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arXiv:1603.07685v1 [math.CA] 23 Mar 2016

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS

EDYTA KANIA AND MARCIN PREISNER

Abstract. Consider the Bessel operator with a potential on L2((0,∞), xαdx), namely

Lf(x) = −f′′(x)−α

xf′(x) + V(x)f(x).

We assume that α > 0and V∈L1

loc((0,∞), xαdx)is a non-negative function. By deﬁnition,

a function f∈L1((0,∞), xαdx)belongs to the Hardy space H1(L)if

sup

t>0

e−tLf

∈L1((0,∞), xαdx).

Under certain assumptions on Vwe characterize the space H1(L)in terms of atomic decompositions

of local type. In the second part we prove that this characterization can be applied to Lfor α∈(0,1)

with no additional assumptions on the potential V.

1. Introduction

1.1. Background. The Schrödinger operator on Rdis given by

e

Lf=−∆f+V · f,

where ∆is the Laplace operator and Vis a function called a potential. If we assume that V ∈ L1

loc(Rd)

and V ≥ 0then one can ﬁnd a densely deﬁned, self-adjoint operator Lon L2(Rd), that corresponds to

e

L. It is well known that Lgenerates the semigroup of contractions Kt= exp(−tL)and Ktadmits an

integral kernel Kt(x, y)such that

Ktf(x) = ZRdKt(x, y)f(y)dy,

0≤Kt(x, y)≤(4πt)−d/2exp −|x−y|2

4t.

There have been wide studies on harmonic analysis related to Schrödinger operators and, more

generally, operators with Gaussian bounds. We refer the reader to [1], [2], [3], [5], [7], [8], [9], [10], [11],

[12], [14], [15], [16], [17], [18], [20], [21], [25], and references therein. In particular, the Hardy spaces

(1.1) H1(L) = (f∈L1(Rd) : kfkH1(L):=

sup

t>0|Ktf|

L1(Rd)

<∞)

related to Lwere intensively studied. At this point let us mention that the classical Hardy space

H1(−∆) has many equivalent deﬁnitions, e.g. in terms of: various maximal functions, singular integrals,

square functions, etc. A particulary useful result is the atomic decomposition theorem (see [6], [22]):

a function f∈H1(−∆) can be decomposed as f(x) = Pkλkak(x), where Pk|λk| ≃ kfkH1(−∆) and

akare classical atoms, that is, there exist balls Bksuch that:

supp ak⊆Bk,

kakk∞≤ |Bk|−1,

Za(x)dx = 0.

In other words atoms satisfy some localization, size, and cancellation conditions.

2010 Mathematics Subject Classiﬁcation. 42B30, 42B25, 35J10 (primary), 47D03, 43A85 (secondary).

Key words and phrases. Hardy space, Schrödinger operator, Bessel operator, maximal function, atomic decomposition.

The second author was supported by Polish funds for sciences grant DEC-2012/05/B/ST1/00672 from Narodowe

Centrum Nauki.

1

2 EDYTA KANIA AND MARCIN PREISNER

Let us mention that Hofmann et al. [21] have found general results (for Vsatisfying 0≤ V ∈ L1

loc(Rd))

saying that H1(L)given above is equal to the Hardy spaces via: square functions, atomic or molecular

decompositions. However, atoms used in [21] are given in terms of L, to be more precise: a function

a∈L2(Rd)is an atom if there exist a ball Band b∈Dom(L), such that: a=Lb,supp b⊆Band b, Lb

satisfy some size condition.

An another approach, started by Dziubański and Zienkiewicz in the 90’s, was to ﬁnd atomic spaces

with simple geometric conditions that characterize H1(L). It appeared that this cannot be done in full

generality, and the properties of atoms depend strictly on the potential Vand the dimension d. For

example, if V ∈ C∞

c(Rd)and V 6≡ 0then atoms have modiﬁed cancellation condition Ra(x)ω(x)dx = 0,

where ωis such that 0< C−1≤ω(x)≤C. For this result and generalizations see [15], [17], [24]. Other

results, see [14], lead to Hardy spaces with local atoms. It was ﬁrst observed by Goldberg [19] that

if we take supremum for 0< t ≤τ2in (1.1), then one obtains atomic space with classical atoms

complemented with the atoms of the form |B|−1B(x), where the ball Bhas radius τ. In [14] the

authors assume that for 0≤ V ∈ L1

loc(Rd)there exists a family of cubes Q={Qk:k∈N}such that

∪kQk=Rd,|Qk∩Qj|= 0 for k6=j, d(Qk)≃d(Qj)if Q∗∗∗

k∩Q∗∗∗

j6=∅.

Here d(Q)is the diameter of Qand Q∗is a cube that has the same center as Qbut with slightly enlarged

diameter. The atomic space H1

at(Q)is built on classical atoms and atoms of the form |Qk|−1Qk(x).

The main result of [14] states that under two additional assumptions on V, Q,Kt(see [14, p.41], con-

ditions: (D),(K)) we have that H1(L) = H1

at(Q), see [14, Thm. 2.2]. In other words the atoms for

H1(L)are either classical atoms or local atoms related to some Qk∈ Q.

Among examples of atoms for which one can ﬁnd a family Qsuch that the assumptions of [14, Thm.

2.2] are satisﬁed, there are potentials Vin the Reverse Hölder class in dimension d≥3. For more

examples see [14]. Later, Czaja and Zienkiewicz [7, Thm. 2.4] proved that in dimension one for any

0≤ V ∈ L1

loc(R)there is a family of intervals such that [14, Thm. 2.2] gives local atomic decompositions

for H1(L).

A question that we are concerned with is: what happens if we replace −∆by the Bessel operator

Bf(x) = −f′′(x)−α/x f ′(x)on L2((0,∞), xαdx)? It is known that if α+ 1 ∈Nthen Bcorresponds

to −∆on radial functions on Rdwith d=α+ 1, however Bexists and generates a semigroup for all

α > −1, which can be considered as the Laplace operator on spaces with non-integer dimensions.

In this paper we prove results similar to [14] and [7] for the Bessel operator with a potential.

1.2. Deﬁnitions. For α > 0let (X, ρ, µ)be a metric-measure space, where X= (0,∞),ρ(x, y) =

|x−y|and dµ(x) = xαdx. Denote B(x, r) = {y∈X:ρ(x, y)< r}and observe that Xis a space of

homogeneous type in the sense of Coifman-Weiss [6], i.e. the doubling condition holds

µ(B(x, 2r)) ≤Cµ(B(x, r)),

where Cdoes not depend on x∈Xand r > 0.

