Báo cáo hóa học: " GENERIC CONVERGENCE OF ITERATES FOR A CLASS OF NONLINEAR MAPPINGS" pdf

ZASLAVSKI Received 4 March 2004 LetK be a nonempty, bounded, closed, and convex subset of a Banach space.. Introduction LetK be a nonempty, bounded, closed, and convex subset of a Banach

OF NONLINEAR MAPPINGS

SIMEON REICH AND ALEXANDER J ZASLAVSKI

Received 4 March 2004

LetK be 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 on K 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 ⊂ X with the relative topology induced by the

norm
·
Set

diam(K) =sup

x − y
:x, y ∈ K

Denote byᏭ the set of all continuous mappings A : K → K which have the following

property:

(P1) for each
> 0, there exists x
∈ K such that

For eachA,B ∈Ꮽ, set

d(A,B) =sup

Ax − Bx
:x ∈ K

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 by B(y,r) the closed ball of center

y ∈ Y and radius r > 0 A subset E ⊂ Y is called porous in (Y,ρ) if there exist α ∈(0, 1) andr0> 0 such that for each r ∈(0,r0] and eachy ∈ Y, there exists z ∈ Y for which

Copyright©2004 Hindawi Publishing Corporation

Fixed Point Theory and Applications 2004:3 (2004) 211–220

2000 Mathematics Subject Classification: 47H09, 47H10, 54E50, 54E52

URL: http://dx.doi.org/10.1155/S1687182004403015

A subset of the spaceY is called σ-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 If Y 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 if E ⊂ Y

is nowhere dense,y ∈ Y, and r > 0, then there are a point z ∈ Y and a number s > 0 such

thatB(z,s) ⊂ B(y,r) \ E If, however, E is also porous, then for small enough r, we can

chooses = αr, where α ∈(0, 1) is a constant which depends only onE.

Our purpose in this paper is to establish the following result

Theorem 1.1 There exists a setᏲ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σ-porous in

(Ꮽ,d) and each A∈ Ᏺ has the following properties:

(i) there exists a unique fixed point x A ∈ K such that

A n x −→ x A as n −→ ∞, uniformly ∀ x ∈ K; (1.5) (ii)
Ax − x A
≤
x − x A
for all x ∈ K;

(iii) for each
> 0, there exist a natural number n and δ > 0 such that for each integer

p ≥ n, each x ∈ K, and each B ∈ Ꮽ satisfying d(B,A) ≤ δ,

2 Auxiliary result

In this section, we present and prove an auxiliary result which will be used in the proof

ofTheorem 1.1inSection 3

Proposition 2.1 Let A ∈ Ꮽ and
∈ (0, 1) Then there exist ¯ x ∈ K and B ∈ Ꮽ such that

d(A,B) ≤
,

¯x − Bx
≤
¯x − x
∀ x ∈ K. (2.1) Proof Choose a positive number

0 < 8 −1
2 

diam(K) + 1−1

SinceA ∈ Ꮽ, there exists ¯x ∈ K such that

Ax − ¯x
≤
x − ¯x
+
0 ∀ x ∈ K. (2.3) Letx ∈ K There are three cases:

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 neighborhoodV xofx in K such that

Defineψ x:V x → K by

Clearly, for ally ∈ V x,

0=ψ

x(y) − ¯x  ≤
y − ¯x
, Ay − ψ

x(y)  =
Ay − ¯x
<
(2.9) Consider now case (2.5) SinceA is continuous, there exists an open neighborhood V xof

x in K such that

Ay − ¯x
<
y − ¯x
∀ y ∈ V x (2.10)

In this case, we defineψ x:V x → K by

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

x − ¯x
≥
Ax − ¯x
−
0 >7

For eachγ ∈[0, 1], set

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



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

γ0



x − ¯x
+
0≥ γ0
Ax − ¯x
=γ0Ax +

1− γ0



¯x − ¯x  =  z

γ0



− ¯x  =
x − ¯x
−
0,

(2.16)

γ0≥
x − ¯x
−
0
x − ¯x
+
0−1=1−2
0
x − ¯x
+
0−1≥1−2
0
x − ¯x
−1.

