A Counterexample to Banach’s Basis Problem

A Counterexample to Banach’s Basis Problem

Viktor Zeh

101.297 Praktikum mit Bachelorarbeit

SS2010

Contents

Introduction 3

1 Coordinate Functionals 4

2 Banach’s Basis Problem 9

2.1 Motivation . . . 9 2.2 A Counterexample to the Basis Problem . . . 9

Bibliography 21

Introduction

Due to the restriction to (finite) linear combinations classical vector space bases are not always suitable for the analysis of infinite dimensional spaces. Therefore, it is natural in some way to consider generalized basis concepts.

Note, that all vector spaces in this paper are spaces over the field F, whereF denotes the field of real numbers R or the field of complex numbers C. Whenever reference is made to some topological property, the norm topology is implied.

Definition 0.1 A sequence(xn)_n∈Nin a Banach space(X,k.k_X)is a called aSchauder basis of Xif for every x∈X there exists a unique sequence(α_n)_n∈N of scalars such that x=P∞

n=1αnxn, i.e. such thatlim_N→∞

x−PN

n=1αnxn

X = 0.

Throughout this paper we make the convention that a basis for a Banach space shall be a Schauder basis, unless explicit reference is made to a vector space basis.

Propostion 0.2 Suppose that (xn)_n∈N is a basis for a Banach space X. Then (xn)_n∈N is linearly independent. In particular, every banach space with a basis is infinite dimen- sional.

Proof:Suppose that an elementxofXcould be written in two different ways as a finite linear combination of the terms of (xn)_n∈N, i.e. forn, m∈N,(αi)ⁿ_i=1 ∈Fⁿ,(βi)^m_i=1∈F^m satisfying α_n 6= 0 6= β_m and (α)ⁿ_i=1 6= (β_i)^m_i=1 we had x = Pn

i=1α_ix_i = Pm i=1β_ix_i. Then we hadP∞

i=1α˜_ix_i=P∞

i=1β˜_ix_i, for ( ˜α_i)i∈N:= (α₁, . . . , α_n,0,0, . . .) and( ˜β_i)i∈N:=

(β1, . . . , βm,0,0, . . .). This is a contradiction to the uniqueness of expansions of vectors in terms of a basis inX.

1 Coordinate Functionals

Our first aim is to show that the continuity of the coordinate functionals - a property that has been required in Schauder’s original defintion - follows from the rest of Definition 0.1.

Definition 1.1 Let X be a Banach space with basis (xn)_n∈N. For each m ∈ N the maps x^∗_m :X → F :P∞

n=1α_nx_n 7→ α_m and P_m : X → X :P∞

n=1α_nx_n 7→ Pm

n=1α_nx_n are called the m^th coordinate functional and the m^th natural projection associated with (xn)_n∈N, respectively.

Remark 1.2 Due to the uniqueness of expansions of each vector in terms of a basis required in definition 0.1 it is instantly verified, that the coordinate functionals actually are linear and the natural projections are projections.

For the sake of convenience we will not work with the original norm of the underlying Banach space, but with the following one.

Lemma 1.3 Let (X,k.k_X) be a Banach space with basis (x_n)_n∈

N. Then the norm k.k defined by the formula kP∞

n=1αnxnk= sup_m∈NkPm

n=1αnxnk_X is a Banach space norm equivalent to the norm k.k_X (i.e. they induce the same topology) satisfying kxk ≥ kxk_X for all x∈X.

Proof: We prove this lemma in four steps. In step one and step two we show, thatk.k actually is a norm and the claimed inequality, respectively. In step three we find a limit in X for a random Cauchy sequence with respect tok.k in X. Finally, in step four we complete the proof by using the open mapping theorem to show the equivalence of the two norms.

Note that in this proof convergence of series in X is always meant to be with respect to k.k_X.

1. Since the other two requirements of the definition of a norm follow instantly from the fact thatk.k_X is a norm, we will confine ourselves to showing that the triangle inequality holds fork.k. For this purpose pick two vectorsxandyinX having the

expansionsx=P∞

n=1α_nx_n andy =P∞

n=1α˜_nx_n. Then by the linearity of P_m

∞

X

n=1

αnxn+

∞

X

n=1

˜ αnxn

= sup

m∈N

m

X

n=1

αnxn+

m

X

n=1

˜ αnxn

X

≤ sup

m∈N

m

X

n=1

α_nx_n X

+

m

X

n=1

˜ α_nx_n

X

!

