Quantifying properties (K) and (μ^{s})
aa r X i v : . [ m a t h . F A ] F e b QUANTIFYING PROPERTIES ( K ) AND ( µ s ) DONGYANG CHEN, TOMASZ KANIA, AND YINGBIN RUAN
Abstract.
A Banach space X has property ( K ), whenever every weak* null sequencein the dual space admits a convex block subsequence ( f n ) ∞ n =1 so that h f n , x n i → n → ∞ for every weakly null sequence ( x n ) ∞ n =1 in X ; X has property ( µ s ) if everyweak ∗ null sequence in X ∗ admits a subsequence so that all of its subsequences areCes`aro convergent to 0 with respect to the Mackey topology. Both property ( µ s ) andreflexivity (or even the Grothendieck property) imply property ( K ). In the present paperwe propose natural ways for quantifying the aforementioned properties in the spirit ofrecent results concerning other familiar properties of Banach spaces. Introduction
The present paper is inspired by many recent results that quantify various famil-iar properties of Banach spaces such as weak sequential completeness [KPS], reciprocalDunford–Pettis property [KS], Schur property [KS1], Dunford–Pettis property [KKS],Banach–Saks property [BKS], property ( V ) [Kr], Grothendieck property [Be], etc. Wecontinue this line of research and investigate possible quantifications of related properties( K ) and ( µ s ) introduced by Kwapie´n and Rodr´ıguez, respectively.Mazur’s lemma (see, e.g. , [D, p. 11]) states that every weakly convergent sequencein a Banach space has a convex block subsequence that is norm convergent to the samelimit. (A sequence ( y n ) ∞ n =1 in a Banach space X is a convex block subsequence of a se-quence ( x n ) ∞ n =1 provided that there exists a strictly increasing sequence of positive integers( k n ) ∞ n =1 so that y n ∈ conv( x i ) k n i = k n − +1 for every n ∈ N , where we set k = 0; we denoteby cbs(( x n ) ∞ n =1 ) the collection of all convex block subsequences of ( x n ) ∞ n =1 .) Kalton andPe lczy´nski [KP, Proposition 2.2] proved that if a Banach space X contains an isomorphiccopy of c , then for every σ -finite measure µ the kernel of any surjection Q from L ( µ )onto X is uncomplemented in its second dual. Consequently, ker Q is not isomorphicto a Banach lattice; the original argument relied on the Lindenstrauss Lifting Principle.Having read a preliminary version of [KP], Kwapie´n introduced property ( K ) to providean alternative proof of [KP, Propositon 2.2] which did not appeal to the Lindenstrauss Date : Version: February 2, 2021.
Keywords:
Property ( K ); Grothendieck property; Reflexivity; Banach–Saks property; Property ( µ s ).The first-named author was supported by the National Natural Science Foundation of China(Grant No.11971403) and the Natural Science Foundation of Fujian Province of China (Grant No.2019J01024). The second-named author acknowledges with thanks funding received from SONATA15 No. 2019/35/D/ST1/01734. Lifting Principle (Kwapie´n’s idea was incorporated in [KP], where it was presented withhis permission). Property ( K ) is central to our considerations: Definition 1.1.
A Banach space X has property ( K ), whenever every weak* null se-quence in X ∗ admits a convex block subsequence ( f n ) ∞ n =1 so that lim n →∞ h f n , x n i = 0 forevery weakly null sequence ( x n ) ∞ n =1 in X .Put equivalently, Definition 1.1 stipulates that the sequence ( f n ) ∞ n =1 converges to 0with respect to the Mackey topology µ ( X ∗ , X ), which is the (locally convex) topology ofuniform convergence in X ∗ on weakly compact subsets of X (see [DDS, Lemma 3.5]).Property ( K ) may be thought of as a counterpart of Mazur’s lemma with respect tothe weak* topology. It was shown in [KP] that the space L ( µ ) for a σ -finite measure hasproperty ( K ), yet c fails to have this property. Schur spaces , i.e. , spaces in which weakconvergence of sequences coincides with norm convergence have property ( K ) for trivialreasons. It follows from Mazur’s lemma that Grothendieck spaces, in particular, reflexivespaces, have property ( K ). (A Banach space X is a Grothendieck space , whenever everyweak* convergent sequence in X ∗ converges weakly.)Figiel, Johnson, and Pe lczy´nski [FJP] refined property ( K ) by introducing a weakerproperty that they call property ( k ); this property appeared implicitly also in [Jo]. Prop-erty ( k ) was used in [FJP] to show that the Separable Complementation Property neednot pass to subspaces.It was proved in [FJP] that property ( k ) is enjoyed by every separable subspaceof a weakly sequentially complete Banach lattice, weakly sequentially complete Banachlattices with weak units, and every separable subspace of the predual of a von Neumannalgebra. Oja [FJP] pointed out that the Radon–Nikod´ym property implies property ( k ).However, it was shown [FJP] that the ℓ -sum of continuum many copies of L [0 ,
1] as wellas Banach spaces containing complemented subspaces isomorphic to c fail property ( k ).Property ( K ) admits a number of characterisations. More precisely, let X be a Ba-nach space. Then the following assertions are equivalent:(a) X has property ( K ).(b) Every weak ∗ null sequence in X ∗ admits a convex block subsequence ( f n ) ∞ n =1 sothat lim n →∞ h f n , x n i = 0 for every weakly null sequence ( x n ) ∞ n =1 in X .(c) Every weak ∗ null sequence in X ∗ admits a convex block subsequence that is µ ( X ∗ , X )-null.(d) Every weak ∗ convergent sequence in X ∗ admits a convex block subsequence thatis µ ( X ∗ , X )-Cauchy.In Section 3 of the present paper, we prove quantitative versions of the aforestatedcharacterisations. In order to do so, we introduce a quantity α that characterises µ ( X ∗ , X )-null sequences and subsequently we introduce a quantity K that characterises property UANTIFYING PROPERTIES ( K ) AND ( µ s ) 3 ( K ). This is a quantitative version of clause (c). In order to quantify (b), we introducea quantity β that turns out to be equivalent to α for weak ∗ null sequences. By using thequantity β , we introduce a further quantity K that we then prove is then equivalent tothe quantity K . Finally, to quantify (d), we introduce a quantity K in terms of ca ρ ∗ defined in [KKS] that measures µ ( X ∗ , X )-Cauchyness and prove that K is equivalent to K . In summary, we quantify property ( K ) by means of the following estimates: K ( X ) K ( X ) K ( X )and K ( X ) K ( X ) K ( X ) . Furthermore, we investigate the values of the quantity K in certain familiar Banachspaces failing property ( K ) and obtain that, in particular, K ( c ) = 1 and the K -valueof the ℓ -sum of c copies of L [0 ,
