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International Journal of Mathematics and Mathematical Sciences Volume 2008, Article ID 287218,13pages

doi:10.1155/2008/287218

Research Article

Image of L

p

R

n

under the Hermite Semigroup

R. Radha and D. Venku Naidu

Department of Mathematics, Indian Institute of Technology Madras, Chennai 600 036, India

Correspondence should be addressed to R. Radha,[email protected] Received 9 June 2008; Revised 8 October 2008; Accepted 9 December 2008 Recommended by Misha Rudnev

It is shown that the Hermitepolynomialsemigroup{e−tH:t >0}mapsLpRn, ρinto the space of holomorphic functions inLrCn, Vt,p/2r/2for each > 0, whereρis the Gaussian measure, Vt,p/2r/2is a scaled version of Gaussian measure withr pif 1< p < 2 andr pif 2< p <∞ with 1/p1/p 1. Conversely ifFis a holomorphic function which is in a “slightly” smaller space, namelyLrCn, Vt,p/2r/2, then it is shown that there is a function fLpRn, ρ such that e−tHf F. However, a single necessary and sufficient condition is obtained for the image of L2Rn, ρp/2undere−tH, 1< p <∞. Further it is shown that ifFis a holomorphic function such that FL1Cn, Vt,p/21/2orFL1,pmR2n, then there exists a functionfLpRn, ρsuch thate−tHfF, wheremx, y e−x2/p−1e4t1e−y2/e4t−1and 1< p <∞.

Copyrightq2008 R. Radha and D. V. Naidu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

It is well known that the Bargmann transform is an isometric isomorphism ofL2Rnonto the Fock spaceFCn, which is associated with the realization of the creation and annihilation operators for Bosons in quantum mechanics. We refer to1,2for further results. Similar type of results are shown for semigroups generated by the Laplace-Beltrami operator on compact spacessee3,4. The image ofL2Hnunder the heat kernel transform is studied for the Heisenberg group in5. Such type of results are also known for Hermite, special Hermite, Bessel, and Laguerre semigroups, see6.

Hall in7studied the problem of characterizing the image ofLpRnunder 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 onRnor Fourier transform onRnto Hermite expansions is not new: to cite a few, summability theorems8,9, multipliers10, Sobolev spaces11, and Hardy’s inequalities12,13.

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In order to prove that a holomorphic function to be in the image ofLpRn, ρunder the Segal-Bargmann transform Hall in7obtained 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 ofLpRn, ρ under Hermite polynomial semigroup. Though we were not completely successful, we could prove the following.

If fLpRn, ρ, then e−tHf is holomorphic and e−tHfLrCn, Vt,p/2r/2 for every >0, but we are able to prove the converse only for the holomorphic functions which are in LrCn, Vt,p/2r/2 withr p, if 1< p <2 andr p, if 2< p <∞with 1/p1/p 1, where Vt,p/2z π2/4−n/2e2ntp−1e4t1e4t−1−n/2e−2x2/p−1e4t1e−2y2/e4t−1andVt,p/2s ,the sth power ofVt,p/2.However, we are able to obtain a single necessary and sufficient condition for the image ofL2Rn, ρp/2undere−tH,1< p <∞,whereρp/2u pπ/2−n/2e−2/pu2.

Notice thatHLrCn, Vt,p/2r/2

>0HLrCn, Vt,p/2r/2 asVt,p/2r/2CVt,p/2r/2 ,where the constant C depends on , t, p, andn. We remark that the Gaussian-type density Vt,p/2z 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 Lp,qm R2n consists of all Lebesgue measurable functions on R2n, such that the norm FLp,qm

Rn

Rn|Fx, w|pmx, wpdxq/pdw1/qis finite. Ifp ∞orq∞,then the corresponding p-norm is replaced by the essential supremum. In this paper, we also prove that if F is holomorphic and ifFL1Cn, Vt,p/21/2orL1,pm R2n,then there exists a functionfLpRn, ρ such thatFis the image off.Heremx, y e−x2/p−1e4t1e−y2/e4t−1.

