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Radial Toeplitz operators on the unit ball

and slowly oscillating sequences

Sergei M. Grudsky∗

Departamento de Matem´aticas, CINVESTAV del I.P.N.,

Apartado Postal 14-740, 07000 M´exico, D.F., M´exico

Egor A. Maximenko†

Escuela Superior de F´ısica y Matem´aticas, Instituto Polit´ecnico Nacional,

C.P. 07730, M´exico, D.F., M´exico

Nikolai L. Vasilevski‡

Departamento de Matem´aticas, CINVESTAV del I.P.N.,

Apartado Postal 14-740, 07000 M´exico, D.F., M´exico

(Communicated by Vladimir Rabinovich)

This is a draft version of a published research article:

Grudsky, Sergei M.; Maximenko, Egor A.; Vasilevski, Nikolai L.

Radial Toeplitz operators on the unit ball and slowly oscillating sequences.

Commun. Math. Anal. 14 (2013), no. 2, 77–94.

http://projecteuclid.org/euclid.cma/1356039033.

We are grateful to Roberto Mois´es Barrera Castel´an who found some errors in the

published article, namely, in the proof of Proposition 4.3. In this version we ﬁxed

the errors.

∗E-mail address: grudsky@math.cinvestav.mx

†E-mail address: maximenko@esfm.ipn.mx

‡E-mail address: nvasilev@math.cinvestav.mx

Abstract

In the paper we deal with Toeplitz operators acting on the Bergman space

A2(Bn) of square integrable analytic functions on the unit ball Bnin Cn. A

bounded linear operator acting on the space A2(Bn) is called radial if it com-

mutes with unitary changes of variables. Zhou, Chen, and Dong [9] showed that

every radial operator Sis diagonal with respect to the standard orthonormal

monomial basis (eα)α∈Nn. Extending their result we prove that the correspond-

ing eigenvalues depend only on the length of multi-index α, i.e. there exists a

bounded sequence (λk)∞

k=0 of complex numbers such that Seα=λ|α|eα.

Toeplitz operator is known to be radial if and only if its generating symbol

gis a radial function, i.e., there exists a function a, deﬁned on [0,1], such that

g(z) = a(|z|) for almost all z∈Bn. In this case Tgeα=γn,a (|α|)eα, where the

eigenvalue sequence γn,a(k)∞

k=0 is given by

γn,a(k) = 2(k+n)Z1

0

a(r)r2k+2n−1dr = (k+n)Z1

0

a(√r)rk+n−1dr.

Denote by Γnthe set {γn,a :a∈L∞([0,1])}. By a result of Su´arez [8], the C∗-

algebra generated by Γ1coincides with the closure of Γ1in `∞and is equal to

the closure of d1in `∞, where d1consists of all bounded sequences x= (xk)∞

k=0

such that

sup

k≥0(k+ 1) |xk+1 −xk|<+∞.

We show that the C∗-algebra generated by Γndoes not actually depend on n,

and coincides with the set of all bounded sequences (xk)∞

k=0 that are slowly

oscillating in the following sense: |xj−xk|tends to 0 uniformly as j+1

k+1 →1

or, in other words, the function x:{0,1,2, . . .} → Cis uniformly continuous

with respect to the distance ρ(j, k) = |ln(j+ 1) −ln(k+ 1)|. At the same

time we give an example of a complex-valued function a∈L1([0,1], r dr ) such

that its eigenvalue sequence γn,a is bounded but is not slowly oscillating in the

indicated sense. This, in particular, implies that a bounded Toeplitz operator

having unbounded deﬁning symbol does not necessarily belong to the C∗-algebra

generated by Toeplitz operators with bounded deﬁning symbols.

AMS (MOS) subject classiﬁcation: Primary 47B35; Secondary 32A36, 44A60.

Keywords. Radial Toeplitz operator, Bergman space, unit ball, slowly oscillating

sequence.

2

1 Introduction and main results

Bergman space on the unit ball

We shall use some notation and well-known facts from Rudin [3] and Zhu [10].

Denote by h·,·i the usual inner product in Cn:hz, wi=Pn

j=1 zjwj. Let |·|be

the Euclidean norm in Cninduced by this inner product, and let Bnbe the unit

ball in Cn. Denote by dv the Lebesgue measure on Cn=R2nnormalized so that

v(Bn) = 1, and denote by dσ the surface measure on the unit sphere S2n−1=∂Bn

normalized so that σ(S2n−1) = 1. Let N={0,1,2, . . .}. Given a multi-index α∈Nn

and a vector z∈Cn, we understand the symbols |α|,α! and zαin the usual sense:

|α|=

n

X

j=1

αj, α! =

n

Y

j=1

αj!, zα=

n

Y

j=1

zαj

j.

Consider the Bergman space A2=A2(Bn, v) of all square integrable analytic func-

tions on Bn. Denote by (eα)α∈Nnthe standard orthonormal monomial basis in A2:

eα(z) = r(n+|α|!)

n!α!zα.