1] is equal to 1. (Curiously, Frankiewicz and Plebanek[FP] proved that under Martin’s Axiom, the ℓ -sum of fewer than c copies of L [0 ,
1] stillhas property ( K ).)The purpose of Section 4 is to quantify the following widely known implications: X is reflexive ⇒ X is a Grothendieck space ⇒ X has property ( K ). ( ⋆ )In order to quantify ( ⋆ ), we first make a slight improvement on characterisationsof weak compactness due to ¨Ulger [U] (see also [DRS]). Using this improvement, weestablish a characterisation of the Grothendieck property, which is used to introduce aquantity G measuring the Grothendieck property. This quantification of the Grothendieckproperty is different from the quantitative Grothendieck property proposed by Kruliˇsov´a(ne´e Bendov´a) in [Be]. Again using the improvement, we introduce a new quantity R measuring reflexivity for Banach spaces. Meanwhile, the relationship between thequantity R and several classical equivalent quantities measuring weak non-compactnessis discussed. We also investigate possible values of the quantity R of some classical Banachspaces. Having introduced G and R , we quantify the implications ( ⋆ ) as follows: K ( X ) G ( X ) R ( X ∗ ) . Avil´es and Rodr´ıguez [AR] studied the implications ( ⋆ ) for Banach spaces not containingisomorphic copies of ℓ and proved that for such space X : X is reflexive ⇔ X is a Grothendieck space ⇔ X has property ( K ). ( ⋆⋆ )Finally, we quantify ( ⋆⋆ ) as follows: K ( X ) G ( X ) R ( X ∗ ) K ( X ) . A bounded subset A of a Banach space X is a Banach–Saks set if each sequence in A has a Ces`aro convergent subsequence. A Banach space X is said to have the Banach–Saks
DONGYANG CHEN, TOMASZ KANIA, AND YINGBIN RUAN property if its closed unit ball B X is a Banach–Saks set. Banach and Saks proved in [BS]that the spaces L p [0 ,
1] and ℓ p (1 < p < ∞ ) enjoy the Banach–Saks property, hence thename. Kakutani [Ka] later showed that uniformly convex spaces have the Banach–Saksproperty (hence so do superreflexive spaces). Any space with the Banach–Saks propertyis reflexive [NW], but there are reflexive spaces without the Banach–Saks property [Ba].A localised version of the result of [NW] says that any Banach–Saks set is relativelyweakly compact [LART].It follows from the Erd˝os–Magidor Theorem (Theorem 2.3) that a Banach space X has the Banach–Saks property if and only if every bounded sequence in X admitsa subsequence such that all of its subsequences are Ces`aro convergent. Property ( µ s ),introduced by Rodr´ıguez [Ro], is a statement refining the Banach–Saks property for weak*null sequences in the dual space. Definition 1.2.
A Banach space X has property ( µ s ), whenever every weak ∗ null sequencein X ∗ admits a subsequence so that all of its subsequences are Ces`aro convergent to 0with respect to µ ( X ∗ , X ).Clearly, if X ∗ has the Banach–Saks property, then X has property ( µ s ). The converseis true for reflexive spaces [Ro, Proposition 2.2]. Moreover, it was pointed out [Ro, Lemma2.1, Remark 2.3] that property ( µ s ) is strictly stronger than property ( K ).The goal of Section 5 is to quantify the following implications ([Ro]): X ∗ has the Banach–Saks property ⇒ X has property ( µ s ) ⇒ X has property ( K ).( ⋆ ⋆ ⋆ )To quantify property ( µ s ), we first introduce a quantity cα by means of α thatmeasures the rate of Ces`aro convergence to 0 with respect to µ ( X ∗ , X ). By using thequantity cα , we introduce a quantity µ s that we then prove characterises property ( µ s ).Furthermore, we introduce a quantity bs( X ) that characterises the Banach–Saks propertyof a Banach space X . This quantity is stronger than the quantity introduced in [BKS]that measures how far a bounded set is from being Banach–Saks. By using the quantities µ s and bs, we quantify ( ⋆ ⋆ ⋆ ) as follows:13 K ( X ) µ s ( X ) bs( X ∗ ) . Finally, we prove that, for a reflexive space X , µ s ( X ) bs( X ∗ ) µ s ( X ) , which is a quantitative version of [Ro, Proposition 2.2].2. Preliminaries
We use standard notation and terminology in-line with [AK] and [LT]. Throughoutthis paper, all Banach spaces are infinite-dimensional over the fixed field of real or complex
UANTIFYING PROPERTIES ( K ) AND ( µ s ) 5 numbers. By a subspace we mean a closed, linear subspace. An operator will always meana bounded linear operator. If X is a Banach space, we denote by B X the closed unitball { x ∈ X : k x k } and by F X the family of all weakly compact subsets in B X . Fora subset A of X , conv( A ) stands for the convex hull of A . For brevity of notation, wedenote by ss(( x n ) ∞ n =1 ) the family of all subsequences of a sequence ( x n ) ∞ n =1 .2.1. Weak compactness.
Let us invoke the following characterisation of weak compact-ness due to ¨Ulger [U].
Lemma 2.1.
A bounded subset A of a Banach space X is relatively weakly compactif and only if given any sequence ( x n ) ∞ n =1 in A , there exists a sequence ( z n ) ∞ n =1 with z n ∈ conv( x i : i > n ) that converges weakly. Diestel, Ruess, and Schachermayer [DRS] improved Lemma 2.1 as follows.
Lemma 2.2.
For a bounded subset A of X the following statements are equivalent:(1) A is relatively weakly compact.(2) For every sequence ( x n ) ∞ n =1 in A , there is a norm-convergent sequence ( z n ) ∞ n =1 such that z n ∈ conv( x i : i > n ) .(3) For every sequence ( x n ) ∞ n =1 in A , there is a weakly convergent sequence ( z n ) ∞ n =1 such that z n ∈ conv( x i : i > n ) . Let A and B are non-empty subsets of a Banach space X , we set • d( A, B ) = inf {k a − b k : a ∈ A, b ∈ B } , • b d( A, B ) = sup { d( a, B ) : a ∈ A } . d( A, B ) is the ordinary distance between A and B , and b d( A, B ) is the (non-symmetrised)Hausdorff distance from A to B . When A is a bounded subset of a Banach space X ,following [KKS], we set • wk X ( A ) = b d (cid:0) A σ ( X ∗∗ ,X ∗ ) , X (cid:1) ; • wck X ( A ) = sup { d(clust X ∗∗ (( x n ) ∞ n =1 ) , X ) : ( x n ) ∞ n =1 is a sequence in A } ,where clust X ∗∗ (( x n ) ∞ n =1 ) is the set of all weak ∗ -cluster points of ( x n ) ∞ n =1 in X ∗∗ . • γ X ( A ) = sup {| lim n lim m h f m , x n i − lim m lim n h f m , x n i| : ( x n ) ∞ n =1 is a sequence in A ,( f m ) ∞ m =1 is a sequence in B X ∗ and all the involved limits exist } .It follows from [AC, Theorem 2.3] thatwck X ( A ) wk X ( A ) γ X ( A ) X ( A ) . Mackey topology.
