# Image ofLp(ℝn) under the Hermite Semigroup

Authors: R. Radha; D. Venku Naidu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

It is shown that the Hermite (polynomial) semigroup{e−tℍ:t>0} maps Lp(ℝn,ρ) into the space of holomorphic functions in Lr(ℂn,Vt,p/2(r+ϵ)/2) for each ϵ>0, where ρ is the Gaussian measure, Vt,p/2(r+ϵ)/2 is a scaled version of Gaussian measure with r=p if 1<p<2 and r=p′ if 2<p<∞ with 1/p+1/p′=1. Conversely if F is a holomorphic function which is in a “slightly” smaller space, namely Lr(ℂn,Vt,p/2r/2), then it is shown that there is a function f∈Lp(ℝn,ρ) such that e−tℍf=F. However, a single necessary and sufficient condition is obtained for the image of L2(ℝn,ρp/2) under e−tℍ, 1<p<∞. Further it is shown that if F is a holomorphic function such that F∈L1(ℂn,Vt,p/21/2) or F∈Lm1,p(ℝ2n), then there exists a function f∈Lp(ℝn,ρ) such that e−tℍf=F, where m(x,y)=e−x2/(p−1)e4t+1e−y2/e4t−1 and 1<p<∞.


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## Body

## 1. Introduction

It is well
known that the Bargmann transform
is an isometric isomorphism ofL2(ℝn) onto the Fock space ℱ(ℂn),
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 L2(ℍn) 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 Lp(ℝn) 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 ℝn or Fourier transform on ℝn 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 ofLp(ℝn,ρ) under the Segal-Bargmann transform Hall in
[7] obtained a
necessary condition for the range 1<p≤2 and the sufficient condition for 2≤p<∞.
In this paper, we tried to obtain a single necessary and sufficient condition
for a holomorphic function to be in the image of Lp(ℝn,ρ) under Hermite (polynomial) semigroup. Though
we were not completely successful, we could prove the following.Iff∈Lp(ℝn,ρ), then e−tℍf is holomorphic and e−tℍf∈Lr(ℂn,Vt,p/2(r+ϵ)/2) for every ϵ>0,
but we are able to prove the converse only for the holomorphic functions which
are in Lr(ℂn,Vt,p/2r/2) (with r=p,
if 1<p<2 and r=p′,
if 2<p<∞ with 1/p+1/p′=1), where Vt,p/2(z)=(π2/4)−n/2e2nt[((p−1)e4t+1)(e4t−1)]−n/2e−2x2/((p−1)e4t+1)e−2y2/(e4t−1) and Vt,p/2s, the sth power of Vt,p/2. However, we are able to obtain a single
necessary and sufficient condition for the image of L2(ℝn,ρp/2) under e−tℍ,1<p<∞, where ρp/2(u)=(pπ/2)−n/2e−2/pu2.Notice thatℋLr(ℂn,Vt,p/2r/2)⊂⋂ϵ>0ℋLr(ℂn,Vt,p/2(r+ϵ)/2) as Vt,p/2(r+ϵ)/2≤CVt,p/2r/2, where the constant C depends on ϵ,t,p,andn. We remark that the Gaussian-type density Vt,p/2(z) defines a finite measure with total mass e2nt when n≥1 and t>0. Let m be a weight function on ℝ2n and let 1≤p,q≤∞. Then the weighted mixed-norm space Lmp,q(ℝ2n) consists of all (Lebesgue) measurable
functions on ℝ2n, such that the norm ∥F∥Lmp,q=(∫ℝn(∫ℝn|F(x,w)|pm(x,w)pdx)q/pdw)1/q is finite. If p=∞ or q=∞, then the corresponding p-norm is replaced by the essential supremum. In
this paper, we also prove that if F is holomorphic and if F∈L1(ℂn,Vt,p/21/2) or Lm1,p(ℝ2n), then there exists a function f∈Lp(ℝn,ρ) such that F is the image of f. Here m(x,y)=e−x2/((p−1)e4t+1)e−y2/(e4t−1).The advantage of takingLp(ℝn,ρ) instead of Lp(ℝn) has been nicely explained by Hall in [7]. We also wish to point out
the following interesting fact, namely, if f∈Lp(ℝn), then the pointwise estimate of e−tℍf is given by |e−tℍf(x+iy)|≤Ct,p,ne(−1/2)tanh2tx2e(1/2)coth2ty2 which is independent of p (except the constant factor), which does not
help in the current problem. Here the constant Ct,p,n=e−nt(2π)−n/2(sinh2t)−n/2(2π/qcoth2t)n/2q, with 1/p+1/q=1. However, if f∈Lp(ℝn,ρ),
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 Lp structure for p≠2 do depend on the measures used in the
particular setup, while questions about L2 structure do not. In Section 2, we discuss the
Hermite (polynomial) semigroup and discuss the image of L2(ℝn,ρ) under the Hermite semigroup. In Section 3, we
prove our main results.


