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COMPACTNESS OF GENERALIZED HANKEL OPERATORS WITH UNBOUNDED

AND ANTI-HOLOMORPHIC SYMBOLS ON GENERALIZED FOCK SPACES

WOLFGANG KNIRSCH and GEORG SCHNEIDER

LetkN\{0}andmR+. Consider the generalized Fock spacesA2(C,|z|m) :=

{fentire: C|f(z)|2 exp{−|z|m}dλ(z)<∞}and denote byP the Bergman pro- jection P : L2(C,|z|m) A2(C,|z|m). In a previous paper ([9]) compactness (boundedness) of Hankel operators

Hzk : (IdP)zk:A2(C,|z|m)A2(C,|z|m)

was proved fork < m/2 (km/2).In this paper we investigate how the mentioned result changes,ifwe replace Hzk by the generalized Hankel operator

Hzk: (IdP1)zk:A2(C,|z|m)A2,1(C,|z|m),

whereA2,1 denotes the closure of the linear span of{zlzn:n, lN, l1}with orthogonal projectionP1 :L2(C,|z|m)−→ A2,1(C,|z|m).The main result is the compactness (boundedness) ofHzk fork < m,(km).

AMS 2000 Subject Classification: Primary 47B35; Secondary 32W05, 32A36.

Key words: Hankel operator, generalized Fock spaces, weighted Bergman space,

-Neumann problem.

1. INTRODUCTION

Letk∈N\ {0}and m∈R+. Consider the spaces of entire functions A2(C,|z|m) :=

f :CCentire :f2m:=

C|f(z)|2exp{−|z|m}dλ(z)<∞ with weight functions exp{−|z|m} and Lebesgue measure dλ in C = R2. We call A2(C,|z|m) generalized Fock spaces and the special case A2(C,|z|2) is the Fock space. Let P = P(m) : L2(C,|z|m) −→ A2(C,|z|m) be the Bergman projection. Here L2(C,|z|m) := {f :C Cmeasurable : f2m :=

C|f(z)|2exp{−|z|m}dλ(z)<∞}.Operators of the form Hψ(f) = (Id−P)(ψf), f ∈A2(C,|z|m)

REV. ROUMAINE MATH. PURES APPL.,52(2007),2, 213–229

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are called (Bergman-)Hankel operators with symbolψ:CCwhile operators of the form

Tψ(f) =P(ψf), f ∈A2(C,|z|m),

are called (Bergman-)Toeplitz operators with symbol ψ. Clearly, for ψ L(C) these operators are bounded with

Tψ ≤ ψ and Hψ ≤ ψ.

Up to now, investigations of Hankel operators have focused on essentially bouned symbols. Operator-theoretic properties of Hankel operators, like com- pactness have already been extensively investigated – see for example [11].

In this paper we investigate Hankel operators with unbounded and anti- holomorphic symbolψ(z) =zk.One motivation for investigating Hankel oper- ators with symbolzkis the following connection with the-Neumann problem:

the canonical solution operatorStois a (Bergman-)Hankel operatorHzwith symbolz,ifS is restricted to the Bergman spaceA2(C,|z|m), i.e.

S = (Id−P)z=Hz :A2(C,|z|m)→A2(C,|z|m).

Clearly, the Hankel operatorHzk with symbol zk corresponds to a solu- tion operator S to k,where k = k

∂zk. For more details on this we refer the reader to [3], [4], [1], [2], [6], [7], [8], [10] and [9].

In the last years the following results about compactness and other operator-theoretic properties ofHz andHzk have been investigated.

Theorem 1.1. The Hankel operator

Hz:A2(C,|z|2)→L2(C,|z|2) is not compact. For m≥3 the Hankel operator

Hz:A2(C,|z|m)→L2(C,|z|m) is compact, but fails to be Hilbert-Schmidt.

Theorem 1.1 is proved in [4]. In [10] is derived an analogue of Theo- rem 1.1 in several dimensions: for n 2 it is shown that for all m 0 the Hankel operator Hzi : A21(Cn,|z|m) L2(Cn,|z|m) is not compact for all i∈ {1, . . . , n}. In [9] we can find the following result about Hankel operators with anti-holomorphic symbolszk, k N.

Theorem 1.2. The following results hold.

