Abstract

It is shown that the Hermite (polynomial) semigroup maps into the space of holomorphic functions in for each , where is the Gaussian measure, is a scaled version of Gaussian measure with if and if with . Conversely if is a holomorphic function which is in a “slightly” smaller space, namely , then it is shown that there is a function such that . However, a single necessary and sufficient condition is obtained for the image of under , . Further it is shown that if is a holomorphic function such that or , then there exists a function such that , where and .

1. Introduction

It is well known that the Bargmann transform is an isometric isomorphism of onto the Fock space , which is associated with the realization of the creation and annihilation operators for Bosons in quantum mechanics. We refer to [1, 2] for further results. Similar type of results are shown for semigroups generated by the Laplace-Beltrami operator on compact spaces (see [3, 4]). The image of under the heat kernel transform is studied for the Heisenberg group in [5]. Such type of results are also known for Hermite, special Hermite, Bessel, and Laguerre semigroups, see [6].

Hall in [7] studied the problem of characterizing the image of under the Segal-Bargmann transform. In this paper, we want to study this problem for Hermite semigroup instead of Segal-Bargmann transform. In fact, the idea of extending the classical results involving the standard Laplacian on or Fourier transform on to Hermite expansions is not new: to cite a few, summability theorems [8, 9], multipliers [10], Sobolev spaces [11], and Hardy's inequalities [12, 13].

In order to prove that a holomorphic function to be in the image of under the Segal-Bargmann transform Hall in [7] obtained a necessary condition for the range and the sufficient condition for . In this paper, we tried to obtain a single necessary and sufficient condition for a holomorphic function to be in the image of under Hermite (polynomial) semigroup. Though we were not completely successful, we could prove the following.

If then is holomorphic and for every , but we are able to prove the converse only for the holomorphic functions which are in (with , if and , if with ), where and the th power of However, we are able to obtain a single necessary and sufficient condition for the image of under where

Notice that as where the constant depends on We remark that the Gaussian-type density defines a finite measure with total mass when and Let be a weight function on and let Then the weighted mixed-norm space consists of all (Lebesgue) measurable functions on such that the norm is finite. If or then the corresponding -norm is replaced by the essential supremum. In this paper, we also prove that if is holomorphic and if or then there exists a function such that is the image of Here

The advantage of taking instead of has been nicely explained by Hall in [7]. We also wish to point out the following interesting fact, namely, if then the pointwise estimate of is given by which is independent of (except the constant factor), which does not help in the current problem. Here the constant with However, if , we get pointwise bounds as in Theorem 3.1. Further, we found the semigroup associated with Hermite polynomials to be more suitable for this problem rather than the semigroup associated with the Hermite functions. This is mainly because questions about structure for do depend on the measures used in the particular setup, while questions about structure do not. In Section 2, we discuss the Hermite (polynomial) semigroup and discuss the image of under the Hermite semigroup. In Section 3, we prove our main results.

2. Hermite (Polynomial) Semigroup

Let denote the Hermite polynomial. For a multi-index define and . This collection forms an orthonormal basis for , where . Further, the functions are eigenfunctions of the operator with eigenvalues . For consider , where the sum converges to in . For each , let denote the orthogonal projection of onto the eigenspace spanned by . Then the spectral decomposition of can be written as The operator defines a semigroup, called the Hermite polynomial semigroup, denoted by for each using the expansion

Before stating our theorem, we will state the following identity which will be repeatedly used in this paper:In fact, the left-hand side of (2.2) can be rewritten asBy using the exponential formula for and the right-hand side of (2.2) can be obtained by straightforward simplification.

We state here Mehler's formula for Hermite functionswhere which will lead to a formula involving Hermite polynomials (for the proof, we refer to [14] or [6]).

Mehler's Formula
For all with, one hasfor all Here and Butfrom which it follows thatIf , then Thus, it follows that whereNotice that, is the kernel of However, is used for computational purpose. Then by using (2.7), we can writeSince extends to an entire function for , can also be extended to as an entire function, where This can be verified by using Morera's theorem.