## 2. Hermite (Polynomial) Semigroup

Letℍk˜(x)=(−1)kex2(dk/dxk)(e−x2) denote the Hermite polynomial. For a
multi-index α∈ℕn,x=(x1,x2,…,xn)∈ℝn, define ℍα˜(x)=ℍα1˜(x1)ℍα2˜(x2)⋯ℍαn˜(xn) and ℍα(x)=(2αα!)−n/2ℍα˜(x).
This collection {ℍα(x):α∈ℕn} forms an orthonormal basis for L2(ℝn,ρ),
where ρ(x)dx=π−n/2e−x2dx.
Further, the functions ℍα are eigenfunctions of the operator ℍ=−Δ+2∑j=1nxj(∂/∂xj)+nI with eigenvalues 2|α|+n.
For f∈L2(ℝn,ρ), consider f=∑α∈ℕn〈f,ℍα〉ℍα,
where the sum converges to f in L2(ℝn,ρ).
For each k∈ℕ,
let ℚk denote the orthogonal projection of L2(ℝn,ρ) onto the eigenspace spanned by {ℍα:|α|=k}.
Then the spectral decomposition of ℍ can be written as ℍf=∑k=0∞(2k+n)ℚkf. The operator ℍ defines a semigroup, called the Hermite
polynomial semigroup, denoted by e−tℍ, for each t>0 using the expansion(2.1)e−tℍf=∑k=0∞e−(2k+n)tℚkf.Before stating our theorem, we will state the
following identity which will be repeatedly used in this paper:(2.2)1(coth2t+1−2/p)sinh22t+1−coth2t=2(p−1)e4t+1.In fact, the left-hand side of
(2.2) can be rewritten as(2.3)1(cosh2t+sinh2t−(2/p)sinh2t)sinh2t+sinh2t−cosh2tsinh2t.By using the exponential
formula for sinh2t and cosh2t, the right-hand side of (2.2) can be obtained
by straightforward simplification.We state here Mehler's formula for Hermite functions(2.4)Φα(x)=∏i=1n(2αiαi!π1/2)−1/2(−1)αidαidxiαi(e−xi2)exi2/2,where α=(α1,α2,…,αn)∈ℕn,x=(x1,x2,…,xn)∈ℝn which will lead to a formula involving Hermite
polynomials (for the proof, we refer to [14] or [6]).Mehler's Formula
For allw∈ℂ with|w|<1,
one has(2.5)∑α∈ℕnΦα(x)Φα(y)w|α|=π−n/2(1−w2)−n/2e(−1/2)((1+w2)/(1−w2))(x2+y2)+(2w/(1−w2))x⋅y,for all x,y∈ℝn. Here x2=∑i=1nxi2 and x⋅y=∑i=1nxi⋅yi. But(2.6)∑α∈ℕnΦα(x)Φα(y)w|α|=π−n/2e−x2/2e−y2/2∑α∈ℕnℍα(x)ℍα(y)w|α|from which it follows
that(2.7)∑α∈ℕnℍα(x)ℍα(y)w|α|=πn/2ex2/2ey2/2∑α∈ℕnΦα(x)Φα(y)w|α|.If f∈L2(ℝn,ρ),
then(2.8)ℚkf(u)=∑|α|=k〈f,ℍα〉ℍα(u)=∑|α|=k{∫ℝnf(x)ℍα(x)ρ(x)dx}ℍα(u).