(1) Fork < m/2 the Hankel operator

Hzk :A2(C,|z|m)→L2(C,|z|m) is compact, but fails to be Hilbert-Schmidt.

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(2) Fork≤m/2 the Hankel operator Hzk :L2(C,|z|m)→L2(C,|z|m) is bounded, while in case k > m/2 unbounded.

In this paper we consider a slightly different type of Hankel operator, which we call the generalized Hankel operator, namely,

Hzk = (Id−P1)zk:A2(C,|z|m)→A2,1(C,|z|m), where

A2,1 :=zlzn:n, l∈N, l1

denotes the closure of the linear span of {zlzn : n, l N, l 1} and P1 : L2 −→ A2,1 is the orthogonal projection. In a former paper ([8]) generalized Hankel operators have already been investigated on the Bergman space of the unit disc and on the Fock space.

Theorem 1.3. On the Bergman space of the unit disc D in C the gen- eralized Hankel operator

Hzk = (Id−P1)zk:A2(D)→A2,1(D) is compact for all k≥1.

A completely different situation appears on the Fock space. We namely have

Theorem 1.4. On the Fock space the generalized Hankel operator Hz2 = (Id−P1)z2 :A2(C,|z|2)→A2,1(C,|z|2)

is bounded, but fails to be compact. Fork≥3 the generalized Hankel operator Hzk = (Id−P1)zk:A2(C,|z|2)→A2,1(C,|z|2)

is unbounded.

2. MAIN RESULTS

In this paper we give a generalization of Theorem 1.4 to generalized Fock spacesA2(C,|z|m). We show that generalized Hankel operatorsHzkare always compact fork < mand bounded fork≤m.So, this is an improvement on the classical (Bergman-)Hankel operator Hzk which only is compact for k < m2 and bounded fork≤ m2.

Consequently, in the special case of the Fock space the generalized Han- kel operatorHz is compact. So, this is an improvement on Hz, which is the canonical solution operator S to ∂, and is not compact (see [4], [10] or Theo- rem 1.1).

The following two theorems are our main results.

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Theorem 2.1. On generalized Fock spaces the generalized Hankel ope- rator

Hzk = (Id−P1)zk:A2(C,|z|m)→A2,1(C,|z|m) is compact fork < m and bounded for k≤m.

The casek≥m is considered in the following theorem.

Theorem 2.2. The following results hold.

(1) Let D := {k, m∈ {1,2, . . .} :k = m}. There exists at most a finite number of (k, m)∈D such that the generalized Hankel operator Hzk = (Id P1)zk:A2(C,|z|m)→A2,1(C,|z|m) is compact.

(2) For k > m we have almost everywhere unboundedness. This means that for every m there exists at most a finite number of k’s such that Hzk = (Id−P1)zk:A2(C,|z|m)→A2,1(C,|z|m) is not unbounded.

The scheme below visualizes Theorem 2.2.

Compactness ) m ( Almost unboundedness Boundedness

k

In the next section we derive some preparatory results. The main theo- rems will be proved at the end of the paper.

3. PRELIMINARIES AND

A COMPLETE ORTHONORMAL SYSTEM

It is well known that{zn/cn,m:n∈N}is a complete orthonormal system for the generalized Fock space

A2(C,|z|m) :=

f entire :f2m :=

C|f(z)|2 exp{−|z|m}dλ(z)<∞

,

wherec2n,m are the so-called moments, which are given by c2n,m=zn2m =

C|zn|2 exp{−|z|m}dλ(z) =

= 2π

0 r2nexp{−rm}rdr= 2π mΓ

2n+ 2

m .

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Frequently, we will abbreviatec2n,mto c2n. Furthermore, Γ is the Gamma function, which is defined for allx >0 by

Γ (x) =

0 tx−1e−tdt.

In the special casem = 2 of the Fock space, the moments are given by c2n=πΓ (n+ 1) =πn!.For the generalized Fock spaces we will use Stirling’s formula

Γ (l+ 1)∼lle−l 2πl

to investigate the asymptotic behavior of the moments c2n=c2n,m= 2π

mΓ

2n+ 2

m

2n+ 2 m 1

2n+2m −1

e2n+2m −1

2n+ 2 m 1. The results below can be found in [8], but we will repeat them for the reader’s convenience.

