Geometric properties of Banach spaces and metric fixed point theory

E extracta mathematicae Vol. 17, N´um. 3, 331 – 349 (2002) IV Curso Espacios de Banach y Operadores. Laredo, Agosto de 2001. Geometric Properties of Banach Spaces and Metric Fixed Point Theory Tom´ as Dom´ ınguez Benavides Facultad de Matem´aticas, P.O. Box 1160, 41080 Sevilla, Spain e-mail: [email protected] AMS Subject Class. (2000): 47H09, 47H10 1. Introduction Let Ta mapping from a set Xinto itself. The mapping Thas a fixed point if there exists x0∈Xsuch that Tx0=x0. The most known fixed point theorem is the Contraction Mapping Principle, due to S. Banach [3]. The classical statement is the following: Theorem. (Banach Fixed Point Theorem [3]) Let Xbe a complete metric space and T:X→Xa contractive mapping, i.e. there exists k∈[0,1) such that d(Tx, Ty)≤kd(x, y) for every x, y ∈X. Then Thas a (unique) fixed point xω. Furthermore xω= limnTnx0for any x0∈X. The proof of the Banach Theorem is rarely simple (probably because the wide diffusion of Banach techniques). Indeed, let x0be an arbitrary point and set xn=Tnx0. We have d(xn+1, xn)≤kd(xn, xn−1)≤... ≤knd(x1, x0). Thus, {xn}is a Cauchy sequence, because d(xn+j, xn)≤d(xn+j, xn+j−1) + ... +d(xn+1, xn) ≤(kn+j−1+... +kn)d(x1, x0)≤kn+j−1 1−k< ² 331 332 t. dom´ ınguez if nis large enough. Obviously, the fixed point is unique, because if x=Tx, y=Ty we have d(x, y) = d(Tx, Ty)≤kd(x, y) which implies d(x, y) = 0. Banach Theorem has proved to be very useful to solve different theoretical and practical problems. For instance, it is used to prove the Picard-Lindel¨of Theorem about existence of solution of differential equations and to prove the Inverse (or Implicit) Function Theorem in infinite dimensional spaces. A big number of generalizations of the Banach Theorem have appeared in the literature, trying to weak some assumptions (see, for instance, [11, Cap. 3]). However, the most natural improvement would be to let the constant kattain the value 1, but in this case the result is not more true. Indeed, a translation in Rsatisfies the assumptions and it is fixed point free. Even the mild assumption d(Tx, Ty)< d(x, y) does not assure the existence of a fixed point. Indeed, consider X= [1,∞) and Tx =x+ 1/x. We have d(Tx, Ty) = ¯¯¯¯ x+1 x−y−1 y¯¯¯¯ =|y−x|−¯¯¯¯ 1 x−1 y¯¯¯¯ <|y−x|. However Tis fixed point free. The fail of the existence of fixed point in this example occurs because, in addition, the interval [1,∞) is unbounded. Indeed, otherwise we could apply Brouwer’s Theorem: Theorem. (Brouwer’s Theorem (1912) [5]) Let Ma bounded convex closed subset of Rnand T:M→Ma continuous mapping. Then Thas a fixed point. Having in mind both Banach and Brouwer Theorems, a question seems to be natural: Assume that Mis a convex, closed bounded subset of an arbitrary Banach space and T:M→Mis a nonexpansive mapping, i.e., kTx −Tyk ≤ kx−ykfor every x, y ∈M. Does Thas a fixed point? We must realize that the assumption on the space has been weakened (the dimension is not necessarily finite) but stronger conditions are assumed on the mapping (nonexpansiveness instead of continuity). The answer in again negative: Example. (Kakutani (1943) [9]). Let Bthe unit ball in c0and T:B→B defined by T(x1, x2, ...) = (1 − kxk, x1, x2, ...). It is easy to check that Tis nonexpansive and fixed point free. As a consequence of these facts, the problems about existence of