Conical square functions associated with Bessel, Laguerre and Schrödinger operators in UMD Banach spaces

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Conical square functions associated with Bessel, Laguerre and Schrödinger operators in UMD Banach spaces ✩ Jorge J. Betancor a , Alejandro J. Castro b , Juan C. Fariña a , L. Rodríguez-Mesa a,∗ a

Departamento de Análisis Matemático, Universidad de La Laguna, Campus de Anchieta, Avda. Astrofísico Francisco Sánchez, s/n, 38271, La Laguna (Sta. Cruz de Tenerife), Spain b Department of Mathematics, Nazarbayev University, 010000 Astana, Kazakhstan

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 15 February 2016 Available online xxxx Submitted by J.A. Ball

In this paper we consider conical square functions in the Bessel, Laguerre and Schrödinger settings where the functions take values in UMD Banach spaces. Following a recent paper of Hytönen, van Neerven and Portal [36], in order to define our conical square functions, we use γ-radonifying operators. We obtain new equivalent norms in the Lebesgue–Bochner spaces Lp ((0, ∞), B) and Lp (Rn , B), 1 < p < ∞, in terms of our square functions, provided that B is a UMD Banach space. Our results can be seen as Banach valued versions of known scalar results for square functions. © 2016 Elsevier Inc. All rights reserved.

Keywords: Conical square functions Vector-valued harmonic analysis UMD Banach spaces Bessel Laguerre Schrödinger

1. Introduction In this paper we obtain equivalent norms in the Lebesgue–Bochner space Lp (Rn , B), 1 < p < ∞, where B is a UMD Banach space, in terms of conical square functions defined via fractional derivatives of Poisson semigroups associated with Bessel, Laguerre and Schrödinger operators. According to the ideas developed by Hytönen, van Neerven and Portal [36] we use appropriate tent spaces using γ-radonifying operators (or, in other words, methods of stochastic analysis in a Banach valued setting). We denote by Pt (z), the classical Poisson kernel in Rn , that is, Pt (z) = cn

(|z|2

t , t > 0 and z ∈ Rn , + t2 )(n+1)/2

where cn = Γ((n + 1)/2)/π (n+1)/2 . ✩ This paper is partially supported by MTM2013-44357-P. The second author was partially supported by Swedish Research Council Grant 621-2011-3629. * Corresponding author. E-mail addresses: [email protected] (J.J. Betancor), [email protected] (A.J. Castro), [email protected] (J.C. Fariña), [email protected] (L. Rodríguez-Mesa).

http://dx.doi.org/10.1016/j.jmaa.2016.10.006 0022-247X/© 2016 Elsevier Inc. All rights reserved.

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Segovia and Wheeden [50] introduced fractional derivatives as follows. Suppose that β > 0 and m ∈ N is such that m − 1 ≤ β < m. If F : Ω × (0, ∞) −→ C is a reasonable nice function, where Ω ⊂ Rn , the β-th derivative with respect to t of F is defined by

∂tβ F (x, t)

e−iπ(m−β) = Γ(m − β)

∞ ∂tm F (x, t + s)sm−β−1 ds t > 0 and x ∈ Ω. 0

In [50] this fractional derivative was used to get characterizations of classical Sobolev spaces. As in [58] we define the β-conical square function Sβ by ⎞1/2   2 dydt ⎟ β β ⎜  Sβ (f )(x) = ⎝ t ∂t Pt (f )(y) n+1 ⎠ , t ⎛

x ∈ Rn ,

Γ(x)

where Pt (f ) denotes the Poisson integral of f , that is,  Pt (x − y)f (y)dy,

Pt (f )(x) =

x ∈ Rn , t > 0,

(1)

Rn

and, for every x ∈ Rn , Γ(x) = {(y, t) ∈ Rn × (0, ∞) : |x − y| < t}. According to [58, Theorems 5.3 and 5.4] the square function Sβ defines an equivalent norm in Lp (Rn ), 1 < p < ∞. Theorem A. Let 1 < p < ∞ and β > 0. Then, there exists C > 0 such that 1 f Lp (Rn ) ≤ Sβ (f )Lp (Rn ) ≤ Cf Lp (Rn ) , C

f ∈ Lp (Rn ).

(2)

The equivalence in Theorem A for β ∈ N can also be seen in [36,42,52]. Coifman, Meyer and Stein [20] introduced a family of spaces called tent spaces. These tent spaces are well adapted to certain questions related to harmonic analysis. Suppose that 1 ≤ p, q < ∞. The tent space Tpq (Rn ) consists of all those measurable functions g on Rn × (0, ∞) such that Aq (g) ∈ Lp (Rn ), where ⎛ ⎜ Aq (g)(x) = ⎝

⎞1/q

 |g(y, t)|q

dydt ⎟ ⎠ tn+1

,

x ∈ Rn .

Γ(x)

The norm  · Tpq (Rn ) in Tpq (Rn ) is defined by gTpq (Rn ) = Aq (g)Lp (Rn ) , g ∈ Tpq (Rn ). More recently Harboure, Torrea and Viviani [32] have simplified some proofs of properties in [20] by using vector valued harmonic analysis techniques. Note that the result in Theorem A can be rewritten in terms of tent spaces as follows. If 1 < p < ∞ and β > 0, then, for every f ∈ Lp (Rn ), tβ ∂tβ Pt (f ) ∈ Tp2 (Rn ) and 1 f Lp (Rn ) ≤ tβ ∂tβ Pt (f )Tp2 (Rn ) ≤ Cf Lp (Rn ) , C where C > 0 does not depend on f . Assume that B is a Banach space. In order to show a version of Theorem A for the Lebesgue–Bochner space Lp (Rn , B), the most natural definition of the β-conical square function Sβ,B is the following

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⎛ ⎜ Sβ,B (f )(x) = ⎝

3

⎞1/2

 tβ ∂tβ Pt (f )(y)2B

dydt ⎟ ⎠ tn+1

f ∈ Lp (Rn , B), 1 < p < ∞.

,

Γ(x)

This type of Banach valued conical function has been considered in [42,58,63]. Since a Banach space B has Lusin type 2 and Lusin cotype 2 if, and only if, B is isomorphic to a Hilbert space, from [58, Theorems 5.3 and 5.4] we can deduce the following result. Theorem B. Assume that B is a Banach space, 1 < p < ∞ and β > 0. The following assertions are equivalent: (i) B is isomorphic to a Hilbert space. (ii) There exists C > 0 such that 1 f Lp (Rn ,B) ≤ Sβ,B (f )Lp (Rn ) ≤ Cf Lp (Rn ,B) , C

f ∈ Lp (Rn , B).

In order to extend the equivalence (2) to Lp (Rn , B), 1 < p < ∞, when B is a Banach space which is not isomorphic to a Hilbert space, Hytönen, van Neerven and Portal [36] introduced new Banach valued tent function spaces. They considered UMD Banach spaces and γ-radonifying operators. As it is well-known, the Hilbert transform H defined by 1 H(f )(x) = lim+ ε→0 π

 |x−y|>ε

f (y) dy, x−y

a.e. x ∈ R,

is a bounded operator from Lp (R) into itself, 1 < p < ∞, and from L1 (R) into L1,∞ (R). Suppose that B is a Banach space. The Hilbert transform can be defined in Lp (R) ⊗ B, 1 < p < ∞, in the obvious way. It is said that B is UMD, provided that H(f )Lp (R,B) ≤ Cp f Lp (R,B) ,

f ∈ Lp (R) ⊗ B.

The main properties of UMD Banach spaces were established by Bourgain [18], Burkholder [19] and Rubio de Francia [49]. Assume that {γj }∞ j=1 is a sequence of independent standard normal variables defined on some probabilistic space (Ω, F, P). Let H be a Hilbert space and let B be a Banach space. We say that a linear operator T : H → B is γ-summing (shortly T ∈ γ ∞ (H, B)) when k 2 1/2

   T γ ∞ (H,B) = sup E γj T (hj ) < ∞, j=1

B

where the supremum is taken over all finite orthonormal family h1 , . . . , hk in H. Here, by E we denote the expectation with respect to P. The space γ ∞ (H, B) becomes a Banach space when it is endowed with the norm  · γ ∞ (H,B) . The space of γ-radonifying operators (shortly, γ(H, B)) is the closure in γ ∞(H, B) of the subspace spanned by the finite rank operators from H into B. According to [59, Proposition 3.15], if H is ∞ separable and {hj }∞ is an orthonormal basis in H, then T ∈ γ(H, B) if, and only if, the series γj T hj j=1 j=1

converges in L2 (Ω, B) and, in this case,

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∞ 2 1/2

   T γ ∞ (H,B) = E γj T hj  . j=1

B

We will write T γ(H,B) = T γ ∞ (H,B) , T ∈ γ(H, B). Hoffman-Jorgensen [33] and Kwapień [40] established that γ(H, B) = γ ∞ (H, B) provided that the Banach space B does not contain any closed subspace isomorphic to c0 . We recall that UMD Banach spaces satisfy this property. Suppose that (M, M, μ) is a measure space and H = L2 (M, M, μ). A function f : M → B is said to be weakly L2 when, for every L ∈ B∗ , the function L ◦ f ∈ H. Then, there exists a bounded and linear operator Tf : H → B (shortly Tf ∈ L (H, B)) such that, for every L ∈ B∗ , 

L, Tf (h)B∗ ,B =

L, f (t)B∗ ,B h(t)dμ(t), h ∈ H, M

provided that f is weakly L2 . We say that f ∈ γ(M, μ; B) provided that Tf ∈ γ(H, B). If B does not contain any closed subspace isomorphic to c0 , then γ(M, μ, B) is a dense subspace of γ(H, B) [40, Remark 2.16]. Assume that B is a UMD Banach space and 1 < p < ∞. Hytönen, van Neerven and Portal [36, Definition 4.1] defined the tent space Tp2 (Rn , B) as the completion of Cc∞ (Rn ×(0, ∞)) ⊗B, where Cc∞ (Rn ×(0, ∞)) denotes the space of smooth functions with compact support in Rn × (0, ∞), with respect to the norm f Tp2 (Rn ,B) = Jf Lp (Rn ,γ(H,B)) , where, from now on,

dydt H = L2 Rn × (0, ∞), n+1 , t the functional J is defined by J : f → [x → [(y, t) → χB(x,t) (y)f (y, t)]] and B(x, t) = {y ∈ Rn : |x − y| < t}, x ∈ Rn and t > 0. By taking into account that γ(H, C) ∼ = H, it is clear that Tp2 (Rn , C) = Tp2 (Rn ). Then, the tent space Tp2 (Rn , B) can be seen as a Banach valued extension of the classical tent space Tp2 (Rn ). The main properties of the space Tp2 (Rn , B) were established in [36]. An alternative and equivalent definition for tent spaces Tp2 (Rn , B) is given in [39]. In [36, Theorem 8.2] (see also the methods in [3] and [39, Example, Section 4]) it was proved a vectorial extension of (2) by using the tent spaces Tp2 (Rn , B). Theorem 1. Let 1 < p < ∞ and β > 0. Assume that B is a UMD Banach space. Then, there exists C > 0 such that 1 f Lp (Rn ,B) ≤ tβ ∂tβ Pt (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) , C

f ∈ Lp (Rn , B).

Since Tp2 (Rn , C) = Tp2 (Rn ) Theorem 1 is an extension of Theorem A to Lebesgue–Bochner spaces L (Rn , B), 1 < p < ∞, provided that B is a UMD. Note that a UMD Banach space is not necessarily isomorphic to a Hilbert space. We present in Section 2 a different proof of this result including some ideas that will be useful in the remainder of the paper. p

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We consider the Schrödinger operator LV in Rn defined by LV = −Δ + V, where Δ represents the usual Laplacian operator, that is, Δ =

n

∂2 j=1 ∂x2j .

We assume that the potential

V ≡ 0 is a nonnegative measurable function for which there exist s > n/2 and C > 0 such that, for every ball B in Rn , ⎛ ⎝ 1 |B|



⎞1/s V (x)s dx⎠

≤

B

C |B|

 V (x)dx.

(3)

B

When V satisfies (3) we say that V verifies the s-reverse Hölder inequality and we write V ∈ RHs (Rn ). In a precise way our Schrödinger operator LV is defined as follows. We consider the sesquilinear form QV given by 

 ∇f (x)∇g(x)dx +

QV [f, g] = Rn

V (x)f (x)g(x)dx,

(f, g) ∈ D(QV ),

Rn

where ∇ denotes the usual gradient. The domain D(QV ) of Q is the product DV × DV , where DV = {f ∈ L2 (Rn ) : |∇f | ∈ L2 (Rn ) and V 1/2 f ∈ L2 (Rn )}. The Schrödinger operator LV is the unique selfadjoint operator such that its domain is DV and

LV f, g = QV [f, g],

f, g ∈ DV .

It is clear that Cc∞ (Rn ), the space of smooth functions with compact support in Rn , is contained in DV and LV = LV on Cc∞ (Rn ). LV is a positive operator. The semigroup of operators {WtLV }t>0 generated by −LV in L2 (Rn ) can be written as  WtLV (f ) =

e−tλ ELV (dλ)f,

f ∈ L2 (Rn ) and t > 0,

[0,∞)

where ELV denotes the spectral measure for LV . For every t > 0 there exists a measurable function WtLV (x, y), x, y ∈ Rn , such that for each f ∈ L2 (Rn ),  WtLV (f )(x) =

WtLV (x, y)f (y)dy.

(4)

Rn

Moreover, according to the Feynman–Kac formula [25, p. 280], we have that 2   e−|x−y| /(4t)  LV  , x, y ∈ Rn and t > 0. Wt (x, y) ≤ C tn/2

Then, the integral in (4) is absolutely convergent for every f ∈ Lp (Rn ), 1 ≤ p ≤ ∞. The family {WtLV }t>0 , where WtLV , t > 0, is defined by (4), is a positive bounded semigroup in Lp (Rn ), 1 ≤ p < ∞, generated by −LV . {WtLV }t>0 is not Markovian because V ≡ 0. The semigroup of operators {PtLV }t>0 subordinated to {WtLV }t>0 , also called Poisson semigroup associated to LV , is defined, for every f ∈ Lp (Rn ), 1 ≤ p < ∞, by

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PtLV

t (f )(x) = √ 2 π

∞

e−t /(4u) LV Wu (f )(x)du, u3/2 2

x ∈ Rn and t > 0.

