Geometric conditions for multiple sampling and interpolation in the Fock space

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Geometric conditions for multiple sampling and interpolation in the Fock space

Alexander Borichev, Andreas Hartmann, Karim Kellay, Xavier Massaneda

To cite this version:

Alexander Borichev, Andreas Hartmann, Karim Kellay, Xavier Massaneda. Geometric conditions for

multiple sampling and interpolation in the Fock space. Advances in Mathematics, Elsevier, 2017, 304,

pp.1262-1295. �10.1016/j.aim.2016.09.019�. �hal-01184573v2�

SAMPLING AND INTERPOLATION IN THE FOCK SPACE

A. BORICHEV, A. HARTMANN, K. KELLAY, X. MASSANEDA

Abstract. We study multiple sampling, interpolation and uni- queness for the classical Fock spaces in the case of unbounded multiplicities. We show that, both in the hilbertian and the uni- form norm case, there are no sequences which are simultaneously sampling and interpolating when the multiplicities tend to infinity.

This answers partially a question posed by Brekke and Seip.

1. Introduction and main results

Sampling and interpolating sequences in Fock spaces were charac- terized by Seip and Seip–Wallst´en in [10, 12] by means of a certain Beurling–type asymptotic uniform density. A consequence of these results is that no sequence can be simultaneously interpolating and sampling, hence there are no unconditional or Riesz bases of reproduc- ing kernels neither. We refer the reader to the monograph [11] by K.

Seip for an account on these problems. For a similar result in several variables see [5].

We would like to consider sampling and interpolation problems in- volving a finite, not necessarily bounded, number of derivatives at each point (Hermite type interpolation). The idea of derivative sampling and interpolation is well-known from the theory of band-limited func- tions.

Band-limited functions appear naturally in the framework of model spaces and their parent Hardy spaces (we refer to [9] for model spaces and their importance in operator theory). Generalized interpolation, a particular instance of which is derivative interpolation, in Hardy spaces

2000 Mathematics Subject Classification. 30D55, 46C07,46E22, 47B32, 47B35.

Key words and phrases. Fock space, multiple interpolation, multiple sampling, Riesz basis.

First author was supported by Russian Science Foundation grant 14-41-00010.

Fourth author was supported by the Generalitat de Catalunya (grant 2014 SGR 289) and the Spanish Ministerio de Ciencia e Innovaci´ on (project MTM2014-51834- P).

1

was thoroughly studied by Vasyunin [14] and Nikolski [8] in the late 1970-s and is characterized in terms of a generalized Carleson condition (see [9, Sections C3.2, C.3.3] for a comprehensive account).

For the case of sequences with uniformly bounded multiplicity in the classical Fock space a complete description of such derivative sampling and interpolating sequences space was given by Brekke and Seip [4].

Again, it turns out that there are no sequences which are simultane- ously sampling and interpolating in this sense.

The natural question which then arises, and was already posed by Brekke and Seip, concerns the case when the multiplicities are un- bounded. We give conditions on sampling and interpolation in that case. The conditions we obtain are of a somewhat different nature, but still imply that there are no sequences which are both interpolating and sampling, at least when the multiplicities tend to infinity.

We should emphasize that our results make sense when the mul- tiplicities are large. For small multiplicities they essentially give no information.

The results of the following subsection concerning the Hilbert space case were earlier announced in the research note [3]. With respect to that research note, observe that the sufficient condition appearing in Theorem 1.1(b) has now a weaker and more natural formulation. We will also discuss the situation in the uniform norm in Subsection 1.2.

1.1. The Hilbert space case. For α > 0, define the Fock space F

by

F

= { f ∈ Hol( C ) : k f k

= k f k

:= α π

Z

C

| f(z) |

e

dm(z) < ∞} . The constant α/π is chosen in such a way that k 1 k

= 1, dm is Lebesgue area measure.

The Fock space F

is a Hilbert space with inner product h f, g i = α

π Z

C

f (z)g(z)e

dm(z).

The orthonormalization of the monomials gives the basis e

(z) =

√ α

√ k! z

, k ≥ 0.

The reproducing kernel is k

(ζ) = P

k≥0

e

(ζ)e

(z) = e

, hence

h f, k

i = f (z), f ∈ F

, z ∈ C .

It’s easy to check that the translations

T

f (ζ) = T

f(ζ) := e

f(ζ − z), f ∈ F

, act isometrically in F

.

Let k

= k

/ k k

k

be the normalized reproducing kernel at z. Note that T

1 = k

.

A sequence Λ ⊂ C is called sampling for F

if k f k

≍ X

λ∈Λ

| f (λ) |

k

(λ) = X

λ∈Λ

|h f, T

1 i|

, f ∈ F

,

and interpolating if for every v = (v

)

∈ ℓ

, there exists f ∈ F

such that

e

f (λ) = h f, T

1 i = v

, λ ∈ Λ.

For numerous results on the Fock space and operators acting thereon see the recent book by Zhu [15].

Let us now define sampling and interpolation for the case of higher multiplicity.

We deal with divisors X given as X = { (λ, m

) }

, where Λ is a sequence of points in C and m

∈ N is the multiplicity associated with λ. We associate with each (λ, m

) the subspace

N

= N

:= { f ∈ F

: f (λ) = f

(λ) = · · · = f

(λ) = 0 } . Definition. The divisor X is called sampling for F

if

k f k

≍ X

λ∈Λ

k f k

= X

λ∈Λ mλ−1

X

k=0

|h f, T

e

i|

, f ∈ F

,

and it is called interpolating for F

if for every sequence v = (v

)

such that

k v k

:= X

λ∈Λ mλ−1

X

k=0

| v

|

< ∞

there exists a function f ∈ F

such that

h f, T

e

i = v

, 0 ≤ k < m

, λ ∈ Λ. (1) Equivalently, X is interpolating if for every sequence (f

)

⊂ F

satisfying

X

λ∈Λ

k f

k

< ∞

there exists a function f ∈ F

such that

f − f

∈ N

, λ ∈ Λ.

This is exactly the way generalized interpolation is defined in Hardy spaces (see [9, Section C3.2] for definitions and results).