≤ sup

m∈N

m

X

n=1

αnxn

X

+ sup

m∈N

m

X

n=1

˜ αnxn

X

=

∞

X

n=1

α_nx_n

+

∞

X

n=1

˜ α_nx_n

= kxk+kyk

2. For a vector x in X with the expansion x = P∞

n=1α_nx_n we obtain from the continuity of the norm k.k_X

kxk=

∞

X

n=1

α_nx_n

= sup

m∈N

m

X

n=1

α_nx_n X

≥ lim

n→∞

m

X

n=1

α_nx_n X

=

∞

X

n=1

α_nx_n X

.

3. We want to show next, that an arbitrary Cauchy sequence(b_i)_i∈

N= (P∞

n=1β_n,ix_n)_i∈

with respect tok.kinXconverges towardsP∞ N

n=1βnxninX, forβn:= limi→∞βn,i, n∈N. In order to see that the sequence (β_n,i)_i∈

Nis Cauchy and hence convergent inFfor each n∈Nletj, k, n be inNand n≥2. Then we have

|β_1,j−β_1,k| kx₁k_X = k(β_1,j−β_1,k)·x₁k_X

≤ sup

m∈N

m

X

l=1

(β_l,j−β_l,k)·x_l X

=

∞

X

l=1

(β_l,j−β_l,k)·x_l and

|β_n,j −β_n,k| kx_nk_X = k(β_n,j −β_n,k)·xnk_X

=

n

X

l=1

(β_l,j−β_l,k)·x_l−

n−1

X

l=1

(β_l,j−β_l,k)·x_l X

≤

n

X

l=1

(βl,j−βl,k)·xl

X

+

n−1

X

l=1

(βl,j−βl,k)·xl

X

≤ sup

m∈N

m

X

l=1

(βl,j−βl,k)·xl

X

+ sup

m∈N

m

X

l=1

(βl,j−βl,k)·xl

X

= 2·

∞

X

l=1

(βl,j−βl,k)·xl

.

As (P∞

n=1β_n,ix_n)_i∈

N is Cauchy so must be (β_n,i)_i∈

N for each n ∈ N. Thus the sequence(βn)_n∈N is well-defined.

Since (P∞

n=1βn,ixn)_i∈

N is Cauchy with respect tok.k we may choose i() ∈N for each fixed >0 such that for i, j, M ∈N, i, j ≥i()

3 >

∞

X

n=1

βn,jxn−

∞

X

n=1

βn,ixn

= sup

m∈N

m

X

n=1

βn,jxn−

m

X

n=1

βn,ixn

X

≥

M

X

n=1

βn,jxn−

M

X

n=1

βn,ixn

X

The inequality holds true for j→ ∞. This way we obtain for alli≥i()

M

X

n=1

βnxn−

M

X

n=1

βn,ixn

X

≤

3. (1.1)

We will complete the proof of this step by showing, that(P∞

n=1β_n,ix_n)_i∈

Nconverges towards P∞

n=1β_nx_n. Therefore, it is necessary to show, that P∞

n=1β_nx_n exists.

Due to the completeness of (X,k.k_X) it is sufficient to proof that P∞

n=1βnxn is Cauchy with respect tok.k_X.

Suppose thatm₁, m₂ ∈N, m₂ ≥m₁ >1. Using (1.1) we have

m2

X

n=m1

βnxn−

m2

X

n=m1

βn,ixn

X

(1.2)

=

m2

X

n=m1

βnxn−

m2

X

n=m1

βn,ixn±

m1−1

X

n=1

βnxn±

m1−1

X

n=1

βn,ixn

X

=

m2

X

n=1

β_nx_n−

m2

X

n=1

β_n,ix_n−

m1−1

X

n=1

β_nx_n+

m1−1

X

n=1

β_n,ix_n X

≤

m2

X

n=1

β_nx_n−

m2

X

n=1

β_n,ix_n X

+

m1−1

X

n=1

β_nx_n−

m1−1

X

n=1

β_n,ix_n X

≤

3 +

3 = 2

3. (1.3)

