On a unified study of relative Chebyshev radii and Hausdorff measures of noncompactness

Houston Journal of Mathematics c 2004 University of Houston Volume 30, No. 1, 2004 ON A UNIFIED STUDY OF RELATIVE CHEBYSHEV RADII AND HAUSDORFF MEASURES OF NONCOMPACTNESS RAFAEL ESP´ INOLA1, ANDRZEJ WI´ SNICKI 2, AND JACEK WO´ SKO 2 Communicated by Gilles Pisier Abstract. The paper is concerned with the notion of the Lifschitz modulus and its relationship with both relative Chebyshev radii and Hausdorff measures of noncompactness. 1. Introduction Throughout the paper we shall use the following notation: Xwill denote a real normed space, Aa nonempty, bounded subset of Xand Ga nonempty subset of X. As usual, B(x, r) will stand for the closed ball centered at x∈Xwith radius r > 0, diam Afor the diameter of A, and dist(x, A) for the distance from xto A. We shall also write B(A, r) = {x∈X: dist (x, A)≤r}, dist (A, G) = inf {dist (x, G) : x∈A}, d(A, G) = inf {r > 0 : A⊂B(G, r)} and d(H, G) = inf {d(F, G) : F∈ H} if His a nonempty family of nonempty bounded subsets of X. Let us recall that the relative Chebyshev radius rG(A) is given by rG(A) = inf {r > 0 : A⊂B(y, r) for some y∈G} 2000 Mathematics Subject Classification. Primary 46B20, 41A65, 47H09. 1Supported by the project PB96-1338-C02-01 from Junta de Andalucia. 2Supported in part by the grant 2 PO3A 029 15 from the State Committee for Scientific Research of Poland. 245 246 RAFAEL ESP´ INOLA, ANDRZEJ WI´ SNICKI, AND JACEK WO´ SKO and the relative Hausdorff measure of noncompactness χG(A) by χG(A) = inf {r > 0 : A⊂B(F, r) for some finite set F⊂G}. If G=Xwe shall abbreviate r(A) and χ(A) respectively. Finally, for ¯≥0, Eε(A) = {y∈X:A⊂B(y, r (A) + ε)}, Hε(A) = {F⊂X:A⊂B(F, χ (A) + ε) and Fis finite}. Note that Hε(A)6=∅and Eε(A)6=∅for every bounded A⊂Xand ε > 0. The central theme of this paper is the concept of the Lifschitz modulus, defined in [16] as follows: Definition 1.1. The Lifschitz modulus of a normed space Xis the function ˜κX(·) defined on [0,+∞) by ˜κX(d) = sup{k > 0 : ∃α∈(0,1) ∀x, y ∈X∀r > 0∃z∈X(kz−yk ≤ αdr ∧B(x, r)∩B(y, kr)⊂B(z, r))}. The origins of this notion come from [13], where slightly similar ideas were used to prove a certain fixed point theorem for uniformly Lipschitzian mappings. Although the definition of the Lifschitz modulus seems to be a little artificial, it is based on very natural geometric intuition (see Section 2 for more details). With the help of this notion we managed [16] to extend the formula rG(A) = r(A) + dist E0(A), G¡ (proved in the case X=C(K) by Smith and Ward in [15], see also [7]), for a wider class of spaces including c0as well as some subspaces of C(K).Moreover, the concept of the Lifschitz modulus let us treat the notions of relative Chebyshev radii and relative Hausdorff measures of noncompactness in a unified way, since Theorem 1.2, stated below, implies that if ˜κX(d) = 1 + dfor all d≥0, then rG(A) = r(A) + lim ε→0+dist (Eε(A), G) and χG(A) = χ(A) + lim ε→0+d(Hε(A), G). The following theorem is a minor modification of the result given in [16]: Theorem 1.2. Let Aand Gbe nonempty subsets of a normed space Xwith A