The advantage of takingLpRn, ρinstead ofLpRnhas been nicely explained by Hall in7. We also wish to point out the following interesting fact, namely, iffLpRn,then the pointwise estimate ofe−tHfis given by|e−tHfxiy| ≤Ct,p,ne−1/2tanh 2t x2e1/2coth 2t y2which is independent ofpexcept the constant factor, which does not help in the current problem.

Here the constantCt,p,n e−nt−n/2sinh 2t−n/22π/qcoth 2tn/2q,with 1/p1/q 1.

However, iffLpRn, 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 aboutLpstructure forp /2 do depend on the measures used in the particular setup, while questions aboutL2structure do not. InSection 2, we discuss the Hermitepolynomial semigroup and discuss the image ofL2Rn, ρunder the Hermite semigroup. InSection 3, we prove our main results.

2. Hermite (polynomial) semigroup

Let Hkx −1kex2dk/dxke−x2 denote the Hermite polynomial. For a multi-index α ∈ Nn, x x1, x2, . . . , xn ∈ Rn, define Hαx Hα1x1Hα2x2· · ·Hαnxn and Hαx 2αα!−n/2Hαx. This collection {Hαx : α ∈ Nn}forms an orthonormal basis for L2Rn, ρ, whereρxdx π−n/2e−x2dx. Further, the functionsHα are eigenfunctions of the operatorH−Δ 2n

j1xj∂/∂xj nIwith eigenvalues 2|α|n. ForfL2Rn, ρ,consider

f

α∈Nn f,HαHα, where the sum converges tofinL2Rn, ρ. For eachk∈N, letQkdenote the orthogonal projection ofL2Rn, ρonto the eigenspace spanned by{Hα :|α| k}. Then the spectral decomposition ofHcan be written as Hf

k02knQkf.The operatorH defines a semigroup, called the Hermite polynomial semigroup, denoted bye−tH,for each

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t >0 using the expansion

e−tHf

k0

e−2kntQkf. 2.1

Before stating our theorem, we will state the following identity which will be repeatedly used in this paper:

1

coth 2t1−2/psinh22t1−coth 2t 2

p−1e4t1. 2.2

In fact, the left-hand side of2.2can be rewritten as 1

cosh 2tsinh 2t−2/psinh 2tsinh 2tsinh 2t−cosh 2t

sinh 2t . 2.3

By using the exponential formula for sinh 2tand cosh 2t,the right-hand side of2.2can be obtained by straightforward simplification.

We state here Mehler’s formula for Hermite functions

Φαx n

i1

2αiαi1/2−1/2−1αi dαi

dxαiie−x2iex2i/2, 2.4

where α α1, α2, . . . , αn ∈ Nn, x x1, x2, . . . , xn ∈ Rn which will lead to a formula involving Hermite polynomialsfor the proof, we refer to14or6.

Mehler’s formula

For allw∈Cwith|w|<1, one has

α∈Nn

Φααyw|α|π−n/21−w2−n/2e−1/21w2/1−w2x2y22w/1−w2x·y, 2.5

for allx, y∈Rn.Herex2 n

i1xi2andx·yn

i1xi·yi.But

α∈Nn

Φααyw|α|π−n/2e−x2/2e−y2/2

α∈Nn

HαxHαyw|α| 2.6

from which it follows that

α∈Nn

HαxHαyw|α|πn/2ex2/2ey2/2

α∈Nn

Φααyw|α|. 2.7

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IffL2Rn, ρ, then

Qkfu

|α|k

f,HαHαu

|α|k RnfxHαxρxdx

Hαu. 2.8

Thus, it follows that

e−tHfx

RnKtx, ufuρudu

RnLtx, ufudu,

2.9

where

Ltx, u π−n/2

α∈Nn

e−2|α|ntHαxHαue−u2 π−n/2e−u2e−nt

α∈Nn

e−2t|α|

HαxHαu. 2.10

Notice that,Ktis the kernel ofe−tH.However,Ltis used for computational purpose. Then by using2.7, we can write