Remark 2.1. The map is one-one. Let for Then
Let denote the Fourier transform of Then it follows from (2.11) thatwhere forcing Then by uniqueness of Fourier transform which in turn implies showing that is one-one. (In fact, the above proof shows that is injective on However, in general, one can show that is injective on the space using the fact that the Fourier transform is injective on the space).

We should call Hermite Bargmann transform. Hereafter, we should write for the class of holomorphic functions in Theorem 2.2. Fix and let Then the Hermite polynomial semigroup is an isometric isomorphism of onto the space of holomorphic functions Proof. Let Then is given by Put ThenMultiplying both sides by and integrating with respect to , we getLet Then it follows thatBy using (2.2), the left-hand side of the above equation can be written aswhich implies that is an isometry from the space into In the above,
It remains to show that defines an onto map. Since is an isometry, the range is closed in It is enough to show that the range is dense in By using exponential formula for and can be written asIt can be easily seen that will take real variable polynomials in to holomorphic polynomials in On the other hand, if we take a holomorphic polynomial in it can be expressed as an image of a real variable polynomial in under In fact, suppose, for instance, if the first one can show thatfor fixed where , Now if we take an th degree holomorphic polynomial in we wish to choose an th degree real variable polynomial in such that This leads to the determination of the coefficients such that Using (2.19) and comparing the coefficients of on both sides for one ends up with a matrix equation with an upper triangular matrix with where Since which in turn gives a unique solution for Thus every holomorphic polynomial in is an image of a real variable polynomial in This idea can be appropriately extended for higher dimensions also.  It remains to show that the set of all holomorphic polynomials are dense in which will force the image of to be dense in Toward this end, we show that any which is orthogonal to all holomorphic polynomials vanishes identically. In particular, is orthogonal to all monomials Now consider the following Fock spaces defined as the space of all entire functions for whichare finite. It is easy to see thatwhere , The assumption that is orthogonal to all leads to the condition that is orthogonal to all in The Taylor expansion leads toSince is orthogonal to all , we get for all and so thus proving our assertion.
In particular, when we obtain the following result in which the function and the measure are independent of , but where which depends on (see also [15]). Corollary 2.3. A holomorphic function on belongs to if and only if for some , where Moreover, one has the equality of normswhenever

3. The Main Results

First, we will obtain a pointwise bound for Hermite Bargmann transform of a function From here onward, in order to define on we will first define on the class of functions Then using standard density argument, we will define on Theorem 3.1. Fix and let Then for all one has Proof. We have for Let , As and by applying Hölder's inequality, it can be shown thatConsiderThuswhere is a constant depending on By using (2.2), it follows thatSince is dense in the result follows.