The reproducing kernel Kzof the space A2at a point z∈Bnsatisﬁes hf, Kzi=f(z)

for all f∈ A2, and is given by the following formula:

Kz(w) = X

α∈Nn

eα(z)eα=1

(1 − hw, zi)n+1 .

The Berezin transform of a bounded linear operator S:A2→ A2is a function

Bn→Cdeﬁned by

(B(S))(z) = hSKz, Kzi

kKzk2= (1 − |z|2)n+1 hSKz, Kzi.

It is well known that the Berezin transform Bis injective: if B(S) is identically zero,

then S= 0. A proof of this fact for the one-dimensional case is given by Stroethoﬀ

[7].

Given a function g∈L1(Bn), the Toeplitz operator Tgis deﬁned on a dense

subset of A2by

(Tg(f))(z) := ZBn

Kzgf dv.

If g∈L∞(Bn), then Tgis bounded and kTgk≤kgk∞.

Radial operators on the unit ball

Following Zhou, Chen and Dong [9] we recall the concept of a radial function on Bn

and of a radial operator acting on A2. The radialization of a measurable function

f:Bn→Cis given by

rad(f)(z) := ZUn

f(Uz)dH (U),

3

where dH is the normalized Haar measure on the compact group Unconsisting of

the unitary matrices of order n.

A function f:Bn→Cis called radial if rad(f) coincides with falmost every-

where. For a continuous function fthis means that f(z) = f(|z|) for all z∈Bn.

Given a unitary matrix U∈ Un, denote by ΨUthe corresponding “change of a

variable operator” acting on A2:

(ΨUf)(z) := f(U∗z).

Here U∗is the conjugated transpose of U. Note that ΨUis a unitary operator on

the space A2, its inverse is ΨU∗, and the formula ΨU1U2= ΨU1ΨU2holds for all

U1, U2∈ Un.

Given a bounded linear operator S:A2→ A2, its radialization Rad(S) is deﬁned

by

Rad(S) := ZUn

ΨUSΨU∗dH(U),

where the integration is understood in the weak sense.

A bounded linear operator Sis called radial if SΨU= ΨUSfor all U∈ Unor,

equivalently, if Rad(S) = S.

Zhou, Chen, and Dong [9] proved that the Berezin transform “commutes with

the radialization” in the following sense: for every bounded linear operator Sacting

in A2

B(Rad(S)) = rad B(S).

It follows that Sis radial if and only if B(S) is radial. In the one-dimensional case

(i.e., for n= 1) these facts were proved by Zorboska [11].

For each α∈Nnwe denote by Pαthe orthogonal projection onto the one-

dimensional space generated by eα:

Pα(x) := hx, eαieα.

Given a bounded sequence λ= (λm)∞

m=0 of complex numbers, denote by Rλthe

following operator (radial operator with eigenvalue sequence λ):

Rλ:= X

α∈Nn

λ|α|Pα,

where the convergence of the series is understood in the strong operator topology.

The Berezin transform of Rλwas computed in [1, 9]:

(B(Rλ))(z) = (1 − |z|2)n+1

∞

X

m=0

2(m+n)!

m! (n−1)! λm|z|2m.(1.1)

Since the function B(Rλ) is radial, the operator Rλis radial.

Theorem 1.1. Let Sbe a bounded linear radial operator in A2. Then there exists

a bounded complex sequence λsuch that S=Rλ.

4

Zhou, Chen, and Dong [9] proved one part of this theorem, namely, that Sis

diagonal with respect to the monomial basis. In Section 2 we prove the remaining

part: the eigenvalues of Sdepend only on the length of the multi-index.

Radial Toeplitz operators on the unit ball

Zhou, Chen, and Dong [9] proved that a Toeplitz operator Tgis radial if and only

if its generating symbol gis radial, i.e., if there exists a function adeﬁned on [0,1]

such that g(z) = a(|z|) for almost all z∈Bn. Then Tgis diagonal with respect

to the orthonormal monomial basis, and the corresponding eigenvalues depend only

on the length of multi-indices. Denote the eigenvalue sequence of such operator by

γn,a:

Tgeα=γn,a(|α|)eα.

An explicit expression of the eigenvalues γn,a (m) in terms of awas found by Grudsky,

Karapetyants and Vasilevski [1] (see also [9]):

γn,a(m)=(m+n)Z1

0

a(√r)rm+n−1dr, (1.2)

or, changing a variable,

γn,a(k) = 2(m+n)Z1

0

a(r)r2m+2n−1dr. (1.3)

Denote by Γn(L∞([0,1])), or Γnin short, the set of all these eigenvalue sequences,

which are generated by the radial Toeplitz operators with bounded generating func-

tions:

Γn:= Γn(L∞([0,1])) = γn,a :a∈L∞([0,1]).(1.4)

Deﬁne γ1,a and Γ1by (1.3) and (1.4) with n= 1:

γ1,a(k) = 2(k+ 1) Z1

0

a(r)r2k+1 dr, (1.5)

Γ1:= Γ1(L∞([0,1])) = γ1,a :a∈L∞([0,1]).(1.6)

Denote by d1(N) the set of all bounded sequences x= (xj)j∈Nsatisfying the condi-

tion

sup

k∈N(k+ 1)(∆x)k<+∞,

where (∆x)k=xk+1 −xk.

