A Komlos Theorem for general Banach lattices ofmeasurable functions

E. Jimenez Fernandez, A. Juan, E. A. Sanchez Perez

Instituto Universitario de Matematica Pura y Aplicada(I.U.M.P.A.),

Universidad Politecnica de Valencia.

Valencia 2010

Consider a Banach function space X(µ) of (classes of) finitely integrable functions overa σ -finite measure space (Ω,Σ,µ) with the Fatou property. In 2010, Day and Lennardproved that the theorem of Komlos on convergence of Cesaro sums holds also in thesespaces; i.e. for every bounded sequence (fn)n in X(µ), there exists a subsequence(fnk )k and a function f ∈ X(µ) such that for any further subsequence (hj )j of (fnk )k , theseries 1

n ∑nj=1 hj converges µ-a.e. to f . In this paper we generalize this result to a more

general class of Banach spaces of classes of measurable functions —spaces L1(ν) ofintegrable functions with respect to a vector measure on a δ -ring— and explore towhich point the Fatou property and the Komlos property are equivalent. In particular weprove that this always holds for sublattices of spaces L1(ν) with the Fatou property, andprovide an example of a Banach space of measurable functions that is Fatou but do notsatisfy the Komlos Theorem.

Definitions

(Ω,Σ,µ) will be a σ -finite measure space, X(µ) a Banach space over µ and BX(µ) theunit ball of X(µ). Let L0(µ) be the space of all measurable real functions on Ω.We will call Banach (or Kothe) function space to any Banach space X ⊆ L0(µ) withnorm ‖ · ‖X satisfying that

1 if f ∈ L0(µ), g ∈ X(µ) with |f | ≤ |g| µ-a.e. then f ∈ X and ‖f‖X ≤ ‖g‖X(µ).

2 For every A ∈Σ of finite measure, the characteristic function χA belongs to X .

(Ω,Σ,µ) will be a σ -finite measure space, X(µ) a Banach space over µ and BX(µ) theunit ball of X(µ). Let L0(µ) be the space of all measurable real functions on Ω.We will call Banach (or Kothe) function space to any Banach space X ⊆ L0(µ) withnorm ‖ · ‖X satisfying that

1 if f ∈ L0(µ), g ∈ X(µ) with |f | ≤ |g| µ-a.e. then f ∈ X and ‖f‖X ≤ ‖g‖X(µ).

2 For every A ∈Σ of finite measure, the characteristic function χA belongs to X .

1 A Banach function space is order continuous if order bounded increasingsequences are convergent in norm.

2 If a Kothe function space X(µ) is order continuous, then X(µ)′ = X(µ)×.3 Consequently, in this case (X(µ)′,‖ · ‖X(µ)× ) is a Kothe function space and for

every f ∈ X(µ),

‖f‖X(µ) = supg∈BX(µ)×

∣∣∣∣∫Ω

fgdµ

∣∣∣∣ .4 If Y (µ) is other Banach function space, we denote by M(X ,Y ) the space of

multiplication operators from X to Y .5 X(µ) has the weak Fatou property if for every increasing sequence in X(µ) fn ↑ f ,

if sup‖fn‖< ∞ then f ∈ X(µ).

In 1967 Komlos proved that for every bounded sequence (fn)n in L1(µ), for a finitemeasure µ, there exists a subsequence (fnk )k and a function f ∈ L1(µ) such that forany further subsequence (hj )j of (fnk )k , the Cesaro sums

1n

n

∑j=1

hj

converge µ-a.e. to f .

Theorem

Let R be a δ -ring and consider a vector measure ν : R→ X, where X is a Banachspace. Let E be an ideal of L1(ν). Then E has the Fatou property if and only if it hasthe Komlos property.

Theorem

Let ν : R→ X be a locally σ−finite vector measure and consider an ideal E of L1w (ν).

Then E has the Komlos property if and only if E has the weak σ -Fatou. In particular,L1

w (ν) has the Komlos property.

