Littlewood-Paley-Stein theory and Banach spaces in the inverse Gaussian setting
aa r X i v : . [ m a t h . C A ] F e b LITTLEWOOD-PALEY-STEIN THEORY AND BANACH SPACES IN THEINVERSE GAUSSIAN SETTING
V´ICTOR ALMEIDA, JORGE J. BETANCOR, JUAN C. FARI ˜NA, AND LOURDES RODR´IGUEZ-MESA
Abstract.
In this paper we consider Littlewood-Paley functions defined by the semigroups asso-ciated with the operator A = − ∆ − x ∇ in the inverse Gaussian setting for Banach valued func-tions. We characterize the uniformly convex and smooth Banach spaces by using L p ( R n , γ − )-properties of the A -Littlewood-Paley functions. We also use Littlewood-Paley functions associ-ated with A to characterize the K¨othe function spaces with the UMD property. ....... But there are those who struggle all their lives: These are the indispensable ones. (Bertolt Brecht) Dedicated to our friend Jos´e M´endez for his retirement Introduction
Let { T t } t> be a semigroup of operators on a measure space (Ω , µ ), k ∈ N and q ∈ (1 , ∞ ). TheLittlewood-Paley function g qk, { T t } t> associated with { T t } t> is defined by g qk, { T t } t> ( f )( x ) = (cid:18)Z ∞ (cid:12)(cid:12) t k ∂ kt T t f ( x ) (cid:12)(cid:12) q dtt (cid:19) /q , x ∈ Ω , for every f ∈ L p (Ω , µ ), 1 ≤ p < ∞ .Littlewood-Paley functions, also called g -functions, allow us to obtain equivalent norms in L p (Ω , µ ), 1 ≤ p < ∞ . This fact makes that L p -boundedness properties of operators connectedwith the semigroup { T t } t> can be obtained in a unified way.Stein developed in [30] the Littlewood-Paley theory for symmetric diffusion semigroups. Heproved that if { T t } t> is a symmetric difussion semigroup then, for every 1 < p < ∞ and k ∈ N there exists C > C k f − E ( f ) k p ≤ k g k, { T t } t> ( f ) k p ≤ C k f k p , f ∈ L p (Ω , µ ) . Actually C depends only on p and k . Here, E ( f ) = lim t →∞ T t ( f ) and E is the projection of L p (Ω , µ )on the fixed point space of { T t } t> .Suppose that X is a Banach space. For every 1 ≤ p < ∞ we denote by L pX (Ω , µ ) the p -thB¨ochner-Lebesgue space. It is wellknown that if L is a bounded operator from L p (Ω , µ ) into itselfwith 1 ≤ p < ∞ such that Lf ≥ ≤ f ∈ L p (Ω , µ ), then the tensor extension of L to L p (Ω , µ ) ⊗ X can be extended to L pX (Ω , µ ) as a bounded operator from L pX (Ω , µ ) into itself thatwill be also denoted by L .If { T t } t> is a positive semigroup of operators on (Ω , µ ), k ∈ N and q ∈ (1 , ∞ ) the Littlewood-Paley function g qk, { T t } t> can be defined on L pX (Ω , µ ) as follows g q,Xk, { T t } t> ( f )( x ) = (cid:18)Z ∞ (cid:13)(cid:13) t k ∂ kt T t ( f )( x ) (cid:13)(cid:13) qX dtt (cid:19) /q , x ∈ R n , Last modification: March 1, 2021.2020
Mathematics Subject Classification.
Key words and phrases.
Littlewood-Paley functions, inverse Gaussian measure, q -uniformly convex, q -uniformlysmooth and UMD Banach spaces, K¨othe function spaces.The authors are partially supported by grant PID2019-106093GB-I00 from the Spanish Government. for every f ∈ L pX (Ω , µ ) with 1 ≤ p < ∞ . Here k · k X denotes the norm in X . The question in thispoint is whether the inequalities in (1.1) are still valid in the Banach valued context.Let T be the unit circle. For every f ∈ L X ( T ) we define G X ( f )( z ) = (cid:18)Z ((1 − r ) k∇ P ( f )( rz ) k X ) dr − r (cid:19) / , z ∈ T . Here P represents the Poisson integral in T and ∇ is the gradient. It is known that there exists C > f ∈ L pX ( T ), with 1 < p < ∞ ,(1.2) 1 C k f k L pX ( T ) ≤ | ˆ f (0) | + k G X ( f ) k L p ( T ) ≤ C k f k L pX ( T ) , where ˆ f (0) = R T f ( w ) dw , if and only if X is isomorphic to a Hilbert space. Xu ([34]) considered g -functions defined as follows. Let q ∈ (1 , ∞ ), we define G q,X ( f )( z ) = (cid:18)Z ((1 − r ) k∇ P ( f )( rz ) k X ) q dr − r (cid:19) /q , z ∈ T , for every f ∈ L X ( T ). Xu characterized the Banach spaces X for which one of the inequalities in(1.2) holds.We recall the martingale type and cotype introduced by Pisier ([26]). A Banach space X issaid to be of martingale cotype (respectively, martingale type) q ∈ (1 , ∞ ) when there exists C > M n ) n ∈ N defined on some probability space with values in X thefollowing inequality holds X n ∈ N E k M n − M n − k qX ≤ C sup n ∈ N E k M n k qX (respectively, sup n ∈ N E k M n k qX ≤ C X n ∈ N E k M n − M n − k qX ) . Here, E denotes the expectation and M − = 0. If X has martingale cotype (respectively, type) q then q ≥ q ∈ (1 , M = ( M n ) n ∈ N with values in X , S q,X ( M ) = X n ∈ N k M n − M n − k qX ! /q , with 1 < q < ∞ . S q,X can be seen as a martingale analogue of G q,X . As it is commented by Xu([34, p. 208]) X if of martingale cotype (respectively, type) q if and only if for every (respectively,for some) 1 < p < ∞ there exists C > L p -martingale M = ( M n ) n ∈ N withvalues in X the following property holds E [ S q,X ( M )] p ≤ C sup n ∈ N E k M n k pX (respectively , sup n ∈ N E k M n k pX ≤ C E [ S q,X ( M )] p ) . The vector valued harmonic analysis is closely connected with the geometry of Banach spaces.The modulus of convexity of X is defined by δ X ( ε ) = inf (cid:26) − (cid:13)(cid:13)(cid:13)(cid:13) a + b (cid:13)(cid:13)(cid:13)(cid:13) X : a, b ∈ X, k a k X = k b k X = 1 , k a − b k X = ε (cid:27) , < ε < , and the modulus of smoothness of X is defined by ρ X ( t ) = sup (cid:26) k a + tb k X + k a − tb k X − a, b ∈ X, k a k X = k b k X = 1 , (cid:27) , t > .X is said to be uniformly convex when δ X ( ε ) >
0, for every 0 < ε <
2, and to be uniformlysmooth when lim t → + ρ X ( t ) t = 0. If q ∈ (1 , ∞ ) we say that X is q -uniformly convex (respectively, q -uniformly smooth) when there exists C > δ X ( ε ) ≥ Cε q , 0 < ε < ρ X ( t ) ≤ Ct q , t > G q,X and S q,X . Theorem A ([27, Chapter 10] and [34, Theorem 3.1 and Corollary 3.2]). Let X be a Banachspace and 1 < q ≤ ≤ q < ∞ . The following assertions are equivalent.(a) There exists a norm (cid:129) · (cid:129) on X that defines the original topology of X and such that( X ; (cid:129) · (cid:129) ) is q -uniformly convex (respectively, q -uniformly smooth).(b) X is of martingale cotype q (respectively, type q ).(c) For every (equivalently, for some) 1 < p < ∞ , there exists C > k G q ,X ( f ) k L p ( T ) ≤ C k f k L pX ( T ) , f ∈ L pX ( T ) , (respectively , k f k L pX ( T ) ≤ C ( k b f (0) k + k G q ,X ( f ) k L p ( T ) ) . (cid:3) This result also holds when the Poisson integral on T is replaced by the Poisson integral in R n × (0 , ∞ ). In this last case the projection E = 0 and the term containing b f (0) does not appear.Theorem A was extended to symmetric diffusion semigroups by Mart´ınez, Torrea and Xu ([21])and Xu ([35]). According to Stein (see [30]) a uniparametric family { T t } t> of operators definedon [ ≤ p ≤∞ L p (Ω , µ ) is called a symmetric diffusion semigroup when the following properties hold(i) T t is a contraction on L p (Ω , µ ), for every 1 ≤ p ≤ ∞ and t > T t + s = T t T s , on L p (Ω , µ ), for every 1 ≤ p ≤ ∞ and t, s > t → T t f = f , in L (Ω , µ ), for every f ∈ L (Ω , µ );(iv) T t is selfadjoint on L (Ω , µ );(v) T t is positive preserving, that is, T t f ≥ t >
0, provided that f ≥ T t is Markovian, that is, T t t > { P t } t> of { T t } t> is defined in the following way P t ( f ) = 1 √ π Z ∞ e − s √ s T t s ( f ) ds, t > . { P t } t> is also a symmetric diffusion semigroup. The Euclidean heat semigroup in R n is thefirst example of symmetric diffusion semigroup. The Euclidean Poisson semigroup is the Poissonsubordinated semigroup of the Euclidean heat semigroup. Theorem B ([21, Theorems 2.1 and 2.2]). Let X be a Banach space. Suppose that { T t } t> is asymmetric diffusion semigroup and that { P t } t> is the Poisson subordinated semigroup of { T t } t> .(a) If q ≥ q ∈ (1 , X is of martingale cotype q (respectively, martingaletype q ) then, for every 1 < p < ∞ , there exists C > (cid:13)(cid:13)(cid:13) g q,X , { P t } t> ( f ) (cid:13)(cid:13)(cid:13) L p (Ω ,µ ) ≤ C k f k L pX (Ω ,µ ) , f ∈ L pX (Ω , µ ) , (1.4)(respectively , k f k L pX (Ω ,µ ) ≤ C (cid:18) k E ( f ) k L pX (Ω ,µ ) + (cid:13)(cid:13)(cid:13) g q,X , { P t } t> ( f ) (cid:13)(cid:13)(cid:13) L p (Ω ,µ ) (cid:19) , f ∈ L pX (Ω , µ )) . (b) If q ≥ q ∈ (1 , { P t } t> = { P t } t> is the Euclidean Poisson semigroup for some 1 < p < ∞ , then X is of martingalecotype q (respectively, martingale type q ). Theorem C ([35, Theorem 2]) Let X be a Banach space. Suppose that { T t } t> is a symmetricdiffusion semigroup. If q ≥ q ∈ (1 , q (respectively,martingale type q ), then, for every 1 < p < ∞ , there exists C > (cid:13)(cid:13)(cid:13) g q,Xk, { T t } t> ( f ) (cid:13)(cid:13)(cid:13) L p (Ω ,µ ) ≤ k f k L pX (Ω ,µ ) , f ∈ L pX (Ω , µ ) , (respectively , k f k L pX (Ω ,µ ) ≤ C (cid:18) k E ( f ) k L pX (Ω ,µ ) + (cid:13)(cid:13)(cid:13) g q,Xk, { T t } t> ( f ) (cid:13)(cid:13)(cid:13) L p (Ω ,µ ) (cid:19) , f ∈ L pX (Ω , µ )) . The result in Theorem C had been previously established for the Euclidean heat semigroup in R n by Hyt¨onen and Naoir ([17]). A version of the Theorem B, (b), when the Euclidean Poissonsemigroup is replaced by Euclidean heat semigroup was proved in [3, Theorems 1.5 and 1.7]. V. ALMEIDA, J.J. BETANCOR, J.C. FARI˜NA, AND L. RODR´IGUEZ-MESA
Markovian property for the semigroup is crucial to prove the above results. Characterizations of q -uniformly convex and smooth Banach spaces as the ones established above but using semigroupsof operators without Markovian property (associated with Laguerre and Hermite operators) wereproved in [3], [4] and [5].Let α >
0. We choose m ∈ N such that m − ≤ α < m . If φ ∈ C m (0 , ∞ ) the Weyl α -derivative ∂ αt φ of φ is defined by ∂ αt φ ( t ) = 1Γ( m − α ) Z ∞ t ( ∂ mu φ )( u )( u − t ) m − α − du, t ∈ (0 , ∞ ) , provided that the last integral exists.Assume that { T t } t> is a symmetric diffusion semigroup on (Ω , µ ). For every α > f ∈ L p (Ω , µ ), 1 < p < ∞ and t >
0, we have that R ∞ t k ∂ mu T u ( f ) k L p (Ω) ( u − t ) m − α − du < ∞ , with m − ≤ α < m ([3, p. 11]). We define, for each f ∈ L p (Ω , µ ), 1 < p < ∞ , ∂ αt T t ( f ) = 1Γ( m − α ) Z ∞ t ∂ mu T u ( f )( u − t ) m − α − du, t > . We now consider the Littlewood-Paley function with fractional derivatives. If α > is defined by g q,Xα, { T t } t> ( f )( x ) = (cid:18)Z ∞ k t α ∂ αt T t ( f )( x ) k qX dtt (cid:19) /q , x ∈ R n , for every f ∈ L p (Ω , µ ), 1 < p < ∞ .Torrea and Zhang ([32, Theorems A and B]) proved a version of Theorem B involving fractionalLittlewood-Paley functions. In [3, Theorems 1.5 and 1.7] Theorem C is generalized by usingfractional g -functions.The measure dγ ( x ) = e −| x | π n/ dx , where dx denotes the Lebesgue measure in R n , is named Gauss-ian measure in R n . Harmonic analysis associated with the Gaussian-measure was began by Muck-enhoupt ([23]). The Ornstein-Uhlenbeck operator L is defined by L φ = − ∆2 φ + x · ∇ φ, φ ∈ C ∞ c ( R n ) . For every k ∈ N we denote by H k the k -th Hermite polynomial given by H k ( x ) = ( − k e x d k dx k e − x , x ∈ R . If k = ( k , . . . , k n ) ∈ N n we define H k ( x ) = ( − | k | n Y i =1 H k i ( x i ) , x = ( x , . . . , x n ) ∈ R n , where | k | = k + . . . + k n .The spectrum of L in L p ( R n , dγ ), 1 < p < ∞ , is the discrete set N and we have that, for every k ∈ N n , L H k = | k | H k . The Ornstein-Uhlenbeck operator −L generates a symmetric diffusion semigroup { T L t } t> . Op-erators defined by { T L t } t> (maximal operators, Riesz transforms, Littlewood-Paley functions,multipliers,. . . ) associated with harmonic analysis in the gaussian setting have been studied byseveral authors (see [13], [14], [15], [16], [24], [25], [33], and the references therein). In [16, The-orem 1.12] and [21, Theorems 6.1 and 6.2] q -uniformly convex and smooth Banach spaces byusing Littlewood-Paley functions defined by the Poisson semigroup associated with the Ornstein-Uhlenbeck operator.We call the inverse gaussian measure on R n to the measure dγ − ( x ) = π n/ e | x | dx . We willwrite γ − to refer us to the inverse Gauss measure. γ − is not doubling. The study of the harmonicanalysis in the inverse Gaussian setting was began by Salogni ([28]).We consider the differential operator A φ = −
12 ∆ φ − x · ∇ φ, for every φ ∈ C ∞ c ( R n ). A is essentially selfadjoint with respect to L ( R n , γ − ). We continuedenoting by A to the closure of this operator. For every k ∈ N n we define e H k ( x ) = e −| x | H k ( x ) , x ∈ R n . We have that A e H k = ( | k | + n ) e H k , k ∈ N n , and the spectrum of A in L p ( R n , γ − ) is the set { k + n, k ∈ N } , for every 1 < p < ∞ . The operator −A generates the semigroup of operators { T A t } t> in L ( R n , γ − ) defined by T A t ( f ) = X k ∈ N n e − ( | k | + n ) t c k ( f ) e H k , f ∈ L ( R n , γ − ) , where c k ( f ) = k e H k k − L ( R n ,γ − ) R R n f ( y ) e H k ( y ) dγ − ( y ), f ∈ L ( R n , γ − ) and k ∈ N n . By using theMehler’s formula we can write, for every t > T A t f ( x ) = Z R n T A t ( x, y ) f ( y ) dy, f ∈ L ( R n , γ − ) , being T A t ( x, y ) = 1 π n/ e − nt e − | x − e − ty | − e − t (1 − e − t ) n , x, y ∈ R n and t > . Thus, { T A t } t> is a symmetric diffusion semigroup and, for every t > ≤ p < ∞ , T A t admitsthe integral representation (1.5) in L p ( R n , γ − ).Salogni ([28, Chapter 3]) studied L p ( R n , γ − )-boundedness properties of the maximal operator T A∗ f = sup t> (cid:12)(cid:12) T A t f (cid:12)(cid:12) and of a certain spectral multiplier associated with the operator A .In [10] Bruno proved L p ( R , γ − )-boundedness and L p ( R , γ − )-unboundedness results for thepurely imaginary powers and the first order Riesz transforms associated with the translate oper-ator A + λI , λ >
0, from certain Hardy type space adaptaded to γ − to L ( R n , γ − ) and from L ( R n , γ − ) into L , ∞ ( R n , γ − ). Recently, Bruno and Sj¨ogren ([11]) have proved that the first andsecond order Riesz transform defined by A are bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) butRiesz transform with order greater than two are not bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).In [1] maximal operators given by T A∗ ,k ( f ) = sup t> (cid:12)(cid:12) t k ∂ kt T A t ( f ) (cid:12)(cid:12) , k ∈ N , are studied in L p ( R n , γ − ), 1 ≤ p < ∞ . Also, the Banach K¨othe function spaces with the Hardy-Littlewood property are characterized by using the maximal operator T A∗ ,k , k ∈ N . The UMDproperty for Banach spaces is characterized by using Riesz transforms and imaginary powers of A in [6, Theorems 1.4, 1.5 and 1.6] where it is established that higher order Riesz transforms in theinverse Gaussian setting are bounded from L p ( R n , γ − ) into itself, for every 1 < p < ∞ .Our objective in this paper is to study the L p ( R n , γ − )-boundedness properties of the Littlewood-Paley functions defined by { T A t } t> and also by the Poisson semigroup { P A t } t> subordinatedto { T A t } t> . We characterize q -uniformly convex and smooth Banach spaces by using theseLittlewood-Paley functions. Also, the UMD property for K¨othe function spaces is characterizedby g -functions in the inverse Gaussian setting.Let β > q ∈ (1 , ∞ ) and k = ( k , . . . , k n ) ∈ N n \ { } . Assume that X is a Banach space. Wedefine the Littlewood-Paley g-functions by g q,Xβ,k, { T A t } t> ( f )( x ) = (cid:18)Z ∞ (cid:13)(cid:13)(cid:13) t | k | + β ∂ kx ∂ βt T A t ( f )( x ) (cid:13)(cid:13)(cid:13) qX dtt (cid:19) /q , x ∈ R n . Here and in the sequel ∂ kx = ∂ | k | ∂x k ...∂x knn , x = ( x , ..., x n ) ∈ R n . When X = C we write g qβ,k, { T A t } t> .In a similar way we define g q,Xβ,k, { P A t } t> . Theorem 1.1.
