2. Preliminaries. Given two metric spaces X and Y with X â Y it is said that a mapping ... Let X be a hyperconvex metric space and A a bounded subset of X. Then. 1. If B â A there exists ..... Nauk. 27(1972), no.1(163), 81-146. (Russian). 11 ...
Remarks on some fixed point theorems for hyperconvex metric spaces and absolute retracts Dariusz Bugajewski and Rafael Esp´ınola Abstract. In this paper we give a review of recently obtained fixed point results for hyperconvex metric spaces and absolute retracts under noncompactness conditions. Moreover, we prove new extensions of some earlier theorems. Our results are illustrated by suitable examples and compared with others of this type. Key words and phrases: absolute retract, fixed point theorem, hyperconvex metric space, Isbell hyperconvex hull, measure of noncompactness, ultimately compact operators. AMS Classification: 47H10
1. Introduction. N. Aronszajn and P. Panitchpakdi introduced hyperconvex metric spaces in 1956 ([1]) while they were studying the problem of extending uniformly continuous mapping between metric spaces. The recent interest of fixed point theory into hyperconvex spaces goes back to results of Sine [16] and Soardi [17] who independently proved that the fixed point property for nonexpansive mappings holds in bounded hyperconvex spaces. Since then many different results have been shown to hold in hyperconvex spaces, and very recently many of them were stated under noncompactness conditions (see for instance [6], [9], [10] and [12]). In the present paper the authors recall some of those results on fixed point theory involving noncompactness conditions at the same time that show suitable examples about these results. We also prove some new extensions of earlier theorems related to perturbation of nonexpansive mappings in a similar way than Krasnoselski [13] did when he considered fixed points of mappings f1 + f2 in Banach spaces, where f1 was contractive and f2 was completely continuous. In Section 2 we introduce some basic definitions and results. In Section 3 we recall some earlier results, prove extensions of these results and show different examples comparing them and others of this type. Finally, in Section 1
4 we set the new results of Section 3 in the more general context of absolute retracts.
2. Preliminaries. Given two metric spaces X and Y with X ⊆ Y it is said that a mapping f : Y → X is a retraction if f (x) = x for all x in X. Hence a metric space X is called an absolute retract if there exists a continuous retraction f : Y → X for all metric spaces Y where X can be isometrically embedded (see [8] for details). Given a metric space X a mapping f : X → X is said to be a nonexpansive mapping if d(f (x), f (y)) ≤ d(x, y) for all x and y in X. The fixed point property for nonexpansive mappings in hyperconvex spaces was first proved by Sine [16] and Soardi [17]. In this paper we will work with mappings of the type f1 + f2 , where f1 is nonexpansive and f2 is completely continuous, previously studied by the first author in [4]. A metric space X is called hyperconvex if \
B(xα , rα ) 6= ∅
α∈A
for any indexed class of closed balls {B(xα , rα ) : α ∈ A} in X, satisfying the condition d(xα , xβ ) ≤ rα + rβ for all α and β in A. N. Aronszajn and P. Panitchpakdi introduced the above definition in [1] giving many different properties of them, in particular they proved that a hyperconvex space is a nonexpansive retract of any metric space in which it is isometrically embedded. Let (X, d) be a metric space and A a bounded subset of X. The Kuratowski measure of noncompactness of A, α(A), is defined by n(ε)
α(A) = inf{ε; A ⊂ ∪i=1 Ai with diam Ai ≤ ε for i = 1, . . . , n(ε)}. For a proper treatment on the Kuratowski measure of noncompactness the reader is refereed to [2]. Now we introduce the concept of a hyperconvex hull of a subset of a hyperconvex space. Let X be a hyperconvex metric space and A ⊂ X. We 2
will say that a set, h(A), is a hyperconvex hull of A if h(A) is a minimal element of the family F(A) = {B ⊆ X ; B is hyperconvex and A ⊂ B}. This concept was first introduced by Isbell in [9]. Isbell proved many different properties about it, in particular he proved that all the hyperconvex hulls of a same set are related by isometries. It is not hard to see (check [10] for details) that every subset of a hyperconvex space has a hyperconvex hull, probably not unique. If given two sets A and B we write A = h(B) we will mean that A is a hyperconvex hull of B. In the following proposition (proof can be found in [9]) we summarize two more properties of the hyperconvex hull. Proposition. Let X be a hyperconvex metric space and A a bounded subset of X. Then 1. If B ⊂ A there exists h(A) and h(B) such that h(B) ⊂ h(A). 2. α(A) = α(h(A)). Given a metric space X and D ⊂ X, a continuous map f : D 7−→ X is said to be α-condensing if, for any bounded A ⊂ D, α(f (A)) ≥ f (A) implies that A is relatively compact. The well known Darbo-Sadovski theorem states that α condensing mappings defined from a bounded, closed and convex subset of a Banach space into itself have a fixed point. In the following section we give two independent generalizations of the hyperconvex version of this result ([9] and [12]). Sadovski [15] introduced the concept of limit compact operators without using the notion of a measure of noncompactness. If X is a linear space and D is a subset of X, then a continuous operator f : D → X is called a limit compact or ultimately compact operator (see [15]) if, for any B ⊂ X, co(f (B ∩ D)) = B implies that B is compact. In a joint paper with G. L´opez [10], the second author introduced the equivalent concept of limit compact operator for mappings between hyperconvex spaces. If X is a hyperconvex space and D a subset of X, then a continuous mapping f : D → X is called an ultimately compact mapping if, for any B ⊂ X, h(f (B ∩ D)) = B implies that B is compact.
