Weak ∗ dentability index of spaces C([0,α])
aa r X i v : . [ m a t h . F A ] J a n WEAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) PETR H ´AJEK, GILLES LANCIEN, AND ANTON´IN PROCH ´AZKA
Abstract.
We compute the weak ∗ -dentability index of the spaces C ( K ) where K is a countable compact space. Namely Dz( C ([0 , ω ω α ])) = ω α +1 , when-ever 0 ≤ α < ω . More generally, Dz( C ( K )) = ω α +1 if K is a scatteredcompact whose height η ( K ) satisfies ω α < η ( K ) ≤ ω α +1 with an α countable. Introduction
The Szlenk index has been introduced in [20] in order to show that there isno universal space for the class of separable reflexive Banach spaces. The generalidea of assigning an isomorphically invariant ordinal index to a class of Banachspaces proved to be extremely fruitful in many situations. We refer to [16] for asurvey with references. In the present note we will give an alternative geometricaldescription of the Szlenk index (equivalent to the original definition whenever X is a separable Banach space not containing any isomorphic copy of ℓ [12]), whichstresses its close relation to the weak ∗ -dentability index. The later index proved tobe very useful in renorming theory ([12], [13], [14]).Let us proceed by giving the precise definitions. Consider a real Banach space X and K a weak ∗ -compact subset of X ∗ . For ε > V be the set of all relativelyweak ∗ -open subsets V of K such that the norm diameter of V is less than ε and s ε K = K \ S { V : V ∈ V} . Then we define inductively s αε K for any ordinal α by s α +1 ε K = s ε ( s αε K ) and s αε K = T β<α s βε K if α is a limit ordinal. We denote by B X ∗ the closed unit ball of X ∗ . We then define Sz( X, ε ) to be the least ordinal α so that s αε B X ∗ = ∅ , if such an ordinal exists. Otherwise we write Sz( X, ε ) = ∞ . The
Szlenkindex of X is finally defined by Sz( X ) = sup ε> Sz(
X, ε ). Next, we introduce thenotion of weak ∗ -dentability index . Denote H ( x, t ) = { x ∗ ∈ K, x ∗ ( x ) > t } , where x ∈ X and t ∈ R . Let K be again a weak ∗ -compact. We introduce a weak ∗ -sliceof K to be any non empty set of the form H ( x, t ) ∩ K where x ∈ X and t ∈ R .Then we denote by S the set of all weak ∗ -slices of K of norm diameter less than ε and d ε K = K \ S { S : S ∈ S} . From this derivation, we define inductively d αε K forany ordinal α by d α +1 ε K = s ε ( d αε K ) and d αε K = T β<α s βε K if α is a limit ordinal.We then define Dz( X, ε ) to be the least ordinal α so that d αε B X ∗ = ∅ , if such anordinal exists. Otherwise we write Dz( X, ε ) = ∞ . The weak ∗ -dentability index isdefined by Dz( X ) = sup ε> Dz(
X, ε ) . Let us now recall that it follows from the classical theory of Asplund spaces (seefor instance [10], [9], [6] and references therein) that for a Banach space X , eachof the following conditions: Dz( X ) = ∞ and Sz( X ) = ∞ is equivalent to X beingan Asplund space. In particular, if X is a separable Banach space, each of theconditions Dz( X ) < ω and Sz( X ) < ω is equivalent to the separability of X ∗ . In Date : July 2008.2000
Mathematics Subject Classification.
Key words and phrases.