As the series P∞

n=1β_n,ix_n is the expansion of a vector in X in terms of (x_n)_n∈

N, it must be convergent and so must be

PN

n=1β_n,ix_n X

N∈N

. Therefore we can choosem()∈Nsuch that

m2

X

n=m1

β_n,ix_n X

<

3 (1.4)

for m₂, m₁ ∈ N, m₂ ≥ m₁ > m(). Finally (1.3) and (1.4) give us the Cauchy criterion for our seriesP∞

n=1βnxn:

m2

X

n=m1

β_nx_n X

=

m2

X

n=m1

β_nx_n±

m2

X

n=m1

β_n,ix_n X

≤

m2

X

n=m1

βnxn−

m2

X

n=m1

βn,ixn

X

+

m2

X

n=m1

βn,ixn

X

≤ 2 3 +

3 = .

and the convergence ofP∞

n=1β_nx_n is proven.

Since inequality (1.1) still holds when taking the supremum over allM ∈Nwe see

∞

X

n=1

βnxn−

∞

X

n=1

βn,ixn

= sup

M∈N

M

X

n=1

βnxn−

M

X

n=1

βn,ixn

X

≤ 3,

which completes the proof of this step.

4. The identity map I : (X,k.k) → (X,k.k_X) is a bijective, linear and due to the inequality, proofed in step 2, continuous operator. The open mapping theorem¹ ensures, that I⁻¹ : (X,k.k_X)→(X,k.k) is continuous too.

Theorem 1.4 Let (X,k.k_X) be a Banach space with basis (xn)_n∈N. Then all natural projections and all coordinate functionals associated with (x_n)_n∈

N are continuous.

Proof: Fix m in N and a member of X having the expansion P∞

n=1α_nx_n. Define a sequence( ˜αn)n∈N by

˜ αn=

(αn, n≤m 0, else.

Then the unique expansion ofP_m(P∞

n=1α_nx_n)in terms of(x_n)_n∈

Nis given byP∞

n=1α˜_nx_n. Now the continuity of the natural projection Pm for(xn)_n∈N follows from

P_m

∞

X

n=1

α_nx_n

!

=

∞

X

n=1

˜ α_nx_n

= sup

M∈N

M

X

n=1

˜ α_nx_n

X

= sup

M=1,...,m

M

X

n=1

˜ α_nx_n

X

= sup

M=1,...,m

M

X

n=1

α_nx_n X

≤ sup

M∈N

M

X

n=1

α_nx_n X

=

∞

X

n=1

α_nx_n .

1see e.g. [1] theorem 12.1 and 12.5

For eachm >1the coordinate functionalx^∗_m associated with(x_n)_n∈

N is continuous as it is the composition of continuous maps

∞

X

n=1

α_nx_n7→(P_m−Pm−1)

∞

X

n=1

α_nx_n

!

=α_mx_m7→α_m

and so it is form= 1

∞

X

n=1

α_nx_n7→P₁

∞

X

n=1

α_nx_n

!

=α₁x₁7→α₁.

Remark 1.5 Consequently each coordinate functional associated with a basis in a Ba- nach spaceX is a member of the continuous dual spaceX⁰ of X. Though it turns out, that in general the sequence of coordinate functionals does not have to be a basis forX⁰, it can be shown, that it is allways a so called basic sequence, i.e. a basis for the closed linear hull of the collection of all coordinate functionals.

2 Banach’s Basis Problem

2.1 Motivation

In the following proposition we will see, that every Banach Space having a basis is sepa- rable. The question whether the converse is true, i.e. whether every infinite dimensional separable Banach space has a basis, is known as the classical basis problem for Banach spaces. It remained open for forty years until Per Enflo found a counterexample in 1973.

Below we will give the construction of a closed subspace of l^p, for 2 < p < ∞, that fails to have a basis, due to A.M. Davie [2], reproduced in [3] and [6].

Propostion 2.1 Every Banach spaceX with a basis(x_n)_n∈

N is separable.

Proof: We want to show, that the countable set A:=

( _m X

n=1

α_nx_n:α₁, . . . , α_n∈Q˜, m∈N )

,

whereQ˜ denotes the rational numbers or respectively the complex numbers with rational real and imaginary part, is dense in X. Since an arbitrary element of X having the expansion P∞

n=1αnxn in terms of (xn)_n∈N can be written as limN→∞PN

n=1αnxn, it follows that X = span({x_n:n∈N}). Therefore, it suffices to show that A is a dense subset of span({x_n:n∈N}).