3
3. Fixed point theorems for hyperconvex metric spaces. We begin this section by comparing two results previously obtained by the two authors in an independent way. In [6] the first author proves the following theorem. Theorem 1 ([6]). Let X be a hyperconvex metric space, x0 ∈ X, α the Kuratowski measure of noncompactness and let f be a continuous mapping of X into itself. If the following implication (V is isometric to h(f (V )) or V = f (V ) ∪ {x0 }) → (α(V ) = 0) holds for every subset V ⊂ X, then f has a fixed point. In [10] the second author obtains the following one. Theorem 2 ([10]). Let X be a bounded hyperconvex metric space, α the Kuratowski measure of noncompactness and let f be a continuous mapping of X into itself. If the following implication (V = h(f (V ))) → (α(V ) = 0) holds for every subset V ⊂ X, i.e. if f is an ultimetely compact mapping, then f has a fixed point. Note that both results extend Darbo-Sadovski fixed point theorem for hyperconvex metric spaces (see [6] and [10] for details). Next we illustrate that both theorems apply in different cases. Example 1. 1. Consider the radial metric defined on IR2 as follows: (
d(v1 , v2 ) =
ρ(v1 , v2 ), if 0 = (0, 0), v1 , v2 are colinear, ρ(v1 , 0) + ρ(v2 , 0), otherwise,
where ρ denotes the usual Euclidean metric in IR2 and v1 = (x1 , y1 ), v2 = (x2 , y2 ) ∈ IR2 . It was proved in [6] that IR2 with this metric is hyperconvex. Let f : IR2+ → IR2+ be the mapping defined as follows: f = f2 ◦ f1 , where (
f1 (v) =
v , ρ(0,v)
if ρ(0, v) > 1, if ρ(0, v) ≤ 1,
v, 4
( ³
f2 (v) =
ϕ+ π
´
r, 2 2 , if v 6= (0, 0), (0, 0), if v = (0, 0),
v ∈ IR2+ and (r, ϕ) denote here the polar coordinates of v. In [6] was shown that this mapping is under conditions of Theorem 1. Hence the existence of its fixed points (for any v = (0, y), where 0 ≤ y ≤ 1 we have f (v) = v) can be deduced from this result while Theorem 2 cannot be applied since the hyperconvex space X is not bounded. 2. In this example we will have the reverse situation to the previous one. Let X = {x = (xn ) ∈ l∞ : 0 ≤ xn ≤ 1 for n ∈ IN }. Recall in this place that the space l∞ is hyperconvex (see [1]). Define f (x) = (
x1 √ √ , x1 , x2 , . . .), 2
x ∈ X.