Let β > , q ∈ [2 , ∞ ) and k ∈ N n .(i) g qβ,k, { T A t } t> is bounded from L p ( R n , γ − ) into itself, for every < p < ∞ . When < β ≤ , g qβ, , { T A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) . V. ALMEIDA, J.J. BETANCOR, J.C. FARI˜NA, AND L. RODR´IGUEZ-MESA (ii) g qβ,k, { P A t } t> is bounded from L p ( R n , γ − ) into itself, for every < p < ∞ . Furthermore, g qβ,k, { P A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) , provided that (respectively,if and only if ) | k | ≤ (respectively, when β ≥ ). We characterize q -uniformly convex and smooth Banach spaces by using g -functions associatedwith A . Theorem 1.2.
Let X be a Banach space, ≤ q < ∞ , < p < ∞ and β > . The followingassertions are equivalent.(i) There exists a norm (cid:129) · (cid:129) on X that defines the original topology of X such that ( X, (cid:129) · (cid:129) ) is q -uniformly convex.(ii) g q,Xβ, , { T A t } t> is bounded from L pX ( R n , γ − ) into L p ( R n , γ − ) .(iii) g q,Xβ, , { P A t } t> is bounded from L pX ( R n , γ − ) into L p ( R n , γ − ) .(iv) g q,Xβ, , { P A t } t> is bounded from L X ( R n , γ − ) into L , ∞ ( R n , γ − ) . Theorem 1.3.
Let X be a Banach space, ≤ q < , < p < ∞ and β > . The followingassertions are equivalent.(i) There exists a norm (cid:129) · (cid:129) on X that defines the original topology of X such that ( X, (cid:129) · (cid:129) ) is q -uniformly smooth.(ii) There exists C > such that k f k L pX ( R n ,γ − ) ≤ C (cid:13)(cid:13)(cid:13) g q,Xβ, , { T A t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . (iii) There exists C > such that k f k L pX ( R n ,γ − ) ≤ C (cid:13)(cid:13)(cid:13) g q,Xβ, , { P A t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . As a consequence of Theorem 1.2 and 1.3 we can deduce the following characterization of theHilbert spaces.
Corollary 1.4.
Let X be a Banach space, < p < ∞ and β > . The following assertions areequivalent.(i) There exists a norm (cid:129) · (cid:129) associated with an inner product in X that defines the originaltopology on X such that ( X, (cid:129) · (cid:129) ) is a Hilbert space.(ii) C k f k L pX ( R n ,γ − ) ≤ (cid:13)(cid:13)(cid:13) g ,Xβ, , { T A t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) .(iii) C k f k L pX ( R n ,γ − ) ≤ (cid:13)(cid:13)(cid:13) g ,Xβ, , { P A t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . Let (Ω , Σ , µ ) be a complete σ -finite measure space. A Banach space X consisting of equivalenceclasses modulus equality almost everywhere with respect to µ of locally integrable real functionson Ω is said to be a K¨othe function space when the following properties hold:(a) If | f ( w ) | ≤ | g ( w ) | , for almost all w ∈ Ω, f is measurable and g ∈ X , then f ∈ X and k f k X ≤ k g k X .(b) For every E ∈ Σ being µ ( E ) < ∞ , the characteristic function X E of E is in X .Each K¨othe function space endowed with the natural order is a Banach lattice. This lattice is σ -order complete. Furthermore, if X is an order continuous Banach lattice having a weak unity,then X is order isometric to a K¨othe function space ([20, Theorem1.b.14]). A Banach lattice E is called order continuous provided that every decreasing net in E whose infimum is zero isnorm-convergent to zero. We refer [20] for the main facts about Banach lattices.We say that a Banach space has the UMD-property when for every (equivalently, for some)1 < p < ∞ there exists C > (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =1 ε k d k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L pX (Ω ,µ ) ≤ C (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =1 d k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L pX (Ω ,µ ) , for every martingale difference sequence ( d k ) nk =1 ∈ ( L pX (Ω , µ )) n and ( ε k ) nk =1 ∈ {− , } n . UMDBanach spaces are very important in harmonic analysis. Burkholder [12] and Bourgain [8] proved that a Banach space X is UMD if and only if the Hilbert transform H defined byH( f )( x ) = lim ε → + π Z | x − y | >ε f ( y ) x − y dy, a.e. x ∈ R , for every f ∈ L p ( R ), 1 < p < ∞ , can be extended from L p ( R ) ⊗ X to L pX ( R ) as a bounded operatorfrom L pX ( R ) into itself, for every (equivalently, for some) 1 < p < ∞ . We refer to [18, Chapters 4and 5] for information on UMD-spaces.Xu ([34, Theorem 4.1]) established the equivalence of the UMD-property of a Banach latticewith a two-sided L p -estimates for g -functions involving Poisson integrals on T . Hyt¨onen ([19, The-orem 1.6]) characterized Banach UMD spaces by Littlewood-Paley function defined by stochasticintegrals. He uses again Poisson integrals on the circle. For Banach lattices, the results obtainedby Hyt¨onen reduce to the ones given by Xu.We now state our results about K¨othe function spaces with the UMD-valued functions and g -functions in the inverse Gaussian setting.Let X be a K¨othe function space and 1 < p < ∞ . Note that if f ∈ L pX ( R n , γ − ), then f can beidentify with a function f defined in R n × Ω. We consider, for every k ∈ N , g k,X { P A t } t> ( f )( x, ω ) = (cid:18)Z ∞ (cid:12)(cid:12) t k ∂ kt P A t ( f ( · , ω ))( x ) (cid:12)(cid:12) dtt (cid:19) / , x ∈ R n and ω ∈ Ω , for every f ∈ L pX ( R n , γ − ).Note that now g -functions are defined in a different way than above. These definitions can begiven for a Banach lattice X where the absolute value is defined but they can not be defined inthis way for general Banach spaces X .Next result can be seen as a version of [19, Theorem 1.6] in the inverse Gauss setting. Theorem D
Let X be a K¨othe function space, < p < ∞ , k ∈ N and { P t } t> a subordinatedsemigroup of a certain symmetric difussion semigroup.(i) If X has the UMD property, then there exists C > such that C k f k L pX ( R n ,γ − ) ≤ k g k,X { P t } t> ( f ) k L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , for every f ∈ L pX ( R n , γ − ) .(ii) If the inequalities in (i) hold for k = 1 and { P t } t> = { P t } t> is the Euclidean Poissonsemigroup, then X has the UMD property. Theorem 1.5.
Let X be an order continuous K¨othe function space and < p < ∞ . The followingassertions are equivalent.(i) X has the UMD-property.(ii) There exists C > such that (cid:13)(cid:13)(cid:13) g ,X { P A t } t> ( f ) (cid:13)(cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) , and (cid:13)(cid:13)(cid:13) g ,X ∗ { P A t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ′ X ∗ ( R n ,γ − ) ≤ C k f k L p ′ X ∗ ( R n ,γ − ) , f ∈ L p ′ X ∗ ( R n , γ − ) , where X ∗ denotes the dual of X and p ′ = pp − . Motivated by [34, Theorem 4.1] we stated the following result.
Theorem 1.6.
Let X be an order continuous K¨othe function space, < p < ∞ and n ∈ N , n ≥ .The following assertions are equivalent.(i) X has the UMD property.(ii) There exists C > such that, for every i = 1 , . . . , n , (cid:13)(cid:13)(cid:13) g Xi, { P A t } t> ( f ) (cid:13)(cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) , and (cid:13)(cid:13)(cid:13) g ,X ∗ { P A− It } t> ( f ) (cid:13)(cid:13)(cid:13) L p ′ X ∗ ( R n ,γ − ) ≤ C k f k L p ′ X ∗ ( R n ,γ − ) , f ∈ L p ′ X ∗ ( R n , γ − ) . Here X ∗ denotes the dual space of X and p ′ = pp − . V. ALMEIDA, J.J. BETANCOR, J.C. FARI˜NA, AND L. RODR´IGUEZ-MESA
Remark . When n = 1 we have that property ( ii ) implies that the space X has the UMD property(see the proof of Theorem 1.6).In the following sections we prove our results. As it is well-known Littlewood-Paley functions canbe seen as vector valued singular integrals. Then, our objective will be to prove L p -boundednessproperties of certain singular integral operators. In order to do this we decompose the operatorunder consideration into two parts: a local part, that is the singular part, and a global one. Thismethod of decomposition of operators in local and global parts was used by Muckenhoupt ([23])on the Gaussian context and then it has been used successfully in other contexts (see [4],[5],[7],[16], [24], and [25], for instance). The global operator is controlled by a positive integral operatorthat must have the L p ( R n , γ − )-boundedness properties that we want. On the other hand wemust prove that the local operator satisfies the same L p ( R n , γ − )-boundedness properties than theoperator appearing when we replace the operator A by the Laplacian operator − ∆.Throughout this paper, m ( x ) = min { , | x | − } , x ∈ R n \ { } , and m (0) = 1, and for every ν > N ν by N ν = n ( x, y ) ∈ R n × R n : | x − y | < νn p m ( x ) o . We name global regions as N cν , ν >
0. Observe that p m ( x ) ∼ (1 + | x | ) − , x ∈ R n . We always use c and C to denote positive constants that can change in each occurrence.2. Proof of Theorem 1.1
Proof of Theorem 1.1 (i), for 1 < p < ∞ . Since { T A t } t> is a symmetric diffusion semigroup,by [3, Theorem 1.5] we deduce that the g -function g qβ, , { T A t } t> is bounded from L p ( R n , γ − ) intoitself.Let f ∈ L p ( R n , γ − ). We have that ∂ kx T A t ( f )( x ) = Z R n ∂ kx T A t ( x, y ) f ( y ) dy, x ∈ R n . As in [32, Proposition 3.1] we obtain g qβ ,k, { T A t } t> ( f ) ≤ Cg qβ ,k, { T A t } t> ( f ) , < β < β . Thus, in order to prove (i) it is sufficient to see that g qm,k ; { T A t } t> is bounded from L p ( R n , γ − )into itself, for every m ∈ N .Let m ∈ N . In [31, Theorem 5.1] Teuwen obtained an explicit expression for ∂ mt T L t ( x, y ), x, y ∈ R n and t >
0, where T L t ( x, y ) = e − | y − e − tx | − e − t (1 − e − t ) n/ , x, y ∈ R n and t > . We think that some signs in the expression of [31, Theorem 5.1] are not correct. The new correctedequality is the following ∂ mt T L t ( x, y ) = ( − m T L t ( x, y ) X | r | = m (cid:18) mr . . . r m (cid:19) n Y i =1 r i X s i =0 s i X ℓ i =0 ( − s i + ℓ i s i (cid:26) r i s i (cid:27)(cid:18) s i ℓ i (cid:19) × (cid:18) e − t √ − e − t (cid:19) s i − ℓ i H l i ( x i ) H s i − ℓ i (cid:18) y i − e − t x i √ − e − t (cid:19) , (2.1)for each x = ( x , . . . , x n ), y = ( y , . . . , y n ) ∈ R n and t >
0. Here, for every
N, ℓ ∈ N , N ≥ ℓ ,the Stirling number of second kind (cid:8) Nℓ (cid:9) is defined as the number of partitions of an N -set into ℓ non-empty subset.Since T A t ( x, y ) = π − n/ e − nt T L t ( y, x ), x, y ∈ R n , t >
0, by using (2.1) we get ∂ mt T A t ( x, y ) = ( − m T A t ( x, y ) m X j =0 (cid:18) mj (cid:19) n m − j X | r | = j (cid:18) jr . . . r n (cid:19) n Y i =1 r i X s i =0 s i X ℓ i =0 ( − s i + ℓ i s i (cid:26) r i s i (cid:27)(cid:18) s i ℓ i (cid:19) × (cid:18) e − t √ − e − t (cid:19) s i − ℓ i H ℓ i ( y i ) H s i − ℓ i (cid:18) x i − e − t y i √ − e − t (cid:19) , for every x = ( x , . . . , x n ), y = ( y , . . . , y n ) ∈ R n and t > By taking into account that ddz e H ℓ ( z ) = − e H ℓ +1 ( z ), z ∈ R and ℓ ∈ N , we have that ∂ kx ∂ mt T A t ( x, y )= ( − m e − nt π n (1 − e − t ) n m X j =0 (cid:18) mj (cid:19) n m − j X | r | = j (cid:18) jr . . . r n (cid:19) n Y i =1 r i X s i =0 s i X ℓ i =0 ( − s i + ℓ i + k i s i (cid:26) r i s i (cid:27)(cid:18) s i ℓ i (cid:19) × (cid:18) e − t √ − e − t (cid:19) s i − ℓ i √ − e − t ) k i H ℓ i ( y i ) e H s i − ℓ i + k i (cid:18) x i − e − t y i √ − e − t (cid:19) = ( − m e − nt π n (1 − e − t ) n m X j =0 (cid:18) mj (cid:19) n m − j X | r | = j (cid:18) jr . . . r n (cid:19) X s ∈ I r X ℓ ∈ I s ( − | s | + | ℓ | + | k | | s | √ − e − t ) | k | (2.2) × (cid:18) e − t √ − e − t (cid:19) | s |−| ℓ | H ℓ ( y ) e H s − ℓ + k (cid:18) x − e − t y √ − e − t (cid:19) n Y i =1 (cid:26) r i s i (cid:27)(cid:18) s i ℓ i (cid:19) , x, y ∈ R n , t > . Here, I r , r = ( r , ..., r n ) ∈ N n , represents the set of all s = ( s , ..., s n ) ∈ N n such that 0 ≤ s j ≤ r j , j = 1 , ..., n .Then, it follows that, for every 0 < η < C > x, y ∈ R n and t > | t m + | k | ∂ kx ∂ mt T A t ( x, y ) | ≤ C m X j =0 X | r | = j X s ∈ I r X ℓ ∈ I s t m + | k | e − t (1 + | y | ) | ℓ | (1 − e − t ) | ℓ | (1 − e − t ) n + | k | + | s | e − η | x − e − ty | − e − t . We now observe that, when s ∈ N n , | s | ≤ m , t m + | k | e − t (1 − e − t ) | k | + | s | ≤ Ce − t , t > . Also, by performance the change of variables t = log s − s , t ∈ (0 , ∞ ), and since | y | ≤ ( | y + x | + | y − x | ), x, y ∈ R n , we obtain for every ℓ ∈ N n ,(1 + | y | ) ℓ (1 − e t ) | ℓ | e − η | x − e − ty | − e − t ≤ C (1 + | y | ) ℓ s | ℓ | e η | y | −| x | e − η (cid:0) s | x + y | + | x − y | s (cid:1) ≤ C (1 + | y | ) ℓ ( | x + y | + | x − y | ) ℓ e η | y | −| x | e − δ (cid:0) s | x + y | + | x − y | s (cid:1) ≤ Ce ( η − δ ) | y | −| x | e − δ | x − e − ty | − e − t , x, y ∈ R n , t > , where 0 < δ < η . As we will see, we choose, for convenience, 0 < δ < η < n > q (1 − δ )and η − δ < p < η + δ .From (2.3) we deduce that(2.4) | t m + | k | ∂ kx ∂ mt T A t ( x, y ) | ≤ C e − t (1 − e − t ) n e ( η − δ ) | y | −| x | e − δ | x − e − ty | − e − t , x, y ∈ R n , t > . We can write g qm,k, { T A t } t> ( f ) ≤ g qm,k, { T A t } t> ; loc ( f ) + g qm,k, { T A t } t> ; glob ( f ) , where g qm,k, { T A t } t> ; loc ( f )( x ) = (cid:18)Z ∞ (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( X N ν ( x, · ) f )( x ) (cid:12)(cid:12)(cid:12) q dtt (cid:19) q , x ∈ R n , and g qm,k, { T A t } t> ; glob ( f )( x ) = (cid:18)Z ∞ (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( X N cν ( x, · ) f )( x ) (cid:12)(cid:12)(cid:12) q dtt (cid:19) q , x ∈ R n . Minkowski inequality leads to g qm,k, { T A t } t> ; glob ( f )( x ) ≤ Z R n X N cν ( x, y ) R A m,k ( x, y ) f ( y ) dy, x ∈ R n , where R A m,k ( x, y ) = (cid:13)(cid:13)(cid:13) t m + | k | ∂ mt ∂ kx T A t ( x, y ) (cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) , x, y ∈ R n . Our objective is to see that the operator R A m,k defined by R A m,k ( f )( x ) = Z R n R A m,k ( x, y ) X N c ( x, y ) f ( y ) dy, x ∈ R n , is bounded from L p ( R n , γ − ) into itself.By (2.4) and since | x − ry | = | y − rx | + (1 − r )( | x | − | y | ), x, y ∈ R n and r ∈ R ,[ R A m,k ( x, y )] q ≤ Ce q ( η − δ ) | y | −| x | Z ∞ e − q t e − δq | x − e − ty | − e − t (1 − e − t ) nq dtt ≤ Ce q ( η + δ ) | y | −| x | Z ∞ e − t e − δq | y − e − tx | − e − t (1 − e − t ) nq dtt , x, y ∈ R n . Let ( x, y ) ∈ N c . As in [25, p. 861] we denote a = | x | + | y | and b = 2 h x, y i . Also we make thechange of variable 1 − s = e − t , t ∈ (0 , ∞ ), and we define u ( s ) = | y − x √ − s | s , s ∈ (0 , R A m,k ( x, y )] q ≤ Ce q ( η + δ ) | y | −| x | Z e − qδu ( s ) s nq ( − log(1 − s )) √ − s ds ≤ Ce q ( η + δ ) | y | −| x | Z e − qδu ( s ) s nq +1 √ − s ds. Assume now b ≤