Fix k∈ N, x1, . . . , x_k ∈ {x_n :n∈ N} and α1, . . . , α_k ∈F. Due to the fact, that Q˜ is dense inF, there is a sequence(α_j,i)i∈N∈Q˜ⁿconverging towardsα_jfor eachj = 1, . . . , k.

From the continuity of the vector space operations it follows, that

i→∞lim

k

X

j=1

αj,ixj =

k

X

j=1

αjxj,

i.e. an arbitrary element of span({x_n:n∈N})can be written as the limit of a sequence inA, which completes the proof of this proposition.

2.2 A Counterexample to the Basis Problem

In our counterexample we will prove the existence of a closed subspace oflp,2< p <∞, lacking a certain property. Therefore, it is essential, that every Banach space with a basis has this property.

Definition 2.2 A Banach space X is said to have the approximation property if for each compact set C ⊂ X and each > 0 there exists a bounded linear Operator AC,

fromX intoX having finite rank such thatkA_C,x−xk_X < for each x inC.

Theorem 2.3 Let (X,k.k_X) be a Banach space with basis (xn)_n∈N. Then X has the approximation property.

Proof: Let C be a compact subset of X and > 0. By the preceding lemma it is sufficient to show that there is anN inNsuch thatkP_Nx−xk_X < for eachx inC.

It follows readily from the uniform boundedness principle¹ that Π := sup_n∈NkP_nk is finite. Due to the compactness ofC we can pick finitely many y1, . . . , y_l inC such that min_i=1,...,lkx−y_ik_X ≤ _2(1+Π) for eachx inC.

Let x0 be in C. Then there is j ∈ N such that kx₀−yjk_X ≤ _2(1+C) and, since it follows from definition 0.1 that limn→∞ky_j−Pnyjk_X = 0, there is N ∈ N such that ky_j−Pnyjk_X ≤ ₂ for eachn≥N. We conclude

kP_nx₀−x₀k_X = kP_nx₀−x₀±y_j ±P_ny_jk_X

≤ ky_j −x₀k_X+ kP_ny_j−P_nx₀k_X

| {z }

≤kPnkkyj−x0k_X≤Πkyj−x0k_X

+ky_j−P_ny_jk_X

| {z }

2

≤ (1 + Π)ky_j −x₀k_X+

2 ≤ (1 + Π)

2(1 + Π) + 2 =

2+ 2 =

In the following lemma we give an equivalent condition for a Banach space to have the approximation property. For a proof we refer e.g. to [3] theorem 1.e.4.

Lemma 2.4 Suppose that (X,k.k_X) is Banach space. Then the following are equivalent.

(i) X has the approximation property.

(ii) P∞

n=1x^∗_n(xn) = 0 for all sequences (xn)_n∈N in X and (x^∗_n)_n∈N in X^∗ satisfying P∞

n=1kx_nk_Xkx^∗_nk<∞ andP∞

n=1x^∗_n(x)x_n= 0 for eachx in X, whenX⁰ denotes the continuous dual space ofX.

In our proof of the existence of a subspace of l_p, 2 < p <∞, lacking the approxima- tion property, we will use an infinite matrix A = (ai,j)i,j∈N of a certain type. For the construction ofA we need two lemmata. Thus we want to recall some simple facts from probability theory and from group theory.

Remark 2.5

(i) If η is a discrete random variable, i.e. a measurable function from a probability space to a discrete subset of the real numbers², taking the valuesη_nwith probabil- itiesP(η=ηn), then the expected value of η is given byE(η) =P

nηnP(η=ηn).

1see e.g. [1] theorem 14.1

2or more general, of a measurable space

(ii) The expected value is linear and monotonic and ifη₁, . . . , η_n are independent ran- dom variables we haveE(Qn

i=1ηi) =Qn

i=1E(ηi).

(iii) Ifγ is a random variable andλ >0we haveP(γ >0) =E(1_(0,∞)(γ)), where1_(0,∞) denotes the characteristic funcion of (0,∞). Since 1_(0,∞)(γ) ≤ e^λγ for all γ and λ >0, we obtain P(γ >0)≤E(e^λγ).

