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Then one can find a subsequence (hn) of (gn) such that x =P∞ 0 hn belongs to infinitely many An’s or to infinitely many Bn’s

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A useful Lemma concerning subseries convergence

The purpose of this note is to show that several classical results in Functional Analysis can be deduced very easily from a simple lemma concerning subseries con- vergence.

Definition. A series P

gn in an abelian topological group G is said to be subseries convergentif all series P

αngn, α∈ {0,1}N, converge in G.

Main Lemma. Let G be an abelian topological group. Let (An), (Bn) be two sequences of Borel subsets of G, and let (gn) be a sequence in G. Assume that An∪(Bn+gn) = G for each n, and that P

gn is subseries convergent. Then one can find a subsequence (hn) of (gn) such that x =P

0 hn belongs to infinitely many An’s or to infinitely many Bn’s.

Proof.

It is easy to check that the convergence of the seriesP

αngn is uniform with respect to α ∈ {0,1}N. Therefore, identifying {0,1}N with the compact abelian group ∆ = (Z/2Z)N, the formulaϕ(α) =P

0 αngn defines a continuous map ϕ: ∆→G.

For each n ∈ N, put An = ϕ−1(An), Bn = ϕ−1(Bn). The sets An, Bn are Borel subsets of ∆, and we have to show that lim supn(An∪Bn) =T

N

S

n>N(An∪Bn) is nonempty. To this end, it is enough to prove thatm(An∪Bn)≥1/4 for alln, where m is the normalized Haar measure on ∆.

Fix n ∈N and put W0(n) ={α∈ ∆ : αn = 0}, W1(n) ={α∈ ∆ : αn = 1}. Observe that

ϕ−1(Bn+gn)∩W1(n)(n)+Bn ∩W0(n), where δ(n) ∈∆ is defined by δn(n) = 1 and δi(n)= 0 if i6=n.

Since ∆ = An∪ϕ−1(Bn+gn), this implies that W1(n) ⊆ An∪(δ(n)+Bn), whence m(An∪Bn)≥ 12 m(An) +m(Bn)

≥m(W1(n))/2 = 1/4.

We now turn to some applications of the Main Lemma. These include the Uniform Boundedness Theorem, a simple result on automatic continuity, Schur’s l1-theorem, the Nikodym boundedness theorem, the Vitali–Hahn–Sacks theorem, and the Orlicz–

Pettis theorem. Of course, all these results are quite well known and very classical.

Yet, we believe that the proofs given in this note may have some interest, mainly because they are almost identical. The Main Lemma is applied each time in exactly the same way, with sets An=Bn arising immediately from the triangle inequality.

Uniform Boundedness Principle. Let X be a Banach space. If Y is a normed space and if F is a pointwise bounded family of continuous linear operators from X into Y, then F is norm-bounded.

1

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Proof.

Assume that sup

T∈F

||T||= +∞. Then one can find two sequences (xn)⊆X, (Tn)⊆X, such that ||xn||<2−n and ||Tn(xn)||> nfor all n. For each n∈N, put

An={x∈X; ||Tn(x)||> n/2}.

Each An is an open subset of X, and by the triangle inequality, one can write X ={x; ||Tn(x)||> n/2} ∪ {x; ||Tn(x−xn)||> n/2}=An∪(An+xn) for all n∈ N. Since the absolutely convergent seriesP

xn is subseries convergent in the Banach space X, the Main Lemma allows us to find a point x ∈ X such that

||Tn(x)||> n/2 for infinitely many n’s. This shows thatF is not pointwise bounded.

Automatic continuity. Let X be Banach space and let Y be a normed space. If T :X→ Y is a Borel linear mapping, then T is continuous

Proof.

IfT is not continuous, then one can find a sequence (xn)⊆X such that ||xn||<2−n and ||T(xn)|| > n (n ∈ N). Put An = {x ∈ X; ||T(x)|| > n/2}. The sets An

are Borel subsets of X, and X = An∪(An+xn) for all n. By the Main Lemma, we get a point x ∈ X such that ||T(x)|| > n/2 for infinitely many n’s, which is a contradiction.

Schur’s l1-Theorem. In the space l1, weakly convergent and norm-convergent se- quences are the same.

Proof.

Let (xn) be a weakly null sequence in l1, and assume that ||xn|| ≥ε for alln and for some ε >0. Using the fact that (xn) is weakly null, one can find a subsequence (yn) of (xn) and a normalized sequence (yn) in l such that the yn have finite, pairwise disjoint supports and hyn, yni> ε/2 for all n. Put An ={y ∈l; |hy, yni| > ε/4}, n ∈ N. The sets An are w-open, and X = An∪(An+yn) for all n. Moreover, since the yn have norm 1 and are disjointely supported, the series P

yn is subseries convergent in G = (l, w). By the Main Lemma, one can find y ∈ l such that

|hy, yni| > ε/2 for infinitely many n’s. This is impossible because (yn) is weakly null.

