0506297

SAMPLING AND INTERPOLATION IN RADIALWEIGHTED SPACES OF ANALYTIC FUNCTIONS

A. BORICHEV, R. DHUEZ, K. KELLAY

Abstract. We obtain sampling and interpolation theorems in ra-dial weighted spaces of analytic functions for weights of arbitrary(more rapid than polynomial) growth. We give an application toinvariant subspaces of arbitrary index in large weighted Bergmanspaces.

1. Introduction

$ Let h : [0, 1) [0,+) be an increasing function such thath(0) = 0, and limr1 h(r) = +. We extend h by h(z) = h(|z|),z D , and call such h a weight function. Denote by Ah(D ) theBanach space of holomorphic functions on the unit disk D with thenorm

fh = supzD|f(z)|eh(z) < +.

A subset of D is called a sampling set for Ah(D ) if there exists > 0 such that for every f Ah(D ) we have

fh fh, = supz|f(z)|eh(z).

A subset of D is called an interpolation set for Ah(D ) if for everyfunction a defined on such that ah, 0 such that for every a withah,

2 A. BORICHEV, R. DHUEZ, K. KELLAY

Next we assume that h C2(D ), and

h(z) =( 2x2

+2

y2

)h(x+ iy) 1, z D .

We consider the weighted Bergman spaces

Aph(D ) ={f Hol (D ) : fpp,h =

D|f(z)|peph(z)dm2(z)

SAMPLING AND INTERPOLATION IN RADIAL WEIGHTED SPACES 3

conjectrure translates into the question on whether any sampling setfor A2h(D ) is a finite union of interpolation sets for A2h(D ). The answeris positive for h we consider in this paper (as follows from Therems 2.2and 2.4).

In the plane case, if h : [0,) [0,+) is an increasing functionsuch that h(0) = 0, limr h(r) = +, we extend h by h(z) = h(|z|),z C , and consider the Banach space Ah(C ) of entire functions withthe norm

fh = supzC|f(z)|eh(z) < +,

and the weighted Fock spaces

Aph(C ) ={f Hol (C ) : fpp,h =

C|f(z)|peph(z)dm2(z)


every z D we find fz Ah(D ) such that|fz(w)| eh(w)|wz|2h(z)/4

in a special neighborhood of z. (For a different type of peak functionsin Ah(D ) see [11].) These peak functions permit us then to reduce ourproblems to those in the standard Fock spaces Aph(C ), h(z) = |z|2.

The construction of peak functions in [16] is based on sharp approx-imation of h by log |f |, f Hol (D ), obtained in the work of Yu. Lyu-barskii and M. Sodin [17]. Here we need a similar construction forradial h of arbitrary (more than polynomial) growth. This is done ina standard way: we atomise the measure h(z)dm2(z) and obtain adiscrete measure

zn . Since our h are radial, we try to get sufficiently

symmetric sequence {zn}:{zn} = k{ske2piim/Nk}, sk 1, Nk .

For approximation of general h see the paper [15] by Yu. Lyubarskiiand E. Malinnikova and the references there.

Our paper is organized as follows. The main results are formulatedin Section 2. We construct peak functions in Section 3. Technicallemmas on sampling and interpolation sets are contained in Section 4.In Section 5 we obtain auxiliary results on asymptotic densities. Thetheorems on sampling sets are proved in Section 6, and the theoremson interpolation sets are proved in Section 7. In Sections 37 we dealwith the disc case. Some changes necessary to treat the plane case arediscussed in Section 8. Finally, in Section 9 we give an application ofour results to subspaces of Aph(D ) invariant under multiplication bythe independent variable.

We do not discuss here the following interesting fact: the familiesof interpolation (sampling) sets are not monotonic with respect to theweight function h. Also, we leave open other questions related to ourresults, including whether our interpolation sets are just sets of freeinterpolation, that is (say, for the spaces Aph(C )) the sets C suchthat

`() Aph(C ) = Aph(C ).

We hope to return to these questions later on.The authors are grateful to Yu. Lyubarskii, N. Nikolski, K. Seip,

M. Sodin, and P. Thomas for helpful discussions.

2. Main results

From now on in the disc case we assume that the function

(r) =[(h)(r)

]1/2, 0 r < 1,


decreases to 0 near the point 1, and

(r) 0, r 1. (2.1)Then for any K > 0, for r (0, 1) sufficiently close to 1, we have[r K(r), r +K(r)] (0, 1), and

(r + x) = (1 + o(1))(r), |x| K(r), r 1. (2.2)Furthermore, we assume that either (ID) the function r 7 (r)(1

r)C increases for some C 0.Given D , we define its lower d-density

D (,D ) = lim infR

lim inf|z|1, zD

Card( D(z,R(z)))R2

,

and its upper d-density

D+ (,D ) = lim supR

lim sup|z|1, zD

Card( D(z, R(z)))R2

.

We remark here that by (2.2), for fixed R, 0 < R


We could compare these densities D to those defined in the case(z) 1 |z| by K. Seip in [21]:

D(,D ) = lim infr1

infAut(D )

z()D(r)\D(1/2) log

1|z|

log 11r

, (2.3)

D+(,D ) = lim supr1

supAut(D )

z()D(r)\D(1/2) log

1|z|

log 11r

, (2.4)

where Aut(D ) is the group of the Mobius authomorphisms of the unitdisc. In contrast to D, the densities D are rather local, and corre-spondingly, it is not difficult to compute D (,D ) for many concrete.

Theorem 2.1. A set D is a sampling set for Ah(D ) if and onlyif it contains a d-separated subset

such that D (,D ) > 1

2.

Theorem 2.2. A set D is a sampling set for Aph(D ), 1 p 1

2.

Theorem 2.3. A set is an interpolation set for Ah(D ) if and onlyif it is d-separated and D

+ (,D ) < 12 .

Theorem 2.4. A set is an interpolation set for Aph(D ), 1 p K(r), and

(r + x) = (1 + o(1))(r), |x| K(r), r .Furthermore, we assume that either (IC) the function r 7 (r)rC in-creases for some C 2, r ;a typical example for (IIC) is

h(r) = exp r, r .