0

An important special case of Schrödinger operator is the Hermite operator H (also called harmonic oscillator) that appears when V (x) = |x|2 , x ∈ Rn . Harmonic analysis associated with Schrödinger and Hermite operators has been developed in the last years by several authors ([2,9,16,17,23–25,27,51,53,56], amongst others). Our second result establishes the equivalence in Theorem 1 when the classical Poisson semigroup is replaced by the Poisson semigroups {PtLV }t>0 or {PtH }t>0 . Theorem 2. Let 1 < p < ∞ and β > 0. Assume that B is a UMD Banach space. (i) If V ∈ RHs (Rn ) for some s > n/2, and n ≥ 3, then there exists C > 0 such that 1 f Lp (Rn ,B) ≤ tβ ∂tβ PtLV (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) , f ∈ Lp (Rn , B). C (ii) For every n ∈ N, there exists C > 0 such that 1 f Lp (Rn ,B) ≤ tβ ∂tβ PtH (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) , f ∈ Lp (Rn , B). C Since Tp2 (Rn , C) = Tp2 (Rn ), 1 < p < ∞, the scalar results can be deduced as special cases of Theorem 2. We now define the space Tp2 ((0, ∞), B), 1 < p < ∞, where B is again a UMD Banach space. Let 1 < p < ∞. The tent space Tp2 ((0, ∞), B) is the completion of Cc∞ ((0, ∞)2 ) ⊗ B with respect to the norm f Tp2 ((0,∞),B) = J+ f Lp ((0,∞),γ(H+ ,B)) , where, from now on,

dydt H+ = L2 (0, ∞)2 , 2 , t the functional J+ is defined by J+ : f → [x → [(y, t) → χB+ (x,t) (y)f (y, t)]] and B+ (x, t) = {y ∈ (0, ∞) : |x − y| < t}, x, t ∈ (0, ∞). Here, by Cc∞ ((0, ∞)2 ) we denote the space of smooth functions with compact support on (0, ∞)2 . We will use the tent space Tp2 ((0, ∞), B) to get equivalent norms in the Lebesgue–Bochner space p L ((0, ∞), B), for every 1 < p < ∞ and every UMD Banach space B. Our new norms (see Theorems 3 and 4 below) involve Poisson semigroups associated with Bessel and Laguerre operators. Harmonic analysis in the Bessel setting began with the deep paper of Muckhenhoupt and Stein [45]. Recently, operators related to the harmonic analysis (Riesz transform, Littlewood–Paley functions, maximal operators, multipliers, . . . ) in the Bessel context have been investigated (see, for instance, [6,10,11,14,61]). We consider the Bessel operator on (0, ∞), Bλ = −xλ Dx2λ Dx−λ = − where λ > 0. The Hankel transform hλ is defined by

d2 λ(λ − 1) + , 2 dx x2

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∞ hλ (f )(x) =

√

xyJλ−1/2 (xy)f (y)dy,

7

f ∈ L1 (0, ∞),

0

where Jα denotes the Bessel function of the first kind and order α. hλ plays with respect to the Bessel operator the same role as the Fourier transform with respect to the classical Laplacian operator. hλ can be extended from L1 (0, ∞) ∩ L2 (0, ∞) to L2 (0, ∞) as an isometry in L2 (0, ∞) [57, Ch. VIII]. By using well-known properties of the Bessel function Jα we can deduce that, for every f ∈ Cc∞ (0, ∞), hλ (Bλ f )(x) = x2 hλ (f )(x),

x ∈ (0, ∞),

[64, Lemma 5.4-1(5)]. We define the operator Bλ by Bλ (f ) = hλ (x2 hλ (f )),

f ∈ D(Bλ ),

where the domain D(Bλ ) of Bλ is given by D(Bλ ) = {f ∈ L2 (0, ∞) : y 2 hλ (f ) ∈ L2 (0, ∞)}. 2 ∞ ∞ Since h−1 λ = hλ in L (0, ∞), Cc (0, ∞) ⊂ D(Bλ ) and Bλ f = Bλ f , f ∈ Cc (0, ∞). The operator −Bλ Bλ p generates a positive and bounded semigroup of operators {Wt }t>0 in L (0, ∞), for every 1 ≤ p < ∞. We can write, for every f ∈ Lp (0, ∞), 1 ≤ p < ∞,

∞ WtBλ (f )(x)

WtBλ (x, y)f (y)dy, x ∈ (0, ∞),

= 0

where √ WtBλ (x, y) =

xy x2 +y2 xy Iλ−1/2 e− 4t , t, x, y ∈ (0, ∞), 2t 2t

and Iν represents the modified Bessel function of the first kind and order ν. Moreover, the Poisson semigroup {PtBλ }t>0 associated with the Bessel operator Bλ can be given by subordination as follows. For every f ∈ Lp (0, ∞), 1 ≤ p < ∞, PtBλ (f )(x)

t = √ 2 π

∞

e−t /(4u) Bλ Wu (f )(x)du, t, x ∈ (0, ∞). u3/2 2

0

Theorem 3. Let 1 < p < ∞ and β, λ > 0. Assume that B is a UMD Banach space. Then, there exists C > 0 such that 1 f Lp ((0,∞),B) ≤ tβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ≤ Cf Lp ((0,∞),B) , C

f ∈ Lp ((0, ∞), B).

Muckenhoupt ([43] and [44]) began the study of harmonic analysis associated to Laguerre operators. Later, Dinger [21] established Lp -boundedness properties for the maximal operator defined by the n-dimensional heat semigroup in the Laguerre context. In the last years a lot of authors have investigated harmonic analysis operators related to Laguerre operators (see, for instance, [12,22,28,31,32,47,54,59]). We consider the Laguerre operator on (0, ∞) Lα = −

d2 α2 − 1/4 + + x2 , dx2 x2

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where α > −1/2. For every k ∈ N, we have that α Lα ϕα k = 2(2k + α + 1)ϕk ,

where  ϕα k (x)

=

2Γ(k + 1) Γ(k + α + 1)

1/2

e−x

2

/2 α+1/2 α 2 x k (x ),

x ∈ (0, ∞),

α and α k represents the k-th Laguerre polynomial of order α [55, pp. 100–102]. The sequence {ϕk }k∈N is an orthonormal basis in L2 (0, ∞). We define the operator Lα as follows

Lα f = 2

∞

α (2k + α + 1)cα k (f )ϕk ,

f ∈ D(Lα ),

k=0

where, for every k ∈ N, ∞ cα k (f )

ϕα k (x)f (x)dx,

=

f ∈ L2 (0, ∞),

0

and the domain D(Lα ) of Lα is given by D(Lα ) = {f ∈ L2 (0, ∞) :

∞

2 (2k + α + 1)2 |cα k (f )| < ∞}.

k=0

The operator −Lα generates a positive and bounded semigroup {WtLα }t>0 in L2 (0, ∞), given by WtLα (f ) =

∞

α e−2(2k+α+1) cα k (f )ϕk ,

f ∈ L2 (0, ∞) and t > 0.

k=0

Mehler’s formula for Laguerre polynomials [56, p. 8] allows us to write, for each f ∈ L2 (0, ∞), ∞

WtLα (f )(x)

=

WtLα (x, y)f (y)dy,

t, x ∈ (0, ∞),

(5)

0

where WtLα (x, y)

 =

2e−2t 1 − e−4t

1/2 

2xye−2t 1 − e−4t

1/2

 Iα

2xye−2t 1 − e−4t



  −4t 1 2 2 1+e exp − (x + y ) . 2 1 − e−4t

(6)

Moreover, if WtLα is defined by (5), for every t > 0, then {WtLα }t>0 is a positive and bounded semigroup of operators in Lp (0, ∞), 1 ≤ p < ∞. As usual the Poisson semigroup {PtLα }t>0 associated with Lα is defined as the one subordinated to {WtLα }t>0 , that is, for every t > 0, PtLα (f )(x)

t = √ 2 π

∞ 0

e−t /(4u) Lα Wu (f )(x)du, u3/2 2

f ∈ Lp (0, ∞) and 1 ≤ p < ∞.

(7)

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Theorem 4. Let 1 < p < ∞, α > −1/2 and β > 0. Assume that B is a UMD Banach space. Then, there exists C > 0 such that 1 f Lp ((0,∞),B) ≤ tβ ∂tβ PtLα (f )Tp2 ((0,∞),B) ≤ Cf Lp ((0,∞),B) , C

f ∈ Lp ((0, ∞), B).

√ √ √ 2 Re z We define, for every β > 0, ϕβ (z) = ( z 2 )β e− z , z ∈ C \ {0}, where z 2 = |Re z| z, z ∈ C \ {0}. Thus, the pair (ϕβ , 0) has sufficient decay in the sense of [36, Definition 7.6] provided that β > n/2. Then, by using [36, Theorem 7.10] we can deduce from Theorem 2 the following result.

Theorem 5. Let B be a U M D Banach space, 1 < p < ∞ and β > n/2. (i) If V ∈ RHs (Rn ) for some s > n/2, and n ≥ 3, then there exists C > 0 such that 1 f Lp (Rn ,B) ≤ tβ ∂tβ WtLV (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) , f ∈ Lp (Rn , B). C (ii) For every n ∈ N, there exists C > 0 such that 1 f Lp (Rn ,B) ≤ tβ ∂tβ WtH (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) , f ∈ Lp (Rn , B). C Actually the results in (i) and (ii) in Theorem 5 hold true for every β > 0. In order to prove them when 0 < β ≤ n/2, by using our comparative procedure we need sufficient good estimates for ∂tβ WtLV (x, y), t > 0 and x, y ∈ Rn , and ∂tβ (WtLV (x, y) − Wt (x − y)), t > 0 and x, y ∈ Rn , playing the role of [23, (2.3)] and (23), respectively. These estimates can be deduced from [23, (2.3)] and (23), by using Cauchy integral formula and by taking into account that {WtLV }t>0 can be extended to an analytic semigroup {WzLV }z∈Ω , where Ω = {z ∈ C : |Arg(z)| < π/4}. However, note that by using subordinated semigroup {PtLV }t>0 , as we make in Theorem 2, the derivative ∂tβ applies on the exponential functions and not on the heat kernel WtLV (x, y) and so, it is sufficient to consider the simpler estimations (20), [23, (2.3)] and (23). In order to prove heat semigroup versions of Theorems 3 and 4 we can proceed in two ways. One of them is to establish the results in [36, Section 7] ([36, Theorem 7.10]) for Tp2 ((0, ∞), B) and then change the square functions. The second method is to use our comparative procedure. In this way the use of the Poisson semigroups has advantages with respect to the one of the heat semigroups. Indeed, by using Poisson semigroups and subordination formulas, the derivatives are not acting on Bessel functions but on the simpler exponential functions. If we work with conical square functions associated with heat semigroups for Bessel and Laguerre operators we need to make hard manipulations involving many terms of the asymptotic expressions for the Bessel functions to get suitable cancellations in the local parts. When conical square functions associated with heat semigroups (with Gaussian bounds) are considered it is usual to introduce a correction motivated by the difference of homogeneities (parabolic setting). For instance, the square function of one order defined by the classical heat semigroup {Wt}t>0 is given usually by

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|t2 (∂s Ws (f )(y))|s=t2 |2

s(f )(x) =

dydt 1/2 . tn+1

Γ(x)

After changing the variable we get



s(f )(x) = √ |x−y|< t

|t∂t Wt (f )(y)|2

dydt 1/2 . 2t1+n/2

√ Note that the square function s is associated with a parabolic tent space and s = ψ(t −Δ), where ψ(z) = 2 z 2 e−z , z ∈ C. Hence, these parabolic square functions are different from the ones that appear in [36, p. 319]. The same words are in order for the conical square functions associated with heat semigroups for Schrödinger (Theorem 5), Bessel and Laguerre (see also, for instance, [34] and [39]). Area Littlewood–Paley functions estimates for Schrödinger, Bessel and Laguerre operators have been studied in the scalar setting (see [7,15,48,54]). But our Theorems 2, 3 and 4 give, when they are reduced to the scalar case, new results (concerning to Poisson semigroups and other orthogonal systems). As it is well known Littlewood–Paley functions are closely connected with H ∞ -functional calculus. In particular, the boundedness of the functional calculus of an operator A ⊗ IB can be deduced from conical square function estimates. From [36, Theorem 7.10] and Theorem 2 it follows that the operator LV ⊗ IB has H ∞ -functional calculus (in the sense of [36]) in Lp (Rn , B), 1 < p < ∞. Also, after establishing a version of [36, Theorem 7.10] for the space Tp2 ((0, ∞), B), by Theorems 3 and 4 we could prove that the operators Bλ ⊗ IB , λ > 0, and Lα ⊗ B, α > −1/2, have H ∞ -functional calculus in Lp ((0, ∞), B), 1 < p < ∞. A Banach space B is U M D if, and only if, the Laplacian Δ ⊗ IB has H ∞ calculus in Lp ((0, ∞), B), 1 < p < ∞. The arguments developed in this paper lead to the same characterization of the U M D property for Banach spaces when the Laplacian operator is replaced by Schrödinger (LV ), Bessel (Bλ ) or Laguerre (Lα ) operators. Note that Banach spaces B having U M D property can be characterized by the Lp (B)-boundedness of the imaginary power Ais , s ∈ R, where A represents Schrödinger, Bessel or Laguerre operators (see [5] and [8]). On the other hand, vertical square functions estimates associated with operators have been widely investigated and used in harmonic analysis. However, in our Banach valued setting conical square functions have some advantages with respect to the vertical ones. In [35], Hytönen studied Littlewood–Paley–Stein theory for semigroups in U M D Banach spaces by using stochastic integration. In order to obtain the equivalence of vertical square function estimates for the Lp norm in U M D Banach spaces he considered subordinated (Poisson) semigroups [35, Theorem 1.6]. This property can be established for symmetric diffusion (not necessarily subordinated) semigroups when the Banach space B is a closed subspace of a complex interpolation space [H, Y ]θ , where Y is a U M D Banach space, H is a Hilbert space, and θ ∈ (0, 1) [35, Theorem 9.7]. The interesting question posed in [49] whether the above interpolation class of Banach spaces (that is contained in the U M D class) exhausts all U M D spaces so far as we know is open. For obtaining conical square function estimates for diffusion semigroups any restriction of the class of U M D spaces is needed (see Theorem 5 and [36]). In [38, Proposition 7.7] Kalton and Weis established the equivalence of the vertical square function estimates associated with operators and Lp -norms. But they impose restrictions on the class of operators under consideration related to Lp (R-) and γ-sectoriality. These properties are not needed to establish the corresponding results in our cases by using conical square functions. In the following sections we prove Theorems 1, 2, 3 and 4. Throughout this paper by C and c we always denote positive constants that can change in each occurrence.

