The convex hull of a Banach-Saks set Pedro Tradacete Universidad - - PowerPoint PPT Presentation

The convex hull of a Banach-Saks set

Pedro Tradacete

Universidad Carlos III de Madrid

Joint work with J. Lopez-Abad (CSIC) and C. Ruiz-Bermejo (UCM)

Congreso de la RSME 22 de enero de 2013

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Convergent sequences

Let (X, · ) be a Banach space. (xn)n ⊂ X xn − → x ⇔ xn − x → 0 xn

Cesaro

− → x ⇔

• 1

n

n

• j=1

xj − x

• → 0

xn

w

− → x ⇔ x∗(xn − x) → 0, ∀x∗ ∈ X ∗ xn − → x ⇒ xn

Cesaro

− → x ⇒ xn

w

− → x

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Definition

A subset A ⊆ X of a Banach space is called Banach-Saks if every sequence in A has a Ces` aro convergent subsequence. Examples:

1

The unit basis of c0, ℓp, p > 1 are Banach-Saks (and weakly-null)

2

The unit basis of ℓ1 is not (and it is not weakly-null)

3

The unit basis of the Schreier space XS is not, but it is weakly-null. Recall XS is the completion of c00 under the norm given by (an)XS = sup

E∈S

• n∈E

|an|, where S is the class of finite sets of the form {n1 < n2 < · · · < nk} with k ≤ n1.

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Convex hulls

A compact ⇒ A Banach-Saks ⇒ A weakly-compact. Given A ⊂ X, co(A) := closure

i

λixi : λi ≥ 0,

• i

λi ≤ 1, xi ∈ A

• A compact ⇒ co(A) compact (Mazur).

A weakly-compact ⇒ co(A) weakly-compact (Krein-Smulian) Question: A Banach-Saks ⇒ co(A) Banach-Saks?

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Positive results

A Banach space has the weak Banach-Saks property if every weakly convergent sequence has a Ces` aro convergent subsequence. Examples: Lp (1 ≤ p < ∞), c0, . . .

Proposition

Let X have the weak Banach-Saks property. A ⊂ X is Banach-Saks if and only if co(A) is Banach-Saks.

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Positive results

A sequence (xn)n in a Banach space X is weakly uniformly convergent to x ∈ X if for every ε > 0, there is n(ε) ∈ N such that for every x∗ ∈ X ∗ #{n ∈ N : |x∗(xn − x)| ≥ ε} ≤ n(ε).

Theorem (Mercourakis)

(xn)n converges uniformly weakly to x ⇔ ∀(xnk)k, xnk

Cesaro

− → x.

Proposition

If (xn)n is uniformly weakly convergent ⇒ co({xn}) is Banach-Saks.

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Schreier spaces

Theorem (Gonz´ alez-Rodr´ ıguez)

A ⊆ XS is Banach-Saks if and only if co(A) is Banach-Saks. The Schreier family S can be extended by induction S2 = S⊗S =

• s1∪· · ·∪sn : si ∈ S, s1 <· · ·<sn, {min(s1), . . . , min(sn)}∈ S
• S3 = S2 ⊗ S

. . . Sα can be defined for any countable ordinal α.

Theorem

A ⊆ XSα is Banach-Saks if and only if co(A) is Banach-Saks.

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Towards a counterexample

Let F ⊆ [N]<ω be a compact family. (an)XF = sup

E∈F

• n∈E

|an|

Definition

A family F is large in M when for every n ∈ N and N ⊆ M there is s ∈ F such that #(s ∩ N) ≥ n.

Definition

A T-family is a compact and hereditary family F on N such that: (1) F is never large in any M ⊆ N. (2) There is a partition

n In = N in finite sets In and δ > 0 such that

Gδ(F) := {t ⊆ N : there is s ∈ F with #(s ∩ In) ≥ δ#In for all n ∈ t} is large.

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Theorem

There is a T-family. In fact, for every ε > 0 there is a family F such that

1

F is not 4-large in any M.

2

G1−ε(F) = S. Therefore, in XF every subsequence of the unit basis (un)n has a subsequence 4-equivalent to the unit basis of c0. In particular, (un)n is Banach-Saks. While, the sequence xn =

1 #In

• j∈In uj ∈ co({un}) is equivalent to the

unit basis of Schreier space XS. Thus, (xn)n is not Banach-Saks.

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Proof of the Theorem:

Please, go to http://arxiv.org/abs/1209.4851

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Proposition (Gillis)

For every ε > 0, δ > 0 and m ∈ N, there is n := n(ε, δ, m) such that for every probability space (Ω, F, µ) and every sequence (Ai)i<n with µ(Ai) ≥ ε for all i < n, there is s ⊂ {1, . . . , n} with ♯s = m such that µ(

• i∈s

Ai) ≥ (1 − δ)εm. A key idea in the proof of our theorem is the following construction by Erd˝

• s and Hajnal:

Let r, n ∈ N. Given i < j < n, let Ai,j := {(ak)k<n ∈ r n : ai = aj}. Clearly #Ai,j = r n−1(r − 1). Now if s ⊆ n has cardinality ≥ r + 1, then

• {i,j}∈[s]2

Ai,j = ∅. This provides a counterexample for double-indexed sequences of the expected generalization of Gillis’ result.

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Thank you very much for your attention.

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