fixed point for nonexpansive mappings were relegated. However in 1965, two surprising theorems appeared: geometric properties of banach spaces 333 Theorem. (Browder’s Theorem [6]) Let Cbe a convex bounded closed subset of a uniformly convex Banach space (a preliminary version was given for Hilbert spaces) and T:C→Ca nonexpansive mapping. Then Thas a fixed point. Theorem. (Kirk’s Theorem [10]) Let Cbe a convex bounded closed subset of a reflexive Banach space with normal structure. If T:C→C is nonexpansive, then Thas a fixed point. In the second section we will include a proof of the latter theorem (which includes Browder’s Theorem). It is noteworthy that these results state a bridge between notions which had usually been considered in Linear Functional Analysis (uniform convexity, reflexivity, normal structure, etc) and problems about existence of fixed point for nonlinear operators. From this starting point a big number of fixed point results have been obtained for different classes of mappings using geometric properties of Banach spaces. In forthcoming sections we will review some of these results. 2. Normal structure. Existence of fixed points for nonexpansive operators We recall some definitions yielding to the notion of normal structure. Definition 1. Let Xbe a Banach space, Aa bounded subset of Xand Ban arbitrary subset of X. The Chebyshev radius of Awith respect to Bis defined by r(A, B) = inf{sup{kx−yk:x∈A}:y∈B} where we write r(A) instead of r(A, co (A)). The Chebyshev center of Awith respect to Bis defined by Z(A, B) = {y∈B: sup{kx−yk:x∈A}=r(A, B)} where we write Z(A) instead of Z(A, co (A)). Remark 1. Roughly speaking, we can say that the Chebyshev radius r(A, B) is the radius of the smallest ball centered at a point in Band covering the set A, the Chebyshev center Z(A, B) being the set formed by all centers of these smallest balls. However, since the infimum appearing in the definition 334 t. dom´ ınguez is not, necessarily attained, the set Z(A, B) can be empty. In opposition, if for every ε > 0 we consider the set Zε(A, B) = {y∈B:r(A, y)≤r(A, B) + ε}, then Zε(A, B) is a nonempty, convex, bounded and closed set if Bsatisfies the same properties. Thus, Zε(A, B) is convex, nonempty and weakly compact if so is B. Since \ ε>0 Zε(A, B) = Z(A, B), the finite intersection property implies that Z(A, B) is nonempty when Bis a convex and weakly compact set. Definition 2. A bounded convex closed subset Aof a Banach space X is said to be diametral if diam (A) = r(A). Equivalently, if Z(A) = A. We say that a Banach space Xhas normal structure (respectively weak normal structure) if every convex closed nonempty (respectively convex weakly compact ) diametral subset of Xis a singleton. Remark 2. According to the above definition, a Banach space has normal structure if every convex set Awhich is not a singleton can be covered by a ball whose radius is less than the diameter of Aand centered at a point in A. We could think that this is the case of every Banach space, and in fact, this occurs for every uniformly convex space (we will see the definition in the next section), for instance, `pand Lp(Ω), 1 < p < +∞. However the sequence space c0fails to have both normal structure and weak normal structure. Indeed, consider the set A= co ({en:n∈N}) where{en}is the standard basis. We have diam (A) = 1 and r(A) = 1 because limn→∞ kx−enk= 1 for every