Modular Spaces and K-widths

In this paper, we show that the ball measure of noncompactness of a modular space Xp is equal to the limit of its K -widths when p is a left continuous, s-convex modular function, without any ~rcondition. We also obtain a similar result for SF-spaces, when the SF-norm N is uniformly continuous. 1991 Mathematics Subject Classification: 46E30, 46A50, 46B99 1. Notation and definitions Throughout the following X is a linear space over a field K (K = IR or K = C). I. A function p : X --+ [0, oo] is called modular if the following hold for arbitrary x,y eX: 1. p(x) = 0 iff x = 0. 2. p(ax) = p(x) if a e K, lal = 1. 3. p(ax + f3y) ~ p(x) + p(y) for a, f3 ;::: 0, a+ f3 = 1. If in place of 3) we have p(ax + f3y) ~as p(x) + f3s p(y) for a, f3 ;::: 0, as+ ps = 1 for an s e (0, 1], then pis called s-convex modular (convex if s = 1). The set is called a modular space. Xp = {x e X : lim p(ax) = 0} a-+0 Each modular space X p may be equipped with an F-norm given by the formula lxlp = inf{u > 0: p(xfu):::; u} for x e Xp. A modular pin X is called left continuous iflimA-+tp(.U) = p(x) for all x e Xp. Offprint from: Functional Analysis, Eds.: Dierolf/Dineen/Domanski © Walter de Gruyter & Co., Berlin· New York 2 A. G. Aksoy and G. Lewicki It is known (see [8]) that, if pis left continuous, s-convex modular in X., then the inequalities lx I~ ::; 1 and p (x) ~ 1 are equivalent for every x E X P· Here by lx 1~ we meM · lxl~ = inf{u > 0: p(u:fs) ~ 1}. A particular class of modular space$ are Orlicz-Musie1ak spaces. To define these spaces, let (E, E, J.£) be a measure space Md let f : E x 1l4 ~ R+ satisfy the following conditions. 1. f(t, ·) : IR+ ~ R+ is anondecreasing, continuous function such that f(t, 0) = 0 and f(t,u) > Oforu > 0. 2. f ( ·, u) : E -+ R+ is a E-measurable function for all u > 0. 3. fA f(t, u) dJ,.£(t) < oo for every u > 0 and A E E, IJ.(A) < oo. Suppose that X is the space of all real (or complex) valued :E-measurable functions defined on E. For x e X, set Pj(X) = £ /(t,ix(t)i) dp.(t). From the above 1) and 2), it is clear that P! is modular. The modular space Xp is called Orlicz-Musielak space (Md Orlicz space, if the function f is independent of the variable t ). We say the function f satisfies a .6.2-condition if f(t, 2u) S Kf(t, u) + h(t) for all u ~ 0, t e E where h e L 1(E, J.L), h ~ 0 and K is a positive cons!Mt independent of the variables t, u. For further theory of modular spaces we refer to [8]. n. Assume a function N : X -+ R+ satisfies the following conditions. 1. N(x) =Oiffx = 0. 2. an ~ a and N (xn x) -+ 0 then N (anXn ax) ~ 0 for all sequences {an} C K and {Xn} C X. 3. If.N'(xn -x)-+ Oand.N'(yny)-+ Othen.N'(xn + Yn -xy)-+ Oforall sequences {xn}. {Yn} C X. 4. N(ax) = N(x) for every x eX and a e K,lal = 1; 5 . .N'(xn x) ~ 0 then N(xn) -+ N(x) for every {xn} C X. 6. The space X is complete with respect to the topology induced by the family C = {B(x; r): x eX, r > 0} where B(x; r) = {y e X.:N(xy) < r}. ....... ________________ __ Modular spaces and K -widths 3 The pair (X, N) Is said to be an SF-space and the function N will be called an SF-norm. Note that each F-space (in particular each Banach space) is an SF-space. If f satisfies a ~2-condition, then each Orlicz-Musielak space (Xp1 • PJ) is an SF-space. However, there exist SF-spaces which are neither F-spaces nor modular spaces as can be seen by the following Example ([7]). Assume f : IR+ -+ JR+ satisfies the following conditions: 1. f is continuous and f(t) = 0 iff t = 0. 