Thus, it follows that
(2.9)e−tℍf(x)=∫ℝnKt(x,u)f(u)ρ(u)du=∫ℝnLt(x,u)f(u)du,
where(2.10)Lt(x,u)=π−n/2∑α∈ℕne−(2|α|+n)tℍα(x)ℍα(u)e−u2=π−n/2e−u2e−nt∑α∈ℕn(e−2t)|α|ℍα(x)ℍα(u).Notice that, Kt is the kernel of e−tℍ. However, Lt is used for computational purpose. Then by
using (2.7), we can write(2.11)Lt(x,u)=(2π)−n/2(sinh2t)−n/2e−1/2(coth2t−1)x2e−1/2(coth2t+1)u2ex⋅u/sinh2t.Since Kt(x,u) extends to an entire function Kt(z,u) for z∈ℂn, F(z)=e−tℍf(z) can also be extended to ℂn as an entire function, where z=x+iy. This can be verified by using Morera's
theorem.Remark 2.1.
The mape−tℍ is one-one.
Let e−tℍf=0 for f∈L2∩Lp(ℝn,ρ). Then e−tℍf(−iξ)=0∀ξ∈ℝn.
Letg^(ξ)=(2π)−n/2∫ℝng(x)e−ix⋅ξdx denote the Fourier transform of g. Then it follows from (2.11) that(2.12)e−tℍf(−iξ)=(sinh2t)−n/2e(1/2)(coth2t−1)ξ2G^(ξsinh2t)=0,where G(u)=f(u)e(−1/2)(coth2t−1)u2 forcing G^(ξ/sinh2t)=0. Then by uniqueness of Fourier transform G=0, which in turn implies f=0, showing that e−tℍ is one-one. (In fact, the above proof shows
that e−tℍ is injective on L2(ℝn,ρ). However, in general, one can show that e−tℍ is injective on the Lp space (1<p<∞) using the fact that the Fourier transform is
injective on the Lp space).We should calle−tℍfHermite Bargmann transform. Hereafter,
we should write ℋLr(ℂn,α(z)) for the class of holomorphic functions in Lr(ℂn,α(z)).Theorem 2.2.
Fixt>0 and let 1<p<∞. Then the Hermite polynomial semigroup e−tℍ is an isometric isomorphism of L2(ℝn,ρp/2) onto the space of holomorphic functions ℋL2(ℂn,Vt,p/2).Proof.
Letf∈L2(ℝn,ρp/2). Then F(z)=e−tℍf(z) is given by  (2.13)F(z)=(2π)−n/2(sinh2t)−n/2e(−1/2)(coth2t−1)z2∫ℝne(−1/2)(coth2t+1)u2e(z⋅u/sinh2t)f(u)du=(2π)−n/2(sinh2t)−n/2e(−1/2)(coth2t−1)z2∫ℝne(−1/2)(coth2t+1)u2e(−i(−y+ix)⋅u)/sinh2tf(u)du.Put g(u)=f(u)e(−1/2)(coth2t+1)u2ex⋅u/sinh2t. Then(2.14)∫ℝn|F(z)|2e(coth2t−1)(x2−y2)dy=(sinh2t)−n∫ℝn|g^(−ysinh2t)|2dy=∫ℝn|g^(y)|2dy(byapplyingchangeofvariables)=∫ℝn|g(y)|2du(usingPlancherelformula)=∫ℝn|f(u)|2e−(coth2t+1)u2e2x⋅u/sinh2tdu=(pπ2)n/2∫ℝn|f(u)|2e−(coth2t+1−2/p)u2e2x⋅u/sinh2tρp/2(u)du.Multiplying both sides by e−x2/(coth2t+1−2/p)sinh22t and integrating with respect to x,
we get(2.15)∫ℂn|F(z)|2e(coth2t−1)(x2−y2)e−x2/(coth2t+1−(2/p))sinh22tdydx=(pπ2)n/2∫ℝn|f(u)|2e−(coth2t+1−2/p)u2∫ℝne−x2/((coth2t+1−2/p)sinh2t)+2x⋅y/sinh2tdxρp/2(u)du=(pπ2)n/2∫ℝn|f(u)|2∫ℝne−[(x/(coth2t+1−2/p)sinh2t)−u(coth2t+1−2/p)]2dxρp/2(u)du.Let 𝒦t,p,n=(pπ2/2)−n/2[(coth2t+1−2/p)sinh22t]−n/2. Then it follows that(2.16)𝒦t,p,n∫ℂn|F(z)|2e(coth2t−1)(x2−y2)e−x2/((coth2t+1−(2/p))sinh22t)dxdy=∥f∥L2(ℝn,ρp/2)2.By using (2.2), the left-hand
side of the above equation can be written as(2.17)(pπ22)−n/2[(coth2t+1−2p)sinh22t]−n/2∫ℂn|F(z)|2e−2x2/((p−1)e4t+1)e−2y2/(e4t−1)dz=(π24)−n/2e2nt[((p−1)e4t+1)(e4t−1)]−n/2∫ℂn|F(z)|2e−2x2/((p−1)e4t+1)e−2y2/(e4t−1)dz,which implies that e−tℍ is an isometry from the space L2(ℝn,ρp/2) into ℋL2(ℂn,Vt,p/2). In the above, dz=dxdy,z=x+iy.