x∈c0. Furthermore, since the sequence {en}is weakly null, Ais a weakly compact set. The same set can be considered in the sequence space `1, giving us that `1fails to have normal structure either. However, we will show in the next section that `1(and any Banach space with the Schur property) has weak normal structure. Theorem 1. Let Xbe a Banach space with weak normal structure, Ca convex weakly compact subset of Xand T:C→Ca nonexpansive mapping. Then Thas a fixed point. Proof. Let Bbe the collection of all convex weakly compact subsets of Cwhich are T-invariant. It is easy to check that B, ordered by inclusion is geometric properties of banach spaces 335 an inductive family. Zorn’s Lemma assures the existence of a minimal set K. Since T(K)⊂Kwe have co (T(K)) ⊂K. Thus, co (T(K)) is a convex weakly compact subset of Kwhich is T−invariant.The minimality of K implies K= co (T(K)). Since Kis a weakly compact convex set, we know from Remark 1 that Z(K) is a nonempty convex weakly compact set. We will prove that Z(K) is T-invariant. Indeed, take x∈Z(K), i.e. r(K, x) = r(K). For every y∈Kwe have kTy−Txk ≤ ky−xk ≤ r(K). Hence T(K) is covered by the closed ball B(Tx, r(K)) which implies co (T(K)) = K⊂B(Tx, r(K)). Therefore r(K, Tx)≤r(K) which implies Tx ∈Z(K). The minimality of Kimplies Z(K) = Kand thus diam (K) = 0 because Xhas weak normal structure. Hence Kis a singleton and contains a fixed point of T. 3. Geometric properties which imply normal structure In order to study some geometric properties implying normal structure we recall a geometric coefficient defined by Bynum [7] in 1980 (a preliminary form had been studied by por J¨ung in 1901). Definition 3. Let Xbe a Banach space. The normal structure coefficient of Xis defined by N(X) = inf ndiam (A) r(A):A⊂Xconvex closed and bounded with diam (A)>0o. It is clear that Xhas normal structure if N(X)>1. However spaces with normal structure exist which satisfy N(X) = 1. This coefficient can be considered as a measure of the “worst” possible relationship between the diameter and the Chebyshev radius of a subset of X. For instance, in the euclidean plane `2 2this “worst” relationship is attained at the equilateral triangle and its value is √3. In the tridimensional euclidean space `3 2it is attained at the tetrahedron with a value equal to 2√2/√3 and, in general, for `n 2the worst value corresponds to the “hipertetrahedron” with a value equal to √2p(n+ 1)/n. It is not easy to evaluate N(X) for a determined space X, and, in fact, its value is unknown in many cases. The following connection between the value of N(X) and the reflexivity of the space is important: Theorem 2. ([12]) Let Xbe a Banach space such that N(X)>1. Then Xis reflexive. 336 t. dom´ ınguez Proof. If Xis not reflexive, for every ε > 0 there exists a sequence {xn} (see [13]) such that 1−ε≤ ku1,n−un,ωk ≤ 1+εfor every u1,n ∈co ({xj}1≤j≤n), un,ω ∈co ({xj}j>n), and for each n. Thus diam ({xn})≤1 + ε. Furthermore, if vbelongs to co ({xn}) and nis large enough we have kxn−vk ≥ 1−ε/2. Since εis arbitrary, we obtain N(X) = 1. The normal structure coefficient is useful to study the stability of the normal structure under renorming. Recall that if Xand Yare isomorphic Banach spaces, the Banach-Mazur distance between Xand Yis defined by d(X, Y ) = inf ©kTkkT−1k:T∈Isom (X, Y )ª. Clearly, d(X, Y ) = 1 when Xand Yare isometric spaces. Theorem 3. Let Xand