Next, we introduce d, and the notion of d-separated subsets of Cas above. Given C , we define its lower d-density

D (,C ) = lim infR

lim inf|z|

Card( D(z,R(z)))R2

,

and its upper d-density

D+ (,C ) = lim supR

lim sup|z|

Card( D(z,R(z)))R2

.

Theorem 2.5. A set C is a sampling set for Ah(C ) if and onlyif it contains a d-separated subset

such that D (,C ) > 1

2.

Theorem 2.6. A set C is a sampling set for Aph(C ), 1 p 1

2.

Theorem 2.7. A set is an interpolation set for Ah(C ) if and onlyif it is d-separated and D

+ (,D ) < 12 .

Theorem 2.8. A set is an interpolation set for Aph(C ), 1 p


(iii) Given s (0, 1) sufficiently close to the point 1, we can define{rk}, {Nk}, {sk} and as above in such a way that s and(3.1) holds (uniformly in s).

Proof. (i) We choose the sequence {rk} in the following way: r0 = 0;given rk, k 0, the number rk is defined by

(rk rk)rk|z|


Set

Wm = log1 zNmsNmm

1 zNmsNmm..

If m > k, then for some constant c > 0,

sm r cm kNm

.

Sincex

y exy, 0 x y 1, (3.4)

we get

rNmsNmm ec(mk).Next we use that if || c < 1, | | 1, thenlog 1

1 c1||, (3.5)

with c1 depending only on c.Therefore,

|Wm| c1ec(mk), m > k.Summing up, we obtain

m>k

|Wm| c1m>k

ec(mk) c2, (3.6)

for some positive constants c1, c2.Thus,

f(z) =m0

1 zNmsNmm1 zNmsNmm

.

Suppose that z D \ , rk r < rk+1, where r = |z|, and for somed, 0 d < Nk, arg z 2pid

Nk

piNk

.

Now we set

A(z) = log |f(z)| h(z) log dist(z,)(z)

.

By Greens formula,

h(r) =

D(r)

h(w) logr

|w|dm2(w)

2pi. (3.7)


Therefore,

A(z) + logdist(z,)

(z)=

0mk1

[log1 zNmsNmm

1 zNmsNmm

rm|w|


with c independent of r. This together with (3.8), (3.9) and (3.6)implies that A is bounded uniformly in z D \ , and (3.1) follows.

Since rm|w|


Since Um(w) = 0, w T , and for t 1, m St, Um are harmonic inthe annulus {w : 1 2t1(1 r) w 1}, we deduce from (3.12) that

mStUm(r)

c2t, t 1,with c independent of r, t, and (3.10) is proved.

Suppose now that satisfies the property (IID). For some constantc > 0,

r sm cNm

, m < k,

and we obtain that

(sm/r)Nm ec,

and again by (3.5),

|Um| c1(sm/r)Nm , m < k,for some positive constant c1.

By (2.2) and (3.4) we obtain thatm


(iii) Given sk s < sk+1, k 0, we may find 0 < rk < s < rk+1 < 1such that

Nk = Nk =

rk|z|


Note that

hr(w) =

D(r)

h(z)(log|w||z|)dm2(z)

2pi, |w| r.

The proof of Proposition 3.1 gives us immediately

Lemma 3.2. In the notations of Proposition 3.1, if sk1 r < sk,and

f() =

D(r)

(1 /1

),

then

|f()| ehrk () min(

1,dist(, rD )

()

), D .

If |z| = r, then we can divide this f by three factors j, j rD D(z, 5(z)), j = 1, 2, 3, and multiply it by (z)3, to obtainLemma 3.3. Given z D such that r = |z| is sufficiently close to1, there exists a function gz analytic and bounded in D and such thatuniformly in z,

|gz(w)|eh(w) 1, |w z| < (z), (3.16)

|gz(w)|eh(w) c(h) min[1,

min[(z), (w)]

|z w|]3, w D . (3.17)

We need only to verify that for some c,

ehr(w)h(w)(z)3 c(w)3, 0 r = |z| |w| < 1.This follows from the inequality h(t) hr(t) and the estimated

dt[h(t)hr(t)] = 1

t

D(t)\D(r)

h(w)dm2(w)

2pi 3(t)

, t r+B(r),

for some B > 0 independent of r.Next, we obtain an asymptotic estimate for partial products of the

Weierstrass -function.

Lemma 3.4. Given R 10, we define = R = (Z + iZ ) D(R2),

PR(z) = z

\{0}

(1 z

)


Then uniformly in R

|PR(z)| dist(z,)e(pi/2)|z|2 , |z| R, (3.18)

|PR(z)| c dist(z,)(e|z|2R2

)(pi/2)R2, |z| > R. (3.19)

Proof. For every denoteQ = {w C : |Re (w )| < 1/2, | Im (w )| < 1/2}.

SetQ =

Q.

If z Q, then we denote by 0 the element of such that z Q0 .For \ {0} we define

B =

Q

log1 z

w

dm2(w) log1 z

=

Q0

[log z + w

z log+ w

] dm2(w).We use thatlog |1 + a| Re (a a2

2) = O(|a|3), a 0.

Since Q0

w dm2(w) = 0,

Q0

w2 dm2(w) = 0,

we conclude that

|B| c[ 1||3 +

1

|z |3], \ {0, 0}.

Furthermore, we define

B0 =

Q0

log1 z

w

dm2(w) log |z|=

Q0

[log

1

|w| + log1 w

z

] dm2(w).If 0 6= 0, then |B0| c for an absolute constant c. Similarly, in thiscase,

B0 + log |z 0| =Q0

[log |w + (0 z)| log

1 + w0

] dm2(w),and, hence,

B0 + log |z 0| c for an absolute constant c. In thesame way, if 0 = 0, then

B0 + log |z| c for an absolute constant c.