2. There exists d > 0 such that flto.dJ is strictly increasing. 3. There exists dt > 0 and M > I such that f(t + s) :5 M(f(t) + f(s)) for s, t E [0, dJ] . 4. There exists lim,~00 f(t) E (0, oo). Let (E, :E, J.L) be a measure space, J.L(E) < oo and let M(E) = {x : E -+ R, xis :E-measurable }. For x E M(E) define N(x) = JE f(lx(t)l) dJ,L(t). By Fatou's lemma, the Riesz and the Lebesgue theorems, one can show that the pair (M(E), N) is an SF-space. It is clear that for non monotonic f the function N is not modular, and if we additionally assume . I ltm f(t) < sup{f(t): t E IR+} r~+oo 2 then the pair (M(E), N) can not be an F-space. The SF-norm N is called nondecreasing iff for any lt, t2 E IR, I t tl :5 lt2l implies N(t1x) :5 N(t2x) and given Din an SF-space (X,N), we say Dis bounded iff .l..n -+ 0, Xn E D implies N(.l..nxn) -+ 0. Analogously D C Xp is p-bounded iff an -+ 0, Xn E D implies p(anxn) -+ 0. The SF-norm N is called uniformly continuous if for every s > 0, there is a 8 > 0 such that if N(fg ) < 8 then IN (f)N (g)i 0 there is V, an open neighbourhood of 0, such that BN (0, r) + V C BN (0, r + s) (*) where BN (x, r) = {y E X: N(xy) :5 r}. The proof is straightforward. 4 A. G. Aksoy and G. Lewicki m. Let (X,.N) be an SF-space. For V c X, V \ {0} ~ 0 put RN(V) = inf{sup{N(tv): t e R+): v e·V \ {0} }. This number which may be equal to +oo, is called the radius of the set V ([7]). As shown in the following example, it m~y also occur that RN(V) = 0. Example. Let X be the space of all complex sequences equipped with an SF~nonn N defined by 00 (1 ) lxtl N(x) = L -2 1 I I for x eX. i=l + Xi Let for n eN, Vn = {x eX: XJ = 0 for j > n}. Then clearly RN(Vn) = 2-n and consequently RN(X) = 0. · In [7] it is shown that if Xp is a modular space with p s-convex modular, then Rp(Xp) = +oo. K -widths are extensively studied in the context of approximation theory [10]. Our aim in this paper is to connect K -widths with measures of noncompactness. Such connections are not only useful in fixed point theory (see [3], [11]) but also in the study of the radius of the essential spectrum (see [6], [9]). Measures of noncompactness for Orlicz spaces are studied in [1], [2] and [4]. In [5] one can find fixed point theorems for Orlicz modular spaces. 2. K-widths in SF-spaces Let (X, N) be an SF-space, D be bounded set in X. ·Then the ball measure of noncompactness of D, a(D), is . n a(D) = inf{r > 0: D C U B"' (XJ:, r)} A:= I and K-th width of D, dA:(D), is defined as dk(D) = inf{r > 0: D c BN (0, r) + Ak dim(AJ:) :;: k}. Theorem 1. Let D be a bounded subset of an SF-space (X, N). Suppose that N is nondecreasing, RN(X) = oo and for every r > 0, e > 0 and E finite dimensional subspace of X, there is an open neighborhood V of 0 in E such that BN (0, r) + V C BN (0, r + e). ( *) Then a(D} = limJ:-..00 dk(D). Modular spaces and K -widths 5 Proof. Let D be a fixed bounded set in X. If there is r > 0 and k e N such that DC U~=l BN(x;, r), then . Therefore, if a(D) < oo then d = lim~edle(D) is finite and d ~ a(D). . To obtain the other inequality, assume that d < +oo, fix k e N, e > 0 and A1e c X with dim A1e = k such that