nonsingleton and bounded. Then rG(A)≥r(A) ˜κX°dr(A,G) r(A)±, RELATIVE CHEBYSHEV RADII AND HAUSDORFF MEASURES 247 where dr(A, G) = (dist E0(A), G¡if E0(A)6=∅, lim ε→0+dist (Eε(A), G)if E0(A) = ∅. If χ(A)6= 0, then χG(A)≥χ(A) ˜κX°dχ(A,G) χ(A)±, where dχ(A, G) = (dH0(A), G¡if H0(A)6=∅, lim ε→0+d(Hε(A), G)if H0(A) = ∅. In [6] we put these results in a more general framework and gave the following characterization of C(K) and C0(Ω) spaces: Theorem 1.3. Let Aand Gbe nonempty subsets of a Banach space Xwith A bounded. The following assertions are equivalent: (1) rG(A) = r(A) + dist E0(A), G¡for every Aand G, as above. (2) rG(A) = r(A) + lim ε→0+dist (Eε(A), G)for every Aand G, as above. (3) Xis isometric to C(K)for some compact Hausdorff topological space K or to C0(Ω) for some locally compact Hausdorff space Ω. (4) ˜κX(d) = 1 + dfor all d≥0. The present paper deals with the following two questions: 1) How “good” are the estimations obtained in Theorem 1.2? 2) Does the characterization of C(K) and C0(Ω) spaces given in Theorem 1.3 still hold if we replace the notion of relative Chebyshev radii by relative Hausdorff measures of noncompactness? In Section 2 we discuss some properties of the Lifschitz modulus. With the use of these properties we show in Section 3 that the estimation of rG(A) given in Theorem 1.2 is, in a sense, optimal. We recall Theorem 1.3 in Section 4 in order to study its counterpart for the relative Hausdorff measure of noncompactness. It clearly follows that if Xis isometric to C(K) or C0(Ω), then (?)χG(A) = χ(A) + lim ε→0+d(Hε(A), G). However, far from stating an equivalent characterizing result we prove that formula (?) does hold in the space `1which is well-known to be nonisometric to the spaces C(K) and C0(Ω). Thus, we discover the rather surprising fact that the notions of relative Chebyshev radii and Hausdorff measures of noncompactness behave in a different way in this problem. Section 5 contains some remarks concerning another problem on relative Chebyshev radii and relative Hausdorff measures of noncompactness. 248 RAFAEL ESP´ INOLA, ANDRZEJ WI´ SNICKI, AND JACEK WO´ SKO 2. Properties of the Lifschitz Modulus It is easily seen that in any normed space X max {1, d −1} ≤ ˜κX(d)≤d+ 1 for all d≥0.The next two propositions show that we may treat the parameter d in the definition of the modulus of Lifschitz as the distance between xand y. Proposition 2.1. For all d≥0 ˜κX(d) = sup{k > 0 : ∃α∈(0,1) ∀x, y ∈X(kx−yk ≤ d⇒ ∃z∈X(kz−yk ≤ αd ∧B(x, 1) ∩B(y, k)⊂B(z, 1)))}. Proof. Let us first notice that we can fix r= 1 in the definition of ˜κX. Put d > 0 and take k≥1 and α∈(0,1) such that for every x, y ∈X, we obtain (2.1) kx−yk ≤ d⇒ ∃z∈X(kz−yk ≤ αd ∧B(x, 1) ∩B(y, k)⊂B(z, 1)) . It is sufficient to prove that k≤˜κX(d). Let x, y ∈Xwith kx−yk> d. We need to construct z∈Xsuch that kz−yk ≤ αd ∧B(x, 1) ∩B(y, k)⊂B(z, 1). Let y1be the element of the segment [x, y] satisfying the equality kx−y1k= d. Therefore there exists z1∈Xwith the property that kz1−y1k ≤ αd and B(x, 1) ∩B(y1, k)⊂B(z1,1).We now consider two cases. Assume first that ky1−yk ≤ (1−α)d. Therefore kz1−yk≤kz1−y1k+ky1−yk ≤ αd + (1 −α)d=dand