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11

2. Proof of Theorem 1 We split the proof of Theorem 1 in the Lemmas 2.1 and 2.3 below. We are going to use the arguments developed in [36, Theorem 8.2]. Lemma 2.1. Let B be a UMD Banach space, 1 < p < ∞ and β > 0. Then, there exists C > 0 such that tβ ∂tβ Pt (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) ,

f ∈ Lp (Rn , B).

(8)

Proof. Let f ∈ Lp (Rn , B). It is not hard to see that, for every k ∈ N,  ∂tk Pt (x − y)f (y)dy,

∂tk Pt (f )(x) =

x ∈ Rn and t > 0.

Rn

Assume that m ∈ N is such that m − 1 ≤ β < m. We have that ∂tβ Pt (f )(x)

e−iπ(m−β) = Γ(m − β) =

e−iπ(m−β) Γ(m − β) 

∞

 sm−β−1 Rn

0



∞ sm−β−1 ∂tm Pt+s (x − y)dsdy

f (y) Rn

0

∂tβ Pt (x − y)f (y)dy,

=

∂tm Pt+s (x − y)f (y)dyds

x ∈ Rn and t > 0.

Rn

The interchange in the order of integration is justified because ∞  f (y)B sm−β−1 |∂tm Pt+s (x − y)|dyds < ∞,

x ∈ Rn and t > 0.

(9)

0 Rn

Indeed, according to Fa di Bruno’s formula [29, (4.6)] we can write, for each n ≥ 2, t > 0 and z ∈ Rn ,  ∂tm

   1 t 1 m+1 = ∂ 1−n t (t2 + |z|2 )(n+1)/2 (t2 + |z|2 )(n−1)/2

1 = 1−n

(m+1)/2

=0

−1 m+1− tm+1−2 (n − 1)(n + 1) · . . . · (n − 1 + 2(m − ))Em, 2 , 2 (t + |z|2 )(n+1+2(m−))/2 (10)

and  ∂tm

 (m+1)/2 1 m+1 1 tm+1−2 t 2 2 = ∂ ln(t + |z| ) = (−1)m− (m − )!Em, 2 , t 2 2 t + |z| 2 2 (t + |z|2 )m+1− =0

where Em, = 2m+1−2 (m + 1)!/(!(m + 1 − 2)!). From (10) and (11) we deduce that      m 1 t ≤C ∂t ,   2 2 (n+1)/2 (t + |z|)m+n (t + |z| )

z ∈ Rn and t > 0.

(11)

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12

Then, ∞ m−β−1

s 0

  ∞  m  t+s sm−β−1 ∂t  ds ≤ C ds  ((t + s)2 + |x − y|2 )(n+1)/2  (t + s + |x − y|)m+n 0

≤C

1 , (t + |x − y|)n+β

x, y ∈ Rn , t > 0,

and Hölder inequality allows us to obtain (9). Also, we have that |tβ ∂tβ Pt (x − y)| ≤ C

tβ , (t + |x − y|)n+β

x, y ∈ Rn , t > 0.

(12)

To simplify we write  (Sf )(t, y) =

k(y, z, t)f (z)dz,

f ∈ Lp (Rn , B),

Rn

where k(y, z, t) = tβ ∂tβ Pt (y − z), y, z ∈ Rn and t > 0. Our objective is to see that S is a bounded operator from Lp (Rn , B) into Tp2 (Rn , B). In order to do this we use [36, Theorem 4.8]. We consider the operator  (Sg)(t, y) =

k(y, z, t)g(z)dz,

g ∈ L2 (Rn ).

Rn

As usual, for every g ∈ L2 (Rn ), we denote by g the Fourier transform of g and by gq the inverse Fourier transform of g. Let g ∈ L2 (Rn ). It is well-known that ∂tk Pt (g)(y) = ((−1)k |z|k e−t|z| g)q,

k ∈ N.

Since (−1)k |z|k e−t|z| g ∈ L1 (Rn ), t > 0 and k ∈ N, we can write ∂tk Pt (g)(y) =

(−1)k (2π)n/2



eiyz |z|k e−t|z| g(z)dz.

Rn

Hence, ∂tβ Pt (g)(y) =

(−1)m e−iπ(m−β) (2π)n/2 Γ(m − β)

eiπβ = (2π)n/2 Γ(m − β) iπβ

=

e (2π)n/2



∞

 sm−β−1 Rn

0

 e Rn

eiyz |z|m e−(t+s)|z| g(z)dzds

iyz

m −t|z|

|z| e

∞ g(z)

e−s|z| sm−β−1 dsdz

0

eiyz |z|β e−t|z| g(z)dz = eiπβ (|z|β e−t|z| g(z))q.

(13)

Rn

The interchange of the order of integration is justified by the absolute convergence of the integral. Thus, Plancherel equality leads to

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J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

  Sg2T22 (Rn )

= Rn Γ(x)

 

=

dydt |(Sg)(y, t)|2 n+1 dx t

  |tβ ∂tβ Pt (g)(y)|2

= Rn Γ(x)

|(e−t|z| (t|z|)β g(z))q(y)|2

Rn Γ(x)

∞ 



0 Rn B(y,t)

∞  |(e

= vn 0 Rn

= vn

Γ(2β) 22β

−t|z|

q

 | g (z)|2 dz = vn Rn

dxdydt tn+1

2 dydt

(t|z|) g(z)) (y)| β

dydt dx tn+1

dydt dx tn+1

|(e−t|z| (t|z|)β g(z))q(y)|2

=

13

t

∞  = vn

|e−t|z| (t|z|)β g(z)|2

0 Rn

dzdt t

Γ(2β) g2L2 (Rn ) , 22β

where vn is the volume of the unit ball in Rn . Hence, S is a bounded operator from L2 (Rn ) into T22 (Rn ). We now prove that |∇z k(y, z, t)| ≤ C

tβ , (t + |y − z|)n+β+1

y, z ∈ Rn and t > 0.

(14)

Let j = 1, . . . , n. According to (10) and (11) we obtain  ∂j ∂tm

 (m+1)/2 tm+1−2l zj t = a , l (t2 + |z|2 )(n+1)/2 (t2 + |z|2 )(n+3+2(m−l))/2 =0

z ∈ Rn and t > 0,

for certain al ∈ R. Then,     β tβ t ≤C t ∂j ∂tm ,  (t + |z|)n+m+1 (t2 + |z|2 )(n+1)/2 

z ∈ Rn and t > 0.

By proceeding as in the proof of (12) we get |∂j k(y, z, t)| ≤ C

tβ , (t + |y − z|)n+1+β

y, z ∈ Rn and t > 0.

Thus (14) is established. From (14) and by using the mean value theorem we can deduce that |k(y, z, t) − k(y, z , t)| ≤ C

tβ |z − z | , (t + |y − z|)n+β+1

provided that t > 0, y, z, z ∈ Rn and |y − z| + t > 2|z − z |. Moreover, we can write 

 k(y, z, t)dz = tβ ∂tβ

Rn

Pt (y − z)dz = 0,

t > 0 and y ∈ Rn .

Rn

According to [36, Theorem 4.8], the operator SB = S ⊗ IB can be extended from Cc∞ (Rn ) ⊗ B to Lp (Rn , B) B from Lp (Rn , B) into T 2 (Rn , B). Also, as a special case, the operator S can be as a bounded operator S p

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14

C from Lp (Rn ) into Tp2 (Rn ). We are going to extended from Cc∞ (Rn ) to Lp (Rn ) as a bounded operator S p n B on L (R , B). show that S = S C = S on Lp (Rn ), where In order to get a better understanding of the proof we see firstly that S  (Sg)(t, y) =

g ∈ Lp (Rn ).

k(y, z, t)g(z)dz, Rn

Let f ∈ Lp (Rn ). We choose a sequence (fk )k∈N in Cc∞ (Rn ) such that fk −→ f,

as k → ∞, in Lp (Rn ).

Then, C f, Sfk −→ S

as k → ∞, in Tp2 (Rn ).

Hence, (J(Sfk ))k∈N converges to a function g ∈ Lp (Rn , H). There exists an increasing sequence (k )∈N ⊂ N and a subset Ω ∈ Rn such that |Rn \ Ω| = 0 and, for every x ∈ Ω, [J(Sfk )](x) −→ g(x), as  → ∞, in H.

(15)

On the other hand, according to (12) we have that, for every  ∈ N,      β S(f − fk )(t, x) ≤ |f (y) − fk (y)||tβ ∂t Pt (y − x)|dy Rn



≤C

|f (y) − fk (y)| Rn

tβ dy (t + |x − y|)n+β ⎛



≤ Cf − fk Lp (Rn ) tβ ⎝

Rn

≤ Ct−n/p f − fk Lp (Rn ) ,

⎞1/p 1 dz ⎠ (t + |z|)(n+β)p x ∈ Rn and t > 0.

(16)

Here p = p/(p − 1). From (16) we deduce that, for each  ∈ N and ε > 0,  2   [JS(fk − f )](x) 2 L

∞  (Rn ×(ε,∞), tdydt n+1

)

≤ Cfk −

f 2Lp (Rn )

t−2n/p

dydt tn+1

ε B(x,t)

≤ Cfk − f 2Lp (Rn ) ε−2n/p ,

x ∈ Rn .

Then, for every ε > 0 and x ∈ Rn , J(Sfk )(x) −→ J(Sf )(x),

dydt as  → ∞, in L2 Rn × (ε, ∞), n+1 . t

(17)

From (15) and (17) it follows that, for every x ∈ Ω, J(Sf )(x) = g(x) as elements of H. We conclude that J(Sfk ) −→ J(Sf ), C f . Hence, Sf = S

as k → ∞, in Lp (Rn , H).

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15

Almost the same proof works for every f ∈ Lp (Rn , B). Let f ∈ Lp (Rn , B). We choose a sequence (fk )k∈N ⊂ ⊗ B such that

Cc∞ (Rn )

fk −→ f,

as k → ∞, in Lp (Rn , B).

Then, B f, SB fk −→ S

as k → ∞, in Tp2 (Rn , B).

Hence, there exists G ∈ Lp (Rn , γ(H, B)) such that J(SB fk ) −→ G,

as k → ∞, in Lp (Rn , γ(H, B)).

There exists an increasing sequence (k )∈N ⊂ N and a set Ω ⊂ Rn such that |Rn \ Ω| = 0 and J(SB fk )(x) −→ G(x),

as  → ∞, in γ(H, B),

for every x ∈ Ω. By proceeding as above we can see that, for every ε > 0 and x ∈ Rn ,

dydt J(SB fk )(x) −→ J(Sf )(x), as  → ∞, in L2 Rn × (ε, ∞), n+1 ; B . t Since γ(H, B) is continuously contained in the space L (H, B) of bounded linear operators from H into B, we have that, for every x ∈ Ω, J(SB fk )(x) −→ G(x),

as  → ∞, in L (H, B).

Assume that h ∈ Cc∞ (Rn × (0, ∞)) and L ∈ B∗ . Then, for every x ∈ Ω,

L, [J(SB fk )(x)](h)B∗ ,B −→ L, [G(x)](h)B∗ ,B ,

as  → ∞,

being  [J(SB fk )(x)](h) =

[J(SB fk )(x)](y, t)h(y, t) Rn ×(0,∞)

dydt , tn+1

 ∈ N.

s Suppose that g = m=1 bm gm , where s ∈ N, bm ∈ B and gm ∈ Cc∞ (Rn ), m = 1, . . . , s. There exists a set W ⊂ Rn such that |Rn \ W | = 0 and, for every x ∈ W , [J(SB g)(x)](y, t) = χB(x,t) (y, t)(SB g)(y, t) =

s

bm χB(x,t) (y, t)(Sgm )(y, t),

(y, t) ∈ Rn × (0, ∞),

m=1



is in L2 Rn × (0, ∞), tdydt and n+1 ; B

L, [J(SB g)(x)](h)

B∗ ,B

=

s



L, bm 

B∗ ,B

m=1



(Sgm )(y, t)h(y, t)

dydt tn+1

Γ(x)

L, [J(SB g)(x)](y, t)B∗ ,B h(y, t)

= Rn ×(0,∞)

dydt , tn+1

x ∈ W.