An application of the open mapping theorem to the restriction op- erator

R (f ) = h f, T

e

i

λ∈Λ,0≤k<mλ

shows that the function f ∈ F

such that (1) holds can always be chosen in such a way that k f k

≤ C k v k

, for some C > 0 depending only on X. The minimal such C is called the interpolation constant of X, and it will be denoted by M

.

Separation between points in Λ plays an important role in our results.

Denote by D(z, r) the disc of radius r centered at z, D(r) = D(0, r).

Definition. A divisor X is said to satisfy the finite overlap condition if

S

= sup

z∈C

X

λ∈Λ

χ

√

mλ/α)

(z) < ∞ . (2)

If Λ is a finite union of subsets Λ

such that for every j, the discs D(λ, p

m

/α), λ ∈ Λ

, are disjoint, then X satisfies the finite overlap condition. It is not clear whether the opposite is true.

The following two results give conditions for sampling and interpola- tion in the case of unbounded multiplicities. They are less precise than the results for the bounded case, in that they do not give characteri- zations. Still, the gap is sufficiently small to show that no divisor can be simultaneously sampling and interpolating when the multiplicities tend to infinity.

Theorem 1.1. (a) If X is a sampling divisor for F

, then X sat- isfies the finite overlap condition and there exists C > 0 such that

[

λ∈Λ

D(λ, p

m

/α + C) = C .

(b) Conversely, let the divisor X satisfy the finite overlap condition.

There exists C = C(S

) > 0 such that if for some compact subset K of C we have

[

λ∈Λ, mλ>αC²

D(λ, p

m

/α − C) = C \ K,

then X is a sampling divisor for F

.

A separation condition stronger than that in Theorem 1.1 (b) permits us to choose a subdivisor with the multiplicities tending to infinity:

Proposition 1.2. Let the divisor X be such that for every C > 0 there exists a compact subset K of C satisfying

[

λ∈Λ, mλ>αC²

D(λ, p

m

/α − C) = C \ K.

Then we can find a subset Λ

of Λ such that for every C > 0 there is a compact subset K of C satisfying

[

λ∈Λ^′, mλ>αC²

D(λ, p

m

/α − C) = C \ K,

and

λ∈Λ^′

lim

m

= + ∞ .

Theorem 1.3. (a) If X is an interpolating divisor for F

, then there exists C = C(M

) > 0 such that the discs

{ D(λ, p

m

/α − C) }

are pairwise disjoint.

(b) Conversely, if for some C > 0 the discs { D(λ, p

m

/α+C) }

are pairwise disjoint, then X is an interpolating divisor for F

. Remark. It is easily seen that if X is a divisor such that for some C > 0 the discs { D(λ, p

m

/α − C) }

are pairwise disjoint, and if lim

|λ|→∞

m

= + ∞ , then X satisfies the finite overlap condition.

Corollary 1.4. Let the divisor X satisfy the condition lim

m

= + ∞ . Then X cannot be simultaneously interpolating and sampling for F

.

Thus, denoting by K

the orthogonal complement of N

in F

, we conclude that the Fock space F

has no unconditional (Riesz) bases of subspaces K

, if lim

m

= + ∞ .

1.2. The F

Fock space. Let F

=

f ∈ Hol( C ) : k f k

= k f k

:= sup

z∈C

| f (z) | e

< ∞ . In order to consider the corresponding L

sampling and interpola- tion problems, we associate to every λ ∈ C the subspace

N

= N

= { f ∈ F

: f (λ) = f

(λ) = · · · = f

(λ) = 0 } . A divisor X is called sampling for F

, if there exists L > 0 such that

k f k

≤ L sup

λ∈Λ

k f k

.

In a similar way we define the generalized interpolation. The divisor X is called interpolating for F

if for every sequence (f

)

with

sup

λ∈Λ

k f

k

< ∞ , there exists a function f ∈ F

such that

f − f

∈ N

, λ ∈ Λ.

As usual, the norm of the interpolating function f is controlled by the ℓ

-norm of the sequence ( k f

k

)

. The minimal constant C such that we can always find an interpolating f with k f k

≤ C sup

k f

k

, is called the interpolation constant of X, and will be denoted by M

.

Theorem 1.5. (a) If X is a sampling divisor for F

then there exists C > 0 such that

[

λ∈Λ

D(λ, p

m

/α + C) = C .

(b) Conversely, if for some C > 0 and a compact set K we have [

λ∈Λ, mλ>αC²

D(λ, p

m

/α − C) = C \ K,

then X is a sampling divisor for F

.

This result is slightly stronger than Theorem 1.1 in that for the p = ∞ case any C > 0 is sufficient for sampling while for p = 2 we need C > C(S

).

Theorem 1.6. (a) If X is an interpolating divisor for F

, then there exists C = C(M

) > 0 such that the discs

{ D(λ, p

m

/α − C) }

are pairwise disjoint.

(b) Conversely, there exists C

> 0 such that if the discs { D(λ, p

m

/α + C

) }

are pairwise disjoint, then X is an interpolating divisor for F

.

This result is slightly weaker than Theorem 1.4 in that for the p = 2 case any C > 0 is sufficient for interpolation while for p = ∞ we need C ≥ C

for some absolute constant C

.

Corollary 1.7. Let the divisor X satisfy the condition lim

m

=

+ ∞ . Then X cannot be simultaneously interpolating and sampling for

F

.

The proof of this Corollary is completely analogous to that of Corol- lary 1.4.

The problems of sampling and interpolation are linked to that of uniqueness, and thus to zero divisors. We say that X is a zero divisor for F

if ∩

N

6 = { 0 } . To our knowledge, there is no characterization of the zero divisors of the Fock space. Some conditions are discussed in [7, 15].

We will establish here a necessary condition for zero divisors which seems sharper than those known so far.

Theorem 1.8. Let X be a divisor. If there exists a compact subset K of C such that

[

λ∈Λ

D(λ, p

m

/α) = C \ K, (3)

then X is not a zero divisor for F

.