Therefore,Q

log1 z

w

dm2(w) log |PR(z)|dist(z,)

= O(1), |z| < R2 + 1,Q

log1 z

w

dm2(w) log |PR(z)| = O(1), |z| R2 + 1. (3.20)

Next we use the identity

2

pi

D(R2+1)

log1 z

w

dm2(w) = 4 min(|z|,R2+1)0

log|z|ss ds

=

{(R2 + 1)2 + 2(R2 + 1)2 log

zR2 + 1

, |z| R2 + 1,|z|2, |z| < R2 + 1,

(3.21)

and the estimatesD(R2+1)\Q log1 z

w

dm2(w)D(R2+1)\Q Re zw dm2(w)

+ c D(R2+1)\Q zw

2dm2(w) c|z|

2

R2, |z| R3/2, (3.22)

andD(R2+1)\Q log1 z

w

dm2(w) m2

(D(R2 + 1) \Q) log[(R2 + 1)(|z|+ (R2 + 1))], z C , (3.23)for an absolute constant c. Now, (3.18) follows from (3.20)(3.22);(3.19) follows from (3.20)(3.23).

Proposition 3.5. Given R 100, there exists (R) > 0 such thatfor every z D with |z| 1 (R), there exists a function g = gz,Ranalytic in D such that uniformly in z,R we have

|g(w)|eh(w) e|zw|2/[4(z)2], w D D(z,R(z)), (3.24)

|g(w)|eh(w) c(h)[ R2(z)2e|z w|2

]R2/4, w D \ D(z,R(z)). (3.25)

Proof. Without loss of generality we may assume that z (0, 1). ByProposition 3.1, we find {rk}, {Nk}, {sk}, =

{ske

2piim/Nk}

, and f


Ah(D ) such that z = sk for some k, Z(f) = , and

|f(w)| eh(w) dist(w,)(w)

, w D . (3.26)

We define = R/

2pi (see Lemma 3.4), and denote

a,b = sk+ae2piib/Nk+a , a+ bi .

Then

maxa+bi

a+ bi a,b 0,02pi(z)

(k),with (k) 0 as k . Denote

Q(w) =w z(z)

a+bi\{0}

(w a,bz a,b

),

and put g = f/Q. Now, estimates (3.24) and (3.25) follow for fixedR when k is sufficiently large, and correspondingly, (k) is sufficientlysmall.

Indeed, by (3.26), for u = geh we have

|u(w)| dist(w,)(w)

(z)|w z|

a+bi\{0}

z a,bw a,b

.If w = z +

2pi(z)w, then

|u(w)| dist(w,)(w)

1|w|

a+bi\{0}

(z a,b)/(2pi(z))w + (z a,b)/(

2pi(z))

.For small (k) and for dist(w,) > 1/10 we have

|u(w)| dist(w,)(w)

1|w|

a+bi\{0}

a+ biw (a+ bi)

= dist(w,)(w)|PR/2pi(w)|

.

Now, for z sufficiently close to 1, by Lemma 3.4 and by the maximumprinciple, we have

|u(w)| dist(w,)PR/2pi(w) exp

[pi

2|w|2

]= exp

[|z w|

2

4(z)2

], |w| R

2pi,


and

|u(w)| c dist(w,)PR/2pi(w) c1

[ R22pie|w|2

](pi/2)(R2/(2pi))= c1

[ R2(z)2e|z w|2

]R2/4, |w| > R

2pi.

Proposition 3.6. Given R 100, there exists (R) > 0 such thatfor every z D with |z| 1 (R), there exists a function g = gz,Ranalytic in D such that uniformly in z,R we have

|g(w)|eh(w) e|zw|2/[4(z)2], w D D(z,R(z)),

|g(w)|eh(w) c(h)[R2 min[(z), (w)]2

e|z w|2]R2/4

, w D \ D(z, R(z)).

Proof. We use the argument from the above proof, and just replaceProposition 3.1 by Lemma 3.2. Furthermore, we use the argumentfrom the proof of Lemma 3.3.

4. d-separated sets

Here we establish several elementary properties of d-separated sets,sets of sampling, and sets of interpolation.

Lemma 4.1. Let 0 < R < , let z be sufficiently close to the unitcircle, (R) < |z| < 1, and let f be bounded and analytic in D =D(z,R(z)). Then

(i)|f(z1)|eh(z1) |f(z2)|eh(z2) c(R, h) d(z1, z2) max

D|feh|,

z1, z2 D(z, R(z)/2),

(ii) |f(z)|eh(z) c(R, h)(z)2

D

|f(w)|eh(w)dm2(w).

Proof. We may assume that () (z), D. We suppose thatmaxD |feh| = 1 and define

H(w) = h(z + wR(z)), |w| 1.Then

H(w) =R2(z)2

(z + wR(z))2 R2, |w| 1.


Set

G(w) =

D

log z w1 zw

H(z) dm2(z), |w| 1.Then |G(w)| + |G(w)| c, w 1, for some c depending only on hand R, and H1 = H G is real and harmonic in D . Denote by H1 theharmonic conjugate of H1, and consider

F (w) = f(z + wR(z))eH1(w)iH1(w).

Then F is analytic and bounded in D , and hence,

|F (w1) F (w2)| c|w1 w2|, w1, w2 D(1/2).Since

|F (w)| = |f(z + wR(z))|eh(z+wR(z))eG(w),we obtain assertion (i). Assertion (ii) follows by the mean value prop-erty for F .

Corollary 4.2. Every set of sampling for Ah(D ) contains a d-separa-ted set of sampling for Ah(D ).Corollary 4.3. Every set of interpolation for Ah(D ) is d-separated.Corollary 4.4. Every set of interpolation for Aph(D ), 1 p < , isd-separated.