It remains to show thate−tℍ defines an onto map. Since e−tℍ is an isometry, the range is closed in ℋL2(ℂn,Vt,p/2). It is enough to show that the range is dense
in ℋL2(ℂn,Vt,p/2). By using exponential formula for sinh2t,cosh2t, and coth2t,e−tℍ can be written as(2.18)e−tℍf(z)=(2π)−n/2(sinh2t)−n/2∫ℝne−(e−tz−etu)2/2sinh2tf(u)du.It can be easily seen that e−tℍ will take real variable polynomials in L2(ℝn,ρp/2) to holomorphic polynomials in ℋL2(ℂn,Vt,p/2). On the other hand, if we take a holomorphic
polynomial in ℋL2(ℂn,Vt,p/2), it can be expressed as an image of a real
variable polynomial in L2(ℝn,ρp/2) under e−tℍ. In fact, suppose, for instance, if n=1, the first one can show that(2.19)e−tℍum(z)=C∑k=0m(−1)kmck(e−tz)m−kwk,for fixed m≥0, where C=−(2π)−n/2(sinh2t)−n/2e−(m+1)t, wk=∫ℝne(−1/2sinh2t)y2ykdy. Now if we take F(z)=∑i=0mcizi, an mth degree holomorphic polynomial in ℋL2(ℂ,Vt,p/2), we wish to choose an mth degree real variable polynomial f(x)=∑i=0maiui in L2(ℝ,ρp/2) such that e−tℍf=F. This leads to the determination of the
coefficients ai such that e−tℍ(∑i=0maiui)=∑i=0mcizi. Using (2.19) and comparing the coefficients of zk on both sides for 0≤k≤m, one ends up with a matrix equation UX=Y with U an upper triangular matrix with Uii=c0e−(i+1)w0, where c0=−(2π)−n/2(sinh2t)−n/2,X=(a0,a1,…,am)t,Y=(c0,c1,…,cm)t. Since w0≠0,detU≠0 which in turn gives a unique solution for a0,a1,…,am. Thus every holomorphic polynomial in ℋL2(ℂn,Vt,p/2) is an image of a real variable polynomial in L2(ℝn,ρp/2). This idea can be appropriately extended for
higher dimensions also.  It remains to show that the set of all holomorphic
polynomials are dense in ℋL2(ℂn,Vt,p/2), which will force the image of e−tℍ to be dense in ℋL2(ℂn,Vt,p/2). Toward this end, we show that any F∈ℋL2(ℂn,Vt,p/2) which is orthogonal to all holomorphic
polynomials vanishes identically. In particular, F is orthogonal to all monomials zα,α∈ℕn. Now consider the following Fock spaces ℱs(t)(ℂn), defined as the space of all entire functions G for which(2.20)∥G∥ℱs(t)2=∫ℂn|G(z)|2e−s(t)|z|2dzare finite. It is easy to see
that(2.21)F∈ℋL2(ℂn,Vt,p/2)⟺G(z)=F(z)ew(t)z2∈ℱs(t),where s(t)=1/((p−1)e4t+1)+1/(e4t−1)(>0), w(t)=(1/2)[1/(e4t−1)−1/((p−1)e4t+1)] The assumption
that F is orthogonal to all zα leads to the condition that G(z)=F(z)ew(t)z2 is orthogonal to all zαew(t)z2 in ℱs(t). The Taylor expansion F(z)=∑α∈ℕnaαzα leads to(2.22)G(z)=∑α∈ℕnaαzαew(t)z2.Since G is orthogonal to all zαew(t)z2,
we get aα=0 for all α and so F=0, thus proving our assertion.In particular, when p=2, we obtain the following result in which the
function f∈L2(ℝn,ρ) and the measure ρ are independent of t,
but e−tℍf∈L2(ℂn,Vt), where Vt=Vt,1, which depends on t (see also [15]).
Corollary 2.3.