1.3. The F

Fock spaces. By analogy to the Fock space F

, we can consider the Fock spaces F

consisting of entire functions f such that f (z) exp( − α | z |

/2) belongs to L

( C ), 1 < p < ∞ , and the correspond- ing sampling and interpolation problems. At the end of the paper we discuss how our results extend to 2 < p < ∞ for which we obtain conditions on sampling (Theorem 8.1) similar to Theorem 1.1 and on interpolation (Theorem 8.2) similar to Theorem 1.6. As a consequence, for 2 < p < ∞ we can obtain an analogue of Corollary 1.7.

It is of interest whether similar results are valid for 1 < p < 2.

The plan of the paper is the following. In the next section we estab- lish two geometrical results including Proposition 1.2 and the proof of Corollary 1.4. In Section 3 we prove the uniqueness Theorem 1.8. In Section 4 we collect some results on local L

estimates and the finite overlap condition. Sections 5 and 6 are devoted to the proofs of The- orems 1.1 and 1.3, respectively. Theorems 1.5 and 1.6 are proved in Section 7. Section 8 is devoted to the discussion of the case 2 < p < ∞ . Let us comment on the techniques used in this paper. One central result is that if a function of norm one on a disc has a small quotient norm with respect to functions vanishing up to a given order (adapted to the radius) at the center of the disc, then it will be small in a disc with a slightly smaller radius. In order to obtain such kind of results, we use the maximum principle and some estimates of the incomplete gamma function. This will indeed be the key for the necessity part of the interpolation result and the sufficiency part of the sampling result.

Another key ingredient for sampling is a uniqueness result based on the

uniform redistribution of the point mass ∆(log | f | ) over discs centered at the zeros of f with radius adapted to the multiplicity, and measuring the radial growth of the total redistributed mass. The sufficiency of the interpolation part for p = 2 is based on ∂ -techniques by H¨ormander requiring some subtle choices of the weight functions in order to fit to arbitrarily big multiplicities. The case p = ∞ adapts a clever trick by Berndtsson allowing to get uniform estimates for the solutions which are optimal in the (weighted) L

norm.

Throughout the remaining part of the paper, we suppose that α = 1.

The general case is dealt with in the same way. We define the measure dµ(z) = (1/π)e

dm(z).

Denote F

= F

, F

= F

. Furthermore, we use the following notations:

• A . B means that there is an absolute constant C such that A ≤ CB.

• A ≍ B if both A . B and B . A.

Acknowledgments. We are thankful to Philippe Charpentier, Daniel Pascuas and St´ephane Rigat for helpful discussions. We are grateful to the referee for interesting comments. The last section of our paper answers one of his questions.

2. Two geometrical results

2.1. An elementary result. We start with a simple geometric lemma that will allow us to deduce Corollary 1.4 from Theorems 1.3 and 1.1, and also play a rˆole in the proof of Theorem 1.8 in the next section.

Lemma 2.1. Suppose that three discs D

= D(Z

, r

), D

= D(Z

, r

), D

= D(Z

, r

) satisfy the property D

∩ D

∩ D

6 = ∅ . Then there exist 1 ≤ i, j ≤ 3, i 6 = j such that

(a) r

+ r

− | Z

− Z

| ≥ c · min(r

, r

) , (b) m(D

∩ D

) ≥ c · min(r

, r

)

;

for some absolute constant c > 0, with m(A) being the area of A.

Proof. The proof is based on elementary geometry. Take Z ∈ D

∩ D

∩ D

. We can assume that each pair of discs intersects at exactly two different points. One of the angles between the vectors −−→ ZZ

, −−→ ZZ

,

−−→ ZZ

is at most 2π/3. Without loss of generality we can assume that

∠Z

ZZ

≤ 2π/3. Denote by W

, W

two points of intersection of ∂D

and ∂D

. Either ∠W

Z

W

or ∠W

Z

W

is at least π/3. Without loss

of generality we can assume that ∠W

Z

W

≥ π/3.

Now, either Z

∈ D

or | W

− W

| ≥ r

. In both situations, (a) and

(b) follow immediately.

We are now ready to prove Corollary 1.4. For this, suppose that there is a divisor X = { (λ, m

) }

which is simultaneously interpolating and sampling and such that lim

m

= ∞ . By Theorem 1.3, there is a constant C

such that the discs D(λ, √ m

− C

) are pairwise disjoint so that for every λ, λ

∈ Λ, λ 6 = λ

, we have

√ m

+ √

m

− | λ − λ

| ≤ 2C

. (4) Next, by Theorem 1.1, the discs D(λ, √

m

+ C

) cover the whole plane, so by Lemma 2.1 (a), we can find pairs (λ, λ

), λ, λ

∈ Λ, λ 6 = λ

such that

2C

+ √ m

+ √ m

−| λ − λ

| ≥ c min( √ m

, √ m

), m

→ ∞ , m

→ ∞ ,

which contradicts (4).

2.2. Proof of Proposition 1.2. Choose an increasing sequence of positive numbers (R

)

, such that

C \ D(R

) ⊂ [

mλ>n²

D(λ, √

m

− n). (U

)

Iteratively, on the step s ≥ 1, we remove from Λ the subsets Λ

= { λ ∈ Λ : | λ | > R

+ s, (s − 1)

≤ m

< s

} . Then

λ∈Λ^′

lim

m

= + ∞ ,

where Λ

= Λ \ ∪

Λ

. It remains only to verify that the conditions (U

) are still valid after each step s. Clearly, (U

), n ≥ s, do not change. Now, for 1 ≤ n < s we have

C \ D(R

) ⊂ [

mλ>n², λ /∈Λs

D(λ, √ m

− n)

∪ [

λ∈Λs

D(λ, √ m

− n) . (5) Observe that if λ ∈ Λ

, then m

< s

and for every z ∈ D(λ, √ m

) we have | z | ≥ | λ | − √ m

> R

+ s − s = R

. Hence,

[

λ∈Λs

D(λ, √ m

− n) ⊂ C \ D(R

). (6)

By (U

), we have C \ D(R

) ⊂ [

mλ>s²

D(λ, √ m

− s)

⊂ [

mλ>s²

D(λ, √ m

− n) ⊂ [

mλ>n², λ /∈Λs

D(λ, √ m

− n). (7) Finally, (5)–(7) yield

C \ D(R

) ⊂ [

mλ>n², λ /∈Λs

D(λ, √ m

− n).

