ON WEAK FILTER CONVERGENCE AND THE RADON-RIESZ TYPE THEOREM?

Lingxin BAO(鮑玲鑫)School of Computer and Information,Fujian Agriculture and Forestry University,Fuzhou 350002,China

?

ON WEAK FILTER CONVERGENCE AND THE RADON-RIESZ TYPE THEOREM?

Lingxin BAO(鮑玲鑫)
School of Computer and Information,Fujian Agriculture and Forestry University,Fuzhou 350002,China

E-mail:bolingxmu@sina.com

AbstractThe author shows a characterization of a(unbounded)weakly filter convergent sequence which is parallel to that every weakly null sequence(xn)in a Banach space admits a norm null sequence(yn)with yn∈co(xk)k≥nfor all n∈N.A version of the Radon-Riesz type theorem is also proved within the frame of the filter convergence.

Key wordsstatistical convergence;filter convergence;Radon-Riesz property;Banach space

2010 MR Subject Classi fi cation40A35;46B15

?Received October 14,2014;revised January 3,2015.The author was partially supported by the Natural Science Foundation of China(11426061,11501108)and the Natural Science Foundation of Fujian province(2015J01579).

1 Introduction

Statistical convergence was independently introduced by Fast[6]and Steinhaus[17]in 1951,which is a generalization of the usual notion of sequential convergence.Over the past 60 years several generalizations and applications of this notion were investigated.As an example among a large amount of literature,we refer the reader to[1-3,5,7,10-12]and[14].A sequence(xn)in a topological space X is said to be statistically convergent to x∈X provided for any neighborhood U of x,we havedenotes the characteristic function of a set A.

The notion of statistical convergence was generalized in different ways.The original notion was introduced for X=R,and there are dozens of its generalizations.Generally speaking,this notion was extended in two directions:one is to discuss statistical convergence in more general spaces,for example,locally convex spaces[13],including Banach spaces with the weak topologies[5,10,11],and general topological spaces[14].The other is to consider generalized notions de fined by various limit processes,for example,A-statistical convergence[4],lacunary statistical convergence[8].The most general notion of statistical convergence is ideal(or filter)convergence[12].The aim of this paper is to study some classical results of Banach space theory within the frame of the filter convergence.

Many meaningful work on statistical convergence started in the frame of Banach space theory.We would like to mention that Connor,Ganichev and Kadets[5]made very nicee ff ort to characterize some properties of Banach spaces by using weakly statistical convergent.Cheng et al.[2,3]presented a measure approach to statistical convergence and to some of its generalizations:A non-negative finitely additive measureμon the measurable space(N,2N)is said to be a statistical measure providedμ(N)=1,andμ({k})=0 for all singletons{k}∈2N.Letbe the set of all statistical measuresμon N.Given a family S of statistical measures,we say that a sequence(xn)in a Banach space X is(measure)S-convergent to x∈X provided μ(A(ε))=0 for allμ∈S and all ε＞0,where A(ε)={n∈N:‖xn?x‖≥ε}.They prove that for each type of target statistical convergence,there is a corresponding family S of statistical measures such that the statistical convergence is just S-convergence.Recently,Bao and Cheng[1]further uni fied every type of generalized statistical convergence via statistical measures:for each ideal I?2N,let XIbe the closure of span{χA:A∈I}in ?∝,pIbe the quotient norm of the quotient space ?∝/XI.Then the I-convergence is equivalent to the measure S-convergence,where S=?pI(e)?χ(·)?,e≡χN.Conversely,for any class S?,then the measure S-convergence coincides with the I-convergence,where I={A?N:μ(A)=0 for allμ∈S}.