Szlenk index, dentability index.Supported by grants: Institutional Research Plan AV0Z10190503, A100190801, GA ˇCR201/07/0394. ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 2 other words, both of these indices measure “quantitatively” the “Asplundness” ofthe space in question. Moreover, these indices are invariant under isomorphism.It is immediate from the definition, that Dz( X ) ≥ Sz( X ) for every Banach space X . Relying on tools from descriptive set theory, Bossard (for the separable case,see [4] and [5]) and the second named author ([14]), proved non-constructively thatthere exists a universal function ψ : ω → ω , such that if X is an Asplund spacewith Sz( X ) < ω , then Dz( X ) ≤ ψ (Sz( X )).Recently, Raja [17] has obtained a concrete example of such a ψ , by showing thatDz( X ) ≤ ω Sz( X ) for every Asplund space. This is a very satisfactory result, butit is not optimal, as we know from [8] that the optimal value ψ ( ω ) = ω . Furtherprogress in this area depends on the exact knowledge of indices for concrete spaces.The Szlenk index has been precisely calculated for several classes of spaces, mostnotably for the class of C ([0 , α ]), α countable (Samuel [19], see also [8]). We haveSz( C ([0 , ω ω α ])) = ω α +1 , so it follows from the Bessaga-Pe lczy´nski ([3]) Theorem 1below, that the value of the Szlenk index characterizes the isomorphism class ([10]).Computations of the Szlenk index for other spaces may be found e.g. in [2], [1],[11]. On the other hand, the precise value of the weak ∗ -dentability index is knownonly for superreflexive Banach spaces, where Dz( X ) = ω ([13], [10]), and for spaceswith an equivalent UKK ∗ renorming ([8]). For a detailed background informationon the Szlenk and dentability indices we refer the reader to [10], [15], [16], [18] andreferences therein.The main result of our note, Theorem 2, is a precise evaluation of the w ∗ -dentability index for the class of C ([0 , α ]), α countable. These spaces have beenclassified isomorphically by C. Bessaga and A. Pe lczy´nski [3] in the following way. Theorem 1. (Bessaga-Pe lczy´nski) Let ω ≤ α ≤ β < ω . Then C ([0 , α ]) is iso-morphic to C ([0 , β ]) if and only if β < α ω . Moreover, for every countable compactspace K there exists a unique α < ω such that C ( K ) is isomorphic to C ([0 , ω ω α ]) . It is also well-known and easy to show that for α ≥ ω , C ([0 , α ]) is isomorphicto C ([0 , α ]) where C ([0 , α ]) = { f ∈ C ([0 , α ]) : f ( α ) = 0 } . The aim of this noteis to prove the next theorem. Note, as a particular consequence, that the weak ∗ -dentability index gives a complete isomorphic characterization of a C ( K ) space,when K is a metrizable compact space (similarly to the case of the Szlenk index). Theorem 2.
Let ≤ α < ω . Then Dz( C ([0 , ω ω α ])) = ω α +1 .Proof. We start by proving the upper estimateDz( C ([0 , ω ω α ])) ≤ ω α +1 , (1)The method of the proof is similar to [8], where a short and direct computationof the Szlenk index of the spaces C ([0 , α ]) is presented. Next lemma is a variantof Lemma 2.2. from [8]. We omit the proof which requires only minor notationalchanges. Lemma 3.
Let X be a Banach space and α an ordinal. Assume that ∀ ε > ∃ δ ( ε ) > d αε ( B X ∗ ) ⊂ (1 − δ ( ε )) B X ∗ . Then
Dz( X ) ≤ α · ω. We shall also use the following Lemma that can be found in [15].
Lemma 4.
Let X be a Banach space and L ( X ) be the Bochner space L ([0 , , X ) .Then Dz( X ) ≤ Sz( L ( X )) . EAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 3 Thus, in order to obtain the desired upper bound we only need to prove thefollowing.
Proposition 5.
Let ≤ α < ω . Then Sz( L ( C ([0 , ω ω α ]))) ≤ ω α +1 .Proof. For a fixed α < ω and γ < ω ω α , let us put Z = L ( ℓ ([0 , ω ω α ))), togetherwith the weak ∗ -topology induced by L ( C ([0 , ω ω α ])) and Z γ = L ( ℓ ([0 , γ ])) withthe weak ∗ -topology induced by L ( C ([0 , γ ])). We recall that for a Banach space X with separable dual, L ( X ∗ ) is canonically isometric to ( L ( X )) ∗ .Let P γ be the canonical projection from ℓ ([0 , ω ω α )) onto ℓ ([0 , γ ]). Then, for f ∈ Z and t ∈ [0 , γ f )( t ) = P γ ( f ( t )). Clearly, Π γ is a norm oneprojection from Z onto Z γ (viewed as a subspace of Z ). We also have that for any f ∈ Z , k Π γ f − f k tends to 0 as γ tends to ω ω α .Next is a variant of Lemma 3.3 in [8]. Lemma 6.