Let X be a Banach space. The Mackey topology, µ ( X ∗ , X ), isthe strongest locally convex topology on X ∗ which is compatible with the dual pairing h X ∗ , X i . In particular, C w ∗ = C µ ( X ∗ ,X ) for every convex subset C of X ∗ . If the dual unitball endowed with the relative Mackey topology, ( B X ∗ , µ ( X ∗ , X )), is metrisable, then X DONGYANG CHEN, TOMASZ KANIA, AND YINGBIN RUAN has property ( K ). Schl¨uchtermann and Wheeler [SW] term Banach spaces X for which( B X ∗ , µ ( X ∗ , X )) is metrisable as strongly weakly compactly generated (SWCG) spaces.As proved in [SW], a Banach space X is SWCG if and only if there exists a weaklycompact subset K of X so that for every weakly compact subset L of X and ε >
0, thereis a positive integer n with L ⊆ nK + εB X . Moreover, reflexive spaces, separable Schurspaces, the space of operators of trace-class on a separable Hilbert space, and L ( µ ) fora σ -finite measure µ are SWCG.2.3. Banach–Saks sets.
Let ( x n ) ∞ n =1 be a bounded sequence in a Banach space. We setca(( x n ) ∞ n =1 ) = inf n ∈ N sup k,l > n k x k − x l k . Clearly ( x n ) ∞ n =1 is norm-Cauchy if and only if ca(( x n ) ∞ n =1 ) = 0. Following [BKS], wedefine cca (cid:0) ( x n ) ∞ n =1 (cid:1) = ca (cid:0) ( 1 n n X i =1 x i ) ∞ n =1 (cid:1) . Clearly, cca(( x n ) ∞ n =1 ) = 0 if and only if ( x n ) ∞ n =1 is Ces`aro convergent.A subset A of a Banach space is Banach–Saks , whenever every sequence in A hasa Ces`aro convergent subsequence.We shall require the well-known 0-1 law by Erd˝os–Magidor [EM]. Theorem 2.3. (Erd˝os–Magidor) Every bounded sequence in a Banach space has a sub-sequence such that either all its further subsequences are Ces`aro convergent, or none ofthem.
For a bounded set A in a Banach space X , Kruliˇsov´a (ne´e Bendov´a), Kalenda, andSpurn´y [BKS] introduced the following quantitybs( A ) = sup ( x n ) ∞ n =1 ⊆ A inf ( y n ) ∞ n =1 ∈ ss(( x n ) ∞ n =1 ) cca(( y n ) ∞ n =1 )measuring how far is A from being a Banach–Saks set. More precisely, they proved that A is a Banach–Saks set if and only if bs( A ) = 0.3. Quantifications of property ( K ) Let ( f n ) ∞ n =1 be a bounded sequence in X ∗ . Following [KKS], we setca ρ ∗ (( f n ) ∞ n =1 ) = sup K ∈F X inf n ∈ N sup k,l > n sup x ∈ K |h f k − f l , x i| , then ca ρ ∗ (( f n ) ∞ n =1 ) = 0 if and only if ( f n ) ∞ n =1 is µ ( X ∗ , X )-Cauchy ( i.e. , µ ( X ∗ , X )-convergentas the Mackey topology µ ( X ∗ , X ) [Ja, Proposition 4 on p. 197] is complete). We set α (( f n ) ∞ n =1 ) = sup K ∈F X lim sup n →∞ sup x ∈ K |h f n , x i| , UANTIFYING PROPERTIES ( K ) AND ( µ s ) 7 then α (( f n ) ∞ n =1 ) = 0 if and only if ( f n ) ∞ n =1 is µ ( X ∗ , X )-null, and β (( f n ) ∞ n =1 ) = sup ( xn ) ∞ n =1 ⊆ BX weakly null lim sup n →∞ |h f n , x n i| . The following result is a quantitative version of [DDS, Lemma 3.3].
Lemma 3.1.
Let ( f n ) ∞ n =1 be a weak ∗ null sequence in X ∗ . Then β (( f n ) ∞ n =1 ) α (( f n ) ∞ n =1 ) β (( f n ) ∞ n =1 ) . Proof.
The former inequality is trivial. It remains to prove the latter one.Let 0 < c < α (cid:0) ( f n ) ∞ n =1 (cid:1) . Then there exist a weakly compact subset K ⊆ B X ,a subsequence ( f k n ) ∞ n =1 of ( f n ) ∞ n =1 , and a sequence ( x n ) ∞ n =1 in K so that |h f k n , x n i| > c for all n . Since K is weakly compact, by the Eberlein–ˇSmulian theorem, ( x n ) ∞ n =1 admitsa subsequence ( x n m ) ∞ m =1 that converges weakly to some x ∈ K . We define a sequence( z n ) ∞ n =1 in X by z k nm = 12 ( x n m − x ) ( m = 1 , , . . . )and z n = 0 for n / ∈ { k n m } ∞ m =1 . Then ( z n ) ∞ n =1 is weakly null in B X . For each m , we get |h f k nm , z k nm i| = 12 |h f k nm , x n m i − h f k nm , x i| >
12 ( c − |h f k nm , x i| ) . Since ( f n ) ∞ n =1 is σ ( X ∗ , X )-null, we getlim sup n →∞ |h f n , z n i| > lim sup m →∞ |h f k nm , z k nm i| > c . Since c is arbitrary, we arrive at β (( f n ) ∞ n =1 ) > α (( f n ) ∞ n =1 ) . (cid:3) Lemma 3.2.
Let ( f n ) ∞ n =1 be a bounded sequence in X ∗ and f ∈ X ∗ . Then ca ρ ∗ (( f n ) ∞ n =1 ) α (( f n − f ) ∞ n =1 ) . Proof.
Let c > α (( f n − f ) ∞ n =1 ) be arbitrary. Let K ∈ F X . Then there exists a positiveinteger n so that sup x ∈ K |h f k − f, x i| < c for all k > n . Hence, for k, l > n , we getsup x ∈ K |h f k − f l , x i| = sup x ∈ K |h ( f k − f ) − ( f l − f ) , x i| c. This implies that ca ρ ∗ (( f n ) ∞ n =1 ) c . As c was arbitrary, the proof is complete. (cid:3) Lemma 3.3.