The classical Bessel operator is given by

Bf(x) = −f′′(x)−α

xf′(x).

Slightly abusing notation, we shall also write Bfor the densely deﬁned, self-adjoint operator on L2(X, µ)

that corresponds to the diﬀerential operator above, see Subsection 2.1 for the semigroup generated by B.

In this paper we consider the Bessel-Schrödinger operator L,

(1.2) Lf=Bf+V·f,

where V∈L1

loc(X, µ), V ≥0. To be more precise, denote hf , giµ=Rfg dµ and deﬁne a quadratic

form

Q(f, g) = hf′, g′iµ+h√V f, √V giµ,

with the domain

Dom(Q) = nf∈L2(X, µ) : f′,√V f ∈L2(X , µ)o.

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 3

The quadratic form Qis positive and closed. Therefore, it deﬁnes a self-adjoint operator Lwith the

domain

Dom(L) = f∈Dom(Q) : ∃h∈L2(X, µ)∀g∈Dom(Q)Q(f, g) = hh, giµ.

For f , h as above we put Lf:= h. Let Kt= exp(−tL)be the semigroup generated by L. Denote by

Bsthe Bessel process on (X, µ). By using the Feynman-Kac formula,

Ktf(x) = Exexp −Zt

0

V(Bs)dsf(Bt),

one gets that Kthas an integral kernel Kt(x, y)and

(1.3) 0≤Kt(x, y)≤Pt(x, y),

where Pt(x, y)is the kernel related to Pt= exp (−tB), see Subsection 2.1.

We deﬁne the Hardy space H1(L)by means of the maximal operator associated with Kt, namely

(1.4) H1(L) = (f∈L1(X, µ) : kfkH1(L)=

sup

t>0|Ktf|

L1(X, µ)

<∞).

The goal of this paper is to give an atomic characterizations of local type for H1(L). Let |I|be the

diameter of I.

Deﬁnition 1.5. Let Ibe a collection of intervals that are closed with respect to the topology on (0,∞).

We call a family Ia proper section of Xif:

(a) for I, J ∈ I,I6=J, the intersection I∩Jis either the empty set or a singleton,

(b) X=SI∈I I,

(c) there exists a constant C0>0such that for I, J ∈ I,I∩J6=∅we have

C−1

0|I| ≤ |J| ≤ C0|I|.

Denote τB(c, r) := B(c, τ r). For an interval I=B(x, r)(if I= (0,2A)we take I=B(A, A)), let

cI := B(x, cr). For a family Ias in Deﬁnition 1.5 we set I∗:= βI for some ﬁxed β > 1, such that

I∗∗∗ ∩J∗∗∗ 6=∅if and only if I∩J6=∅.

We say that a function a:X→Cis an (I, µ)-atom if:

(i)there exist I∈ I and an interval J⊂I∗∗,such that: supp(a)⊂J, kak∞≤µ(J)−1,Za dµ = 0,

or

(ii)there exists I∈ I, such that a(x) = µ(I)−1I(x).

The atoms as in (ii)are called local atoms.

The atomic Hardy space H1

at(I, µ)associated with the collection Iis deﬁned in the following way.

We say that f∈ H1

at(I, µ)if

(1.6) f(x) = X

n

λnan(x),

where λn∈C,anare (I, µ)-atoms, and Pn|λn|<∞. Set

(1.7) kfkH1

at(I,µ):= inf X

n|λn|,

where the inﬁmum is taken over all possible representations of fas in (1.6).

For a collection Ias above and V≥0,V∈L1

loc(X, µ)we consider the following two conditions:

- there exist constans C, ε > 0such that

(D) sup

y∈I∗∗ ZX

K2k|I|2(x, y)dµ(x)≤Ck−1−εfor I∈ I, k ∈N,

- there exist constans C, δ > 0such that

(K) Z2t

0ZX

Ps(x, y)I∗∗∗ (y)V(y)dµ(y)ds ≤Ct

|I|2δ

for x∈X, I ∈ I, t ≤ |I|2.

4 EDYTA KANIA AND MARCIN PREISNER

1.3. Statement of results. Our ﬁrst main result is the following, cf. [14, Thm. 2.2]

Theorem 1.8. Assume that a proper section Iand 0≤V∈L1

loc(X, µ)are given, so that (D)and

(K)hold. Then H1(L) = H1

at(I, µ)and there exists a constant C > 0, such that

C−1kfkH1

at(I,µ)≤ kfkH1(L)≤CkfkH1

at(I,µ).

In the second part we give an important application of Theorem 1.8. Let us restrict ourselves to

α∈(0,1). We prove that for any 0≤V∈L1

loc(X, µ)we can ﬁnd a family I(V)such that the

assumptions of Theorem 1.8 hold, cf. [7, Thm. 2.4] for the case α= 0. To be more precise, let Dbe a

family of dyadic intervals on (0,∞), that is D={[k2n,(k+ 1)2n] : k∈N∪ {0}, n ∈Z}. Consider the

family I(V)that consists of maximal dyadic closed intervals Ithat satisfy

(S) |2I|2

µ(2I)Z2I

V(y)dµ(y)≤1.

In Section 4we prove that I(V)is a well deﬁned proper section. The second main result is the

following.

Theorem 1.9. Let α∈(0,1) and 0≤V∈L1

loc(X, µ). Then the family I(V)satisﬁes the assumptions

of Theorem 1.8.

Corollary 1.10. Let α∈(0,1) and 0≤V∈L1

loc(X, µ). Then there is C > 0, such that

C−1kfkH1

at(I(V),µ)≤ kfkH1(L)≤CkfkH1

at(I(V),µ).

The paper is organized as follows. In Section 2we study the atomic Hardy spaces related to Band

its local versions. This is used in a proof of Theorem 1.8, which is provided in Section 3. Finally, in

Section 4a proof of Theorem 1.9 is given.