Let α < ω , γ < ω ω α , β < ω and ε > . If z ∈ s β ε ( B Z ) and k Π γ z k > − ε , then Π γ z ∈ s βε ( B Z γ ) .Proof. We will proceed by transfinite induction in β . The cases β = 0 and β a limitordinal are clear. Next we assume that β = µ +1 and the statement has been provedfor all ordinals less than or equal to µ . Consider f ∈ B Z with k Π γ f k > − ε andΠ γ f / ∈ s βε ( B Z γ ). Assuming f / ∈ s µ ε ( B Z ) ⊃ s β ε ( B Z ) finishes the proof, so we maysuppose that f ∈ s µ ε ( B Z ). By the inductive hypothesis, Π γ f ∈ s µε ( B Z γ ). Thusthere exists a weak ∗ -neighborhood V of f such that the diameter of V ∩ s µε ( B Z γ )is less than ε . We may assume that V can be written V = T ki =1 H ( ϕ i , a i ), where a i ∈ R and ϕ i ∈ L ( C ([0 , γ ])). We may also assume, using Hahn-Banach theorem,that V ∩ (1 − ε ) / B Z γ = ∅ .Define Φ i ∈ L ( C ([0 , ω ω α )) by Φ i ( t )( σ ) = ϕ i ( t )( σ ) if σ ≤ γ and Φ i ( t )( σ ) = 0otherwise. Then define W = T ki =1 H (Φ i , a i ). Note that for f in Z , f ∈ W ifand only if Π γ f ∈ V . In particular W is a weak ∗ -neighborhood of f . Considernow g, g ′ ∈ W ∩ s µ ε ( B Z ). Then Π γ g and Π γ g ′ belong to V and therefore theyhave norms greater than (1 − ε ) / . It follows from the induction hypothesis thatΠ γ g, Π γ g ′ ∈ s µε ( B Z γ ) thus k Π γ g − Π γ g ′ k ≤ ε . Since k Π γ g k > − ε and k g k ≤ k g − Π γ g k < ε . The same is true for g ′ and therefore k g − g ′ k < ε .This finishes the proof of the Lemma. (cid:3) We are now in position to prove Proposition 5. For that purpose it is enough toshow that for all α < ω : ∀ γ < ω ω α ∀ ε > s ω α ε ( B Z γ ) = ∅ . (2)We will prove this by transfinite induction on α < ω .For α = 0, γ is finite and the space Z γ is isomorphic to L and therefore s ωε ( B Z γ )is empty. So (2) is true for α = 0.Assume that (2) holds for α < ω . Let Z = L ( C ([0 , ω ω α ])). It follows fromLemma 6 and the fact that for all f ∈ Z k Π γ f − f k tends to 0 as γ tends to ω ω α ,that ∀ ε > s ω α ε ( B Z ) ⊂ (1 − ε ) / B Z . From this and Lemma 3 it follows that ∀ ε > s ω α +1 ε ( B Z ) = ∅ . By Theorem 1 we know that the spaces C ([0 , γ ]), C ([0 , ω ω α ]), and also C ([0 , ω ω α ])are isomorphic, whenever ω ω α ≤ γ < ω ω α +1 . Thus s ω α +1 ε ( B Z γ ) = ∅ for any ε > γ < ω ω α +1 , i.e. (2) holds for α + 1.Finally, the induction is clear for limit ordinals. (cid:3) EAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 4 In the rest of the note, we will focus on proving the converse inequality. Notethat it suffices to deal with the spaces C ([0 , ω ω α ]) where α < ω . Indeed, in case α ≥ ω , our inequality (1) implies thatDz( C ([0 , ω ω α ])) = Sz( C ([0 , ω ω α ])) = ω α +1 . Proposition 7.
Let
X, Z be Banach spaces and let Y ⊂ X ∗ be a closed subspace.Let there be T ∈ B ( X, Z ) such that T ∗ is an isometric isomorphism from Z ∗ onto Y . Let ε > , α be an ordinal such that B X ∗ ∩ Y ⊂ d αε ( B X ∗ ) , and z ∈ Z ∗ . If z ∈ d βε ( B Z ∗ ) , then T ∗ z ∈ d α + βε ( B X ∗ ) .Proof. By induction with respect to β . The cases when β = 0 or β is a limit ordinalare clear. Let β = µ + 1 and suppose that T ∗ z / ∈ d α + βε ( B X ∗ ). If z / ∈ d µε ( B Z ∗ ),then the proof is finished. So we proceed assuming that z ∈ d µε ( B Z ∗ ), which bythe inductive hypothesis implies that T ∗ z ∈ d α + µε ( B X ∗ ). There exist x ∈ X , t >
0, such that T ∗ z ∈ H ( x, t ) ∩ d α + µε ( B X ∗ ) = S and diam S < ε . Consider theslice S ′ = H ( T x, t ) ∩ d µε ( B Z ∗ ). We have h T x, z i = h x, T ∗ z i , so z ∈ S ′ . Also,diam S ′ ≤ diam S < ε as T ∗ is an isometry. We conclude that z / ∈ d βε ( B Z ∗ ), whichfinishes the argument. (cid:3) Let us introduce a shift operator τ m : ℓ ([0 , ω ]) → ℓ ([0 , ω ]), m ∈ N , by letting τ m h ( n ) = h ( n − m ) for n ≥ m , τ m h ( n ) = 0 for n < m and τ m h ( ω ) = h ( ω ). Corollary 8.
Let h ∈ d αε ( B ℓ ([0 ,ω ]) ) . Then τ m h ∈ d αε ( B ℓ ([0 ,ω ]) ) for every m ∈ N .Proof. Indeed, consider the mapping T : C ([0 , ω ]) → C ([0 , ω ]) defined as T (( x (0) , x (1) , . . . , x ( ω ))) = ( x (1) , x (2) , . . . , x ( ω )). Clearly, T ∗ = τ and the as-sertion for m = 1 follows by the previous proposition. For m > (cid:3) Definition 9.
Let α be an ordinal and ε >
0. We will say that a subset M of X ∗ is an ε - α -obstacle for f ∈ B X ∗ if(i) dist( f, M ) ≥ ε ,(ii) for every β < α and every w ∗ -slice S of d βε ( B X ∗ ) with f ∈ S we have S ∩ M = ∅ .It follows by transfinite induction that if f has an ε - α -obstacle, then f ∈ d αε ( B X ∗ ).An ( n, ε ) -tree in a Banach space X is a finite sequence ( x i ) n +1 − i =0 ⊂ X such that x i = x i + x i +1 k x i − x i +1 k ≥ ε for i = 0 , . . . , n −
1. The element x is called the root of the tree ( x i ) n +1 − i =0 . Notethat if ( h i ) n +1 − i =0 ⊂ B X ∗ is an ( n, ε )-tree in X ∗ , then h ∈ d nε ( B X ∗ ).Define f β ∈ ℓ ([0 , α ]), for α ≥ β , by f β ( ξ ) = 1 if ξ = β and f β ( ξ ) = 0 otherwise. Lemma 10. f ω ∈ d ω / ( B ℓ ([0 ,ω ]) ) Proof.
In [7, Exercise 9.20] a sequence is constructed of ( n, B ℓ ([0 ,ω ]) withroots r n = ( 12 n , . . . , n | {z } n − times , , . . . )whose elements belong to P = (cid:8) h ∈ B ℓ ([0 ,ω ]) : k h k = 1 , h ( n ) ≥ , h ( ω ) = 0 (cid:9) . Wehave r n ∈ d n / ( B ℓ ([0 ,ω ]) ), and dist( f ω , P ) = 2. Finally, for every h ∈ P , every x ∈ C ([0 , ω ]) and every t ∈ R such that f ω ∈ H ( x, t ), there exists m ∈ N such that τ m h ∈ H ( x, t ). Therefore the set (cid:8) τ m r n : ( m, n ) ∈ N (cid:9) is an - ω -obstacle for f ω .Thus f ω ∈ d ω / ( B ℓ ([0 ,ω ]) ). (cid:3) EAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 5 Proposition 11.
For every α < ω , f ω ωα ∈ d ω α / ( B ℓ ([0 ,ω ωα ]) ) (3) Proof.
The case α = 0 is contained in Lemma 10. Let us suppose that we haveproved the assertion (3) for all ordinals (natural numbers, in fact) less than or equalto α . It is enough to show, for every n ∈ N , that f ( ω ωα ) n ∈ d ω α n / ( B ℓ ([0 , ( ω ωα ) n ]) ) . (4)Indeed, (4) implies f ( ω ωα ) n ∈ d ω α n / ( B ℓ ([0 ,ω ωα +1 ]) ) . Since f ( ω ωα ) n w ∗ −→ f ω ωα +1 and (cid:13)(cid:13) f ( ω ωα ) n − f ω ωα +1 (cid:13)(cid:13) = 2, we see that { f ( ω ωα ) n : n ∈ N } is an - ω α +1 -obstacle for f ω ωα +1 . That implies (3) for α + 1.In order to prove (4) we will proceed by induction. The case n = 1 follows fromthe inductive hypothesis as indicated above, so let us suppose that n = m + 1 and(4) holds for m .Define the mapping T : C ([0 , ( ω ω α ) n ]) → C ([0 , ω ω α ]) by( T x )( γ ) = x (( ω ω α ) m (1 + γ )) , γ ≤ ω ω α A simple computation shows that the dual map T ∗ is given by( T ∗ g )( γ ) = ( g ( ξ ) , if γ = ( ω ω α ) m (1 + ξ ) , ξ ≤ ω ω α T ∗ is an isometric isomorphism of ℓ ([0 , ω ω α ]) onto rng T ∗ . We claim that B ℓ ([0 , ( ω ωα ) n ]) ∩ rng T ∗ ⊂ d ω α m / ( B ℓ ([0 , ( ω ωα ) n ]) ) . (5)Note that the set of extremal points of B ℓ ([0 , ( ω ωα ) n ]) ∩ rng T ∗ satisfiesext( B ℓ ([0 , ( ω ωα ) n ]) ∩ rng T ∗ ) ⊂ { f γ , − f γ : γ = ( ω ω α ) m (1 + ξ ) , ξ ≤ ω ω α } By the inductive assumption and by symmetry, f ( ω ωα ) m and − f ( ω ωα ) m belong to d ω α m / ( B ℓ ([0 , ( ω ωα ) n ]) ). It is easy to see that more generally, f γ and − f γ belongto d ω α m / ( B ℓ ([0 , ( ω ωα ) n ]) ), whenever γ = ( ω ω α ) m (1 + ξ ) , ξ ≤ ω ω α . Thus we haveverified that ext( B ℓ ([0 , ( ω ωα ) n ]) ∩ rng T ∗ ) ⊂ d ω α m / ( B ℓ ([0 , ( ω ωα ) n ]) ) , and the claim (5) follows using the Krein-Milman theorem.This together with the inductive assumption (3) allows us to apply Proposition 7(with ℓ ([0 , ( ω ω α ) n ]) as X ∗ , C ([0 , ω ω α ]) as Z , and rng T ∗ as Y ) to get f ( ω ωα ) n = T ∗ f ω ωα ∈ d ω α n / ( B ℓ ([0 , ( ω ωα ) n ]) ) . (cid:3) To finish the proof of Theorem 2, we use that for every Asplund space X ,Dz( X ) = ω ξ for some ordinal ξ (see [15, Proposition 3.3], [10]). Combining Propo-sition 11 with (1) we obtain Dz( C ([0 , ω ω α ])) = ω α +1 for α < ω . For ω ≤ α < ω , we use that ω α +1 = ω α +1 = Sz( C ([0 , ω ω α ])) =Dz( C ([0 , ω ω α ])), which finishes the proof. (cid:3) Our next proposition is a direct consequence of Theorem 2, Lemma 4 and Propo-sition 5.
EAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 6 Proposition 12.
Let ≤ α < ω . Then Sz( L ( C ([0 , ω ω α ]))) = ω α +1 . Our main result can be extended to the non separable case as follows.
Theorem 13.