(2.17) Inequalities (2.17) and (2.12) imply that

γ0≥1−2
0



7

8

−1

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

z

γ0



− Ax  =  γ0Ax +

1− γ0



¯x − Ax  = 1− γ0



Ax − ¯x

≤1− γ0



diam(K) ≤16
0(7
)−1diam(K)

≤3
0diam(K)
−1≤3

8
,

(2.19)

z

γ0



− Ax  ≤3

Relations (2.15) and (2.20) imply that there exists an open neighborhoodV x ofx in K

such that for eachy ∈ V x,

z

γ0 

− Ay<
, z

γ0 

− ¯x<
y − ¯x
. (2.21)

Defineψ x:V x → K by

ψ x(y) = z

γ0



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

ofx in K and a continuous mapping ψ x:V x → K such that for each y ∈ V x,

Ay − ψ x(y)<
, ¯x − ψ x(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{ φ i } i∈IonK subordinated to { V x } x∈K, where eachφ i:K →[0, 1],i ∈ I, is a continuous function such that for each y ∈ K, there

is a neighborhoodU of y in K such that

U ∩supp

φ i

(2.24) only for a finite number ofi ∈ I;



i∈I

and for eachi ∈ I, there is x i ∈ K such that

supp

φ i

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

Bz =

i∈I

φ i(z)ψ x i(z), z ∈ K. (2.27)

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

Letz ∈ K There are a neighborhood U of z in K and i1, ,i n ∈ I such that

U ∩supp

φ i



= ∅ for anyi ∈ I \
i1, ,i n



We may assume, without loss of generality, that

z ∈supp

φ i p



Then

n



p=1

φ i p(z) =1, Bz =n

p=1

φ i p(z)ψ x ip(z). (2.30)

Relations (2.26), (2.29), and (2.23) imply that forp =1, ,n, the following relations also

hold:z ∈ V x ip,

Az − ψ x

ip(z)<
, ¯x − ψ x

ip(z)  ≤
¯x − z
(2.31)

By (2.31) and (2.30),

Bz − Az
=





n



p=1

φ i p(z)ψ x ip(z) − Az





 ≤

n



p=1

φ i p(z)ψ x

ip(z) − Az<
,

¯x − Bz  =¯x −n

p=1

φ i p(z)ψ x ip(z)



 ≤

n



p=1

φ i p(z)¯x − ψ x

ip(z)  ≤
¯x − z
,

Bz − Az
<
,
¯x − Bz
≤
¯x − z

(2.32)

3 Proof of Theorem 1.1

For eachC ∈ Ꮽ and x ∈ K, set C0x = x For each natural number n, denote by Ᏺ nthe set

of allA ∈Ꮽ which have the following property:

(P2) there exist ¯x ∈ K, a natural number q, and a positive number δ > 0 such that

¯x − Ax
≤
¯x − x
+n −1 ∀ x ∈ K, (3.1) and such that, for eachB ∈ Ꮽ satisfying d(B,A) ≤ δ, and each x ∈ K,

Define

Ᏺ= ∩ ∞

Lemma 3.1 Let A ∈ Ᏺ Then there exists a unique fixed point x A ∈ K such that

(i)A n x → x A as n → ∞ , uniformly on K;

(ii)

(iii) for each
> 0, there exist a natural number q and δ > 0 such that, for each B ∈Ꮽ

satisfying d(B,A) ≤ δ, each x ∈ K, and each integer i ≥ q,

Proof Let n be a natural number Since A ∈Ᏺ⊂Ᏺn, it follows from (P2) that there exist

x n ∈ K, an integer q n ≥1, and a numberδ n ≥0 such that

x n − Ax  ≤  x n − x+n −1 ∀ x ∈ K, (3.6) and we have the following property:

(P3) for eachB ∈ Ꮽ satisfying d(B,A) ≤ δ n, and eachx ∈ K,

B q n x − x n  ≤ 1

Property (P3) implies that for eachx ∈ K,
A q n x − x n
≤1/n This fact implies, in turn,

that for eachx ∈ K,

A i x − x n  ≤1

n for any integeri ≥ q n (3.8) Sincen is any natural number, we conclude that for each x ∈ K, { A i x } ∞

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



lim

i→∞ A i x − x n

Sincen is again an arbitrary natural number, we conclude further that lim i→∞ A i x does

not depend onx Hence, there is x A ∈ K such that

x A =lim

By (3.9) and (3.10),

x A − x n  ≤1

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

Ax − x A  ≤  Ax − x n+x n − x A  ≤1

n+Ax − x n  ≤1

n+x − x n+1

n

≤2

n+x − x

A+x

A − x n  ≤  x − x A+3

n,

(3.12)

so that

Ax − x A  ≤  x − x A+3

Sincen is an arbitrary natural number, we conclude that

Ax − x A  ≤  x − x A for eachx ∈ K. (3.14)

Let
> 0 Choose a natural number

n >8

Property (P3) implies that

B i x − x n  ≤1

for eachx ∈ K, each integer i ≥ q n, and eachB ∈ Ꮽ satisfying d(B,A) ≤ δ n Inequalities (3.16), (3.11), and (3.15) imply that for eachB ∈ Ꮽ satisfying d(B,A) ≤ δ n, eachx ∈ K,

and each integeri ≥ q n,

B i x − x A  ≤  B i x − x n+x n − x A  ≤1

n+

1

n <
(3.17)

Completion of the proof of Theorem 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 set Ꮽ \Ᏺnis porous in (Ꮽ,d).