The notion of weak convergence in a Banach space di ff er signi fi cantly from weak filter convergence.In particular a weakly convergent sequence must be bounded,but,a weakly statistical(fi lter)convergent sequence can tent to in finity in norm[5].The purpose of this paper is to characterize a sequence(xn)in a Banach space which is weakly filter convergence and present a Radon-Riesz type theorem in the frame of the filter convergence.We show that a sequence(xn)in Banach space X is weakly F-convergent to x∈X if and only if for each S≡{nk}∈F?,there exists a sequence(yk)?X with yk∈co(xnj)j≥kfor all k∈N such that(yk)is convergent to x in the norm topology.Let X be a Banach space with a separable dual X?.We also show that X has Radon-Riesz property if and only if X has Radon-Riesz property within the frame of F-convergence.

2 Main Results

For a real Banach space X,SXstands for its unit sphere,and X?its dual.For a set A∈X,(co(A),resp.)presents the closure(convex hull,resp.)of A.

The notions of filter and ideal of subsets are two complementary concepts.A filter F is a collection of subsets of a set ? satisfying i);ii)A,B∈F implies A∩B∈F;iii)A∈F and A?B?? entail B∈F.A filter F is said to be free if∩{F∈F}=?;otherwise,it is called principle.An ideal I on the set ? is a collection of subsets of ? satisfying i)A,B∈I implies A∪B∈I;iii)A∈I and B?A entail B∈I.For a filter F,we denote by the ideal IFcorresponding to F,i.e.,IF≡{NF:F∈F}.In this paper,we shall assume ?=N,and every filter contains the Fréchet filter FF≡{A∈N:(NA)?＜∞},where A?denotes the cardinality of A.Then any such filter F is free.

A subset of A is called stationary with respect to a filter F if it has nonempty intersection with each member of the filter.Denote the collection of all F-stationary sets by F?.Note the assumption that F?FF,every A∈F?contains in finitely elements.Any subset of A is either a member of F or the complement of a member of F or the set and its complement are both F-stationary sets.

More about filters and their applications one can find in every advanced general topologytextbook,for example in[18].

Recall that sequence(xn),n∈N,in a topological space X is said to be F-convergent to x∈X(and we write F-limxn=x)if for every neighborhood U of x the set B(U)∈F;or,equivalently,A(U)∈IF.Where A(U)={n∈N:xnU}and B(U)={n∈N:xn∈U}.A sequence(xn)which is F-convergent to x is also said to be ideal IF-convergent to x.Denote Fst={A?N:and remark that Fstis a filter.Then statistical convergence is the same as convergence with respect to the filter Fst.The notion of weak filter convergence can be traced back to Connor,Ganichev and Kadets[5],in which they introduce the notion of weakly statistical convergence and show a number of characterizations of a separable Banach space being finite dimensional,an Asplund space,not containing ?1.

De finition 2.1Let X be a Banach space,and F be a free filter on N.We say that a sequence(xn)?X is weakly F-convergent to x∈X(and we write w-F-limxn=x)if for every ε＞0,and every x?∈X?,A(x?;ε)={n∈N:|〈x?,xn?x〉|≥ε}lies in IF;and that(xn)is a weakly F-null sequence if it is weakly F-convergent to 0.

Remark 2.2It was shown in[5]that dimX＜∞if and only if every weakly statistically null sequence contains a bounded subsequence.

Though a direct consequence of Remark 2.2 is that every in finite dimensional Banach space has a weakly statistically null sequence not containing any bounded subsequence,the following theorem allows us to see that a weakly statistically null sequence admits again“convex hull selection convergent property”.

Theorem 2.3Suppose that F is a free filter on N.Then a sequence(xn)in Banach space X is weakly F-convergent to x∈X if and only if for each S≡{nj}∈F?,there exists a sequence(yk)?X with yk∈co(xnj)j≥kfor all k∈N such that(yk)is convergent to x in the norm topology.

ProofWithout loss of generality,we can assume that x=0.