(iv) Let (α_i)^N_i=1 be a sequence of real numbers. Suppose, that the sequence (p_i)^N_i=1 ⊂ [0,1] satisfies PN

i=1p_i = 1. Then there exists a sequence of independent random variables(ρn)_n∈N, such that eachρntakes the valueαi with probabilitypi, for each i= 1, . . . , N.

Let be Gbe a finite abelian Group of order k, i.e. an abelian group having kelements, then

(v) G has exactly k characters, i.e. homomorphism from G into the multiplicative group({z∈C:|z|= 1},·),

(vi) if ω is a character of Gwe have ω(g) =ω(g⁻¹) for all g ∈G and ω(e) = 1 for the identity elementeof Gand

(vii) any two different charactersω₁andω₂ofGare orthogonal, i.e. P

g∈Gω₁(g)ω₂(g⁻¹) = P

g∈Gω₁(g)ω₂(g) = 0.

.

Lemma 2.6 Let (ρ_n)^N_n=1 and (α_n)^N_n=1 be finite sequences of independent random vari- ables and complex numbers, respectively. If

(i) each ρ_n takes the value 2 with probability ¹₃ and −1 with probability ²₃ or (ii) each ρn takes the values 1 and−1 with probability ¹₂,

then there exists an absolute constant L such that

P







N

X

n=1

αnρn

> L log(N)

N

X

n=1

|α_n|²

!¹₂





< L

N³. (2.1)

One possible choice of L isL= 3√ 3.

Proof: First we consider real α_n’s. If α_n = 0 for all n = 1, . . . , N, the assertion of this lemma is trivial. Thus, let(αn)^N_n=1 be so thatPN

n=1|α_n|² 6= 0. Furthermore we can assume without loss of generality thatPN

n=1|α_n|² = 1, because otherwise we have

P







N

X

n=1

α_nρ_n

> L log(N)

N

X

n=1

|α_n|²

!¹₂





=P







N

X

n=1

α_n q

PN j=1|α_j|²

ρ_n

> L(log(N)·1)¹²





 .

ρ₁, . . . , ρ_N are independent and e^|t| ≤e^|t|+e^−|t| =e^t+e^−t for all t∈ R. Thus ,taking remark 2.5 into account we obtain for anyλ >0

E

e^λ|^PN_n=1αnρn|

≤ E

e^{λPNn=1αnρn}

+E

e^{−λPNn=1αnρn}

= E

N

Y

n=1

e^λαnρn

! +E

N

Y

n=1

e^−λαnρn

!

=

N

Y

n=1

E(e^λαnρn) +

N

Y

n=1

E(e^−λαnρn)(2.2) Suppose first that (i) holds. Then (2.2) yields

E

e^λ|^PN_n=1αnρn|

≤

N

Y

n=1

E(e^λαnρn) +

N

Y

n=1

E(e^−λαnρn)

=

N

Y

n=1

1

3e^2λαn+2 3e^−λαn

+

N

Y

n=1

1

3e^−2λαn+2 3e^λαn

. (2.3) In order to see that

1

3e^2t+2

3e^−t≤e^2t2 for all t∈R, (2.4) consider

e^2t+ 2e^−t =

∞

X

n=0

(2t)ⁿ+ 2(−t)ⁿ

n! =

∞

X

n=0

(2t)²ⁿ+ 2(−t)²ⁿ

(2n)! + (2t)²ⁿ⁺¹+ 2(−t)²ⁿ⁺¹ (2n+ 1)!

= 3 + 3t²+t³+

∞

X

n=2

2²ⁿ+ 2

(2n)! +t2²ⁿ⁺¹−2 (2n+ 1)!

. (2.5)

It is instantly verified, that (2.4) holds fort≥1. Ift <1and n≥2 then 2²ⁿ+ 2

(2n)! +t2²ⁿ⁺¹−2

(2n+ 1)! < 2²ⁿ⁺¹

(2n)! + 2²ⁿ⁺¹

(2n+ 1)! = 2²ⁿ⁺¹(2n+ 2) (2n+ 1)!

< 3·2²ⁿ·2(n+ 1)

n!(n+ 1)(n+ 2). . .(2n+ 1) = 3·2²ⁿ

n! · 2

(n+ 2)(n+ 3). . .(2n+ 1) < 3·2²ⁿ n! .