Then the C∗-algebra generated by radial Toeplitz operators with bounded gen-

erated symbols is isometrically isomorphic to the C∗-algebra generated by Γn.

Theorem 1.2 (Su´arez [8]).The C∗-algebra generated by Γ1coincides with the topo-

logical closure of Γ1in `∞(N), being the topological closure of d1(N)in `∞(N).

5

Slowly oscillating sequences

Denote by SO(N) the set of all bounded sequences that slowly oscillate in the sense

of Schmidt [5] (see also Landau [2] and Stanojevi´c and Stanojevi´c [6]):

SO(N) := nx∈`∞: lim

j+1

k+1 →1|xj−xk|= 0o.

In other words, SO(N) consists of all bounded functions N→Cthat are uniformly

continuous with respect to the “logarithmic metric” ρ(j, k) := |ln(j+ 1) −ln(k+ 1)|.

In Section 3 we give some properties and equivalent deﬁnitions of the C∗-algebra

SO(N).

In Section 4 we prove that the C∗-algebra generated by Γndoes not actually

depend on n. Applying Theorem 1.2 and some standard approximation techniques

(de la Vall´ee-Poussin means) we obtain the main result of the paper.

Theorem 1.3. For each nthe C∗-algebra generated by Γncoincides with the topo-

logical closure of Γnin `∞and is equal to SO(N).

As shown by Grudsky, Karapetyants and Vasilevski [1], if a∈L1([0,1], r2n−1dr)

and the sequence γn,a is bounded, then γn,a (m+ 1) −γn,a(m)→0. At the same

time, in this situation γn,a does not necessarily belong to SO(N). The next result is

proved in Section 5.

Theorem 1.4. There exists a function a∈L1([0,1], r dr)such that γn,a ∈`∞(N)\

SO(N).

That is, a bounded Toeplitz operator having unbounded deﬁning symbol does not

necessarily belong to the C∗-algebra generated by Toeplitz operators with bounded

deﬁning symbols.

2 Diagonalization of radial operators in the monomial

basis

Lemma 2.1 (Zhou, Chen, and Dong [9]).Let S:A2→ A2be a bounded radial

operator and αbe a multi-index. Then eαis an eigenfunction of S, i.e., hSeα, eβi= 0

for every multi-index βdiﬀerent from α.

Proof. For a reader convenience we give here a proof, slightly diﬀerent from [9].

Choose an index j∈ {1, . . . , n}such that αj6=βjand a complex number tsuch

that |t|= 1 and tαj6=tβj. For example, put

t=eiφ where φ=π

|αj−βj|.

Denote by Uthe diagonal matrix with (j, j)st entry equal to t−1and all other

diagonal entries equal to 1:

U= diag(1,...,1, t−1

|{z}

jst position

,1,...,1).

6

Then Uis a unitary matrix, ΨUeα=tαjeα, and

tαjhSeα, eβi=hSΨUeα, eβi=hΨUSeα, eβi=hSeα,ΨU∗eβi=tβjhSeα, eβi.

Since tαj6=tβj, it follows that hSeα, eβi= 0.

Lemma 2.2 (Berezin transform of basic projections).Let α∈Nnand z∈B. Then

B(Pα)(z) = (1 − |z|2)n+1 qα(z),

where qα:B→Cis the square of the absolute value of eα:

qα(z) = |eα(z)|2=(n+|α|)!

n!α!|zα|2.

Proof. We calculate PαKzfor an arbitrary z∈B:

PαKz=Pα

X

β∈Nn

eβ(z)eβ

=eα(z)eα.

The reproducing property of Kzimplies that heα, Kzi=eα(z). Therefore

B(Pα)(z) = 1

Kz(z)hPαKz, Kzi= (1 − |z|2)n+1 heα(z)eα, Kzi

= (1 − |z|2)n+1 |eα(z)|2.

Lemma 2.3. For each m∈N, the function z7→ |z|2mis n

m+ntimes the arithmetic

mean of the functions qαwith |α|=m:

|z|2m=m!n!

(m+n)! X

|α|=m

qα(z) = n

m+n

m! (n−1)!

(m+n−1)! X

|α|=m

qα(z).

Proof. Apply the multinomial theorem and the deﬁnition of qα:

|z|2m=

n

X

j=1 |zj|2

m

=X

|α|=m

m!

α!

n

Y

j=1 |zj|2αj

=X

|α|=m

m!

α!|zα|2=m!n!

(m+n)! X

|α|=m

qα(z).

Lemma 2.4. Let α∈Nn. Then for all z∈B,

rad(qα)(z) = n+|α|

n|z|2|α|.

7

Proof. Express the integration over Unthrough the integration over S2n−1:

rad(qα)(z) = ZUn

n+|α|!

n!α!|(Uz)α|2dH (U) = n+|α|!

n!α!|z|2|α|ZS2n−1|ζα|2dσ(ζ).