A
holomorphic functionF on ℂn belongs to ℋL2(ℂn,Vt) if and only if F(z)=e−tℍf(z) for some f∈L2(ℝn,ρ),
where Vt(z)=(π2/2)−n/2(sinh4t)−n/2e−2x2/(e4t+1)e−2y2/(e4t−1). Moreover, one has the equality of norms(2.23)∥F(z)∥ℋL2(ℂn,Vt)=∥f∥L2(ℝn,ρ),whenever F=e−tℍf.


## 3. The Main Results

First, we will
obtain a pointwise bound for Hermite Bargmann transform of a functionf∈Lp(ℝn,ρ). From here onward, in order to define e−tℍ on Lp(ℝn,ρ),1<p<∞, we will first define e−tℍ on the class of functions L2∩Lp(ℝn,ρ). Then using standard density argument, we will
define e−tℍ on Lp(ℝn,ρ).Theorem 3.1.
Fixt>0 and let 1<p<∞. Then for all f∈Lp(ℝn,ρ), one has (3.1)|e−tℍf(x+iy)|≤Kt,p,n∥f∥Lp(ℝn,ρ)ex2/((p−1)e4t+1)ey2/(e4t−1).Proof.
We havee−tℍf(z)=(2π)−n/2(sinh2t)−ne(−1/2)(coth2t−1)z2∫ℝne(−1/2)(coth2t−1)u2ez⋅u/sinh2tf(u)e−u2du, for f∈L2∩Lp(ℝn,ρ). Let h(u)=e(−1/2)(coth2t−1)u2ez⋅u/sinh2t, Ct(x,y)=(2sinh2t)−n/2e(−1/2)(coth2t−1)(x2−y2). As f∈Lp(ℝn,ρ) and h∈Lp′(ℝn,ρ) by applying Hölder's inequality, it can be
shown that(3.2)|e−tℍf(z)|≤Ct(x,y)∥f∥Lp(ℝn,ρ)∥e(−1/2)(coth2t−1)u2ez⋅u/sinh2t∥Lp′(ℝn,ρ).Consider(3.3)∥e(−1/2)(coth2t−1)u2ez⋅u/sinh2t∥Lp′(ℝn,ρ)p′=∫ℝne(−p′/2)(coth2t−1)u2ep′x⋅u/sinh2tπ−n/2e−u2du=π−n/2∫ℝne(−p′/2)[(coth2t−1+2/p′)u−(x/sinh2tcoth2t−1+2/p′)]2ep′x2/2sinh22t(coth2t−1+2/p′)du=(2p′)n/2ep′x2/2sinh22t(coth2t+1−2/p)(coth2t−1+2p′)−n/2.Thus(3.4)|e−tℍf(z)|≤Kt,p,n∥f∥Lp(ℝn,ρ)e(1/2)(coth2t−1)y2e(x2/2)[1/(sinh22t(coth2t+1−2/p))+1−coth2t],where Kt,p,n is a constant depending on t,p,n. By using (2.2), it follows that(3.5)|e−tℍf(z)|≤Kt,p,n∥f∥Lp(ℝn,ρ)ex2/((p−1)e4t+1)ey2/(e4t−1),z=x+iy.Since L2∩Lp(ℝn,ρ) is dense in Lp(ℝn,ρ),1<p<∞, the result follows.The next theorem follows from Theorem3.1, by a
straightforward computation.Theorem 3.2.
Fixt>0 and let 1<p≤2.
If f∈Lp(ℝn,ρ), then e−tℍf∈ℋLp(ℂn,Vt,p/2(p+ϵ)/2) for every fixed ϵ>0.
In particular e−tℍf∈⋂ϵ>0ℋLp(ℂn,Vt,p/2(p+ϵ)/2).Remark 3.3.
The above theorem is valid for1<p<∞. 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, Theorem3.8.Theorem 3.4.
IfF is holomorphic and F∈Lp(ℂn,Vt,p/2p/2),
where 1<p≤2 and t is a fixed number greater than zero,
then there exists a unique function f∈Lp(ℝn,ρ) such that e−tℍf=F.Proof.