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16

For every  ∈ N, we denote by W the set in Rn associated with fk as above. We define Ω0 = Ω∩(∩∈N W ). We have that |Rn \ Ω0 | = 0. Let x ∈ Ω0 . We can write

L, [G(x)](h)B∗ ,B = lim L, [J(SB fk )(x)](h)B∗ ,B →∞  dydt = lim

L, [J(SB fk )(x)](y, t)B∗ ,B h(y, t) n+1 →∞ t Rn ×(0,∞)



L, [J(Sf )(x)](y, t)B∗ ,B h(y, t)

= Rn ×(0,∞)

dydt . tn+1

We conclude that L, [J(Sf )(x)](·, ·)B∗ ,B ∈ H, x ∈ Ω0 . Then, for every x ∈ Ω0 , G(x) = [J(Sf )(x)](·, ·), as elements of γ(H, B). B f . Thus, (8) is proved. 2 Hence, Sf = S Before establishing the first inequality in Theorem 1 we need the following polarization formula. Lemma 2.2. Let f ∈ Cc∞ (Rn ) ⊗ B and g ∈ Cc∞ (Rn ) ⊗ B∗ . Then,  

tβ ∂tβ Pt (g)(y), tβ ∂tβ Pt (f )(y)B∗ ,B Rn

dydt Γ(2β) dx = vn 2β n+1 t 2

Γ(x)



g(y), f (y)B∗ ,B dy, Rn

where vn denotes the volume of the unit ball in Rn . Proof. Suppose firstly that f, g ∈ Cc∞ (Rn ). According to (13) and by using Plancherel equality we get  

dydt tβ ∂tβ Pt (f )(y)tβ ∂tβ Pt (g)(y) n+1 dx

Rn Γ(x)

t

 

=

  tβ ∂tβ Pt (f )(y)tβ ∂tβ Pt (g)(y)

= Rn Γ(x)

[(|z|t)β e−t|z| f]q(y)[(|z|t)β e−t|z| g]q(y)

Rn Γ(x)

∞ 

2β −2t|z|

(|z|t) e

= vn 0 Rn

dydt dx tn+1

dydt dx tn+1

dzdt Γ(2β) = vn 2β g (z) f(z) t 2

 f (y)g(y)dy.

(18)

Rn

The interchanges in the order of integration are justified because the integrals are absolutely convergent. In order to see this it is sufficient recall that the Littlewood–Paley–Stein function defined by ⎛

⎞1/2

∞

gβ (f )(x) = ⎝

|tβ ∂tβ Pt (f )(x)|2 0

is bounded from L2 (Rn ) into itself. From (18) we easily conclude the proof of this lemma. 2

dt ⎠ t

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17

Lemma 2.3. Let B be a UMD Banach space, 1 < p < ∞ and β > 0. Then, there exists C > 0 such that f Lp (Rn ,B) ≤ Ctβ ∂tβ Pt (f )Tp2 (Rn ,B) ,

f ∈ Lp (Rn , B).

Proof. Let f ∈ Cc∞ (Rn ) ⊗ B and g ∈ Cc∞ (Rn ) ⊗ B∗ . Since B is a UMD Banach space, B∗ has also the UMD property. According to Lemma 2.1, Lemma 2.2 and [37, Proposition 2.4] we can write        g(y), f (y)B∗ ,B dy     n  R     dydt   β β ≤C  χB(x,t) (y, t)tβ ∂t Pt (g)(y), χB(x,t) (y, t)tβ ∂t Pt (f )(y)B∗ ,B  n+1 dx t Rn Rn ×(0,∞)



χB(x,t) (y, t)tβ ∂tβ Pt (f )(y)γ(H,B) χB(x,t) (y, t)tβ ∂tβ Pt (g)(y)γ(H,B∗ ) dx

≤C Rn

≤ Ctβ ∂tβ Pt (f )(y)Tp2 (Rn ,B) tβ ∂tβ Pt (g)(y)T 2 (Rn ,B∗ ) p

≤

Ctβ ∂tβ Pt (f )(y)Tp2 (Rn ,B) gLp (Rn ,B∗ ) .

Then, since Cc∞ (Rn ) ⊗ B is a dense subspace of Lp (Rn , B), by taking into account Lemma 2.1 we conclude the proof of this lemma. 2 3. Proof of Theorem 2, (i) 3.1. Suppose that V ∈ RHs (Rn ), where s > n/2 and n ≥ 3. We are going to show that, for every f ∈ Lp (Rn , B), tβ ∂tβ PtLV (f )Tp2 (Rn ,B) ≤ Cf Lp (Rn ,B) . In order to see this we cannot proceed as in the proof of Lemma 2.1. In this Schrödinger case [36, Theorem 4.8] does not apply because ∂t PtLV (1) ≡ 0. Note that, since lim+ PtLV (1)(x) = 1, x ∈ Rn , if ∂t PtLV (1)(x) = 0, t→0

t > 0 and x ∈ Rn , then PtLV (1)(x) = 1, t > 0 and x ∈ Rn , and we could infer that 0 = (∂t − LV )PtLV (1)(x) = −V (x)PtLV (1)(x) = −V (x). Hence, since V ≡ 0, we have that ∂t PtLV (1)(x) ≡ 0. We define, for every x ∈ Rn , ⎧ ⎪ ⎨ 1 ρ(x) = sup r > 0 : n−2 ⎪ r ⎩

⎫ ⎪ ⎬

 V (y)dy ≤ 1 B(x,r)

⎪ ⎭

.

The function ρ (usually called critical radius) plays an important role in the development of the harmonic analysis associated with LV (see, for instance, [23,25,27,51]). The main properties of ρ are described in [51, Section 1]. m Suppose that f = j=1 bj fj , where bj ∈ B and fj ∈ Cc∞ (Rn ), j = 1, . . . , m. It is clear that tβ ∂tβ PtLV (f )(y) =

m j=1

bj tβ ∂tβ PtLV (fj )(y),

t > 0 and y ∈ Rn .

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J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

18

Moreover, for almost every x ∈ Rn , we have that χΓ(x) (y, t)tβ ∂tβ PtLV (f )(y) ∈ H ⊗ B. Indeed, let g ∈ L2 (Rn ), we can write     ∞  2 dtdy 2 dtdy  β β LV  β β LV   , t ∂t Pt (g)(y) n+1 = vn t ∂t Pt (g)(y) t t

Rn Γ(x)

Rn 0

where once again vn denotes the volume of the unit ball in Rn . Then, according to [1, Theorem A], it follows that    2 dydt  β β LV  t ∂t Pt (g)(y) n+1 dx ≤ Cg2L2 (Rn ) . t Rn Γ(x)

Hence, for almost x ∈ Rn , tβ ∂tβ PtLV (g)(y)χΓ(x) (t, y) ∈ H ⊗ B. To simplify, we write K LV (f )(x; y, t) = tβ ∂tβ PtLV (f )(y)χΓ(x) (y, t),

x, y ∈ Rn and t > 0.

We decompose the operator K LV as follows: LV LV K LV (f )(x; y, t) = Kloc (f )(x; y, t) + Kglob (f )(x; y, t),

x, y ∈ Rn and t > 0,

where LV Kloc (f )(x; y, t) = K LV (f χB(x,ρ(x)) )(x; y, t),

x, y ∈ Rn and t > 0.

LV LV The operators Kloc and Kglob are usually called local and global part of K LV , respectively. We also consider the operators

K(f )(x; y, t) = tβ ∂tβ Pt (f )(y, t)χΓ(x) (y, t),

x, y ∈ Rn and t > 0,

Kloc (f )(x; y, t) = K(f χB(x,ρ(x)) )(x; y, t),

x, y ∈ Rn and t > 0.

and

We can write LV K LV (f )(x; y, t) = DLV (f )(x; y, t) + Kloc (f )(x; y, t) + Kglob (f )(x; y, t),

x, y ∈ Rn and t > 0,

(19)

where LV DLV (f )(x; y, t) = Kloc (f )(x; y, t) − Kloc (f )(x; y, t),

x, y ∈ Rn and t > 0.

The three terms in the right hand side of (19) are studied separately in the following lemmas. Lemma 3.1. Let B be a Banach space and 1 < p < ∞. Then, LV Kglob (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) ,

f ∈ Cc∞ (Rn ) ⊗ B.

m Proof. Let f = j=1 bj fj , where bj ∈ B and fj ∈ Cc∞ (Rn ), j = 1, . . . , m. Firstly, we show that for every LV x ∈ Rn , Kglob (f )(x; ·, ·) ∈ γ(H, B). Fix x ∈ Rn . According to the subordination formula, we have that

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J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

PtLV

t (y, z) = √ 4π

∞

e−t /(4u) LV Wu (y, z)du, u3/2

19

2

y, z ∈ Rn and t > 0.

0

By [4, Lemma 4] |∂tβ [te−t

2

/(4u)

]| ≤ Ce−t

2

/(8u) (1−β)/2

u

,

t, u ∈ (0, ∞).

(20)

Also, by taking into account the Feynman–Kac formula [23, (2.2)] and (20) we can interchange the order of integration and differentiate under the integral sign to get ∂tβ PtLV

1 (y, z) = √ 4π

∞

∂tβ [te−t /(4u) ] LV Wu (y, z)du, u3/2 2

y, z ∈ Rn and t > 0.

0

Then, by using [23, (2.3)] and (20) we obtain |tβ ∂tβ PtLV

∞ (y, z)| ≤ Ct ρ(y) β

e−t

2

/(8u) −(3+β+n)/2 −c|y−z|2 /u

u

e

du

0

≤C

tβ ρ(y) , (t + |y − z|)n+β+1

y, z ∈ Rn and t > 0.

(21)

Estimation (21) justifies the differentiation under the integral sign, and we get 

   LV  Kglob (f )(x; y, t) ≤ Ctβ ρ(y)χΓ(x) (y, t) B

Rn \B(x,ρ(x))

f (z)B dz, (t + |y − z|)n+β+1

x, y ∈ Rn and t > 0.

We observe that, if x ∈ Rn and (y, t) ∈ Γ(x), then, for certain C > 0, independent of x, y and t we have that C 1 ≤ . n+β+1 (t + |y − z|) (t + 2|x − y| + |y − z|)n+β+1 Thus, we can write,    LV  Kglob (f )(x; y, t)

B

⎛



⎜ ≤ Ctβ ρ(y)χΓ(x) (y, t)f Lp (Rn ,B) ⎝

Rn \B(x,ρ(x))

⎛



⎜ ≤ Ctβ ρ(y)χΓ(x) (y, t)f Lp (Rn ,B) ⎝

Rn \B(x,ρ(x))

⎛ ⎜ ≤ Ctβ ρ(y)χΓ(x) (y, t)f Lp (Rn ,B) ⎝

∞

⎞1/p dz ⎟ ⎠ (t + 2|x − y| + |y − z|)(n+β+1)p ⎞1/p dz ⎟ ⎠ (t + |x − y| + |z − x|)(n+β+1)p ⎞1/p

rn−1 dr ⎟ ⎠ (t + |x − y| + r)(n+β+1)p

ρ(x)

≤C

tβ ρ(y)χΓ(x) (y, t) f Lp (Rn ,B) , (t + |x − y| + ρ(x))n+β+1−n/p

x, y ∈ Rn and t > 0.

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J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

20

According to [23, Proposition 1] we deduce that  ρ(y) ≤ Cρ(x) 1 +

t ρ(x)

γ ,

|x − y| < t,

x, y ∈ Rn and t > 0,

(22)

for some 1/2 ≤ γ < 1. Then,  dtdy LV Kglob (f )(x; y, t)2B n+1 t Rn ×(0,∞)

∞ ≤ Cρ(x)

2



f 2Lp (Rn ,B) 0 |x−y|
∞ ≤ Cρ(x)

2

f 2Lp (Rn ,B)

t2β−n−1 (t + |x − y| + ρ(x))2n+2β+2−2n/p

t2β−1 (t + ρ(x))2n+2β+2−2n/p



t 1+ ρ(x)

 1+

t ρ(x)

2γ dydt

2γ dt

0

≤ Cρ(x)

−2n/p

f 2Lp (Rn ,B) ,

x ∈ Rn .

Here we have used that 0 < γ < 1. Since 0 < ρ(z) < ∞, z ∈ Rn [51, p. 519], we conclude that

LV dydt Kglob (f )(x; ·, ·) ∈ L2 Rn × (0, ∞), tn+1 ; B , x ∈ Rn . LV LV To simplify we write G(x) = Kglob (f )(x; ·, ·) and Gj (x) = Kglob (fj )(x; ·, ·), x ∈ Rn and j = 1, . . . , m. We denote by TG(x) the linear and bounded operator from H into B defined by

 LV Kglob (f )(x; y, t)h(y, t)

TG(x) (h) = Rn ×(0,∞)

dydt , tn+1

h ∈ H.

For every j = 1, . . . , m, TGj is defined in a similar way. For every S ∈ B∗ , we have that  dydt LV

S, TG(x) (h)B∗ ,B =

S, Kglob (f )(x; y, t)B∗ ,B h(y, t) n+1 , t

h ∈ H.

Rn ×(0,∞)

Moreover, it is clear that TG(x) =

m

bj TGj (x) .

j=1

Then, for every h ∈ H and S ∈ B∗ the function Rn

−→ C

x

−→ S, TG(x) (h)B∗ ,B =

n

S, bj B∗ ,B TGj (x) (h) j=1

is measurable. According to [60, Lemma 2.5] and Pettis measurability theorem [60, Proposition 2.1] the function Rn x is strongly measurable.

−→ γ(H, B) −→ G(x)(≡ TG(x) )

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.21 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

21

∞ Suppose that {hj }∞ j=1 is an orthonormal basis in H and {γj }j=1 is a sequence of independent standard normal variables on a probability space (Ω, F, P ). We have that ⎛ ⎞1/2 ∞  2   G(x)γ(H,B) = ⎝E γj TG(x) (hj ) ⎠ B

j=1

2 ⎞1/2 ⎛      ∞  dydt   ⎟ ⎜  β β LV = ⎝E  γj f (z) χΓ(x) (y, t)t ∂t Pt (y, z)hj (y, t) n+1 dz  ⎠ ,   t  j=1 Rn \B(x,ρ(x))  Rn ×(0,∞)

x ∈ Rn .