3. Zero divisors

Lemma 3.1. If X = { (λ

, m

) }

is a zero divisor for F

, then 2

Z

D(R)

X

k≥1

χ

(z) log R

| z | dm(z) ≤ πR

+ O(1), R → ∞ . Proof. Assume that f ∈ F

vanishes at λ

of order m

, k ≥ 1. Then log | f | is a subharmonic function, not identically −∞ and bounded above by

s(z) := log k f k

+ 1 2 | z |

.

In an inductive argument, we will construct a new subharmonic func- tion h such that log | f | ≤ h ≤ s by redistributing the mass of ∆(log | f | ) of each D(λ

, √ m

) uniformly on this disc.

Set h

:= log | f | . Then

h

= m

log | z − λ

|

√ m

+ u

,

where u

is subharmonic on C and harmonic in some small neighbour- hood of λ

(since the zeros of the entire function f are isolated). Define

v

(z) =



 

 

 

 

| z − λ

|

− m

2 if z ∈ D(λ

, √ m

) m

log | z − λ

|

√ m

otherwise.

Then v

∈ C

( C ), it is harmonic outside D(λ

, √ m

) and it has constant Laplacian 2 in D(λ

, √ m

). Hence the total mass of the measure ∆v

on C is equal to 2m(D(λ

, √ m

)) = 2πm

, which corresponds to the

total mass of the measure ∆(m

(log | z − λ

| / √ m

)) = 2πm

δ

.

Now set h

= v

+ u

, and restart the procedure for λ

, i.e. write h

= m

log | z − λ

|

√ m

+ u

. We construct v

as above and obtain h

= v

+ u

.

Iterating this procedure we obtain a sequence of subharmonic func- tions (h

)

. We claim that, for every z ∈ C , (h

(z))

is increasing and

log | f (z) | ≤ h

(z) ≤ s(z).

As above we will give the argument just for the first step. The rest will follow by induction.

Let us begin by showing that h

(z) ≤ s(z) for every z ∈ C . This is clear when z / ∈ D(λ

, √ m

), since for these z we have not changed the function (h

= h

outside D(λ

, √ m

)). This estimate holds also on

∂D(λ

, √ m

).

Consider the function w

:= v

+ u

− s. Then ∆(v

+ u

− | z |

/2) =

∆u

≥ 0 on D(λ

, √ m

), since u

is subharmonic. Hence w

is subhar- monic and it is non-positive on the boundary of D(λ

, √ m

). There- fore, it is non-positive throughout this disc, yielding h

= v

+ u

≤ s also in D(λ

, √ m

).

The fact that h

≤ h

is almost obvious. Again, there is nothing to prove for z outside D(λ

, √ m

). Inside the disc it remains to estimate m

log( | z − λ

| / √ m

) from above by ( | z − λ

|

− m

)/2, which is clear, since r → m

log(r/ √ m

) is concave, r → (r

− m

)/2 is convex, and these functions touch smoothly at r = m

.

The pointwise limit h of the sequence (h

)

is still subharmonic (because the sequence is locally eventually stable) and by construction it is comprised between log | f | and s(z). We also know that it is not identically equal to −∞ , since log | f | is not.

Since h(z) ≤ s(z), by Green’s formula, Z

D(R)

∆h(z) log R

| z | dm(z) = − 2πh(0) + 1 R

Z

|z|=R

h(z) d | z |

≤ 2π

− h(0) + log k f k

+ R

2

. On the other hand,

Z

D(R)

∆h(z) log R

| z | dm(z) ≥ 2 Z

D(R)

X

k≥1

χ

(z) log R

| z | dm(z),

and the proof is complete.

3.1. Proof of Theorem 1.8. Let X = { (λ, m

) }

be a zero divisor for F

satisfying (3). Denote E = { z ∈ C : P

λ∈Λ

χ

(z) > 1 } . Choose R > 0 such that K ⊂ D(R). Then, by Lemma 3.1, there exists A > 0 such that

πR

2 + A ≥ Z

D(R)

X

λ∈Λ

χ

(z) log R

| z | dm(z)

= Z

D(R)

log R

| z | dm(z) + Z

D(R)∩E

h X

λ∈Λ

χ

(z) − 1 i log R

| z | dm(z)+

+ Z

D(R)\E

h X

λ∈Λ

χ

(z) − 1 i log R

| z | dm(z)

≥ Z

D(R)

log R

| z | dm(z) + Z

D(R)∩E

log R

| z | dm(z) − Z

K

log R

| z | dm(z)

≥ πR

2 +

Z

D(R)∩E

log R

| z | dm(z) − m(K) log R − c

, (8) where c

depends only on the compact set K.

By Lemma 2.1 (b), the area of E is infinite. Choose R

such that m(E ∩ D(R

)) ≥ m(K ) + 1. For R ≥ R

we have

Z

D(R)∩E

log R

| z | dm(z) ≥ Z

D(R0)∩E

log R

| z | dm(z)

≥ (m(K) + 1) log R

R

≥ (m(K) + 1) log R − c

, with c

depending only on R

. By (8) we get a contradiction, and the

proof is completed.

4. Local L

estimates and the finite overlap condition Given x ≥ 0 and k ∈ Z

we define

σ

(x) = 1 k!

Z

y

e

dy,

ω

(x) = e

k

X

s=0

x

s! .

Then lim

σ

(x) = lim

ω

(x) = 1. Integration by parts gives σ

(x) + ω

(x) = 1.

The following estimates on partial sums of exponentials (or the in-

complete Gamma function) will be useful in what follows.

Lemma 4.1. (a) Given t ≥ 0 , there exist ε > 0 and k

> 0 such that for every k ≥ k

σ

(k − t √

k) ≥ ε.

(b) Given t ≥ 0 , there exists ε > 0 such that for every k ≥ 0 ω

(k + t √

k) ≥ ε.

(c) Given ε > 0, there exist t > 0 such that for every k ≥ t

σ

(m − t √

m) ≤ εσ

(m), t

≤ m ≤ k.

