Some geometric properties of a new modular space defined by Zweier operator

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Et et al. Fixed Point Theory and Applications 013, 013:165 R E S E A R C H Open Access Some geometric properties of a new modular space defined by Zweier operator Mikail Et 1,MuratKarakaş * and Muhammed Çınar 3 Dedicated to Professor Hari M Srivastava * Correspondence: m.karakas33@hotmail.com Department of Statistics, Bitlis Eren University, Bitlis, 13000, Turkey Full list of author information is available at the end of the article Abstract In this paper, we define the modular space Z σ s, p by using the Zweier operator and a modular. Then, we consider it equipped with the Luxemburg norm and also examine the uniform Opial property and property β. Finally, we show that this space has the fixed point property. MSC: 0A05; 6A5; 6B0 Keywords: Zweier operator; Luxemburg norm; modular space; uniform Opial property; property β 1 Introduction In literature, there are many papers about the geometrical properties of different sequence spaces such as [1 9]. Opial [10] introduced the Opial property and proved that the sequence spaces l p 1 < p < havethispropertybutl p [0, π] p,1<p < doesnot have it. Franchetti [11] showed that any infinite dimensional Banach space has an equivalent norm that satisfies the Opial property. Later, Prus [1] introduced and investigated the uniform Opial property for Banach spaces. The Opial property is important because Banach spaces with this property have the weak fixed point property. Definition and preliminaries Let X, be a real Banach space and let SXresp. BX betheunitsphereresp. the unit ball of X. ABanachspaceX has the Opial property if for any weakly null sequence {x n } in X and any x in X \{0}, theinequalitylim n inf x < lim n inf x n + x holds. We say that X has the uniform Opial property if for any ε >0thereexistsr >0suchthat for any x X with x ε and any weakly null sequence {x n } in the unit sphere of X,the inequality 1 + r lim n inf x n + x holds. For a bounded set A X, the ball-measure of noncompactness was defined by βa= inf{ε >0:A can be covered by finitely many balls with diameter ε}. The function defined by ε =inf{1 inf x : x A :A is closed convex subset of BXwithβA ε} is called the modulus of noncompact convexity. A Banach space X is said to have property L, if lim ε 1 ε = 1. This property is an important concept in the fixed point theory and a Banach space X possesses property L if and only if it is reflexive and has the uniform Opial property. 013 Et et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License http://creativecommons.org/licenses/by/.0, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Et et al. Fixed Point Theory and Applications 013, 013:165 Page of 10 A Banach space X is said to satisfy the weak fixed point property if every nonempty weakly compact convex subset C and every nonexpansive mapping T : C C Tx Ty x y, x, y C haveafixedpoint,thatis,thereexistsx C such that Tx=x. Property L and the fixed point property were also studied by Goebel and Kirk [13], Toledano et al. [1], Benavides [15], Benavides and Phothi [16]. A Banach space X is said to have property H if every weakly convergent sequence on the unit sphere is convergent in norm. Clarkson [17] introduced the uniform convexity, and it is known that the uniform convexity implies the reflexivity of Banach spaces. Huff [18] introduced the concept of nearly uniform convexity of Banach spaces. A Banach space X is called uniformly convex UC if for each ε >0,thereisδ > 0 such that for x, y SX, the inequality x y > ε implies that 1 x + y <1 δ. For any x / BX, the drop determined by x is the set Dx, BX = conv{x} BX. A Banach space X has the drop property D ifforevery closed set C disjoint with BX, there exists an element x C such that Dx, BX C = {x}. Rolewicz [19] showed that the Banach space X is reflexive if X has the drop property. Later, Montesinos [0]extendedthisresultandprovedthatX has the drop property if and only if X is reflexive and has property H. A sequence {x n } is said to be ε-separated sequence for some ε >0if sepx n =inf { x n x m : n m } > ε. A Banach space X is called nearly uniformly convex NUC if for every ε >0,thereexists δ 0, 1 such that for every sequence x n BX withsepx n >ε, wehaveconvx n 1 δbx. Huff[18] proved that every NUC Banach space is reflexive and has property H. ABanachspace X hasproperty β ifandonlyifforeach ε >0,thereexists δ > 0 such that for each element x BXandeachsequencex n inbxwithsepx n ε, there is an index k for which x+x k <1 δ. For a real vector space X, a function ρ : X [0, ] is called a modular if it satisfies the following conditions: i ρx=0if and only if x =0, ii ραx=ρx for all scalar α with α =1, iii ραx + βy ρx+ρy for all x, y X and all α, β 0 with α + β =1. The modular ρ iscalledconvex if iv ραx + βy αρx+βρy for all x, y X and all α, β 0 with α + β =1. For any modular ρ on X,thespace X ρ = { x X : ρσ x< for some σ >0 } is called a modular space. In general, the modular is not subadditive and thus it does not behave as a norm or a distance. But we can associate the modular with an F-norm. A functional : X [0, ]definesanf-norm if and only if i x =0 x =0, ii αx = x whenever α = 1, iii x + y x + y, iv if α n α and x n x 0,then α n x n αx 0.