0. Then, u ( s ) ≥ as − | x | , s ∈ (0 , Z e − qδu ( s ) s nq +1 √ − s ds ≤ e − qδ | y | Z e − qδ ( as − a ) s nq +1 √ − s ds ≤ Ce − qδ | y | Z ∞ e − qδr ( r + a ) nq − a nq dr √ r ≤ Ce − qδ | y | Z ∞ e − qδr ( r + 1) nq − dr √ r ≤ Ce − qδ | y | , because a ≥ n . We obtain R A m,k ( x, y ) ≤ Ce − ( η + δ ) | x | +( η − δ ) | y | . Suppose that b >
0. Since n > q (1 − δ ), we can write (see [25, p. 862]) Z e − qδu ( s ) s nq +1 √ − s ds = Z (cid:18) e − u ( s ) s n (cid:19) nq − n e − ( q ( δ − n ) u ( s ) s √ − s ds ≤ C e − u s n ! nq − n Z e − ( q ( δ − n ) u ( s ) s √ − s ds, where s = 2 √ a − b a + √ a − b and u = u ( s ) = | y | −| x | + | x + y || x − y | .We now proceed as in the proof of [25, Lemma 2.3]. Since a − b ≥
1, we get Z e − ( q ( δ − n ) u ( s ) s √ − s ds ≤ C e − ( q ( δ − n ) u √ s Z ∞ e − ( q ( δ − n ) s ( a − b ) (cid:18) √ s + 1 √ s (cid:19) ds ≤ C e − ( q ( δ − n ) u √ s . Then, Z e − qδu ( s ) s nq +1 √ − s ds ≤ C e − qδu s nq . According to [25, p. 863], since | x − y || x + y | ≥ n and s ∼ √ a − b a , we obtain R A m,k ( x, y ) ≤ C e ( η + δ ) | y | −| x | − δu s n/ = C e η ( | y | −| x | ) − δ | x + y || x − y | s n/ ≤ C (cid:16) | x + y || x − y | (cid:17) n/ e η ( | y | −| x | ) − δ | x + y || x − y | ≤ C | x + y | n e η ( | y | −| x | ) − δ | x + y || x − y | . By putting together the above estimates we get, when ( x, y ) ∈ N c ,(2.5) R A m,k ( x, y ) ≤ C e − ( η + δ ) | x | +( η − δ ) | y | , h x, y i ≤ , | x + y | n e η ( | y | −| x | ) − δ | x + y || x − y | , h x, y i > . Since (cid:12)(cid:12) | y | − | x | (cid:12)(cid:12) ≤ | x + y || x − y | , x, y ∈ R n , and that we have chosen η − δ < p < η + δ , we obtain(see [24, p. 501]) Z R n e | x | p − | y | p R A m,k ( x, y ) X N c ( x, y ) dy ≤ C (cid:18)Z R n e ( p − η + δ ) | x | e −| y | ( p + η − δ ) dy + Z R n | x + y | n e − ( δ − | η − p | ) | x − y || x + y | dy (cid:19) ≤ C, x ∈ R n . In a similar way we can see thatsup y ∈ R n Z R n e | x | −| y | p R A m,k ( x, y ) X N c ( x, y ) dx < ∞ . It follows that the operator R m,k ; p defined by R A m,k ; p ( f )( x ) = Z R n e | x | −| y | p R A m,k ( x, y ) X N c ( x, y ) f ( y ) dy, x ∈ R n , is bounded from L r ( R n , dx ) into itself, for every 1 ≤ r < ∞ . Then, the operator R A m,k is boundedfrom L p ( R n , γ − ) into itself. Hence, g qm,k, { T A t } t> ; glob is bounded from L p ( R n , γ − ) into itself.We now deal with g qm,k, { T t } t> ; loc . We define the operator G m,k, { T A t } t> by G m,k, { T A t } t> ( f )( x ) = t m + | k | ∂ mt ∂ kx T A t ( f )( x ) , x ∈ R n and t > . We are going to see that(a) G m,k, { T t } t> defines a bounded operator from L q ( R n , γ − ) into L qL q ((0 , ∞ ) dtt ) ( R n , γ − ).(b) R A m,k ( x, y ) ≤ C | x − y | n , ( x, y ) ∈ N .(c) For every i = 1 , . . . , n , (cid:18)Z ∞ (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ x i ∂ kx T A t ( x, y ) (cid:12)(cid:12)(cid:12) q dtt (cid:19) q ≤ C | x − y | n +1 , ( x, y ) ∈ N . This means that G m,k, { T A t } t> is a local Calder´on-Zygmund operator with respect to the Banachspace L q ((0 , ∞ ) , dtt ) in the inverse Gaussian setting. Salogni [28, § x, y ) ∈ N . Then, || x | − | y | | ≤ C and according to (2.4) weobtain that R A m,k ( x, y ) q ≤ C Z ∞ e − δq | x − e − ty | − e − t (1 − e − t ) nq e − qt t dt. The argument in the proof of [28, Lemma 3.3.1] leads to Z m ( x )0 e − δq | x − e − ty | − e − t (1 − e − t ) nq e − qt t dt ≤ C Z m ( x )0 e − c | x − y | t t nq +1 dt. Also, we have Z ∞ m ( x ) e − δq | x − e − ty | − e − t (1 − e − t ) nq e − qt t dt ≤ C Z ∞ m ( x ) e − qt (1 − e − t ) nq dtt ≤ C Z ∞ m ( x ) dtt nq +1 ≤ C ( m ( x )) nq ≤ C | x − y | nq . We conclude that R A m,k ( x, y ) ≤ C | x − y | n , ( x, y ) ∈ N . Thus (b) is proved. The estimation in (c) can be seen in a similar way.
Next we handle property (a). Let f ∈ L q ( R n , γ − ) and write B q = L q ((0 , ∞ ) , dtt ). Since q ≥ (cid:13)(cid:13)(cid:13) G qm,k, { T A t } t> ( f ) (cid:13)(cid:13)(cid:13) qL q B q ( R n ,γ − ) = Z R n Z ∞ (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12)(cid:12) q dtt dγ − ( x )= Z R n Z ∞ (cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12) sup t> (cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12) q − dtt dγ − ( x ) ≤ (cid:18)Z R n (cid:12)(cid:12) g m,k, { T A t } t> ( f )( x ) (cid:12)(cid:12) q dγ − ( x ) (cid:19) q (cid:18)Z R n (cid:12)(cid:12) T A m,k, ∗ ( f )( x ) (cid:12)(cid:12) q dγ − ( x ) (cid:19) − q . Here T A m,k, ∗ is the maximal operator defined by T A m,k, ∗ f ( x ) = sup t> (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12)(cid:12) , x ∈ R n . Thus, to get (a) it is sufficient to see that g m,k, { T A t } t> and T A m,k, ∗ are bounded from L q ( R n , γ − )into itself, because these properties allow us to conclude that (cid:13)(cid:13)(cid:13) G m,k, { T A t } t> ( f ) (cid:13)(cid:13)(cid:13) qL q B q ( R n ,γ − )) ≤ C k f k L q ( R n ,γ − ) k f k q − L q ( R n ,γ − ) = C k f k qL q ( R n ,γ − ) . In order to see that g m,k, { T A t } t> is bounded from L q ( R n , γ − ) into itself we are going to establishthe following properties:(a’) G m,k, { T A t } t> is bounded from L ( R n , γ − ) into L B ( R n , γ − ).(b’) (cid:13)(cid:13) t m + k ∂ mt ∂ kx T A t ( x, y ) (cid:13)(cid:13) B ≤ C | x − y | n , ( x, y ) ∈ N ,(c’) For every i = 1 . . . , n , (cid:13)(cid:13)(cid:13) t m + k ∂ mt ∂ x i ∂ kx T A t ( x, y ) (cid:13)(cid:13)(cid:13) B ≤ C | x − y | n +1 , ( x, y ) ∈ N . Note that (b’) and (c’) coincide with (b) and (c), respectively, when q = 2. So, we only need toprove (a’). Let f ∈ L ( R n , γ − ). We have that T A t ( f )( x ) = X ℓ ∈ N n c ℓ ( f ) e − ( n + | ℓ | ) t e H ℓ ( x ) , x ∈ R n . According to [29, p. 324], for every j ∈ N , | e H j ( u ) | ≤ p j j ! e − u , u ∈ R . Also, k e H j k L ( R n ,γ − ) = p π j j ! , j ∈ N . Then, for every ℓ = ( ℓ , . . . , ℓ n ) ∈ N n , | c ℓ ( f ) | ≤ C k e H ℓ k L ( R n ,γ − ) Z R n | f ( x ) || e H ℓ ( x ) | e | x | dx ≤ C k f k L ( R n ,γ − ) k e H l k L ( R n ,γ − . We get, for every ℓ = ( ℓ , . . . , ℓ n ) ∈ N n , | c ℓ ( f ) || e H ℓ ( x ) | ≤ Ce − | x | , x ∈ R n . Hence, for every t >
0, the series defining T A t ( f ) is absolute and uniformly convergent in R n . Also,since ddu e H j ( u ) = − e H j +1 ( u ), u ∈ R and j ∈ N , we can derivate under the sum sign to obtain ∂ mt ∂ kx T A t ( f )( x ) = X ℓ ∈ N n ( − | k | + n + | ℓ | c ℓ ( f ) e − t ( n + | ℓ | ) ( n + | ℓ | ) m e H ℓ + k ( x ) , x ∈ R n , t > . By using Plancherel identity for Hermite polynomials it follows that (cid:13)(cid:13)(cid:13) t m + | k | ∂ mt ∂ kx T A t ( f ) (cid:13)(cid:13)(cid:13) L B ( R n ,γ − ) = Z ∞ t m + | k | Z R n (cid:12)(cid:12) ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12) dγ − ( x ) dtt = Z ∞ t m + | k | X ℓ ∈ N n | c ℓ ( f ) | e − t ( n + | ℓ | ) ( n + | ℓ | ) m k e H ℓ + k k L ( R n ,γ − ) dtt = X ℓ ∈ N n | c ℓ ( f ) | k e H ℓ k L ( R n ,γ − ) k e H ℓ + k k L ( R n ,γ − ) k e H ℓ k L ( R n ,γ − ) ( n + | ℓ | ) m Z ∞ e − t ( n + | ℓ | ) t m + | k | dtt ≤ C X ℓ ∈ N n | c ℓ ( f ) | k e H ℓ k L ( R n ,γ − ) k e H ℓ + k k L ( R n ,γ − ) k e H ℓ k L ( R n ,γ − ) n + | ℓ | ) | k | ≤ C X ℓ ∈ N n | c ℓ ( f ) | k e H ℓ k L ( R n ,γ − ) ( n + | ℓ | ) | k | n Y j =1 ( ℓ j + k j )! ℓ j ! ≤ C X ℓ ∈ N n | c ℓ ( f ) | k e H ℓ k L ( R n ,γ − ) ( n + | l | ) | k | n Y j =1 ℓ k j j ≤ C X ℓ ∈ N n | c ℓ ( f ) | k e H ℓ k L ( R n ,γ − ) = C k f k L ( R n ,γ − ) . Thus, we obtain that G m,k, { T A t } t> is bounded from L ( R n , γ − ) into L B ( R n , γ − ). Accordingto a vector valued L ((0 , ∞ ) , dtt )) version of [28, Proposition 3.2.5] we conclude that the operator G m,k, { T A t } t> ; loc defined by G m,k, { T A t } t> ; loc ( f )( x ) = t m + | k | ∂ mt ∂ kx T A t ( X N ( x, · ) f )( x ) , x ∈ R n , t > , is bounded from L r ( R n , γ − ) into L r B ( R n , γ − ), for every 1 < r < ∞ . In other words, we have seenthat the operator g m,k, { T A t } t> ; loc is bounded from L r ( R n , γ − ) into itself, for every 1 < r < ∞ . Aswe had proved, g m,k, { T A t } t> ; glob is also bounded from L r ( R n , γ − ) into itself, for every 1 < r < ∞ .Hence, g m,k, { T A t } t> is bounded from L q ( R n , γ − ) into itself.We now consider the maximal operator T A m,k, ∗ ( f )( x ) = sup t> (cid:12)(cid:12)(cid:12) t m + | k | ∂ mt ∂ kx T A t ( f )( x ) (cid:12)(cid:12)(cid:12) , x ∈ R n . We are going to see that T A m,k, ∗ ( f )( x ) := sup t> (cid:12)(cid:12)(cid:12) M ( f )( t, x ) | , x ∈ R n , is bounded from L q ( R n , γ − ) into itself, where M ( f )( t, x ) = Z R n M ( t, x, y ) f ( y ) dy, x ∈ R n , t > , and M ( t, x, y ) = e − t (1 − e − t ) n e ( η + δ ) | y | −| x | e − δ | y − e − tx | − e − t , x, y ∈ R n , t > . Here, 0 < δ < η <
1. We observe that, once we prove this fact, according to (2.4) and since | x − ry | = | y − rx | + (1 − r )( | x | + | y | ), x, y ∈ R n , r >
0, we can deduce that T A m,k, ∗ is boundedfrom L q ( R n , γ − ) into itself.Consider M loc and M glob defined by M loc ( f )( t, x ) = M ( X N ν ( x, · ) f )( t, x ) and M glob ( f )( t, x ) = M ( X N cν ( x, · ) f )( t, x ) , x ∈ R n , t > . The operators T m,k, ∗ ; loc and T m,k, ∗ ; glob are defined in the obvious way. Here ν will be fixed later.We first study T A m,k, ∗ ; glob . Let λ, α ∈ (0 , ∞ ). Suppose that ( u, v ) ∈ N λ , that is, | u − v | <λn p m ( u ). Then, | αu − αv | < αλn p m ( u ) ≤ αλn ( p m ( αu ) , < α < ,α p m ( αu ) , α ≥ . Hence, ( αu, αv ) ∈ N αλ , when 0 < α <
1, and ( αu, αv ) ∈ N α λ , if α ≥