The value of the latter integral is well known (e.g., see [3, Proposition 1.4.9]):

ZS2n−1|ζα|2dσ(ζ) = (n−1)! α!

(n−1 + |α|)!.

Lemma 2.5 (radialization of basic projections).Let α∈Nn. Then the radialization

of Pαis the arithmetic mean of all Pβwith |β|=|α|:

Rad(Pα) = (n−1)! |α|!

(n−1 + |α|)! X

β∈Nn

|β|=|α|

Pβ.(2.1)

Proof. We shall prove that both sides of (2.1) have the same Berezin transform, then

(2.1) will follow from the injectivity of the Berezin transform. We use the fact the

Berezin transform “commutes with the radialization” [9], and apply then Lemmas

2.2 and 2.4:

B(Rad(Pα))(z) = rad(B(Pα))(z) = (1 − |z|2)n+1 rad(qα)(z)

=n+|α|

n|z|2|α|(1 − |z|2)n+1.

On the other hand, by Lemmas 2.4 and 2.3,

(n−1)! |α|!

(n−1 + |α|)! X

|β|=|α|B(Pβ)(z) = (1 − |z|2)n+1 (n−1)! |α|!

(n−1 + |α|)! X

|β|=|α|

qβ(z)

=n+|α|

n|z|2|α|(1 − |z|2)n+1.

Lemma 2.6 (radialization of a diagonal operator).Let (cα)α∈Nnbe a bounded family

of complex numbers. Consider the operator S:A2→ A2given by

S=X

α∈Nn

cαPα.

Then

Rad(S) =

∞

X

m=0

m! (n−1)!

(m+n−1)! X

|β|=m

cβ

X

|α|=m

Pα

.

Proof. Follows from Lemma 2.5 and the fact that the sum of a converging serie of

mutually orthogonal vectors does not depend on the order of summands.

8

Proof of Theorem 1.1. Let Sbe a bounded linear radial operator in A2. By Lemma

2.1,

S=X

α∈Nn

cαPα.

Since Rad(S) = S, it follows from Lemma 2.6 that the coeﬃcients cαdepend only

on |α|. Deﬁning λmequal to cαfor some αwith |α|=m, we obtain

S=

∞

X

m=0

λm

X

|α|=m

Pα

=Rλ.

3 Slowly oscillating sequences

Deﬁnition 3.1 (logarithmic metric on N).Deﬁne ρ:N×N→[0,+∞) by

ρ(j, k) := ln(j+ 1) −ln(k+ 1).

The function ρis a metric on Nbecause it is obtained from the usual metric

d:R×R→[0,+∞), d(t, u) := |t−u|, via the injective function N→R,j7→ ln(j+1).

Deﬁnition 3.2 (modulus of continuity of a sequence with respect to the logarithmic

metric).Given a complex sequence x= (xj)j∈N, deﬁne ωρ,x : [0,+∞)→[0,+∞] by

ωρ,x(δ) := sup|xj−xk|:j, k ∈N, ρ(j, k)≤δ.

Deﬁnition 3.3 (slowly oscillating sequences).Denote by SO(N) the set of the

bounded sequences that are uniformly continuous with respect to the logarithmic

metric:

SO(N) = λ∈`∞(N): lim

δ→0+ωρ,λ(δ) = 0.

Note that the class SO(N) plays an important role in Tauberian theory, see

Landau [2], Schmidt [5, §9], Stanojevi´c and Stanojevi´c [6].

For every sequence xthe function ωρ,x : [0,+∞)→[0,+∞] is increasing (in the

non-strict sense). Therefore the condition limδ→0+ωρ,x(δ) = 0 is equivalent to the

following one: for all ε > 0 there exists a δ > 0 such that ωρ,x(δ)< ε.

The same class SO(N) can be deﬁned using another special metric ρ1on N:

Deﬁnition 3.4. Deﬁne ρ1:N×N→[0,+∞) by

ρ1(j, k) = |j−k|

max(j+ 1, k + 1) = 1 −min(j+ 1, k + 1)

max(j+ 1, k + 1).

Proposition 3.5. ρ1is a metric on N.

Proof. Clearly ρ1is non-negative, symmetric, and ρ1(j, k) = 0 only if j=k. We

have to prove that for all j, k, p ∈N

ρ1(j, p) + ρ1(p, k)−ρ1(j, k)≥0.(3.1)

9

Denote the left-hand side of (3.1) by Λ(j, k, p). Since Λ(j, k, p) is symmetric with

respect to jand k, assume without loss of generality that j≤k. If j≤p≤k, then

Λ(j, k, p) = 1−j+ 1

p+ 1+1−p+ 1

k+ 1−1−j+ 1

k+ 1

=p−j

p+ 1 −p−j

k+ 1 =(p−j)(k−p)

(k+ 1)(p+ 1) ≥0.

If j≤k < p, then

Λ(j, k, p) = (p−k)(j+k+ 2)

(k+ 1)(p+ 1) ≥0.

