

(3.16), (3.11), and (3.15) imply that for eachB∈Ꮽsatisfyingd(B,A)≤δn, eachx∈K,

and each integeri≥qn,

_Bi_x−x

A≤Bix−xn+xn−xA≤1

n+ 1

n<. (3.17)

This completes the proof ofLemma 3.1.

Completion of the proof ofTheorem 1.1. In order to complete the proof of this theorem, it is sufficient, byLemma 3.1, to show that for each natural numbern, the setᏭ\Ᏺnis

porous in (Ꮽ,d).

Letnbe a natural number. Choose a positive number

α <(16n)−1₂−1 _{diam(K) + 1}2_16·8n−1_{. (3.18)}

Let

A∈Ꮽ, r∈(0, 1]. (3.19)

ByProposition 2.1, there existA0∈Ꮽand ¯x∈Ksuch that

dA,A0≤r

8, (3.20)

_{A0x−x¯≤ x−x¯ for eachx∈K. (3.21)}

Set

γ=8−1_{rdiam(K) + 1}−1 _(3.22)

and choose a natural numberqfor which

1≤q diam(K) + 1216n·8r−1−1_{≤2. (3.23)}

Define ¯A:K→Kby

¯

Clearly, the mapping ¯Ais continuous and, for eachx∈K,

_Ax¯ _{−x =¯ (1−γ)A0x+γx¯−x¯=(1−γ)A0x−x¯≤(1−γ)x−x.¯ (3.25)}

Thus, ¯A∈Ꮽ. Relations (1.3), (3.24), (1.1), (3.22), and (3.25) imply that

dA¯,A0=supAx¯ −A0x:x∈K=supγx¯−A0x:x∈K

≤γdiam(K)≤r 8.

(3.26)

Together with (3.20), this implies that

d( ¯A,A)≤dA¯,A0+dA0,A≤r

4. (3.27)

Assume now that

B∈Ꮽ, dB, ¯A≤αr. (3.28)

Then (3.28), (3.18), and (3.25) imply that for eachx∈K,

_{Bx−x¯≤Bx−Ax}¯ _+Ax¯ _{−x¯≤ x−x¯ +αr≤ x−x¯ +}1

n. (3.29)

In addition, (3.28), (3.27), and (3.18) imply that

d(B,A)≤d(B, ¯A) +d( ¯A,A)≤αr+r 4≤

r

2. (3.30)

Assume that x∈K. We will show that there exists an integer j∈[0,q] such that Bj_{x−x ≤¯ (8n)}−1_{. Assume the contrary. Then}

_Bi_x−x¯>(8n)−1_{, i=0,...,q. (3.31)}

Let an integeri∈ {0,...,q−1}. By (3.28) and (3.25),

_Bi+1_x−x¯=BBi_x−x¯≤BBi_x−A¯_Bi_x+A¯_Bi_x−x¯

≤d(B, ¯A) +A¯Bix−x¯≤αr+ (1−γ)Bix−x¯, _Bi+1_{x−x¯≤αr+ (1−γ)B}i_x−x¯.

(3.32)

When combined with (3.31), (3.18), and (3.22), this inequality implies that _Bi_x−x¯−Bi+1_x−x¯≥Bi_{x−x¯−αr−(1−γ)B}i_x−x¯

=γBi_{x−x¯−αr >(8n)}−1_{γ−αr≥(16n)}−1_γ, (3.33)

so that

_Bi_x−x¯−Bi+1

When combined with (1.1), this inequality implies that

diam(K)≥ x−x −¯ Bqx−x¯≥

q−1

i=0

_Bi_x−x¯−Bi+1_{x−x¯≥q(16n)}−1_γ,

q≤diam(K)16n γ ,

(3.35)

a contradiction (see (3.22) and (3.23)). The contradiction we have reached shows that there exists an integer j∈ {0,...,q−1}such that

_Bj_{x−x¯≤(8n)}−1

. (3.36)

It follows from (3.28) and (3.25) that for each integeri∈ {0,...,q−1}, _Bi+1_x−x¯=BBi_x−x¯≤BBi_x−A¯_Bi_x+A¯_Bi_x−x¯

≤d( ¯A,B) +A¯Bi_{x−x¯≤αr+B}i_x−x¯,

_Bi+1

x−x¯≤Bix−x¯+αr.

(3.37)

This implies that for each integerssatisfying j < s≤q,

_Bs_x−x¯≤Bj_{x−x¯+αr(s−j)≤B}j_{x−x¯+αrq. (3.38)}

It follows from (3.36), (3.38), (3.23), and (3.18) that

_Bq_{x−x¯≤αrq+ (8n)}−1_≤(2n)−1_{. (3.39)}

Thus, we have shown that the following property holds: for eachBsatisfying (3.28) and eachx∈K,

_Bq_{x−x¯≤(2n)}−1_{, Bx−x¯≤ x−x¯ +}1

n (3.40)

(see (3.29)). Thus

B∈Ꮽ:d(B, ¯A)≤αr 2

⊂Ᏺn∩

B∈Ꮽ:d(B,A)≤r. (3.41)

In other words, we have shown that the setᏭ\Ᏺnis porous in (Ꮽ,d). This completes the

proof ofTheorem 1.1.

Acknowledgments

References

[1] F. S. De Blasi and J. Myjak,Sur la porosit´e de l’ensemble des contractions sans point fixe[On the porosity of the set of contractions without fixed points], C. R. Acad. Sci. Paris S´er. I Math.308

(1989), no. 2, 51–54 (French).

[2] ,On a generalized best approximation problem, J. Approx. Theory94(1998), no. 1, 54– 72.

[3] F. S. De Blasi, J. Myjak, and P. L. Papini,Porous sets in best approximation theory, J. London Math. Soc. (2)44(1991), no. 1, 135–142.

[4] S. Reich and A. J. Zaslavski,The set of divergent descent methods in a Banach space isσ-porous, SIAM J. Optim.11(2001), no. 4, 1003–1018.

[5] ,The set of noncontractive mappings isσ-porous in the space of all nonexpansive map-pings, C. R. Acad. Sci. Paris S´er. I Math.333(2001), no. 6, 539–544.

[6] ,Well-posedness and porosity in best approximation problems, Topol. Methods Nonlinear Anal.18(2001), no. 2, 395–408.

Simeon Reich: Department of Mathematics, The Technion – Israel Institute of Technology, 32000 Haifa, Israel

E-mail address:[email protected]

Current address: Department of Mathematics, University of California, Santa Barbara, CA 93106-3080, USA

E-mail address:[email protected]

Alexander J. Zaslavski: Department of Mathematics, The Technion – Israel Institute of Technology, 32000 Haifa, Israel