B

The interchange of the order of integration is justified because by proceeding as above we can show that the integrals are norm convergent. Then, from (21) and (22) it follows that   ∞

 dydt 2 1/2  G(x)γ(H,B) ≤ f (z)B E γj tβ ∂tβ PtLV (y, z)hj (y, t) n+1  dz t j=1 Rn \B(x,ρ(x))

Γ(x)



    f (z)B tβ ∂tβ PtLV (y, z)χΓ(x) (y, t) dz

≤C

H

Rn \B(x,ρ(x))

 f (z)B

≤C



2γ ρ(x)2 1 + t/ρ(x) t2β−n−1



Rn \B(x,ρ(x))

(t + |y − z|)2(n+β+1)

dz

Γ(x)

 ≤ Cρ(x)

f (z)B

Rn \B(x,ρ(x))

f (z)B

≤ Cρ(x)





1/2

t2β−n−1

dtdy (t + 2|x − y| + |y − z|)2(n+β+1)

2γ

∞ 1 + t/ρ(x) t2β−1

dz

0



ρ(x)



f (z)B

Rn \B(x,ρ(x))

0

1/2

dt (t + |x − z|)2(n+β+1)

Rn \B(x,ρ(x))

≤ Cρ(x)

2γ 1 + t/ρ(x)

Γ(x)



∞

1/2 dtdy

dz

t2β−1 dt (t + |x − z|)2(n+β+1)

t2γ+2β−1 dt 1/2 dz (t + |x − z|)2(n+β+1) ρ(x)2γ

+ ρ(x)



≤ Cρ(x)

f (z)B

Rn \B(x,ρ(x))



≤ C ρ(x)β+1

Rn \B(x,ρ(x))

≤ C ρ(x)β+1

∞



1/2 1 ρ(x)2β + dz ρ(x)2γ (ρ(x) + |x − z|)2n+2−2γ |x − z|2(n+β+1)

f (z)B dz + ρ(x)1−γ |x − z|n+β+1 

k=0 k 2 ρ(x)≤|x−z|<2k+1 ρ(x) 1−γ

+ ρ(x)

∞



k=0 k 2 ρ(x)≤|x−z|<2k+1 ρ(x)

 Rn \B(x,ρ(x))

f (z)B dz |x − z|n+β+1

f (z)B dz |x − z|n+1−γ

f (z)B dz n+1−γ |x − z|

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.22 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

22

≤C

∞

k=0

+

∞ k=0



ρ(x)β+1 k (2 ρ(x))n+β+1

f (z)B dz B(x,2k+1 ρ(x))



ρ(x)1−γ (2k ρ(x))n+1−γ

f (z)B dz

B(x,2k+1 ρ(x))

≤ C M (f B )(x),

x ∈ Rn ,

where M denotes the Hardy–Littlewood maximal operator. Note that we have used that 1/2 ≤ γ < 1. As it is well-known the maximal operator M is bounded from Lp (Rn ) into itself. Then LV Kglob (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) ,

where C > 0 does not depend on f . 2 Lemma 3.2. Let B be a Banach space and 1 < p < ∞. Then, DLV (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) , Proof. Let f =

m

j=1 bj fj ,

∂tβ (PtLV

f ∈ Cc∞ (Rn ) ⊗ B.

where bj ∈ B and fj ∈ Cc∞ (Rn ), j = 1, . . . , m. We can write

1 (y, z) − Pt (y, z)) = √ 4π

∞

 ∂tβ [te−t /(4u) ]  LV Wu (y, z) − Wu (y − z) du, 3/2 u 2

0

where e−|y| /(4u) , (4πu)n/2 2

Wu (y) =

y ∈ Rn and u > 0.

According to Kato–Trotter’s formula [26, (2.10)] we have that Wu (y − z) − WuLV (y, z) =

u 

LV Ws (y − v)V (v)Wu−s (v, z)dvds

0 Rn

u/2 =

LV Ws (y − v)V (v)Wu−s (v, z)dvds

0 Rn

u/2 +

Wu−s (y − v)V (v)WuLV (v, z)dvds,

y, z ∈ Rn and u > 0.

0 Rn

From [23, (2.2)] and [23, (2.8)] we infer that    u/2  u/2 2 2   e−c|v−z| /u e−c|y−v| /s   LV Ws (y − v)V (v)Wu−s (v, z)dvds ≤ V (v) dvds    sn/2 (u − s)n/2  0 Rn  0 Rn u/2 ≤C 0 Rn

e−c(|y−v| +|v−z| (su)n/2 2

2

)/u

V (v)e−c|y−v|

2

/s

dvds

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.23 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

e−c|y−z| ≤C un/2

2

/u

u/2 0 Rn

−c|y−z| /u

e−c|y−v| V (v) sn/2

2

/s

e−c|y−z| dvds ≤ C un/2

2

/u

23

δ u/2  1 s ds s ρ(y)2 0

2

=C

e , u−δ+n/2 ρ(y)2δ

y, z ∈ Rn ,

0 < u < ρ(y)2 ,

for a certain δ > 0. Also, we get     u/2 2   e−c|y−z| /u   LV Wu−s (y − v)V (v)Ws (v, z)dvds ≤ C −δ+n/2 ,    u ρ(z)2δ   0 Rn

0 < u < ρ(z)2 ,

y, z ∈ Rn .

Then, according to [51, Lemma 1.4, (a)] if |x − z| < ρ(x) and |y − z| < 2ρ(x), then ρ(y) ∼ ρ(x) ∼ ρ(z). Hence, we conclude that e−c|y−z| /u (y, z)| ≤ C −δ+n/2 , u ρ(z)2δ 2

|Wu (y − z) −

WuLV

(23)

for 0 < u < ρ(z)2 and y, z ∈ Rn such that |y − z| ≤ 2ρ(x) and |x − z| < ρ(x). Estimates (20) and (23) lead to ρ(z)   2 2 2  ρ(z)

 ∂tβ [te−t /(4u) ] LV C e−c(t +|y−z| )/u   du, Wu (y, z) − Wu (y − z) du ≤  ρ(x)2δ u3/2 u1−δ+(n+β)/2 2

2

0

(24)

0

for y, z ∈ Rn such that |y − z| ≤ 2ρ(x) and |x − z| < ρ(x). On the other hand, [23, (2.2)] and (20) imply that ∞ −c(t2 +|y−z|2 )/u  ∞ ∂ β [te−t2 /(4u) ]

 e   t LV du, Wu (y, z) − Wu (y − z) du ≤ C  u3/2 u1+(n+β)/2 ρ(z)2

y, z ∈ Rn , t > 0,

(25)

ρ(z)2

and also  2  ρ(z)

 1 ∂tβ [te−t /(4u) ] LV   (y, z) − W (y − z) du ≤ C , W  u u (t + |y − z|)n+β u3/2 2

y, z ∈ Rn and t > 0.

(26)

0

We are going to see that DLV (f )(x; ·, ·) ∈ H, for every x ∈ Rn . Fix x ∈ Rn . We can write  D

LV

tβ f (z) √ 4π

(f )(x; y, t) = χΓ(x) (y, t) B(x,ρ(x))

 + χΓ(x) (y, t) B(x,ρ(x))

2 ρ(z) 

∂tβ [te−t /(4u) ] LV (y, z) − W (y − z) dudz W u u u3/2

0

tβ f (z) √ 4π

∞

∂tβ [te−t /(4u) ] LV (y, z) − W (y − z) dudz W u u u3/2 2

ρ(z)2

= D1LV (f )(x; y, t) + D2LV (f )(x; y, t), Minkowski’s inequality and (25) lead to

2

y ∈ Rn and t > 0.

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.24 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

24

D2LV (f )(x; ·, ·)L2 Rn ×(0,∞), dydt ;B tn+1



 ∞ ∂ β [te−t2 /(4u) ]

   t LV f (z)B tβ (y, z) − W (y − z) du dz W u u H u3/2

≤C

ρ(z)2

B(x,ρ(x))



∞

≤C

f (z)B

u1+(n+β)/2

ρ(z)2

B(x,ρ(x))



∞

≤C

u1+(n+β)/2



∞

1

f (z)B

e−ct

u

dudz ≤ 1+n/2

2

/u

2

+|y−z|2 )/u

dt 1/2 t1−2β

1/2

dydt

dudz

t−2β+n+1

dudz



C ρ(x)n

ρ(z)2

B(x,ρ(x))

e−c(t

0

ρ(z)2

B(x,ρ(x))



0 |x−y|
∞

1

f (z)B

≤C

∞

1

f (z)B dz ≤ C M (f B )(x). B(x,ρ(x))

We have taken into account that ρ(z) ∼ ρ(x) because |x − z| < ρ(x). We now decompose D1LV (f ) as follows LV LV D1LV (f ) = D1,1 (f ) + D1,2 (f ),

where LV D1,1 (f )(x; y, t) = D1LV (f χB(y,2ρ(x)) )(x; y, t),

y ∈ Rn , t > 0.

By using again the Minkowski’s inequality and (26) we get LV D1,2 (f )(x; ·, ·)L2 Rn ×(0,∞), dydt ;B tn+1

⎛ ∞ ⎜ f (z)B ⎝

 ≤C

0 |x−y|
B(x,ρ(x))

⎛∞  ⎝ f (z)B

 ≤C



⎞1/2 2β−n−1

t ⎟ χRn \B(y,2ρ(x)) (z)dydt⎠ (t + |x − y|)2n+2β

⎞1/2 C t2β−1 dt⎠ dz ≤ (t + ρ(x))2n+2β ρ(x)n

0

B(x,ρ(x))

 f (z)B dz B(x,ρ(x))

≤ C M (f B )(x). Now, estimations (20) and (24) imply that LV D1,1 (f )(x; ·, ·)L2 Rn ×(0,∞), dydt ;B tn+1

 ∞







≤C

f (z)B 0 |x−y|
B(x,ρ(x))∩B(y,2ρ(x))

 2  ρ(z)

 2 dydt 1/2 ∂tβ [te−t /(4u) ] LV   × tβ (y, z) − W (y − z) dudz W u u tn+1 u3/2 2

0

dz

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Doctopic: Functional Analysis

[m3L; v1.190; Prn:13/10/2016; 10:18] P.25 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

 ∞





≤C 0 |x−y|
 ∞



Rn



≤C 0 |x−y|
 ∞



Rn



≤C 0 |x−y|
 ∞



≤C 0 |x−y|
 ≤ C sup

u>0 Rn

 ≤ C sup

u>0 Rn

 ≤ C sup

u>0 Rn

Rn

tβ f (z)B ρ(x)2δ

2 cρ(x) 

tβ f (z)B ρ(x)2δ

2 cρ(x) 

2

e−c(t +|x−y| +|y−z| u1−δ+(n+β)/2 2

2

tβ f (z)B ρ(x)2δ

2 cρ(x) 

2

2

2

dydt tn+1

1/2

2

)/u

dudz

2

dydt tn+1

1/2

2

dydt tn+1

1/2

0

 cρ(x) 

2

e−ct

2

/u

u1−δ+(n+β)/2

/u

2 cρ(x) 

f (z)B dz

 e

−c|x−z|2 /u

/u

cρ(x) 

uδ−1−β/2 ρ(x)2δ

2

uδ−1−β/2 ρ(x)2δ

f (z)B dz

/u

f (z)B dzdu

 ∞



t2β−n−1 e−ct

2

dydt tn+1

1/2

1/2 /u

dydt

du

0 |x−y|
 ∞

t2β−1 e−ct

2

1/2 /u

dt

du

0

cρ(x) 

f (z)B dz

2

Rn

0

e−c|x−z| un/2

2

e−c(t +|x−z| )/u dudz u1−δ+(n+β)/2

0

e−c|x−z| un/2

2

0

0 2

2

0

t2β ρ(x)4δ

e−c|x−z| un/2

e−c(t +|y−z| )/u dudz u1−δ+(n+β)/2

25

2

uδ−1 du ≤ CW∗ (f B )(x), ρ(x)2δ

0

where W∗ represents the maximal operator associated with the heat semigroup {Wt }t>0 defined by W∗ (g) = sup |Wt (g)|, t>0

g ∈ Lq (Rn ),

1 ≤ q ≤ ∞.

We conclude that

DLV (f )(x; ·, ·)L2 Rn ×(0,∞), dydt ;B ≤ C W∗ (f B )(x) + M (f B )(x) . tn+1

Also from the above estimations we get DLV (f )(x; ·, ·)L2 Rn ×(0,∞), dydt ;B ≤ Cf L∞ (Rn ,B) , tn+1

x ∈ Rn .

LV We can now proceed as in the study of Kglob to obtain that

DLV (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) , where C does not depend on f , because M and W∗ are bounded operators from Lp (Rn ) into itself. Lemma 3.3. Let B be a UMD Banach space and 1 < p < ∞. Then, Kloc (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) ,

f ∈ Cc∞ (Rn ) ⊗ B.

2

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26

m Proof. Let f = j=1 bj fj , where bj ∈ B and fj ∈ Cc∞ (Rn ), j = 1, . . . , m. We are going to use the ideas developed in the proof of [1, Theorem 3.7].

n Our first goal is to see that Kloc (f )(x; ·, ·) ∈ L2 Rn × (0, ∞), tdydt n+1 ; B , for every x ∈ R . According to n [25, Lemma 2.3] we consider a sequence (xk )∞ k=1 in R such that if Qk = B(xk , ρ(xk )), k ∈ N, we have that (i)

∞ 

Qk = Rn ;

k=1

∗∗ (ii) There exists N = N (ρ) such that, for every k ∈ N, card{j ∈ N : Q∗∗ j ∩ Qk = ∅} ≤ N , where ∗∗ Q = B(x , 4ρ(x )),  ∈ N.