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 3 of 10 F-norm defines a distance on X by dx, y= x y. If the linear metric space X, d is complete, then it is called an F-space. The modular space X ρ can be equipped with the following F-norm: { } x x = inf α >0:ρ α. α If the modular ρ is convex, then the equality x = inf{α >0:ρ x 1} defines a norm α whichiscalledtheluxemburgnorm. Amodularρ is said to satisfy the δ -condition if for any ε > 0, there exist constants K, a >0suchthatρu Kρu +ε for all u X ρ with ρu a. Ifρ provides the δ -condition for any a >0withK dependentona, thenρ provides the strong δ -condition briefly ρ δ s. Let us denote by l 0 the space of all real sequences. The Cesàro sequence spaces { Ces p = x l 0 : n=1 n 1 p } n x i <, 1 p <, i=1 and { } n Ces = x l 0 : sup n 1 x i <, n i=1 were introduced by Shiue [1]. Jagers [] determined the Köthe duals of the sequence space Ces p 1 < p <. It can be shown that the inclusion l p Ces p is strict for 1 < p < although it does not hold for p = 1. Also, Suantai [3] defined the generalized Cesàro sequence space by cesp= { x l 0 : ρλx< for some λ >0 }, where ρx= n=1 1 n n i=1 xi p n.ifp =p n isbounded,then { cesp= x =x k : n=1 n 1 n xi pn } <. i=1 The sequence space Cs, p was defined by Bilgin [] asfollows: { Cs, p= x =x k : r r k s x k pr <, s 0 } for p =p r withinf p r >0,where r denotes a sum over the ranges r k < r+1.the special case of Cs, pfors =0isthespace { Cesp= x =x k : r r x k pr < }

Et et al. Fixed Point Theory and Applications 013, 013:165 Page of 10 which was introduced by Lim [5]. Also, the inclusion Cesp Cs, pholds.aparanorm on Cs, pisgivenby ρx= r r pr 1/M k s x k for M = max1, HandH = sup p r <. The Z-transform of a sequence x =x k isdefinedbyzx n = y n = αx n +1 αx n 1 by using the Zweier operator α, k = n, Z =z nk = 1 α, k = n 1, 0, otherwise for n, k N and α F\{0}, where F is the field of all complex or real numbers. The Zweier operator was studied by Şengönül and Kayaduman [6]. Now we introduce a new modular sequence space Z σ s, pby Z σ s, p= { x l 0 : σ tx<, forsomet >0 }, where σ x= r r k s αx k +1 αx k 1 p r < and s 0. If we take α =1,then Z σ s, p =Cs, p; if α =1ands =0,thenZ σ s, p =Cesp. It can be easily seen that σ : Z σ s, p [0, ] isamodularonz σ s, p. We define the Luxemburg norm on the sequence space Z σ s, p as follows: { } x x = inf t >0:σ 1, x Z σ s, p. t It is easy to see that the space Z σ s, p is a Banach space with respect to the Luxemburg norm. Throughout the paper, suppose that p =p r isboundedwithp r >1forallr N and e i = i 1 {}}{ 0,0,...,0,1,0,0,0,..., x i = x1, x, x3,...,xi,0,0,0,..., x N i = 0,0,0,...,xi +1,xi +,..., for i N and x l 0. In addition, we will require the following inequalities: a k + b k p k C a k p k + b k p k, a k + b k t k a k t k + b k t k, where t k = p k M 1andC = max{1, H 1 } with H = sup p k. 3 Main results Since l p is reflexive and convex, lp-type spaces have many useful applications, and it is natural to consider a geometric structure of these spaces. From this point of view, we