2. Hardy spaces for the Bessel operator

2.1. Global Hardy space for B.In this section we consider the case α > 0. Let Pt= exp(−tB)be

the Bessel semigroup given by

Ptf(x) = ZX

Pt(x, y)f(y)dµ(y),

Pt(x, y) = (2t)−1exp −x2+y2

4tIα−1

2xy

2t(xy)−α−1

2,

where Iα(x) = P∞

m=0 1

m!Γ(m+α+1) x

22m+αis the modiﬁed Bessel function of the ﬁrst kind. It is clear

that Pt(x, y) = Pt(y, x)and, since B(0,∞)(x) = 0, we have that

(2.1) ZX

Pt(x, y)dµ(x) = 1.

Let us recall that

(2.2) µ(B(x, t)) ≃t(x+t)α.

It is known that the kernel Pt(x, y)satisﬁes the two-side Gaussian estimates (see, e.g. [13, Lem. 4.2]),

(2.3) C−1µ(B(x, √t))−1exp −|x−y|2

c1t!≤Pt(x, y)≤Cµ(B(x, √t))−1exp −|x−y|2

c2t!,

while the derivative satisﬁes

(2.4)

∂

∂x Pt(x, y)≤Ct−1/2µ(B(x, √t))−1exp −|x−y|2

ct !.

Let H1(B)be the Hardy space related to Pt, i.e. the space deﬁned as in (1.4) with Land Ktreplaced

by Band Pt, respectively. We call a function aan µ-atom if

(iii)there exists an interval J⊆X, such that: supp(a)⊂J, kak∞≤µ(J)−1,Za dµ = 0.

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 5

The atomic Hardy space H1

at(µ)is deﬁned as in (1.6) and (1.7) with anbeing µ-atoms.

Theorem 2.5. [4, Thm. 1.7] Let α > 0. There is C > 0such that

C−1kfkH1

at(µ)≤ kfkH1(B)≤CkfkH1

at(µ).

2.2. Local Hardy space for B.For τ > 0we deﬁne h1

τ(B),the local Hardy space related to B, as

the set of L1(X, µ)functions for which the norm

kfkh1

τ(B)=

sup

t≤τ2|Ptf|

L1(X, µ)

is ﬁnite.

Let Iτbe a proper section of Xthat consists of closed intervals of length τ.

Theorem 2.6. (a)There exists C > 0such that for τ > 0and an (Iτ, µ)-atom awe have

kakh1

τ(B)≤C.

(b)Let Ibe an interval such that supp(f)⊂I∗and f∈h1

|I|(B). Then

f=

∞

X

n=0

λnan,

∞

X

n=0 |λn| ≤ Ckfkh1

|I|(B),

where a0(x) = µ(I)−1I(x)and anare µ-atoms supported in I∗∗ for n≥1.

Let us remark that another characterization of h1

τ(B), by mean of a local Riesz transform, was given

in [23, Thm. 2.11]. Theorem 2.6 will be used to prove Theorem 1.8.

Proof. (a)Obviously, if ais µ-atom, then the statement follows from Theorem 2.5. Assume then that

a(x) = µ(I)−1I(x)and |I|=τ. It is well known that (2.3) implies the boundedness of the maximal

operator supt>0|Ptf|on L2(X, µ). Using this fact and the Schwarz inequality,

sup

t≤τ2|Pta|

L1(I∗∗,µ)≤Cµ(I)1/2

sup

t>0|Pta|

L2(X,µ)≤Cµ(I)1/2kakL2(X,µ)≤C.

Denote by cIthe center of Iand notice that |x−y| ≃ |x−cI|when y∈Iand x∈(I∗∗)c. By (2.3),

sup

t≤τ2|Pta|

L1((I∗∗)c,µ)≤CZ(I∗∗ )c

sup

t≤τ2ZI

µ(B(x, √t))−1exp −|x−cI|2

c3tµ(I)−1dµ(y)dµ(x)

≤CZ(I∗∗)c

sup

t≤τ2

t−1/2exp −|x−cI|2

c3tdx

≤CZ(I∗∗)c|I|−1exp −|x−cI|2

c4|I|2dx ≤C.

(b)Deﬁne λ0:= Rfdµ and g(x) = f(x)−λ0µ(I)−1I(x). Notice that

|λ0| ≤ kfkL1(X,µ)≤ kfkh1

|I|(B).

Therefeore, kgkL1(X,µ)≤2kfkL1(X,µ). Our goal is to prove that

(2.7) kgkH1(B)≤Ckfkh1

|I|(B).

For t≤ |I|2using (a)we obtain

sup

t≤|I|2|Ptg|

L1(X,µ)≤ kfkh1

|I|(B)+|λ0|

µ(I)−1I

h1

|I|(B)≤Ckfkh1

|I|(B)

6 EDYTA KANIA AND MARCIN PREISNER

As for t≥ |I|2notice that Rgdµ = 0. Therefore, using (2.4),

sup

t≥|I|2|Ptg|

L1((I∗∗)c,µ)≤Z(I∗∗ )c

sup

t≥|I|2ZI∗

(Pt(x, y)−Pt(x, cI)) g(y)dµ(y)dµ(x)

≤Z(I∗∗)cZI∗

sup

t≥|I|2

|y−cI|

√tµ(B(x, √t))−1exp −|x−cI|2

c2t|g(y)|dµ(y)dµ(x)

≤CkgkL1(X,µ)|I|Z(I∗∗)c

sup

t≥|I|2

t−1exp −|x−cI|2

c2tdx

≤CkfkL1(X,µ)|I|Z(I∗∗)c|x−cI|−2dx ≤Ckfkh1

|I|(B).

Likewise,

sup

t≥|I|2|Ptg|

L1(I∗∗,µ)≤CZI∗∗ ZI∗

sup

t≥|I|2

µ(B(x, √t))−1|g(y)|dµ(y)dµ(x)

≤CkgkL1(X,µ)ZI∗∗

sup

t≥|I|2

t−1/2dx ≤Ckfkh1

|I|(B).

From (2.7) we have that g∈H1(B), so using Theorem 2.5 we obtain λkand µ-atoms aksuch that

g=P∞

k=1 λkak. Consequently,

f=

∞

X

k=0

λkak,

∞

X

k=0 |λk| ≤ Ckfkh1

|I|(B),

where a0(x) = µ(I)−1I(x).