Proof. This lemma is essentially contained in [13]; alternatively, one can use the Stirling formula to evaluate the sums ω

(x). Yet another proof of parts (a) and (b) uses the Poisson law and its approximation by the standard normal law via the central limit theorem.

To prove (c), we will check that for every ε > 0, there exist t > 0 satisfying

(y − t √ y)

e

≤ εy

e

, t

≤ y ≤ k. (9) Assuming for the moment (9), and integrating from t

to x ∈ [t

, k], we get

Z

u

e

du

1 − t/(2 √ y) = Z

t²

(y − t √ y)

e

dy

≤ ε Z

t²

y

e

dy,

where u = y − t √ y. Since y ≥ t

, we have 1 − t/(2 √ y) ≥

, and hence, Z

x 0

u

e

du ≤ 2ε Z

t²

y

e

dy, which completes the proof.

To verify (9), we use that t √

y + k log y − t √ y y

= t √

y − X

j≥1

kt

jy

= − (k − y) t

√ y − kt

2y − X

j≥3

kt

jy

≤ − t

2

≤ log ε for t

≤ y ≤ k and t ≥ p

2 log(1/ε).

Lemma 4.2. For every A ≥ 0 there exists C(A) > 0 , n(A) > 0 such that for every f ∈ F

, n ≥ n(A) , λ ∈ C , we have

n−1

X

k=0

|h f, T

e

i|

≤ C(A) Z

D(λ,√n−A)

| f(z) |

dµ(z).

Proof. Let g = T

f . Our statement is equivalent to

n−1

X

k=0

|h g, e

i|

. Z

D(√

n−A)

| g(z) |

dµ(z), (10) Let

g = X

k≥0

a

e

.

We use that e

are mutually orthogonal with respect to χ

dµ and Z

D(R)

| e

(z) |

dµ(z) = σ

(R

). (11) Therefore, (10) can be rewritten as

n−1

X

k=0

| a

|

. X

k≥0

σ

(( √

n − A)

) | a

|

. (12) By Lemma 4.1 (a), for n > k ≥ k

(A), we have

σ

(( √

n − A)

) ≥ σ

(k − 2A √

k) ≥ ε(A)

> 0.

For every k such that 0 ≤ k < k

(A), n ≥ n(A, k) we have σ

(( √

n − A)

) ≥ 1 2 . Therefore, for n ≥ max

[n(A, k)] we obtain

σ

(( √

n − A)

) ≥ min 1

2 , ε(A)

> 0, k ≥ 0,

and (12) follows.

Lemma 4.3. Let X = { (λ, m

) }

. The divisor X satisfies the finite overlap condition if and only if there exists C > 0 satisfying

X

λ∈Λ mλ−1

X

k=0

|h f, T

e

i|

≤ C k f k

, f ∈ F

. (13)

Proof. Suppose that (13) holds. Given z ∈ C , set f = T

1; we have k f k

= 1. Next,

|h f, T

e

i| = |h T

1, T

e

i| = |h e

, T

e

i| = | T

e

(0) |

= 1

√ k! | λ − z |

e

.

Denote ρ

= | λ − z | . Then (13) implies that 1 = k f k

& X

λ∈Λ mλ−1

X

k=0

ρ

k! e

≥ X

λ∈Λ, ρ²_λ<mλ

e

X

0≤k≤ρ²_λ

ρ

k!

= X

λ∈Λ

χ

(z)ω

λ]

(ρ

),

where [ρ

] is the integer part of ρ

. By Lemma 4.1 (b) with t = 0 we conclude that

1 & X

λ∈Λ

χ

(z).

In the opposite direction, if X satisfies the finite overlap condition,

we just apply Lemma 4.2.

5. Proof of Theorem 1.1

5.1. Necessary condition. Let X = { (λ, m

) }

be a sampling di- visor. By Lemma 4.3 it satisfies the finite overlap condition. Suppose that for every C > 0

[

λ∈Λ

D(λ, √

m

+ C) 6 = C . Then there exists a sequence (z

)

⊂ C such that

ρ

:= dist(z

, ∪

D(λ, √ m

)) −→ ∞ as n → ∞ . (14) Set f

= T

1. By Lemma 4.2, we have

X

λ∈Λ mλ−1

X

k=0

|h f

, T

e

i|

. X

λ∈Λ

Z

D(λ,√mλ)

e

dm(w)

. Z

∪^λ∈ΛD(λ,√mλ)

e

dm(w) . Z

|ζ|≥ρn

e

dm(ζ) → 0, (15)

and X cannot be sampling for F

, which contradicts the hypothesis.

5.2. Sufficient condition. Let us start with a local estimate.

Lemma 5.1. Given 0 < η ≤ 1 there exists a(η) < ∞ such that if f ∈ F

, m ≥ a(η)

, and if

X

0≤k<m

|h f, e

i|

≤ η/2,

Z

D(√

m)

| f(z) |

dµ(z) ≤ 1, then

Z

D(√

m−a(η))

| f (z) |

dµ(z) ≤ η.

Proof. Let

f = X

k≥0

a

e

.

By (11), (e

/ p

σ

(R

))

is an orthonormal system with respect to the scalar product R

D(R)

f gdµ, and it remains only to verify that X

k≥m

| a

|

σ

(( √

m − a(η))

) ≤ (η/2) X

k≥m

| a

|

σ

(m).

This inequality follows from Lemma 4.1 (c).

Now we are ready to complete the sufficiency part of the proof of Theorem 1.1. Suppose that there exists a sequence (f

)

such that k f

k

= 1 and

X

λ∈Λ mλ−1

X

k=0

|h f

, T

e

i|

→ 0 as n → ∞ .

Passing to a weakly convergent subsequence denoted again by (f

)

we have two possibilities: either (A) f

converge weakly to f 6 = 0 or (B) f

converge weakly to 0.

(A): In this case X is a zero divisor for f ∈ F

. Since F

⊂ F

, we obtain, by Theorem 1.8, that C \ ∪

D(λ, √ m

) cannot be compact, thus contradicting the hypothesis.