Suppose that ( f n ) ∞ n =1 converges to f ∈ X ∗ in the weak* topology. Then α (( f n − f ) ∞ n =1 ) ca ρ ∗ (( f n ) ∞ n =1 ) . Proof.
Let c > ca ρ ∗ (( f n ) ∞ n =1 ) be arbitrary. Let K ∈ F X . Then there exists a positiveinteger n so that sup x ∈ K |h f k − f l , x i| < c for all k, l > n . Hence, for each x ∈ K , we get |h f k − f l , x i| < c for all k, l > n . Letting l → ∞ , we get |h f k − f, x i| c . This meansthat sup x ∈ K |h f k − f, x i| c for all k > n and, consequently, α (cid:0) ( f n − f ) ∞ n =1 (cid:1) c . As c wasarbitrary, the proof is finished. (cid:3) DONGYANG CHEN, TOMASZ KANIA, AND YINGBIN RUAN
Definition 3.4.
Let X be a Banach space. We set K ( X ) = sup ( fn ) ∞ n =1 ⊆ BX ∗ weak ∗ null inf ( g n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) α (cid:0) ( g n ) ∞ n =1 (cid:1) ,K ( X ) = sup ( fn ) ∞ n =1 ⊆ BX ∗ weak ∗ null inf ( g n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) β (cid:0) ( g n ) ∞ n =1 (cid:1) , and K ( X ) = sup ( fn ) ∞ n =1 ⊆ BX ∗ weak ∗ Cauchy inf ( g n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) ca ρ ∗ (cid:0) ( g n ) ∞ n =1 (cid:1) . The three quantities K , K , and K are actually equivalent. Proposition 3.5.
Let X be a Banach space. Then(i) K ( X ) K ( X ) K ( X ) , (ii) K ( X ) K ( X ) K ( X ) . Proof.
The statement (i) follows from Lemma 3.1. It suffices to prove (ii).It follows from Lemma 3.3 that K ( X ) K ( X ). Let 0 < c < K ( X ). Then thereexists a weak ∗ -Cauchy sequence ( f n ) ∞ n =1 in B X ∗ so that for every ( h n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 )we have ca ρ ∗ (( h n ) ∞ n =1 ) > c . Clearly, ( f n ) ∞ n =1 converges to some f ∈ B X ∗ in the weak*topology. Take any ( g n ) ∞ n =1 ∈ cbs(( ( f n − f )) ∞ n =1 ). Then 2 g n = h n − f ( n ∈ N ), where( h n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ). By Lemma 3.2, we get c < ca ρ ∗ (cid:0) ( h n ) ∞ n =1 (cid:1) = 2 ca ρ ∗ (cid:0) ( g n ) ∞ n =1 (cid:1) α (cid:0) ( g n ) ∞ n =1 (cid:1) . Hence c K ( X ). Since c is arbitrary, we get K ( X ) K ( X ) . (cid:3) The subsequent result implies that the quantities K , K , and K do characteriseproperty ( K ). Theorem 3.6.
A Banach space X has property ( K ) if and only if K ( X ) = 0 . To prove Theorem 3.6, we require two elementary lemmata whose proofs are omitted.
Lemma 3.7. If ( y n ) ∞ n =1 ∈ cbs(( x n ) ∞ n =1 ) , then cbs(( y n ) ∞ n =1 ) ⊆ cbs(( x n ) ∞ n =1 ) . More pre-cisely, if y n ∈ conv( x i ) k n i = k n − +1 and z n ∈ conv( y j ) m n j = m n − +1 , then z n ∈ conv( x i ) k mn i = k mn − +1 . Lemma 3.8.
Let ( f n ) ∞ n =1 be a bounded sequence in X ∗ . Then α (cid:0) ( g n ) ∞ n =1 (cid:1) α (cid:0) ( f n ) ∞ n =1 (cid:1) , (cid:0) ( g n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) (cid:1) . Proof of Theorem 3.6.
The necessity is trivial, so it suffices to prove the sufficiency.Let ( f n ) ∞ n =1 be a weak*-null sequence in B X ∗ . Since K ( X ) = 0, for each k we getinductively a sequence ( f ( k ) n ) ∞ n =1 in X ∗ so that for k ∈ N . • ( f (1) n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) , UANTIFYING PROPERTIES ( K ) AND ( µ s ) 9 • ( f ( k +1) n ) ∞ n =1 ∈ cbs(( f ( k ) n ) ∞ n =1 ) , • α (( f ( k ) n ) ∞ n =1 ) < k . For n ∈ N , we set g n = f ( n ) n . By Lemma 3.7, we get ( g n ) n > k ∈ cbs(( f ( k ) n ) ∞ n =1 ) for each k .By Lemma 3.8, we get α (cid:0) ( g n ) ∞ n =1 (cid:1) = α (cid:0) ( g n ) n > k (cid:1) α (cid:0) ( f ( k ) n ) ∞ n =1 (cid:1) < k ( k = 1 , , . . . ) . This means that α (cid:0) ( g n ) ∞ n =1 (cid:1) = 0 and ( g n ) ∞ n =1 is µ ( X ∗ , X )-null. Consequently, X hasproperty ( K ). (cid:3) Example 3.9. (a) Let X be a Banach space so that B X ∗ is σ ( X ∗ , X )-sequentially compact or ℓ doesnot embed into X . If X contains a subspace isomorphic to c , then K ( X ) = 1.In particular, K ( c ) = K ( c ) = K ( C [0 , . (b) K (cid:0) ℓ ( R , L [0 , (cid:1) = 1; here ℓ ( R , L [0 , ℓ -sum of c copies of L [0 , Proof. (a). Let ε >
0. It follows from [DRT, Theorem 6] (respectively, [DF, Theorem 2.2])that there exists a subspace Z of X so that Z is (1 + ε )-isomorphic to c and a projection P from X onto Z with k P k ε . Let T : c → Z be an operator so that11 + ε k z k k T z k k z k ( z ∈ c ) . Let S = T − P . Then ST = I c and k S k (1 + ε ) . For each n , we set f n = S ∗ e ∗ n (1+ ε ) ,where ( e ∗ n ) n is the unit vector basis of ℓ . Then ( f n ) ∞ n =1 is weak ∗ null in B X ∗ . Take any( y ∗ n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) and write y ∗ n = k n X i = k n − +1 λ i f i , where k n P i = k n − +1 λ i = 1 , and λ i > . For every n , let z n = k n P i = k n − +1 e i , where ( e n ) ∞ n =1 is theunit vector unit basis of c . Clearly, ( T z n ) ∞ n =1 is weakly null in B X . Moreover, for every n , we get |h y ∗ n , T z n i| = 1(1 + ε ) |h k n X i = k n − +1 λ i e ∗ i , k n X j = k n − +1 e j i| = 1(1 + ε ) . This means that β (( g n ) ∞ n =1 ) > ε ) and so K ( X ) > ε ) . Letting ε →
0, we get K ( X ) = 1.(b). Let Λ be the set of all strictly increasing sequences ( k n ) ∞ n =1 of positive integerswith k = 1. Set X = ℓ (Λ , L [0 , . Let ( r j ) ∞ j =1 be a sequence of Rademacher functions.Define ( g ∗ n ) ∞ n =1 ⊆ X ∗ by g ∗ n ( t ) = r j ( n,t ) , where t = ( k m ) ∞ m =1 ∈ Λ, k j ( n,t ) n < k j ( n,t )+1 . Since g ∗ n ( t ) weak ∗ −→ L ∞ [0 ,
1] ( t ∈ Λ) and k g ∗ n k = 1 ( n ∈ N ), we get g ∗ n weak ∗ −→
0. Given( h ∗ m ) ∞ m =1 ∈ cbs(( g ∗ n ) n ), we write h ∗ m = k ◦ m +1 − X j = k ◦ m λ j g ∗ j ( t = ( k ◦ m ) m ∈ Λ) . For each m , define h m ∈ X by h m ( t ) = r m if t = t and h m ( t ) = 0 otherwise. Then( h m ) ∞ m =1 is weakly null in B X . Moreover, h h ∗ m , h m i = 1 for each m . This implies that β (( h ∗ m ) ∞ m =1 ) = 1. Consequently, K ( X ) = 1. (cid:3) Quantifying the Grothendieck property and reflexivity
The following result is a slight improvement on Lemma 2.2. For the sake of com-pleteness, we include the proof here.