The only problem we have to deal with is that akare not necessarily supported in I∗∗ . Let ψbe

a function such that ψ≡1on I∗,ψ≡0on (I∗∗ )c, and kψ′k∞≤C|I|−1. To complete the proof we

will show, that for every µ-atom akthere exist a sequence aj

k, such that each aj

kis either µ-atom or

aj

k=µ(I)−1I. Moreover, ak=Pjλj

kaj

kand Pj|λj

k|< C, with Cthat not depend on k. Fix a=ak

and an interval J⊆Xsuch that: supp(a)⊂J,kak∞≤µ(J)−1. Obviously supp(ψa)⊆I∗∗ ∩J,

and if J⊂I∗, then ψa =ais an µ-atom. Furthermore, if J⊂(I∗∗ )c, then ψa = 0, so it suﬃces

to consider the case that J∩(I∗)c∩I∗∗ 6=∅. Observe that |J| ≤ |I|and let N∈Nbe such that

1/2N+1|I| ≤ |J| ≤ 1/2N|I|.

Deﬁne λ:= Rψa dµ and notice that

|λ|=ZJ

ψ(x)a(x)dµ(x)=ZJ

a(x)(ψ(x)−ψ(cJ)) dµ(x)

≤µ(J)−1ZJ|x−cJ||ψ′(ξ)|dµ(x)

≤C|J||I|−1≤C2−N.

Let us choose intervals Ij, such that J=: I0⊂I1⊂ ··· ⊂ IN⊂I∗∗, where |Ij+1 |/|Ij|= 2 and

|IN| ≃ |I|. Then

ψa =ψa −λµ(I0)−1I0+

N

X

j=1

λµ(Ij−1)−1Ij−1−µ(Ij)−1Ij

+λ(µ(IN)−1IN−µ(I)−1I) + λµ(I)−1I=

N+2

X

j=0

bj.

Observe that:

(1) for j= 0, ..., N we have: supp(bj)⊆Ij,Rbjdµ = 0, and

kbjk∞≤C|λ|µ(Ij)−1≤C2−Nµ(Ij)−1,

(2) supp(bN+1)⊆I∗∗ ,kbN+1k∞≤C|λ|µ(I)−1and RbN+1 dµ = 0.

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 7

We conclude that bjare multiples of (I|I|, µ)-atoms and

N+2

X

j=0

bj

H1

at(I|I|,µ)

≤C

N+2

X

j=0

2−N≤C.

Corollary 2.8. There exists a constant C > 0such that for τ > 0we have

C−1kfkH1

at(Iτ,µ)≤ kfkh1

τ(B)≤CkfkH1

at(Iτ,µ).

The right inequality in Corolarry 2.8 follows easily from Theorem 2.6(a). For the left inequality one

uses Theorem 2.6(b)with a suitable partition of unity and methods as in Lemmas 3.2 and 3.5 below.

We omit the details.

3. Proof of Theorem 1.8

3.1. Auxiliary estimates. For a proper section Ilet {φI}I∈I be a partition of unity associated with

I, that is a family of C∞functions on X, such that supp(φI)⊂I∗,0≤φI≤1,kφ′

Ik∞≤ |I|−1and

PI∈I φI(x) = 1 for all x∈X.

The perturbation formula states that

(3.1) Pt(x, y)−Kt(x, y) = Zt

0ZX

Pt−s(x, z)V(z)Ks(z, y)dµ(z)ds.

To prove Theorem 1.8 we closely follow the proof of [14, Thm. 2.2]. However, in our weighted space,

some technical diﬃculties appear. Therefore we present all the details for convenience of the reader.

Lemma 3.2. For I∈ I and f∈L1(X, µ),

sup

t≤4|I|2|Pt(φIf)|

L1((I∗∗)c, µ)≤CkφIfkL1(X, µ).

Proof. Denote by cIthe center of the interval I∈ I. For x∈(I∗∗)cand y∈I∗we have |x−y| ≃

|x−cI|. Notice that (2.2) implies

(3.3) µ(B(x, √t))−1xα≤t−1/2.

Using (2.3) the left-hand side is bounded by

Z(I∗∗)c

sup

t≤4|I|2ZI∗

µ(B(x, √t))−1e−|x−y|2

c2t|φI(y)f(y)|dµ(y)dµ(x)

≤CkφIfkL1(X,µ)Z(I∗∗)c

sup

t≤4|I|2

x−αt−1

2e−|x−cI|2

c3tdµ(x)

≤CkφIfkL1(X,µ)Z(I∗∗)c|I|−1e−|x−cI|2

c4|I|2dx

≤CkφIfkL1(X, µ).

Corollary 3.4. For I∈ I and f∈L1(X, µ),

kφIfkh1

|I|(B)≃

sup

t≤|I|2|Pt(φIf)|

L1(I∗∗, µ)

.

Denote

e

fI:= X

J:I∩J6=∅

φJf, ¯

fI:= f−e

fI=X

J:I∩J=∅

φJf.

8 EDYTA KANIA AND MARCIN PREISNER

Lemma 3.5. For I∈ I and f∈L1(X, µ),

sup

t≤|I|2KtφIe

fI−φIKte

fI

L1(I∗∗, µ)≤CX

J:I∩J6=∅kφJfkL1(X, µ)

Proof. Denote e

I=∪J:I∩J6=∅J∗. Note that for x∈I∗∗ and y∈e

I, there is |x−y| ≤ C|I|and

(3.6) sup

t≤|I|2

t−1

2e−|x−y|2

c2t≤C|x−y|−1.

Using (1.3), (2.3), (3.3), and (3.6),

sup

t≤|I|2KtφIe

fI−φIKte

fI

L1(I∗∗, µ)≤ZI∗∗

sup

t≤|I|2Ze

I|φI(y)−φI(x)|Kt(x, y)e

fI(y)dµ(y)dµ(x)

≤CZI∗∗ Ze

I|x−y|kφ′

Ik∞sup

t≤|I|2

t−1

2e−|x−y|2

c2te

fI(y)dµ(y)dx

≤CZI∗∗ |I|−1dx

e

fI

L1(X,µ).

Lemma 3.7. Assume that Vand Iare given, so that (D)holds. Then

X

I∈I

sup

t>0Kt¯

fI

L1(I∗∗∗, µ)≤CkfkL1(X, µ).