hence using (2.1), we obtain B(z1,1) ∩B(y, k)⊂B(z, 1) for some zwith kz−yk ≤ αd. Thus, since k≥1, B(x, 1) ∩B(y, k) = B(x, 1) ∩B(y1, k)∩B(y, k)⊂B(z1,1) ∩B(y, k)⊂B(z, 1) and we have the desired result. Assume now that ky1−yk>(1 −α)d. Therefore ky1−yk ≤ n(1 −α)dfor some n∈Nand we can find the elements y2, y3, ..., ynof the segment [y1, y] such that ky1−y2k= (1 −α)d,ky1−y3k= 2(1 −α)d, ..., ky1−ynk= (n−1)(1 −α)d and kyn−yk ≤ (1 −α)d. Using similar arguments as before we find the elements z2, z3, ..., znsuch that B(x, 1) ∩B(y2, k)⊂B(z2,1),kz2−y2k ≤ αd, B(x, 1) ∩B(y3, k)⊂B(z3,1),kz3−y3k ≤ αd, ...... B(x, 1) ∩B(yn, k)⊂B(zn,1),kzn−ynk ≤ αd. Finally, B(x, 1) ∩B(y, k)⊂B(z, 1) for some zsatisfying kz−yk ≤ αd and the proof is complete. £ RELATIVE CHEBYSHEV RADII AND HAUSDORFF MEASURES 249 Proposition 2.1 leads to the aforementioned desired result. Proposition 2.2. For all d≥0 ˜κX(d) = sup{k > 0 : ∃α∈(0,1) ∀x, y ∈X(kx−yk=d⇒ ∃z∈X(kz−yk ≤ αd ∧B(x, 1) ∩B(y, k)⊂B(z, 1)))}. Proof. Fixing d > 0 and taking k > 0 and α0∈(0,1) such that for every x, y ∈X, we obtain kx−yk=d⇒ ∃z∈X(kz−yk ≤ α0d∧B(x, 1) ∩B(y, k)⊂B(z, 1)) . It is sufficient to prove that k≤˜κX(d). Choose an arbitrary α∈(α0,1) and x, y ∈Xwith kx−yk< d. Let y1be the element of the half-line xy such that kx−y1k=d. Therefore there exists z∈Xwith the property that kz−y1k ≤ α0dand B(x, 1) ∩B(y1, k)⊂B(z, 1). If kx−yk ≤ αd we have B(x, 1)∩B(y, k)⊂B(x, 1) and xitself is the required center. If αd < kx−yk< d, then ky−y1k<(1 −α)dand B(x, 1) ∩B(y, k −(1 −α)d)⊂B(x, 1) ∩B(y1, k)⊂B(z, 1). Moreover kz−yk ≤ (1 + α0−α)d. Combining these two cases and bearing Proposition 2.1 in mind we conclude that k−(1 −α)d≤˜κX(d). Taking αclose to 1 we complete the proof. £ From Proposition 2.1 we also obtain the following proposition. Proposition 2.3. The function ˜κXis nondecreasing on [0,∞). Nonexpansivity follows in a less trivial way. Proposition 2.4. The function ˜κXis nonexpansive, i.e. ˜κX(d2)−˜κX(d1)≤d2−d1 for every 0≤d1≤d2. Proof. Fix 0 < d1< d2and choose k < ˜κX(d2). Therefore there exists α∈(0,1) such that ∀x, y ∈X∃z∈X(kz−yk ≤ αd2∧B(x, 1) ∩B(y, k)⊂B(z, 1)) . We will now prove that ˜κX(d1)≥k−(d2−d1).By Proposition 2.2 it is sufficient to consider x, y1∈Xwith kx−y1k=d1. Let y2be the element on the half-line xy1such that kx−y2k=d2. Hence B(x, 1) ∩B(y1, k −(d2−d1)) ⊂B(x, 1) ∩B(y2, k)⊂B(z, 1) 250 RAFAEL ESP´ INOLA, ANDRZEJ WI´ SNICKI, AND JACEK WO´ SKO for some z∈Xwith kz−y2k ≤ αd2. We notice that y1=d2−d1 d2x+d1 d2y2and set p=d2−d1 d2x+d1 d2z. Therefore kp−y1k=d1 d2kz−y2k ≤ αd1. Moreover it is not difficult to see that B(x, 1) ∩B(y1, k −(d2−d1)) ⊂B(p, 1). Thus ˜κX(d1)≥k−(d2−d1) and, taking supremum over k, we can deduce ˜κX(d1)≥˜κX(d2)−(d2−d1). £ We shall use Propositions 2.2, 2.3 and 2.4 in the next section. 