Ltx, u 2π−n/2sinh 2t−n/2e−1/2coth 2t−1x2e−1/2coth 2t1u2ex·u/sinh 2t. 2.11

SinceKtx, uextends to an entire functionKtz, uforz ∈Cn,Fz e−tHfzcan also be extended toCnas an entire function, wherezxiy.This can be verified by using Morera’s theorem.

Remark 2.1. The mape−tHis one-one . Lete−tHf 0 forfL2LpRn, ρ.Thene−tHf−iξ 0∀ξ∈Rn.

Letgξ 2π −n/2

Rngxe−ix·ξdxdenote the Fourier transform ofg.Then it follows from2.11that

e−tHf−iξ sinh 2t−n/2e1/2coth 2t−1ξ2G ξ

sinh 2t

0, 2.12

whereGu fue−1/2coth 2t−1u2forcingGξ/ sinh 2t 0.Then by uniqueness of Fourier transformG0,which in turn impliesf0,showing thate−tHis one-one.In fact, the above proof shows thate−tHis injective onL2Rn, ρ.However, in general, one can show thate−tHis injective on theLpspace1 < p <∞using the fact that the Fourier transform is injective on theLpspace.

We should call e−tHf Hermite Bargmann transform. Hereafter, we should write HLrCn, αzfor the class of holomorphic functions inLrCn, αz.

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Theorem 2.2. Fixt > 0 and let 1 < p < ∞.Then the Hermite polynomial semigroupe−tH is an isometric isomorphism ofL2Rn, ρp/2onto the space of holomorphic functionsHL2Cn, Vt,p/2. Proof. LetfL2Rn, ρp/2.ThenFz e−tHfzis given by

Fz 2π−n/2sinh 2t−n/2e−1/2coth 2t−1z2

Rne−1/2coth 2t1u2ez·u/sinh 2tfudu−n/2sinh 2t−n/2e−1/2coth 2t−1z2

Rne−1/2coth 2t1u2e−i−yix·u/sinh 2tfudu.

2.13

Putgu fue−1/2coth 2t1u2ex·u/sinh 2t.Then

Rn|Fz|2ecoth 2t−1x2−y2dy sinh 2t−n

Rn

g −y

sinh 2t 2dy

Rn|gy|2dy by applying change of variables

Rn|gy|2du using Plancherel formula

Rn|fu|2e−coth 2t1u2e2x·u/sinh 2tdu

2

n/2

Rn|fu|2e−coth 2t1−2/pu2e2x·u/sinh 2tρp/2udu.

2.14

Multiplying both sides bye−x2/coth 2t1−2/psinh22tand integrating with respect tox, we get

Cn|Fz|2ecoth 2t−1x2−y2e−x2/coth 2t1−2/psinh22tdy dx

2

n/2

Rn|fu|2e−coth 2t1−2/pu2

Rne−x2/coth 2t1−2/psinh2t2x·y/sinh 2tdx ρp/2udu

2

n/2

Rn|fu|2

Rne−x/

coth 2t1−2/psinh 2t−u

coth 2t1−2/p2dx ρp/2udu.

2.15

LetKt,p,n2/2−n/2coth 2t1−2/psinh22t−n/2.Then it follows that

Kt,p,n

Cn|Fz|2ecoth 2t−1x2−y2e−x2/coth 2t1−2/psinh22tdx dyf2L2Rnp/2. 2.16

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By using2.2, the left-hand side of the above equation can be written as 2

2

−n/2

coth 2t1− 2 p

sinh22t

−n/2

Cn|Fz|2e−2x2/p−1e4t1e−2y2/e4t−1 dz

π2 4

−n/2 e2nt

p−1e4t1e4t−1−n/2

Cn|Fz|2e−2x2/p−1e4t1e−2y2/e4t−1 dz, 2.17

which implies thate−tHis an isometry from the spaceL2Rn, ρp/2intoHL2Cn, Vt,p/2.In the above,dzdx dy, zxiy.