Proof. Denote sm= 2m|I|2and let m≥2. By the semigroup property, (1.3), (2.3),

sup

sm≤t≤sm+1

Kt(|φJf|)

L1(X, µ)

≤ZX

sup

sm≤t≤sm+1 ZXZX

Pt−sm−1(x, z)Ksm−1(z, y)|φJ(y)f(y)|dµ(z)dµ(y)dµ(x)

≤CZX|φJ(y)f(y)|ZX

Ksm−1(z, y)ZX

s−1/2

mexp −|x−z|2

c3smdx dµ(z)dµ(y)

≤C(m−1)−1−ε· kφJfkL1(X, µ).

In the last inequality we have applied (D). Using the above estimate and Lemma 3.2,

X

I∈I

sup

t>0Kt X

J:I∩J=∅

φJf!

L1(I∗∗∗, µ)

≤X

J∈I X

I:J∩I=∅

sup

t>0

Kt(|φJf|)

L1(I∗∗∗, µ)≤CX

J∈I

sup

t>0

Kt(|φJf|)

L1((J∗∗)c, µ)

≤CX

J∈I

sup

t≤4|J|2

Kt(|φJf|)

L1((J∗∗)c, µ)

+CX

J∈I

∞

X

m=2

sup

sm≤t≤sm+1

Kt(|φJf|)

L1((J∗∗)c, µ)

≤CX

J∈I kφJfkL1(X, µ)+CX

J∈I

∞

X

m=2

m−1−ε· kφJfkL1(X, µ)≤CkfkL1(X, µ).

Lemma 3.8. Z∞

0ZX

V(z)Ks(|f|)(z)dµ(z)ds ≤CkfkL1(X, µ)

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 9

Proof. By integrating (3.1) and using (2.1) we obtain

Zt

0ZX

V(z)Ks(z, y)dµ(z)ds ≤C.

The proof is ﬁnished by setting t→ ∞.

Lemma 3.9. Assume that Vand Iare given, so that (K)holds. Then for I∈ I,

sup

t≤|I|2|(Pt−Kt)(φIf)|

L1(X, µ)≤CkφIfkL1(X, µ).

Proof. By (1.3) and Lemma 3.2,

sup

t≤|I|2|(Pt−Kt)(φIf)|

L1((I∗∗)c, µ)≤CkφIfkL1(X, µ).

Consider the integral on I∗∗ . From (3.1),

(Pt−Kt)(φIf)(x) = Zt

0

Pt−sV′′Ks(φIf)(x)ds +Zt

0

Pt−sV′Ks(φIf)(x)ds

where V=I∗∗∗ V+(I∗∗∗)cV=V′+V′′ . Repeating the argument from the proof of Lemma 3.2 and

using Lemma 3.8 we obtain

sup

t≤|I|2Zt

0

Pt−sV′′Ks(φIf)ds

L1(I∗∗

,µ)≤CkφIfkL1(X, µ).

To estimate the integral that contains V′write

Zt

0

Pt−sV′Ks(φjf))(x)ds≤Zt/2

0

Pt−sV′Ps(|φjf|)(x)ds +Zt

t/2

Pt−sV′Ps(|φjf|)(x)ds

=A1,t(x) + A2,t (x).

Let tm= 2−m|I|2. Similarly as in proof of the Lemma 3.7 we obtain

sup

t≤|I|2

A1,t

L1(X, µ)≤

∞

X

m=0 ZX

sup

tm+1≤t≤tmZt/2

0ZX

Pt−s(x, y)V′(y)Ps(|φIf|)(y)dµ(y)ds dµ(x)

≤C

∞

X

m=0 ZXZtm

0

V′(y)Ps(|φIf|)(y)ZX

t−1

2

me−|x−y|2

c3tmdx ds dµ(y)

≤C

∞

X

m=0 Ztm

0ZX

I∗∗∗ (y)V(y)Ps(|φIf|)(y)ds dµ(y)

≤C

∞

X

m=0 2−m|I|2

|I|2δ

kφIfkL1(X, µ).

In the last inequality we have used (K). To estimate A2,t we proceed similarly noticing that for

t∈[tm+1, tm]and s∈[t/2, t]we have s∈[tm+2, tm]. The details are left to the reader.

3.2. Proof of Theorem 1.8. First inequality. Observe that φIf=φIe

fIand

Kt(φIf) = KtφIe

fI−φIKte

fI−φIKt¯

fI+φIKt(f).

From Lemmas 3.5 and 3.7 we deduce that

10 EDYTA KANIA AND MARCIN PREISNER

X

I∈I

sup

t≤|I|2|Kt(φIf)|

L1(I∗∗, µ)≤X

I∈I

sup

t≤|I|2KtφIe

fI−φIKte

fI

L1(I∗∗, µ)

+X

I∈I

sup

t≤|I|2φIKt¯

fI

L1(I∗∗, µ)

+X

I∈I

sup

t≤|I|2|φIKtf|

L1(I∗∗, µ)

≤CkfkL1(X, µ)+kfkH1(L)≤CkfkH1(L).

The above estimate, together with Corollary 3.4 and Lemma 3.9, lead to

X

I∈I kφIfkh1

|I|(B)≤CX

I∈I

sup

t≤|I|2|(Pt−Kt)(φIf)|

L1(I∗∗, µ)

+CX

I∈I

sup

t≤|I|2|Kt(φIf)|

L1(I∗∗, µ)

≤CkfkH1(L).

Now we use Theorem 2.6(b)for each φIfgetting (λI

n)nand (I, µ)-atoms (aI

n)n, so that

φIf(x) = X

n

λI

naI

n(x),and X

nλI

n≤CkφIfkh1

|I|(B).

Summing up for all I∈ I we ﬁnish the ﬁrst part of the proof.

Second inequality. Let abe an (I, µ)-atom, such that supp(a)⊂I∗∗ . There exists an integer

m≥0, independent of I, such that

inf |I|2:J∩I6=∅≥2−m|I|2.

Denote tn= 2−n|I|2. Observe that

sup

t≤tm|Kt(a)|

L1(X, µ)≤

sup

t≤tm|(Kt−Pt)(a)|

L1(X, µ)

+

sup

t≤tm|Pt(a)|

L1(X, µ)≤ kakL1(X, µ).

In the last inequality we applied Lemma 3.9 and Theorem 2.6(b), since ais also an (I|I|, µ)-atom.