Ybe isomorphic Banach spaces. Then N(X)≤d(X, Y )N(Y). Proof. Let Cbe a bounded convex closed subset of Y. If U:Y→Xis an isomorphism we have r(C)≤ kU−1kr(U(C)) ≤ kU−1kdiam(U(C))/N(X) ≤ kU−1kkUkdiam(C)/N(X). Thus r(C)≤d(X, Y ) diam(C)/N(X) and this inequality implies the result. The first geometric property, related to normal structure, which will be considered is the uniform convexity. Let us recall that a Banach space is said to be strictly convex if the unit sphere does not contain a segment. Equivalently: Definition 4. We say that a Banach space Xis strictly convex if for every vectors xand yin Xwhich are not collinear, we have kx+yk<kxk+kyk. A stronger notion appears if we assume this property in a uniform sense, that is, roughly speaking, assuming that there is no segment with a predetermined length as close to the unit sphere as wanted. geometric properties of banach spaces 337 Definition 5. We say that a Banach space Xis uniformly convex, if for every ε∈(0,2] there exists δ > 0 such that for x,y∈Xwith kx−yk ≥ ε x, y ∈B(0,1))⇒1−° ° ° ° x+y 2° ° ° ° > δ. Example 1. Hilbert spaces are uniformly convex as a consequence of the parallelogram identity. Indeed, if x, y ∈B(0,1) and kx−yk ≥ ε, we have ° ° ° ° x+y 2° ° ° °≤r1−³ε 2´2. Considering δ=q1−(ε/2)2we obtain the uniform convexity of the space. To prove that `pspaces are uniformly convex is more technical (see, for instance, [4]). On the other hand, recall that a Banach space Xis finitely representable in another Banach space Yif for every finite dimensional subspace Eof X and every ε > 0 there exists a subspace Fof Ysuch that d(E, F)<1 + ε. It is not difficult to prove that Lp(Ω) is finitely representable in `p. Indeed, if Eis an n-dimensional subspace of Lp(Ω) and {f1, f2, ..., fn}is a normalized basis of E, with basic constant c, for every ε > 0 we can find simple functions {s1, s2, ..., sn}such that kfk−skk< ε/nc(2 + ε) for k= 1, ..., n. If f=Pn k=1 akfk, we define Tf =Pn k=1 aksk. Then Tis an isomorphism from Eonto span{s1, ..., sn}and kTkkT−1k<1 + ε. Since span{s1, ..., sn} can be isometrically embedded in `p(by discretization of the measure), there exists a subspace Fof `psuch that d(E, F)<1 + ε. Since the definition of the uniform convexity only depends on 2-dimensional subspaces and Lp([0,1]) contains isometrically to `p, we can assure that Lp([0,1]) spaces are uniformly convex for the same choice of δas in `p. We will need a measure of the uniform convexity of the space: Definition 6. Let Xbe a Banach space. The modulus of convexity of X,δX(ε), is defined by δX(ε) = inf ½1−° ° ° ° x+y 2° ° ° ° :x, y ∈B(0,1),kx−yk ≥ ε¾. Theorem 4. Let Xbe a Banach space with modulus of convexity δX. Then N(X)≥(1 −δX(1))−1. 338 t. dom´ ınguez Proof. Let Abe a closed convex bounded subset of Xwhich is not a singleton and choose ε > 0. Denote d= diam (A) and r=r(A). Choose x and yin Asuch that kx−yk ≥ d−ε. Write w= (x+y)/2, and choose zin Asuch that kz−wk ≥ r−ε. Since k(z−x)/dk ≤ 1,k(z−y)/dk ≤ 1 and k(z−x)/d −(z−y)/dk>(d−ε)/d from the definition of δXwe obtain kz−wk ≤ dµ1−δXµd−ε d¶¶. Thus r≤ε+dµ1−δXµd−ε d¶¶ which implies the result using the continuity of the norm. Remark 3. Notice that uniform convexity implies normal structure, but this property is also shared by every space such that segments with length equal to 1 are separated from the unit sphere. On the other hand, it is well known that