Lemma 4.1.
For a bounded subset A of X the following are equivalent:(1) A is relatively weakly compact.(2) Every sequence in A admits a convex block subsequence that is norm convergent.(3) Every sequence in A admits a convex block subsequence that is weakly convergent.Proof. (1) ⇒ (2). Given a sequence ( x n ) ∞ n =1 in A . Then ( x n ) ∞ n =1 admits a subsequence( y n ) ∞ n =1 that is weakly convergent. By Mazur’s lemma, ( y n ) ∞ n =1 admits a convex blocksubsequence ( z n ) ∞ n =1 that is norm convergent. It follows from Lemma 3.7 that ( z n ) ∞ n =1 isa convex block subsequence of ( x n ) ∞ n =1 .(2) ⇒ (3) is trivial. It remains to prove (3) ⇒ (1).Let K = conv( A ). Given any f ∈ X ∗ . We let c = sup x ∈ K h f, x i = sup x ∈ A h f, x i . Choosea sequence ( x n ) ∞ n =1 in A so that h f, x n i → c . By the assumption, there exists a sequence( z n ) ∞ n =1 ∈ cbs(( x n ) ∞ n =1 ) so that ( z n ) ∞ n =1 converges weakly to some x ∈ K . It is easyto see that h f, z n i → c . Hence c = h f, x i . It follows from James’ characterisation ofweak compactness via norm-attaining functionals that K is weakly compact and so A isrelatively weakly compact. (cid:3) Proposition 4.2.
A Banach space X has the Grothendieck property if and only if everyweak ∗ null sequence in X ∗ admits a convex block subsequence that is norm null.Proof. The necessity follows from Mazur’s lemma. It remains to prove the sufficiency.Given a weak ∗ null sequence ( f n ) ∞ n =1 in X ∗ and any subsequence ( h n ) ∞ n =1 of ( f n ) ∞ n =1 .By the hypothesis, ( h n ) ∞ n =1 admits a convex block subsequence ( g n ) ∞ n =1 that is norm null.By Lemma 4.1, the sequence ( f n ) ∞ n =1 is relatively weakly compact and hence is weaklynull. Thus X has the Grothendieck property. (cid:3) UANTIFYING PROPERTIES ( K ) AND ( µ s ) 11 Definition 4.3.
Let X be a Banach space. We set G ( X ) = sup ( fn ) ∞ n =1 ⊆ BX ∗ weak ∗ null inf ( g n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) lim sup n →∞ k g n k . The above-defined quantity measures, in a certain sense, how far is a given Banachspace from being a Grothendieck space. This quantification of the Grothendieck propertyis very different from the one proposed by Kruliˇsov´a ([Be]) who introduced the so-called λ - Grothendieck spaces parametrised by λ >
1. Every λ -Grothendieck space is Grothendieckbut not every Grothendieck space is λ -Grothendieck for some λ > Example 4.4. (1) G ( c ) = 1,(2) G ( ℓ ) = 1,(3) G ( C [0 , Proof. (1) is clear.For (2), let ( s n ) ∞ n =1 be the summing basis of c , that is, s n = n P k =1 e k ( n ∈ N ). Then( s ω − s n ) ∞ n =1 is a weak ∗ null sequence in B ℓ ∞ , where s ω is the sequence constantly equalto 1. It is easy to see that for any ( g n ) ∞ n =1 ∈ cbs(( s ω − s n ) ∞ n =1 ) we have k g n k = 1 ( n ∈ N ).Consequently, G ( ℓ ) = 1.In order to prove (3), for the sake of convenience, we consider C [ − ,
1] instead. Foreach n , we define h n ( t ) = − n , − n t < n , t n , otherwiseand ϕ ( t ) = (cid:26) − , − t < , t ν be the Lebesgue measure. A routine argument shows that lim n →∞ R f h n d ν = 0 for all f ∈ C [ − , h n ) ∞ n =1 is a weak ∗ null sequence in B C [ − , ∗ if we vieweach h n ∈ L [ − ,
1] as an element of C [ − , ∗ . Clearly, R ϕ · h n dν = 1 for each n . Takeany ( ν n ) ∞ n =1 ∈ cbs(( h n ) ∞ n =1 ) and write ν n = k n P i = k n − +1 λ i h i . Then h ϕ, ν n i = k n X i = k n − +1 λ i h ϕ, h i i = 1 ( n ∈ N ) , which implies that k ν n k = 1 if we regard ϕ as an element of B C [ − , ∗∗ . We have thusproved that G ( C [ − , (cid:3) We are going to use G to quantify how far is a given Banach space from beinga Grothendieck space. Theorem 4.5.