It remains to show thate−tHdefines an onto map. Sincee−tHis an isometry, the range is closed inHL2Cn, Vt,p/2.It is enough to show that the range is dense inHL2Cn, Vt,p/2.By using exponential formula for sinh 2t,cosh 2t,and coth 2t, e−tHcan be written as

e−tHfz 2π−n/2sinh 2t−n/2

Rne−e−tz−etu2/2 sinh 2tfudu. 2.18

It can be easily seen that e−tH will take real variable polynomials in L2Rn, ρp/2 to holomorphic polynomials inHL2Cn, Vt,p/2.On the other hand, if we take a holomorphic polynomial inHL2Cn, Vt,p/2,it can be expressed as an image of a real variable polynomial inL2Rn, ρp/2undere−tH.In fact, suppose, for instance, ifn1,the first one can show that

e−tHumz C m k0

−1kmcke−tzm−kwk, 2.19

for fixedm≥0,whereC−2π−n/2sinh 2t−n/2e−m1t,wk

Rne−1/2 sinh 2ty2ykdy.Now if we takeFz m

i0cizi,anmth degree holomorphic polynomial inHL2C, Vt,p/2,we wish to choose anmth degree real variable polynomialfx m

i0aiui inL2R, ρp/2such that e−tHf F. This leads to the determination of the coefficientsai such thate−tHm

i0aiui m

i0cizi.Using 2.19 and comparing the coefficients of zk on both sides for 0 ≤ km, one ends up with a matrix equation UX Y with U an upper triangular matrix with Uiic0e−i1w0,wherec0−2π−n/2sinh 2t−n/2, X a0, a1, . . . , amt, Y c0, c1, . . . , cmt. Sincew0/0, detU /0 which in turn gives a unique solution fora0, a1, . . . , am.Thus every holomorphic polynomial in HL2Cn, Vt,p/2 is an image of a real variable polynomial in L2Rn, 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 inHL2Cn, Vt,p/2,which will force the image ofe−tH to be dense inHL2Cn, Vt,p/2.Toward this end, we show that any F∈ HL2Cn, Vt,p/2which is orthogonal to all holomorphic polynomials vanishes identically.

In particular,Fis orthogonal to all monomialszα, α∈Nn.Now consider the following Fock spacesFstCn,defined as the space of all entire functionsGfor which

G2Fst

Cn|Gz|2e−st|z|2dz 2.20

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are finite. It is easy to see that

F∈ HL2Cn, Vt,p/2⇐⇒Gz Fzewtz2 ∈ Fst, 2.21

wherest 1/p−1e4t1 1/e4t−1>0,wt 1/21/e4t−1−1/p−1e4t1 The assumption thatFis orthogonal to allzαleads to the condition thatGz Fzewtz2is orthogonal to allzαewtz2inFst.The Taylor expansionFz

α∈Nnaαzαleads to

Gz

α∈Nn

aαzαewtz2. 2.22

SinceGis orthogonal to allzαewtz2, we getaα 0 for allαand soF 0,thus proving our assertion.

In particular, whenp 2,we obtain the following result in which the functionfL2Rn, ρand the measureρ are independent oft, bute−tHfL2Cn, Vt,whereVt Vt,1, which depends ontsee also15.

Corollary 2.3. A holomorphic function F on Cn belongs to HL2Cn, Vt if and only if Fz e−tHfzfor somefL2Rn, ρ, whereVtz π2/2−n/2sinh 4t−n/2e−2x2/e4t1e−2y2/e4t−1. Moreover, one has the equality of norms

FzHL2Cn,VtfL2Rn, 2.23

wheneverFe−tHf.