Note that f (X) is itself hyperconvex and isometric to X but not relatively compact. Nevertheless Theorem 2 works on this example since the only set A such that A = h(f (A)) is the singleton A = {(0, 0, · · · , 0, · · ·)} and the rest of its hypothesis are also verified. 3. Consider again the space IR2 with the radial metric. Let B = {(x, y) ∈ IR2 : x2 + y 2 ≤ 1} and let f : B → B be defined by the formula f ((x, y)) = (x3 , y 3 ). Note that f is continuous, α(B) = α(f (B)) = 2 and the following implication is verified: if V is isometric to h(f (V )) or V = f (V ) ∪ {(0, 0)} then V ⊂ {(x, y) ∈ B : x = 0 or y = 0}; hence V is relatively compact. How, in addition, B is bounded we are in the hypothesis of both theorems while Darbo-Sadovski theorem cannot. 4. Finally, consider the space IR2 with the metric “river” defined as follows: (
d(v1 , v2 ) =
| y1 − y2 |, if x1 = x2 , | y1 | + | y2 | + | x1 − x2 |, if x1 6= x2 ,
where v1 = (x1 , y1 ), v2 = (x2 , y2 ) ∈ IR2 . The space IR2 with the metric “river” is also hyperconvex (see [6], Lemma 3). Let X = {(x, y) ∈ IR2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} and let f : X → X be defined by the formula x+1 , y). Note that f is continuous, α(A) = α(f (A)) = 2, f ((x, y)) = ( 2 where A = {(x, y) ∈ X : y = 1} and the following implication is verified: if 5
V is isometric to hf (V ) or V = f (V )∪{(1, 1)} then V ⊂ {(x, y) ∈ X : x = 1}, so V is relatively compact. And again Theorem 1 and 2 work while DarboSadovski one does not. Now we extend Sine and Soardi’s fixed point theorem for nonexpansive mappings in hyperconvex metric spaces ([3], [4], [16], [17]) to mappings f1 +f2 , where f1 is nonexpansive and f2 is completely continuous. Theorem 3 [4]. Let K be a bounded hyperconvex subset of a normed space (X, k·k). Assume that 10 f1 : K → X is nonexpansive, 20 f2 : K → X is completely continuous, 30 f1 (x) + f2 (y) ∈ K for any x, y ∈ K, 40 every sequence (xn ) such that xn ∈ K for n ∈ IN and lim (xn − f (xn )) = 0,
n→∞
where f = f1 + f2 has a limit point. Then f has at least one fixed point in K. Remark 1. Originally Theorem 3 was proved in [4] with the additional assumption that λK ⊂ K for every 0 < λ ≤ 1. Now we present the new proof of this theorem which allows to omit this assumption. Proof of Theorem 3. For any y ∈ K consider the following equation (1)
1 x = r( (f1 (x) + f2 (y))), q
q > 1,
where r : X → K is a nonexpansive retraction (the existence of such a retraction follows from [1, Th.8]). In view of the Banach contraction principle the above equation has exactly one solution x = uq (y) for every y ∈ K. In this way we obtain a function uq : K → K defined by uq (y) = x, where x is the unique solution of (1) for every y ∈ K. It is clear that (2)
1 uq (y) = r( (f1 (uq (y)) + f2 (y))), q 6
q > 1.
Further, arguing similarly as in [4, Th.2] we infer that there exists y q ∈ K such that (3)
uq (y q ) = y q .
Let (qn )n∈IN be a sequence such that qn > 1, n ∈ IN and qn → 1 as n → ∞. In view of (3), (2) we obtain 1 (f1 (y qn ) + f2 (y qn ))) − r(f1 (y qn ) + f2 (y qn ))k ≤ qn qn − 1 ≤ kf1 (y qn ) + f2 (y qn )k, qn
ky qn − f (y qn )k = kr(
so lim ky qn − f (y qn )k = 0, because K is bounded. By 40 , the sequence (y qn ) n→∞ has a convergent subsequence. So by passing to a subsequence, if necessary, one can claim that there exists y0 ∈ K such that lim y qn = y0 . In view of (2) n→∞ it is clear that y0 = f1 (y0 ) + f2 (y0 ), which completes the proof of Theorem 3. We can illustrate Theorem 3 by the following Example 2 [4]. Let K = {x = (x1 , x2 , . . . , xn , . . .) ∈ l∞ : kxk ≤ 1, xn ≥ 0 for n ∈ IN } and let (qn )n≥2 be a sequence such that qn ∈ (0, 1) for n ≥ 2 and qn → 1. Define f1 (x) = (0, q2 x2 , q3 x3 , q4 x4 , . . .),
√ f2 (x) = ( x1 , 0, 0, . . .) for x ∈ K.
It can be shown that the mappings f1 , f2 satisfy the assumptions 10 − 40 of Theorem 3. Note also that the function f1 + f2 does not satisfy the assumptions of Darbo-Sadovski’s theorem for hyperconvex metric spaces. Finally note that above presented Theorem 1 and Theorem 2 are independent from Theorem 3 what follows from quite simple examples.