Su ffi ciency.Suppose that(xn)is not weakly F-null.Then there exists x?∈SX?such that real sequence{〈x?,xn〉}is not F-null,i.e.,there exists ε＞0,such that A(ε)≡{n∈N:|〈x?,xn〉|≥ε}IF.Then A(ε)∈F?.Set A±(ε)≡{n∈N:±〈x?,xn〉≥ε}.Then at least one of the mutually complementary sets,say A+(ε)and A?(ε),belongs to F?.We can assume that A+(ε)∈F?,and let S=A+(ε)≡(nj).Then for everyhaveThis entails that(yk)is not a norm null sequence.

Necessity.Let S≡(nj)∈F?.Clearly,it suffices to show thatfor all k∈N.Otherwise,there exists N∈N such that for all k≥N,By separation theorem(see,for instance,[9]),there exists x?∈X?such that 0＜δk=infy∈Ck〈x?,y〉for all k≥N.This implies that〈x?,xnj〉≥δk≥δN＞0 for all j≥k≥N.On the other hand,since(xn)is weakly F-null,{〈x?,xn〉}?R is F-null.For any 0＜ε＜δN,let B(ε)={n∈N:|〈x?,xn〉|＜ε}.Then B(ε)∈F.since S∈F?,(S∩A)?=∞.Thus,{〈x?,xn〉}n∈S∩Ais again an in finite null subsequence.But this contradicts to that〈x?,xnj〉≥δN＞0 for all but finitely many j∈N.

Remark 2.4This result is achieved with the collaborative e ff ort made by professor LixinCheng.

Note that many classical results may not valid again within the frame of the filter convergence.Recall the Schur theorem states that weak and strong convergence of sequences in ?1coincide.However,Salinas,Kadets,Leonov and Lopez[16]showed that the Schur theorem does not valid again when the ordinary convergence of sequences is substituted by filter convergence.For instance,Fstdoes not have the Schur property([16],Theorem 5.3).In the following,we start with the Radon-Riesz property(a weakened version of the Schur property)for a Banach space with a separable dual.

De finition 2.5(see[15])A normed space X is said to have Radon-Riesz property or the Kadets-Klee property or property(H)if it satisfies the following condition:whenever(xn)is a sequence in X and x an element of X such that(xn)weakly convergent to x and‖xn‖→‖x‖,it follows that xn→x.In this case,we say X have RRP for short.

De finition 2.6Suppose that F is a free filter on N.A normed space X is said to have F-RRP if it satisfies the following condition:whenever(xn)is a sequence in X and x an element of X such that w-F-limxn=x and F-lim‖xn‖=‖x‖,it follows that F-limxn=x.

Now we have the following theorem.

Theorem 2.7Suppose that X is a Banach space with a separable dual X?,and F is a free filter on N.Then X has RRP if and only if X has F-RRP.

ProofNecessity.Suppose that(xn)is a sequence in X and x an element in X such that w-F-limxn=x and F-lim‖xn‖=‖x‖.To show F-limxn=x.Otherwise,there exists ε0＞0 such that A(ε0)={n∈N:‖xn?x‖≥ε0}∈F?.Let{x?m}be a countable dense subset of X?.For each ε＞0,de fi ne

Note any set in F?is an in finite subset of N,we can select a subsequence(xnk)of the sequence(xn)n∈A(ε0)such that for each k∈N,

Therefore,it is not difficult to see that(xnk)weakly convergent to x and(‖xnk‖)convergent to‖x‖as k→∞.Note that X has RRP,(xnk)convergent to x in norm topology.But this is a contradiction to‖xnk?x‖≥ε0for all k∈N.

Su ffi ciency.Suppose,to the contrary,that there exist a sequence(xn)?X,a vector x∈X and a real positive number t such that(xn)weakly convergent to x and(‖xn‖)convergent to ‖x‖but‖xn?x‖＞t for all n∈N.By assumption,w-F-limxn=x and F-lim‖xn‖=‖x‖.Since X has F-RRP,F-limxn=x.Consequently,for any 0＜ε＜t,B(ε)={n∈N:‖xn?x‖＜ε}∈F.This is a contradiction.

Remark 2.8The question is unknown whether Theorem 2.7 is valid while the assumption that X has separable dual was removed?

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