3. The main results

First, we will obtain a pointwise bound for Hermite Bargmann transform of a functionfLpRn, ρ.From here onward, in order to definee−tH onLpRn, ρ,1 < p < ∞,we will first definee−tH on the class of functionsL2LpRn, ρ.Then using standard density argument, we will definee−tHonLpRn, ρ.

Theorem 3.1. Fixt >0 and let 1< p <∞.Then for allfLpRn, ρ,one has

e−tHfxiyKt,p,nfLpRnex2/p−1e4t1ey2/e4t−1. 3.1 Proof. We have e−tHfz 2π−n/2sinh 2t−ne−1/2coth 2t−1z2

Rne−1/2coth 2t−1u2ez·u/sinh 2t fue−u2du, for fL2LpRn, ρ. Let hu e−1/2coth 2t−1u2ez·u/sinh 2t, Ctx, y 2 sinh 2t−n/2e−1/2coth 2t−1x2−y2.AsfLpRn, ρandhLpRn, ρby applying H ¨older’s inequality, it can be shown that

e−tHfzCtx, yfLpRne−1/2coth 2t−1u2ez·u/sinh 2t

LpRn. 3.2

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Consider

e−1/2coth 2t−1u2ez·u/sinh 2tp

LpRn

Rne−p/2coth 2t−1u2epx·u/sinh 2tπ−n/2e−u2du π−n/2

Rne−p/2

coth 2t−12/pu−x/sinh 2t

coth 2t−12/p2epx2/2sinh22tcoth 2t−12/pdu

2

p n/2

epx2/2sinh22tcoth 2t1−2/p

coth 2t−1 2 p

−n/2 .

3.3

Thus

|e−tHfz| ≤Kt,p,nfLpRne1/2coth 2t−1y2ex2/21/sinh22tcoth 2t1−2/p1−coth 2t, 3.4

whereKt,p,nis a constant depending ont, p, n.By using2.2, it follows that

|e−tHfz| ≤Kt,p,nfLpRnex2/p−1e4t1ey2/e4t−1, zxiy. 3.5

SinceL2LpRn, ρis dense inLpRn, ρ,1< p <∞,the result follows.

The next theorem follows fromTheorem 3.1, by a straightforward computation.

Theorem 3.2. Fixt > 0 and let 1 < p2. IffLpRn, ρ,thene−tHf ∈ HLpCn, Vt,p/2p/2for every fixed >0. In particulare−tHf

>0HLpCn, Vt,p/2p/2.

Remark 3.3. The above theorem is valid for 1 < p < ∞. We will see in Theorem 3.8 that Theorems 3.2 and 3.7can 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. IfFis holomorphic andFLpCn, Vt,p/2p/2, where 1< p2 and t is a fixed number greater than zero, then there exists a unique functionfLpRn, ρsuch thate−tHfF.

Proof. Let Gf e−tHf. Then it follows from Theorem 2.2, that G is an isometric isomorphism fromL2Rn, ρp/2ontoHL2Cn, Vt,p/2.Gcan be explicitly written asGfz

RnLtz, u/ρp/2ufuρp/2udu, whereLtis given in2.11. LetG∗,pdenote the adjoint of G,whereGis an operator from Hilbert spaceL2Rn, ρp/2into the Hilbert spaceL2Cn, Vt,p/2. Note that,G∗,pFwill coincide withG−1FifF is a holomorphic function onCn.Thus we can write

G∗,pFu

Cn

Ltz, u

ρp/2 FzVt,p/2dx dy. 3.6

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In order to change the measuresVt,p/2andρp/2into Lebesgue measure, construct a mapG∗,p: L2Cn, dxdyL2Rn, dudefined by

G∗,pFu ρ1/2p/2G∗,pVt,p/2−1/2Fz. 3.7

An explicit computation shows thatG∗,pcan be written as

G∗,pFu C

Cneix·ycoth 2t−1e−iy·u/sinh 2t

×e{−1/2coth 2t1−2/pu−x/sinh 2tcoth 2t1−2/p2}Fzdx dy.