It suﬃces to estimate

supt>tm|Kta|

L1(X, µ). This is done by using (1.3) and (K). Indeed, using

similar methods as in the proof of Lemma 3.2,

sup

t>tm|Kta|

L1(X, µ)≤X

n≤m

sup

tn≤t≤tn−1|Kta|

L1(X, µ)

≤X

n≤m

sup

tn+1≤t≤3tn+1

KtKtn+1 a

L1(X, µ)

≤CX

n≤m+1 ZXZX

K2−n|I|2(x, y)|a(y)|dµ(y)dµ(x)

≤CkakL1(X, µ)≤C,

since the operator suptn+1 ≤t≤3tn+1 Ktis bounded on L1(X, µ), see (2.3).

4. Proof of Theorem 1.9.

In the whole section we assume that α∈(0,1). Recall that for 0≤a < b,

(4.1) µ((a, b)) = b1+α−a1+α

1 + α≃(bα+1 2a≤b

(b−a)aα2a≥b.

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 11

By (4.1) and the Mean-Value Theorem we easily get

(4.2) |I|2

µ(I)≃b1−α−a1−α.

Lemma 4.3. For α∈(0,1) and I⊂J⊂(0,∞),

|I|2

µ(I)ZI

V(y)dµ(y)≤|J|2

µ(J)ZJ

V(y)dµ(y).

Proof. Obviously, since V≥0, it is enough to prove that

(4.4) |I|2

µ(I)≤|J|2

µ(J).

Let a≤b≤c≤dand I= (b, c)⊂(a, d) = J. Denote Γ(x, y) = (y−x)2/(yα+1 −xα+1 ). Now, (4.4) is

equivalent to Γ(b, c)≤Γ(a, d). This is done in two steps.

Step 1: Γ(b, c)≤Γ(a, c). Denote a=sc and b=tc, where 0< s ≤t < 1. It is enough to prove

(1 −t)2

1−tα+1 ≤(1 −s)2

1−sα+1 .

By a simple calculus argument, the function F1(x) = (1 −x)2/(1 −xα+1 )is monotonically decreasing

for x∈(0,1).

Step 2: Γ(a, c)≤Γ(a, d). Similarly, let c=t′a,d=s′a,1< t′≤s′. The function F2(x) =

(x−1)2/(xα+1 −1) is monotonically increasing in (1,∞), thus

(t′−1)2

(t′)α+1 −1≤(s′−1)2

(s′)α+1 −1.

Proposition 4.5. Let V∈L1

loc(X, µ),V≥0. Then the family I(V)of maximal dyadic intervals

satisfying (S)is a well-deﬁned proper section (see Deﬁnition 1.5).

Proof. For a closed dyadic interval Iconsider F(I) = |2I|2µ(2I)−1R2IV dµ and denote by Idthe

smallest dyadic interval containing I. Notice that 2I⊆2Idand, by Lemma 4.3, we have F(I)≤F(Id).

Also, for an increasing sequence of dyadic intervals In⊂In+1 we have limn→∞ F(In) = ∞, see (4.2).

This justiﬁes the choice of I(V)as maximal dyadic intervals such that (S) holds. What is left to

prove is that I(V)is a proper section, namely we need to show that for I , J ∈ I(V),I∩J6=∅we have

|I| ≃ |J|.

By contradiction, suppose that there exist Ik,Jksuch that Ik∩Jk6=∅and |Ik|/|Jk| → ∞. We can

assume that 2Jd

k⊆2Ikfor all k. Denote, ak=|Ik|2µ(Ik)−1|Jk|−2µ(Jk). By the choice of I,

1≥|2Ik|2

µ(2Ik)Z2Ik

V(y)dµ(y)≥|2Ik|2µ(2Jd

k)

µ(2Ik)2Jd

k22Jd

k2

µ(2Jd

k)Z2Jd

k

V(y)dµ(y)≥C−1ak.

The proof will be ﬁnished when we show that ak→ ∞. This follows from (4.1) by considering several

cases. Let a, b, c be such that 0≤a < b < c.

Case 1: Jk= [a, b],Ik= [b, c].

Subcase 1: 4a≤2b≤c. Using (4.1) we have

ak≃(c/b)2(b/c)1+α= (c/b)1−α≃(|Ik|/|Jk|)1−α.

Subcase 2: 4a≤2b≥c. This subcase can hold only for ﬁnite k.

Subcase 3: 4a≥2b≤c. Using (4.1) we have

ak≃c2

(b−a)2

(b−a)aα

cα+1 =aαc

cα(b−a)≥c1−α

(b−a)1−α≃(|Ik|/|Jk|)1−α.

12 EDYTA KANIA AND MARCIN PREISNER

Subcase 4: 4a≥2b≥c. Using (4.1) we have

ak≃c−b

b−a2(b−a)aα

(c−b)bα≃ |Ik|/|Jk|.

Case 2: Jk= [b, c],Ik= [a, b]. Then

ak≥C|Ik|

|Jk|2|Jk|bα

|Ik|bα≃ |Ik|/|Jk|.

Recall that hf, g iµ=RXfg dµ, so that h−Bφ, ψ iµ=hφ, −Bψiµfor appropriate ψ , φ. For y > 0the

distributional equation

−Bφ=δy

has the solution given by φy(x) = 1

2(1−α)x1−α−y1−α. We shall use φyto construct superharmonic

functions that will be crucial in the proof of (D).

Lemma 4.6. Let α∈(0,1),0≤V∈L1

loc(X, µ), and I(V)is as in Proposition 4.5. Then

ZX

K2n|I|2(x, y)dµ(x)≤C2−1−α

2n

for y∈I∗∗,I∈ I(V), and n≥0.

Proof. Let Ibe a dyadic interval such that

|2I|2

µ(2I)Z2I

V dµ ≤1,|2Id|2

µ(2Id)Z2Id

V dµ > 1.

By continuity argument there exists Jsuch that I∗∗ ⊂2I⊂J⊂2Idand |J|2

µ(J)RJV dµ = 1. Let

J= (a, b)and observe that |J| ≃ |I|. Deﬁne

φI(x) = 1 + 1

2(1 −α)ZJ

V(y)x1−α−y1−αdµ(y).

Fix z∈I∗∗. By (4.2) and the doubling condition,

(4.7) φI(z)≤1 + Csup

y,y′∈J|y1−α−(y′)1−α|ZJ

V dµ ≃C.

Also, we claim that for x∈X,

(4.8) φI(x)≃1 + µ(J)x1−α−z1−α

|J|2.