(B): In this case we define η = (S

+ 1)

. We set the constant

C = C(S

) from the formulation of the theorem to be equal to a(η)

defined in Lemma 5.1. Denote Λ

= { λ ∈ Λ : m

> a(η)

} . Then take

B = B(η) such that

C \ D(B) ⊂ [

λ∈Λ1

D(λ, √

m

− a(η))

and obtain

1 = Z

C

| f

(z) |

dµ(z)

≤ Z

D(B)

| f

(z) |

dµ(z) + X

λ∈Λ1

Z

D(λ,√mλ−a(η))

| f

(z) |

dµ(z).

Denote by Λ

the set of λ ∈ Λ

such that

mλ−1

X

k=0

|h f

, T

e

i|

≤ (η/2) Z

D(λ,√mλ)

| f

(z) |

dµ(z).

By Lemma 5.1 we obtain that 1 ≤ o(1) + X

λ∈Λ2

Z

D(λ,√mλ−a(η))

| f

(z) |

dµ(z)

+ X

λ∈Λ1\Λ2

Z

D(λ,√mλ−a(η))

| f

(z) |

dµ(z)

≤ o(1) + η X

λ∈Λ2

Z

D(λ,√mλ)

| f

(z) |

dµ(z) + (2/η) X

λ∈Λ1\Λ2

mλ−1

X

k=0

|h f

, T

e

i|

≤ o(1) + η X

λ∈Λ

Z

D(λ,√mλ)

| f

(z) |

dµ(z)

≤ o(1) + ηS

Z

C

| f

(z) |

dµ(z) = o(1) + ηS

, n → ∞ .

This contradiction completes the proof.

6. Proof of Theorem 1.3

6.1. Sufficient condition. Suppose that the discs D(λ, √ m

+ C

), λ ∈ Λ are pairwise disjoint for some C

> 0.

Let v = (v

)

be a sequence with k v k

< ∞ and let (p

)

be a sequence of polynomials such that

h p

, e

i = v

, λ ∈ Λ, 0 ≤ k < m

, and

k p

k

= k p

k

, X

λ∈Λ

k p

k

< ∞ .

Set Q

= T

p

, r

= √ m

+ C

, D

= D(λ, √ m

), and D

= D(λ, r

), so that the discs (D

)

are pairwise disjoint. Consider first the interpolating function

F (z) = X

λ∈Λ

Q

(z)η( | z − λ | − r

), where η is a smooth cut-off function on R with

• supp η ⊂ ( −∞ , 0],

• η ≡ 1 on ( −∞ , − C

],

• | η

| . 1.

Note that supp F ⊂ ∪

D

and Z

C

| F (z) |

e

dm(z) . X

λ

Z

D^′_λ

| Q

(z) |

e

dm(z) ≤ X

λ

k Q

k

< ∞ .

Furthermore, F interpolates (v

)

on X, that is for every λ ∈ Λ, the function

F −

mλ−1

X

k=0

v

T

e

vanishes at λ of order at least m

.

We search for a holomorphic interpolating function of the form f = F − u. This leads to the ¯ ∂-equation ∂u = ∂F , which we solve using H¨ormander’s result [6, Theorem 4.2.1]: if ψ is a subharmonic function, then there exists u such that ∂F = ∂u and

Z

C

| u |

e

dm ≤ 4 Z

C

| ∂F |

e

dm

∆ψ . (16)

We set ψ(z) = | z |

+ v(z), where v(z) = X

λ∈Λ

m

h log | z − λ |

m

+ 1 − | z − λ |

m

i χ

(z).

Since

log(1 − t) + t ≤ 0, 0 ≤ t ≤ 1,

we have v(z) ≤ 0. Next, supp ∂F ⊂ ∪

(D

\ D

). Furthermore, for z ∈ D

\ D

, λ ∈ Λ, we have v (z) = 0 and

| ∂F (z) | . | Q

(z) | . (17) A direct computation yields

∆v = − 4 + 4πm

δ

on D

, where δ

is the unit mass at λ. Therefore,

∆ψ = 4πm

δ

on D

. Next,

∆ψ = 4 on C \ [

λ∈Λ

D

. (18)

By (16)–(18), we obtain Z

C

| u |

e

dm ≤ Z

C

| u |

e

dm ≤ 4 Z

C

| ∂F |

e

dm

∆ψ . X

λ∈Λ

Z

D_λ^′\Dλ

| Q

(z) |

e

dm(z) . X

λ

k Q

k

.

Thus, f = F − u ∈ F

.

Finally, since in a neighbourhood of λ holds the estimate e

≍ e

≍ | z − λ |

and u is analytic, the bound

Z

C

| u(z) |

e

dm(z) < ∞

forces u to vanish at order m

at λ, so that f interpolates the values (v

)

, i.e.:

h f, T

e

i = h Q

, T

e

i = v

, λ ∈ Λ, 0 ≤ k < m

.

6.2. Necessary condition. Let X = { (λ, m

) }

be an interpolat- ing divisor and assume that the discs { D(λ, √ m

− A) }

are not separated for some A > 0 to be chosen later on. Then there exist λ, λ

∈ Λ and w ∈ C such that

D(w, 1) ⊂ D(λ, √ m

− A + 1) ∩ D(λ

, √ m

− A + 1).

Since X is an interpolating divisor, there exists f ∈ F

such that f ∈ N

,

f − T

1 ∈ N

λ′

, k f k

≤ M

. By Lemma 5.1,

Z

D(λ,√mλ−A+1)

| f (z) |

dµ(z) + Z

D(λ^′,√m_λ^′−A+1)

| (f − T

1)(z) |

dµ(z)

= o(1) · M

, A → ∞ , and hence,

Z

D(w,1)

| f(z) |

dµ(z) + Z

D(w,1)

| (f − T

1)(z) |

dµ(z)

= o(1) · M

, A → ∞ .

On the other hand, Z

D(w,1)

| (T

1)(z) |

dµ(z)

is a positive constant. This gives a contradiction when A > A(M

).