LetGf=e−tℍf. Then it follows from Theorem 2.2, that G is an isometric isomorphism from L2(ℝn,ρp/2) onto ℋL2(ℂn,Vt,p/2). G can be explicitly written as Gf(z)=∫ℝn(𝕃t(z,u)/ρp/2(u))f(u)ρp/2(u)du,
where 𝕃t is given in (2.11). Let G*,p denote the adjoint of G, where G is an operator from Hilbert space L2(ℝn,ρp/2) into the Hilbert space L2(ℂn,Vt,p/2). Note that, G*,pF will coincide with G−1F if F is a holomorphic function on ℂn. Thus we can write (3.6)G*,pF(u)=∫ℂn𝕃t(z,u)¯ρp/2F(z)Vt,p/2dxdy.In order to change the measures Vt,p/2 and ρp/2 into Lebesgue measure, construct a map G*,p¯:L2(ℂn,dxdy)→L2(ℝn,du) defined by(3.7)G*,p¯F(u)=ρp/21/2G*,p(Vt,p/2−1/2F(z)).An explicit computation shows
that G*,p¯ can be written as (3.8)G*,p¯F(u)=C∫ℂneix⋅y(coth2t−1)e−iy⋅u/sinh2t×e{−1/2(coth2t+1−2/p)[u−(x/sinh2t(coth2t+1−2/p))]2}F(z)dxdy.It can be easily verified that G*,p¯ defines a bounded linear map of L1(ℂn,dxdy) into L1(ℝn,du). By applying interpolation theorem ([16], M. Riesz convexity theorem), it follows that G*,p¯ is a bounded map of Lq(ℂn,dxdy) into Lq(ℝn,du) for q satisfying 1≤q≤2. In particular if we take p=q,
then G*,p¯ will be bounded from Lp(ℂn,dxdy) into Lp(ℝn,du). Again, we wish to change Lebesgue measure on ℝn to ρ,
and Lebesgue measure on ℂn to Vt,p/2p/2.
Toward this end, we define(3.9)G*,p¯¯:Lp(ℂn,Vt,p/2p/2)→Lp(ℝn,ρ)byG*,p¯¯F(u)=ρ−1/p(u)G*,p¯(Vt,p/21/2F(z)).Note that, ρ(u)−1/p=ceu2/p=c′ρp/2−1/2 for some constants c and c′.
Then it follows from (3.7) that G*,p¯¯F(u)=CG*,pF(u). This shows that G*,p is bounded from Lp(ℂn,Vt,p/2p/2) into Lp(ℝn,ρ).
We claim that ifF is in the holomorphic subspace of Lp(ℂn,Vt,p/2p/2), then GG*,pF=F.
LetPs(z) be the “polydisk” of radius s, centered at z, namely, Ps(z)={w∈ℂn:|wk−zk|<s,k=1,2,…,n, and wk=uk+ivk,zk=xk+iyk} (see [17]). Then F(z) can be written as(3.10)F(z)=(πs2)−n∫ℂnχPs(z)1α(w)F(w)α(w)dudv,where χPs(z) denotes the characteristic function on Ps(z).
Defineα(w)=e−pu2/((p−1)e4t+1)e−pv2/(e4t−1),w=u+iv,β=p′−1/p′. By using Hölder's inequality,(3.11)∫ℂn|Fm(z)|2e−2x2/((p−1)e4t+1)e−2y2/(e4t−1)dxdy=∫ℂn|F(λmz)|2e−2x2/((p−1)e4t+1)e−2y2/(e4t−1)dxdy=C∫ℂn|F(z)|2e−2x2/((p−1)e4t+1)e−2y2/λm2(e4t−1)dxdy.From this we can see
that(3.12)|F(z)|≤Ce(x+s)2/((p−1)e4t+1)e(y+s)2/(e4t−1).Now define Fm(z)=F(λmz), where λm is an increasing  sequence of numbers tending to 1. Then Fm∈Lp(ℂn,Vt,p/2p/2) and Fm will converge to F in the norm of Lp(ℂn,Vt,p/2p/2). Consider(3.13)|F(z)|≤C∥χPs(z)1|α(w)|∥Lp′(ℂn,α(w))∥F∥Lp(ℂn,α(w))≤Csupw∈Ps(z)1|α(w)|βm(Ps(z))1/p′∥F∥Lp(ℂn,α(w))=Csupw∈Ps(z)1|α(w)|β=Csupw∈Ps(z)eu2/((p−1)e4t+1)ev2/(e4t−1)sincepβ=1.Then by using (3.12),
we get(3.14)(3.11)≤C∫ℂne(x+s)2/((p−1)e4t+1)e(y+s)2/(e4t−1)e−2x2/λm2((p−1)e4t+1)e−2y2/λm2(e4t−1)dxdy<∞(as2<2λm2foreachm).This shows that Fm∈ℋL2(ℂn,Vt,p/2) for each m which in turn implies that GG*,pFm=Fm for each m. Since G*,p is bounded from Lp(ℂn,Vt,p/2p/2) into Lp(ℝn,ρ),G*,pFm converges to G*,pF. Then GG*,pFm=Fm will converge uniformly to GG*,pF on compact sets. Since Fm also converges to F in the norm of Lp(ℂn,Vt,p/2p/2),
the pointwise limit and Lp limit must coincide, showing GG*,pF=F. Then taking f=G*,pF proves our existence assertion. The uniqueness
follows from Remark 2.1.Remark 3.5.
