On subsets of deciding the norm convergence of sequences in 1 - PowerPoint PPT Presentation

On subsets of ℓ ∞ deciding the norm convergence of sequences in ℓ 1 Damian Sobota Institute of Mathematics, Polish Academy of Sciences SE | = OP 2016, Fruˇ ska Gora 2016

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R So, ℓ ∞ ∼ = ℓ ∗ 1 (the space of continuous linear functionals on ℓ 1 )

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R So, ℓ ∞ ∼ = ℓ ∗ 1 (the space of continuous linear functionals on ℓ 1 ) �� � � So, � x � 1 = sup � � x , y � � : y ∈ S ℓ ∞

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R So, ℓ ∞ ∼ = ℓ ∗ 1 (the space of continuous linear functionals on ℓ 1 ) �� � � So, � x � 1 = sup � � x , y � � : y ∈ S ℓ ∞ � � So, for ( x n ) n ∈ ω ⊆ ℓ 1 we have 1 → 0 ( the norm convergence ) � x n �

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R So, ℓ ∞ ∼ = ℓ ∗ 1 (the space of continuous linear functionals on ℓ 1 ) �� � � So, � x � 1 = sup � � x , y � � : y ∈ S ℓ ∞ � � So, for ( x n ) n ∈ ω ⊆ ℓ 1 we have 1 → 0 ( the norm convergence ) � x n � only if � x n , y � → 0 for every y ∈ S ℓ ∞ ( the weak convergence )

Little ℓ -spaces � x ∈ R ω : � x � 1 = � � n | x ( n ) | < ∞ ℓ 1 = y ∈ R ω : � y � ∞ = sup n | y ( n ) | < ∞ � � ℓ ∞ = � � S ℓ ∞ = y ∈ ℓ ∞ : � y � ∞ = 1 The action of ℓ ∞ on ℓ 1 x ∈ ℓ 1 , y ∈ ℓ ∞ − → � x , y � = � n x ( n ) · y ( n ) ∈ R So, ℓ ∞ ∼ = ℓ ∗ 1 (the space of continuous linear functionals on ℓ 1 ) �� � � So, � x � 1 = sup � � x , y � � : y ∈ S ℓ ∞ � � So, for ( x n ) n ∈ ω ⊆ ℓ 1 we have 1 → 0 ( the norm convergence ) � x n � only if � x n , y � → 0 for every y ∈ S ℓ ∞ ( the weak convergence ) What about if ? Does the weak convergence imply the norm one?

Schur’s theorem Theorem (Schur, 1921) Every weakly convergent sequence in ℓ 1 is norm convergent.

Schur’s theorem Theorem (Schur, 1921) Every weakly convergent sequence in ℓ 1 is norm convergent. � � x ∈ ℓ ∞ : c 0 = lim n x ( n ) = 0 with the sup norm 0 ∼ c ∗ = ℓ 1

Schur’s theorem Theorem (Schur, 1921) Every weakly convergent sequence in ℓ 1 is norm convergent. � � x ∈ ℓ ∞ : c 0 = lim n x ( n ) = 0 with the sup norm 0 ∼ c ∗ = ℓ 1 � ∈ c 0 � 0 , . . . , 0 , 1 , 0 , . . . Let e n = Then, ( e n ) n ∈ ω doesn’t converge in norm But for every f ∈ ℓ 1 , � e n , f � = f ( n ) → 0 So, ( e n ) n ∈ ω converges weakly

Measures on ω A signed bounded finitely additive function µ : ℘ ( ω ) → R is a measure on ω ba denotes the Banach space of all measures on ω with the variation norm

Measures on ω A signed bounded finitely additive function µ : ℘ ( ω ) → R is a measure on ω ba denotes the Banach space of all measures on ω with the variation norm 1 ∼ → ℓ ∗∗ = ℓ ∗ ℓ 1 ֒ ∞

Measures on ω A signed bounded finitely additive function µ : ℘ ( ω ) → R is a measure on ω ba denotes the Banach space of all measures on ω with the variation norm 1 ∼ ∞ ∼ → ℓ ∗∗ = ℓ ∗ ℓ 1 ֒ = ba (by Riesz’s representation theorem)

Measures on ω A signed bounded finitely additive function µ : ℘ ( ω ) → R is a measure on ω ba denotes the Banach space of all measures on ω with the variation norm 1 ∼ ∞ ∼ → ℓ ∗∗ = ℓ ∗ ℓ 1 ֒ = ba (by Riesz’s representation theorem) ℓ 1 ∋ x �→ µ x ∈ ba by the formula: µ x ( A ) = � x , χ A � = � n ∈ A x ( n ) for every A ∈ ℘ ( ω ) Note that χ A ∈ S ℓ ∞ !