uniformly convex spaces are reflexive. From theorems 2 and 4 a stronger result is obtained: the condition δX(1) >0 implies reflexivity. Next we will review another coefficient which implies weak normal structure. We will use the notion of asymptotically equilateral sequence Definition 7. Let Xbe a metric space. A sequence {xn}in Xis said to be asymptotically equilateral if limn,m ;n6=md(xn, xm) exists, i.e., a number d exists such that for every ε > 0 there is a nonnegative integer n0such that d−ε < d(xn, xm)< d +εif n, m > n0and n6=m. To study the existence of asymptotically equilateral sequence we will use a simple version of Ramsey Lemma. Let us fix the notation: By Nwe denote the set of nonnegative integers, [N] the collection of its infinite subsets and for every set Cin [N], [C]2will denote the set formed by all ordered pair formed with numbers in C. Lemma. (Ramsey Lemma) Let f: [N]2→ {1,2}be a function. Then, there exists C∈[N]such that the restriction of fto [C]2is a constant. Theorem 5. Let {xn}be a bounded sequence in a metric space. Then {xn}contains an asymptotically equilateral subsequence. geometric properties of banach spaces 339 Proof. For every subsequence {yn}of {xn}we denote φ({yn}) = inf{ε > 0 : {yn}can be covered by finitely many sets with diameter ≤ε}. Claim.“There exists a subsequence {yn}of {xn}such that φ({zn}) = φ({yn}) for every subsequence {zn}of {yn}.” To prove the claim, define by induction {z0 n}={xn}and φm+1 = inf{φ({zn}) : {zn}subsequence of {zm n}}. Let {zm+1 n}be a subsequence of {zm n}such that φ({zm+1 n)< φm+1 +1 m+ 1. Consider the diagonal subsequence {zn n}. We will show that this sequence satisfies the required condition. Since {zn n}is a subsequence of {zm n}for n > m we have φ({zn n})≤φ({zm n}) for each m. Assume that {zn}is a subsequence of {zn n}. Hence {zn}is a subsequence of {zm n}for n > m. Thus φ({zn n})≤φ({zm n})< φm+1 m≤φ({zn}) + 1 m. Since mis arbitrary we obtain φ({zn n})≤φ({zn})≤φ({zn n}) and the claim is proved. Choose now an arbitrary ε > 0 and a subsequence {yn}of {xn}satisfying the property in the claim. Taking a subsequence (which “a fortiori” satisfies the same property) we can assume φ({yn})+ε≥ kyn−ymk for every n, m. Define the following function from [N]2into {1,2}:f(n, m) = 1 if kyn−ymk> φ({yn})−εand f(n, m) = 2 if kyn−ymk ≤ φ({yn})−ε. By Ramsey’s Lemma, there exists a subsequence {zn}of {yn}satisfying either kzn−zmk> φ({yn})−εfor every n, m ;n6=mor kzn−zmk ≤ φ({yn})−ε for every n, m. Since the second possibility is a contradiction according to the property satisfied by {yn}, we deduce that the first possibility always holds and we have φ({yn})−ε≤ kzn−zmk< φ({yn}) + ε for every n, m ;n6=m. Choosing ε= 1/n we can conclude the proof by a diagonal argument. To introduce a coefficient for weak normal structure we need some previous definitions. 346 t. dom´ ınguez where x+and x−are vectors whose coordinates are (x+)i= max{xi,0}=|xi|+xi 2 (x−)i= max{−xi,0}=|xi|−xi 2. It is not difficult to prove that WCS(`2,k·k2,1) = √2 (see [2, Theorem VI.3.11]) and so this space has normal structure. Furthermore, its conjugate space is (`2,k·k2,∞). We will show that this space fails to have normal structure. Indeed, the standard basic sequence is a diametral sequence because ken−emkp,∞= 1 if n6=mand for every point u=Pn i=1 αiei, αi≥ 0,Pn i=1 αi= 1 we have ken+1 −ukp p,∞= sup{1,Pαp i}= 1. Remark 6. Many other geometric properties of Banach spaces implying normal structure can be found in the books [8] and [2, Chapter VI]. 4. Existence of fixed point in absence of normal structure Until now, we have studied the existence of fixed point for nonexpansive mappings as a consequence of the normal structure. In this section we will show some geometric properties which imply the existence of fixed points in absence of normal structure. The most important case corresponds to the space c0which fails to have weak normal structure. However, we will see that Theorem 1 still holds for this space. We first recall a “classical” result in Metric Fixed Point Theory. Lemma. (Goebel-Karlovitz Lemma) Let Kbe a convex weakly compact subset of a Banach space X, and T:K→Ka nonexpansive mapping. Assume that Kis minimal with these properties and let {xn}be an approximated fixed point sequence for Tin K, i.e. limnkxn−Txnk= 0. Then lim n→∞ ky−xnk= diam (K) for every y∈K. Proof. From the proof of Theorem 1 we know that K=Z(K) which implies that Kis a diametral set. We will prove that Za({xn}, K) = K. Let Za,ε({xn}, K) = {y∈K: lim sup n→∞ kxn−yk ≤ ra({xn}, K) + ε}. geometric properties of banach spaces 347 It is easy to check that Za,ε({xn}, K) is a nonempty closed convex and T-invariant set. Thus Za,ε({xn}, K) = Kand Za({xn}, K) = Tε>0Za,ε({xn}, K) = K. We claim that lim supn→∞ ky−xnk= diam (K) for every y∈K. Indeed, assume that there exists y∈Ksuch that lim supn→∞ ky−xnk<diam (K). Denote r= lim supn→∞ ky−xnk,d= diam (K) and consider the collection {B(z, (r+d)/2) ∩K:z∈K}. Choose an arbitrary positive number εsuch that ε < (d−r)/2. From the first part in the proof we know that lim supn→∞ kxn−zk=rfor every z∈K. Thus, for every finite subset {z1, ..., zk}of Kthere exists a nonnegative integer N such that kxN−zik ≤ r+ε= (r+d)/2 for i= 1, ..., k. Hence xNbelongs to Tk i=1 B(zi,(r+d)/2). The weak compactness of Kimplies the existence of x0∈Tz∈KB(z, (r+d)/2) ∩Kand this point is not diametral because sup z∈Kkz−x0k<r+d 2< d = diam (K). This contradiction proves the claim. If lim infn→∞ ky−xnk<diam (K) for some y∈Kthere exists a subsequence {yn}of {xn}such that lim supn→∞ kyn−yk= lim infn→∞ kxn−yk<diam (K), which is a contradiction according to the claim applied to the sequence {yn}which is again an approximated fixed point sequence. Theorem 12. Let Kbe a convex weakly compact subset of a Banach space X, and T:K→Ka nonexpansive mapping. Assume that Kis minimal for these conditions, diam(K)=1and {xn}is an approximated fixed point sequence which is weakly null. Then, for every ε > 0and t∈[0,1], there exists a sequence {zn}in Ksuch that: (i) {zn}is weakly convergent to a point z∈K.(ii) kznk>1−εfor every n∈N.(iii) kzn−zmk ≤ tfor every n, m ∈N.(iv) lim supnkzn−xnk ≤ 1−t. Proof. Since {wn}is an approximated fixed point sequence in K, diam(K) = 1 and 0 lies in K, from Goebel-Karlovitz Lemma we deduce limnkwnk= 1. Hence, for every ε > 0 there exists δ(ε)>0 such that kxk>1−εif x∈Kand kTx −xk< δ(ε). Indeed, otherwise, there exists ε0>0 such that we can find wn∈Ksatisfying kTwn−wnk<1/n and kwnk ≤ 1−ε0for every n∈N. Therefore, the sequence {wn}is an approximated fixed point sequence in Ksatisfying lim supnkwnk ≤ 1−ε0. Let ε > 0 and t∈[0,1]. Choose γ < min{1, δ(ε)}and for any n∈Ndefine the contraction Sn:K→Kby Sn(x) = (1 −γ)T(x) + γtxn. 348 t. dom´ ınguez The Contractive Mapping Principle assures that there exists a (unique) fixed point znof Sn. Since Kis a weakly compact set, we can assume, taking a subsequence if necessary, that {zn}satisfies (i). Since kzn−Tznk< γ we know that {zn}satisfies (ii). Condition (iii) is easily obtained and (iv) is a consequence of the inequalities kzn−xnk ≤ k(1 −γ)Tzn+γtxn−xnk ≤(1 −γ)kTzn−Txnk+ (1 −γ)kTxn−xnk+γ(1 −t)kxnk. Thus, kzn−xnk ≤ 1−t+1−γ γkTxn−xnk. Taking limsup as ntends to infinity, we obtain (iv). Definition 12. Let Xbe a Banach space. We define the coefficient R(X) = sup{lim inf n→∞ kxn+xk} where the supremum is taken over all x∈Xwith kxk ≤ 1 and over all weakly null sequences in B(0,1). Theorem 13. Let Xbe a Banach space with R(X)<2. If Cis a convex weakly compact subset of Xand T:C→Cis a nonexpansive mapping, then Thas a fixed point. Proof. Otherwise, we can find a convex weakly compact T-invariant subset of Xwhich is not a singleton and which is minimal for these conditions. By multiplication we assume that its diameter is 1. Furthermore, from the Contractive Mapping Principle it is easy to construct an approximated fixed point sequence {xn}in K. We can assume that {xn}is weakly convergent and, by translation, that {xn}is weakly null. Consider a sequence {zn}satisfying (i)-(iv) in Theorem 12 for t= 1/2. Taking again a subsequence, if necessary, we can assume that limnkzn−zkexists. Furthermore, limnkzn−zk ≤ limnlimmkzn−zmk ≤ 1/2. We can choose η > 0 such that ηR(X)< 1−R(X)/2. For nlarge enough, we have kzn−zk ≤ 1/2 + η. Furthermore kzk ≤ lim infn→∞ kzn−xnk ≤ 1/2. Thus ° ° ° ° zn 1/2 + η° ° ° ° =° ° ° ° zn−z 1/2 + η+z 1/2 + η° ° ° °≤R(X). Therefore lim supn→∞ kznk ≤ R(X)(1/2 + η)<1 which is a contradiction because 0 ∈K. geometric properties of banach spaces 349 Remark 7. For X=c0it is easy to check that R(c0) = 1. Thus Theorem 1 holds for c0even though this space fails to have weak normal structure. References [1] Alspach, D.E., A fixed point free nonexpansive map, Proc. Amer. Math. Soc., 82 (1981), 423 – 424. [2] Ayerbe, J.M., Dom´ ınguez, T., L´ opez, G., “Measures of Noncompactnees in Metric Fixed Point Theory”, Birkh¨auser, 1997. [3] Banach, S., Sur les op´erations dans les ensembles abstraits et leurs applications, Fund. Math., 3(1922), 133 – 181. [4] Beauzamy, B., “Introduction to Banach Spaces and Their Geometry”, North-Holland, 1986. [5] Brouwer, L.E.J., ¨ Uber Abbildungen von Mannigfaltigkeiten, Math. Ann., 71 (1912), 97 – 115. [6] Browder, F.E., Nonexpansive nonlinear operators in a Banach space, Proc. Nat. Acad. Sci. USA, 54 (1965), 1041 – 1044. [7] Bynum, W.L., Normal structure coefficients for Banach spaces, Pacific J. Math., 86 (1980), 427 – 436. [8] Goebel,K., Kirk, W.A., “Topics in Metric Fixed Point Theory”, Cambridge University Press, 1990. [9] Kakutani, S., Topological properties of the unit sphere of a Hilbert space, Proc. Imp. Acad. Tokyo, 14 (1943, 242 – 245. [10] Kirk, W.A., A fixed point theorem for mappings which do not increase distances, Amer. Math. Monthly, 72 (1965), 1004 – 1006. [11] Istr˘ at¸escu, V.I., “Fixed Point Theory”, Reidel Pub. Co., 1981. [12] Maluta, E., Uniformly normal structure and related coefficients, Pacific J. Math., 111 (2) (1984), 357 – 369. [13] Milman, D.P., Milman, V.D., The geometry of nested families with empty intersection-structure of the unit sphere of a nonreflexive space, Trans. Amer. Math. Soc., 85 (2) (1969), 233 – 243.