As mentioned in the introduction,⋂ϵ>0ℋLp(ℂn,Vt,p/2(p+ϵ)/2) is larger than ℋLp(ℂn,Vt,p/2p/2). In fact, if f(z)=ez2/((p−1)e4t+1),
then f∈⋂ϵ>0ℋLp(ℂn,Vt,p/2(p+ϵ)/2) but f∉ℋLp(ℂn,Vt,p/2p/2). But we are able to show that the transform e−tℍ is only onto the functions in ℋLp(ℂn,Vt,p/2p/2).The following
theorem shows that the image of Lp(ℝn,ρ) under Hermite polynomial semigroup will be
contained in ℋLp′(ℂn,Vt,p/2p′/2) also.Theorem 3.6.
Fixt>0 and let 1<p≤2.
If f∈Lp(ℝn,ρ), then e−tℍf∈ℋLp′(ℂn,Vt,p/2p′/2), where p′ is such that 1/p+1/p′=1.Proof.
LetGf=e−tℍf. Then it follows from Theorem 2.2 that G is an
isometric isomorphism from L2(ℝn,ρp/2) onto ℋL2(ℂn,Vt,p/2). In order to change the weighted measure into
Lebesgue measures, construct a map Gp:L2(ℝn,du)→L2(ℂn,dxdy) defined by Gpf(z)=Vt,p/21/2G(ρp/2(u)−1/2f(u)). An explicit computation shows that Gp can be written as(3.15)Gpf(x+iy)=C∫ℝne−i(coth2t−1)x⋅yeiy⋅u/sinh2te{(−1/2)(coth2t+1−2/p)[u−x/sinh2t(coth2t+1−2/p)]2}f(u)du,where z=x+iy and C is a constant depending on t,p,n. It can be easily verified that Gp defines a bounded operator from L1(ℝn) into L∞(ℂn). By the interpolation theorem, Gp is also bounded from Lq(ℝn,du) into Lq′(ℂn,dxdy), for q satisfying 1≤q≤2. In particular we take p=q, then Gp will be bounded from Lp(ℝn,du) into Lp′(ℂn,dxdy). Again, to change measures, we define Gp¯:Lp(ℝn,ρ(u))→Lp′(ℂn,Vt,p/2p′/2) by Gp¯f(z)=Vt,p/2−1/2Gp(ρ(u)1/pf(u)) we see that the operators G and Gp¯ turn out to be the same up to a constant
multiple. Thus G is bounded from Lp(ℝn,ρ(u)) into Lp′(ℂn,Vt,p/2p′/2), proving our assertion.Using pointwise estimate in Theorem 3.1, one can obtain the
following result.Theorem 3.7.
Fixt>0 and let 2≤p<∞. If f∈Lp(ℝn,ρ), then e−tℍf∈ℋLp′(ℂn,Vt,p/2(p+ϵ)/2) for any fixed ϵ>0. In particular e−tℍf∈⋂ϵ>0ℋLp′(ℂn,Vt,p/2(p+ϵ)/2).As mentioned earlier, Theorems3.2 and 3.7 are
special cases of the following theorem.Theorem 3.8.
Suppose thatf∈Lp(ℝn,ρ) and that 1<p<∞,t>0 and ϵ>0. Then e−tℍf∈ℋLs(ℂn,Vt,p/2(s+ϵ)/2) for any 1≤s<∞.The proof is
simply an application of the pointwise estimate proved in Theorem 3.1. Then one
gets Theorem 3.2 by taking s=p and Theorem 3.7 by taking s=p'.As in the case of Theorem3.4, we prove the following
result.Theorem 3.9.
IfF is holomorphic and F∈Lp′(ℂn,Vt,p/2p′/2),
where 2≤p<∞ and t is a fixed number greater than zero,
then there exists a unique function f∈Lp(ℝn,ρ) such that e−tℍf=F,
where p′ is such that 1/p+1/p′=1.Proof.