If p<j≤k, then

Λ(j, k, p) = (j−p)(j+k+ 2)

(j+ 1)(k+ 1) ≥0.

Proposition 3.6 (relations between ρand ρ1).

1. For all j, k ∈N,

ρ1(j, k)≤ρ(j, k).(3.2)

2. For all j, k ∈Nsatisfying ρ1(j, k)≤1

2,

ρ(j, k)≤2 ln(2)ρ1(j, k).(3.3)

Proof. Since the functions ρand ρ1are symmetric and vanish on the diagonal

(ρ(j, j) = ρ1(j, j ) = 0), consider only the case j < k. Denote k+1

j+1 −1 by t, then

ρ(j, k) = ln(1 + t), ρ1(j, k) = 1 −1

1 + t=t

1 + t.

Deﬁne f: (0,+∞)→(0,+∞) by

f(t) := ln(1 + t)

1−1

1+t

.

Then

f0(t) = t−ln(1 + t)

t2>0,

and thus fis strictly increasing on (0,+∞). Since limt→0+f(t) = 1 and f(1) =

2 ln(2), we see that f(t)>1 for all t > 0 and f(t)≤2 ln(2) for all t∈(0,1].

Substituting tby k+1

j+1 −1 we obtain (3.2) and (3.3).

Corollary 3.7. The set SO(N)can be deﬁned using the metric ρ1instead of ρ:

SO(N) = nλ∈`∞(N): lim

δ→0+sup

ρ1(j,k)≤δ|λj−λk|= 0o.

Let us mention some simple properties of SO(N).

10

Proposition 3.8. SO(N)is a closed subalgebra of the C∗-algebra `∞(N).

Proof. It is a general fact that the set of the uniformly continuous functions on some

metric space Mis a closed subalgebra of the C∗-algebra of the bounded continuous

functions on M. In our case M= (N, ρ). Since

ωρ,f+g≤ωρ,f +ωρ,g, ωρ,λf =|λ|ωρ,f ,

ωρ,fg ≤ωρ,f kgk∞+ωρ,gkfk∞, ωρ,f=ωρ,f ,

the set SO(N) is closed with respect to the algebraic operations. The topological

closeness of SO(N) in `∞(N) follows from the inequality

ωρ,f (δ)≤2kf−gk∞+ωρ,g(δ).

Proposition 3.9 (comparison of SO(N) to c(N)).The set of the converging se-

quences c(N)is a proper subset of SO(N).

Proof. 1. Denote by N:= N∪ {∞} the one-point compactiﬁcation (Alexandroﬀ

compactiﬁcation) of N. The topology on Ncan be induced by the metric

dN(j, k) :=

j

j+ 1 −k

k+ 1.

If σ∈c(N), then σis uniformly continuous with respect to the metric dN, but dNis

less or equal than ρ:

dN(j, k) = |j−k|

(j+ 1)(k+ 1) ≤|j−k|

max(j+ 1, k + 1) =ρ1(j, k)≤ρ(j, k).

2. The sequence x= (xj)j∈Nwith xj= cos(ln(j+ 1)) does not converge but belongs

to SO(N) since

xj−xk=cos(ln(j+ 1)) −cos(ln(k+ 1))≤ln(j+ 1) −ln(k+ 1)=ρ(j, k).

We deﬁne now the left and right shifts of a sequence. Given a complex sequence

x= (xj)j∈N, deﬁne the sequences τL(x) and τR(x) as follows:

τL(x) := (x1, x2, x3, . . .), τR(x) := (0, x0, x1, . . .).

More formally,

τL(x)j:= xj+1;τR(x)j:= (0, j = 0;

xj−1, j ∈ {1,2,3, . . .}.

Note that τL(τR(x)) = xfor every sequence x.

Both τLand τRare bounded linear operators on `∞(N). In the following two

propositions we show that SO(N) is an invariant subspace of each one of these

operators.

11

Proposition 3.10. For every x∈SO(N),τL(x)∈SO(N).

Proof. The image of τL(x) is a subset of the image of x, therefore kτL(x)k≤kxk.

If δ > 0, j, k ∈N,j < k and ρ(j, k)≤δ, then

ρ(j+ 1, k + 1) = ln k+ 2

j+ 2 = ln k+ 1

j+ 1 + ln 1 + 1

k+ 1−ln 1 + 1

j+ 1

<ln k+ 1

j+ 1 =ρ(j, k)≤δ.

It follows that ωρ,τL(x)(δ)≤ωρ,x(δ) and lim

δ→0+ωρ,τL(x)(δ) = 0.

Proposition 3.11. For every x∈SO(N),τR(x)∈SO(N).

Proof. The sequences xand τR(x) have the same image up to one element zero:

{τR(x)j:j∈N}={xj:j∈N}∪{0}.

Therefore kτR(x)k∞=kxk∞.

2. Let δ∈0,1

3,j, k ∈N,j < k and ρ(j, k)≤δ. Then j≥1, k≥2, and

ρ1(j−1, k −1) = k−j

k=k+ 1

k·(k+ 1) −(j+ 1)

k+ 1 ≤3

2ρ1(j, k).