Nikodym Boundedness Theorem. Let (X,T) be a measurable space. If M is a family of countably additive set functions defined on T such that sup

µ∈M

|µ(E)|<+∞

for all E ∈ T, then sup{|µ(E)|; E ∈ T, µ∈ M} <+∞.

Proof.

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Assume that ME = supM|µ(E)| < +∞ for all E ∈ T, and that supT ME = +∞.

Choose E1 ∈ T and µ1 ∈ M such that |µ1(E1)| >1 +MX; then |µ1(E1)| >1, and

1(X \E1)| ≥ |µ1(E1)| − |µ1(X)| > 1. Moreover, at least one of sup{MF; F ⊆ E1}, sup{MF; F ⊆ X\E1} is infinite; say supF⊆X\E1MF = +∞. Repeating this argument, one can construct a sequence (µn)⊆ Mand a sequence (En) of pairwise disjoint sets in T such that|µn(En)|> n for all n.

Denote by ca(T) the family of all countably additive set-functions defined on T, and let Gbe the group of all bounded, scalar-valued, T-measurable functions on X, endowed with the topology of pointwise convergence on ca(T). Since the En’s are pairwise disjoint, the series P

1En is subseries convergent in G; moreover, the limit of any subseries ofP

1En is the characteristic function of some setE ∈ T. Now, put An ={f ∈G; |R

f dµn|> n/2}. Each setAnis open in G, andG=An∪(An+1En) for all n. Applying the Main Lemma, we get that Mis not pointwise bounded.

Vitali–Hahn–Sacks Theorem. Let (X,T, µ) be a measure space (µ≥0), and let (µn) be a sequence of countably additive measures defined on T. Assume that each µn isµ-continuous, and that limnµn(E) exists for all setsE ∈ T. Then the sequence (µn) is uniformly µ-continuous.

Proof.

If (µn) is not uniformly µ-continuous, then one can find a subsequence (νn) of (µn) and a sequence (En) ⊆ T such that µ(En) → 0 and |νn(En)| > ε > 0 for all n.

Using the µ-continuity of the νn’s and extracting further subsequences if necessary, we may assume that |νn(E)| < ε/3 for all n and for every set E ∈ T contained in

k>nEk. Thus, putting Fn =En\ ∪k>nEk

, we get a sequence of pairwise disjoint sets in T such that |νn(Fn)| > 2ε/3 and |νn(Fn+1)| < ε/3, n ∈ N; in particular,

n+1(Fn+1) −νn(Fn+1)| > ε/3 for all n. The proof now proceeds exactly in the same way as for the Nikodym Boundedness Theorem: applying the Main Lemma, we get a set F ∈ T such that |νn+1(F)−νn(F)|> ε/6 for infinitely manyn’s, which contradicts the convergence of the sequence (νn(F)).

Orlicz–Pettis Theorem. If X is a normed space, then a series P

xn in X is subseries convergent for the weak topology if and only if it is subseries convergent for the norm topology.

Proof.

It is enough to prove that ifP

xn is subseries convergent inX for the weak topology, then ||xn|| →0. Indeed, when applied to series of the form P

P

k∈Fnαkxk

, where (Fk) is a sequence of pairwise disjoint intervals of N, this result yields that for each α ∈ {0; 1}N, the partial sums of P

αnxn form a norm-Cauchy sequence, and since they converge weakly, it follows that P

xn is norm subseries convergent. Moreover,

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it follows from Mazur’s theorem that if P

xn is weakly subseries convergent in X, then it is weakly subseries convergent in the norm-closed linear span of the xn’s, hence it is enough to consider the case of aseparable normed space. So, assume that X is separable, that P

xn is subseries convergent for the weak topology, and that

||xn||> ε >0 for all n.

First, we observe that for any point a ∈ X, the set {n; ||xn−a|| ≤ ε/2} is finite.

Indeed, it is even empty if ||a|| ≤ε/2 (by the triangle inequality), and if ||a||> ε/2, the result follows because the set {x; ||x|| > ε/2} is weakly open and a−xn → a weakly.

Now, let (Fn) be an increasing sequence of finite subsets of X such that S

nFn is norm-dense in X. Using the preceding remark, one can construct by induction a subsequence (zn) of (xn) such that dist(zn, Fn − Fn) > ε/2 for all n. The sets An = {x; dist(x, Fn) > ε/4} are weakly Borel (actually weakly open), and X = An∪(zn+An) for all n. By the Main Lemma, one can find a point x∈X such that dist(x, Fn) > ε/4 for infinitely many n, hence for all n because (Fn) is increasing.

Since S

nFn is dense in X, this is the required contradiction.

Remark. Obvious modifications of the above proof give the following Orlicz-Pettis type result due to N. Kalton: Let G be an abelian group endowed with two (Haus- dorff) group topologies τ1 ⊆ τ2. Assume that τ2 is separable and admits a neigh- bourhood basis consisting of τ1-closed sets. Then subseries convergence in (G, τ1) is equivalent to subseries convergence in (G, τ2).

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