Lemma 4.5. For every > 0, 1 p < , we have Aph(D ) A(1+)h(D ).Proof. By (2.1) and (3.7),

|(r)|(r)

= o( 1(r)

)= o(h(r)

), r 1,

and hence,

eh(z)(z)2 , |z| 1. (4.1)Applying Holders inequality and Lemma 4.1 (ii) with R = 1, we obtainour assertion:

|f(z)|e(1+)h(z) c eh(z)

(z)2

D(z,(z))

|f(z)|eh(z)dm2(z)

c eh(z)

(z)2/p

(D(z,(z))

|f(z)|peph(z)dm2(z))1/p c, z D .


Lemma 4.6. Let be a d-separated (with constant ) subset of D .If R > 0, (R,) = {w : minz d(w, z) R}, and if f is analytic in(R,), then

fpp,h, c(,R, h, p)

(R,)

|f(w)|peph(w)dm2(w).

Proof. The assertion follows from Lemma 4.1 (ii). Lemma 4.7. Let D . Then

fpp,h, c()fpp,h, f Aph(D ), (4.2)if and only if is a finite union of d-separated subsets.

Proof. For every z D with |z| close to 1, we apply Lemma 3.3 toobtain the function f = gz such that

|f(w)|eh(w) 1, |w z| < (z), (4.3)

|f(w)|eh(w) c(h) (z)3

|z w|3 , w D . (4.4)

By (4.3),

fpp,h,

wD(z,(z))|f(w)|peph(w)(w)2

cCard( D(z, (z)))(z)2. (4.5)Furthermore, by (4.3)(4.4),

|wz|


Proof. Let be a set of sampling for Aph(D ). For every > 0 we canfind a d-separated subset

of such that

supw

minz

d(z, w) .

Suppose that there exists f Aph(D ) such thatf2,h 1, fp,h, 1, fp,h, .

By Lemma 4.7, for some N,K independent of , both and areunions of N subsets d-separated with constant K. Without loss ofregularity we can assume that |z|+ (z) < 1, z . For every zk we choose wk D(zk, (zk)) such that

2|f(wk)|peph(wk) upk = supD(zk,(zk))

|f |peph.

Then the sequence {wk} is the union of c(N,K) subsets d-separatedwith constant c1(N,K). By Lemma 4.6,

zkupk(zk)

2 Cfpp,h,

with C independent of . Furthermore, by Lemma 4.1 (i), for every kand for every w D(zk, (zk)),|f(w)|eh(w) |f(zk)|eh(zk) Cuk.Therefore,

fpp,h, =w|f(w)|peph(w)(w)2

Czk

(|f(zk)|peph(zk) + pupk)(zk)2 c(fpp,h, + pfpp,h) Cp,

with C independent of . This contradiction implies our assertion.

5. Asymptotic densities

Given a set D such that D (,D )


Lemma 5.1. (i) If R > 0, and 0 < < (R), then for R R(R, ), and for z D such that |z| 1(R) we have

Card (z,R)R2

q(R) ;

(ii) if > 0, R > 0, R R(R, ), |z| 2(R),

E ={w D(z,R(z)/2) : Card (w,R)

R2 q(R) +

},

and

m2E m2D(z,R(z)/2),then

Card (z,R)R2

q(R) + 2

5.

Proof. By (2.2),

maxwD(z,R(z))

log (z)(w)

= o(1), |z| 1,and hence, for small , for fixed R, and for |z| close to 1 we have

Card( D(w, (R + 3)(z))) Card (w,R)

(q(R) 3)R2, w D(z,R(z)). (5.1)In the same way,

E E = {w D(z, R(z)/2) :Card

( D(w, (R + 3)(z))) (q(R) + )R2}. (5.2)

We use that by the Fubini theorem, for 0 < r1 < r2 and for F D(r2 r1),

CardF =1

pir21

D(r2)

Card(F D(w, r1)) dm2(w). (5.3)

(i) By (5.3),

Card (z, R)R2

D(z,(RR3)(z))

Card( D(w, (R + 3)(z)))pi(R + 3)2(z)2R2

dm2(w). (5.4)


Therefore, by (5.1), for |z| close to 1,Card (z, R)

R2

(q(R) 3)(R R 3

R

)2( RR + 3

)2 q(R) ,

for < (R), R R(R, ).(ii) By (5.1)(5.2), for small > 0, fixed R and |z| close to 1 we

have D(z,(RR3)(z))

Card( D(w, (R + 3)(z)))pi(R + 3)2(z)2R2

dm2(w)

=

E. . .+

D(z,(RR3)(z))\E

R2m2E

pi(R + 3)2(z)2R2(q(R) + )

+R2(pi(R R 3)2(z)2 m2E )

pi(R + 3)2(z)2R2(q(R) 3

).

If m2E m2D(z,R(z)/2), then by (5.4) we obtain

Card (z,R)R2

D(z,(RR3)(z))

Card( D(w, (R + 3)(z)))pi(R + 3)2(z)2R2

dm2(w)

( RR + 3

)2[(R R 3R

)2(q(R) 3) +

4( + 3)

] q(R) +

2

5

for = (), R R(R, ), |z| 2(R). By Lemma 5.1(i), for every R0 and such that 0 < < (R0), we

haveD (,D ) = lim inf

Rq(R) q(R0) .

Therefore, we obtain

Corollary 5.2.limR

q(R) = D (,D ), (5.5)

andq(R) D (,D ), R > 0. (5.6)


Given closed subsets A and B of C , the Frechet distance [A,B] isthe smallest t > 0 such that A B + tD , B A + tD . A sequence{An}, An C , converges weakly to A C if for every R > 0,[

(An D(R)) RT , (A D(R)) RT] 0, n.

In this case we use the notation An A. Given any sequence {An},An C , we can choose a weakly convergent subsequence {Ank}.Lemma 5.3. If D , and D (,D ) 12 , then there exists a se-quence of points zj D , |zj| 1, a sequence Rj , j , and asubset 0 of C such that

#(zj, Rj) 0, j , (5.7)

lim infR

Card(0 D(R))R2

12, (5.8)

where

#(z, R) ={w C : z + w(z) (z,R)}

={w D(R) : z + w(z) }.