Indeed, if xis such that |x1−α−z1−α| ≤ C|J|2µ(J)−1, with Clarge enough, this follows exactly as in

(4.7). In the opposite case |x1−α−z1−α| ≥ C|J|2µ(J)−1, we have |x1−α−z1−α| ≃ |x1−α−y1−α|for

y∈Jand the claim follows.

Now we proceed to a crucial argument that uses superharmonicity. Observe that formal calculation

gives

−LφI(x) = −BφI(x)−V(x)φI(x) = V(x)( J(x)−φI(x)) ≤0

and, consequently,

∂

∂t KtφI(z) = ZX

Kt(z, x)(−LφI(x)) dµ(x)≤0.

This leads to

(4.9) KtφI(z)≤φI(z), t > 0.

However, φIis not in Dom(L)(or even in L2(X, µ)), thus we provide a detailed proof of (4.9) in

Appendix.

Denote

θ(t) = ZX

Kt(z, x)dµ(x).

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 13

Our goal is prove that

(4.10) θ(2n|I|2)≤c02−1−α

2n.

This will follow by induction argument. By (1.3) and (2.3),

(4.11) K2t(z, x) = ZX

Kt(z, y)Kt(y, x)dµ(y)≤Cθ(t)µ(B(x, √t))−1

and

θ(2t) = Z|x1−α−z1−α|<R

K2t(z, x)dµ(x) + Z|x1−α−z1−α|>R

K2t(z, x)dµ(x) = A1+A2.

By (4.8), (4.9), and (4.7) we have

A2≤|J|2

Rµ(J)Z|x1−α−z1−α|>R

K2t(z, x)µ(J)x1−α−z1−α

|J|2dµ(x)

≤CR−1|J|2µ(J)−1ZK2t(z, x)φI(x)dµ(x)

≤CR−1|J|2µ(J)−1.

To estmate A1we use (4.11) and (2.2),

A1≤Cθ(t)Z|x1−α−z1−α|<R

µ(B(x, √t))−1dµ(x)

≤Cθ(t)t−1/2Z|x1−α−z1−α|<R

dx.

Case A: z1−α≥2R, then

Z|x1−α−z1−α|<R

dx =1

1−α(z1−α+R)1

1−α−(z1−α−R)1

1−α≤CRzα.

In this case

(4.12) θ(2t)≤c1θ(t)t−1/2Rzα+R−1|J|2µ(J)−1.

Case B: z1−α<2R,

Z|x1−α−z1−α|<R

dx ≤C(z1−α+R)1

1−α≤CR 1

1−α.

In this case

(4.13) θ(2t)≤c2θ(t)t−1/2R1

1−α+R−1|J|2µ(J)−1.

Now we proceed to the proof of (4.10). The ﬁrst step, θ(|I|2)≤C, follows simply by (1.3). Assume

that (4.10) holds for some n. Consider the following cases:

Case 1: ρ(0, I )≥2|I|. In this case µ(I)≃ |I|zα.

Subcase 1.1: θ(2n|I|2)≥2n/2|I|2z−2. Observe that

R1:= 2−1θ(2n|I|2)−1/2(2n|I|2)1/4|I|1/2z−α< z1−α/2.

Putting R=R1and t= 2n|I|2into (4.12) and using the induction hypothesis,

θ(2n+1|I|2)≤c3θ(2n|I|2)1/2(2n|I|2)−1/4|I|1/2

≤c3c1/2

02−n1−α

42−n

4≤c02−(n+1) 1−α

2.

The last inequality holds if we choose c0such that c0≥c2

321−α.

Subcase 1.2: θ(2n|I|2)≤2n/2|I|2z−2. One easily checks that

R2:= 2−1θ(2n|I|2)−1(2n|I|2)1/2|I|z−α1−α

2−α> z1−α/2.

14 EDYTA KANIA AND MARCIN PREISNER

Putting R=R2and t= 2n|I|2into (4.13) and using the induction hypothesis,

θ(2n+1|I|2)≤c4θ(2n|I|2)(2n|I|2)−1/2|I|1

1−αz−α

1−α1−α

2−α

≤c4θ(2n|I|2)(2n|I|2)−1/2|I|1−α

2−α

≤c4c0(2−n)1−α

2+1

21−α

2−α≤c02−(n+1) 1−α

2.

In the last inequality we choose c0such that c0≥c2−α

42(1−α)(2−α)/2. Notice that we have used z≥ |I|,

which follows from ρ(0, I)≥2|I|.

Case 2: ρ(0, I )≤2|I|. In this case µ(I)≃ |I|α+1 . Notice that z≤4|I|.

Subcase 2.1 θ(2n|I|2)≥2n/2|I|2−αzα−2. Observe that

R3:= 2−1θ(2n|I|2)−1/2(2n|I|2)1/4|I|1−α

2z−α/2< z1−α/2.

Putting R=R3and t= 2n|I|2into (4.12) and using the induction hypothesis,

θ(2n+1|I|2)≤c5θ(2n|I|2)1/22−n/4zα/2|I|−α/2

≤c5c1/2

02−n1−α

42−n

4zα/2|I|−α/2

≤c5c1/2

02α2−n1−α

42−n

4

≤c02−(n+1) 1−α

2.

The last inequality holds if we choose c0such that c0≥c2

521+α.

Subcase 2.2 θ(2n|I|2)<2n/2|I|2−αzα−2. One easily checks that

R4:= 2−1θ(2n|I|2)−1(2n|I|2)1/2|I|1−α1−α

2−α> z1−α/2.

Putting R4and t= 2n|I|2into (4.13) we obtain

θ(2n+1|I|2)≤c6θ(2n|I|2)(2n|I|2)−1/2|I|1−α

2−α

≤c6c0(2−n)1−α

2+1

21−α

2−α≤c02−(n+1) 1−α

2,

similarly as in subcase 1.2.

Lemma 4.14. Let α∈(0,1),0≤V∈L1

loc(X, µ), and I(V)is as in Proposition 4.5. Then the pair

(V, I(V)) satisﬁes (K).

Proof. Case 1: ρ(0, I )≤2|I|. In this case µ(I)≃ |I|1+α. Using (2.3) and (2.2),

Z2t

0ZI∗∗∗

Ps(x, y)V(y)dµ(y)ds ≤CZ2t

0

s−1+α

2ds ·ZI∗∗∗

V dµ ≤C t 1−α

2µ(I)

|I|2≤Ct

|I|21−α

2

.