3. The Optimality Theorem In this section we show that the evaluation of rG(A) given in Theorem 1.2 is in a sense optimal. Theorem 3.1. Let Xbe a normed space. Then for every d≥0and ε > 0there exist sets A, G ⊂Xsuch that (3.1) |dr(A,G) r(A)−d|< ε and rG(A) r(A)<˜κX°dr(A,G) r(A)±+ε. Proof. Fix d≥0, ε > 0, and take ksatisfying the inequality ˜κX(d)< k < ˜κX(d) + ε 2and choose α∈(0,1) (notice that k > 1). Then there exist x, y ∈X with kx−yk=dsuch that (3.2) ∀z∈X(B(x, 1) ∩B(y, k)⊂B(z, 1) ⇒ kz−yk> αd). We set A=B(x, 1) ∩B(y, k), G ={y}. It is clear that rG(A) = k, r(A)≤1 and dr(A, G)≤d. We claim that (3.3) dr(A, G)≥αd. Indeed, if r(A) = 1, then x∈E0(A) and E0(A) is nonempty. Using this fact and (3.2), we can deduce dr(A, G) = dist E0(A), y¡≥αd. If r(A)<1, then we set 0< δ < 1−r(A) and, again, dr(A, G)≥dist Eδ(A), y¡≥αd which proves (3.3). Moreover (3.4) r(A)≥1−(1 −α)d. Assume conversely, that there exist z∈Xand r0<1−(1 −α)dsuch that A⊂B(z, r0). In the case 0 ≤d≤1 we may use the fact that kz−yk ≥ αd and that B(y, 1−d)⊆A. If we take the point u∈Ain the line zy which is the farthest from zin B(y, 1−d), it can be seen that ku−zk ≥ 1−d+αd = 1 −(1 −α)d, which is a contradiction with u∈A. In the case d > 1, let p∈Abe the element of the half-line xy such that kx−pk= 1. Then ky−pk=d−1 and it follows RELATIVE CHEBYSHEV RADII AND HAUSDORFF MEASURES 251 from (3.2) that kz−pk≥kz−yk−ky−pk>1−(1 −α)dwhich contradicts p∈A. Thus (3.4) is proved. Summarizing we have (3.5) 1 −(1 −α)d≤r(A)≤1 and αd ≤dr(A, G)≤d. From this fact and Propositions 2.3 and 2.4 rG(A) = k≤˜κX(d) + ε 2≤˜κX°dr(A,G) αr(A)±+ε 2≤ ˜κX°dr(A,G) r(A)±+dr(A,G) r(A)²1 α−1³+ε 2. Finally rG(A) r(A)≤˜κX(d) + ε 2 1−(1 −α)d≤ ˜κX°dr(A,G) r(A)±+dr(A,G) r(A)1 α−1¡+ε 2 1−(1 −α)d and from (3.5) (3.6) αd ≤dr(A, G) r(A)≤d 1−(1 −α)d. Taking αsufficiently close to 1 the formula (3.1) holds. £ Theorem 1.2 also states ˜κX(d)≤inf {rG(A) : A, G ⊂X, r(A) = 1 and d=dr(A, G)}. Using slightly more subtle arguments than in the proof of Theorem 3.1 we may improve the previous formulae. We shall need the following observation, which is easy to check. Lemma 3.2. Let Ebe a convex subset of a normed space X,y∈Xand dist(y, E) = d > 0. Then for every ε > 0there exist y1, y2∈B(y, ε)such that dist(y1, E)≤max{d−ε 2,0}, dist(y2, E)≥d+ε 2. Theorem 3.3. In any normed space X ˜κX(d) = inf {rG(A) : A, G ⊂X, r(A) = 1 and d=dr(A, G)}. Proof. Fix d, ε > 0. As in the proof of Theorem 3.1, for sufficiently large α < 1, there exist A⊂Xand y∈Xsuch that ˜κX(d)≥r{y}(A) r(A)−εand αd ≤ dr(A,{y}) r(A)≤d 1−(1−α)d(see (3.6)). Moreover, it follows from Lemma 3.3 that we can fix αclose to 1 for which there exist y1, y2∈B(y, ε) with dr(A, {y1})< dr(A) and dr(A, {y2})> dr(A). Since the function x→dr(A, {x}) is continuous, there exists z∈B(y, ε) with dr(A, {z}) = dr(A).Put ¯ A=A r(A)and ¯z=z r(A). Then 252 RAFAEL ESP´ INOLA, ANDRZEJ WI´ SNICKI, AND JACEK WO´ SKO dr¯ A, {¯z}¡=d,r¯ A¡= 1 and ˜κX(d)≥r{¯z}¯ A¡−2ε. This completes the proof. £ Notice that Theorem 3.3 gives a geometric description of the Lifschitz modulus. 