The next theorem follows from Theorem 3.1, by a straightforward computation.Theorem 3.2. Fix and let . If then for every fixed . In particular Remark 3.3. The above theorem is valid for We will see in Theorem 3.8 that Theorems 3.2 and 3.7 can be put together in a general form. At this point, we thank one of the referees for suggesting us this general form, namely, Theorem 3.8.Theorem 3.4. If is holomorphic and , where and t is a fixed number greater than zero, then there exists a unique function such that .Proof. Let Then it follows from Theorem 2.2, that is an isometric isomorphism from onto . can be explicitly written as , where is given in (2.11). Let denote the adjoint of where is an operator from Hilbert space into the Hilbert space Note that, will coincide with if is a holomorphic function on Thus we can write In order to change the measures and into Lebesgue measure, construct a map defined byAn explicit computation shows that can be written as It can be easily verified that defines a bounded linear map of into By applying interpolation theorem ([16], M. Riesz convexity theorem), it follows that is a bounded map of into for satisfying In particular if we take , then will be bounded from into Again, we wish to change Lebesgue measure on to , and Lebesgue measure on to . Toward this end, we defineNote that, for some constants and . Then it follows from (3.7) that This shows that is bounded from into
We claim that if is in the holomorphic subspace of then
Let be the “polydisk” of radius centered at namely, and (see [17]). Then can be written aswhere denotes the characteristic function on
Define By using Hölder's inequality,From this we can see thatNow define where is an increasing sequence of numbers tending to Then and will converge to in the norm of ConsiderThen by using (3.12), we getThis shows that for each which in turn implies that for each Since is bounded from into converges to Then will converge uniformly to on compact sets. Since also converges to in the norm of , the pointwise limit and limit must coincide, showing Then taking proves our existence assertion. The uniqueness follows from Remark 2.1.
Remark 3.5. As mentioned in the introduction, is larger than In fact, if , then but But we are able to show that the transform is only onto the functions in The following theorem shows that the image of under Hermite polynomial semigroup will be contained in also.Theorem 3.6. Fix and let . If then where is such that Proof. Let Then it follows from Theorem 2.2 that is an isometric isomorphism from onto In order to change the weighted measure into Lebesgue measures, construct a map defined by An explicit computation shows that can be written aswhere and is a constant depending on It can be easily verified that defines a bounded operator from into By the interpolation theorem, is also bounded from into for satisfying In particular we take then will be bounded from into Again, to change measures, we define by we see that the operators and turn out to be the same up to a constant multiple. Thus is bounded from into proving our assertion.Using pointwise estimate in Theorem 3.1, one can obtain the following result.Theorem 3.7. Fix and let If then for any fixed In particular

As mentioned earlier, Theorems 3.2 and 3.7 are special cases of the following theorem.Theorem 3.8. Suppose that and that and Then for any The proof is simply an application of the pointwise estimate proved in Theorem 3.1. Then one gets Theorem 3.2 by taking and Theorem 3.7 by taking

As in the case of Theorem 3.4, we prove the following result.Theorem 3.9. If is holomorphic and , where and is a fixed number greater than zero, then there exists a unique function such that , where is such that Proof. In the proof of Theorem 3.4, we have noticed that is bounded from into Instead, one can also verify that is bounded from into In this case the interpolation theorem will show that will be bounded from into So will also be bounded from into In this case also, we can show that , for Now, let The uniqueness follows from Remark 2.1.Remark 3.10. As mentioned earlier, we are able to show that the transform is only onto the functions in instead of

The following theorem gives a sufficient condition for a holomorphic function to be in the image of As we will now see, this condition is a certain type of integrability of Theorem 3.11. If is holomorphic and , where and t is a fixed number greater than zero, then there exists a unique function such that Proof. By proceeding as in Theorem 3.4, can be rewritten asFurther, by considering norm with respect to the variable , we can show that where is independent of and Since we can show by Minkowski's integral inequality thatThus is bounded from into Again as in Theorem 3.4, by taking an increasing sequence of real numbers converging to one can show that for any holomorphic function in Then taking proves our existence assertion. The uniqueness follows from Remark 2.1.Theorem 3.12. Fix and let Suppose is holomorphic and where then there exists a unique with Proof. We have as in (3.16). ThenThen, by applying Minkowski's integral inequality, it follows from hypothesis that which means that Thus is bounded from into Again, by showing for holomorphic and , we obtain the required result. The uniqueness follows from Remark 2.1.

Acknowledgments

The authors wish to express their sincere gratitude to the referees for spending their valuable time in reading the manuscript very carefully, pointing out many inaccuracies, gaps, and errors and giving them several useful suggestions for improving the earlier version of the manuscript. The authors thank Professor S. Thangavelu for initiating them into this work and giving them valuable suggestions. They also wish to thank Professor B. C. Hall for giving clarifications to their questions related to Segal-Bargmann transform. D. Venku Naidu thanks the Council of Scientific and Industrial Research, India, for the financial support.