Case 2: ρ(0, I)>2|I|. In this case µ(I)≃ |I|cα

I, where cIdenotes the center of I. Using (2.3) and

the doubling condition,

Z2t

0ZI∗∗∗

Ps(x, y)V(y)dµ(y)ds ≤CZ2t

0ZI∗∗∗

µ(B(y, √s))−1V(y)dµ(y)ds

≤CZ2t

0

s−1

2c−α

Ids ·ZI∗∗∗

V(y)dµ(y)

≤Ct 1

2µ(I)

|I|2c−α

I≤Ct

|I|21

2

.

Combining Lemmas 4.6 and 4.14 we obtain Theorem 1.9.

HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 15

Appendix

The goal of this Appendix is to give a precise proof of the formula (4.9). Recall that J= (a, b). By

the deﬁnition of φIwe have

(4.15)

φI(x) = 1 + 1

2(1 −α)

RJV(y)y1−αdµ(y)−x1−αRJV(y)dµ(y), x < a

Rx

aV(y)(x1−α−y1−α)dµ(y)−Rb

xV(y)(y1−α−x1−α)dµ(y), x ∈(a, b)

x1−αRJV(y)dµ(y)−RJV(y)y1−αdµ(y), x > b.

Recall that RJV=µ(J)|J|−2. The formula (4.15) easily implies that

(4.16) |φ′

I(x)| ≤ Cµ(J)|J|−2x−α

for all x∈X.

Lemma 4.17. Let ψ∈Dom(Q),ψ≥0,η∈C∞

c(X, µ),η≡1on supp(ψ). Then φIη∈Dom(Q)and

Q(ψ, φIη) = ZX

ψ(x)V(x)(φI(x)−J(x)) dµ(x)≥0.

Proof. Observe that using (4.15) we obtain that

−φ′′

I(x)−α

xφ′

I(x) = −J(x)V(x).

Since supp(η)is compact, using facts (4.8), (4.16) we deduce that φIη∈Dom(Q). Since η≡1on

supp (ψ),ZX

ψ′(x)(φI(x)η(x))′dµ(x) = ZX

ψ′(x)φ′

I(x)dµ(x)

=ZX

ψ(x)−φ′′

I(x)−α

xφ′

I(x)dµ(x)

=−ZX

ψ(x)J(x)V(x)dµ(x).

The lemma follows, since φI≥1.

Recall that z∈Jis ﬁxed and denote ϑ(u) = KuφI(z).Our goal is to prove that ϑ(t+s)≤ϑ(t)for

t, s > 0.

Denote k(x) := Kt(x, z) = Kt/2(Kt/2(·, z ))(x). Since the semigroup Ktis analytic and Kt/2(·, z )∈

L2(X, µ)(see (1.3) and (2.3)), we have k∈Dom(L)⊂Dom(Q). Let ηn→X,ηn∈C∞

c(X, µ),

supp(ηn)⊂(0, n + 1],ηn≡1on (0, n], and kη′

nk∞+kη′′

nk∞≤C.

First, observe that

(Ku(k)ηn)′(φIηn+1)′= (Ku(k)ηn)′φ′

I=Ku(k)′φ′

Iηn+Ku(k)φ′

Iη′

n.

Using this,

Q(Ku(k), φIηn) = ZKu(k)′(φIηn)′dµ +ZV Ku(k)φIηndµ

=ZKu(k)′φ′

Iηndµ +ZKu(k)′φIη′

ndµ +ZV Ku(k)φIηndµ

=Z(Ku(k)ηn)′(φIηn+1)′dµ −ZKu(k)φ′

Iη′

ndµ +ZKu(k)′φIη′

ndµ

+ZV(Ku(k)ηn)(φIηn+1)dµ

=Q(Ku(k)ηn, φIηn+1)−2ZKu(k)φ′

Iη′

ndµ −ZKu(k)φIη′′

ndµ −ZKu(k)φIη′

n

α

xdµ

=Q(Ku(k)ηn, φIηn+1)−2B1−B2−B3.

(4.18)

16 EDYTA KANIA AND MARCIN PREISNER

Proposition 4.19. The function ϑis non-increasing.

Proof. Using (4.18),

ϑ(t+s)−ϑ(t) = ZX

(Ks(k)−k) (x)φI(x)dµ(x)

= lim

n→∞ ZX

(Ks(k)−k)(x)φI(x)ηn(x)dµ(x)

= lim

n→∞ ZXZs

0

(−L)Ku(k)(x)duφI(x)ηn(x)dµ(x)

=−lim

n→∞ Zs

0ZX

LKu(k)(x)φI(x)ηn(x)dµ(x)du

=−lim

n→∞ Zs

0

Q(Ku(k), φIηn)du

=−lim

n→∞ Zs

0

(Q(Ku(k)ηn, φIηn+1)−2B1−B2−B3)du.

Having in mind Lemma 4.17 it is enough to show that Rs

0Bidu →0as n→ ∞ for i= 1,2,3.

This follows from Lebesgue’s Dominated Convergence Theorem and the estimates we have already

established. For example, for B2observe that |φI(x)η′′

n(x)| ≤ Cµ(J)|J|−2|x|1−α[n,n+1](x)for n≥N

with Nlarge enough. Then the majorant is

Zs

0ZX

sup

n≥N|Ku(k)(x)φI(x)η′′

n(x)|dµ(x)du ≤CZs

0ZX

|x|1−α

µ(B(x, √t+u)) exp −|x−z|2

c2(t+u)xαdx du

≤CZs

0ZX

|x|1−α

(t+u)1/2x

x+√t+uα

exp −|x−z|2

c2(t+u)dx du

≤Ct−1/2sZX|x|1−αexp −|x−z|2

c2(t+s)dx

≤C(I, t, s).

The integrals with B3and B1goes similarly. For the latter one we use (4.16).

Acknowledgments: The authors would like to thank Jacek Dziubański for his helpful remarks.

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Edyta Kania

Instytut Matematyczny, Uniwersytet Wrocławski

pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland

E-mail address :edyta.kania@math.uni.wroc.pl

Marcin Preisner

Instytut Matematyczny, Uniwersytet Wrocławski

pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland

E-mail address :marcin.preisner@math.uni.wroc.pl