Fix x ∈ Rn . We define the operators S(f )(x; y, t) =

∞

χQk (x)K(f χQ∗k )(x; y, t),

y ∈ Rn and t > 0,

k=1

where Q∗k = B(xk , 2ρ(xk )), k ∈ N, and S(f )(x; y, t) =

∞

χQk (x)Kloc (f )(x; y, t) − S(f )(x; y, t),

y ∈ Rn and t > 0.

k=1

We can write for every y ∈ Rn and t > 0,  ∞

S(f )(x; y, t) = χΓ(x) (y, t) χQk (x)tβ ∂tβ Pt (y − z) χB(x,ρ(x)) (z) − χQ∗k (z) f (z)dz. Rn k=1

  According to [51, Lemma 1.4, (a)] we deduce that, if χQk (x) χB(x,ρ(x)) (z) − χQ∗k (z) = 0, for some k ∈ N; then C11 ρ(x) ≤ |x − z| ≤ C1 ρ(x), for a certain C1 > 0. By (12) it follows that ⎛ ⎞1/2 ∞    2β−n−1 t β β  ⎜ ⎟ dydt⎠ t ∂t Pt (y − z)χΓ(x) (y, t) ≤ C ⎝ H (t + |y − z|)2(n+β) 0 |x−y|
⎛ ∞ ⎜ ≤C⎝

⎞1/2



2(n+β) 2β−n−1

0 |x−y|
⎛ ∞ ⎜ ≤C⎝

⎞1/2



2β−n−1

0 |x−y|
≤

C , |x − z|n

2 t ⎟ dydt⎠ 2(n+β) (t + |x − y| + |y − z|)

t ⎟ dydt⎠ (t + |x − z|)2(n+β)

⎛∞  ≤C⎝ 0

⎞1/2 2β−1

t dt⎠ (t + |x − z|)2(n+β)

x, z ∈ Rn .

∞ Minkowski’s inequality and the property (ii) of the sequence {Q∗∗ k }k=1 lead to  f (z)B S(f )(x; ·, ·)L2 Rn ×(0,∞), dydt ,B ≤ C dz |x − z|n tn+1 ρ(x)/C1 ≤|x−z|≤C1 ρ(x)

≤C

1 ρ(x)n



f (z)B dz ≤ Cf L∞ (Rn ,B) . |x−z|≤C1 ρ(x)

Note that by virtue of (ii), C does not depend on x ∈ Rn .

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27

LV By proceeding as in the study of Kglob we conclude that

S(f )Lp (Rn ,γ(H,B)) ≤ CM (f B )Lp (Rn ) ≤ Cf Lp (Rn ) . On the other hand, according to Lemma 2.1 and by taking into account the properties of the sequence (xk )k∈N we get Sf Lp (Rn ,γ(H,B)) ≤ C

 ∞

1/p χQk K(f χQ∗k )pLp (Rn ,γ(H,B))

≤C

 ∞

k=1

1/p f χQ∗k pLp (Rn ,B)

k=1

≤ Cf Lp (Rn ,B) . Also, we have that  ∞     Kloc (f )Lp (Rn ,γ(H,B)) ∼  χ K (f ) Qk loc  

.

Lp (Rn ,γ(H,B))

k=1

Then, we conclude that Kloc (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) , where C does not depend on f . 2 By combining Lemmas 3.1, 3.2 and 3.3 we obtain K LV (f )Lp (Rn ,γ(H,B)) ≤ Cf Lp (Rn ,B) ,

f ∈ Cc∞ (Rn ) ⊗ B,

   β β LV  t ∂t Pt (f )

f ∈ Cc∞ (Rn ) ⊗ B,

or, in other words,

Tp2 (Rn ,B)

≤ Cf Lp (Rn ,B) ,

provided that B is a UMD Banach space. Here C > 0 does not depend on f . We define the operator T (f ) = tβ ∂tβ PtLV (f ),

f ∈ Cc∞ (Rn ) ⊗ B.

Since Cc∞ (Rn ) ⊗ B is a dense subspace of Lp (Rn ; B), T can be extended from Cc∞ (Rn ) ⊗ B to Lp (Rn , B) as a bounded operator T from Lp (Rn , B) into Tp2 (Rn , B). The same argument developed in Lemma 2.1 allows us to obtain that Tf = tβ ∂tβ PtLV (f ),

f ∈ Lp (Rn , B),

and then,    β β LV  t ∂t Pt (f )

Tp2 (Rn ,B)

≤ Cf Lp (Rn ,B) , f ∈ Lp (Rn , B).

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28

3.2.

We now prove that, there exists C > 0 for which     f Lp (Rn ,B) ≤ C tβ ∂tβ PtLV (f )

Tp2 (Rn ,B)

, f ∈ Lp (Rn , B).

According to [13, Proposition 2.1, (ii)] we have that, for every f, g ∈ L2 (Rn ),  ∞ tβ ∂tβ PtLV Rn

(f )(x)tβ ∂tβ PtLV

Γ(2β) dtdx = 2β (g)(x) t 2

0

 f (x)g(x) dx. Rn

Then, for every f, g ∈ L2 (Rn ),   tβ ∂tβ PtLV

(f )(y)tβ ∂tβ PtLV

Rn Γ(x)

dydt (g)(y) n+1 = vn t

 ∞ tβ ∂tβ PtLV (f )(y)tβ ∂tβ PtLV (g)(y)

Rn 0

Γ(2β) = vn 2β 2

dydt t

 f (y)g(y)dy. Rn

Hence, for every f ∈ L2 (Rn ) ⊗ B and g ∈ L2 (Rn ) ⊗ B∗ , we get  

tβ ∂tβ PtLV (g)(y), tβ ∂tβ PtLV (f )(y)B∗ ,B Rn

dydt Γ(2β) = vn 2β n+1 t 2

Γ(x)



g(y)f (y)B∗ ,B dy. Rn

Now our objective can be established by proceeding as in the proof of Lemma 2.3. 4. Proof of Theorem 2, (ii) For every k ∈ N we consider the k-th Hermite function defined by √ 2 hk (x) = ( π2k k!)−1/2 e−x /2 Hk (x),

x ∈ R,

where Hk represents the k-th Hermite polynomial [55, pp. 105–106]. If k = (k1 , . . . , kn ) ∈ Nn , the k-th Hermite function hk in Rn is defined by n

hk (x) =

hkj (xj ),

x = (x1 , . . . , xn ) ∈ Rn .

j=1

For every k ∈ Nn , hk is an eigenfunction of the Hermite operator H satisfying H hk = (2|k| + n)hk ,

where |k| = k1 + . . . + kn . The system {hk }k∈Nn is an orthonormal basis in L2 (Rn ). The heat semigroup associated with {hk }k∈Nn is defined by WtH (f )(x) =

k∈Nn

where

e−t(2|k|+n) ck (f )hk (x),

f ∈ L2 (Rn ),

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.29 (1-44)

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29

 ck (f ) =

hk (y)f (y)dy,

k ∈ Nn .

Rn

According to the Mehler’s formula [56, (1.1.36)] we can write, for every t > 0, WtH



WtH (x, y)f (y)dy,

(f )(x) =

f ∈ L2 (Rn ),

(27)

Rn

where, for every x, y ∈ Rn and t > 0, WtH (x, y) =



1 π n/2

e−2t 1 − e−4t

n/2

   −2t 1 1 + e−2t 21−e . |x − y|2 exp − + |x + y| 4 1 − e−2t 1 + e−2t

We define, for each t > 0, and 1 ≤ p ≤ ∞, the operator WtH on Lp (Rn ) by (27). Then, {WtH }t>0 is a positive semigroup of contractions in Lp (Rn ), for every 1 < p < ∞. WtH can be extended to Lp (Rn , B) with the same boundedness properties, for every t > 0 and 1 < p < ∞. In the Hermite setting the critical radius ρ(x), x ∈ Rn , satisfies that

ρ(x) ∼

⎧ 1 ⎪ ⎪ , ⎪ ⎨ 1 + |x|

|x| ≥ 1

⎪ ⎪ ⎪ ⎩ 1, 2

|x| < 1.

We are going to see that in this context we can establish properties that allow us to prove Theorem 2, (ii), by proceeding as in the proof of Theorem 2, (i). Note that now n can be any nonnegative integer. Firstly, according to Feynman–Kac formula we have that e−|x−y| /(4t) , x, y ∈ Rn and t > 0. tn/2 2

|WtH (x, y)| ≤ C On the other hand, we have that

• If x, y ∈ Rn , x · y > 0, then |x + y| ≥ |y| and |WtH

 (x, y)| ≤ C

e−2t 1 − e−4t

≤C

e−c|x−y| tn/2

/t

≤C

e−c|x−y| tn/2

/t

2

2

  −2t −2t 1 21 + e 21 −e |x − y| exp − + |y| 4 1 − e−2t 1 + e−2t   2 1/2 e−c|x−y| /t 1 1 1 + e−2t √ ≤ C 1 − e−2t |y| tn/2 t|y|

n/2

ρ(y) √ , t



|y| > 1 and t > 0.

• If x, y ∈ Rn , x · y < 0, then |x − y| ≥ |y| and n/2    −2t −2t 1 e−2t 21 + e 21 +e |x − y| exp − + |y| 1 − e−4t 8 1 − e−2t 1 − e−2t  1/2  1 − e−2t 1 e−ct 1 + e−2t 1 ≤ C n/2 exp − |x − y|2 8 1 − e−2t 1 + e−2t |y| t √ −ct −c|x−y|2 /t 2 e te e−c|x−y| /t ρ(y) √ , |y| > 1 and t > 0. ≤C ≤ C |y| tn/2 tn/2 t

|WtH (x, y)| ≤ C



(28)

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30

• If x, y ∈ Rn and |y| < 1, then e−c|x−y| tn/2

2

|WtH (x, y)| ≤ C

/t

e−c|x−y| 1 √ ≤C tn/2 t

2

/t

ρ(y) √ , t

|y| ≤ 1 and t > 0.

We conclude that −c|x−y|2 /t   H ρ(y) Wt (x, y) ≤ C e √ , n/2 t t

x, y ∈ Rn and t > 0.

(29)

The Poisson semigroup {PtH }t>0 associated with the Hermite operator is defined, as usual, by subordination PtH

t (f )(x) = √ 4π

∞

e−t /(4u) H Wu (f )(x)du, u3/2 2

f ∈ Lp (Rn ) and t > 0.

0

The Poisson kernel PtH (x, y) can be written as t PtH (x, y) = √ 4π

∞

e−t /(4u) H Wu (x, y)du, u3/2 2

x, y ∈ Rn and t > 0.

0

By using (20) and (29) we obtain   β β H  t ∂t Pt (z, y) ≤ C

ρ(y)tβ , (t + |y − z|)β+n+1

z, y ∈ Rn and t > 0.

(30)

We also have that  |z|2 Rn

e−c|y−z| sn/2

2



/s

dz =

√ 2 |y − w s|2 e−c|w| dw =

Rn

≤

C s s ρ(y)2





√ 2 (|y|2 + |w|2 s + 2|y||w| s)e−c|w| dw

Rn

(1 + |w|2 + |w|)e−c|w| dw ≤

Rn

2

C s , s ρ(y)2

0 < s < ρ(y)2 .

(31)

Note that ρ(y) ≤ 1, y ∈ Rn . By proceeding as in the proof of [51, Lemma 1.4] we can see that there exists 1/2 ≤ γ < 1 such that γ  |x − y| ρ(y) ≤ Cρ(x) 1 + , ρ(x)

x, y ∈ Rn .

(32)

Also, we can find a sequence (xk )k∈N in Rd such that if Qk = B(xk , ρ(xk )) and Q∗∗ k = B(xk , 4ρ(xk )), k ∈ N, the following properties hold [25, Lemma 2.3] (a) ∪k∈N Qk = Rn ; ∗∗ (b) There exists N ∈ N such that, for every k ∈ N, card{j ∈ N : Q∗∗ j ∩ Qk = ∅} ≤ N . These properties of the sequence of balls {Qk }k∈N and the estimates (28)–(32) allow us to show Theorem 2 (ii), by proceeding as in the proof of Theorem 2, (i).

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31

5. Proof of Theorem 3 5.1.

Our first objective is to show that tβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ≤ Cf Lp ((0,∞),B) ,

f ∈ Lp ((0, ∞), B).

(33)

We can write Bλ = −

d2 + V (x), dx2

x ∈ (0, ∞),

where V (x) = λ(λ − 1)/x2 , x ∈ (0, ∞). Then, ∂t PtBλ (1) = 0, and in order to prove (33) we cannot use [36, Theorem 4.8]. We are going to proceed as in Section 3, by comparing the operator tβ ∂tβ PtBλ (f ) with the one related to the one-dimensional classical Poisson semigroup given by (1). The Bessel–Poisson semigroup PtBλ , t > 0, is defined by using subordination with respect to the Besselheat semigroup WtBλ , t > 0. We recall that the Bessel-heat kernel WtBλ (x, y), t, x, y ∈ (0, ∞), is given by √ WtBλ (x, y) =

xy x2 +y2 xy Iλ−1/2 e− 4t , t, x, y ∈ (0, ∞), 2t 2t

where λ > 0. We will use the following properties of Iα , α > −1, that can be encountered in [41, Ch. 5], zα

Iα (z) ∼ √ ez zIα (z) = √ 2π

, + 1)

1 



2α Γ(α 1+O

z

,

as z → 0+ ,

(34)

z ∈ (0, ∞).

(35)

From (34) and (35) we easily deduce that 0≤

WtBλ (x, y)

e−c(x−y) √ ≤C t

2

/t

,

t, x, y ∈ (0, ∞).

(36)

By (20) and (36) we also deduce that ∞ |tβ ∂tβ (PtBλ (y, z)|

≤C

|tβ ∂tβ [te−t /(4u) ]| Bλ Wu (y, z)du u3/2 2

0

∞ ≤ Ctβ

e−c(t +(y−z) u(β+3)/2 2

2

)/u

0

du ≤ C

tβ , (t + |y − z|)β+1

t, y, z ∈ (0, ∞).

(37)

Then, by using Hölder’s inequality, it can be justified that, for every f ∈ Lp ((0, ∞), B), ∞ tβ ∂tβ PtBλ (f )(y) =

tβ ∂tβ PtBλ (y, z)f (z)dz,

t, y ∈ (0, ∞).

0

Let f ∈ Cc∞ (0, ∞) ⊗ B. We define the function f0 ∈ Cc∞ (R) ⊗ B by ! f (x), x > 0 f0 (x) = 0, x ≤ 0.