1. Since 0 < δ < √ δx, √ δy ) N provided that ( x, y ) N ν , with ν = δ − .Let ν = δ − . By [22, Proposition 2.1] we can write, for every ( x, y ) ∈ N cν ,sup t> M ( t, x, y ) ≤ Ce ( η + δ ) | y | −| x | ( e − δ | y | , h x, y i ≤ , | x + y | n e − δ ( | y | −| x | + | x + y || x − y | ) , h x, y i > , ≤ C e − ( η + δ ) | x | +( η − δ ) | y | , h x, y i ≤ , | x + y | n e η ( | y | −| x | ) − δ | x + y || x − y | , h x, y i > . By using that || y | − | x | | ≤ | x + y || x − y | , we obtain e | x | p − | y | p sup t> M ( t, x, y ) ≤ C e | x | ( p − η + δ ) e | y | ( η − δ − p ) , h x, y i ≤ , | x + y | n e − ( δ −| η − p | ) | x + y || x − y | , h x, y i > , ( x, y ) ∈ N cν . We choose 0 < δ < η < η − δ < p < η + δ . It follows that (see [24, p. 501])sup x ∈ R n Z R n e | x | p − | y | p sup t> M ( t, x, y ) X N cν ( x, y ) dy < ∞ , and, in a similar way, sup y ∈ R n Z R n e | x | p − | y | p sup t> M ( t, x, y ) X N cν ( x, y ) dx < ∞ . Then, the operator T A m,k, ∗ ; glob is bounded from L r ( R n , γ − ) into itself, for every 1 < r < ∞ . Inparticular, T A m,k, ∗ ; glob is bounded from L q ( R n , γ − ) into itself.On the other hand since || y | − | x | | ≤ | x − y || x + y | ≤ C when ( x, y ) ∈ N ν , by proceeding as inthe proof of [28, Lemma 3.3.1] M ( t, x, y ) ≤ C e − t (1 − e − t ) n e ( η − δ ) | y | −| x | e − δ | x − e − ty | − e − t ≤ C e − t e − c | x − y | − e − t (1 − e − t ) n ≤ C e − c | y − x | t t n , ( x, y ) ∈ N ν , t > . Then T A m,k, ∗ ; loc ( f )( x ) ≤ sup t> Z R n e − c | y − x | t t n | f ( y ) | dy, x ∈ R n . Hence, T A m,k, ∗ ; loc is bounded from L r ( R n , dx ) into itself, for every 1 < r < ∞ and by [28, Propo-sition 3.2.5] we deduce that T A m,k, ∗ ; loc is bounded from L q ( R n , γ − ) into itself. It follows that T A m,k, ∗ ; loc is bounded from L q ( R n , γ − ) into itself and property (a) is established.Since (a), (b) and (c) holds, by [28, Theorem 3.2.8], the operator G m,k, { T A t } t> ; loc given by G m,k, { T A t } t> ; loc ( f )( x ) = G m,k, { T A t } t> ( X N ( x, · ) f )( x ) , x ∈ R n , is bounded from L p ( R n , γ − ) into L p B q ( R n , γ − ). In other words, we have shown that g qm,k, { T A t } t> ; loc is bounded from L p ( R n , γ − ) into itself.Thus, we conclude that g m,k, { T A t } t> is bounded from L p ( R n , γ − ) into itself.2.2. Proof of Theorem 1.1 (i), for p = In order to see that g qβ, , { T A t } t> , with 0 < β ≤ L ( R n , γ − ) into L , ∞ ( R n , γ − ), it is sufficient to prove that g q , , { T A t } t> is boundedfrom L ( R n , γ − ) into L , ∞ ( R n , γ − ).We consider the operator G , , { T A t } t> ( f )( t, x ) = t∂ t T A t ( f )( x ) , x ∈ R n , t > . We proved that this operator is a local Calder´on-Zygmund one in the L q ((0 , ∞ ) , dtt )-setting. Then, G , , { T A t } t> ; loc is bounded from L ( R n , γ − ) into L , ∞ B q ( R n , γ − ), where B q = L q ((0 , ∞ ) , dtt ). Inother words, g q , , { T A t } t> ; loc is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).In order to finish the proof of our objective we have to see that g q , , { T A t } t> ; glob is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ). To establish this result we use some ideas developed in [11].We have that, for every x, y ∈ R n and t > t∂ t T A t ( x, y ) = − te | y | −| x | T A t ( y, x )1 − e − t n + 2 e − t n X i =1 ( y i − x i e − t ) x i − e − t | y − e − t x | − e − t ! . Then, | t∂ t T A t ( x, y ) | ≤ Cte − nt e | y | −| x | e − | y − e − tx | − e − t (1 − e − t ) n +1 (cid:16) e − t | x || y − e − t x | + e − t | y − e − t x | − e − t (cid:17) ≤ Cte − nt e | y | −| x | e − c | y − e − tx | − e − t (1 − e − t ) n +1 (cid:16) e − t | x | (1 − e − t ) (cid:17) , x, y ∈ R n , t > . (2.6)Let f ∈ L ( R n , γ − ). Minkowski inequality leads to g q , , { T A t } t> ; glob ( f )( x ) ≤ C Z R n f ( y ) X N c ( x, y ) (cid:13)(cid:13) t∂ t T A t ( x, y ) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) dy, x ∈ R n . Next we analyze the kernel (cid:13)(cid:13) t∂ t T A t ( x, y ) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) , when ( x, y ) ∈ N c . From (2.6) and byperfoming the change of variables r = e − t , t ∈ (0 , ∞ ) we get (cid:13)(cid:13) t∂ t T A t ( x, y ) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ≤ Ce | y | −| x | Z ∞ t q − e − nqt e − c | y − e − tx | − e − t (1 − e − t ) ( n +1) q (cid:16) | x | q (1 − e − t ) q (cid:17) dt q ≤ Ce | y | −| x | Z ( − log r ) q − r nq − e − c | y − rx | − r (1 − r ) ( n +1) q (cid:16) | x | q (1 − r ) q (cid:17) dr q ≤ Ce | y | −| x | Z ( − log r ) q − r nq − e − c | y − rx | (1 + | x | q ) dr ! q + (cid:16) Z e − c | y − rx | − r (1 − r ) n q +1 (cid:16) | x | q (1 − r ) q (cid:17) dr (cid:17) q = K ( x, y ) + K ( x, y ) , x, y ∈ R n . Then, g q , , { T A t } t> ; glob ( f )( x ) ≤ K ( | f | )( x ) + K ( | f | )( x ) , x ∈ R n , where K j ( f )( x ) = Z R n f ( y ) X N c ( x, y ) K j ( x, y ) dy, x ∈ R n , j = 0 , . We study K j , j = 0 ,
1. For every x, y ∈ R n with x = 0 we write y = y x + y ⊥ , where y x isparallel to x and y ⊥ is orthogonal to x . We define r ( x, y ) = | y || x | cos θ ( x, y ), where θ ( x, y ) ∈ [0 , π )represents the angle between x and y .First we show that(2.7) K ( x, y ) ≤ Ce | y | −| x | (cid:18) | x | − n + | x | − n + e − c | y ⊥ | | x | (cid:16) | y || x | (cid:17) n − X | y |≤ | x | ( x, y ) (cid:19) , ( x, y ) ∈ N c , and thus, by virtue of [11, Lemmas 4.2 and 4.3], we can conclude that K is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).We observe that | y − rx | ≥ | x | , when r ∈ (0 , ) and | y | ≥ | x | . Then, K ( x, y ) ≤ Ce | y | −| x | (1 + | x | ) e − c | x | Z ( − log r ) q − r nq − dr ! q ≤ Ce | y | −| x | | x | − n , | y | ≥ | x | . Consider now ( x, y ) ∈ N c , with | y | < | x | . Since | y − rx | = ( r − r ) | x | + | y ⊥ | , we have that K ( x, y ) ≤ Ce | y | −| x | (1 + | x | ) e − c | y ⊥ | × Z ( − log r ) q − ( | r | n − + | r − r | n − ) q e − c ( r − r ) | x | r q − dr ! q ≤ Ce | y | −| x | (1 + | x | ) e − c | y ⊥ | (cid:18)(cid:16) | y || x | (cid:17) n − + | x | − n (cid:19) Z ( − log r ) q − r q − dr ! q ≤ Ce | y | −| x | e − c | y ⊥ | (cid:18) | x | (cid:16) | y || x | (cid:17) n − + | x | − n (cid:19) . In the last inequality we have taken into account that | x | ≥ C , when ( x, y ) ∈ N c and | y | < | x | .Property (2.7) is then established. Next we deal with K ( x, y ), ( x, y ) ∈ N c . We make the following decomposition (see [11, p.12])( K ( x, y )) q = Ce ( | y | −| x | ) q × Z e − c ( r − r | x | | y ⊥| − r (1 − r ) nq +1 (cid:16) | x | q (1 − r ) q (cid:17)(cid:16) X { r ( , } ( x, y ) + X { r ∈ ( , } ( x, y ) (cid:17) dr = ( K , ( x, y )) q + ( K , ( x, y )) q , ( x, y ) ∈ N c . Let ( x, y ) ∈ N c . Assume first that r = r ( x, y ) ≤ or r ≥
2. In these cases, | r − r | ≥ c , foreach r ∈ ( , K , ( x, y ) ≤ Ce | y | −| x | Z e − c | x | − r (1 − r ) nq +1 (1 + | x | q (1 − r ) q ) dr q ≤ Ce | y | −| x | Z e − c | x | − r (1 − r ) nq +1 dr q ≤ Ce | y | −| x | | x | − n . Suppose now that r ∈ ( , | y x | ∼ | x | and | y ⊥ | ≥ | x | sin θ ( x, y ). As in [11, p.12] we splitthe interval ( ,
1) in the following three parts according to the regions in Figure 2.2:(1) I = (cid:8) r ∈ ( ,
1) : 1 − r ≤ max { (1 − r ) , ( r − } (cid:9) ;(2) I = (cid:8) r ∈ ( ,
1) : | − r | < (1 − r ) (cid:9) ;(3) I = (cid:8) r ∈ ( ,
1) : | r − r | ≤ (1 − r ) , r ∈ ( , (cid:9) . r r
13 23 R R R R When ( r , r ) ∈ R , we have that | r − r | ≥ c (1 − r ) and | r − r | ∼ | − r | = | x − y x || x | . Then, Z I e − c ( r − r | x | | y ⊥| − r (1 − r ) nq +1 (cid:16) | x | q (1 − r ) q (cid:17) dr ≤ C Z I e − c (1 − r ) | x | e − c | x − yx | | y ⊥| − r (1 − r ) nq +1 (cid:16) | x | q (1 − r ) q (cid:17) dr ≤ C Z e − c | x − yx | | y ⊥| − r (1 − r ) nq +1 dr ≤ C ( | x − y x | + | y ⊥ | ) nq ≤ C min (cid:8) | x − y | − nq , | y ⊥ | − nq (cid:9) ≤ C min (cid:8) (1 + | x | ) nq , | x sin θ ( x, y ) | − nq (cid:9) . If ( r , r ) ∈ R , then | r − r | ∼ − r and we can write Z I e − c ( r − r | x | | y ⊥| − r (1 − r ) nq +1 (cid:16) | x | q (1 − r ) q (cid:17) dr ≤ C Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 (cid:16) | x | q (1 − r ) q (cid:17) dr ≤ C Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 dr. We have that Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 dr ≤ C min (cid:8) (1 + | x | ) nq , | x sin θ ( x, y ) | − nq (cid:9) . Indeed, Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 dr ≤ C Z e − c | y ⊥| − r (1 − r ) nq +1 dr ≤ C | y ⊥ | nq ≤ C | x sin θ ( x, y ) | − nq . On the other hand, by making the change of variables s = (1 − r ) | x | , r ∈ I , we obtain Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 dr = C | x | nq Z s> | x || x − y x | e − cs e − c | y ⊥| | x | s s nq +1 ds. Since ( x, y ) ∈ N c , it follows that | x || x − y | ≥ c and then(2.8) | x || x − y x | ≥ c, when | y ⊥ | ≤ | x − y x | , and | y ⊥ || x | ≥ c, when | x − y x | ≤ | y ⊥ | . We get Z I e − c (1 − r ) | x | e − c | y ⊥| − r (1 − r ) nq +1 dr ≤ C | x | nq × Z ∞ e − cs ds, if | y ⊥ | ≤ | x − y x | Z ∞ e − cs s nq +1 ds, if | x − y x | ≤ | y ⊥ | ≤ C (1 + | x | ) nq . Finally, we analyze the integral when ( r , r ) ∈ R . In this case, 1 − r ∼ − r = | x − y x || x | . So, byusing [11, (5.3)] we get Z I e − c ( r − r | x | | y ⊥| − r (1 − r ) nq +1 (cid:0) | x | q (1 − r ) q (cid:1) dr ≤ C | x | q (1 − r ) q (1 − r ) nq +1 e − c | y ⊥| − r Z I e − c ( r − r | x | − r dr ≤ C | x | q (1 − r ) q (1 − r ) nq +1 e − c | y ⊥| − r min n − r , √ − r | x | o ≤ C | x | q (1 − r ) q (1 − r ) nq e − c | y ⊥| − r = C | x | nq (1 + ( | x || x − y x | ) q ) | x − y x | nq e − c | y ⊥| | x || x − yx | . When | x − y x | ≤ | y ⊥ | , we have that | x − y | ≤ | y ⊥ | , and since ( x, y ) ∈ N c we get | x | nq (1 + ( | x || x − y x | ) q ) | x − y x | nq e − c | y ⊥| | x || x − yx | ≤ C | x | nq | x − y x | nq e − c | y ⊥| | x || x − yx | ≤ C | y ⊥ | nq ≤ min n | x − y | nq , | x sin θ ( x, y ) | nq o ≤ C min n (1 + | x | ) nq , | x sin θ ( x, y ) | − nq o . Assume now that | y ⊥ | ≤ | x − y x | . From (2.8), we obtain that | x | nq (1 + ( | x || x − y x | ) q ) | x − y x | nq e − c | y ⊥| | x || x − yx | ≤ C | x | nq | x − y x | nq + | x | ( n +1) q | x − y x | ( n − q ! ≤ C | x | nq . Also, we can write | x | nq (1 + ( | x || x − y x | ) q ) | x − y x | nq e − c | y ⊥| | x || x − yx | ≤ C | y ⊥ | nq + | x | ( n +1) q | x − y x | ( n − q e − c | y ⊥| | x || x − yx | ! ≤ C | x sin θ ( x, y ) | nq + | x | ( n +1) q | x − y x | ( n − q e − c | y ⊥| | x || x − yx | X {| x || x − y x |≥ } ( x, y ) ! . In the last inequality we have taken into account that if | x || x − y x | ≤
1, then | x | ( n +1) q | x − y x | ( n − q e − c | y ⊥| | x || x − yx | ≤ C ( | x || x − y x | ) q | y ⊥ | nq ≤ C | y ⊥ | nq . By collecting all the estimates above whe have obtained that Z I e − c ( r − r | x | | y ⊥| − r (1 − r ) nq +1 (cid:0) | x | q (1 − r ) q (cid:1) dr ≤ C min n (1 + | x | ) nq , | x sin θ ( x, y ) | − nq o + | x | ( n +1) q | x − y x | ( n − q e − c | y ⊥| | x || x − yx | X {| x || x − y x |≥ } ( x, y ) ! , and we deduce that K ( x, y ) ≤ Ce | y | −| x | (cid:16) | x | − n + min n (1 + | x | ) n , | x sin θ ( x, y ) | − n o + | x | n +12 | x − y x | n − e − c | y ⊥| | x || x − yx | X {| x || x − y x |≥ } ( x, y ) ! , ( x, y ) ∈ N c . By taking into account [28, Lemma 3.3.4] and [11, Lemmas 4.2 and 4.4] we conclude that theoperator K is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).Hence, g , , { T A t } t> ; glob defines a bounded operator from L ( R n , γ − ) into L , ∞ ( R n , γ − ). Wehave proved that g , , { T A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).2.3. Proof of Theorem 1.1 (ii), for 1 < p < ∞ . We recall that P A t ( f ) = 1 √ π Z ∞ e − s √ s T A t / (4 s ) ( f ) ds, t > . As in ( i ) it is sufficient to prove ( ii ) with β ∈ N . Let f ∈ L p ( R n , γ − ) and m ∈ N . We have that ∂ mt ∂ kx P A t ( f )( x ) = 1 √ π Z ∞ e − s √ s ∂ kx ∂ mt T A t / (4 s ) ( f )( x ) ds, x ∈ R n , t > . By using Fa`a di Bruno’s formula we get ∂ mt T A t / (4 s ) ( f )( x ) = 12 m s m ∂ mv T A v ( f )( x ) (cid:12)(cid:12) v = t √ s = 1 s m X ℓ ∈ N , ≤ ℓ ≤ m c ℓ,m (cid:16) t √ s (cid:17) m − ℓ ∂ m − ℓv T A v ( f )( x ) (cid:12)(cid:12) v = t s , x ∈ R n , t, s > . Here c ℓ,m ∈ R , ℓ ∈ N , 0 ≤ ℓ ≤ m .We can write ∂ mt ∂ kx P A t ( f )( x ) = X ℓ ∈ N , ≤ ℓ ≤ m c ℓ,m Z ∞ e − s s m +12 (cid:16) t √ s (cid:17) m − ℓ ∂ kx ∂ m − ℓv T A v ( f )( x ) (cid:12)(cid:12) v = t s ds, x ∈ R n , t > . By using Minkowski inequality we obtain g qm,k, { P A t } t> ( f )( x ) = (cid:13)(cid:13)(cid:13) t m + | k | ∂ mt ∂ kx P A t ( f )( x ) (cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ≤ C X ℓ ∈ N , ≤ ℓ ≤ m Z ∞ e − s s m − ℓ + (cid:13)(cid:13)(cid:13)(cid:13) t m − ℓ )+ | k | ∂ kx ∂ m − ℓv T A v ( f )( x ) (cid:12)(cid:12) v = t s (cid:13)(cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ds ≤ C X ℓ ∈ N , ≤ ℓ ≤ m Z ∞ e − s s | k |− (cid:13)(cid:13) v m − ℓ + | k | ∂ kx ∂ m − ℓv T A v ( f )( x ) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ds ≤ C X ℓ ∈ N , ≤ ℓ ≤ m g qm − ℓ,k, { T A t } t> ( f )( x ) , x ∈ R n . (2.9)By using the property established in Section 2.1 we conclude that g qm,k, { P A t } t> is bounded from L p ( R n , γ − ) into itself, for every 1 < p < ∞ .2.4. Proof of Theorem 1.1 (ii), for p = Let m ∈ N and k ∈ N n . Our next objective is to seethat g qm,k, { P A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) when | k | ≤ G m,k, { T A t } t> satisfies the properties (a), (b) and (c). Then, fromthe L q ((0 , ∞ ) , dtt )-version of [28, Theorem 3.2.8] we deduce that g qm,k, { T A t } t> ; loc is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).By (2.9) it follows that g qm,k, { P A t } t> ; loc is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ).We now study the operator g qm,k, { P A t } t> ; glob . We can write P A t ( f ) = t √ π Z ∞ e − t s s T A s ( f ) ds = − √ π Z ∞ ∂ t ( e − t s ) T A s f ds √ s , t > . Let f ∈ L ( R n , γ − ). By using Minkowski inequality we get (cid:13)(cid:13) t m + | k | ∂ mt ∂ kx P A t ( f )( x ) (cid:13)(cid:13) L q (0 , ∞ ) , dtt ) ≤ C Z ∞ (cid:13)(cid:13) t m + | k | ∂ m +1 t ( e − t s ) (cid:13)(cid:13) L q (0 , ∞ ) , dtt ) | ∂ kx T A s f ( x ) | ds √ s . Formula de Fa`a di Bruno leads to ∂ m +1 t ( e − t s ) = e − t s s m +12 X ℓ ∈ N , ≤ ℓ ≤ m +12 c m,ℓ (cid:16) t s (cid:17) m +12 − ℓ , t, s > , for certain c m,ℓ ∈ R , 0 ≤ ℓ ≤ ( m + 1) /