Proof. Choose a sequence of positive numbers k,

k1 k 1, setrk = 2

k, k 1, and apply Lemma 5.1(ii) to find 1 > 0, 0 < 1 < 1, R1such that for 1 |w| < 1, if

Card (w,R1)

R21 q(R1) + 1,

then there exists z = z1(w) D(w,R1(w)/2) such thatCard (z, r1)

r21 q(r1) + 1.

Applying Lemma 5.1(ii) repeatedly, we find m 0, m 1, Rm , m, such that for m 1, m |w| < 1, if

Card (w,Rm)

R2m q(Rm) + m,

then there exists z = zm(w) D(w,Rm(w)/2), such thatCard (z, rk)

r2k q(rk) + k, 1 k m.

Next, by the definition of q(Rm), we can find wm D , m |wm| < 1,such that

Card (wm, Rm)

R2m q(Rm) + m,


and define zm = zm(wm). We obtain

lim supm

Card((zm, rk))

r2k D (,D )

1

2, k 1. (5.9)

Finally, we choose a sequence {mk} and a set 0 C such that#(zmk , rmk) 0, mk .

The property (5.8) follows from (5.9). Analogously, we have

Lemma 5.4. If D , and D+ (,D ) 12 , then there exists a se-quence of points zj D , |zj| 1, a sequence Rj , j , and asubset 0 of C such that

#(zj, Rj) 0, j , (5.10)

lim supR

Card(0 D(R))R2

12. (5.11)

6. Sampling theorems

We set (z) = |z|2/4.Proof of Theorem 2.1. By Corollary 4.2, every sampling set for Ah(D )contains a d-separated subset which is also a sampling set for Ah(D ).

(A) Suppose that is d-separated, and D (,D ) 12 . We follow

the scheme proposed in [16]. We apply Lemma 5.3 to obtain zj, Rj,and 0 satisfying (5.7)(5.8). Fix > 0. By the theorem of Seip onsampling in Fock type spaces [20, Theorem 2.3], there exists f A(C )such that

f = 1, f,0 .For K > 1 we set fK(z) = f((1K3/2)z). Then|fK(z)|e(z) |fK(z)|e(1K3/2)2(z)

|f(z)|e(z) + |f(z)|e(z) |f((1K3/2)z)|e(1K3/2)2(z)= |f(z)|e(z) + o(1), |z| K, K ,

where in the last relation we use [20, Lemma 3.1] (for a similar estimatesee Lemma 4.1 (i)). Furthermore,

|fK(z)|e(z) = o(1), |z| > K, K .Therefore, for sufficiently large K we get

fK 1, fK,0 2.


We fix such K and for N 0 setTNfK(z) =

0nN

cnzn,

wherefK(z) =

n0

cnzn.

As in [16, page 169], by the Cauchy formula,

|cn| c infr

exp[(1K3/2)2r2/4]rn

,

and hence,n0|cnzn|e(z) c(1 + |z|)4e[(1K3/2)21]|z|2/4, z C . (6.1)

Therefore, for sufficiently large N we have

TNfK 1, TNfK,0 3.We fix such N , set P = TNfK , and choose a C such that

|P (a)|e(a) 1.By (5.7), we can find large R > |a| and z close to the unit circle such

that

|P (w)| |w|R, |w| R,P,#(z,R) 4.

We set z = z+a(z), apply Proposition 3.5 to get g = gz,R, and define

f(w) = g(w)P(w z(z)

).

Then f Ah(D ),|f(z)|eh(z) |g(z)|eh(z) |P (a)| e|a|2/4e(a) = 1,

|f(w)|eh(w) e|wz|2/(4(z)2) 4e|wz|2/(4(z)2) = 4, w (z, R),

|f(w)|eh(w) c[ R2(z)2e|z w|2

]R2/4 [ |z w|(z)

]R c, w D \ D(z, R(z)),

with c independent of , R. Since can be chosen arbitrarily small,this shows that is not a sampling set for Ah(D ).

(B) Now we assume that is a d-separated subset of D , D (,D ) >12, and is not a sampling set for Ah(D ). Then there exist functionsfn Ah(D ) such that fnh = 1 and fnh, 0, n . By


the normal function argument, either (B1) fn tend to 0 uniformly oncompact subsets of the unit disc or (B2) there exists a subsequence fnkconverging uniformly on compact subsets of the unit disc to f Ah(D ),f 6= 0, with f = 0.

In case (B1), using Proposition 3.5 we can find zn D , Rn ,n, such that the functions

Fn(w) =f(zn + w(zn))

gzn,Rn(zn + w(zn))

satisfy the conditions:

|Fn(0)| 1,|Fn(w)| e|w|2/4, |w| Rn,

sup#(zn,Rn)

|Fn| 0, n.

By Corollary 5.2, we can find q, 12< q < D (,D ), and 0 < C < Rn 0 and large R > R(). ThenwkD(r)

logr

|wk|

=

wkD(r)log

r

|wk| 1

piR2(wk)2

D(wk,R(wk))

dm2(w)

O(1) + (1 )D(rR2(r))

1

piR2(w)2 log r|w|

Card[wk : w D(wk, R(wk))] dm2(w) O(1) + (1 )

2q(R )pi

D(rR2(r))

1

(w)2log

r

|w| dm2(w)

= O(1) +(1 )2q(R )

pi2pih(r), r 1,

that contradicts to (6.2) for small > 0 and R > R(). This provesour assertion.

Proof of Theorem 2.2. By Lemmas 4.7 and 4.8, every sampling set forAph(D ) is a finite union of d-separated subsets and contains a d-separated subset which is also a sampling set for Aph(D ).

(A) Suppose that is a d-separated subset of the unit disc andD (,D ) 12 . As in part (A) of the proof of Theorem 2.1, we applyLemma 5.3 to obtain zj, Rj, and 0 satisfying (5.7)(5.8). Furthermore,0 is uniformly separated, that is

inf{|z1 z2| : z1, z2 0, z1 6= z2} > 0.Then, by a version of a result of Seip [20, Lemma 7.1] (see also Lem-ma 4.7),

z0ep(z)|g(z)|p cgpp,, g Ap(C ). (6.3)

Fix > 0. By the theorem of Seip on sampling in Fock spaces [20,Theorem 2.1], there exists f Ap(C ) such that

fp, = 1, fp,,0 .