Now (2.5) yields for t <1

e^2t+ 2e^−t≤3 + 6t²+

∞

X

n=2

3·2ⁿ

n! t²ⁿ= 3e^2t2.

Thus, (2.4) is proven. Inserting (2.4) back in (2.3) for t= λαn and t= −λα_n, respec- tively, for n= 1, . . . , N we obtain

E

e^λ|^PN_n=1αnρn|

≤2

N

Y

n=1

e^2(λαn)2 = 2e^{2λ2PNn=1|αn|2} = 2e^2λ2

If(ii) holds, (2.2) shows that

E

e^λ|^PN_n=1αnρn|

≤ 2

N

Y

n=1

e^−λαn+e^λαn 2

(2.6) and because of

e^t+e^−t

2 = cosht =

∞

X

n=0

t²ⁿ (2n)! ≤

∞

X

n=0

t²ⁿ

n! = e^t2, for all t∈R, furthermore

E

e^λ|^PNn=1αnρn|

≤ 2

N

Y

n=1

e^(λαn)2 = 2e^{λ2PNn=1|αn|2} = 2e^λ2 ≤ 2e^2λ2.

So in either case we have E

e^λ|^PN_n=1αnρn|

≤2e^2λ2.

Consequently E

e^λ|^PN_n=1αnρn|−2λ²−3 logN

=e^{−2λ2−3 logN}E

e^λ|^PN_n=1αnρn|

≤2e^{−3 logN} = 2 N³.

Now putting λ=√

3 logN gives E

e

√3 logN|^PN_n=1αnρn|^{−9 log}N

≤ 2 N³.

Applying remark 2.5 (iii) to γ =

PN

n=1αnρn

−3√

3(logN)¹² and λ = √

3 logN we obtain

P (

N

X

n=1

α_nρ_n

−3√

3(logN)¹² >0 )

≤E

e

√3 logN|^PNn=1αnρn|^{−9 log}N

≤ 2

N³ ≤ 3√ 3 N³ ,

(2.7) i.e. (2.1) is proven for real αn’s. In order to see that (2.1) still holds for complex αn = un+ivn, n = 1, . . . , N, observe that |α_n|² = |u_n|² +|v_n|², n = 1, . . . , N, and

PN

n=1αnρn

2

=

PN

n=1unρn

2

+

PN

n=1vnρn

2

. By (2.7) we have

P







N

X

n=1

α_nρ_n

>3√

3 log(N)

N

X

n=1

|α_n|²

!¹₂





= P







N

X

n=1

α_nρ_n

2

>27 log(N)

N

X

n=1

|α_n|²







≤ P











N

X

n=1

u_nρ_n

2

>27 log(N)

N

X

n=1

|u_n|²



∨





N

X

n=1

v_nρ_n

2

>27 log(N)

N

X

n=1

|v_n|²











≤ P







N

X

n=1

u_nρ_n

2

>27 log(N)

N

X

n=1

|u_n|²





 + P







N

X

n=1

v_nρ_n

2

>27 log(N)

N

X

n=1

|v_n|²







≤ 2 N³ + 2

N³ = 4

N³ ≤ 3√ 3 N³ ,

which completes the proof of this lemma.

Corollary 2.7 LetG_kandGk−1be abelian groups³of order3·2^kand3·2^k−1, respectively.

Denote the characters ofGkandGk−1 byωn, n= 1, . . . ,3·2^k, andγn, n= 1, . . . ,3·2^k−1. Let(j_n)^N_n=1 be in{1, . . . ,3·2^k}^N and let(l_n)^N_n=1 be in{1, . . . ,3·2^k−1}^N, for anN ≥3·2^k. Then there exist sequences (ρ¹_n)^N_n=1 ∈ {−1,2}^N and (ρ²_n)^N_n=1 ∈ {−1,1}^N such that for some absolute constantM and for all g∈Gk and for all h∈Gk−1

N

X

n=1

ρⁱ_nωjn(g)γln(h)

≤M k¹²2^k−12 , i∈ {1,2}.

Proof: Let theρ˜¹_n’s be random variables⁴ as in lemma 2.6(i) and the ρ˜²_n as in lemma 2.6(ii). Recall that|ω_n(g)|=|γ_n(h)|= 1for all n,g∈G_k,h∈Gk−1.