2. Then, (cid:13)(cid:13) t m + | k | ∂ m +1 t ( e − t s ) (cid:13)(cid:13) qL q ((0 , ∞ ) , dtt ) ≤ C X ℓ ∈ N , ≤ ℓ ≤ m +12 Z ∞ e − qt s t (2 m + | k | +1 − ℓ ) q − s ( m +1 − ℓ ) q dt ≤ Cs ( | k |− q , s > , and we obtain (cid:13)(cid:13) t m + | k | ∂ mt ∂ kx P A t ( f )( x ) (cid:13)(cid:13) L q (0 , ∞ ) , dtt ) ≤ C Z ∞ s | k | − | ∂ kx T A s f ( x ) | ds ≤ C Z R n | f ( y ) | Z ∞ s | k | − | ∂ kx T A s ( x, y ) | dsdy, x ∈ R n . According to [11, Proposition 5.1] we deduce that the operator(2.10) L ( f )( x ) = Z R n f ( y ) X N c ( x, y ) Z ∞ s | k | − | ∂ kx T A s ( x, y ) | dsdy, x ∈ R n , is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) when | k | ≤
2. Note that the proof of [11, Propo-sition 5.1] also works for | k | = 0.Hence, the operator g qm,k, { P A t } t> ; glob is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) providedthat | k | ≤
2. We conclude that g qm,k, { P A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) when | k | ≤ g qm,k, { P A t } t> is not bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) when | k | ≥
3. We use some ideas developed in [11, § η >
0. We define z = ( η, ..., η ). We will choose η large. For every x ∈ R n we write x = x z + x ⊥ , where x z is parallel to z and x ⊥ is orthogonal to z . Let us consider the set J ( z ) = n x ∈ R n : | x ⊥ | < , | z | < | x z | < | z | o . We have that (see [11, § η > η > η we have that, for every r ∈ (0 , y ∈ B ( z,
1) and x ∈ J ( z ), x i − ry i √ − r ≥ C | z | , i = 1 , ..., n, and hence H k (cid:18) x − ry √ − r (cid:19) ≥ C | z | k . Also we get e − | y − rx | − r ≥ Ce − c | r | x z |−| z || , x ∈ J ( z ) , y ∈ B ( z,
1) and r ∈ (cid:16) , (cid:17) . We choose 0 < t < t < ∞ such that for a certain C > (cid:12)(cid:12)(cid:12) H m +1 (cid:16) t − log r ) (cid:17)(cid:12)(cid:12)(cid:12) ≥ C, t ∈ ( t , t ) and r ∈ (cid:16) , (cid:17) , and the sign of H m +1 (cid:16) t − log r ) (cid:17) is constant when t ∈ ( t , t ) and r ∈ ( , ). We write a = − sign H m +1 (cid:16) t − log r ) (cid:17) , t ∈ ( t , t ) and r ∈ (cid:16) , (cid:17) . We now take a function 0 ≤ f ∈ L ( R n , γ − ) such that supp f ⊂ B ( z,
1) and k f k L ( R n ,γ − ) = 1.We have that g qm,k, { P A t } t> ( f )( x ) = (cid:13)(cid:13)(cid:13) t m + | k | ∂ mt ∂ kx P A t ( f )( x ) (cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) = (cid:13)(cid:13)(cid:13) Z R n f ( y ) H k,m ( t, x, y ) dy (cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) , where, for each x, y ∈ R n and t > H k,m ( t, x, y ) = − t m + | k | √ π Z ∞ ∂ m +1 t ( e − t s ) ∂ kx T A s ( x, y ) dss = ( − t ) m + | k | m +1 √ π Z ∞ H m +1 (cid:16) t √ s (cid:17) e − t s s m + e − ns H k (cid:16) x − e − s y √ − e − s (cid:17) e − | x − e − sy | − e − s (1 − e − s ) n + | k | ds. By making r = e − s , s ∈ (0 , ∞ ), it follows that, when t ∈ ( t , t ), y ∈ B ( z, η > η and x ∈ J ( z ), a H k,m ( t, x, y ) = ( − t ) m + | k | m +1 √ π Z H m +1 (cid:16) t − log r ) (cid:17) e − t − log r ) ( − log r ) m + r n − H k (cid:16) x − ry √ − r (cid:17) e − | x − ry | − r (1 − r ) n + | k | dr ≥ Ct m + | k | η | k | e − ct e | y | −| x | Z / / e − c | r | x z |−| z || dr. Then, g qm,k, { P A t } t> ( f )( x ) ≥ C (cid:18)Z t t t ( m + | k | ) q − e − ct dt (cid:19) q η | k | e −| x | Z / / e − c | r | x z |−| z || dr ≥ Cη | k |− e −| x | ≥ C | z | k − e − ( | z | ) , x ∈ J ( z ) , η > η . Moreover, γ − ( J ( z )) ≥ e ( | z | ) | z | − . We getsup s> sγ − ( { x ∈ R n : g qm,k, { P A t } t> ( f )( x ) > s } ) ≥ Ce − ( | z | ) | z | | k |− γ − ( J ( z )) ≥ C | z | | k |− , η > η . We conclude that g qm,k, { P A t } t> is not bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) when | k | > Remark . As it was mentioned during the last proof we have seen that the operator g qm,k, { T A t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ) except when m = 1 and k = 0. As far as we knowthis question is also open even when q = 2, k = 0 and m = 1 in the Gaussian setting, that is, it isnot known whether the Littlewood-Paley function g , , { T L t } t> is bounded from L ( R n , γ − ) into L , ∞ ( R n , γ − ). We recall that L denotes the Ornstein-Uhlenbeck operator.3. Proof of Theorems 1.2 and 1.3 and Corollary 1.4
Proof of Theorem 1.2.
According to [35, Theorem 2] and [3, Theorem 1.5], since { T A t } t> is a symmetric diffusion semigroup, (i) ⇒ (ii) holds. By using [32, Theorem A] (see also [21,Theorem 2.1]) we can see that (i) ⇒ (iii).Suppose that (ii) holds. We are going to prove (i). According to [3, Theorem 1.5 (b)], thereexists a norm (cid:129) · (cid:129) on X that defines the original topology of X such that ( X, (cid:129) · (cid:129) ) is q -uniformlyconvex, provided that g q,Xα, , { W t } t> is bounded from L pX ( R n , dx ) into L p ( R n , dx ), for some α > { W t } t> denotes the classical heat semigroup being, for every t > W t ( f )( x ) = 1 π n/ Z R n e −| x − y | / t (2 t ) n/ f ( y ) dy, x ∈ R n . Since g q,Xβ , , { W t } t> ( f ) ≤ g q,Xβ , , { W t } t> ( f ) ([32, Proposition 3.1]), 0 < β < β , we can assumethat β ∈ (0 , g q,Xβ, , { W t } t> is bounded from L pX ( R n , dx ) into L p ( R n , dx )We define the local and global operators in the usual way associated to N . Minkowski inequalityleads to g q,Xβ, , { W t } t> ; loc ( f )( x ) ≤ (cid:12)(cid:12)(cid:12) g q,Xβ, , { W t } t> ; loc ( f )( x ) − g q,Xβ, , { T A t } t> ; loc ( f )( x ) (cid:12)(cid:12)(cid:12) + g q,Xβ, , { T A t } t> ; loc ( f )( x ) ≤ H β ( f )( x ) + g q,Xβ, , { T A t } t> ; loc ( f )( x ) , x ∈ R n . where H β ( f )( x ) = Z R n k f ( y ) k X (cid:18)Z ∞ | t β ∂ βt ( W t ( x − y ) − T A t ( x, y )) | q dtt (cid:19) /q X N ( x, y ) dy, x ∈ R n . Since β ∈ (0 , ∂ βt ( W t ( x − y ) − T A t ( x, y )) = 1Γ(1 − β ) Z ∞ t ∂ u ( W u ( x − y ) − T A u ( x, y ))( u − t ) − β du, x, y ∈ R n . By proceeding as in the proof [32, Proposition 3.1] we get (cid:18)Z ∞ | t β ∂ βt ( W t ( x − y ) − T A t ( x, y )) | q dtt (cid:19) /q ≤ C (cid:18)Z ∞ | t∂ t ( W t ( x − y ) − T A t ( x, y )) | q dtt (cid:19) /q , x, y ∈ R n . Then, g q,Xβ, , { W t } t> ; loc ( f ) ≤ C (cid:16) H ( f ) + g q,Xβ, , { T A t } t> ; loc ( f ) (cid:17) . We consider the operator G Xβ, { T A t } t> ( f )( t, x ) = t β ∂ βt T A t ( f )( x ) , x ∈ R n and t > . Since (ii) holds, G Xβ, { T A t } t> is bounded from L pX ( R n , γ − ) into L pL qX ((0 , ∞ ) , dtt ) ( R n , γ − ). By theproperty (b) in the proof of Theorem 1.1 we have that (cid:18)Z ∞ | t β ∂ βt T A t ( x, y ) | q dtt (cid:19) /q ≤ C | x − y | n , ( x, y ) ∈ N . A vector valued version of [28, Proposition 3.2.7] (see [16, Proposition 2.3]) allow us to obtainthat G Xβ, { T A t } t> ; loc is bounded from L pX ( R n , dx ) into L pL qX ((0 , ∞ ) , dtt ) ( R n , dx ). or, in other words g q,Xβ, , { T A t } t> ; loc is bounded from L pX ( R n , dx ) into L p ( R n , dx ).We now study the operator H . We have that t∂ t ( W t ( x − y ) − T A t ( x, y )) = − nt (cid:18) W t ( x − y )2 t − T A t ( x, y )1 − e − t (cid:19) + 2 t (cid:18) | x − y | (2 t ) W t ( x − y ) − e − t | x − e − t y | (1 − e − t ) T A t ( x, y ) (cid:19) + 2 te − t h y, x − e − t y i − e − t T A t ( x, y ) =: X j =1 K j ( t, x, y ) , x, y ∈ R n , t > . We are going to estimate k K j ( · , x, y ) k L q ((0 , ∞ ) , dtt ) , j = 1 , ,
3, for every ( x, y ) ∈ N . By perform-ing the change of variable t = log s − s , s ∈ (0 , ∞ ), we obtain k K ( · , x, y ) k qL q ((0 , ∞ ) , dtt ) ≤ C | y | q Z ∞ t q e − q ( n +1) t e − c | x − e − ty | − e − t (1 − e − t ) n +12 q dtt ≤ C | y | q e − c ( | x | −| y | ) Z ∞ e − qnt e − c | y − e − tx | − e − t (1 − e − t ) n − q dtt ≤ C | y | q e − c ( | x | −| y | ) Z (cid:18) − s s (cid:19) qn e − c ( s | x + y | + s | x − y | ) s n − q ds (1 − s ) log s − s ≤ C | y | q e − c ( | x | −| y | ) Z e − c | x − y | s s n − q e − cs ( | x + y | + | x − y | ) ds − log(1 − s ) , x, y ∈ R n . Since 2 | y | ≤ ( | x + y | + | x − y | ), x, y ∈ R n , − log(1 − s ) ≥ cs , s ∈ (0 , ∞ ) and || y | − | x | | ≤ C ,( x, y ) ∈ N , it follows that k K ( · , x, y ) k qL q ((0 , ∞ ) , dtt ) ≤ C | y | q/ Z e − c | x − y | s s ( n − + ) q +1 ds ≤ C | y | q/ | x − y | ( n − / q , ( x, y ) ∈ N . Then k K ( · , x, y ) k L q ((0 , ∞ ) , dtt ) ≤ C p | x || x − y | n − / , ( x, y ) ∈ N . Next we deal with k K j ( · , x, y ) k L q ((0 , ∞ ) , dtt ) , j = 1 ,
2. Let us first observe that | K ( t, x, y ) | + | K ( t, x, y ) | ≤ C (cid:18)(cid:16) | x − y | t (cid:17) W t ( x − y ) + t − e − t (cid:16) | x − e − t y | − e − t (cid:17) T A t ( x, y ) (cid:19) ≤ C (cid:18) t n + te − nt (1 − e − t ) n +1 (cid:19) ≤ C t n , x, y ∈ R n , t > . Then, by using that | x − y | ≤ n p m ( x ) when ( x, y ) ∈ N , and that p m ( x ) ∼ (1 + | x | ) − , x ∈ R n ,we get that Z ∞ m ( x ) ( | K ( t, x, y ) | q + | K ( t, x, y ) | q ) dtt ≤ C Z ∞ m ( x ) dtt n q +1 = Cm ( x ) n q ≤ Cm ( x ) q | x − y | ( n − ) q ≤ C (cid:16) p | x || x − y | n − (cid:17) q , ( x, y ) ∈ N . (3.1)On the other hand, we can write K ( t, x, y ) = − nt (cid:18) t (cid:0) W t ( x − y ) − T A t ( x, y ) (cid:1) + (cid:16) t − − e − t (cid:17) T A t ( x, y ) (cid:19) , x, y ∈ R n , t > , and K ( t, x, y ) = | x − y | t (cid:0) W t ( x − y ) − T A t ( x, y ) (cid:1) + 2 t (cid:16) | x − y | (2 t ) − | x − e − t y | (1 − e − t ) (cid:17) T A t ( x, y )+ 2 t | x − e − t y | − e − t T A t ( x, y ) x, y ∈ R n , t > . We have that, for each x, y ∈ R n and t ∈ (0 , (cid:12)(cid:12) | x − y | − | x − e − t y | (cid:12)(cid:12) = | − (1 − e − t ) | y | + 2(1 − e − t ) h x − y, y i| ≤ t | y | + 2 t | x − y || y | , and (cid:12)(cid:12)(cid:12)(cid:12) | x − y | (2 t ) − | x − e − t y | (1 − e − t ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) | x − y | (cid:18) t ) − − e − t ) (cid:19) − | y | + 2 h x − y, y i − e − t (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) | x − y | t + | y | + | x − y || y | t (cid:19) . We also get that, for every x, y ∈ R n and t ∈ (0 , (cid:12)(cid:12)(cid:12)(cid:12) e − | x − y | t − e − | x − e − ty | − e − t (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12)(cid:12) | x − y | t − | x − e − t y | − e − t (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:16) − min n | x − y | t , | x − e − t y | − e − t o(cid:17) ≤ C ( | x − y | + t | y | + | x − y || y | ) exp (cid:16) − min n | x − y | t , | x − e − t y | − e − t o(cid:17) , and since e − | x − e − ty | − e − t ≤ Ce − c | x − y | t , ( x, y ) ∈ N , t ∈ (0 , , we obtain that(3.2) (cid:12)(cid:12)(cid:12)(cid:12) e − | x − y | t − e − | x − e − ty | − e − t (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:16) t (1 + | y | ) + | x − y || y | (cid:17) e − c | x − y | t , ( x, y ) ∈ N , t ∈ (0 , . Hence, (cid:12)(cid:12) ( W t ( x − y ) − T A t ( x, y ) (cid:12)(cid:12) ≤ π n (cid:18) t ) n (cid:12)(cid:12)(cid:12)(cid:12) e − | x − y | t − e − | x − e − ty | − e − t (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) t ) n − − e − t ) n (cid:12)(cid:12)(cid:12)(cid:12) e − | x − e − ty | − e − t + (1 − e − nt ) e − | x − e − ty | − e − t (1 − e − t ) n ≤ C e − c | x − y | t t n − (cid:18) | y | + | x − y || y | t (cid:19) , ( x, y ) ∈ N , t ∈ (0 , . (3.3)We have used that | (2 t ) r − (1 − e − t ) r | ≤ Ct r , when t ∈ (0 ,
1) and r ∈ R . By considering all these estimations we obtain | K ( t, x, y ) | + | K ( t, x, y ) | ≤ C " e − c | x − y | t t n − (cid:18) | y | + | x − y || y | t (cid:19) + tT A t ( x, y ) (cid:18) | x − y | t + | y | + | x − y || y | t + | x − e − t y | − e − t (cid:19)(cid:21) ≤ C e − c | x − y | t t n − (cid:18) | y | + | x − y || y | t (cid:19) ≤ C e − c | x − y | t t n − (cid:18) | y | + | y |√ t (cid:19) , ( x, y ) ∈ N , t ∈ (0 , . By taking into account that m ( x ) | y | ≤ C , ( x, y ) ∈ N , we can write t | y | + t | y | ≤ t | y | (cid:0) m ( x ) | y | + m ( x ) | y | (cid:1) ≤ Ct | y | , ( x, y ) ∈ N , t ∈ (0 , m ( x )) . (3.4)Then, we get, for j = 1 , Z m ( x )0 | K j ( t, x, y ) | q dtt ≤ C Z m ( x )0 e − c | x − y | t t ( n − q +1 (cid:18) | y | + | y |√ t (cid:19) q dt ≤ C Z e − c | x − y | t t ( n − q +1 dt + | y | q Z m ( x )0 e − c | x − y | t t ( n − ) q +1 dt ! ≤ C | x − y | ( n − ) q Z t q − dt + | y | q Z ∞ e − c | x − y | t t ( n − ) q +1 dt ! ≤ C | y | q | x − y | ( n − ) q ≤ C p | x || x − y | n − ! q , ( x, y ) ∈ N . (3.5)We obtain k K ( · , x, y ) k L q ((0 , ∞ ) , dtt ) + k K ( · , x, y ) k L q ((0 , ∞ ) , dtt ) ≤ C p | x || x − y | n − / , ( x, y ) ∈ N . By putting together all the above estimates we conclude that(3.6) (cid:13)(cid:13)(cid:13) t∂ t ( W t ( x − y ) − T A t ( x, y )) (cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ≤ C p | x || x − y | n − , ( x, y ) ∈ N . We have that Z R n p | x || x − y | n − X N ( x, y ) dy ≤ C p | x | Z √ m ( x )0 dρ √ ρ ≤ Cm ( x ) p | x | ≤ C, x ∈ R n . Since 1 + | x | ∼ | y | when ( x, y ) ∈ N we can also see thatsup y ∈ R n Z R n p | x || x − y | n − X N ( x, y ) dx < ∞ . Then, the operator H is bounded from L q ( R n , dx ) into itself, for every 1 ≤ q < ∞ .We conclude that g q,Xβ, , { W t } t> ; loc is bounded from L pX ( R n , dx ) into L p ( R n , dx ).The properties of invariance of the operator g q,Xβ, , { W t } t> allow us to prove (see [16, p. 21])that g q,Xβ, , { W t } t> can be extended to L pX ( R n , dx ) as a bounded operator from L pX ( R n , dx ) into L p ( R n , dx ). Our objective is established and (i) is proved.We are going to see that (iii) ⇒ (i) and (iv) ⇒ (i). Assume that (iii) holds. As in the proof of(ii) ⇒ (i) we can assume that 0 < β <