We approximate f by a polynomial P in the norm ofAp(C ) and obtain,using (6.3), that

1 Pp, 1 + , Pp,,0 2.For some M > 0 we have

D(M)|P (z)|pep(z)dm2(z) 1

2. (6.4)

By (5.7), we can find large R > M and z close to the unit circle suchthat

|P (w)| |w|R, |w| R. (6.5)Pp,,#(z,R) 3, (6.6)

We apply Proposition 3.5 to get g = gz,R, and define

f(w) =g(w)

(z)2/pP(w z(z)

).

Then, by (3.24) and (6.4),

fpp,h |wz|R(z)

|f(w)|peph(w)dm2(w)

|wz|R(z)

1

(z)2

P(w z(z)

)pep|wz|2/[4(z)2]dm2(w)=

D(R)|P (w)|pep|w|2/4dm2(w) 1

2. (6.7)

On the other hand,

fpp,h, =w|f(w)|peph(w)(w)2 =

w(z,R)

. . .+

w\(z,R). . . .

By (3.24) and (6.6),w(z,R)

(w)2

(z)2|g(w)|peph(w)

P(w z(z)

)p c

w#(z,R)

ep|w|2/4|P (w)|p Cp.

Since is d-separated, by Lemma 4.6 we havew\(z,R)

|f(w)|peph(w)(w)2 cD \D(z,(R1)(z))

|f(w)|peph(w)dm2(w).


Furthermore, by (3.24), (3.25), and (6.5),D \D(z,(R1)(z))

1

(z)2|g(w)|peph(w)

P(w z(z)

)pdm2(w) cp

D \D(z,R(z))

1

(z)2

[ |w z|(z)

]pR[ R2(z)2e|w z|2

]pR2/4dm2(w)

+cpD(z,R(z))\D(z,(R1)(z))

1

(z)2RpRep|wz|

2/[4(z)2]dm2(w)

= cp|w|>R

|w|pR[ R2e|w|2

]pR2/4dm2(w)

+cpR1 0 such that is a

sampling set for A(1+)h(D ). Following the method of [7, Section 6],we are going to prove that is a sampling set for Aph(D ).

We set

A(1+)h,0(D ) ={F Hol (D ) : lim

|z|1F (z)e(1+)h(z) = 0

},

R : F A(1+)h,0(D ) 7{F (zk)e

(1+)h(zk)}zk c0.

Since is a sampling set for A(1+)h(D ), R is an invertible linearoperator onto a closed subspace V of the space c0. Therefore, linearfunctionals Ez : v V 7 (R1 v)(z)e(1+)h(z), Ez 1, z D , (seealso Lemma 3.2) extend to linear functionals on c0 bounded uniformlyin z. Thus, for every z D , there exist bk(z), k 1, such that

k1|bk(z)| C, (6.8)

with C independent of z, and

F (z)e(1+)h(z) =k1

bk(z)F (zk)e(1+)h(zk). (6.9)

Let f Aph(D ). By Lemma 4.5, f A(1+)h,0(D ). For every z D ,we use Lemma 3.3 (with h replaced by h) to get gz satisfying (3.16)(3.17) (with replaced by 1/2). Now, we apply (6.9) to F = fgz.


By (3.16) we obtain

|f(z)|eh(z) Ck1|bk(z)||f(zk)|eh(zk)|gz(zk)|eh(zk).

By (6.8),

|f(z)|peph(z) Ck1|f(zk)|peph(zk)|gz(zk)|peph(zk)

and hence,

fpp,h Ck1|f(zk)|peph(zk)

D|gz(zk)|peph(zk)dm2(z).

It remains to note that by (3.17),D|gz(zk)|peph(zk)dm2(z)

c()D

min[1,

(zk)3p

|z zk|3p]dm2(z) C()(zk)2.

Then fp,h Cfp,h,, and using Lemma 4.7, we conclude that isa sampling set for Aph(D ).

7. Interpolation theorems

Proof of Theorem 2.3. Corollary 4.3 claims that every interpolation setfor Ah(D ) is d-separated.

(A) Let be a d-separated subset of D , and let D+ (,D ) 12 .Suppose that is an interpolation set for Ah(D ). We apply Lemma 5.4to obtain zj, Rj, and 0 = {z0k}k1 satisfying (5.10), (5.11). Supposethat ak C satisfy the estimate

|ak|e(z0k) 1.By (5.10), we can choose a subsequence {zj} of {zj} and Rj satisfying the following properties: (zj, R

j) are disjoint,

Bj = Card (zj, R

j) = Card[0 D(Rj)],

and we can enumerate (zj, Rj) = {wjk}1kBj in such a way that

max1kBj

z0k wjk zj(zj) 0, j . (7.1)

Without loss of generality, we can assume that |zj| > 1(Rj). There-fore, by Proposition 3.5, there exist gj = gzj ,Rj satisfying (3.24) and

(3.25).


For j 1 we consider the following interpolation problem:

fj(w) =

{akgj(wjk), w = wjk (zj, Rj),0, w \ (zj, Rj).

(7.2)

By our assumption on , we can find fj Ah(D ) satisfying (7.2). Thenthe functions

Fj(w) =fj(z

j + w(z

j))

gj(zj + w(zj))

satisfy the properties

Fj

(wjk zj(zj)

)= ak, 1 k Bj,

|Fj(w)| ce(w), |w| Rj.By a normal families argument and by (7.1), we conclude that there

exists an entire function F A(C ) such thatF (z0k) = ak, k 1.

Thus, 0 is a set of interpolation for A(C ). However, by the theoremof Seip on interpolation in the Fock type spaces [20, Theorem 2.4], thisis impossible for 0 satisfying (5.11). This contradiction proves ourassertion.