Letn be a natural number Choose a positive number

α < (16n) −12−1 

diam(K) + 1 2

16·8n −1. (3.18) Let

ByProposition 2.1, there existA0∈ Ꮽ and ¯x ∈ K such that

d

A,A0



≤ r

Set

γ =8−1r

diam(K) + 1−1

(3.22) and choose a natural numberq for which

1≤ q 

diam(K) + 1 2

16n ·8r −1 −1

Define ¯A : K → K by

¯

Clearly, the mapping ¯A is continuous and, for each x ∈ K,

Ax¯ − ¯x
=(1− γ)A0x + γ ¯x − ¯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

d¯

A,A0



=sup
Ax¯ − A0x:x ∈ K

=sup

γ¯x − A0x:x ∈ K

≤ γ diam(K) ≤ r

Together with (3.20), this implies that

d( ¯ A,A) ≤ d¯

A,A0



+d

A0,A

≤ r

Assume now that

B, ¯ A

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

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

d(B,A) ≤ d(B, ¯ A) + d( ¯ A,A) ≤ αr + r

4≤ r

Assume that x ∈ K We will show that there exists an integer j ∈[0,q] such that

B j x − ¯x
≤(8n) −1 Assume the contrary Then

B i x − ¯x> (8n) −1, i =0, ,q. (3.31) Let an integeri ∈ {0, ,q −1} By (3.28) and (3.25),

B i+1 x − ¯x  =  B

B i x

− ¯x  ≤  B

B i x

− A¯

B i x+A¯

B i x

− ¯x

≤ d(B, ¯ A) +A¯

B i x

− ¯x  ≤ αr + (1 − γ)B i x − ¯x,

When combined with (3.31), (3.18), and (3.22), this inequality implies that

B i x − ¯x − B i+1 x − ¯x  ≥  B i x − ¯x − αr −(1− γ)B i x − ¯x

= γB i x − ¯x − αr > (8n) −1γ − αr ≥(16n) −1γ, (3.33)

so that

B i x − ¯x − B i+1 x − ¯x  ≥(16n) −1γ. (3.34)

When combined with (1.1), this inequality implies that

diam(K) ≥
x − ¯x
−B q x − ¯x  ≥ q− 1

i=0

B i x − ¯x − B i+1 x − ¯x  ≥ q(16n) −1γ,

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

B j x − ¯x  ≤(8n) −1. (3.36)

It follows from (3.28) and (3.25) that for each integeri ∈ {0, ,q −1},

B i+1 x − ¯x  =  B

B i x

− ¯x  ≤  B

B i x

− A¯

B i x+A¯

B i x

− ¯x

≤ d( ¯ A,B) +A¯

B i x

− ¯x  ≤ αr +B i x − ¯x,

This implies that for each integers satisfying j < s ≤ q,

B s x − ¯x  ≤  B j 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

B q x − ¯x  ≤ αrq + (8n) −1≤(2n) −1. (3.39) Thus, we have shown that the following property holds: for eachB satisfying (3.28) and eachx ∈ K,

B q x − ¯x  ≤(2n) −1, Bx − ¯x  ≤
x − ¯x
+1

(see (3.29)) Thus

B ∈ Ꮽ : d(B, ¯A) ≤ αr

2

⊂Ᏺn ∩
B ∈ Ꮽ : d(B,A) ≤ r

In other words, we have shown that the setᏭ\Ᏺnis porous in (Ꮽ,d) This completes the proof ofTheorem 1.1

Acknowledgments

The work of the first author was partially supported by the Israel Science Foundation founded by the Israel Academy of Sciences and Humanities (Grant 592/00), by the Fund for the Promotion of Research at the Technion, and by the Technion VPR Fund

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

E-mail address:sreich@tx.technion.ac.il

Current address: Department of Mathematics, University of California, Santa Barbara, CA

93106-3080, USA

E-mail address:sreich@math.ucsb.edu

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

32000 Haifa, Israel

E-mail address:ajzasl@tx.technion.ac.il

[1] F S De Blasi and J Myjak, Sur la porosit´e de l’ensemble des contractions...

E-mail address:sreich@tx.technion.ac.il

Current address: Department of Mathematics, University of California, Santa Barbara, CA ...

93106-3080, USA

E-mail address:sreich@math.ucsb.edu

Alexander J Zaslavski: Department of Mathematics, The Technion – Israel