4. The Case of Relative Hausdorff Measures of Noncompactness It is not difficult to see that in any normed space X, χG(A)≤χ(A) + lim ¯→0+d(H¯(A), G). If Xis isometric to C(K) or C0(Ω), then by Theorem 1.3 ˜κX(d) = 1 + dand by Theorem 1.2 χG(A) = χ(A) + lim ε→0+d(Hε(A), G). In this section we extend this result to the space `1. Let us first recall the notion of minimal sets introduced by Dom´ınguez Benavides in [5]. Definition 4.1. Let Mbe a metric space and ϕa measure of noncompactness. A bounded, infinite set A⊂Mis said to be minimal for the measure ϕ(or, in short, ϕ-minimal) if ϕ(A) = ϕ(B) for every infinite subset Bof A. Definition 4.2. A measure of noncompactness ϕis said to be strictly minimalizing for a metric space Mif for every bounded A⊂Mthere exists a ϕminimal set B⊂Asuch that ϕ(B) = ϕ(A). It is known [3] that the Hausdorff measure of noncompactness χis strictly minimalizing for a wide class of spaces including separable as well as reflexive Banach spaces. In particular, χis strictly minimalizing in `1space. Thus, for any bounded set A⊂`1, there exists a χ-minimal sequence for A, that is, a sequence {xn}in Asuch that χ(A) = χ({x1, x2, ...}). We shall also need the following result concerning `1space. We say that a bounded sequence {xn}of points of `1is coordinate-wise convergent to x∈`1if for any i= 1,2, ..., limn→∞ xi n=xi. For a fixed sequence {xn}in `1and arbitrary y∈`1,we set ra(y, {xn}) = lim supn→∞ kxn−yk. Proposition 4.3. ([9]). If {xn}is a bounded sequence in `1converging coordinatewise to x∈`1, then for any y∈`1 ra(y, {xn}) = ra(x, {xn}) + kx−yk. To establish the main result of this section we shall use the following lemmas. RELATIVE CHEBYSHEV RADII AND HAUSDORFF MEASURES 253 Lemma 4.4. Fix x∈`1, r > 0, ε ≥0and let {xn}be a sequence of points of the ball B(x, r), which is coordinate-wise convergent to x0.If χ({x1, x2, x3, ...})≥ r−ε, then kx−x0k ≤ ε. Proof. Notice that ra(x0,{xn})≥r−εand it follows from Proposition 4.3 that kx−x0k=ra(x, {xn})−ra(x0,{xn})≤r−(r−ε) = ε. £ Lemma 4.5. Let Abe a bounded subset of a Banach space Xand ε > 0. Then there exist x1, ..., xn∈X,r1, ...., rn>0such that A⊂Sn i=1 B(xi, ri) and χ(A∩B(xi, ri)) ≥ri−ε, i = 1, ..., n. Proof. Let A⊂Sm i=1 B(yi, ai), where y1, ..., ym∈Xand a1, ..., am>0.If χ(A∩B(yj, aj)) < aj−εfor some j∈ {1, ..., m},there exist z1, ..., zl∈X, b1, ...., bl< aj−εsuch that A∩B(yj, aj)⊂Sl i=1 B(zi, bi).If, again, χ(A∩B(yj, aj)∩B(zk, bk)) < bk−ε for some k∈ {1, ..., l}we have A∩B(yj, aj)∩B(zk, bk)⊂Sp i=1 B(ui, ci), where u1, ..., up∈Xand c1, ..., cp< bk−ε<aj−2ε. After a finite number of steps we obtain the desired cover. £ Lemma 4.6. Let x, y ∈`1,0< r ≤kand ε > 0. Therefore there exists a finite set F⊂`1with the properties B(x, r)∩B(y, k)⊂B(F, r)and F⊂B(y, k−r+ε). Proof. It follows from Lemma 4.5 that there exist f1, ..., fm∈`1and r1, ...., rm≤ rsuch that B(x, r)∩B(y, k)⊂Sm i=1 B(fi, ri) and χ(B(x, r)∩B(y, k)∩B(fi, ri)) ≥ri−ε 2, i = 1, ..., m. For each set Ui=B(x, r)∩B(y, k)∩B(fi, ri) we select a χminimal sequence {xi 1, xi 2, ...} ⊂ Ui. We can assume, by taking subsequences if necessary, that the sequences are coordinate wise convergent to some w1, ..., wm∈`1, respectively. Since χ{xi 1, xi 2, ...}¡≥ri−ε 2, it follows from Lemma 4.4 that kfi−wik ≤ ε 2. Moreover kwi−yk=ra(y, ¨xi n©)−ra(wi,¨xi n©)≤k−ri+ε 2 and hence kfi−yk≤kfi−wik+kwi−yk ≤ k−ri+ε, i = 1, ..., m. Let us construct the set F={¯ f1, ..., ¯ fm}in the following way: For i= 1, ..., m, if kfi−yk ≥ r−ri,¯ fiis denoted as the point of the segment [fi, y] satisfying the equality kfi−¯ fik=r−ri,r−ri≥0. If kfi−yk< r −riwe set ¯ fi=y. It is not