Lemma 3.1. The set 0,k, ψ1,k : k N} is a complete orthonormal system forA2,1(C,|z|m). Here

ψ0,k = zk ck

and ψ1,k = zzk+1c2k+1c2 k zk

c2k+2c4k+1c2 k

.

Next, we express the action of the orthogonal projection P1 on the com- plete orthonormal system 0,k, ψ1,k : k N} in terms of moments. We thus have

Proposition3.2. LetP1 :L2(C,|z|m)−→ A2,1(C,|z|m) be the orthogo- nal projection. ThenP1(zkzn+k) =an,0zn+an,1zzn+1,where the coefficients an,0, an,1 are given by

an,1= c2nc2n+k+1−c2n+1c2n+k

c2nc2n+2−c4n+1 , an,0= c2n+k

c2n −c2n+1 c2n an,1. Proof. This can be shown as in [8].

After these preparations we can calculate a convenient form of the norm ofHzk(zn+k/cn+k).

Proposition 3.3. For all n∈N we have Hzk

zn+k

cn+k 2=an−bnhn,

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with

an= c2n+2k

c2n+k −c2n+k

c2n , bn= c2n+k+1

c2n+1 −c2n+k

c2n and hn =

c2n+k+1 c2n+k c2n+1c2n

c2n+2

c2n+1 c2n+1c2n

.

Proof. Clearly,

an,1 = c2nc2n+k+1−c2n+1c2n+k

c2nc2n+2−c4n+1 = c2n+k c2n+1

c2n+k+1 c2n+k c2n+1c2

n

c2n+2 c2n+1 c2n+1c2

n

=: c2n+k c2n+1hn

and

an,0 = c2n+k

c2n c2n+1

c2n an,1 = c2n+k

c2n −c2n+k

c2n hn= c2n+k

c2n (1−hn). Calculating the norm ofHzk

zn+k yields

Hzk(zn+k)2=zkzn+k−an,0zn−an,1zzn+12=zkzn+k2+a2n,0zn2+ +a2n,1zzn+122an,0

zkzn+k|zn

2an,1

zkzn+k|zzn+1

+2an,0an,1

zn|zzn+1

=c2n+2k2an,1c2n+k+12an,0c2n+k+a2n,1c2n+2+ 2an,0an,1c2n+1+a2n,0c2n. Consequently,

Hzk

zn+k

cn+k 2 =h2n

c2n+kc2n+2

c4n+1 c2n+k c2n

2hn

c2n+k+1

c2n+1 c2n+k c2n

+ +c2n+2k

c2n+k −c2n+k c2n =h2n

c2n+kc2n+2

c4n+1 −c2n+k c2n

2hnbn+an. With

c2n+kc2n+2

c4n+1 c2n+k

c2n = c2n+k c2n+1

c2n+2

c2n+1 −c2n+1 c2n

=

= c2n+k c2n+1

c2n+2

c2n+1 c2n+1c2n

c2n+k+1 c2n+k c2n+1c2n

c2n+k+1

c2n+k −c2n+1 c2n

=h−1n

c2n+k+1

c2n+1 −c2n+k c2n

=h−1n bn

we get

Hzk

zn+k

cn+k 2=an−hnbn. This completes the proof.

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4. PROOF OF THE MAIN RESULT

It follows from Proposition 3.3 that we have to investigate the asymptotic behaviour of expressions of the form

c2n+l

c2n+j −c2n+r c2n+s,

where, fortunately,l−j =! r−s, that we will essentially use later on. First, we investigate the asymptotic behavior of quotients of moments.

Proposition 4.1. Let l, j N with l > j. The asymptotic behavior of quotients of moments is given by

c2n+l c2n+j

2

m(n+l+ 1−m/2)

m2(l−j)

em2(l−j)γn(l, j), where

γn=γn(l, j) =

1 +l+1−m/2n 1 + j+1−m/2n

2n

m

1 +l+1−m/2n 1 +j+1−m/2n

4(j+1)−m

2m

.