In the
proof of Theorem3.4, we have noticed that G*,p¯ is bounded from L1(ℂn,dxdy) into L1(ℝn,du). Instead, one can also verify that G*,p¯ is bounded from L1(ℂn,dxdy) into L∞(ℝn,du). In this case the interpolation theorem will
show that G*,p¯ will be bounded from Lp′(ℂn,dxdy) into Lp(ℝn,du). So G*,p¯¯ will also be bounded from Lp′(ℂn,Vt,p/2p′/2) into Lp(ℝn,ρ). In this case also, we can show that GG*,pF=F,
for F∈ℋLp′(ℂn,Vt,p/2p′/2). Now, let f=G*,pF. The uniqueness follows from Remark 2.1.Remark 3.10.
As mentioned earlier, we are able
to show that the transforme−tℍ is only onto the functions in ℋLp′(ℂn,Vt,p/2p′/2), instead of ⋂ϵ>0ℋLp′(ℂn,Vt,p/2(p′+ϵ)/2).The following theorem gives a sufficient condition for
a holomorphic functionF to be in the image of e−tℍ. As we will now see, this condition is a
certain type of integrability of F.Theorem 3.11.
IfF is holomorphic and F∈L1(ℂn,Vt,p/21/2),
where 1<p<∞ and t is a fixed number greater than zero,
then there exists a unique function f∈Lp(ℝn,ρ) such that e−tℍf=F.Proof.
By
proceeding as in Theorem3.4, G*,p can be rewritten as(3.16)G*,pF(u)=C∬ℝne2u2/pe(−1/2)(coth2t−1)(x−iy)2e(−1/2)(coth2t+1)u2e(x−iy)⋅u/sinh2t×e−x2/((p−1)e4t+1)e−y2/e4t−1[F(z)e−x2/((p−1)e4t+1)e−y2/e4t−1]dxdy.Further, by considering Lp norm with respect to the variable u,
we can show that (3.17)∥e2u2/pe(−1/2)(coth2t−1)(x2−y2)e(−1/2)(coth2t+1)u2ex⋅u/sinh2te−x2/((p−1)e4t+1)e−y2/(e4t−1)∥Lp(ℝn,ρ)=C,
where C=(p/2(coth2t+1)−1)−n/2 is independent of x and y. Since F∈L1(ℂn,Vt,p/21/2), we can show by Minkowski's integral inequality
that(3.18)∥G*,pF∥Lp(ℝn,ρ)≤C∫ℂn|F(x+iy)|e−x2/((p−1)e4t+1)e−y2/(e4t−1)dxdy.Thus G*,p is bounded from L1(ℂn,e−x2/((p−1)e4t+1)e−y2/(e4t−1)) into Lp(ℝn,ρ). Again as in Theorem 3.4, by taking an
increasing sequence λm of real numbers converging to 1, one can show that GG*,pF=F, for any holomorphic function F in L1(ℂn,e−x2/((p−1)e4t+1)e−y2/(e4t−1)). Then taking f=G*,pF proves our existence assertion. The uniqueness
follows from Remark 2.1.Theorem 3.12.
Fixt>0 and let 1<p<∞. Suppose F is holomorphic and F∈Lm1,p(ℝ2n), where m(x,y)=e−x2/((p−1)e4t+1)e−y2/(e4t−1) then there exists a unique f∈Lp(ℝn,ρ) with e−tℍf=F.Proof.
We haveG*,pF(u) as in (3.16). Then(3.19)|G*,pF(u)|≤Ceu2/p∫ℝne(−1/2)(coth2t+1−2/p)[u−x/sinh2t(coth2t+1−2/p)]2×(∫ℝn|F(x+iy)|e−x2/((p−1)e4t+1)e−y2/(e4t−1)dy)dx.Then, by applying Minkowski's
integral inequality, it follows from hypothesis that e−u2/pG*,pF∈Lp(ℝn,du), which means that G*,pF∈Lp(ℝn,ρ(u)du). Thus G*,p is bounded from Lm1,p(ℝ2n) into Lp(ℝn,ρ). Again, by showing GG*,pF=F, for F holomorphic and F∈Lm1,p(ℝ2n),
we obtain the required result. The uniqueness follows from Remark 2.1.



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*Source: 287218-2009-02-26.xml*