A Banach space X has the Grothendieck property if and only if G ( X ) = 0 .Proof. The necessary implication follows from Proposition 4.2.Suppose that G ( X ) = 0. Given a weak ∗ null sequence ( f n ) ∞ n =1 in B X ∗ , by induction,for each k , we get a sequence ( f ( k ) n ) ∞ n =1 so that for all k = 1 , , . . . , • ( f (1) n ) ∞ n =1 ∈ cbs (cid:0) ( f n ) ∞ n =1 (cid:1) , • ( f ( k +1) n ) ∞ n =1 ∈ cbs (cid:0) ( f ( k ) n ) ∞ n =1 (cid:1) , • lim sup n →∞ k f ( k ) n k < k . For each n , we set h n = f ( n ) n . By Lemma 3.7, ( h n ) n > k ∈ cbs(( f ( k ) n ) ∞ n =1 ) for each k .Hence lim sup n →∞ k h n k lim sup n →∞ k f ( k ) n k < k ( k ∈ N ) . This implies that ( h n ) ∞ n =1 is a convex block subsequence of ( f n ) ∞ n =1 that converges to 0 innorm. Again by Proposition 4.2, X enjoys the Grothendieck property. (cid:3) Definition 4.6.
Let X be a Banach space. We set R ( X ) = sup ( x n ) ∞ n =1 ⊆ B X inf ( z n ) ∞ n =1 ∈ cbs(( x n ) ∞ n =1 ) ca(( z n ) ∞ n =1 ) . Theorem 4.7.
A Banach space X is reflexive if and only if R ( X ) = 0 .Proof. The necessity follows from Lemma 4.1. To prove the sufficiency, we need [BF, Fact1]: an operator T from a Banach space X to a Banach space Y is weakly compact if andonly if the image under T of every normalised basic sequence in X does not dominatethe summing basis ( s n ) ∞ n =1 of c . In particular, a Banach space X is reflexive if and onlyif every normalised basic sequence in X does not dominate the summing basis ( s n ) ∞ n =1 of c . Assume that X is non-reflexive. Then there exists a normalised basic sequence( x n ) ∞ n =1 in X that dominates the summing basis ( s n ) ∞ n =1 in c . That is, for some constant C >
0, we get k n X i =1 a i x i k > C k n X i =1 a i s i k = C max k n | n X i = k a i | , for all n and all scalars a , a , . . . , a n . By the hypothesis, there exists a sequence ( z n ) ∞ n =1 in cbs(( x n ) ∞ n =1 ), z n = k n P i = k n − +1 λ i x i , so that ca(( z n ) ∞ n =1 ) < C/
2. Thus, for n = m we have k z n − z m k < C , yet k z n − z m k = k k n X i = k n − +1 λ i x i − k m X i = k m − +1 λ i x i k > C k k n X i = k n − +1 λ i s i − k m X i = k m − +1 λ i s i k > C. This contradiction completes the proof. (cid:3)
We discuss the relationship between the quantity R and several commonly usedequivalent quantities measuring weak non-compactness. UANTIFYING PROPERTIES ( K ) AND ( µ s ) 13 Theorem 4.8.
Let X be a Banach space. Then wck X ( B X ) R ( X ) . Proof. Case 1. X is separable.Let 0 < c < wck X ( B X ) be arbitrary. Then there exists a sequence ( x n ) ∞ n =1 in B X so that d(clust X ∗∗ (( x n ) ∞ n =1 ) , X ) > c . Let ε >
0. Take any x ∗∗ ∈ clust X ∗∗ (( x n ) ∞ n =1 ) andlet d = d( x ∗∗ , X ). By the Hahn–Banach theorem, there exists x ∗∗∗ ∈ S X ∗∗∗ so that h x ∗∗∗ , x ∗∗ i = d and h x ∗∗∗ , x i = 0 for all x ∈ X . We let C = B X ∗ ∩ { x ∗∗∗ ∈ X ∗∗∗ : |h x ∗∗∗ , x ∗∗ i − d | < ε } . By Goldstine’s theorem, x ∗∗∗ ∈ C σ ( X ∗∗∗ ,X ∗∗ ) . Since h x ∗∗∗ , x i = 0 for all x ∈ X , we get0 ∈ C σ ( X ∗ ,X ) . Since X is separable, there exists a weak ∗ null sequence ( f m ) ∞ m =1 in C .By passing to a subsequence, we may assume that the limit lim m h x ∗∗ , f m i exists, which isdenoted by a . By the definition of C , | a − d | ε . Since x ∗∗ ∈ clust X ∗∗ (( x n ) ∞ n =1 ), we geta subsequence ( y n ) ∞ n =1 of ( x n ) ∞ n =1 so that |h x ∗∗ − y n , f m i| < n for m = 1 , , . . . , n . Thisimplies that lim n →∞ h f m , y n i = h x ∗∗ , f m i for each m and then lim m →∞ lim n →∞ h f m , y n i = a . Givenany ( z n ) ∞ n =1 ∈ cbs(( y n ) ∞ n =1 ). It is easy to see that lim m →∞ lim n →∞ h f m , z n i = a .We claim that | a | ca(( z n ) ∞ n =1 ). Indeed, for any δ >
0, we may choose a N ∈ N sothat k z n − z N k < ca(( z n ) ∞ n =1 ) + δ for all n > N . Then for each m and n > N , we get |h f m , z n i| ca(( z n ) ∞ n =1 ) + δ + |h f m , z N i| . Since ( f m ) ∞ m =1 is weak ∗ null, we get, by letting n → ∞ and m → ∞ , | a | ca(( z n ) ∞ n =1 ) + δ. As δ was arbitrary, the proof of the claim is complete.It follows that c < d | a | + ε R ( X ) + ε. As c and ε are arbitrary, we get wck X ( B X ) R ( X ) . Case 2 . X is possibly non-separable.Let 0 < c < wck X ( B X ) be arbitrary. Then there exists a sequence ( x n ) ∞ n =1 in B X sothat d(clust X ∗∗ (( x n ) ∞ n =1 ) , X ) > c . Let Y = span { x n : n = 1 , , . . . } and i Y : Y → X bethe inclusion map. Since i ∗∗ Y : Y ∗∗ → X ∗∗ is an isometric embedding, we getd(clust Y ∗∗ (( x n ) ∞ n =1 ) , Y ) > d(clust X ∗∗ (( x n ) ∞ n =1 ) , X ) > c. Indeed, let y ∗∗ ∈ clust Y ∗∗ (( x n ) ∞ n =1 ) and y ∈ Y be arbitrary. Then i ∗∗ Y y ∗∗ ∈ clust X ∗∗ (( x n ) ∞ n =1 )and k y ∗∗ − y k = k i ∗∗ Y y ∗∗ − y k > d(clust X ∗∗ (( x n ) ∞ n =1 ) , X ) . Finally, by Case 1, we get c wck Y ( B Y ) R ( Y ) R ( X ) . As c was arbitrary, the proof is complete. (cid:3) Example 4.9. (1) Let X be a Banach space containing a subspace isomorphic to ℓ . Then R ( X ) = 2.In particular, R ( ℓ ) = R ( C [0 , R ( c ) = 2, where c denotes the space of all convergent scalar sequences equippedwith the supremum norm.(3) 1 R ( c ) . Proof. (1). Let ε >