Proof. We have c2n+l=c2n+l,m

2(n+l) + 2

m 1

2(n+l)+2 m −1

e2(n+ml)+2−1

2(n+l) + 2

m 1

and so c2n+l c2n+j

1+l+1−m/2n 1+j+1−m/2n

m2(n+1)−1

1+l+1−m/2n 1+j+1−m/2n

2jm 2

m(n+l+1−m/2)

m2(l−j)

×em2(l−j)

1 +l+1−m/2n 1 + j+1−m/2n . Now, the proposition follows from

γn(l, j) =

1+l+1−m/2n 1+j+1−m/2n

2(n+1)−m

m

1+l+1−m/2n 1+j+1−m/2n

2j

m

1+l+1−m/2n 1 +j+1−m/2n =

=

1 +l+1−m/2n 1 + j+1−m/2n

2nm

1 + l+1−m/2n 1 +j+1−m/2n

4(j+1)−m2m

.

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Clearly, γn(l, j)const(l, j) forn→ ∞,but for our purpose this is not enough. We also have to investigate how fast γn tends to const(l, j). To de- scribe the limiting behaviour we use the so-called Landau symbols Ok :=O 1

nk

.

Proposition 4.2. Let l, j N withl > j. Then γn =γn(l, j) =

= exp

(l−j)

c1+d1 n

1+c2(d2−c2)1 n+

K+d1d2−c2 2

1 n2+O

1 n3

, where c1 = m2, c2 = m1, d1 = d1(j) = 4(j+1)−m2m , c = c(l) := l+ 1−m/2, d=d(j) :=j+ 1−m/2 andK =K(l, j) = 2(c3m3−d3) +(d22m−c22)2.

Proof. Note that γn= exp{lnγn}=

= exp

2n m ln

1+l+1nm/2 1+j+1nm/2

(I)

·exp

4(j+1)−m 2m ln

1+l+1−m/2n 1+j+1−m/2n

=:(II)

.

Setc=c(l) :=l+ 1−m/2,d=d(j) :=j+ 1−m/2. Using the equation ln

1 + 1

n = 1 n− 1

2n2 + 1 3n3 +O

1 n4 one can write

(I) = exp 2n

m

ln

1 + c n

ln

1 + d n

= exp

2(c−d)

m + d2−c2

m n + 2c3−d3 3m n2 +O

1 n3

= exp

2(c−d) m

×

×

1+

d2−c2 m

1

n+2c3−d3 3m n2

+ 1 2

d2−c2 m

1

n+ 2c3−d3 3m n2

2 +O

1 n3

= exp

2(l−j) m

1+d2−c2 m

1 n+

2(c3−d3)

3m +(d2−c2)2 2m2

1 n2+O

1 n3

=: exp

2(l−j)

m 1 +d2−c2 m

1

n+K 1 n2 +O

1 n3

.

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Setd1 :=d1(j) = 4(j+1)−m2m .Then it is easy to see that (II) = exp

d1

ln

1 + c

n ln

1 +d

n

= exp

d1

c n− c2

2n2 +O 1

n3 d

n− d2 2n2 +O

1 n3

= exp

d1

c−d

n +d2−c2 2n2 +O

1 n3

= exp

d1(c−d) n

exp

d1d2−c2 2n2 +O

1 n3

= exp

d1(l−j)

n 1 +d1d2−c2 2n2 +O

1 n3

.

Consequently, we deduce that (I) (II) = exp

2(l−j)

m +d1(l−j) n

×

×

1 +d2−c2 m

1

n+K 1 n2 +O

1 n3

1 +d1d2−c2 2n2 +O

1 n3

= exp

(l−j)

c1+d1

n 1+d2−c2 m

1 n+

K+d1d2−c2 2

1 n2+O

1 n3

.

This completes the proof.

For further calculations of the moments we use the formula below, which we call the exact Stirling’s formula. This formula consists of an asymptotic termA, which corresponds to Stirling’s formula, and the so-called error term E,which we have to simplify.

EXACT STIRLING’S FORMULA ([5]). We have

c2n+l=

2(n+l) + 2

m 1

2(n+l)+2 m −1

e2(n+l)+2m −1

2(n+l) + 2

m

×

1 + 1 12

1

2(n+l)+2

m 1+ 1

288

$ 1

2(n+l)+2

m 1

%2 +O 1

n3



=: [Asymptotic term]l×[Error term]l=:Al×El=A×E.

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Calculation of the error termE. Letc=c(l) := 1 +η:=l+ 1−m/2.