Applying lemma 2.6(i) and (ii) to (αn)^N_n=1 = (ωjn(g)γln(h))^N_n=1, respectively, yields

P (

N

X

n=1

ωjn(g)γln(h) ˜ρⁱ_n

>3

√

3 (log(N)N)¹² )

< 3√ 3

N³ , for allg∈Gk, h∈Gk−1, i∈ {1,2}.

3for a givenm∈None example of an abelian group of ordermisZ/mZ, the factor group of Zover mZ

4the existence of these random variables follows from remark 2.5 [iv]

We conclude P

(

∃(g, h)∈Gk×Gk−1 :

N

X

n=1

ωjn(g)γln(h) ˜ρⁱ_n

>3

√

3 (log(N)N)¹² )

≤ X

(g,h)∈G_k×Gk−1

P (

N

X

n=1

ωjn(g)γ_ln(h) ˜ρⁱ_n

>3√

3 (log(N)N)¹² )

< 3·2^k3·2^k−13√ 3

N³ .

SinceN ≥2^k, we obtain fork large enough (k > k0)

P (

∃(g, h)∈G_k×Gk−1 :

N

X

n=1

ω_jn(g)γ_ln(h) ˜ρⁱ_n

>3√

3 (log(N)N)¹² )

< 1.

Consequently P

(

∀(g, h)∈Gk×Gk−1 :

N

X

n=1

ωjn(g)γln(h) ˜ρⁱ_n

≤3

√

3 (log(N)N)¹² )

>0.

Thus there exist sequences (ρ¹_n)^3·2_n=1^k ∈ {−1,2}^N and (ρ²_n)^3·2_n=1^k ∈ {−1,1}^N for k large enough such that for allg∈G_k and for allh∈Gk−1

N

X

n=1

ω_jn(g)γ_ln(h)ρⁱ_n

≤3√

3 (log(N)N)¹² ≤3√

3(log(3·2^k−1)3·2^k−1)¹² ≤M k¹²2^k−12 .

Since for eachk ∈N∪ {0} the product Gk×Gk−1 is a finite set, the same inequalities hold for k = 0, . . . , k₀, by increasing M if necessary, which completes the proof of this corollary.

Lemma 2.8 Let G_k be an abelian group of order 3·2^k, for each k in N∪ {0}. Then there exists a partition of the set of all charactersH_k={ω_n}^3·2_n=1^k of G_k into two disjoint sets H_k⁺ = {σ_n}²_n=1^k and H_k⁻ = {τ_n}²_n=1^k+1 with cardinalities satisfying |H_k⁺| = 2^k and

|H_k⁻|= 2^k+1 such that for some absolute constant K and for each g∈Gk

2

2^k

X

n=1

σ_n(g)−

2^k+1

X

n=1

τ_n(g)

≤K(k+ 1)¹²2^k2. (2.8)

Proof:It suffices to show that there exists a sequence of numbers(ρ_n)^3·2_n=1^k ∈ {−1,2}^3·2k such that

3·2^k

X

n=1

ρnωn(g)

≤K(n+ 1)¹²2ⁿ² for allg∈Gk (2.9)

and

3·2^k

X

n=1

ρ_n= 0, (2.10)

since then (2.9) proofs (2.8), while (2.10) ensures that the requirements concerning the cardinalities of our subsetsH_k⁺ and H_k⁻ are satisfied.

Let Gk+1 be any abelian group of order 3·2^k+1 and let I be the trivial character⁵ of G_k+1. Applying corollary 2.7 toG_k,G_k+1,{ω_n}^3·2_n=1^k and {I}^3·2_n=1^k shows, that there exists a sequence(ρ_n)^3·2_n=1^k ∈ {−1,2}^3·2k such that,

3·2^k

X

n=1

ρ_nω_n(g)

≤ K2^k2(k+ 1)¹²,

for some absolute constantK, i.e. (2.9) holds for(ρ_n)^3·2_n=1^k. Put Sj :=

ρn:n∈ {1, . . . ,3·2^k}, ρ_n=j for j ∈ {−1,2}. For the cardinalities of this to sets we obviously have |S₋₁|+|S₂|= 3·2^k. Therefore

3·2^k

X

n=1

ρn = 2|S₂| − |S₋₁| = 2|S₂| −3·2^k+|S₂| = 3(|S₂| −2^k). (2.11) If |S₂|= 2^k our sequence (ρn)^3·2_n=1^k satisfies (2.10) in addition to (2.9) and we are done.