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 5 of 10 generalized the space Cs, p by using the Zweier operator and then obtained the equality Z σ s, p=cesp, that is, it was seen that the structure of the space Cesp was preserved. In this section, our goal is to investigate a geometric structure of the modular space Z σ s, p related to the fixed point theory. For this, we will examine property β and the uniform Opial property for Z σ s, p. Finally, we will give some fixed point results. To do this, we need some results which are important in our opinion. Lemma 3.1 [] If σ δ s, then for any L >0and ε >0,there exists δ >0such that σ u + v σ u < ε, where u, v X σ with σ u Landσ v δ. Lemma 3. [] If σ δ s, convergence in norm and in modular are equivalent in X σ. Lemma 3.3 [] If σ δ s, then for any ε >0,there exists δ = δε>0such that x 1+δ implies σ x 1+ε. Now we give the following two lemmas without proof. Lemma 3. If x L <1for any x Z σ s, p, then σ x x L. Lemma 3.5 For any x Z σ s, p, x L =1if and only if σ x=1. Lemma 3.6 If lim inf p r >1,then for any x Z σ s, p, there exist k 0 N and μ 0, 1 such that x k σ 1 μ σ x k k 1 {}}{ for all k N with k k 0, where x k = 0,0,...,0, r i k xi, xk +1,xk +,... and r k < r+1. Proof Let k N be fixed. Then there exists r k N such that k I rk.letγ be a real number 1<γ lim inf p r,andsothereexistsk 0 N such that γ < p rk for all k k 0.Chooseμ 0, 1 such that 1 γ 1 μ. Therefore, we have x k σ = = r r 1 1 γ < 1 μ σ x k k s αxk+1 αxk 1 pr r r r r pr k s αxk+1 αxk 1 pr k s αxk+1 αxk 1 pr for each x Z σ s, pandk k 0.

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 6 of 10 Lemma 3.7 If σ δ s, then for any ε 0, 1, there exists δ 0, 1 such that σ x 1 ε implies x 1 δ. Proof Suppose that lemma does not hold. So, there exist ε >0andx n Z σ s, p such that σ x n <1 ε and 1 x 1 n 1. Take s n = x n 1,andsos n 0asn.Let P = sup{σ x n :n N}.ThereexistsD suchthat σ u Dσ u+1 3.1 for every u Z σ s, pwithσ u<1,sinceσ δ s.wehave σ x n Dσ x n +1<D +1 for all n N by 3.1. Therefore, 0 < P < and from Lemma 3.5 we have xn 1=σ = σ s n x n +1 s n x n x n s n σ x n +1 s n σ x n s n P +1 ε 1 ε. This is a contradiction. So, the proof is complete. Theorem 3.8 The space Z σ s, p has property β. Proof Let ε >0andx n BZ σ s, p with sepx n ε and x BZ σ s, p. For each l N, we can find r k N such that r k l < r k+1.let l 1 x l n = {}}{ 0,0,...,0, r k i l xi, xn l +1,x n l +,.... Since for each i N,x n i i=1 is bounded, by using the diagonal method, we can find a subsequence x nj ofx n suchthatx nj i converges for each i N with 1 i l. Therefore, there exists an increasing sequence of positive integers t l such that sepx l n j j tl ε.thus, there exists a sequence of positive integers r l l=1 with r 1 < r < such that x l r l ε for all l N.Sinceσ δ s,thereisη >0suchthat σ x l r l η for all l N 3. from Lemma 3.3.However,thereexistk 0 N and μ 0, 1 such that v k σ 1 μ σ v k 3.3 for all v Z σ s, pandk k 0 by Lemma 3.6.Thereexistsδ >0suchthat σ y 1 μη y 1 δ 3. for any y Z σ s, p by Lemma 3.7.