Applying Proposition 3.6 we see that

ρ1(j−1, k −1) ≤3

2ρ(j, k) = 3

2δ≤1

2

and

ρ(j−1, k −1) ≤2 ln(2)ρ1(j−1, k −1) ≤2 ln(2) 3

2δ= 3 ln(2)δ.

Thus for every δ∈0,1

3,

ωρ,τR(x)(δ)≤ωρ,x(3 ln(2)δ).

Therefore lim

δ→0+ωρ,τR(x)(δ) = 0.

4 Γnis a dense subset of SO(N)

First we prove that Γnis contained in SO(N).

Proposition 4.1. Let a∈L∞([0,1]). Then γn,a ∈SO(N). More precisely,

kγn,ak∞≤ kak∞,(4.1)

and for all j, k ∈N,

γn,a(j)−γn,a(k)≤2kak∞ρ(j, k).(4.2)

12

Proof. The inequality (4.1) follows directly from (1.3):

|γn,a(j)| ≤ 2(n+j)Z1

0

r2n+2j−1kak∞dr =kak∞.

The proof of (4.2) is based on an idea communicated to us by K. M. Esmeral Garc´ıa.

Since both sides of (4.2) are symmetric with respect to the indices jand k, without

loss of generality we consider the case j < k. First factorize a(r) and bound it by

kak∞:

γn,a(j)−γn,a(k)=Z1

0(n+j)r2n+2j−1−(n+k)r2n+2k−1a(r)dr(4.3)

≤ kak∞Z1

0(n+j)r2n+2j−1−(n+k)r2n+2k−1dr. (4.4)

Denote by r0the unique solution of the equation (n+j)r2n+2j−1−(n+k)r2n+2k−1= 0

on the interval (0,1):

r0=n+j

n+k1

2(k−j).

The function r7→ (n+j)r2n+2j−1−(n+k)r2n+2k−1takes positive values on the

interval (0, r0) and negative values on the interval (r0,1). Dividing the integral (4.4)

on two parts by the point r0, we obtain:

γn,a(j)−γn,a(k)≤2kak∞r2n+2j

0−r2n+2k

0= 2kak∞r2n+2j

0ρ1(j, k).

Since r0<1 and ρ1(j, k)≤ρ(j, k), the inequality (4.2) follows.

Deﬁnition 4.2. Denote by d1(N) the set of the bounded sequences xsuch that

sup

j∈N(j+ 1)|xj+1 −xj|<+∞.

Proposition 4.3. d1(N)is a proper subset of SO(N).

Proof. 1. Let x∈d1(N) and

M= sup

j∈N(j+ 1)|xj+1 −xj|.

Then for all j, k ∈Nwith j < k we have

|xk−xj| ≤

k−1

X

q=j|xq+1 −xq| ≤ M

k−1

X

q=j

1

q+ 1 ≤2M

k−1

X

q=j

1

q+ 2

= 2M

k−1

X

q=j

ρ1(q, q + 1) ≤2M

k−1

X

q=j

ρ(q, q + 1) = 2Mρ(j, k ).

Therefore d1(N) is contained in SO(N).

13

2. Consider the sequence

xj:= sin πblog2(j+ 2)c

plog2(j+ 2) .

For every jand kwith k > j,

|xk−xj| ≤ πblog2(k+ 2)c

plog2(k+ 2) −πblog2(j+ 2)c

plog2(j+ 2)

≤πlog2(k+ 2)

plog2(k+ 2) −π(log2(j+ 2) −1)

plog2(j+ 2)

=πplog2(k+ 2) −plog2(j+ 2)+π

plog2(j+ 2)

=πlog2k+2

j+2

plog2(k+ 2) + plog2(j+ 2) +π

plog2(j+ 2).

Thus x∈SO(N). On the other hand, if j= 2k2−3, then

|xj+1 −xj|=|xj|=

sin π(k2−1)

plog2(2k2−1)!

=

sin kπ −π(k2−1)

plog2(2k2−1)!

.

Appying the inequality |sin(t)| ≥ 2|t|

π, which holds for all twith |t| ≤ π

2, we obtain:

|xj+1 −xj| ≥ 2 k−(k2−1)

plog2(2k2−1)!≥2k−pk2−1≥1

k=1

plog2(j+ 3).

Therefore x /∈d1(N).

Lemma 4.4. Let x∈`∞(N)and δ∈(0,1). Denote by ythe sequence of the de la

Vall´ee-Poussin means of x:

yj=1

1 + bjδc

j+bjδc

X

k=j

xk.(4.5)

Then y∈d1(N)and

ky−xk∞≤ωρ,x(δ).(4.6)

Proof. Note that for all j∈N, the sum in the right-hand side of (4.5) contains

1 + bjδcterms. Therefore

|yj| ≤ 1

1 + bjδc

j+bjδc

X

k=jkxk∞=kxk∞.