Let ≤ α < ω . Let K be a compact space whose Cantor derivedsets satisfy K ω α = ∅ and K ω α +1 = ∅ . Then Dz( C ( K )) = ω α +1 .Proof. The upper estimate follows from the separable determination of the weak ∗ -dentability index when it is countable and from Theorem 2 (the argument is iden-tical to the one given for the computation of Sz( C ( K )) in [14]).On the other hand, since K ω α = ∅ , we have that Sz( C ( K )) ≥ ω α +1 (see [14] orProposition 7 in [15]). Therefore there is a separable subspace X of C ( K ) such thatSz( X ) ≥ ω α +1 . By considering the closed subalgebra of C ( K ) generated by X , wemay as well assume that X is isometric to C ( L ), where L is a compact metrizablespace. Since Sz( C ( L )) ≥ ω α +1 , it follows from Theorem 2 that Dz( C ( L )) ≥ ω α +1 and finally that Dz( C ( K )) ≥ ω α +1 . (cid:3) References [1] D. Alspach,
The dual of the Bourgain-Delbaen space , Isr. J. Math. 117 (2000), 239–259.[2] D.E. Alspach, R. Judd and E. Odell,
The Szlenk index and local ℓ -indices , Positivity, (2005).[3] C. Bessaga and A. Pe lczy´nski, Spaces of continuous functions (IV) (on isomorphical classi-fication of spaces of continuous functions) , Studia Math., (1960), 53-62.[4] B. Bossard, Codage des espaces de Banach s´eparables. Familles analytiques ou coanalytiquesd’espaces de Banach , C. R. Acad. Sci. Paris S´er. I Math., (1993), 1005-1010.[5] B. Bossard,
Th´eorie descriptive des ensembles et g´eom´etrie des espaces de Banach , Th`ese,Universit´e Paris VI (1994).[6] R. Deville, G. Godefroy and V. Zizler,
Smoothness and renormings in Banach spaces , PitmanMonographs and Surveys , Longman Ed (1993).[7] M. Fabian, P. Habala, P. H´ajek, V. Montesinos, J. Pelant and V. Zizler, Functional analysisand infinite dimensional geometry , CMS Books in Mathematics (Springer-Verlag) (2001).[8] P. H´ajek and G. Lancien,
Various slicing indices on Banach spaces , Mediterranean J. Math.4 (2007) 179–190.[9] P. H´ajek, G. Lancien and V. Montesinos,
Universality of Asplund spaces , Proc. AMS 135,no.7 (2007), 2031–2035.[10] P. H´ajek, V. Montesinos, J. Vanderwerff and V. Zizler,
Biorthogonal systems in Banachspaces , CMS Books in Mathematics (Springer-Verlag) (2007).[11] H. Knaust, E. Odell and T. Schlumprecht,
On asymptotic stucture, the Szlenk index andUKK properties in Banach spaces , Positivity, (1999), 173-199.[12] G. Lancien, Dentability indices and locally uniformly convex renormings , Rocky MountainJ. Math., (Spring 1993), 2, 635-647.[13] G. Lancien, On uniformly convex and uniformly Kadec-Klee renormings , Serdica Math. J., (1995), 1-18.[14] G. Lancien, On the Szlenk index and the weak ∗ -dentability index , Quart. J. Math. Oxford(2), (1996), 59-71.[15] G. Lancien, A survey on the Szlenk index and some of its applications , Revista R. Acad.Cien. Serie A Math., (2006), 209-235.[16] E. Odell,
Ordinal indices in Banach spaces , Extracta Math. 19 (2004), 93–125.[17] M. Raja,
Dentability indices with respect to measures of non-compactness , Journal of Func-tional Analysis 253 (2007), 273-286[18] H.P. Rosenthal,
The Banach spaces C ( K ), Handbook of the Geometry of Banach spaces Vol.2, W.B. Johnson and J. Lindenstrauss editors, Elsevier Amsterdam (2003), 1547-1602.[19] C. Samuel, Indice de Szlenk des C ( K ), S´eminaire de G´eom´etrie des espaces de Banach, Vol.I-II, Publications Math´ematiques de l’Universit´e Paris VII, Paris (1983), 81-91.[20] W. Szlenk, The non existence of a separable reflexive Banach space universal for all separablereflexive Banach spaces , Studia Math., (1968), 53-61. EAK ∗ DENTABILITY INDEX OF SPACES C ([0 , α ]) 7 Mathematical Institute, Czech Academy of Science, ˇZitn´a 25, 115 67 Praha 1, CzechRepublic
E-mail address : [email protected] Universit´e de Franche Comt´e, Besanc¸on, 16, Route de Gray, 25030 Besanc¸on Cedex,France
E-mail address : [email protected] Charles University, Sokolovsk´a 83, 186 75 Praha 8, Czech Republic and Universit´eBordeaux 1, 351 cours de la liberation, 33405, Talence, France.
E-mail address ::