0. By James’ distortion theorem, there is a sequence ( x n ) ∞ n =1 in B X so that k n P i =1 a i x i k > (1 − ε ) n P i =1 | a i | for all n and all scalars a , a , . . . , a n . For each( z n ) ∞ n =1 ∈ cbs(( x n ) ∞ n =1 ) we write z n = P k n i = k n − +1 λ i x i . Then, for n < m , we get k z n − z m k = k k n X i = k n − +1 λ i x i − k m X i = k m − +1 λ i x i k > − ε ) . This implies that ca(( z n ) ∞ n =1 ) > − ε ) and hence R ( X ) > − ε ). As ε was arbitrary,we proved (1).(2). For each n , let x n ( i ) = (cid:26) , i n − , i > n Given ( z n ) ∞ n =1 ∈ cbs(( x n ) ∞ n =1 ), we write z n = k n P i = k n − +1 λ i x i . Then, for n < m k n X i = k n − +1 λ i x i ( k m − + 1) = − , k m X i = k m − +1 λ i x i ( k m − + 1) = 1 . This implies that k z n − z m k = 2 and so ca(( z n ) ∞ n =1 ) = 2. Thus, we obtain R ( c ) = 2.(3). The inequality R ( c ) > X one has wck X ( B X ) = 1, which follows for example from [GHP, Theorem 1] and[CKS, Proposition 2.2]. The inequality R ( c ) / x n ) ∞ n =1 is a sequence in B c . By passing to a subsequence, we mayassume that ( x n ) ∞ n =1 converges coordinate-wise to some x ∈ B ℓ ∞ . By passing to furthersubsequence and making a small perturbation we may assume that there are k < k < . . . so that x n is supported on { , , . . . , k n } and x n +1 ( i ) = x ( i ) , i = 1 , , . . . , k n . We define z n = x n + x n +1 ( n ∈ N ).We claim that k z n − z m k for all n, m , m > n . Indeed, | z n ( i ) − z m ( i ) | = | x n ( i ) + x ( i ) − x m ( i ) − x ( i ) | , i k n | x n +1 ( i ) − x m ( i ) − x ( i ) | , k n < i k n +1 | − x m ( i ) − x ( i ) | , k n +1 < i k m | − x m +1 ( i ) | , k m < i k m +1 Consequently, ca(( z n ) ∞ n =1 ) and the proof is completed. (cid:3) UANTIFYING PROPERTIES ( K ) AND ( µ s ) 15 We require an elementary lemma whose proof is straightforward.
Lemma 4.10.
Suppose that ( f n ) ∞ n =1 is a weak ∗ null sequence in X ∗ . Then lim sup n →∞ k f n k ca(( f n ) ∞ n =1 ) n →∞ k f n k . An immediate consequence of Lemma 4.10 is the following quantification of implica-tions ( ⋆ ). Theorem 4.11.
Let X be a Banach space. Then K ( X ) G ( X ) R ( X ∗ ) . In order to quantify ( ⋆⋆ ), we require a lemma. Lemma 4.12.
Let X be a Banach space containing no subspaces isomorphic to ℓ . Sup-pose that f n weak ∗ −→ in X ∗ . Then lim sup n →∞ k f n k β (cid:0) ( f n ) ∞ n =1 (cid:1) . Proof.
Let 0 < c < lim sup n →∞ k f n k . By passing to a subsequence, we may assume that k f n k > c for all n . Choose x n ∈ B X with h f n , x n i > c ( n ∈ N ). Passing to a furthersubsequence, by Rosenthal’s ℓ -theorem, we may assume that ( x n ) ∞ n =1 is weakly Cauchy.Let ε >
0. Since f n weak ∗ −→
0, we obtain, by induction, a strictly increasing sequence ( k n ) ∞ n =1 of even integers so that h f k n , x k n − x n − i > c − ε for all n . We set y n = ( x k n − x n − ).Then ( y n ) ∞ n =1 is weakly null in B X . Let us define a weakly null sequence ( z n ) ∞ n =1 in B X by z k n = y n and 0 otherwise. Then β (cid:0) ( f n ) ∞ n =1 (cid:1) > lim sup n →∞ |h f n , z n i| > lim sup n →∞ |h f k n , z k n i| > c − ε . Letting ε →
0, we get β (cid:0) ( f n ) ∞ n =1 (cid:1) > c . As c was arbitrary, the proof is complete. (cid:3) Theorem 4.13.
Let X be a Banach space containing no subspaces isomorphic to ℓ .Then K ( X ) G ( X ) R ( X ∗ ) K ( X ) . Proof.
By Theorem 4.11, it suffices to prove the inequality R ( X ∗ ) K ( X ) . Let 0 < c < R ( X ∗ ). Then there exists a sequence ( f n ) ∞ n =1 in B X ∗ so thatca (cid:0) ( g n ) ∞ n =1 (cid:1) > c (cid:16) ( g n ) ∞ n =1 ∈ cbs (cid:0) ( f n ) ∞ n =1 (cid:1)(cid:17) . Since X contains no isomorphic copy of ℓ , it follows from [Bo, Proposition 3.11] ( cf .[Pf, Proposition 11]) that B X ∗ is weak ∗ convex block compact , that is, every sequence in B X ∗ admits a weak ∗ convergent convex block subsequence. By passing to a convex blocksubsequence, by Lemma 3.7 we may assume that f n weak ∗ −→ f for some f ∈ B X ∗ . Hence, weget ca(( g n ) ∞ n =1 ) > c (cid:16) ( g n ) ∞ n =1 ∈ cbs (cid:0) ( f n − f ) ∞ n =1 (cid:1)(cid:17) . Rescaling if necessary, we may assume that ( f n ) ∞ n =1 is a weak ∗ null sequence in B X ∗ andca (cid:0) ( g n ) ∞ n =1 (cid:1) > c (cid:16) ( g n ) ∞ n =1 ∈ cbs (cid:0) ( f n ) ∞ n =1 (cid:1)(cid:17) . By Lemma 4.10 and Lemma 4.12, we arrive at c < ca (cid:0) ( g n ) ∞ n =1 (cid:1) n k g n k β (cid:0) ( g n ) ∞ n =1 (cid:1) (cid:16) ( g n ) ∞ n =1 ∈ cbs (cid:0) ( f n ) ∞ n =1 (cid:1)(cid:17) . This implies that K ( X ) > c . Since c was arbitrary, the proof is complete. (cid:3) Quantifying property ( µ s ) For a bounded sequence ( f n ) ∞ n =1 in X ∗ , we define cα (cid:0) ( f n ) ∞ n =1 (cid:1) = α (cid:0) ( 1 n n X i =1 f i ) ∞ n =1 (cid:1) . Then cα (( f n ) ∞ n =1 ) = 0 if and only if ( f n ) ∞ n =1 is Ces`aro convergent to 0 with respect to µ ( X ∗ , X ). A direct argument shows that cα (( f n ) ∞ n =1 ) = cα (( f n ) n > N +1 ) for every positiveinteger N . Definition 5.1.