3.8

It can be easily verified that G∗,p defines a bounded linear map of L1Cn, dx dy into L1Rn, du.By applying interpolation theorem16, M. Riesz convexity theorem, it follows thatG∗,p is a bounded map ofLqCn, dx dy intoLqRn, du forqsatisfying 1 ≤ q ≤ 2.In particular if we take p q, thenG∗,p will be bounded fromLpCn, dx dyintoLpRn, du.

Again, we wish to change Lebesgue measure onRn to ρ, and Lebesgue measure onCn to Vt,p/2p/2. Toward this end, we define

G∗,p:Lp

Cn, Vt,p/2p/2

−→LpRn, ρ by G∗,pFu ρ−1/puG∗,p

Vt,p/21/2Fz

. 3.9

Note that,ρu−1/p ceu2/p cρ−1/2p/2 for some constantscandc. Then it follows from3.7 thatG∗,pFu CG∗,pFu.This shows thatG∗,pis bounded fromLpCn, Vt,p/2p/2intoLpRn, ρ.

We claim that ifFis in the holomorphic subspace ofLpCn, Vt,p/2p/2,thenGG∗,pFF.

Let Psz be the “polydisk” of radius s, centered atz, namely, Psz {w ∈ Cn :

|wkzk|< s, k 1,2, . . . , n,andwk ukivk, zk xkiyk}see17. ThenFzcan be written as

Fz πs2−n

CnχPsz 1

αwFwαwdu dv, 3.10

whereχPszdenotes the characteristic function onPsz.

Defineαw e−pu2/p−1e4t1e−pv2/e4t−1, wuiv, β p−1/p.By using H ¨older’s inequality,

Cn|Fmz|2e−2x2/p−1e4t1e−2y2/e4t−1dx dy

Cn|Fλmz|2e−2x2/p−1e4t1e−2y2/e4t−1dx dy C

Cn|Fz|2e−2x2/p−1e4t1e−2y22me4t−1dx dy.

3.11

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From this we can see that

|Fz| ≤Cexs2/p−1e4t1eys2/e4t−1. 3.12

Now defineFmz mz,whereλmis an increasing sequence of numbers tending to 1.

ThenFmLpCn, Vt,p/2p/2andFmwill converge toFin the norm ofLpCn, Vt,p/2p/2.Consider

|Fz| ≤C

χPsz 1

|αw|

LpCn,αwFLpCn,αw

C sup

w∈Psz

1

|αw|βmPsz1/pFLpCn,αw C sup

w∈Psz

1

|αw|β C sup

w∈Psz

eu2/p−1e4t1ev2/e4t−1 since1.

3.13

Then by using3.12, we get

3.11≤C

Cnexs2/p−1e4t1eys2/e4t−1e−2x22mp−1e4t1e−2y22me4t−1dx dy

<

as 2< 2

λ2m for eachm

.

3.14

This shows thatFm ∈ HL2Cn, Vt,p/2for eachmwhich in turn implies thatGG∗,pFm Fm for each m. Since G∗,p is bounded from LpCn, Vt,p/2p/2 into LpRn, ρ, G∗,pFm converges to G∗,pF.ThenGG∗,pFm Fmwill converge uniformly toGG∗,pFon compact sets. SinceFmalso converges toFin the norm ofLpCn, Vt,p/2p/2, the pointwise limit andLplimit must coincide, showingGG∗,pF F.Then takingf G∗,pFproves our existence assertion. The uniqueness follows fromRemark 2.1.