14

For j∈N, let us estimate the diﬀerence |yj+1 −yj|:

|yj+1 −yj|=

1

1 + b(j+ 1)δc

j+b(j+1)δc

X

k=j

xk−1

1 + bjδc

j+bjδc

X

k=j

xk

≤b(j+ 1)δc − bjδc

(1 + b(j+ 1)δ)(1 + bjδc)

j+b(j+1)δc

X

k=j|xk|+|xj+b(j+1)δc|

1 + b(j+ 1)δc

≤kxk∞(bjδc+ 1)

(j+ 1)δ(1 + bjδc)+kxk∞

(j+ 1)δ

=kxk∞

(j+ 1)δ.

Thus y∈d1(N). Let us prove (4.6). If j≤k≤j+bjδc, then

ρ(j, k) = ln k+ 1

j+ 1 ≤ln k

j≤ln(1 + δ)≤δ.

Therefore

|yj−xj| ≤ 1

1 + bjδc

j+bjδc

X

k=j|xk−xj| ≤ ωρ,x(δ).

Proposition 4.5. d1(N)is a dense subset of SO(N).

Proof. Let ε > 0. Using the fact that ωρ,x(δ)→0 as δ→0, choose a δ > 0 such

that ωρ,x(δ)< ε. Deﬁne yby (4.5). Then y∈d1(N) and kx−yk∞< ε by Lemma

4.4.

Theorem 1.3 follows from Proposition 4.5 and Theorem 1.2:

Proposition 4.6. Γ1is a dense subset of SO(N).

Proof. Proposition 4.1 implies that Γ1is contained in SO(N). Let x∈SO(N) and

ε > 0. Applying Proposition 4.5 ﬁnd a sequence y∈d1(N) such that

ky−xk∞<ε

2.

Using Theorem 1.2 we ﬁnd a function a∈L∞([0,1]) such that kγ1,a −yk∞<ε

2.

Then

kγ1,a −xk∞≤ kγ1,a −yk∞+ky−xk∞< ε.

Lemma 4.7. Let a∈L∞([0,1]). Then γn,a =τn−1

L(γ1,a).

Proof. Follows directly from the deﬁnitions of γn,a and γ1,a, see (1.3) and (1.5).

Proposition 4.8. Γnis a dense subset of SO(N).

15

Proof. By Proposition 4.1, Γnis a subset of SO(N).

Let x∈SO(N) and ε > 0. Denote τn−1

R(x) by y. By Proposition 3.11, y∈SO(N).

Using Proposition 4.6 ﬁnd a function a∈L∞([0,1]) such that ky−γ1,ak∞< ε. Then

apply Lemma 4.7:

kx−γn,ak∞=kτn−1

L(y)−τn−1

L(γ1,a)k∞=kτn−1

L(y−γ1,a)k∞≤ ky−γ1,ak∞< ε.

We ﬁnish this section with an important observation. The results stated up to

this moment do not take into account the multiplicities of the eigenvalues. In this

connection we recall that for each bounded radial operator Rλon A2(Bn) with the

eigenvalue sequence λ∈`∞(N), the equality

Rλeα=λpeα

holds for all multi-indices α∈Nnsatisfying |α|=p, and there are n+p−1

n−1such

multi-indices.

As was mentioned, for each natural number nthe C∗-algebra generated by

Toeplitz operators on A2(Bn) with bounded radial symbols is isomorphic and iso-

metric to the C∗-algebra of multiplication operators Rλon `2(N) whose eigenvalue

sequences belong to SO(N), and thus its C∗structure does not depend on n. At the

same time these algebras, when nis varied, are quite diﬀerent if we count multiplic-

ities of eigenvalues, that is when the operators forming the algebra are considered

by their action on the basis elements of the corresponding Hilbert space A2(Bn).

Let us consider in more detail sequences of eigenvalues with multiplicities. For-

mula for the rising sum of binomial coeﬃcients states that

p

X

m=0 n+m−1

n−1=n+p−1

n.

Now, for every j∈Nthere exists a unique pin Nsuch that

n+p−1

n≤j < n+p

n.

Denote this pby πn(j), and say that the index jis located on the p-st “level”.

Given a sequence λ∈`∞, deﬁne Φn(λ) as the sequence obtained from λby

repeating each λpaccording to its multiplicity. That is,

Φn(λ) :=

(n+p−1

n)elements

z }| {

λ0

|{z}

(n−1

n−1)

times

, λ1

|{z}

(n

n−1)

times

, λ2

|{z}

(n+1

n−1)

times

, λ3

|{z}

(n+2

n−1)

times

, . . . , λp

|{z}

(n+p−1

n−1)

times

, . . ..

Since the isometric homomorphism Φnof `∞(N) is injective, the C∗-algebra gener-

ated by the set {Φn(γn,a) : a∈L∞[0,1]}coincides with Φn(SO(N)), that is, with

the C∗-algebra obtained from SO(N) by applying the mapping Φn.