4. Fixed point theorems for absolute retracts. In this section we extend Theorem 3 to absolute retracts. Let D be an absolute retract (shortly: AR). In view of well known Arens-Eells’ theorem 7
we infer that D is isometric to a closed subset K of a normed space (E, k·k). We will denote by i such isometry. Let K c =convK. In view of [8, Th.10.6(i), p.93] there exists a retraction r : K c → K. Now we formulate the general fixed point principle for absolute retracts, namely Theorem 4 [5]. Let D be an AR, x0 ∈ D and let f : D → D be a continuous mapping of D into itself. If V is relatively compact whenever V = conv(i ◦ f ◦ i−1 ◦ r(V )) or V = f (V ) ∪ {x0 }, then f has a fixed point. As a corollary from the above principle we obtain the following DarboSadovski fixed point theorem for absolute retracts. For simplicity we shall denote the Kuratowski measure of noncompactness by the same letter α in different spaces. Theorem 5 [5]. Let D be an AR and let f : D → D be a continuous mapping such that (4)
α(f (A)) < α(A) for each A ⊂ D such that α(A) > 0.
If α(r(V )) ≤ α(V ) for any V ⊂ K c (shortly: if r is α-nonexpansive), then f has a fixed point (r, E, K, K c denote the same as at the beginning of this section). P r o o f. First, let A ⊂ D be such that A = f (A) ∪ {x0 }. If α(A) > 0, then α(A) = α(f (A)) < α(A), what gives a contradiction. Now assume that V ⊂ K c is such that V =conv (i ◦ f ◦ i−1 ◦ r(V )). If α(r(V )) = 0, then α(V ) = α(f ◦ i−1 ◦ r(V )) ≤ α(f ◦ i−1 (r(V ))) = 0, so V is relatively compact. On the other hand, if α(r(V )) > 0, then by the assumptions we obtain α(V ) = α(f ◦ i−1 ◦ r(V )) < α(i−1 ◦ r(V )) ≤ α(V ), what gives a contradiction. Hence f satisfies the assumptions of Theorem 4 and therefore it has a fixed point. 8
Now we illustrate Theorem 4 by the following Example 3 [5]. Let D = {x ∈ c0 : kxk ≤ 1, | x1 |=| x2 |} and let q
q
f (x) = ( | x1 |, | x2 |, q3 x3 , q4 x4 , . . .),
x ∈ D,
where (qn )n≥2 is the sequence from Example 2. One can show that this mapping f satisfies all assumptions of Theorem 4. Note also that f does not satisfy the condition (4) in Theorem 5. Now we apply Theorem 5 to prove the following Theorem 6. Let E be a Banach space, K ⊂ E be a bounded AR and let r : E → K be a retraction of E on K. Further, for any ε > 0 let K ε = {x ∈ E : dist(x, K) ≤ ε}. Suppose that α(r(V )) ≤ α(V )
for any V ⊂ K
and for some ε > 0 the restriction r |K ε is uniformly continuous. If the following conditions: i) f1 : K → E is nonexpansive; ii) f2 : K → E is completely continuous; iii) f1 (x) + f2 (y) ∈ K for any x, y ∈ K; iv) every sequence (xn ) such that xn ∈ K for n ∈ IN and lim (xn − f (xn )) = 0,
n→∞
where f = f1 + f2 has a limit point; hold, then f has at least one fixed point in K. P r o o f. For any q > 1 define the mapping fq : K → K by the formula 1 fq (x) = r( (f1 (x) + f2 (x))), q
9
x ∈ K.
For any subset A of K we have 1 1 α(fq (A)) = α(r( (f1 (A) + f2 (A)))) ≤ α( (f1 (A) + f2 (A))) ≤ q q 1 ≤ α(A) < α(A), if α(A) > 0. q In view of Theorem 5 there exists xq ∈ K such that xq = fq (xq ). For this xq we have 1 kxq − f (xq )k = kr( (f1 (xq ) + f2 (xq ))) − (f1 (xq ) + f2 (xq ))k = q 1 = kr( (f1 (xq ) + f2 (xq ))) − r(f1 (xq ) + f2 (xq ))k. q By the above inequality and by the assumption that the restriction r |K ε is uniformly continuous we infer that there exists a sequence (xqn ) such that kxqn − f (xqn )k