Let X be a Banach space. We set µ s ( X ) = sup ( fn ) ∞ n =1 ⊆ BX ∗ weak ∗ null inf ( g n ) ∞ n =1 ∈ ss(( f n ) ∞ n =1 ) sup ( h n ) ∞ n =1 ∈ ss(( g n ) ∞ n =1 ) cα (cid:0) ( h n ) ∞ n =1 (cid:1) . Theorem 5.2.
A Banach space X has property ( µ s ) if and only if µ s ( X ) = 0 .Proof. The sufficient part is trivial. We only prove the necessary part.Given a weak ∗ null sequence ( f n ) ∞ n =1 in B X ∗ , by induction, for each k we may finda sequence (( g n ) ( k ) ) ∞ n =1 in X ∗ such that • (( g n ) (1) ) ∞ n =1 ∈ ss(( f n ) ∞ n =1 ) , • (( g n ) ( k +1) ) n ∈ ss((( g n ) ( k ) ) n ) , • cα (cid:0) ( g n ) ∞ n =1 (cid:1) < k (cid:0) ( g n ) ∞ n =1 ∈ ss((( g n ) ( k ) ) n ) (cid:1) . Let g n = ( g n ) ( n ) ( n = 1 , , . . . ). Then ( g n ) ∞ n =1 is a subsequence of ( f n ) ∞ n =1 . Take anysubsequence ( h n ) ∞ n =1 of ( g n ) ∞ n =1 . By construction, for each k , there exists N k ∈ N so that( h n ) n > N k +1 ∈ ss((( g n ) ( k ) ) n ) . Consequently, cα (cid:0) ( h n ) ∞ n =1 (cid:1) = cα (( h n ) n > N k +1 ) < k . As k was arbitrary, cα (cid:0) ( h n ) ∞ n =1 (cid:1) = 0. Thus the sequence ( h n ) ∞ n =1 is Ces`aro convergent to0 with respect to µ ( X ∗ , X ), which completes the proof. (cid:3) Definition 5.3.
For a Banach space X , we setbs( X ) = sup ( x n ) ∞ n =1 ⊆ B X inf ( y n ) ∞ n =1 ∈ ss(( x n ) ∞ n =1 ) sup ( z n ) ∞ n =1 ∈ ss(( y n ) ∞ n =1 ) cca(( z n ) ∞ n =1 ) . Clearly, bs( B X ) bs( X ). Combining Theorem 2.3 with [BKS, Corollary 4.3], we seethat bs( X ) = 0 if and only if X has the Banach–Saks property. UANTIFYING PROPERTIES ( K ) AND ( µ s ) 17 Theorem 5.4.
Let X be a Banach space. Then K ( X ) µ s ( X ) bs( X ∗ ) . Proof.
The latter inequality follows from Lemma 3.3, so it remains to prove only theformer one.Let 0 < c < K ( X ) and let ε >
0. Then there exist a weak ∗ null sequence ( f n ) ∞ n =1 in B X ∗ and a subsequence ( g n ) ∞ n =1 of ( f n ) ∞ n =1 such that • α (cid:0) ( h n ) ∞ n =1 (cid:1) > c (cid:0) ( h n ) ∞ n =1 ∈ cbs(( f n ) ∞ n =1 ) (cid:1) , • cα (( g n ) ∞ n =1 ) < µ s ( X ) + ε. It follows from Lemma 3.7 that c < α (cid:0) ( 12 n − n X i =2 n − +1 g i ) ∞ n =1 (cid:1) α (cid:0) ( 12 n n X i =1 g i ) ∞ n =1 (cid:1) + α (cid:0) ( 12 n − n − X i =1 g i ) ∞ n =1 (cid:1) α (cid:0) ( 1 n n X i =1 g i ) ∞ n =1 (cid:1) µ s ( X ) + 3 ε. As c and ε are arbitrary, we arrive at K ( X ) µ s ( X ), which completes the proof. (cid:3) Finally, we present a result that directly quantifies [Ro, Proposition 2.2].
Theorem 5.5.
Let X be a reflexive space. Then µ s ( X ) bs( X ∗ ) µ s ( X ) . Proof.
The former inequality follows from Lemma 3.3, so we need to prove the latter one.Let 0 < c < bs( X ∗ ). Then there exists a sequence ( f n ) ∞ n =1 in B X ∗ so thatsup ( h n ) ∞ n =1 ∈ ss(( g n ) ∞ n =1 ) cca(( h n ) ∞ n =1 ) > c (cid:0) ( g n ) ∞ n =1 ∈ ss(( f n ) ∞ n =1 ) (cid:1) . (5.1)Due to reflexivity, we may assume that f n weak ∗ −→ f for some f ∈ B X ∗ .Given any ( g n ) ∞ n =1 ∈ ss(( f n − f ) ∞ n =1 ) , by (5.1), there exists a subsequence ( h n ) ∞ n =1 of(2 g n + f ) ∞ n =1 such that cca(( h n ) ∞ n =1 ) > c. Again, by reflexivity of X , we get2 cα (cid:0) ( h n − f ) ∞ n =1 (cid:1) > cca(( h n − f ) ∞ n =1 ) = cca(( g n ) ∞ n =1 ) > c. As ( h n − f ) ∞ n =1 is a subsequence of ( g n ) ∞ n =1 , µ s ( X ) > c . Since c was arbitrary, the proof iscomplete. (cid:3) Acknowledgements.
The first-named author would like to thank W. B. Johnson andB. Wallis for helpful discussions and comments. The second-named author is indebtedto G. Plebanek for making available [FP] to us.
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Email address : [email protected] Mathematical Institute, Czech Academy of Sciences, ˇZitn´a 25, 115 67 Praha 1, CzechRepublic, and, Institute of Mathematics and Computer Science, Jagiellonian University, Lojasiewicza 6, 30-348 Krak´ow, Poland
Email address : [email protected], [email protected] College of Mathematics and informatics, Fujian Normal University, Fuzhou, 350007,China
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