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 7 of 10 By Lemma 3.1,thereexistsδ 0 such that σ u + v σ u < μη, 3.5 where σ u 1andσ v δ 0.Hence,wegetthatσx 1sincex BZ σ s, p. Then there exists k k 0 such that σ x k δ 0.Letu = x l r l and v = x l.then σ u v <1 and σ < δ 0. We obtain from 3.3and3.5that u + v σ σ u + μη 1 μ σ u+μη. 3.6 Choose s i = r li.bytheinequalities3., 3.3, 3.6 and the convexity of the function f u= u p r,wehave x + xsk σ r r r k 1 = r r + r r = r=r k k s αxk+x si k + 1 αxk 1+x si k 1 k s αxk+x si k + 1 αxk 1+x si k 1 k s αxk+x si k + 1 αxk 1+x si k 1 1 r k 1 r k s pr αxk+1 αxk 1 r + 1 r k 1 r k s αxsi k+1 αx si k 1 pr r + r k s αx si k+1 αx si k 1 pr + μη r r=r k 1 r k 1 r k s pr αxk+1 αxk 1 r + 1 r k 1 r k s αxsi k+1 αx si k 1 pr r + 1 μ r k s αx si k+1 αx si k 1 pr + μη r r=r k 1 r k 1 r k s pr αxk+1 αxk 1 r + 1 r k s αx si k+1 αx si k 1 pr r pr pr pr

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 8 of 10 μ r r r=r k 1 + 1 μη + μη =1 μη. k s αx si k+1 αx si k 1 pr + μη So, the inequality 3. impliesthat x+x s k 1 δ. Consequently, the space Z σ s, p possesses property β. Since property β implies NUC, NUC implies property D and property D implies reflexivity, we can give the following result from Theorem 3.8. Corollary 3.9 The space Z σ s, p is nearly uniform convex, reflexive and also it has property D. Theorem 3.10 The space Z σ s, p has the uniform Opial property. Proof Let ε >0andx Z σ s, p besuchthat x ε and x n be a weakly null sequence in SZ σ s, p. By σ δ s,thereexistsζ 0, 1 independent of x such that σ x >ζ by Lemma 3..Alsosinceσ δ s, by Lemma 3.1,thereisζ 1 0, ζ suchthat σ y + z σ y < ζ 3.7 whenever σ y 1andσ z ζ 1.Taker 0 N such that r r k s αxk+1 αxk 1 pr < ζ 1. 3.8 Hence, we have ζ < r 0 r=1 + r k s pr αxk+1 αxk 1 r r k s pr αxk+1 αxk 1 r r 0 r=1 r r k s αxk+1 αxk 1 pr + ζ 1 3.9 and this implies that r 0 r=1 r r k s αxk+1 αxk 1 pr > ζ ζ 1 > ζ ζ = 3ζ. 3.10

Et et al. Fixed Point Theory and Applications 013, 013:165 Page 9 of 10 Since x n w 0, by the inequality 3.10, there exists r 0 N such that r 0 r=1 r r k s α x n k+xk +1 α x n k 1+xk 1 pr > 3ζ. 3.11 Again, by x n w 0, there is r 1 > r 0 such that for all r > r 1 x n r0 <1 1 ζ 1/M, 3.1 where p r M N for all r N. Therefore, we obtain that x n N r0 > 1 ζ 1/M 3.13 by the triangle inequality of the norm. It follows from the definition of the Luxemburg norm that xn N r0 1 σ 1 ζ 1/M r r = k s αx n k+1 αx n k 1 1 ζ 1/M 1 M 1 ζ 1/M and this implies that r r pr k s αxn k+1 αx n k 1 pr 3.1 r r k s αxn k+1 αx n k 1 pr 1 ζ. 3.15 By 3.7, 3.8, 3.11, 3.15 and since x n w 0 x n 0 coordinatewise, we have for any r > r 1 that σ x n + x= r 0 r=1 + r r r r r r k s α xn k+xk +1 α x n k 1+xk 1 pr k s α xn k+xk +1 α x n k 1+xk 1 pr k s α x n k+xk +1 α x n k 1+xk 1 pr ζ + 3ζ 3ζ + 1 ζ ζ =1+ ζ.

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