(38)

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.32 (1-44)

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32

By taking into account (38) and also that (see proof of Lemma 2.1) ∞ tβ ∂tβ Pt (f0 )(y)

tβ ∂tβ Pt (y − z)f (z)dz,

=

t, y ∈ (0, ∞),

0

for every x ∈ (0, ∞), we can write the following decomposition tβ ∂tβ [PtBλ (f )(y) − Pt (f0 )(y)]χΓ+ (x) (y, t) ⎞ ⎛ x/2 ∞ 2x ⎟ ⎜ = χΓ+ (x) (y, t) ⎝ + + ⎠ tβ ∂tβ [PtBλ (y, z) − Pt (y − z)]f (z)dz 0

2x

x/2

t, y ∈ (0, ∞).

1 2 (f )(x, y, t) + Kglob (f )(x, y, t) + Kloc (f )(x, y, t), = Kglob

Here Γ+ (x) = {(y, t) ∈ (0, ∞)2 : |x − y| < t}, x ∈ (0, ∞). We are going to establish the boundedness properties of each of the above operators. Lemma 5.1. Let B be a Banach space and 1 < p < ∞. Then, for each j = 1, 2, f ∈ Cc∞ (0, ∞) ⊗ B.

j Kglob (f )Lp ((0,∞),γ(H+ ,B)) ≤ Cf Lp ((0,∞),B) ,

Proof. Let f ∈ Cc∞ (0, ∞) ⊗ B. We can write

tβ ∂tβ [PtBλ (y, z)

1 − Pt (y − z)] = √ 2 π

∞

tβ ∂tβ [te−t /(4u) ] (WuBλ (y, z) − Wu (y − z))du, t, y, z ∈ (0, ∞). u3/2 2

0

By using (20) and (36) we deduce that ∞ |tβ ∂tβ [PtBλ (y, z)

− Pt (y − z)]| ≤ C

|tβ ∂tβ [te−t /(4u) ]| (WuBλ (y, z) + Wu (y − z))du u3/2 2

0

≤C

tβ , (t + |y − z|)β+1

t, y, z ∈ (0, ∞).

Hence we obtain    β β Bλ  t ∂t (Pt (y, z) − Pt (y − z))χΓ+ (x) (y, t)

H+

⎛ ∞ ⎜ ≤C⎝



0 |x−y|
⎛ ∞ ⎜ ≤C⎝

0 |x−y|
⎞1/2 2β+2 2β−2



2 t ⎟ dydt⎠ (t + |x − y| + |y − z|)2β+2

⎞1/2 2β−2

t ⎟ dydt⎠ (t + |y − z|)2β+2

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.33 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

⎛ ∞ ⎜ ≤C⎝

⎞1/2



2β−2

t ⎟ dydt⎠ 2β+2 (t + |x − z|)

0 |x−y|
⎧ 1 ⎪ ⎪ ⎪ ⎨ z,

C ≤C ≤ ⎪ |x − z| ⎪ 1 ⎪ ⎩ , x

⎛∞  ⎝ ≤C

⎞1/2 t2β−1 dt⎠ (t + |x − z|)2β+2

0

0 < 2x < z, (39) x 0
Then, since γ(H+ , C) = H+ , we get 1 2 Kglob (f )(x, y, t)γ(H+ ,B) + Kglob (f )(x, y, t)γ(H+ ,B)



∞    β β Bλ  + t ∂t [Pt (y, z) − Pt (y − z)]χΓ+ (x) (y, t)

x/2

≤C

H+

0

f (z)B dz

2x

x/2 ∞

1  f (z)B dz , f (z)B dz + ≤C x z 0

x ∈ (0, ∞).

2x

By using Hardy inequalities [65, p. 20] we conclude that j Kglob (f )Lp ((0,∞),γ(H+ ,B)) ≤ Cf Lp ((0,∞),B) ,

j = 1, 2.

2

Lemma 5.2. Let B be a Banach space and 1 < p < ∞. Then, Kloc (f )Lp ((0,∞),γ(H+ ,B)) ≤ Cf Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Proof. Let f ∈ Cc∞ (0, ∞) ⊗ B. We decompose tβ ∂tβ [PtBλ (y, z) − Pt (y − z)], t, y, z ∈ (0, ∞), as follows tβ ∂tβ [PtBλ (y, z) − Pt (y − z)] ⎛ yz ∞⎞   2 tβ ⎝ ∂tβ [te−t /(4u) ] Bλ ⎠ = √ + [Wu (y, z) − Wu (y − z)]du 2 π u3/2 yz

0

= I1 (y, z, t) + I2 (y, z, t),

t, y, z ∈ (0, ∞).

According to (20) and (34) we get |I2 (y, z, t)| ∞ ≤ Ct

β

|∂tβ [te−t /(4u) ]| Bλ [Wu (y, z) + Wu (y − z)]du u3/2 2

yz

∞ ≤ Ctβ

(y−z)2 1 e−t /(8u) (yz)λ − y2 +z2 e 4u + √ e− 4u du (β+2)/2 λ+1/2 u u u 2

yz

∞ ≤ Ct

β

e−c(t +y +z )/u du u(β+3)/2 2

2

2

yz

∞ ≤ Ct

β 0

33

tβ e−c(t +y +z )/u du ≤ C , (t + y + z)β+1 u(β+3)/2 2

2

2

t, z, y ∈ (0, ∞).

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.34 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

34

By proceeding as in (39) we obtain I2 (y, z, t)χΓ+ (x) (y, t)H+ ≤

C C ≤ , z+x x

0<

x < z < 2x. 2

(40)

On the other hand, (20) and (35) lead to yz |I1 (y, z, t)| ≤ Ct

β

e−c(t +(y−z) u(β+3)/2

2

e−c(t +(y−z) u(β+3)/2

2

2

)/u

u du yz

0

yz ≤ Ctβ

2

)/u

u 1/3 du yz

0

tβ ≤C (yz)1/3

∞

e−c(t +(y−z) )/u du u(3β+7)/6 2

2

0

≤C

(yz)1/3 (t

tβ , + |y − z|)β+1/3

t, z, y ∈ (0, ∞).

Then, by choosing 0 < ε < min{2/3, 2β} we can write ⎛ ⎜ I1 (y, z, t)χΓ+ (x) (y, t)H+ ≤ C ⎝

⎞1/2



2β−2

(yz)2/3 (t

t ⎟ dtdy ⎠ + |y − z|)2β+2/3

Γ+ (x)

⎫ ⎛⎧ ⎞1/2 ⎪ ⎬ ∞ ⎨ z/4 ∞⎪ 2β−2 t ⎜ ⎟ ≤C⎝ + dtdy ⎠ 2/3 (t + |y − z|)2β+2/3 ⎪ ⎪ (yz) ⎭ ⎩ 0

⎛ ⎜ ≤C⎝

z/4

z/4 ∞

|x−y|

dtdy + (yz)2/3 t8/3

0 x/2

⎛ ⎜ ≤C⎝

z/4

∞

1 (yz)2/3 |x − z|ε+2/3

z/4

dy (yz)2/3 x5/3

+

0

z 2/3 |x

1 − z|ε+2/3

 ≤C

|x−y|

⎞1/2 dt ⎟ dy ⎠ t2−ε ⎞1/2

∞

dy ⎟ ⎠ − y|1−ε

y 2/3 |x x/8

⎛ ⎜ ≤C⎝

∞

1 z 1/3 x5/3

+

1 z 2/3 |x

−

xε+2/3 1 + x2 x2 |x − z|ε+2/3

⎞1/2

∞

z|ε+2/3 x−ε+2/3

1/2

C ≤ x

du ⎟ ⎠ 1−ε − u|

u2/3 |1 1/8

"

 1+

x |x − z|

(3ε+2)/6 # ,

0<

x < z < 2x. 2

By combining (40) and (41) we get    β β Bλ  t ∂t [Pt (y, z) − Pt (y − z)]χΓ+ (x) (y, t)

H+

For every x ∈ (0, ∞),

C ≤ x

"

 1+

x |x − z|

(3ε+2)/6 # ,

0<

x < z < 2x. 2

(41)

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.35 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

1 dμx (z) = C0



35

 x (3ε+2)/6 dz 1+ χ(x/2,2x) (z) |x − z| z

is a probability measure on (0, ∞) when 2  C0 =

1 1+ |1 − u|(3ε+2)/6

 du.

1/2

Then, since γ(H+ , C) = H+ , Jensen’s inequality leads to ⎛ ⎜ Kloc (f )(x, ·, ·)pγ(H+ ,B) ≤ C ⎝

2x

⎞p

    f (z)B tβ ∂tβ [PtBλ (y, z) − Pt (y − z)]χΓ+ (x) (y, t)

H+

x/2

⎛

⎜1 ≤C⎝ x



2x

f (z)B 1 +



x |x − z|

(3ε+2)/6 

⎟ dz ⎠

⎞p ⎟ dz ⎠

x/2

2x

1 x

f (z)pB

≤C



1+

 x (3ε+2)/6 dz, |x − z|

x ∈ (0, ∞).

x/2

Hence, ⎛ ⎜ Kloc (f )Lp ((0,∞),γ(H+ ,B)) ≤ C ⎝

∞ 2x f (z)pB

1 x



1+

0 x/2

x |x − z|

(3ε+2)/6 

⎞1/p ⎟ dzdx⎠

⎞1/p ⎛  ∞ 2z 

(3ε+2)/6 x 1 ⎟ ⎜ dxdz ⎠ 1+ ≤ C ⎝ f (z)pB ≤ Cf Lp ((0,∞),B) . x |x − z| 0

2

z/2

Putting together Lemmas 5.1 and 5.2 we conclude that tβ ∂tβ PtBλ (f ) − tβ ∂tβ Pt (f0 )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

It is clear that if {hj }j=1 is an orthonormal system in H+ and, for every j ∈ N, we define hj by ! hj (y, t) =

hj (y, t), y, t > 0, 0, t > 0, y ≤ 0,

then, {hj }j=1 is an orthonormal system in L2 (R × (0, ∞), dydt t2 ). Hence, since B is UMD we have that   β β  t ∂t Pt (f0 )(y)χΓ+ (x) (y, t)

γ(H+ ,B)

   2 1/2   β β = sup E γj [t ∂t Pt (f0 )(y)χΓ+ (x) (y, t)](hj ) B

j=1

   2 1/2   β β  = sup E γj [t ∂t Pt (f0 )(y)](χΓ(x) (y, t)hj ) j=1

B

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.36 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

36

   2 1/2   β β ≤ sup E γj [t ∂t Pt (f0 )(y)](χΓ(x) (y, t)ej ) B

j=1

    = tβ ∂tβ Pt (f0 )(y)χΓ(x) (y, t)



, γ L2 (R×(0,∞), dydt ),B t2

a.e. x ∈ (0, ∞),

where the two first suprema are taken over all orthonormal systems {hj }j=1 in H+ and the last one over all orthonormal systems {ej }j=1 in L2 (R × (0, ∞), dydt t2 ). According to Theorem 1, since B is a UMD Banach space, we have that tβ ∂tβ Pt (f0 )Tp2 ((0,∞),B) ≤ tβ ∂tβ Pt (f0 )Tp2 (R,B) ≤ Cf Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Hence, tβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Finally, using (37), and proceeding as in the proof of Lemma 2.1, we can deduce that tβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) , 5.2.

f ∈ Lp ((0, ∞), B).

Now we are going to show that f Lp ((0,∞),B) ≤ Ctβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ,

f ∈ Lp ((0, ∞), B).

If f ∈ Cc∞ (0, ∞) then, according to [7, Lemma 3.1] we have that hλ (tβ ∂tβ PtBλ (f ))(x) = eiπβ (tx)β e−tx hλ (f )(x),

t, x ∈ (0, ∞).

Suppose that f, g ∈ Cc∞ (0, ∞). We denote by Jt (y) = {x ∈ (0, ∞) : |x − y| < t}, t, y ∈ (0, ∞). Plancherel equality for Hankel transform leads to ∞  tβ ∂tβ PtBλ (f )(y)tβ ∂tβ PtBλ (g)(y)

dydt dx t|Jt (y)|

0 Γ+ (x)

∞ ∞ iπβ

hλ [e

= 0

(tz) e

iπβ

hλ (f )](y)hλ [e

=

(tz) e

 hλ (g)](y)

dx

dy dt t|Jt (y)|

Jt (y)

e2iπβ (ty)2β e−2ty hλ (f )(y)hλ (g)(y)

dydt t

0

∞ =e

β −tz

0

∞ ∞ 0

β −tz

2iπβ

∞ hλ (f )(y)hλ (g)(y)y

0

e2iπβ Γ(2β) = 22β

2β

e−2ty t2β−1 dtdy

0

∞ 0

e2iπβ Γ(2β) hλ (f )(y)hλ (g)(y)dy = 22β

∞ f (x)g(x)dx. 0

Suppose now that f ∈ Cc∞ (0, ∞) ⊗ B and g ∈ Cc∞ (0, ∞) ⊗ B∗ . From (42) it follows that

(42)

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.37 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

∞ ∞  e−2iπβ 22β

g(x), f (x)B∗ ,B dx = Γ(2β) 0

37

tβ ∂tβ PtBλ (g)(y), tβ ∂tβ PtBλ (f )(y)B∗ ,B

dydt dx. t|Jt (y)|

0 Γ+ (x)

Then, since |Jt (y)| ≥ t, for every t, y ∈ (0, ∞), according to [37, Proposition 2.2] we get ∞    ∞    dydt  g(x), f (x)B∗ ,B dx ≤ C | tβ ∂tβ PtBλ (g)(y), tβ ∂tβ PtBλ (f )(y)B∗ ,B | 2 dx   t   0

0 Γ+ (x)

∞ tβ ∂tβ PtBλ (f )(y)χΓ+ (x) (y, t)γ(H+ ,B) tβ ∂tβ PtBλ (g)(y)χΓ+ (x) (y, t)γ(H+ ,B∗ ) dx

≤C 0

≤ Ctβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) tβ ∂tβ PtBλ (g)T 2 ((0,∞),B∗ ) , p

where p = p/(p − 1). Since B∗ is UMD, by using (33) where p is replaced by p , we obtain ∞       g(x), f (x)B∗ ,B dx ≤ Ctβ ∂tβ PtBλ (f )T 2 ((0,∞),B) )||g|| p L ((0,∞),B∗ ) .   p   0



From [30, Lemma 2.3] and by taking into account that Cc∞ (0, ∞) ⊗ B∗ is dense in Lp ((0, ∞), B∗ ) we get f Lp ((0,∞),B) ≤ Ctβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

By using again (33) and the fact that Cc∞ (0, ∞) ⊗ B is dense in Lp ((0, ∞), B) we conclude that f Lp ((0,∞),B) ≤ Ctβ ∂tβ PtBλ (f )Tp2 ((0,∞),B) ,

f ∈ Lp ((0, ∞, B).