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then:

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω Let ( x n ) n ∈ ω ⊆ ℓ 1 be weakly convergent

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω Let ( x n ) n ∈ ω ⊆ ℓ 1 be weakly convergent Then, � x n , χ A � → 0 for every A ∈ ℘ ( ω )

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω Let ( x n ) n ∈ ω ⊆ ℓ 1 be weakly convergent Then, � x n , χ A � → 0 for every A ∈ ℘ ( ω ) Hence, µ x n ( A ) → 0 for every A ∈ ℘ ( ω )

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω Let ( x n ) n ∈ ω ⊆ ℓ 1 be weakly convergent Then, � x n , χ A � → 0 for every A ∈ ℘ ( ω ) Hence, µ x n ( A ) → 0 for every A ∈ ℘ ( ω ) � = 0 � � { j } �� So by Phillips: lim n � � µ x n j ∈ ω

Phillips’s lemma Theorem (Phillips, 1948) Let ( µ n ) n ∈ ω ⊆ ba . If µ n ( A ) → 0 for every A ∈ ℘ ( ω ) , then: � = 0 . � � �� lim � { j } � µ n n j ∈ ω Let ( x n ) n ∈ ω ⊆ ℓ 1 be weakly convergent Then, � x n , χ A � → 0 for every A ∈ ℘ ( ω ) Hence, µ x n ( A ) → 0 for every A ∈ ℘ ( ω ) � = 0 � � { j } �� So by Phillips: lim n � � µ x n j ∈ ω � � But this means exactly that: lim n � x n 1 = 0! �

Phillips and Schur families Definition A family F ⊆ ℘ ( ω ) is Phillips if for every sequence ( µ n ) n ∈ ω ⊆ ba such that µ n ( A ) → 0 for every A ∈ F , we have � = 0 . � lim � � { j } �� � µ n n j ∈ ω

Phillips and Schur families Definition A family F ⊆ ℘ ( ω ) is Phillips if for every sequence ( µ n ) n ∈ ω ⊆ ba such that µ n ( A ) → 0 for every A ∈ F , we have � = 0 . � lim � � { j } �� � µ n n j ∈ ω Definition A family F ⊆ ℘ ( ω ) is Schur if for every sequence ( x n ) n ∈ ω ⊆ ℓ 1 such that � x n , χ A � → 0 for every A ∈ F , we have � � lim � x n 1 = 0 . � n

Phillips and Schur families Definition A family F ⊆ ℘ ( ω ) is Phillips if for every sequence ( µ n ) n ∈ ω ⊆ ba such that µ n ( A ) → 0 for every A ∈ F , we have � = 0 . � lim � � { j } �� � µ n n j ∈ ω Definition A family F ⊆ ℘ ( ω ) is Schur if for every sequence ( x n ) n ∈ ω ⊆ ℓ 1 such that � x n , χ A � → 0 for every A ∈ F , we have � � lim � x n 1 = 0 . � n Every Phillips family is Schur

Phillips and Schur families Definition A family F ⊆ ℘ ( ω ) is Phillips if for every sequence ( µ n ) n ∈ ω ⊆ ba such that µ n ( A ) → 0 for every A ∈ F , we have � = 0 . � lim � � { j } �� � µ n n j ∈ ω Definition A family F ⊆ ℘ ( ω ) is Schur if for every sequence ( x n ) n ∈ ω ⊆ ℓ 1 such that � x n , χ A � → 0 for every A ∈ F , we have � � lim � x n 1 = 0 . � n Every Phillips family is Schur ℘ ( ω ) is Phillips

Quest for small Phillips families Question Is it consistent that there exists a Phillips family of cardinality strictly smaller than c ?

Martin’s axiom and Schur families Theorem Assume MA κ ( σ - centered ) for some cardinal number κ . Let F ⊆ ℘ ( ω ) be such that |F| � κ .