Otherwise we have to change an appropriate number of theρn’s

Suppose first, that |S₂|<2^k and letI₁ ⊂ {1, . . . ,3·2k} be such that|I₁|= 2^k− |S₂| and ρ_n=−1 for alln∈I₁. We define a sequence {ρ_n,1}^3·2_n=1^k by

ρn,1=

(2, forn∈I₁, ρn, else.

If |S₂|> 2^k, we fix I2 ⊂ {1, . . . ,3·2k} such that |I₂|= |S₂| −2^k and ρn = 2 for all n∈I2. The sequence {ρ_n,2}^3·2_n=1^k is defined by

ρ_n,2=

(−1, forn∈I2, ρ_n, else.

In both cases (2.9) and (2.11) give

3·2^k

X

n=1

ρ_nω_n(g)−

3·2^k

X

n=1

ρ_n,jω_n(g)

=

X

i∈I_j

3·ω_i(g)

≤3|I_j| = 3

|S₂| −2^k =

3·2^k

X

n=1

ρ_n

=

3·2^k

X

n=1

ρ_nω_n(e)

≤ K(n+ 1)¹²2ⁿ²

5i.e. the maph7→1, h∈Gk+1

for all g ∈ G_k, j ∈ {1,2} and the identity element e of G_k. Using this, the triangle inequality and (2.9) we obtain

3·2^k

X

n=1

ρn,j

≤2K(k+ 1)¹²2ⁿ², for allg∈G_k, j ∈ {1,2},

and thus (2.9) holds for K and {ρ_n}^3·2_n=1^k replaced by 2K and {ρ_n,j}^3·2_n=1^k (j ∈ {1,2}), respectively. In addition, by (2.11), we have

3·2^k

X

n=1

ρn−

3·2^k

X

n=1

ρn,j =X

i∈I_j

3 = 3|I_j|= 3

|S₂| −2^k =

3·2^k

X

n=1

ρn j ∈ {1,2}

and consequentlyP3·2^k

n=1ρ_n,j = 0 for j ∈ {1,2}, i.e. we have (2.10) for {ρ_n}^3·2_n=1^k replaced by {ρ_n,j}^3·2_n=1^k.

Now we are able to pass the construction of the infinite matrix A needed for our coun- terexample.

Lemma 2.9 There exists an infinite matrixA= (ai,j)i,j∈Nof complex numbers satisfying A² = 0,trA=P∞

i=1a_i,i 6= 0andP∞

i=1(maxj∈N|a_i,j|)^r<∞ for eachr > ²₃. Furthermore each row and each column contains only finitely many nonzero entries.

Proof: For each kinN∪ {0} letGk= ({g₁, . . . , g_3·2^k},∗)be an abelian group of order 3·2^k. Furthermore let H_k⁺ = {σ_n,k}²_n=1^k and H_k⁻ = {τ_n,k}²_n=1^k+1 be as in the preceding lemma.

We define complex matrices Pk =

3⁻¹²2^−2k+12 τi,k(gj)

i=1,...,2^k+1;j=1,...,3·2^k, Qk=

3⁻¹²2^−kρi,kσi,k(gj)

i=1,...,2^k;j=1,...,3·2^k,

for eachk inN∪ {0}, where(ρ_i,k)²_i=1^k ∈ {−1,1}^2k will be determined below. We have PkP_k^∗ =



3⁻¹2^−2k−1

3·2^k

X

l=1

τi,k(gl)τj,k(gl)





i,j=1,...2^k+1

, (2.12)

where P_k^∗ denotes the conjugate transpose of P_k. Taking remark 2.5 into account we obtain

3·2^k

X

l=1

τ_i,k(g_l)τ_j,k(g_l) =

(P3·2^k

l=1 |τ_i,k(gl)|²= 3·2^k, j =i= 1, . . . ,2^k

0, i, j ∈ {1, . . . ,2^k}, i6=j.

Hence, we have

P_kP_k^∗ = 2^−k−1I₂k+1, k∈N∪ {0}, (2.13)