Remark. The Bessel–Poisson kernel PtBλ (x, y), t, x, y ∈ (0, ∞), also can be written as ([45, (16.4)], [62]) PtBλ (x, y)

2λ(xy)λ t = π

π 0

(sin θ)2λ−1 dθ, (t2 + (x − y)2 + 2xy(1 − cos θ))λ+1

t, x, y ∈ (0, ∞).

(43)

We prefer to use in the proof of Theorem 3 the expression for the Bessel–Poisson kernel given by subordination formula instead of (43) directly because the manipulations are easier in the former case. 6. Proof of Theorem 4 6.1.

We are going to show that tβ ∂tβ PtLα (f )Tp2 ((0,∞),B) ≤ Cf Lp ((0,∞),B) ,

f ∈ Lp ((0, ∞), B).

(44)

We have that Lα = −

d2 + V (x), dx2

x ∈ (0, ∞),

where V (x) = x2 + (α2 − 1/4)/x2 . Then, ∂t PtLα (1) = 0 and [36, Theorem 4.8] can not be applied to prove (44). The strategy will be the same as in previous sections: comparing with an operator whose

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.38 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

38

boundedness property is already known. This time we are going to relate the operator tβ ∂tβ PtLα with tβ ∂tβ PtH , studied in Section 4. Recall (7) for the definition of the Poisson Laguerre semigroup. It is given via subordination with respect to the heat Laguerre semigroup WtLα , in which it is involved the modified Bessel function Iα (see (6)). From (34) and (35) we can infer that 0 ≤ WtLα (x, y) ≤ C

e−c(x−y) √ t

2

/t

,

t, x, y ∈ (0, ∞).

(45)

Let f ∈ Cc∞ (0, ∞) ⊗ B. We define the function f0 ∈ Cc∞ (R) ⊗ B by ! f (x), x > 0 0, x ≤ 0.

f0 (x) =

By (20), (28), (45) and applying Hölder’s inequality, we obtain tβ ∂tβ PtLα (f )(y) =

∞

tβ ∂tβ PtLα (y, z)f (z)dz,

t, y ∈ (0, ∞),

0

and tβ ∂tβ PtH

∞ (f0 )(y) =

tβ ∂tβ PtH (y, z)f (z)dz,

t, y ∈ (0, ∞).

0

Here, PtH (x, y) denotes the Poisson kernel associated with the Hermite operator on R. For every x ∈ (0, ∞), in a similar way as in the previous section we write the decomposition that follows tβ ∂tβ [PtLα (f )(y) − PtH (f0 )(y)]χΓ+ (x) (y, t) ⎞ ⎛ x/2 ∞ 2x ⎟ ⎜ = χΓ+ (x) (y, t) ⎝ + + ⎠ tβ ∂tβ [PtLα (y, z) − PtH (y, z)]f (z)dz 0

=

1 Kglob (f )(x, y, t)

2x

+

x/2

2 Kglob (f )(x, y, t)

+ Kloc (f )(x, y, t),

t, y ∈ (0, ∞).

Next, we analyze the boundedness properties of each of the above operators. Lemma 6.1. Let B be a Banach space and 1 < p < ∞. Then, for each j = 1, 2, j Kglob (f )Lp ((0,∞),γ(H+ ,B)) ≤ Cf Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Proof. Let f ∈ Cc∞ (0, ∞) ⊗ B. By using (20), (28) and (29) we deduce that |tβ ∂tβ [PtLα (y, z)

−

PtH

∞ (y, z)]| ≤ C

|tβ ∂tβ [te−t /(4u) ]| Lα |Wu (y, z) − WuH (y, z)|du u3/2 2

0

∞ ≤ Ctβ 0

e−c(t +(y−z) u(β+3)/2 2

2

)/u

du ≤ C

tβ , (t + |y − z|)β+1

t, y, z ∈ (0, ∞).

The proof of this lemma can be finished by proceeding as in the proof of Lemma 5.1.

(46) 2

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.39 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

39

Lemma 6.2. Let B be a Banach space and 1 < p < ∞. Then, Kloc (f )Lp ((0,∞),γ(H+ ,B)) ≤ Cf Lp ((0,∞),B) , f ∈ Cc∞ (0, ∞) ⊗ B. Proof. Let f ∈ Cc∞ (0, ∞) ⊗ B. To simplify notation, we call ξ = ξ(u, y, z) =

2yze−2u , 1 − e−4u

u, y, z ∈ (0, ∞).

We make the following decomposition tβ ∂tβ [PtLα (y, z) − PtH (y, z)] ⎛  β t ⎜ √ + = ⎝ 2 π {u∈(0,∞) : ξ≤1}

⎞



⎟ ∂t [te−t /(4u) ] Lα [Wu (y, z) − WuH (y, z)]du ⎠ u3/2 2

β

{u∈(0,∞) : ξ≥1}

t, y, z ∈ (0, ∞).

= I1 (y, z, t) + I2 (y, z, t), According to (20) and (34) we get |I1 (y, z, t)|  {u∈(0,∞) : ξ≤1}

∞ ≤ Ct

β

e−t /(8u) u(β+2)/2 2

≤ Ctβ



e−2u 1 − e−4u

1/2

  % 1 1 + e−4u $ α+1/2 ξ exp − (y 2 + z 2 ) + e du ξ 2 1 − e−4u

tβ e−c(t +y +z )/u du ≤ C , (β+3)/2 (t + y + z)β+1 u 2

2

2

t, z, y ∈ (0, ∞).

0

As in (39) we obtain I1 (y, z, t)χΓ+ (x) (y, t)H+ ≤

C C ≤ , z+x x

0 < x/2 < z < 2x.

(47)

On the other hand, (20) and (35) lead to 

e−t /(8u) H du ≤ Ctβ Wu (y, z) (β+2)/2 ξ u 2

|I2 (y, z, t)| ≤ Ctβ {u∈(0,∞) : ξ≥1}



{u∈(0,∞) : ξ≥1}

∞ 0

{u∈(0,∞) : ξ≥1}

e−t /(8u) H du Wu (y, z) 1/3 (β+2)/2 u ξ 2

e−t /(8u) e−u e−c|y−z| /u du √ u(β+2)/2 1 − e−4u ξ 1/3 2

≤ Ctβ

tβ ≤C (yz)1/3



2

tβ e−c(t +|y−z| )/u du ≤ C , (3β+7)/6 1/3 u (yz) (t + |y − z|)β+1/3 2

2

t, z, y ∈ (0, ∞).

Then, by choosing 0 < ε < min{2/3, 2β}, by proceeding as in the proof of Lemma 5.2 we can write ⎛ ⎞1/2  2β−2 t ⎜ ⎟ I2 (y, z, t)χΓ+ (x) (y, t)H+ ≤ C ⎝ dtdy ⎠ 2/3 (yz) (t + |y − z|)2β+2/3 ≤

C x

Γ+ (x)

"

 1+

x |x − z|

(3ε+2)/6 #

,

0 < x/2 < z < 2x.

(48)

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.40 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

40

By combining (47) and (48) we get    β β Lα  t ∂t [Pt (y, z) − PtH (y, z)]χΓ+ (x) (y, t)

H+

C ≤ x

"

 1+

x |x − z|

(3ε+2)/6 # ,

0 < x/2 < z < 2x.

Then, since γ(H+ , C) = H+ , Jensen’s inequality leads to ⎛ ⎜ Kloc (f )(x, ·, ·)pγ(H+ ,B) ≤ C ⎝

2x

    f (z)B tβ ∂tβ [PtLα (y, z) − PtH (y, z)]χΓ+ (x) (y, t)

H+

x/2

⎛

⎜1 ≤C⎝ x



2x

f (z)B 1 +



x |x − z|

(3ε+2)/6 

⎞p ⎟ dz ⎠

⎞p ⎟ dz ⎠

x/2

2x ≤C

1 f (z)pB

x

 x (3ε+2)/6 dz, 1+ |x − z|





x ∈ (0, ∞).

x/2

Hence,

Kloc (f )Lp ((0,∞),γ(H+ ,B))

⎞1/p ⎛   ∞ 2x

(3ε+2)/6 x 1 ⎟ ⎜ dzdx⎠ 1+ ≤C⎝ f (z)pB x |x − z| 0 x/2

⎞1/p ⎛  ∞ 2z 

(3ε+2)/6 x 1 ⎟ ⎜ dxdz ⎠ 1+ ≤ C ⎝ f (z)pB ≤ Cf (z)Lp ((0,∞),B) . x |x − z| 0

z/2

Putting together Lemmas 6.1 and 6.2 we conclude that tβ ∂tβ PtLα (f ) − tβ ∂tβ PtH (f0 )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Moreover, according to Theorem 2, (ii), since B is a UMD Banach space, we have that tβ ∂tβ PtH (f0 )Tp2 ((0,∞),B) ≤ tβ ∂tβ PtH (f0 )Tp2 (R,B) ≤ Cf Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

Hence, tβ ∂tβ PtLα (f )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) ,

f ∈ Cc∞ (0, ∞) ⊗ B.

We get |tβ ∂tβ PtLα (y, z)| ≤ C

tβ , (t + |y − z|)β+1

t, y, z ∈ (0, ∞),

and by proceeding as in the proof of Lemma 2.1, we can deduce that tβ ∂tβ PtLα (f )Tp2 ((0,∞),B) ≤ f Lp ((0,∞),B) ,

f ∈ Lp ((0, ∞), B).

2

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[m3L; v1.190; Prn:13/10/2016; 10:18] P.41 (1-44)

J.J. Betancor et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

6.2.

41

In this paragraph we establish that f Lp ((0,∞),B) ≤ Ctβ ∂tβ PtLα (f )Tp2 ((0,∞),B) ,

f ∈ Lp ((0, ∞), B).

As in Section 5.2, by duality and density arguments, it is enough to have the following identity. Lemma 6.3. Let α, β > 0 and f, g ∈ Cc∞ (0, ∞). Then, ∞ 

tβ ∂tβ PtLα (f )(y)tβ ∂tβ PtLα (g)(y)

e2πiβ Γ(2β) dydt dx = t|Jt (y)| 22β

0 Γ+ (x)

∞ f (x)g(x)dx, 0

where Jt (y) = {x ∈ (0, ∞) : |x − y| < t}. Proof. We can write PtLα (f )(x)

=

∞

√ 2k+α+1 α ck (f )ϕα k (x),

e−t

t, x ∈ (0, ∞).

(49)

k=0

According to [46, p. 1124] there exists C > 0 such that |ϕα k (x)| ≤ C, x ∈ (0, ∞). Moreover, for every m ∈ N −m there exists C > 0 such that |cα (f )| ≤ C(1 + k) , k ∈ N. Hence, the series in (49) converges in L2 (0, ∞) k and uniformly in (t, x) ∈ (0, ∞) × (0, ∞). Also, for every m ∈ N, we get ∂tm PtLα (f )(x) =

∞

√ 2k+α+1 α ck (f )ϕα k (x),

(−1)m (2k + α + 1)m/2 e−t

x, t ∈ (0, ∞).

k=0

If m ∈ N is such that m − 1 ≤ β < m, then ∂tβ PtLα (f )(x)

e−iπ(m−β) = Γ(m − β) =

e−iπ(m−β) Γ(m − β)

∞

Lα ∂tm Pt+s (f )(x)sm−β−1 ds

0

∞ sm−β−1 0

∞

(−1)m (2k + α + 1)m/2 e−(t+s)

√ 2k+α+1 α ck (f )ϕα k (x)ds

k=0

 ∞ √ √ e−iπ(m−β) α = (−1)m (2k + α + 1)m/2 e−t 2k+α+1 cα (f )ϕ (x) sm−β−1 e−s 2k+α+1 ds k k Γ(m − β) ∞

k=0

= eiπβ

∞

(2k + α +

0

√ α 1)β/2 e−t 2k+α+1 cα k (f )ϕk (x),

t, x ∈ (0, ∞).

k=0

The interchange between the series and the integral is justified because (cα k (f ))k∈N is rapidly decreasing as k → ∞. We have that ∞ 

tβ ∂tβ PtLα (f )(y)tβ ∂tβ PtLα (g)(y)

dydt dx t|Jt (y)|

0 Γ+ (x)

∞ ∞ = 0

0

tβ ∂tβ PtLα (f )(y)tβ ∂tβ PtLα (g)(y)



Jt (y)

dx

dydt t|Jt (y)|

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∞ ∞ = 0

0

= e2iπβ

tβ ∂tβ PtLα (f )(y)tβ ∂tβ PtLα (g)(y) ∞ ∞ & ∞ 0

&

0

dydt t '

√

(t 2k + α +

√ α 1)β e−t 2k+α+1 cα k (f )ϕk (y)

k=0

' ∞ √ √ dydt β −t 2m+α+1 α α × (t 2m + α + 1) e cm (g)ϕm (y) t m=0 =e

2iπβ

∞

α cα k (f )ck (g)

k=0

= e2iπβ

Γ(2β) 22β

∞ √ √ dt (t 2k + α + 1)2β e−2t 2k+α+1 t 0

∞ k=0

α 2iπβ cα k (f )ck (g) = e

Γ(2β) 22β

∞ f (x)g(x)dx.

2

0

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