16

Note that for all p, q with p<qthe following estimates hold:

ln q+ 1

p+ 1 ≤ln n+q

n−ln n+p

n≤nln q+ 1

p+ 1,

which implies that Φn(SO(N)) coincides with the C∗-algebra SOrep,n(N), a subal-

gebra SO(N), which consists of all sequences having the same elements on each

“level”:

SOrep,n(N) := nµ∈SO(N) : if πn(j) = πn(k),then µj=µko.

That is, the described above eigenvalue repetitions do not change in essence a slowly

oscillating behavior of sequences.

5 Example

In this section we construct a bounded sequence λ= (λj)j∈Nsuch that λ=γn,a

for a certain function a∈L1([0,1], r dr) but λ /∈SO(N). This implies that the

corresponding radial Toeplitz operator is bounded, but it does not belong to the

C∗-algebra generated by radial Toeplitz operators with bounded symbols.

Proposition 5.1. Deﬁne f:{z∈C:<(z)≥0} → Cby

f(z) := 1

z+nexp i

3πln2(z+n),(5.1)

where ln is the principal value of the natural logarithm (with imaginary part in

(−π, π]). Then there exists a unique function A∈L1(R+,e−udu)such that fis the

Laplace transform of A:

f(z) = Z+∞

0

A(u)e−zu dz.

Proof. For every z∈Cwith <(z)≥0 we can write ln(z+n) as ln |z+n|+iarg(z+n)

with −π

2<arg(z+n)<π

2. Then

|f(z)|=1

|z+n|exp i

3πln |z+n|+iarg(z+n)2

=1

|z+n|exp −2 arg(z+n)

3πln |z+n|

=1

|z+n|1+ 2 arg(z+n)

3π

.

Since |z+n| ≥ 1 and −1

3<−2 arg(z+n)

3π<1

3,

|f(z)| ≤ 1

|z+n|2/3.

17

Therefore for every x > 0,

ZR|f(x+iy)|2dy ≤ZR

dy

((x+n)2+y2)2/3<ZR

dy

(1 + y2)2/3<+∞,

and fbelongs to the Hardy class H2on the half-plane {z∈C:<(z)>0}. By

Paley–Wiener theorem (see, for example, Rudin [4, Theorem 19.2]), there exists a

function A∈L2(0,+∞) such that for all x > 0

f(x) = Z+∞

0

A(u)e−ux du.

The uniqueness of Afollows from the injective property of the Laplace transform.

Applying H¨older’s inequality we easily see that A∈L1(R+,e−udu):

Z+∞

0|A(u)|e−udu ≤ kAk2Z+∞

0

e−2udu1/2

=kAk2

√2.

Proposition 5.2. The sequence λ= (λj)j∈N, where

λj:= exp i

3πln2(j+n),(5.2)

belongs to `∞(N)\SO(N). Moreover there exists a function a∈L1([0,1], r dr)such

that λ=γn,a.

Proof. Since |λj|= 1 for all j∈N, the sequence λis bounded. Let Abe the function

from Proposition 5.1. Deﬁne a: [0,1] →Cby

a(r) = A(−2 ln r).

Then

Z1

0|a(r)|r dr =1

2Z1

0|a(√t)|dt =1

2Z1

0|A(−ln(t))|dt

=1

2Z+∞

0|A(u)|e−udu < +∞,

and

γn,a(j)=(j+n)Z1

0

a(√r)rj+n−1dr = (j+n)Z1

0

A(−ln r)rj+n−1dr

= (j+n)Z+∞

0

A(u) e−(j+n)udu = (j+n)f(j+n) = λj.

Let us prove that λ /∈SO(N). For every j, k ∈Nwe have

|λj−λk|=exp i

3πln2(j+n)−ln2(k+n)−1.

18

Replace jby the following function of k:

j(k) := k+$k+n

ln1/2(k+n)%.

Then

j(k)−k

k+n=1

ln1/2(k+n)+O1

k+n

and

ln(j(k) + n) = ln(k+n) + ln 1 + j(k)−k

k+n

= ln(k+n) + 1

ln1/2(k+n)−1

2 ln(k+n)+O 1

ln3/2(k+n)!.

Denote ln2(j(k) + n)−ln2(k+n) by Lkand consider the asymptotic behavior of Lk

as k→ ∞:

Lk:= ln2(j(k) + n)−ln2(k+n) = −1 + 2 ln1/2(k+n) + O1

ln(k+n).

Since Lkincreases slowly for large k, for every K > 0 there exists an integer k≥K

such that Lk+ 1 is close enough to an integer multiple of 6π2, say to 6mπ2:

Lk+ 1 ≈6mπ2.

For such k,

|λj(k)−λk|=exp i

3π(Lk+ 1 −6mπ2)exp −i

3π−1

≈exp −i

3π−16= 0.

It means that |λj(k)−λk|does not converge to 0 as kgoes to inﬁnity. On the other

hand,

ρ(j(k), k) = ln j(k)+1

k+ 1 ≤(k+n)

(k+ 1) ln1/2(k+n)→0.

It follows that λ /∈SO(N).

Acknowledgments

This research was partially supported by the projects CONACyT 102800, IPN-SIP

20120730, and by PROMEP via “Proyecto de Redes”.

19

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