Then forn large enough we have 1

2(n+l)+2

m 1 = m

2

1

n+l+ 1−m/2 = m 2n

1

1 +l+1−m/2n =: m 2n

1 1 + nc

= m 2n

) j=0

−c n

j

= m 2n

1 c

n+ c2 n2 +O

1 n3 and

$ 1

2(n+l)+2

m 1

%2 = m2 4n2

12c

n +3c2 n2 +O

1 n3 . Consequently,

E= 1 + m 24n

1 c

n+ c2 n2 +O

1

n3 + m2 1152n2

12c

n +3c2 n2 +O

1 n3

= 1 + m 24

1 n +

m2

1152 −mc 24

1 n2 +

mc2

24 −m2c 576

1 n3 +O

1 n4

= 1 + m 24

1 n +

m2

1152 −mc 24

1 n2 +O

1 n3 .

Now, we shall use the exact Stirling’s formula to determine the limiting behavior ofc2n+l/c2n+j. First, a preparatory result.

Proposition 4.3. Forl > j we have c2n+l

c2n+j = 2

m(n+l+η)

m2(l−j)

×

×

1 +α1+ (l−j)d1

n +α1(l−j)d1+α2+ ((l−j)d1)2

n2 +O

1 n3 , where

α1 =α1(l, j) = j2−l2+ 2(j−l)η and m

α2 =α2(l, j) = m

24(j−l) +K+d1j2−l2+ 2(j−l)η

2 .

Furthermore, d1 =d1(j) and K = K(l, j) are defined in Proposition 4.2 and η=η(m) = 1−m/2.

Proof. From Proposition 4.1 we have () c2n+l

c2n+j 2

m(n+l+η) 2

m(l−j)

em2(l−j)γn(l, j) Al

Aj.

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Now, we use Proposition 4.2, the exact Stirling’s formula and the fact thatc1 = m2 to see that

c2n+l

c2n+j = Al×El

Aj×Ej (†)=

2

m(n+l+η)

m2(l−j)

em2(l−j)γn(l, j) × El

Ej Prop.4.2

=

2

m(n+l+η)

m2(l−j)

em2(l−j)exp

(l−j)

c1+d1

n ×

×

1 +c2(d2−c2)

n +

K+d1d2−c2 2

1 n2+O

1

n3 ×El

Ej Exact Stirling

=

2

m(n+l+η) 2

m(l−j)

exp

(l−j)d1 n

×

×

1 + c2(d2−c2)

n +

K+d1d2−c2 2

1 n2 +O

1 n3 ×

× 1 +24m1n+

$ m2

1152 mc(l)24 %

1

n2 +O 1

n3

1 + m241n+

$ m2

1152 mc(j)24 %

1

n2 +O 1

n3

2

m(n+l+η) 2

m(l−j)

exp

(l−j)d1 n

× P.

It is routine to check that P =α0+α11

n+α2 1 n2 +O

1

n3 = 1 +

j2−l2+ 2(j−l)η m

1 n+ +

m

24(j−l) +K+d1j2−l2+ 2(j−l)η 2

1 n2 +O

1 n3 . Next, we only have to compute

exp

(l−j)d1

n

× P =

1 +(l−j)d1

n +

(l−j)d1

n

2

+O3

×

×

1 +α11

n+α2 1 n2 +O

1 n3 ,

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which yields

exp

(l−j)d1

n

× P= 1 +α1+ (l−j)d1

n +

+α1(l−j)d1+α2+ ((l−j)d1)2

n2 +O

1 n3 . This completes the proof.

Remark. In the following we will use the generalized binomial formula:

forα∈Rand 0≤y <1,we have (1 +y)α =

) k=0

α k yk =

α

0 +

α 1 y+

α

2 y2+· · · , where

α

k := α(α−1)· · ·(α−k+ 1)

k! and

α

0 := 1.

Now, we can finally determine the limiting behavior of c2n+l/c2n+j. The following proposition yields the desired result.

Proposition 4.4. Let l > j and let n be large enough. Then we have c2n+l

c2n+j =C(l−j, m)n2(l−j)m ×

1 +β1(l, j)

n +β2(l, j) n2 +O

1 n3 , where C(l−j, m) = 2

m

2(l−j)

m . Furthermore,

β1(l, j) =

2(l−j)

m

1 (l+η) +α1(l, j) + (l−j)d1(j) and

β2(l, j) =

2(l−j)

m

2 (l+η)2+α1(l, j)(l−j)d1(j) +α2(l, j) + ((l−j)d1)2+ +

2(l−j)

m

1 (l+η) [α1(l, j) + (l−j)d1(j)],

whereα1(l, j),andα2(l, j)are defined in Proposition 4.3whiled1(j)is defined in Proposition 4.2with η=η(m) = 1−m/2.