1. We can also see that g q,Xβ, , { P t } t> ; loc ( f ) ≤ (cid:16) H ( f ) + g q,Xβ, , { P A t } t> ; loc ( f ) (cid:17) , where H ( f )( x ) = Z R n k f ( y ) k (cid:13)(cid:13) t∂ t ( P t ( x − y ) − P A t ( x, y )) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) dy, x ∈ R n . Here, { P t } t> denotes the Euclidean semigroup in R n . We consider the operator G Xβ, { P A t } t> ( f )( t, x ) = t β ∂ βt P A t ( f )( x ) , x ∈ R n and t > . Since (iii) holds, G Xβ, { P A t } t> is bounded from L pX ( R n , γ − ) into L pL qX ((0 , ∞ ) , dtt ) ( R n , γ − ).According to [2, Lemma 4] we obtain k t β ∂ βt P A t ( x, y ) k L q ((0 , ∞ ) , dtt ) ≤ C Z ∞ (cid:18)Z ∞ | t β ∂ βt ( te − t / u ) | q dtt (cid:19) /q T A u ( x, y ) duu / ≤ C Z ∞ Z ∞ t β e − t / u u β − ! q dtt ! /q T A u ( x, y ) duu / ≤ C Z ∞ T A u ( x, y ) u du, x, y ∈ R n . Since e − | x − e − ty | − e − t ≤ Ce − | x − y | − e − t , ( x, y ) ∈ N , t >
0, we get k t β ∂ βt P A t ( x, y ) k L q ((0 , ∞ ) , dtt ) ≤ C Z e − | x − y | u u n +1 du + Z ∞ m ( x ) u n +1 du ! ≤ C (cid:18) | x − y | n + 1 m ( x ) n (cid:19) ≤ C | x − y | n , ( x, y ) ∈ N . In an analogous way we can see that, for every i = 1 , ..., n , k ∂ x i t β ∂ βt P A t ( x, y ) k L q ((0 , ∞ ) , dtt ) ≤ C | x − y | n +1 , ( x, y ) ∈ N . From a L q ((0 , ∞ ) , dtt )-version of [28, Proposition 3.2.7] (see [16, Proposition 2.3]) we deduce that G Xβ, { P A t } t> ; loc is bounded from L pX ( R n , γ − ) into L pL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) and from L pX ( R n , dx ) into L pL qX ((0 , ∞ ) , dtt ) ( R n , dx ). Hence, g q,Xβ, , { P A t } t> ; loc is bounded from L pX ( R n , dx ) into L p ( R n , dx ).On the other hand, we can write (cid:13)(cid:13) t∂ t ( P t ( x − y ) − P A t ( x, y )) (cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) = 1 √ π (cid:13)(cid:13)(cid:13)(cid:13)Z ∞ e − u t∂ t (cid:0) W t / u ( x − y ) − T A t / u ( x, y ) (cid:1) du √ u (cid:13)(cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) ≤ C Z ∞ e − u (cid:13)(cid:13)(cid:13) t∂ t (cid:0) W t / u ( x − y ) − T A t / u ( x, y ) (cid:1)(cid:13)(cid:13)(cid:13) L q ((0 , ∞ ) , dtt ) du √ u ≤ C (cid:13)(cid:13) s∂ s (cid:0) W s ( x − y ) − T A s ( x, y ) (cid:1)(cid:13)(cid:13) L q ((0 , ∞ ) , dss ) Z ∞ e − u du √ u ≤ C p | x || x − y | n − , ( x, y ) ∈ N . (3.7)In the last inequality we have taken into account the estimation (3.6).We deduce that the operator H is bounded from L qX ( R n , dx ) into L q ( R n , dx ), for every 1 ≤ q ≤ ∞ .It is concluded that g q,Xβ, , { P t } t> ; loc is bounded from L pX ( R n , dx ) into L p ( R n , dx ). As above, theproperties of invariance of the operator g q,Xβ, , { P t } t> allow us to deduce that g q,Xβ, , { P t } t> defines abounded operator from L pX ( R n , dx ) into L p ( R n , dx ). By [32, Theorem C] (i) holds.If (iv) holds the same arguments lead to that (i) is true (see [32, Theorem 5.2]).Suppose now that (iii) holds. We are going to prove that (iv) is true.As we have just proved we deduce that g q,Xβ, , { P t } t> is bounded from L pX ( R n , dx ) into L p ( R n , dx ).According to [32, Theorem 5.2], g q,Xβ, , { P t } t> is bounded from L X ( R n , dx ) into L , ∞ ( R n , dx ).We can write g q,Xβ, , { P A t } t> ( f ) ≤ H ( f ) + g q,Xβ, , { P t } t> ; loc ( f ) + g q,Xβ, , { P A t } t> ; glob ( f ) . According to [28, Propositions 3.2.5 and 3.2.7] in the L q ((0 , ∞ ) , dtt )-setting, g q,Xβ, , { P t } t> ; loc isbounded from L X ( R n , γ − ) into L , ∞ ( R n , γ − ). We proved that H is a bounded operator from L X ( R n , dx ) into L ( R n , dx ). Since H is local, [28, Proposition 3.2.5] says us that H is boundedfrom L X ( R n , γ − ) into L ( R n , γ − ). By taking into account (2.10) we have that g q,X , , { P A t } t> ; glob ( f )( x ) ≤ Z R n k f ( y ) k L ( x, y ) dy, f ∈ L X ( R n , γ − ) , where L ≥ L ( f )( x ) = Z R n k f ( y ) k L ( x, y ) dy, f ∈ L X ( R n , γ − ) , bounded from L X ( R n , γ − ) into L , ∞ ( R n , γ − ). Since 0 < β < g q,Xβ, , { P A t } t> ; glob is bounded from L X ( R n , γ − ) into L , ∞ ( R n , γ − ).We conclude that g q,Xβ, , { P A t } t> is bounded from L X ( R n , γ − ) into L , ∞ ( R n , γ − ) and (iv) holds.The proof is complete.3.2. Proof of Theorem 1.3. { T A t } t> is a symmetric diffusion semigroup. According to [35,Theorem 2, (ii)] and [3, Theorem 1.7] we deduce that the property (i) ⇒ (ii) holds. From [32,Theorem B] (see also [21, Theorem 2.2]) it deduces that the property (i) ⇒ (iii) is true.We are going to see that (ii) ⇒ (i). Assume that (ii) is true. According to [32, Proposition 3.1]we have that, for every δ ≥ β , k f k L pX ( R n ,γ − ) ≤ C k g q,Xδ, , { T A t } t> ( f ) k L p ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . We can assume that β ∈ N \ { } .Since X is a Banach space, X is a closed subspace of ( X ∗ ) ∗ . Then according to [18, Proposition1.3.1] L pX ( R n , γ − ) is norming for L p ′ X ∗ ( R n , γ − ), where p ′ = pp − . This means that, for every g ∈ L p ′ X ∗ ( R n , γ − ) k g k L p ′ X ∗ ( R n ,γ − ) = sup f ∈ L pX ( R n , γ − ) k f k L pX ( R n ,γ − ) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z R n h f ( x ) , g ( x ) i dγ − ( x ) (cid:12)(cid:12)(cid:12)(cid:12) . We now define the operator Q A β ( h )( x ) = Z ∞ s β − ( ∂ βs T A s )( h ( s, · ))( x ) ds, x ∈ R n . Our objective is to prove that the operator L A q,β defined by L A q,β ( h ) = g q,Xβ, , { T A t } t> ( Q A β ( h ))is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into L r ( R n , γ − ), 1 < r < ∞ .When this objective is established we can see that (i) holds following the procedure devel-oped in the proof of [21, Theorem 3.1]. Indeed, let f ∈ C ∞ c ( R n ) ⊗ X ∗ . According to Theorem1.1, g q ′ ,X ∗ β, , { T A t } t> ( f ) ∈ L p ( R n , γ − ), or in other words, t β ∂ βt T A t ( f ) ∈ L pL q ′ X ∗ ((0 , ∞ ) , dtt ) ( R n , γ − ) ⊆ L p ( L qX ((0 , ∞ ) , dtt )) ∗ ( R n , γ − ). Then, without loss of generality we can suppose that there exists h ∈ L p ′ L qX ((0 , ∞ ) , dtt ) ( R n , γ − ), that can be chosen smooth enough, such that k h k L p ′ LqX ((0 , ∞ ) , dtt ) ( R n ,γ − ) ≤ k g q ′ ,X ∗ β, , { T A t } t> ( f ) k L p ( R n ,γ − ) = k t β ∂ βt T A t ( f ) k L pLq ′ X ∗ ((0 , ∞ ) , dtt ) ( R n ,γ − ) = Z R n Z ∞ h t β ∂ βt T A t ( f )( x ) , h ( t, x ) i X ∗ ,X dtt dγ − ( x ) . By interchanging the order of integration that is justified by the smoothness of f and h we obtain k g q ′ ,X ∗ β, , { T A t } t> ( f ) k L p ( R n ,γ − ) = Z R n Z ∞ h t β ∂ βt T A t ( f )( x ) , h ( t, x ) i X ∗ ,X dtt dγ − ( x ) = Z R n h f ( x ) , Q A β ( h )( x ) i X ∗ ,X dγ − ( x ) . By using H¨older’s inequality, and our objective, that is assumed proved, we get k g q ′ ,X ∗ β, , { T A t } t> ( f ) k L p ( R n ,γ − ) ≤ k f k L pX ∗ ( R n ,γ − ) k Q A β ( h ) k L p ′ X ( R n ,γ − ) ≤ C k f k L pX ∗ ( R n ,γ − ) kL A q,β ( h ) k L p ′ ( R n ,γ − ) ≤ C k f k L pX ∗ ( R n ,γ − ) k h k L p ′ LqX ((0 , ∞ ) , dtt ) ( R n ,γ − ) ≤ C k f k L pX ∗ ( R n ,γ − ) . By Theorem 1.2, there exists a norm (cid:129) · (cid:129) X ∗ in X ∗ defining the original topology of X ∗ and suchthat ( X ∗ , (cid:129) · (cid:129) X ∗ ) is q ′ -uniformly convex. This is equivalent to that X ∗ has q ′ martingale cotype.Then, X has q -martingale type (see [21, Theorem 3.1]). Hence, there exists a norm (cid:129) · (cid:129) X in X that defines the topology of X being ( X, (cid:129) · (cid:129) X ) q -uniformly smooth.We are going to prove our objective. Suppose that h ∈ C ∞ c ( R n ) ⊗ C c (0 , ∞ ) ⊗ X . Note that C ∞ c ( R n ) ⊗ C c (0 , ∞ ) ⊗ X is a dense subspace of L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ). We can write ∂ βt T A t ( Q A β ( h ))( x ) = Z R n ∂ βt T A t ( x, y ) Q A β ( h )( y ) dy = Z R n ∂ βt T A t ( x, y ) Z ∞ s β − Z R n ∂ βs T A s ( y, z ) h ( s, z ) dzdsdy = Z ∞ Z R n h ( s, z ) Z R n s β − ∂ βs T A s ( y, z ) ∂ βt T A t ( x, y ) dydzds, x ∈ R n and t > . By using the semigroup property we get Z R n ∂ βs T A s ( y, z ) ∂ βt T A t ( x, y ) dy = ∂ βt ∂ βs Z R n T A s ( y, z ) T A t ( x, y ) dy = ∂ βt ∂ βs T A t + s ( x, z ) = ∂ βu T A u ( x, z ) (cid:12)(cid:12) u = t + s , x, z ∈ R n and t, s ∈ (0 , ∞ ) . We consider the operator L A β defined by L A β ( h )( t, x ) = Z ∞ Z R n h ( s, z ) K A β ( t, x ; s, z ) dz dss , x ∈ R n and t > , where K A β ( t, x ; s, z ) = ( st ) β ∂ βu T A u ( x, z ) (cid:12)(cid:12) u = t + s , x, z ∈ R n and t, s > . Note that L A q,β ( h )( x ) = k L A β ( h )( · , x ) k L qX ((0 , ∞ ) , dtt ) , x ∈ R n . We are going to see that L A β is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into itself, for every 1 < r < ∞ .We consider L A β, loc and L A β, glob defined in the usual way.We first observe that, since q ′ /q ≥ Z ∞ ( st ) βq ( s + t ) βq dtt = Z ∞ z βq − (1 + z ) βq dz = B ( βq, βq ), s >
0, by using Jensen’s inequality we obtain that, if F is a function defined on (0 , ∞ ) such that u β F ∈ L q ′ ((0 , ∞ ) , duu ), then (cid:13)(cid:13)(cid:13) ( st ) β F ( t + s ) (cid:13)(cid:13)(cid:13) L q ′ Lq ((0 , ∞ ) , dtt ) ((0 , ∞ ) , dss ) = C Z ∞ (cid:18)Z ∞ ( st ) βq | F ( t + s ) | q dtt (cid:19) q ′ /q dss ! /q ′ ≤ C (cid:18)Z ∞ Z ∞ ( st ) βq − ( s + t ) βq | ( t + s ) β F ( t + s ) | q ′ dtds (cid:19) /q ′ = C (cid:18)Z ∞ u β ( q ′ − q ) | F ( u ) | q ′ Z u (( u − t ) t ) βq − dtdu (cid:19) /q ′ = C k u β F k L q ′ ((0 , ∞ ) , duu ) . (3.8)We now study the operator L A β, glob .According to (2.4) we have that, for 0 < δ < η < | u β ∂ βu T A u ( x, z ) | ≤ C e − u (1 − e − u ) n e ( η − δ ) | z | −| x | e − δ | z − e − ux | − e − u , x, z ∈ R n and u > . By using Minkowski and H¨older’s inequalities we get k L A β, glob ( h )( · , x ) k L qX ((0 , ∞ ) , dtt ) ≤ Z R n Z ∞ k K A β ( · , x ; s, z ) k L q ((0 , ∞ ) , dtt ) k h ( s, z ) k X X N c ( x, z ) dss dz ≤ Z R n (cid:13)(cid:13)(cid:13) K A β ( · , x ; · , z ) (cid:13)(cid:13)(cid:13) L q ′ Lq ((0 , ∞ ) , dtt ) ((0 , ∞ ) , dss ) k h ( · , z ) k L qX ((0 , ∞ ) , dss ) X N c ( x, z ) dz, x ∈ R n . By considering (3.8) and (3.9) it follows that (cid:13)(cid:13)(cid:13) K A β ( · , x ; · , z ) (cid:13)(cid:13)(cid:13) L q ′ Lq ((0 , ∞ ) , dtt ) ((0 , ∞ ) , dss ) = (cid:13)(cid:13)(cid:13) ( st ) β ∂ βu T A u ( x, z ) | u = t + s | (cid:13)(cid:13)(cid:13) L q ′ Lq ((0 , ∞ ) , dtt ) ((0 , ∞ ) , dss ) ≤ C k u β ∂ βu T A u ( x, z ) k L q ′ ((0 , ∞ ) , duu ) ≤ Ce ( η − δ ) | z | −| x | (cid:13)(cid:13)(cid:13) e − u e − δ | z − e − ux | − e − u (1 − e − u ) n (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) ≤ Ce ( η − δ ) | z | −| x | (cid:13)(cid:13)(cid:13) e − u e − δ | z − e − ux | − e − u (1 − e − u ) n (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) , x, z ∈ R n . Then, k L A β, glob ( h )( · , x ) k L qX ((0 , ∞ ) , dtt ) ≤ C Z R n H ( x, z ) k h ( · , z ) k L qX ((0 , ∞ ) , dss ) X N c ( x, z ) dz, x ∈ R n , where H ( x, z ) = e ( η − δ ) | z | −| x | (cid:13)(cid:13)(cid:13) e − u e − δ | z − e − ux | − e − u (1 − e − u ) n (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) , x, z ∈ R n . By proceeding as in the estimation (2.5) we get that, H ( x, z ) ≤ C e − ( η + δ ) | x | +( η − δ ) | z | , h x, z i ≤ , | x + z | n e η ( | z | −| x | ) − δ | x + z || x − z | , h x, z i > . , ( x, z ) ∈ N c . Let 1 < r < ∞ . By choosing 0 < η < δ < η − δ < r < η + δ , we deduce thatsup x ∈ R n Z R n X N c ( x, z ) H ( x, z ) e | x | −| z | r dz < ∞ , and sup z ∈ R n Z R n X N c ( x, z ) H ( x, z ) e | x | −| z | r dx < ∞ . We conclude that the operator defined by H ( g )( x ) = Z R n H ( x, z ) g ( z ) dz, x ∈ R n , is bounded from L r ( R n , γ − ) into itself. Hence, L A β, glob is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − )into itself.We now study the local operator L A β, loc . We consider the operator L β defined by L β ( h )( t, x ) = Z ∞ Z R n h ( s, z ) K β ( t, x ; s, z ) dz dss , x ∈ R n , t > , where K β ( t, x ; s, z ) = ( st ) β ∂ βu W u ( x − z ) | u = t + s , x, z ∈ R n and s, t > ℓ ∈ N ,(3.10) ∂ ℓu W u ( z ) = 12 ℓ √ π u ) + ℓ e H ℓ (cid:18) z √ u (cid:19) = 1(4 u ) ℓ H ℓ (cid:16) z √ u (cid:17) W u ( z ) , z ∈ R , u > . We prove this equality by induction on ℓ . Since e H ′ n ( z ) = − e H n +1 ( z ) and H n +1 ( z ) = 2 zH n ( z ) − nH n − ( z ), z ∈ R and n ∈ N , we have that ∂ u W u ( z ) = ∂ u (cid:18) √ πu e H (cid:16) z √ u (cid:17)(cid:19) = 1 √ π (2 u ) (cid:18) − e H (cid:16) z √ u (cid:17) + z √ u e H (cid:16) z √ u (cid:17)(cid:19) = 12 √ π u ) / e H (cid:16) z √ u (cid:17) , z ∈ R , u > . Suppose that (3.10) is true for certain ℓ ∈ N . Then, ∂ ℓ +1 u W u ( z ) = 12 ℓ + √ π (cid:18) − (cid:16)
12 + ℓ (cid:17) u + ℓ e H ℓ (cid:16) z √ u (cid:17) + z √ u ℓ e H ℓ +1 (cid:16) z √ u (cid:17)(cid:19) = 12 ℓ √ π u ) + ℓ (cid:18) − (2 ℓ + 1) e H ℓ (cid:16) z √ u (cid:17) + z √ u e H ℓ +1 (cid:16) z √ u (cid:17)(cid:19) = 12 ℓ +1 √ π u ) + ℓ e H ℓ +2 (cid:16) z √ u (cid:17) , z ∈ R , u > . From (3.10) it follows that, for every m ∈ N and x = ( x , ..., x n ) ∈ R n , ∂ mu W u ( x ) = ∂ u (cid:16) n Y i =1 W u ( x i ) (cid:17) = W u ( x ) X | r | = m (cid:18) mr . . . r n (cid:19) n Y i =1 u ) r i H r i (cid:18) x i √ u (cid:19) = W u ( x )(4 u ) m X | r | = m (cid:18) mr . . . r n (cid:19) H r (cid:18) x √ u (cid:19) . We define M β, loc = L A β, loc − L β, loc . Let us see that M β, loc is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − )into itself, for every 1 ≤ r < ∞ .By Minkowski integral and H¨older inequalities and (3.8) we get, for every x ∈ R n , kM A β, loc ( h )( · , x ) k L qX ((0 , ∞ ) , dtt ) ≤ Z R n (cid:13)(cid:13)(cid:13) K A β ( · , x ; · , z ) − K β ( · , x ; · , z ) (cid:13)(cid:13)(cid:13) L q ′ Lq ((0 , ∞ ) , dtt ) ((0 , ∞ ) , dss ) k h ( · , z ) k L qX ((0 , ∞ ) , dss ) X N ( x, z ) dz ≤ C Z R n (cid:13)(cid:13)(cid:13) u β ∂ βu ( T A u ( x, z ) − W u ( x − z )) (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) k h ( · , z ) k L qX ((0 , ∞ ) , dss ) X N ( x, z ) dz. Our objective is to established that(3.11) (cid:13)(cid:13)(cid:13) u β ∂ βu ( T A u ( x, z ) − W u ( x − z )) (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) ≤ C p | x || x − z | n − / , ( x, z ) ∈ N . Then, since M β, loc is a local operator, by virtue of [28, Proposition 3.2.5] we can conclude that M β, loc is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into itself, for every 1 ≤ r < ∞ . Recall that theoperator T ( g )( x ) = Z R n p | x || x − z | n − / X N ( x, y ) g ( y ) dy, x ∈ R n , is bounded from L r ( R n , dx ) into itself, for every 1 ≤ r < ∞ (see the proof of Theorem 1.2).Let us show (3.11). By considering (2.2) and (3.10) we have that ∂ βu ( T A u ( x, z ) − W u ( x − z ))= T A u ( x, z ) X ( j,r,s,ℓ ) ∈ J c j,r,s,ℓ (cid:18) e − u √ − e − u (cid:19) | s |−| ℓ | H ℓ ( z ) H s − ℓ (cid:18) x − e − u z √ − e − u (cid:19) − W u ( x − z )(4 u ) β X | r | =2 β (cid:18) βr . . . r n (cid:19) H r (cid:18) x − z √ u (cid:19) , x, z ∈ R n , u > . Here we understand that ( j, r, s, ℓ ) ∈ J when j = 0 , ..., β , and r, s, ℓ ∈ N n satisfy | r | = j and ℓ i ≤ s i ≤ r i , i = 1 , ..., n . For every ( j, r, s, ℓ ) ∈ J the constant c j,r,s,ℓ is given by c j,r,s,ℓ = ( − | s | + | ℓ | | s | (cid:18) βj (cid:19) n β − j (cid:18) jr . . . r n (cid:19) n Y i =1 (cid:26) r i s i (cid:27)(cid:18) s i ℓ i (cid:19) . Denoting by K = J \ { (2 β, r, r, ℓ ) : r, ℓ ∈ N n , | r | = 2 β, ℓ = (0 , ..., } , we can write ∂ βu ( T A u ( x, z ) − W u ( x − z ))= T A u ( x, z ) X ( j,r,s,ℓ ) ∈ K c j,r,s,ℓ (cid:18) e − u √ − e − u (cid:19) | s |−| ℓ | H ℓ ( z ) H s − ℓ (cid:18) x − e − u z √ − e − u (cid:19) + T A u ( x, z ) e − βu β (1 − e − u ) β X | r | =2 β (cid:18) βr . . . r n (cid:19) H r (cid:18) x − e − u z √ − e − u (cid:19) − W u ( x − z )(4 u ) β X | r | =2 β (cid:18) βr . . . r n (cid:19) H r (cid:18) x − z √ u (cid:19) = I ( u, x, z ) + I ( u, x, z ) , x, z ∈ R n , u > . We have that | I ( u, x, z ) | ≤ Ce − nu X ( j,r,s,ℓ ) ∈ K (1 + | z | ) | ℓ | e − c | x − e − uz | − e − u (1 − e − u ) n + | s |− | ℓ | , x, z ∈ R n , u > . Then, by making the change of variables u = log v − v , v ∈ (0 , ∞ ), we get (cid:13)(cid:13)(cid:13) u β I ( u, x, z ) (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) ≤ C X ( j,r,s,ℓ ) ∈ K Z ∞ (cid:16) u β e − nu (1 + | z | ) | ℓ | e − c | x − e − uz | − e − u (1 − e − u ) n + | s |− | ℓ | (cid:17) q ′ duu /q ′ ≤ Ce − c ( | x | −| z | ) X ( j,r,s,ℓ ) ∈ K Z (cid:16) log 1 + v − v (cid:17) βq ′ − (1 − v ) nq ′ − (1 + | z | ) q ′ | ℓ | e − c ( v | x + z | + | x − z | v ) v ( n + | s |− | ℓ | ) q ′ dv ! /q ′ ≤ C X ( j,r,s,ℓ ) ∈ K Z (1 + | z | ) q ′ | ℓ | e − c ( v | x + z | + | x − z | v ) v ( n + | s |− | ℓ | − β ) q ′ +1 dv ! /q ′ , ( x, z ) ∈ N . In the last inequality we have taken into account that || x | − | z | | ≤ C , ( x, z ) ∈ N , that log v − v ≤ Cv , v ∈ (0 , ), and that ( − log(1 − v )) βq ′ − (1 − v ) nq ′ − ≤ C , v ∈ ( , j, r, s, ℓ ) ∈ K . In the case that j ≤ β − j = 2 β and s = r we have that | s | ≤ β − | z | ) q ′ | ℓ | e − c ( v | x + z | + | x − z | v ) v ( n + | s |− | ℓ | − β ) q ′ +1 ≤ C (1 + | x + z | q ′ | ℓ | + | x − z | q ′ | ℓ | ) e − c ( v | x + z | + | x − z | v ) v ( n − − | ℓ | ) q ′ +1 ≤ C e − c | x − z | v v ( n − q ′ +1 ≤ C e − c | x − z | v v ( n − ) q ′ +1 , x, z ∈ R n , v ∈ (0 , . If j = 2 β and s = r , then | s | = 2 β and | ℓ | ≥
1. In this case we have that, for every x, z ∈ R n and v ∈ (0 , | z | ) q ′ | ℓ | e − c ( v | x + z | + | x − z | v ) v ( n + | s |− | ℓ | − β ) q ′ +1 ≤ C (1 + | z | ) q ′ e − c ( v | x + z | + | x − z | v ) v ( n − ) q ′ +1 ≤ (1 + | z | ) q ′ e − c | x − z | v v ( n − ) q ′ +1 . Hence, we deduce that (cid:13)(cid:13)(cid:13) u β I ( u, x, z ) (cid:13)(cid:13)(cid:13) L q ′ ((0 , ∞ ) , duu ) ≤ C p | z | Z e − c | x − z | v v ( n − ) q ′ +1 dv ! /q ′ ≤ C p | z || x − z | n − , ( x, z ) ∈ N . On the other hand, we can write, for each x, z ∈ R n and u > | I ( u, x, z ) | = 12 β π n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − (4 β + n ) u (1 − e − u ) n +2 β X | r | =2 β (cid:18) βr . . . r n (cid:19) (cid:20) e H r (cid:16) x − e − u z √ − e − u (cid:17) − e H r (cid:16) x − z √ u (cid:17)(cid:21) + X | r | =2 β (cid:18) βr . . . r n (cid:19) e H r (cid:16) x − z √ u (cid:17) (cid:20) e − (4 β + n ) u (1 − e − u ) n +2 β − u ) n +2 β (cid:21)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C u n +2 β X | r | =2 β (cid:12)(cid:12)(cid:12)(cid:12) e H r (cid:16) x − e − u z √ − e − u (cid:17) − e H r (cid:16) x − z √ u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) + e − c | x − z | u (cid:12)(cid:12)(cid:12)(cid:12) e − (4 β + n ) u (1 − e − u ) n +2 β − u ) n +2 β (cid:12)(cid:12)(cid:12)(cid:12) . We observe that, since e H ′ n ( s ) = − e H n +1 ( s ), s ∈ R , n ∈ N , and e − | x − e − uz | − e − u ≤ Ce − c | x − z | u , ( x, z ) ∈ N , u ∈ (0 , r = ( r , ..., r n ) ∈ N n , (cid:12)(cid:12)(cid:12)(cid:12) e H r (cid:16) x − e − u z √ − e − u (cid:17) − e H r (cid:16) x − z √ u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X k =1 k − Y i =1 (cid:12)(cid:12)(cid:12)(cid:12) e H r i (cid:16) x i − z i √ u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) e H r k (cid:16) x k − e − u z k √ − e − u (cid:17) − e H r k (cid:16) x k − z k √ u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) n Y i = k +1 (cid:12)(cid:12)(cid:12)(cid:12) e H r i (cid:16) x i − e − u z i √ − e − u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Ce − c | x − z | u n X k =1 (cid:12)(cid:12)(cid:12)(cid:12) x k − e − u z k √ − e − u − x k − z k √ u (cid:12)(cid:12)(cid:12)(cid:12) ≤ Ce − c | x − z | u n X k =1 (cid:18) | z k | p − e − u + | x k − z k | (cid:12)(cid:12)(cid:12) √ − e − u − √ u (cid:12)(cid:12)(cid:12)(cid:19) ≤ Ce − c | x − z | u √ u (1 + | z | ) , ( x, z ) ∈ N , u ∈ (0 , . We also have that, when u ∈ (0 , (cid:12)(cid:12)(cid:12)(cid:12) e − (4 β + n ) u (1 − e − u ) n +2 β − u ) n +2 β (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) e − (4 β + n ) u − − e − u ) n +2 β (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) − e − u ) n +2 β − u ) n +2 β (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cu n +2 β − . Thus, we conclude that | I ( u, x, z ) | ≤ C e − c | x − z | u u n +2 β ( √ u (1 + | z | ) + u ) ≤ C (1 + | x | ) e − c | x − z | u u n +2 β − , ( x, z ) ∈ N , u ∈ (0 , . Also we have that | I ( u, x, z ) | ≤ C (cid:18) e − (4 β + n ) u (1 − e − u ) n +2 β + 1 u n +2 β (cid:19) ≤ Cu n +2 β , x, z ∈ R n , u ∈ (0 , ∞ ) . By these estimations we get, since p m ( x ) ∼ (1 + | x | ) − , x ∈ R n , (cid:13)(cid:13)(cid:13) u β I ( u, x, z ) (cid:13)(cid:13)(cid:13) q ′ L q ′ ((0 , ∞ ) , duu ) ≤ C (1 + | x | ) q ′ Z m ( x )0 e − c | x − z | u u ( n − ) q ′ +1 du + Z ∞ m ( x ) duu n q ′ +1 ! ≤ C (1 + | x | ) q ′ m ( x ) q ′ Z m ( x )0 e − c | x − z | u u ( n − ) q ′ +1 du + 1 m ( x ) n q ′ ! ≤ C p | x || x − y | n − ! q ′ , ( x, z ) ∈ N . Hence, (3.11) is established.We have seen that L A β is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into itself, with 1 < r < ∞ ,provided that so is L β, loc .We are going to prove that L β, loc is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into itself, for every1 < r < ∞ . We use vector valued Calder´on-Zygmund theory.We have that | ∂ βt W t ( z ) | ≤ C e − c | z | /t t n/ β , z ∈ R n and t > . We get Z R n Z ∞ | K β ( t, x ; s, z ) | dss dz ≤ C Z R n Z ∞ ( st ) β e − c | x − z | s + t ( s + t ) n +2 β dss dz ≤ C Z R n Z ∞ ( st ) β (cid:16) | x − z | s + t (cid:17) n +2 β ( s + t ) n +2 β dss dz ≤ C Z R n Z ∞ ( st ) β ( s + t + | x − z | ) n +2 β dss dz ≤ C Z ∞ ( st ) β Z ∞ ρ n − ( s + t + ρ ) n +2 β dρ dss ≤ C Z ∞ ( st ) β ( s + t ) β dss = C Z ∞ u β − (1 + u ) β du = C, x ∈ R n and t > . It follows that k L β ( h ) k qL qLqX ((0 , ∞ ) , dtt ) ( R n ,dx ) = Z R n Z ∞ (cid:18)Z R n Z ∞ | K β ( t, x ; s, z ) |k h ( s, z ) k X dss dz (cid:19) q dtt dx ≤ Z R n Z ∞ Z R n Z ∞ | K β ( t, x ; s, z ) |k h ( s, z ) k qX dss dz (cid:18)Z R n Z ∞ | K β ( t, x ; s, z ) | dss dz (cid:19) q/q ′ dtt dx ≤ C k h k qL qLqX ((0 , ∞ ) , dtt ) ( R n ,dx ) , h ∈ L qL qX ((0 , ∞ ) , dtt ) ( R n , dx ) . For every x, z ∈ R n , x = z , we define the operator K β ( x, z ) by K β ( x, z )( g )( t ) = Z ∞ K β ( t, x ; s, z ) g ( s ) dss , t ∈ (0 , ∞ ) . As before we have that Z ∞ | K β ( t, x ; s, z ) | dss ≤ C Z ∞ ( st ) β ( s + t + | x − z | ) n + β dss ≤ C | x − z | n , x, z ∈ R n , x = z. Then, for every g ∈ L qX ((0 , ∞ ) , dtt ), k K β ( x, z )( g ) k qL qX ((0 , ∞ ) , dtt ) = Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ K β ( t, x ; s, z ) g ( s ) dss (cid:12)(cid:12)(cid:12)(cid:12) q dtt ≤ Z ∞ (cid:18)Z ∞ K ( t, x ; s, z ) dss (cid:19) q/q ′ Z ∞ K β ( t, x ; s, z ) k g ( s ) k q dss dt ≤ C | x − z | ( qq ′ +1) n k g k qL qX ((0 , ∞ ) , dtt ) ≤ C | x − z | qn k g k qL qX ((0 , ∞ ) , dtt ) , x, z ∈ R n , x = z. Hence, k K β ( x, z ) k L qX ((0 , ∞ ) , dtt ) → L qX ((0 , ∞ ) , dtt ) ≤ C | x − z | n , x, z ∈ R n , x = z. In a similar way we can see that, for every i = 1 , ..., n , k ∂ x i K β ( x, z ) k L qX ((0 , ∞ ) , dtt ) → L qX ((0 , ∞ ) , dtt ) ≤ C | x − z | n +1 , x, z ∈ R n , x = z, and k ∂ z i K β ( x, z ) k L qX ((0 , ∞ ) , dtt ) → L qX ((0 , ∞ ) , dtt ) ≤ C | x − z | n +1 , x, z ∈ R n , x = z, where ∂ x i K β ( x, z ) and ∂ z i K β ( x, z ) are understood as the integral operators defined by the kernels ∂ x i K β ( t, x ; s, z ) and ∂ z i K β ( t, x ; s, z ), respectively.We also have that, for every h ∈ C c ( R n ) ⊗ C c (0 , ∞ ) ⊗ X , K β ( h )( t, x ) = (cid:18)Z R n K β ( x, z )[ h ( · , x )] dz (cid:19) ( t ) , for almost all ( t, x ) / ∈ supp h . Here the integral is understood in the L ((0 , ∞ ) , dtt )-Bochnersense. According to vector valued Calder´on-Zygmund theory , K β defines a bounded operatorfrom L rL qX ((0 , ∞ ) , dtt ) ( R n , dx ) into itself, for every 1 < r < ∞ .By [28, Propositions 3.2.5 and 3.2.7] K β, loc is bounded from L rL qX ((0 , ∞ ) , dtt ) ( R n , γ − ) into itself,for every 1 < r < ∞ .The proof of (ii) ⇒ (i) is thus finished.In order to prove that (iii) ⇒ (i) we can proceed as in the proof of (ii) ⇒ (i) with the semigroup { T A t } t> replaced by { P A t } t> . We can also argue as follows. Suppose that (iii) is true. Accordingto (2.9) and by taking into account that g q,Xβ , , { T A t } t> ( f ) ≤ g q,Xβ , , { T A t } t> ( f ) when 0 < β ≤ β , wehave that k f k L pX ( R n ,γ − ) ≤ C k g q,Xm, , { T A t } t> ( f ) k L p ( R n ,γ − ) , f ∈ L p ( R n , γ − ) , where m ∈ N and m − ≤ β < m . Then by using the property we have just proved, (i) holds. Proof of Corollary 1.4.
The equivalence stated in Corollary 1.4 follows by using Theorems1.2 and 1.3 and by taking into account that X is isomorphic to a Hilbert space if and only if X has 2-martingale type and 2-martingale cotype.4. Proofs of Theorems 1.5 and 1.6
Proof of Theorem 1.5.
Since { P A t } t> is a subordinated Poisson semigroup of the symmet-ric diffusion semigroup { T A t } t> the property ( i ) ⇒ ( ii ) can be deduced from [19, Theorem 1.6].We recall that X ∗ is UMD provided that X is UMD.Assume now that ( ii ) is true. We are going to see that X is UMD. Our first objective is to seethat there exists C > k g ,X { P t } t> ( f ) k L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . We define the global and local operators in the usual way and consider the operator G ( f )( t, x, w ) = t (cid:2) ∂ t P A t, loc ( f ( · , w ))( x ) − ∂ t P t, loc ( f ( · , w ))( x ) (cid:3) = Z R n X N ( x, y ) t∂ t [ P A t ( x, y ) − P t ( x − y )] f ( y, w ) dy, x ∈ R n , t > , w ∈ Ω . According to (3.7) we have that H ( x, y ) = (cid:13)(cid:13) t∂ t [ P A t ( x, y ) − P t ( x − y )] (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C p | x || x − y | n − , ( x, y ) ∈ N . Then, the operator H defined by H ( f )( x, w ) = Z R n H ( x, y ) f ( y, w ) X N ( x, y ) dy, x ∈ R n , w ∈ Ω , is bounded from L rX ( R n , dx ) into itself, for every 1 ≤ r ≤ ∞ .By using Minkowski inequality we deduce that (cid:13)(cid:13)(cid:13) k G ( f )( · , x, w ) k L ((0 , ∞ ) , dtt ) (cid:13)(cid:13)(cid:13) L rX ( R n ,dx ) ≤ C k f k L rX ( R n ,dx ) , for every f ∈ L rX ( R n , dx ), 1 ≤ r < ∞ . Then, since the operators are local (see [28, Proposition3.2.5]) (cid:13)(cid:13) g ,X { P A t } t> ; loc ( f ) − g ,X { P t } t> ; loc ( f ) (cid:13)(cid:13) L rX ( R n ,γ − ) ≤ C k f k L rX ( R n ,γ − ) , for every f ∈ L rX ( R n , γ − ), 1 ≤ r < ∞ . We now consider the operator M glob defined by M glob ( f )( t, x, w ) = Z R n t∂ t P A t ( x, y ) f ( y, w ) X N c ( x, y ) dy, x ∈ R n , t > , w ∈ Ω . As in (2.9) we obtain (cid:13)(cid:13) t∂ t P A t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C (cid:13)(cid:13) t∂ t T A t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) , x, y ∈ R n . According to (2.5) when m = 1, k = 0 and q = 2 it follows that, for every r ∈ (1 , ∞ ) thereexists a positive function F r defined in R n × R n such that k t∂ t T A t ( x, y ) k L ((0 , ∞ ) , dtt ) ≤ F r ( x, y ) , ( x, y ) ∈ N c , being sup x ∈ R n Z R n e | x | −| y | r F r ( x, y ) X N c ( x, y ) dy < ∞ , and sup y ∈ R n Z R n e | x | −| y | r F r ( x, y ) X N c ( x, y ) dx < ∞ . Then, (cid:13)(cid:13)(cid:13) kM glob ( f )( · , x, w ) k L ((0 , ∞ ) , dtt ) (cid:13)(cid:13)(cid:13) L rX ( R n ,γ − ) ≤ C k f k L rX ( R n ,γ − ) , for every f ∈ L r ( R n , γ − ) and r ∈ (1 , ∞ ).It follows that k g ,X { P A t } t> ; glob ( f ) k L rX ( R n ,γ − ) ≤ C k f k L rX ( R n ,γ − ) , for each f ∈ L r ( R n , γ − ) and r ∈ (1 , ∞ ). Since ( ii ) holds we have that k g ,X { P A t } t> ; loc ( f ) k L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . We can write k g ,X { P t } t> ; loc ( f ) k L pX ( R n ,γ − ) ≤ k g ,X { P A t } t> ; loc ( f ) − g ,X { P t } t> ; loc ( f ) k L pX ( R n ,γ − ) + k g ,X { P A t } t> ; loc ( f ) k L pX ( R n ,γ − ) . Hence, g ,X { P t } t> ; loc is bounded from L pX ( R n , γ − ) into itself, and also from L pX ( R n , dx ) into itself(see [28, Proposition 3.2.5]).By using now the corresponding properties of invariance (see [16, p. 21]) we deduce that k g ,X { P t } t> ( f ) k L pX ( R n ,dx ) ≤ C k f k L pX ( R n ,dx ) , f ∈ L pX ( R n , dx ) . By changing X by X ∗ and p by p ′ we also infer from ( ii ) that k g ,X ∗ { P t } t> ( f ) k L p ′ X ∗ ( R n ,dx ) ≤ C k f k L p ′ X ∗ ( R n ,dx ) , f ∈ L p ′ X ∗ ( R n , dx ) . Let now f ∈ L ( R n , γ − ) ⊗ X and h ∈ L ( R n , γ − ) ⊗ X ∗ . Since X is order continuous the K¨otheand topological dual coincides. We can write the following polarization equality Z R n h f ( x, · ) , h ( x, · ) i X,X ∗ dx = 4 Z R n Z ∞ h t∂ t P t ( f ( · , w ))( x ) , t∂ t P t ( h ( · , w ))( x ) i X,X ∗ dtt dx = 4 Z R n Z ∞ Z Ω t∂ t P t ( f ( · , w ))( x ) t∂ t P t ( h ( · , w ))( x ) dµ ( w ) dtt dx. By using H¨older inequality it follows that (cid:12)(cid:12)(cid:12)(cid:12)Z R n h f ( x, · ) , h ( x, · ) i X,X ∗ dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ C Z R n Z Ω (cid:13)(cid:13) t∂ t P t ( f ( · , w ))( x ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) (cid:13)(cid:13) t∂ t P t ( h ( · , w ))( x ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) dµ ( w ) dx ≤ C Z R n (cid:13)(cid:13) k t∂ t P t ( f ( · , w ))( x ) k L ((0 , ∞ ) , dtt ) k X (cid:13)(cid:13) k t∂ t P t ( h ( · , w ))( x ) k L ((0 , ∞ ) , dtt ) (cid:13)(cid:13) X ∗ dx ≤ C (cid:13)(cid:13)(cid:13) g ,X { P t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,dx ) (cid:13)(cid:13)(cid:13) g ,X ∗ { P t } t> ( h ) (cid:13)(cid:13)(cid:13) L p ′ X ∗ ( R n ,dx ) ≤ C (cid:13)(cid:13)(cid:13) g ,X { P t } t> ( f ) (cid:13)(cid:13)(cid:13) L p ( R n ,dx ) k h k L p ′ X ∗ ( R n ,dx ) . Since L p ′ X ∗ ( R n , dx ) is norming in L pX ( R n , dx ) it follows that k f k L pX ( R n ,dx ) ≤ C (cid:13)(cid:13) g ,X { P t } t> ( f ) (cid:13)(cid:13) L pX ( R n ,dx ) . From [19, Theorem 1.6] we deduce that X is UMD and the proof is complete. Remark . The property established in [19, Theorem 1.6] suggests to ask whether in Theorem 1.5( ii ) we can replace the condition involving the dual space X ∗ by this other one k f k L pX ( R n ,γ − ) ≤ C (cid:13)(cid:13) g ,X { P A t } t> ( f ) (cid:13)(cid:13) L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) , and then the UMD property for X can be again deduced from ( ii ). At this moment we do notknow what is the answer for this question.4.2. Proof of Theorem 1.6.
In [6, Theorems 1.4 and 1.5] the Banach spaces with the UMD-property were characterized by using L p ( R n , γ − )-boundedness properties of the Riesz transformsassociate with the operator A . For every i = 1 , ..., n , the Riesz transform R A i in the A -setting isdefined by R A i ( f )( x ) = lim ε → + Z | x − y | >ε R A i ( x, y ) f ( y ) dy, for almost all x ∈ R n , for every f ∈ L p ( R n , γ − ), 1 ≤ p < ∞ , where R A i ( x, y ) = 1 √ π Z ∞ ∂ x i T A t ( x, y ) dt √ t , x, y ∈ R n . For every i = 1 , ..., n , R A i is bounded from L p ( R n , γ − ) into itself, when 1 < p < ∞ , and from L ( R n , γ − ) into L , ∞ ( R n , γ − ); moreover, R A i can be extended, in the usual way, to L p ( R n , γ − ) ⊗ Y , 1 ≤ p < ∞ , where Y is a Banach space. In [6, Theorem 1.4] it was established that Y has the UMD property if and only if R A i , i = 1 , ..., n , can be extended to L pY ( R n , γ − ) for some(equivalently, for any) 1 < p < ∞ as a bounded operator from L pY ( R n , γ − ) into itself.Suppose that ( ii ) holds. Let i = 1 , ..., n . We have that, for every k ∈ N n , ∂ x i e H k = − e H k + e i ,where e i = ( e ij ) nj =1 , being e ij = 1, i = j , and e ij = 0, if i = j . Then, since A − / f = X ℓ ∈ N n p n + | ℓ | c ℓ ( f ) e H ℓ , f ∈ L ( R n , γ − ) , we have that R A i e H k = − p n + | k | e H k + e i , k ∈ N n . On the other hand, we can write P A−I t ( f ) = X ℓ ∈ N n e − t √ n + | ℓ |− c ℓ ( f ) e H ℓ , f ∈ L ( R n , γ − ) . Then we get ∂ t P A−I t ( R A i e H k ) = − p n + | k | ∂ t P A−I t ( e H k + e i ) = e − t √ n + | k | e H k + e i , k ∈ N n . and also, by considering that P A t ( f ) = X ℓ ∈ N n e − t √ n + | ℓ | c ℓ ( f ) e H ℓ , f ∈ L ( R n , γ − ) , it follows that ∂ x i P A t ( e H k ) = − e − t √ n + | k | e H k + e i , k ∈ N n . Let us denote by F = span { e H k } k ∈ N n the linear space generated by { e H k } k ∈ N n . For every f ∈ F we have that(4.1) ∂ x i P A t ( f ) = − ∂ t P A−I t ( R A i ( f )) . We recall that F is a dense subspace of L p ( R n , γ − ), for every 1 ≤ p < ∞ . It is clear that R A i f ∈ F ,for every f ∈ F and i = 1 , ..., n .Since X is order continuous, and by taking into account that, when n ≥
2, 0 is not an eigenvalueof
A − I , it follows that, for every f ∈ L ( R n , γ − ) ⊗ X and g ∈ L ( R n , γ − ) ⊗ X ∗ , Z R n h f ( x, · ) , g ( x, · ) i X,X ∗ dγ − ( x ) = 4 Z R n Z ∞ h t∂ t P A−I t ( f ( · , w ))( x ) , t∂ t P A−I t ( g ( · , w ))( x ) i dtt dγ − ( x ) . As above, by taking into account that (cid:13)(cid:13) g ,X ∗ { P A−I t } t> ( h ) (cid:13)(cid:13) L p ′ X ∗ ( R n ,γ − ) ≤ C k h k L p ′ X ∗ ( R n ,γ − ) , h ∈ L p ′ X ∗ ( R n , γ − ) , we obtain that, for every f ∈ ( L ( R n , γ − ) ∩ L p ( R n , γ − )) ⊗ X ,(4.2) k f k L pX ( R n ,γ − ) ≤ C (cid:13)(cid:13) g ,X { P A− It } t> ( f ) (cid:13)(cid:13) L pX ( R n ,γ − ) . According to (4.1) and (4.2) we get (cid:13)(cid:13) R A i ( f ) (cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C (cid:13)(cid:13) g ,X { P A−I t } t> ( R A i ( f )) (cid:13)(cid:13) L pX ( R n ,γ − ) = C (cid:13)(cid:13) g Xi, { P A t } t> ( f ) (cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ F ⊗ X. By [6, Theorem 1.4] we conclude that the space X has the UMD property.We observe that when n = 1, we have that Z R h ( f − E ( f ))( x, · ) , ( g − E ( g ))( x, · ) i X,X ∗ dγ − ( x )= 4 Z R Z ∞ h t∂ t P A−I t ( f ( · , w ))( x ) , t∂ t P A−I t ( g ( · , w ))( x ) i dtt dγ − ( x ) , for every f ∈ L ( R , γ − ) ⊗ X and g ∈ L ( R , γ − ) ⊗ X ∗ , where E ( f ) is the projection of f intothe subspace generated by e H . Property (4.2) can be written in the following way k f − E ( f ) k L pX ( R ,γ − ) ≤ C (cid:13)(cid:13) g ,X { P A− It } t> ( f ) (cid:13)(cid:13) L pX ( R ,γ − ) , for every f ∈ ( L ( R , γ − ) ∩ L p ( R , γ − )) ⊗ X . Since E ( R A ( f )) = 0, for f ∈ F ⊗ X , with the sameargument as before we can deduce that X is UMD when n = 1.Suppose now that the Banach space X has the UMD property. Let i = 1 , ..., n . Our objectiveis to see that, for certain C > (cid:13)(cid:13) g ,X ∗ { P A− It } t> ( f ) (cid:13)(cid:13) L p ′ X ∗ ( R n ,γ − ) ≤ C k f k L p ′ X ∗ ( R n ,γ − ) , f ∈ L p ′ X ∗ ( R n , γ − ) , and(4.4) (cid:13)(cid:13) g Xi, { P t } t> ( f ) (cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , f ∈ L pX ( R n , γ − ) . Let us consider first the estimation for g ,X ∗ { P A− It } t> . We define the local and global operators in theusual way and write g ,X ∗ { P A−I t } t> ( f ) ≤ g ,X ∗ { P A−I t } t> ; glob ( f ) + g ,X ∗ { P A−I t } t> ; loc ( f ) − g ,X ∗ { P t } t> ; loc ( f ) + g ,X ∗ { P t } t> ; loc ( f ) , for every f ∈ L pX ( R n , γ − ).By using Minkowski inequality we get, for each x ∈ R n and w ∈ Ω, g ,X ∗ { P A−I t } t> ; glob ( f ( · , w ))( x ) ≤ C Z R n X N c ( x, y ) | f ( y, w ) | (cid:13)(cid:13) t∂ t P A−I t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) dtt ) dy, and (cid:12)(cid:12) g ,X ∗ { P A−I t } t> ; loc ( f ( · , w ))( x ) − g ,X ∗ { P t } t> ; loc ( f ( · , w ))( x ) (cid:12)(cid:12) ≤ C Z R n X N ( x, y ) | f ( y, w ) | (cid:13)(cid:13) t∂ t (cid:2) P A−I t ( x, y ) − P t ( x − y ) (cid:3)(cid:13)(cid:13) L ((0 , ∞ ) dtt ) dy. We recall that P A−I t ( x, y ) = 1 √ π Z ∞ e − u T A−I t / (4 u ) ( x, y ) du √ u , x, y ∈ R n , t > . We obtain, for every x, y ∈ R n and t > t∂ t P A−I t ( x, y ) = t √ π Z ∞ e − u ∂ v T A− Iv ( x, y ) | v = t u duu = t √ π Z ∞ e − t v ∂ v T A− Iv ( x, y ) dv √ v . Then, Minkowski inequality leads to (cid:13)(cid:13) t∂ t P A−I t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C Z ∞ (cid:13)(cid:13) te − t v (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) | ∂ v T A−I v ( x, y ) | dv √ v ≤ C Z ∞ | ∂ v T A−I v ( x, y ) | dv = (cid:13)(cid:13) v∂ v T A− Iv ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dvv ) , x, y ∈ R n . We now observe that, since n ≥
2, the estimation (2.5) is also valid when m = 1, k = 0, q = 1 and A is replaced by A − I . Then, for 0 < δ < η <
1, we have that (cid:13)(cid:13) t∂ t P A−I t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C e − ( η + δ ) | x | +( η − δ ) | y | , h x, y i ≤ , | x + y | n e η ( | y | −| x | ) − δ | x + y || x − y | , h x, y i > . , ( x, y ) ∈ N c . By choosing η − δ < p ′ < η + δ we deduce thatsup x ∈ R n Z R n e | x | −| y | p ′ X N c ( x, y ) (cid:13)(cid:13) t∂ t P A−I t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) dy < ∞ , and sup y ