Báo cáo hóa học: "HYPERBOLIC MONOTONICITY IN THE HILBERT BALL" docx

EVA KOPECK ´A AND SIMEON REICHReceived 17 August 2005; Accepted 22 August 2005 We first characterizeρ-monotone mappings on the Hilbert ball by using their resolvents and then study the a

EVA KOPECK ´A AND SIMEON REICH

Received 17 August 2005; Accepted 22 August 2005

We first characterizeρ-monotone mappings on the Hilbert ball by using their resolvents

and then study the asymptotic behavior of compositions and convex combinations of these resolvents

Copyright © 2006 E Kopeck´a and S Reich This is an open access article distributed un-der the Creative Commons Attribution License, which permits unrestricted use, distri-bution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Monotone operator theory has been intensively developed with many applications to Convex and Nonlinear Analysis, Partial Differential Equations, and Optimization In this note we intend to apply the concept of (hyperbolic) monotonicity to Complex Analysis

As we will see, this application involves the generation theory of one-parameter continu-ous semigroups of holomorphic mappings

Let (H, ·,·) be a complex Hilbert space with inner product·,·and norm| · |, and letB:= { x ∈ H : | x | < 1 } be its open unit ball The hyperbolic metric ρ on B × B[5, page 98] is defined by

ρ(x, y) : =argtanh

1− σ(x, y) 1/2

where

σ(x, y) : =



1− | x |2 

1− | y |2 

1−  x, y  2 , x, y ∈ B (1.2)

A mappingg : B → Bis said to beρ-nonexpansive if

ρ

g(x), g(y)

for allx, y ∈ B It is known (see, for instance, [5, page 91]) that each holomorphic self-mapping ofBisρ-nonexpansive.

Hindawi Publishing Corporation

Fixed Point Theory and Applications

Volume 2006, Article ID 78104, Pages 1 15

DOI 10.1155/FPTA/2006/78104

Recall that ifC is a subset of H, then a (single-valued) mapping f : C → H is said to be monotone if

Re

x − y, f (x) − f (y)

Equivalently, f is monotone if

Re

x, f (x)

+

y, f (y)

≥Re

y, f (x)

+

x, f (y)

, x, y ∈ C. (1.5)

It is also not difficult to see that f is monotone if and only if

| x − y | ≤x + r f (x) −

y + r f (y), x, y ∈ C, (1.6) for all (small enough)r > 0.

LetI denote the identity operator A mapping f : C → H is said to satisfy the range condition if

If f is monotone and satisfies the range condition, then the mapping J r:C → C,

well-defined for positiver by J r:=(I + r f ) −1, is called a (nonlinear) resolvent of f It is clearly nonexpansive, that is, 1-Lipschitz:

J r x − J r y  ≤ | x − y |, x, y ∈ C. (1.8)

As a matter of fact, this resolvent is even firmly nonexpansive:

J r x − J r y  ≤  J r x − J r y + s

x − J r x −y − J r y (1.9) for allx and y in C and for all positive s.

This is a direct consequence of (1.6) becausex − J r x = r f (J r x) and y − J r y = r f (J r y)

for allx and y in C We remark in passing that, conversely, each firmly nonexpansive

mapping is a resolvent of a (possibly set-valued) monotone operator To see this, letT :

C → C be firmly nonexpansive Then the operator

M : = [Tx, x − Tx] : x ∈ C

(1.10)

is monotone becauseT satisfies (1.9)

We now turn to the concept of hyperbolic monotonicity which was introduced in [19, page 244]; there it was calledρ-monotonicity In the present paper we will use both terms

interchangeably

We say that a mapping f : B → H is ρ-monotone onBif for each pair of points (x, y) ∈

B × B,

ρ(x, y) ≤ ρ

x + r f (x), y + r f (y)

(1.11) for allr > 0 such that the points x + r f (x) and y + r f (y) belong toB.

We say that f : B → H satisfies the range condition if

If aρ-monotone f satisfies the range condition (1.12), then for eachr > 0, the resolvent

J r:=(I + r f ) −1is a single-valued,ρ-nonexpansive self-mapping ofB As a matter of fact,

this resolvent is firmly nonexpansive of the second kind in the sense of [5, page 129] (see

Lemma 4.2 below) We remark in passing that this resolvent is different from the one introduced in [17] which is firmly nonexpansive of the first kind [5, page 124]

Our first aim in this note is to establish the following characterization ofρ-monotone

mappings Recall that a subset ofBis said to lie strictly insideBif its distance from the boundary ofB(the unit sphere ofH) is positive.

Theorem 1.1 LetBbe the open unit ball in a complex Hilbert space H, and let f : B → H

be a continuous mapping which is bounded on each subset strictly insideB(equivalently, on each ρ-ball) Then f is ρ-monotone if and only if for each r > 0, its resolvent J r:=(I + r f ) −1

is a single-valued, ρ-nonexpansive self-mapping ofB.

This result shows that in some cases the hyperbolic monotonicity off : B → H already

implies the range condition (1.12) This is in analogy with the Euclidean Hilbert space case, where it is known that if f : H → H is continuous and monotone, then the range R(I + r f ) = H for all r > 0 To see this, we may first note that a continuous and

mono-tone f : H → H is maximal monotone and then invoke Minty’s classical theorem [11] to conclude thatR(I + r f ) is indeed all of H for all positive r.

However, as pointed out on [14, page 393], Minty’s theorem is equivalent to the Kirszbraun-Valentine extension theorem which is no longer valid, generally speaking, outside Hilbert space, or for the Hilbert ball of dimension larger than 1 [8,9] On the other hand, it is known [10] that ifE is any Banach space and f : E → E is continuous and

accretive, then f is m-accretive, that is, R(I + r f ) = E for all r > 0.

Our proof ofTheorem 1.1uses finite dimensional projections The separable case is due to Itai Shafrir (see [19, Theorem 2.3]) This proof is presented inSection 3, which also contains a discussion of continuous semigroups of holomorphic mappings and their (infinitesimal) generators (seeCorollary 3.2) It is preceded by three preliminary results

inSection 2 InSection 4, the last section of our note, we study the asymptotic behavior

of compositions and convex combinations of resolvents ofρ-monotone mappings (see

Theorems4.14and4.15).Theorem 4.14, in particular, provides two methods for finding

a common null point of finitely many (continuous)ρ-monotone mappings.

2 Preliminaries

We precede the proof ofTheorem 1.1with the following three preliminary results Givenz ∈ B, let{ u α:α ∈Ꮽ}be a complete orthonormal system inH which contains z/ | z |ifz =0 LetΓ be the set of all finite dimensional subspaces of H which contain z and

are spanned by vectors from{ u α:α ∈Ꮽ}, ordered by containment For eachF ∈Γ, let

P F:H → F be the orthogonal projection of H onto F.

Lemma 2.1 For each y ∈ H, the net { P F y } F ∈Γconverges to y.

Proof Let y =∞ i =1 y, u αi  u αiand let > 0 There is N = N( ) such that ifn ≥ N, then







n

i =1



y, u αi



u αi − y







2

=





∞

i = n+1



y, u αi



u αi







2

=

∞

i = n+1

y, u αi 2

< 2. (2.1)

LetF0:=span{ u α1,u α2, , u αN }

If F ∈ Γ, F ⊃ F0, and F =span{ u α1, , u αN,v1, , v m }, then | P F y − y |2= |N

i =1 y,

u αi  u αi+m

j =1 y, v j  v j − y |2 If  y, v j  =0, then v j ∈ { u αi: ≥ N + 1 } and therefore

Next, we recall a characterization ([19, Theorem 2.1]) of ρ-monotone mappings in

terms of the inner product ofH.

Proposition 2.2 A mapping f : B → H is ρ-monotone if and only if for each x, y ∈ B ,

Re

x, f (x)

1− | x |2 +Re



y, f (y)

1− | y |2 ≥Re



y, f (x)

+

x, f (y)

1−  x, y 



Note that (2.2) is the hyperbolic analog of the Euclidean (1.5)

Finally, we recall a fixed point theorem which will be used in the proof ofTheorem 1.1 LetC be a subset of a vector space E and let the point x belong to C Recall that the

inward setI C(x) of x with respect to C is defined by

I C(x) : = z ∈ E : z = x + a(y − x) for some y ∈ C, a ≥0

IfE is a topological vector space, then a mapping f : C → E is said to be weakly inward

if f (x) belongs to the closure of I C(x) for each x ∈ C.

Theorem 2.3 Let C be a nonempty, compact and convex subset of a locally convex, Haus-dor ff topological vector space E If a continuous f : C → E is weakly inward, then it has a fixed point.

This theorem is due to Halpern and Bergman [6] A simple proof can be found in [13]

3 The range condition

We begin this section with the proof ofTheorem 1.1

Proof of Theorem 1.1 One direction is clear: if J risρ-nonexpansive, and the points x, y,

x + r f (x), y + r f (y) all belong toB, then

ρ(x, y) = ρ

J r

x + r f (x)

,J r

y + r f (y)

≤ ρ

x + r f (x), y + r f (y)

Thus, it is enough to prove that if f is ρ-monotone, then for each z ∈ Bandr > 0, there

exists a solutionx ∈ Bto the equationx + r f (x) = z Fix z ∈ Band consider the corre-sponding directed setΓ of finite dimensional subspaces of H.

For eachF ∈Γ, letBF:= B ∩ F and denote the composition P F ◦ f by f F The (re-stricted) mapping f F:BF → F is also ρ-monotone because when x, y ∈ B F, we have

 x, P F f (x)  =  x, f (x)  and  y, P F f (x)  =  y, f (x) , and f F is seen to be ρ-monotone

by the characterization (2.2)

Now we want to show that there is a pointw F ∈ B Fsuch that

w F+r f F

w F

Indeed, consider the mappingh :BF → F defined by

h F(x) : = z − r f F(x), x ∈ B F (3.3) Using (2.2) withy =0, we get

Re

f F(x), x

≥1− | x |2 

Re

x, f F(0)

(3.4) for allx ∈ B F Hence

Re

h F(x), x

=Re z, x  − r Re

f F(x), x

≤ | z || x | − r

1− | x |2 

Re

x, f F(0)

. (3.5) Since| f F(0)| = | P F f (0) | ≤ | f (0) |, it follows that there is| z | < s < 1 (independent of F)

such that Re h F(x), x  ≤ | x |2for allx ∈ F with | x | = s Thus h Fis weakly inward on{ x ∈

F : | x | ≤ s }by [12, Proposition 2] (alternatively, it satisfies the Leray-Schauder condition

on{ x ∈ F : | x | = s }) and therefore has a fixed point byTheorem 2.3 This fixed point

w F ∈ B F(0,s) ⊂ B(0,s) is a solution of (3.2)

Let{ v E:E ∈Δ}be a subnet of{ w F:F ∈Γ}which converges weakly tov ∈ B(0,s).

We can assume that{| v E |} E ∈Δconverges tot, with | v | ≤ t ≤ s < 1 Since f is bounded on

B(0,s), we can also assume that { f (v E)} E ∈Δconverges weakly top ∈ H.

Our next claim is that| v | = t.

To see this, note first that



v E,y +

rg E

v E ,y

for allE ∈ Δ and y ∈ H, where { g E } E ∈Δis a subnet of{ f F } F ∈Γ

Also, ifϕ :Δ→Γ is the mapping associated with the subnet{ v E:E ∈Δ}, theng E =

f ϕ(E) and  g E(v E),y  =  f ϕ(E)(v E),y  =  P ϕ(E) f (v E),y  =  f (v E),P ϕ(E) y  =  f (v E),y +

 f (v E),P ϕ(E) y − y  →  p, y because{ f (v E)} E ∈Δis bounded and{ P E y } E ∈Δconverges to

y byLemma 2.1 Hence v, y +r  p, y  =  z, y for ally ∈ H, and v + r p = z.

Writing (2.2) withx : = v and y : = v E, we see that

Re

v, f (v)

/

1− | v |2 

+

v E, 

v E

/

1−v E 2 

≥Re v, f

v E

+

f (v), v E

/

1−v, v E

.

(3.7)

Also, v E,v E +r  g E(v E),v E  =  z, v E  Hence (lettingQ F = I − P F),



v E, 

v E



=w ϕ(E),P ϕ(E) f

v E

 +Q ϕ(E) f

v E



=w ϕ(E),P ϕ(E) f

v E



=v E,g E

v E

=

v E,z

−v E 2 

/r,

Rer

v E, 

v E



=Re

v E,z

−v E 2 

.

(3.8)

Re

r

v, f (v)

/

1− | v |2 

+

v E,z

−v E 2 

/

1−v E 2 

≥Re r

v, f

v E

+r

f (v), v E

/

1−v, v E

Taking limits, we get

Re r

v, f (v)

/

1− | v |2 

+

 v, z  − t2 

/

1− t2 

≥Re r  v, p +r

f (v), v

/

1− | v |2 

Now v, v +r  p, v  =  z, v  Therefore,

Re v, z  /

1− t2 

− t2

1− t2 ≥Re

 z, v  − | v |2

1− | v |2

 ,

Re v, z 

 1

1− t2− 1

1− | v |2



≥ t2

1− t2− | v |2

1− | v |2.

(3.11)

If| v | < t, then this inequality yields Re  v, z  ≥1 But Re v, z  ≤ | v || z | ≤ t ≤ s < 1, a

con-tradiction Hence| v | = t, as claimed.

Since{ v E } E ∈Δconverges weakly tov and {| v E |} E ∈Δconverges tot = | v |,{ v E } E ∈Δ con-verges strongly tov Since f is continuous, f (v E)→ f (v) and p = f (v) Hence v + r f (v) =

Why is it important to know that in certain cases aρ-monotone mapping already

sat-isfies the range condition? To answer this question, letD be a domain (open, connected

subset) in a complex Banach spaceX, and recall that a holomorphic mapping f : D → X

is said to be a semi-complete vector field onD if the Cauchy problem

∂u(t, z)

∂t +f



u(t, z)

=0

u(0, z) = z

(3.12)

has a unique global solution{ u(t, z) : t ≥0} ⊂ D for each z ∈ D It is known (see, e.g.,

[1,18]) that if a holomorphic f : D → X is semi-complete, then the family S f = { F t } t ≥0

defined by

F t(z) : = u(t, z), t ≥0,z ∈ D, (3.13)

is a one-parameter (nonlinear) semigroup (semiflow) of holomorphic self-mappings of

D, that is,

F t+s = F t ◦ F s, t, s ≥0,

whereI denotes the restriction of the identity operator on X to D In addition,

lim

uniformly on each ball which is strictly insideD.

A semigroup{ F t } t ≥0is said to be generated if, for eachz ∈ D, there exists the strong

limit

f (z) : =lim

t →0 +



z − F t(z)

This mapping f is called the (infinitesimal) generator of the semigroup It is, of course,

a semi-complete vector field Analogous definitions apply to (continuous) semigroups of

ρ-nonexpansive mappings, where ρ is a pseudometric assigned to D by a Schwarz-Pick

system [5, page 91]

When is a mapping f : D → X a generator? An answer to this question is provided

by the following result [19, page 239] Recall that ifD is a convex domain, then all the

pseudometrics assigned to D by Schwarz-Pick systems coincide If D is also bounded,

then this common pseudometric is, in fact, a metric, which we call the hyperbolic metric

ofD.

Theorem 3.1 Let D be a bounded convex domain in a complex Banach space X, and let ρ denote its hyperbolic metric Suppose that f : D → X is bounded and uniformly continuous

on each ρ-ball in D Then f is a generator of a ρ-nonexpansive semigroup on D if and only if, for each r > 0, the mapping J r:=(I + r f ) −1is a well-defined ρ-nonexpansive self-mapping

of D.

If, in the setting of this theorem, f : D → X is a generator of a ρ-nonexpansive

semi-group{ F t } t ≥0, then the following exponential formula holds:

F t(z) =lim

n →∞



I + t

n f

− n

Combining Theorems1.1and3.1, we obtain the following corollary

Corollary 3.2 Let f : B → H be bounded and uniformly continuous on each ρ-ball inB Then f is the generator of a ρ-nonexpansive semigroup onBif and only if f is ρ-monotone.

If follows from the Cauchy inequalities that this corollary applies, in particular, to holomorphic mappings which are bounded on eachρ-ball.

Note that all the mappings of the form f = I − T, where I is the identity operator

andT : B → Bisρ-nonexpansive (in particular, holomorphic), are generators of

semi-groups ofρ-nonexpansive (resp., holomorphic) mappings More applications of

hyper-bolic monotonicity and, in particular, of the characterizations provided byProposition 2.2andCorollary 3.2, can be found in [2]

4 Asymptotic behavior

In this section we study the asymptotic behavior of compositions and convex combina-tions of resolvents ofρ-monotone mappings.

Consider the functionψ : [0, δ] →[0,∞) defined by

wherex, y, u and v are any four points inBandδ > 0 is sufficiently small We begin by recalling [19, Lemma 2.2] Note thatψ is differentiable at the origin byLemma 2.1there See also [20, Proposition 4.3]

Lemma 4.1 Let the function ψ be defined by ( 4.1 ) Then the following are equivalent:

(a)ψ(t) ≤ ψ(0), 0 ≤ t ≤ δ;

(b)ψ decreases on [0, δ];

(c)ψ (0)≤ 0.

LetD be a subset of the Hilbert ballB Recall that a mappingT : D → Bis said to

be firmly nonexpansive of the second kind [5, page 129] if the functionϕ : [0, 1] →[0,∞) defined by

ϕ(s) : = ρ

(1− s)x + sTx, (1 − s)y + sT y

is decreasing for all pointsx and y in D.

We denote the family of firmly nonexpansive mappings of the second kind byFN2

Lemma 4.2 Any resolvent of a ρ-monotone mapping is firmly nonexpansive of the second kind.

Proof Fix a positive r and let J rbe a resolvent of aρ-monotone mapping f : B → H Let

x and y be any two points in the domain of J r To show that the functionρ(tx + (1 −

t)J r x, t y + (1 − t)J r y) increases on [0, 1], we have to show that the function ψ : [0, 1] →

[0,∞) defined by

ψ(t) : = σ

J r x + t

x − J r x ,J r y + t

y − J r y

decreases on [0, 1] To this end, it suffices, according toLemma 4.1, to check thatψ(t) ≤

ψ(0) for all 0 ≤ t ≤1

Indeed, since f is ρ-monotone, x − J r x = r f (J r x), and y − J r y = r f (J r y), we know

that, by (1.11),

ρ

J r x, J r y

≤ ρ

J r x + s f

J r x ,J r y + s f

J r x

= ρ

J r x + (s/r)

x − J r x ,J r y + (s/r)

y − J r y (4.4) for all 0≤ s ≤ r In other words,

ψ(0) = σ

J r x, J r y

≥ σ

J r x + t

x − J r x ,J r y + t

y − J r y

We now turn to the class of strongly nonexpansive mappings

LetT : D → Bbe aρ-nonexpansive mapping with a nonempty fixed point set F(T) Re-call that such a mapping is Re-called strongly nonexpansive ([4,16]) if for anyρ-bounded

se-quence{ x n:n =1, 2, 3, } ⊂ D and every y ∈ F(T), the condition ρ(x n,y) − ρ(Tx n,y) →

0 implies thatρ(x n,Tx n)→0

To define this concept for fixed point free mappings, we first recall two notations

If the pointb belongs to the boundary ofB, let the functionϕ b:B →(0,∞) be defined by

ϕ b(x) : =1−  x, b  2

/

1− | x |2 

and for positiver consider the ellipsoids E(b, r) : = { x ∈ B:ϕ b(x) < r }

Now we recall [5, page 126] that if aρ-nonexpansive mapping T : B → Bis fixed point free, then there exists a unique pointe = e(T) of norm one (the sink point of T) such

that all the ellipsoidsE(e, r), r > 0, are invariant under T We say that such a mapping

is strongly nonexpansive if for any sequence { x n:n =1, 2, } ⊂ Bsuch that{ ϕ e(x n)}is bounded, the conditionϕ e(x n)− ϕ e(Tx n)→0 implies thatx n − Tx n →0

Proofs of the following two lemmas can be found in [15]

Lemma 4.3 Let { x n } and { z n } be two sequences inB Suppose that for some y inB,

lim supn →∞ ρ(x n,y) ≤ M, lim sup n →∞ ρ(z n,y) ≤ M, and lim inf n →∞ ρ((x n+z n)/2, y) ≥ M Then lim n →∞ | x n − z n | = 0.

Lemma 4.4 Let the point b belong to the boundary ofB, and let { x n } and { z n } be two sequences in B Suppose that lim sup n →∞ ϕ b(x n)≤ M, lim sup n →∞ ϕ b( n)≤ M, and

lim infn →∞ ϕ b((x n+z n)/2) ≥ M Then lim n →∞ | x n − z n | = 0.

Our interest in strongly nonexpansive mappings stems from the following two facts

Lemma 4.5 If a mapping T ∈ FN2has a fixed point, then it is strongly nonexpansive Proof Suppose that the sequence { x n }isρ-bounded, y ∈ F(T), and ρ(x n,y) − ρ(Tx n,

y) →0 In order to prove thatρ(x n,Tx n)→0, we may assume without loss of generality that limn →∞ ρ(x n,y) =limn →∞ ρ(Tx n,y) = d > 0 Since T ∈ FN2, we also have

ρ

Tx n,y

≤ ρ

x n+Tx n

/2, y

≤ ρ

x n,y

Hence limn →∞ ρ((x n+Tx n)/2, y) = d, too Now we can invoke Lemma 4.3to conclude thatx n − Tx n →0 Since{ x n }isρ-bounded, it follows that ρ(x n,Tx n)→0 as well 

Lemma 4.6 If a mapping T : B → B belongs to FN2and is fixed point free, then it is strongly nonexpansive.

Proof Let e be the sink point of T and let { x n:n =1, 2, } ⊂ Bbe a sequence such that

{ ϕ e(x n)}is bounded andϕ e(x n)− ϕ e(Tx n)→0 In order to prove thatx n − Tx n →0, we may assume thatϕ e(x n)→ M Hence ϕ e(Tx n)→ M, too Since T ∈ FN2, we know by [5, Lemma 30.7 on page 142] that the functiong : [0, 1] →(0,∞) defined by

g(s) : = ϕ e

(1− s)x + sTx

is decreasing for eachx ∈ B Hence

ϕ e

Tx n

≤ ϕ e

x

n+Tx n

2



≤ ϕ e

x n

(4.9) for eachn =1, 2, .

Thus limn →∞ ϕ e((x n+Tx n)/2) = M, too, and hence lim n →∞(x n − Tx n)=0 byLemma

Next, we recall [16] the following weak convergence result

Proposition 4.7 If T : B → B has a fixed point and is strongly nonexpansive, then for each point x inB, the sequence of iterates { T n x } converges weakly to a fixed point of T.

In view ofLemma 4.5, this result applies, in particular, to all those mappingsT : B → B

inFN2which have a fixed point

It follows from [8,9] that in the setting ofProposition 4.7, strong convergence does not hold in general However, our next result shows that when a strongly nonexpansive mapping is fixed point free, its iterates do converge strongly

Proposition 4.8 If T : B → B is strongly nonexpansive and fixed point free, then for each point x inB, the sequence of iterates { T n x } converges strongly to the sink point of T Proof Let e be the sink point of T and denote T n x by x n,n =1, 2, Since ϕ e(Tx) ≤

ϕ e(x) for all x ∈ B, the sequences{ ϕ e(x n)} and{ ϕ e(Tx n)}decrease to the same limit

M Since T is strongly nonexpansive, it follows that x n − Tx n →0 SinceT is fixed point

free, this implies that{ x n }cannot have aρ-bounded subsequence Thus lim n →∞ | x n | =1,

Now we consider compositions and convex combinations of strongly nonexpansive mappings

The following result is proved in [16]

Lemma 4.9 Let the mappings T j:B → B , 1 ≤ j ≤ m, be strongly nonexpansive, and let

T = T m T m −1··· T1 If F = ∩{ F(T j) : 1≤ j ≤ m } is not empty, then F = F(T) and T is also strongly nonexpansive.

Here is an analog of this result for the fixed point free case

Lemma 4.10 If the fixed point free mappings T j:B → B , 1 ≤ j ≤ m, have a common sink point and are strongly nonexpansive, then T = T m T m −1··· T1is also strongly nonexpansive Proof Let T1andT2be two fixed point free and strongly nonexpansive mappings with

a common sink pointe = e(T1)= e(T2) We first note that the compositionT = T2T1is also fixed point free Indeed, letx ∈ B and consider the iteratesx n = T n x, n =1, 2, .

Since the decreasing sequence{ ϕ e(x n)}converges, we see that

0≤ ϕ e

x n

− ϕ e

T1x n

≤ ϕ e

x n

− ϕ e

Tx n

and thereforex n − T1x n →0

If{ x n }wereρ-bounded, then its asymptotic center [5, page 116] would be a fixed point

ofT1 Hence{ x n }isρ-unbounded and T is fixed point free, as claimed Thus e = e(T)

is also the sink point ofT To show that T is strongly nonexpansive, let { x n } ⊂ Bbe a

Finally, we recall a fixed point theorem which will be used in the proof ofTheorem 1.1 LetC be a subset of a vector space E and let the. ..

Combining Theorems1.1and3.1, we obtain the following corollary

Corollary 3.2 Let f : B → H be bounded and uniformly continuous on each ρ-ball in< /i>B Then... aρ-nonexpansive mapping T : B → Bis fixed point free, then there exists a unique pointe = e(T) of norm one (the sink point of T) such

that all the ellipsoidsE(e,