Lemma

Let R be a δ -ring and consider a sequence (An)n ⊂ R. Take U = ∪n≥1An. Then:

(1) The class of subsets R∩U is a δ -ring of subsets of U such that R∩U ⊆ R.

(2) (R∩U)loc = U ∩Rloc as δ -rings of subsets of U.

(3) If ν is a vector measure on R and νU is its restriction to R∩U, thenL1(ν)∩M ((R∩U)loc) = L1(νU ) and L1

w (ν)∩M ((R∩U)loc) = L1w (νU ).

Lemma

Let (Ω,Σ,µ) be a finite measure space and let (Ω,Σ,η) be a σ -finite measure spacesuch that µ and η are equivalent measures. Let E(η) be a Banach function space overη with the weak σ -Fatou property, and let Mg : E(η)→ L1(µ) be a multiplicationoperator given by a Σ-measurable a function g > 0. Then E(η) has the Komlosproperty.

Proof. Let (fn)n be a sequence of functions of E(η) such that ‖fn‖ ≤M for all n ∈ N.Consider the sequence (fng)n in L1(µ). We can apply Komlos theorem in the spaceL1(µ): there exists a subsequence (gfni )i such that for any further subsequence (hnij

)j

of (fni )i , 1k ∑

kj=1 ghnij

converges µ−a.e. to a function h0 ∈ L1(µ). We need to show that

h := h0g ∈ E , since η is equivalent to µ, and then 1

k ∑kj=1 hnij

converges η-a.e. to h. For

each n ∈ N, consider the measurable function kn(w) := ınfk≥n | 1k ∑ki=1 fni (w)|, w ∈Ω. It

is clearly an increasing sequence kn ↑ |h| and ‖kn‖E(η) ≤ 1n ∑

ni=1 ‖fni ‖ ≤M.

The case of L1w (ν)

Definition

Let R be a δ−ring of subsets of Ω, X a Banach space and ν : R→ X a vectormeasure. We say that ν is locally σ−finite with respect to a δ−ring R if given a setB ∈R loc with ‖ν‖(B) < ∞, B can be written as B = (∪n≥1An)∪N, with An ∈R andN ∈R loc a ν−null set.

Theorem

Let ν : R→ X be a locally σ−finite vector measure and consider an ideal E of L1w (ν).

Then E has the Komlos property if and only if E has the weak σ -Fatou. In particular,L1

w (ν) has the Komlos property.

Example. A Fatou Banach lattice of measurable functions that is not Komlos. Considerthe space `∞(I), where I = 2N. Take the sequence (fn) defined as follows. Let w ⊂ 2N.Let us order w increasingly and write it as w = (nk )m

k=1, where m = |w |. Put fn(w) := 0if n is not in w . In other case, put fn1 (w) = 1, fn2 (w) =−1, fn3 (w) = 1, fn4 (w) = 1,fn5 (w) =−1, fn6 (w) =−1, fn7 (w) = 1, fn8 (w) = 1, fn9 (w) = 1, fn10 (w) = 1,fn11 (w) = 1,fn12 (w) = 1, fn13 (w) =−1, fn14 (w) =−1... . This sequence is defined as follows. Take asequence defined as a1 = 3 and ar = 3ar−1−2 for r ≥ 2. We define fn1 (w) = 1,fnk (w) = 1 if k ∈ ar , ...,2ar −2 for any r ≥ 1, and fnk =−1 in other case.Clearly, this sequence is norm bounded by 1. Let us show that 1

m ∑mk=1 fnk do not

converge pointwise —and so n-almost everywhere for the counting measure n— forany subsequence (fnk ) of (fn). If (fnk ) is such a subsequence, take the setw0 := nk : k ∈ N. A direct computation shows that 1

2ar−2 ∑2ar−2k=1 fnk (w0) = 1

2 and1

ar−1 ∑ar−1k=1 fnk (w0) = 0, for every r ∈ N; so the sequence ( 1

m ∑mk=1 fnk (w0))∞

k=1 do notconverge.