Remark 3.5. As mentioned in the introduction,

>0HLpCn, Vt,p/2p/2 is larger than HLpCn, Vt,p/2p/2. In fact, if fz ez2/p−1e4t1, then f

>0HLpCn, Vt,p/2p/2 but f/∈HLpCn, Vt,p/2p/2.But we are able to show that the transforme−tHis only onto the functions inHLpCn, Vt,p/2p/2.

The following theorem shows that the image ofLpRn, ρunder Hermite polynomial semigroup will be contained inHLpCn, Vt,p/2p/2also.

Theorem 3.6. Fixt >0 and let 1< p2. IffLpRn, ρ,thene−tHf ∈ HLpCn, Vt,p/2p/2,where pis such that 1/p1/p1.

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Proof. LetGf e−tHf.Then it follows fromTheorem 2.2thatGis an isometric isomorphism from L2Rn, ρp/2 onto HL2Cn, Vt,p/2. In order to change the weighted measure into Lebesgue measures, construct a mapGp :L2Rn, duL2Cn, dx dydefined byGpfz Vt,p/21/2 p/2u−1/2fu.An explicit computation shows thatGpcan be written as

Gpfxiy C

Rne−icoth 2t−1x·yeiy·u/sinh 2te{−1/2coth 2t1−2/pu−x/sinh 2tcoth 2t1−2/p2}fudu, 3.15

wherez xiy andCis a constant depending ont, p, n. It can be easily verified thatGp defines a bounded operator fromL1Rn intoLCn.By the interpolation theorem, Gp is also bounded fromLqRn, duintoLqCn, dx dy,forqsatisfying 1 ≤q≤2.In particular we takep q,thenGpwill be bounded fromLpRn, duintoLpCn, dx dy.Again, to change measures, we defineGp :LpRn, ρuLpCn, Vt,p/2p/2byGpfz Vt,p/2−1/2Gpρu1/pfu we see that the operatorsGandGpturn out to be the same up to a constant multiple. ThusG is bounded fromLpRn, ρuintoLpCn, Vt,p/2p/2,proving our assertion.

Using pointwise estimate inTheorem 3.1, one can obtain the following result.

Theorem 3.7. Fixt >0 and let 2p <∞.IffLpRn, ρ,thene−tHf ∈ HLpCn, Vt,p/2p/2for any fixed >0.In particulare−tHf

>0HLpCn, Vt,p/2p/2.

As mentioned earlier, Theorems3.2and3.7are special cases of the following theorem.

Theorem 3.8. Suppose thatfLpRn, ρand that 1< p < ∞, t > 0 and >0.Thene−tHf ∈ HLsCn, Vt,p/2s/2for any 1s <∞.

The proof is simply an application of the pointwise estimate proved inTheorem 3.1.

Then one getsTheorem 3.2by takingspandTheorem 3.7by takingsp. As in the case ofTheorem 3.4, we prove the following result.

Theorem 3.9. IfFis holomorphic andFLpCn, Vt,p/2p/2, where 2≤p <andtis a fixed number greater than zero, then there exists a unique functionfLpRn, ρsuch thate−tHf F, wherepis such that 1/p1/p1.

Proof. In the proof ofTheorem 3.4, we have noticed thatG∗,p is bounded fromL1Cn, dx dy into L1Rn, du. Instead, one can also verify that G∗,p is bounded from L1Cn, dx dy into LRn, du.In this case the interpolation theorem will show thatG∗,pwill be bounded from LpCn, dx dyintoLpRn, du.SoG∗,pwill also be bounded fromLpCn, Vt,p/2p/2intoLpRn, ρ.

In this case also, we can show thatGG∗,pF F, forF ∈ HLpCn, Vt,p/2p/2.Now, letf G∗,pF.

The uniqueness follows fromRemark 2.1.

Remark 3.10. As mentioned earlier, we are able to show that the transforme−tH is only onto the functions inHLpCn, Vt,p/2p/2,instead of

>0HLpCn, Vt,p/2p/2.

参照

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