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Proof. SetC:=C(l−j, m) :=2

m

2(l−j)m

and letnbe large enough. Then from Proposition 4.3 we get

c2n+l c2n+j =

2

m(n+l+η) 2

m(l−j)

×

×

1 +α1+ (l−j)d1

n +α1(l−j)d1+α2+ ((l−j)d1)2

n2 +O

1 n3

=C

n

1 +l+η n

2(l−j)

m ×

×

1 +α1+ (l−j)d1

n +α1(l−j)d1+α2+ ((l−j)d1)2

n2 +O

1 n3

=C(l−j, m)n2(l−j)m

1+

2(l−j)

m

1

l+η n +

2(l−j)

m

2

l+η n

2

+O 1

n3 ×

×

1 +α1+ (l−j)d1

n +α1(l−j)d1+α2+ ((l−j)d1)2

n2 +O

1 n3 . An elementary computation completes the proof.

We will use the following convenient definition for power series coeffi- cients.

Definition 4.1. Let P = *

kckxk be a formal power series. Then [P]k denotes the coefficientck of xk.In the following we consider x= 1n.

Now, we shall apply the formula from Proposition 4.4 to compute

an= c2n+2k

c2n+k −c2n+k

c2n , bn= c2n+k+1

c2n+1 −c2n+k

c2n , hn= h[ab]n

h[bel]n

:=

c2n+k+1 c2n+k c2n+1c2

n

c2n+2

c2n+1 c2n+1c2 n

.

from Proposition 3.3. We thus have

Corollary 4.5. Let n be large enough and an, bn, h[ab]n and h[bel]n as above. Then we have

an=C(k, m)n2km ×

β1(2k, k)−β1(k,0)

n + β2(2k, k)−β2(k,0)

n2 +O

1 n3

=:C(k, m)n2km × [P]1

n +[P]2 n2 +O

1

n3 =:C(k, m)n2km ×P,

(14)

bn=C(k, m)n2km ×

β1(k+1,1)−β1(k,0)

n +β2(k+1,1)−β2(k,0)

n2 +O

1 n3

=:C(k, m)n2km × [T]1

n +[T]2 n2 +O

1

n3 =:C(k, m)n2km ×T, h[ab]n =C(1, m)nm2 ×

β1(k+1, k)−β1(1,0)

n +β2(k+1, k)−β2(1,0)

n2 +O

1 n3

=:C(1, m)n2km × [H]1

n + [H]2 n2 +O

1

n3 =:C(1, m)n2km ×H, and

h[bel]n =C(1, m)nm2 ×

β1(2,1)−β1(1,0)

n +β2(2,1)−β2(1,0)

n2 +O

1 n3

=:C(1, m)n2km × [S]1

n +[S]2 n2 +O

1

n3 =:C(1, m)n2km ×S.

Finally, we can now determine the limiting behavior ofHzk(zn+k/cn+k)2. The following result holds.

Corollary 4.6. Let C(k, m) :=2

m

2km 2k2

m . Then Hzk

zn+k

cn+k 2 =an−bnhn=C(k, m)n2km ×γ,

whereγ is a power series in1/nwith[γ]0 = [γ]1 = 0and[γ]2 = [P]2[T][S]2[H]2 2. Proof. We recall that

β1(l, j) =

2(l−j)

m

1 (l+η) +α1(l, j) + (l−j)d1(j) and so forl, j, r, s with l−j=τ =r−swe get

β1(l, j)−β1(r, s) = 2τ

m(l−r) + (α1(l, j)−α1(r, s)) +τ(d1(j)−d1(s)). With

α1(l, j) = j2−l2+ 2(j−l)η

m = j2−l2+ 2τ η m and

d1(j) = 4(j+ 1)−m 2m we get

β1(l, j)−β1(r, s) = 2τ

m[(l−r) + (j−s)] +(j2−l2)(s2−r2)

m ,

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