7. Proofs of Theorems 1.5 and 1.6

Let us start with some technical lemmas. First of all, given m ≥ 1, the function ϕ(t) = ϕ

(t) = t

/2 − m log t decreases on (0, √

m) and increases on ( √

m, + ∞ ).

Lemma 7.1. Given m ≥ 1, (a) for every a > 0,

ϕ( √

m + a) ≤ ϕ( √

m) + a

, (b) for every a > 0,

ϕ( √

m − a) ≥ ϕ( √

m) + a

, m ≥ a

. Proof. Since ϕ

( √

m) = 0, ϕ

(t) < 2, t > √

m, and ϕ

(t) > 2, 0 < t <

√ m, we have

ϕ( √

m + a) − ϕ( √ m) =

Z

√m

ϕ

(t) dt < 2 Z

0

s ds = a

and

ϕ( √

m − a) − ϕ( √

m) = − Z

√m−a

ϕ

(t) dt > 2 Z

0

s ds = a

, m ≥ a

. Lemma 7.2. Given δ > 0 there exist ε, η > 0 such that if f ∈ F

, k f k

≤ 1, m ≥ δ

, and k f k

< ε, then

| f (z) | ≤ (1 − η)e

, | z | ≤ √ m − δ.

Proof. Let h be an entire function such that the function g : z → z

h(z) belongs to F

and

k f − z

h k

< ε.

Then, using that k f k

≤ 1, we obtain

| z

h(z) | ≤ | f (z) − z

h(z) | + | f(z) | ≤ (1 + ε)e

, z ∈ C . Hence,

| h(z) | ≤ (1 + ε) exp ϕ( | z | ), z ∈ C ,

where, as above, ϕ = ϕ

. Set | z | = √

m and use Lemma 7.1 (b) to get

| h(z) | ≤ (1 + ε)e

e

≤ (1 + ε)e

e

. We will have

| h(z) | ≤ (1 − ε − η)e

, | z | = √ m, as soon as we choose ε, η small enough so that

(1 + ε)e

≤ 1 − ε − η.

By the maximum principle, the above estimate holds then for all | z | ≤

√ m. Using that ϕ( | z | ) ≥ ϕ( √

m − δ) for | z | ≤ √

m − δ we finally see that

| h(z) | ≤ (1 − ε − η) exp ϕ( | z | ), | z | ≤ √ m − δ, and

| f(z) | ≤ | z

h(z) | + εe

≤ (1 − η)e

, | z | ≤ √ m − δ.

Lemma 7.3. Given β > 0 , there exist ε, δ > 0 such that if f ∈ F

, k f k

≤ 1 , m ≥ δ

, and k f k

< ε , then

| f(z) | ≤ βe

, | z | ≤ √ m − δ.

Proof. We argue as in the proof of Lemma 7.2.

Lemma 7.4. Let δ > 0 . If m ≥ 1 and f is a function analytic in D(R) and continuous up to ∂D(R) , R = √

m + δ , satisfying f(0) = f

(0) = . . . = f

(0) = 0 and | f (Re

) | ≤ e

, t ∈ [0, 2π], then

| f(z) | ≤ e

e

, | z | ≤ R.

Proof. Let f(z) = z

g (z). The function g is holomorphic in D(R) and continuous up to ∂D(R). By the maximum principle,

log | g(z) | ≤ ϕ(R), | z | ≤ R.

By Lemma 7.1 (a), and since √ m is the point of minimum for ϕ, we have

log | g(z) | ≤ ϕ( √

m) + δ

≤ ϕ( | z | ) + δ

, | z | ≤ R, and hence,

| f (z) | ≤ e

e

, | z | ≤ R.

7.1. Proof of Theorem 1.5. Sufficient condition. Suppose that there exists a sequence (f

)

such that k f

k

= 1, n ∈ N and

sup

λ∈Λ

k f

k

→ 0 as n → ∞ .

Passing to a subsequence converging uniformly on compact subsets and denoted again by (f

) we have two possibilities: either (A) the se- quence (f

)

converges to f 6 = 0 or (B) the sequence (f

) converges to 0.

(A): In this case X is a zero divisor for F

. Then, by Theorem 1.8, the set C \ ∪

D(λ, √ m

) cannot be compact, thus contradicting the hypothesis.

(B): In this case we recall our assumption that for a compact set K we have

Ω = [

λ∈Λ, mλ>C²

D(λ, √ m

− C) = C \ K.

Next, since (f

)

tends to 0,

| f

(z) | < 1

2 e

, z ∈ K, n ≥ n

.

On the other hand, applying Lemma 7.2 to T

f

on D( √ m

), λ ∈ Λ, we obtain that

| f

(z) | < (1 − η(C))e

, z ∈ Ω, n ≥ n

, Hence, k f

k

< 1, n ≥ n

, and we arrive at a contradiction.

7.2. Proof of Theorem 1.5. Necessary condition. The argument is completely analogous to that in the proof of the necessity part of Theorem 1.1. Once again, we choose (z

)

satisfying (14) and set f

= T

1,

g

=

mλ−1

X

k=0

h f

, T

e

i T

e

, λ ∈ Λ.

By (15),

X

λ∈Λ

k g

k

→ 0, n → ∞ , and since k h k

≤ C k h k

for h ∈ F

, we have

max

k g

k

→ 0, n → ∞ .

Since g

− f

∈ N

, λ ∈ Λ, we obtain that X cannot be sampling for

F

and our proof is completed.

7.3. Proof of Theorem 1.6. Here, we need an additional construc- tion.

Lemma 7.5. Given q, a ≥ 1, there exists a function y

∈ C

( C \ { 0 } ), y

(z) = y

( | z | ) such that z 7→ y

(z) − 2q

log | z | is C

-smooth at 0,

y

(z) = | z |

, | z | ≥ q + a, and

∆y

(z) ≥ 4a

(q + 2a − | z | )

, | z | < q + a. (19) Proof. Consider

γ(t) = a (q + 2a − t)

and set

b = Z

D(q+a)

γ( | z | ) dm(z) ≤ 2π(q + a)

q + 2a . (20)

Solve the Dirichlet problem







∆g(z) = 4γ( | z | ) − 4b

π(q + a)

, | z | < q + a, g(z) = 0, | z | = q + a.