(B) Now we assume that is a d-separated subset of D , D+ (,D ) 0, and for every R R0, z \ D(1(R)), there


exists Fz,R A(1)(C ), such thatFz,R(0) = 1,

Fz,R #(z, R) \ {0} = 0,Fz,R(1) c.

(7.4)To continue, we need a simple estimate similar to (6.1).

Lemma 7.1. If F (z) =

n0 cnzn, |F (z)| exp |z|2, z C , N Z +,

and

(TNF )(z) =

0nNcnz

n,

then

|(F TNF )(z)| 2(N3)/2, |z| N/(4e),

|(TNF )(z)| (N + 1) exp |z|2,N/(4e) < |z|

N/2,

|(TNF )(z)| (N + 1)|z|N(2e/N)N/2, |z| >N/2.

Proof. By the Cauchy formula, we have

|cn| infr>0

[rn exp r2

]= exp

[n

2log

n

2e

],

and hence,n>N

|cn|rn n>N

( n2e 4eN

)n/2 2(N3)/2, r

N/(4e),

and0nN

|cn|rn (N + 1) max0nN

exp[n

2log

n

2e+ n log r

], r 0.

Furthermore,

n2

logn

2e+ n log r r2, r 0,

and

n2

logn

2e+ n log r N log r N

2log

N

2e, r

N/2, 0 n N.


Corollary 7.2. If > 0, F A(1)(C ), F(1) 1, N Z +,TNF is defined as above, and R =

2N/(1 ) is sufficiently large,

R > R(), then for some c = c() > 0 independent of z,R we have

|(F TNF )(z)|e|z|2/4 ecR2 , |z| R,|(TNF )(z)|e|z|2/4 2ec|z|2 , |z| R,

|(TNF )(z)|(e|z|2R2

)R2/4(e|z|2R2

)R2/5, |z| > R.

For large N we set R =

2N/(1 ), and for z sufficientlyclose to the unit circle, define, using gz,R from Proposition 3.6 and Fz,Rfrom (7.4),

Uz(w) = gz,R(w) (TNFz,R)(w z(z)

). (7.5)

If = {zn}n1, an C , n 1, andsupn1|an|eh(zn) 1, (7.6)

then for |zn| = max((R), 1(R)) ((R) is introduced in Proposi-tion 3.6) we put

Vn =anUznUzn(zn)

. (7.7)

ThenVn(zn) = an, (7.8)

and by the estimates in Proposition 3.6 and in Corollary 7.2 we obtainfor |zn| that|Vn(zk)|eh(zk) c0ecR2 , |zk zn| R(zn), k 6= n, (7.9)|Vn(z)|eh(z) c0ec|zzn|2/[(zn)2], |z zn| R(zn), (7.10)

|Vn(z)|eh(z) c0(R2 min[(zn), (z)]2

e|z zn|2)R2/5

,

|z zn| > R(zn), (7.11)for some c0 independent of zn, z, R.

For R > 1 we define

AR(z) =

zn, |zzn|>R(zn)

(R2 min[(zn), (z)]2e|z zn|2

)R2/5.

Suppose that for every > 0, we can find arbitrarily large R such that

supD

AR . (7.12)


Then for 0 < < 1/(2c0), for sufficiently large R, and for = max((R), 1(R)) we can define

f1 =

zn\D()Vn,

and obtain that for some B independent of {an} satisfying (7.6),f1h B, (7.13)

supzn\D()

|f1(zn) an|eh(zn) 12. (7.14)

Indeed, by (7.10), (7.11), and (7.12), for z D we have

|f1(z)|eh(z)

zn\D(), |zzn|R(zn)|Vn(z)|eh(z) +

zn\D(), |zzn|>R(zn)

|Vn(z)|eh(z)

c+ c0.By (7.8), (7.9), (7.11), and (7.12), for zk \ D() we have

|f1(zk) ak|eh(z)

|zkzn|R(zn), k 6=n|Vn(z)|eh(z) +

zn\D(), |zkzn|>R(zn)

|Vn(z)|eh(z)

cR2ecR2 + c0 12

for sufficiently large R. We fix such , R, .Iterating the approximation construction and using (7.13) and (7.14),

we obtain f2 Ah(D ) such thatf2h B/2,

supzn\D()

|f2(zn) + f1(zn) an|eh(zn) 14.

Continuing this process, we arrive at f =

n1 fn such that

fh 2B,f(zn) = an, zn \ D().

Thus, \D() is a set of interpolation for Ah(D ), and hence, is aset of interpolation for Ah(D ).


It remains to estimate AR for large R. Since is d-separated, using(2.2) we obtain

zn, |zzn|>R(zn)

(R2 min[(zn), (z)]2e|z zn|2

)R2/5 C(, h)

D \D(z,R(z))

(R2 min[(w), (z)]2e|z w|2

)R2/5dm2(w)(w)2

R2||>1

( 1e||2

)R2/5dm2() = o(1), R, (7.15)

because for any z, w D , R 5/,(min[(w), (z)](z)

)2R2/5( (z)(w)

)2 1.

This completes the proof of our assertion. Proof of Theorem 2.4. By Corollary 4.4, every set of interpolation forAph(D ) is d-separated.

(A) The argument is analogous to that in the part (A) of the proofof Theorem 2.3. We just use [20, Theorem 2.2] instead of [20, Theo-rem 2.4].

(B) Now we assume that is a d-separated subset of D , D+ (,D ) 1, such that for R R0, z \D(1(R)), the sets #(z,R)satisfy (7.3), and there exist Fz,R Ap(12)(C ), such that

Fz,R(0) = 1,

Fz,R #(z, R) \ {0} = 0,Fz,Rp,(12) c.

(7.16)Instead of Lemma 7.1 and Corollary 7.2 we use

Lemma 7.3. If F Ap,(12)(C ), Fp,(12) 1, ifF (z) =

n0 cnz

n, TNF is defined as in Lemma 7.1, and if R =2N/(1 ) is sufficiently large, R > R(), then for some c = c() >

0 independent of z,R we have

|(F TNF )(z)|e|z|2/4 ecR2 , |z| R, (7.17)

|(TNF )(z)|(e|z|2R2

)R2/4(e|z|2R2

)R2/5, |z| > R, (7.18)

D(R)|(TNF )(z)|pep|z|2/4dm2(z) 1. (7.19)


Proof. We just use Lemma 4.5 and Corollary 7.2 to deduce (7.17)(7.18). Inequality (7.19) is evident.

Next, we define Uz as in (7.5) using Fz,R from (7.16). If = {zn}n1,an C , n 1, and

n1|an|peph(zn)(zn)2 1, (7.20)

then for |zn| = max((R), 1(R)) ((R) is introduced in Propo-sition 3.6) we define Vn by (7.7). Set n = |an|eh(zn). As above, weobtain

Vn(zn) = an,

|Vn(zk)|eh(zk) c1n ecR2 , |zk zn| R(zn), k 6= n,D(z,R(zn))

|Vn(z)|peph(z)dm2(z) c1pn (zn)2,

|Vn(z)|eh(z) c1n (R2 min[(zn), (z)]2

e|z zn|2)R2/5

, |z zn| > R(zn),

for some c, c1 independent of zn, z, R.For z, D we set

W (z, ) =(R2 min[(), (z)]2

e|z |2)R2/5

(1 D(,R())(z)).

Now, to complete the proof as in part (B) of Theorem 2.3, we needonly to verify that for every > 0 there exists arbitrarily large R suchthat

k1

(n1

nW (zk, zn))p(zk)

2 ,D

(n1

nW (z, zn))pdm2(z) 1.

Furthermore, using (2.2), we can deduce these inequalities from theinequality

D

(n1

nW (z, zn))pdm2(z) 1, (7.21)

with small 1. By (7.15), for any small 2 > 0 we can find large R suchthat

n1W (z, zn) 2, z D . (7.22)


An estimate analogous to (7.15) gives us for large RDW (z, zn) dm2(z) c(zn)2.

Therefore,D

(n1

nW (z, zn))pdm2(z)

D

(n1

W (z, zn))p1

(n1

pnW (z, zn))dm2(z)

p12 n1

pn

DW (z, zn) dm2(z) cp12

n1

pn(zn)2 = cp12 .

This completes the proof of our assertion in the case p > 1. If p = 1,then (7.21) follows from (7.20) and (7.22).

8. The plane case

The results of Sections 37 easily extend to the plane case. First, wecan approximate h by log |f | for a special infinite product f .Proposition 8.1. There exist sequences {rk}, {sk}, 0 = r0 < s0 0, r 1, andmS0

eNm(rsm) k0

eckr(r)/(r) c1,

for some c, c1 independent of r 1.Using Proposition 8.1, we arrive at analogs of Propositions 3.5 and

3.6. For example, we have

Proposition 8.2. Given R 100, there exists (R)


9. Subspaces of large index

Let X be a Banach space of analytic functions in the unit disc, suchthat the operator Mz of multiplication by the independent variablef 7 zf acts continuously on X. Examples of such spaces are theHardy spaces Hp and the weighted Bergman spaces Ah(D ), Aph(D ),1 p


satisfying the same conditions as h and such that

h(r) = (1 + o(1))h(r), r 1,(r) = (1 + o(1))(r), r 1,

log1

(r)= o(h(r) h(r)), r 1,

where (r) =[(h)(r)

]1/2, 0 r < 1. Then we apply Proposition 3.1

to h to obtain ={ske

2piim/Nk}k0, 0m


Now we set

d ={ske

2piim/Nk , k = 2d+1(2v + 1), v 0, 0 m < Nk},

d = \ d, d 0.It is clear that

D+ (,D ) = D (,D ) =1

2,

D+ (d,D ) = D (d,D ) =1 2d1

2, d 0.

Using Theorem 2.3, we obtain that I(d) 6= {0}, d 0. To completeour proof we apply the following criterion from [14, Theorem 2.1].

Suppose that for every d 0 there exists cd > 0 such thatcd|g(0)| g + g1p,h, g I(d), g1 l0, l 6=dI(l). (9.3)

Thenindd0I(d) = +.

It remains to verify (9.3). Since is d-separated and d are pairwise

disjoint, by Lemma 4.7 and by (9.2) we obtain

g + g1p,h cg + g1p,h, cgp,h,d = cgp,h, c1|g(0)|.This proves our theorem.

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[18] N. Marco, X. Massaneda, J. Ortega-Cerda`, Interpolating and sampling se-quences for entire functions, Geom. and Funct. Anal. 13 (2003) 862914.

[19] J. Ortega-Cerda`, K. Seip, Beurling-type density theorems for weighted Lp

spaces of entire functions, J. dAnal. Math. 75 (1998) 247266.[20] K. Seip, Density theorems for sampling and interpolation in the Bargmann

Fock space, I, J. Reine Angew. Math. 429 (1992) 91106.[21] K. Seip, Beurling type density theorems in the unit disk, Inv. Math. 113 (1993)

2131.[22] K. Seip, Developments from nonharmonic Fourier series, Documenta Math. J.

DMV, Extra Volume ICM (1998) 713722.[23] K. Seip, Interpolating and sampling in spaces of analytic functions, American

Mathematical Society, Providence, 2004.[24] K. Seip, R. Wallsten, Density theorems for sampling and interpolation in the

BargmannFock space, II, J. Reine Angew. Math. 429 (1992) 107113.

Alexander Borichev, Department of Mathematics, Univer-sity of Bordeaux I, 351, cours de la Liberation, 33405 Tal-ence, FranceE-mail: [email protected]

Remi Dhuez, LATPCMI, Universite de Provence, 39 rue F.Joliot-Curie, 13453 Marseille, FranceE-mail: [email protected]

Karim Kellay, LATPCMI, Universite de Provence, 39 rueF. Joliot-Curie, 13453 Marseille, FranceE-mail: [email protected]

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