Then g(z) = g( | z | ), Z

D(q+a)

∆g(z) dm(z) = 0, and by Green’s formula, ∇ g = 0 on ∂D(q + a).

Next, set h(z) = q

log

z q + a

2

+ 1 −

z q + a

2

, | z | ≤ q + a.

Then h = ∇ h = 0 on ∂D(q + a), and

∆h = − 4q

(q + a)

+ 4πq

δ

on D(q + a).

Finally, we set y

(z) =

( | z |

+ g (z) + h(z), | z | ≤ q + a,

| z |

, | z | > q + a.

It is clear that y

∈ C

( C \ { 0 } ) and z 7→ y

(z) − 2q

log | z | is C

-smooth at 0. It remains to verify (19). In fact, using (20) and the fact that q, a ≥ 1, we obtain

1

4 ∆y

(z) ≥ 1 + γ( | z | ) − b

π(q + a)

− q

(q + a)

≥ γ( | z | ) = a

(q + 2a − | z | )

, | z | < q + a.

Proof of Theorem 1.6. We start with the sufficiency part. Suppose that the discs D(λ, √ m

+ 6), λ ∈ Λ, are pairwise disjoint. Denote Λ = (λ

)

. Let (ρ

)

be a sequence of data with sup

k ρ

k

≤ 1.

We want to find functions f

such that

f

− ρ

∈ N

, 1 ≤ n ≤ N,

and k f

k

≤ M , N ≥ 1. A normal family argument will then complete the proof.

Set a = 3, m

= m

, r

= √ m

+ a, D

= D(λ

, r

), D

= D(λ

, r

+ a), D

= D(λ

, r

+ a + 1),

F

(z) = X

1≤n≤N

ρ

(z)˜ η( | z − λ

| − r

), with ˜ η is a smooth cut-off function on R with

• supp ˜ η ⊂ ( −∞ , a],

• η ˜ ≡ 1 on ( −∞ , 0],

• | η ˜

| . 1.

Then

supp ∂F

⊂ Q

= [

1≤n≤N

D

\ D

.

Next, using Lemma 7.5, we define ψ

(z) = | z |

+ X

1≤n≤N

h y

(z − λ

) − | z − λ

|

i ,

where q(n) = √ m

(by construction, the function y

(z − λ

) − | z − λ

|

vanishes outside D

). Then







∆ψ

(z) = 4, z ∈ C \ ∪

D

,

∆ψ

(z) ≥ 4a

( √ m

+ 2a − | z − λ

| )

, z ∈ D

, 1 ≤ n ≤ N.

Given ζ ∈ Y

= C \ S

1≤n≤N

D

, we define w

(z) = 1

| z − ζ |

+ 2a , z ∈ C . Furthermore, we set

Ω

= 1 2 χ

. We have

∆w

(z)

w

(z) = 9 | z − ζ | ( | z − ζ |

− 2a) ( | z − ζ |

+ 2a)

.

An elementary calculation (consider separately the cases t ≤ 8/3 and t > 8/3) shows that

9 t(t

− 2a)

(t

+ 2a)

≤ 2, t ≥ 0.

Next,

9 t(t

− 2a) (t

+ 2a)

≤ 9

t

, t ≥ 0.

Therefore,

∆w

(z)

w

(z) ≤ min

2, 4a

| z − ζ |

,

and we obtain that on the whole plane,

∆w

≤ w

(∆ψ

− 4Ω

). (21) Indeed, for z ∈ C \∪

D

we have ∆ψ

− 4Ω

≥ 2 and for z ∈ D

, 1 ≤ n ≤ N , we have

4a

| z − ζ |

≤ 4a

( √ m

+ 2a − | z − λ

| )

because | ζ − λ

| ≥ √ m

+ 2a + 1, | z − λ

| ≤ √ m

+ 2a.

Now, we use a remarkable result by Berndtsson [1, Theorem 4] (see also [2]): the solution u

of the equation ∂u

= ∂F

minimizing the integral

Z

C

| u

(z) |

e

dm(z) satisfies the inequality

Z

C

| u

(z) |

e

w

(z) dm(z) ≤ Z

C

| ∂F

(z) |

e

w

(z)

Ω

(z) dm(z)

under condition (21), for every ζ ∈ Y

. Since supp ∂F

⊂ Q

and Ω

=

χ

we have

Z

C

| u

(z) |

e

w

(z) dm(z) ≤ 2 Z

QN

| ∂F

(z) |

e

w

(z) dm(z).

Hence, we obtain that Z

D(ζ,1)

| u

(z) |

e

dm(z)

≤ 2 X

1≤n≤N

Z

D^′n\Dn

| ∂F

(z) |

e

dm(z)

| z − ζ |

+ 2a ≤ C for an absolute constant C. Next, ∂F

vanishes on D(ζ, 1), and, hence, u

is analytic in D(ζ, 1). By the mean value property applied to the function u

(z) exp( | ζ |

/2 − zζ ), we have

| u

(ζ) |

e

≤ C

Z

D(ζ,1)

| u

(z) |

e

dm(z).

Thus,

| u

(ζ) |

e

≤ C

. To extend this estimate to ζ ∈ S

1≤n≤N

D

, we just apply Lemma 7.4 to the functions u

+ ρ

− F

analytic in D

and use that u

and, hence, u

+ ρ

− F

vanish of order m

at λ

, 1 ≤ n ≤ N .

It remains to set f

= F

− u

. Then

| f

(z) | ≤ Ce

, z ∈ C , with C independent of N ≥ 1 and

f

− ρ

∈ N

, 1 ≤ n ≤ N.

Finally, for the necessity part, the argument is completely analogous to that in the proof of the necessity part of Theorem 1.3. Instead of

Lemma 5.1 we use Lemma 7.3.

8. The case F

, 2 < p < ∞

In this section, we formulate and sketch the proofs of the following two results. Let 2 < p < ∞ .

Theorem 8.1. (a) If X is a sampling divisor for F

, then X sat- isfies the finite overlap condition and there exists C > 0 such that

[

λ∈Λ

D(λ, p

m

/α + C) = C .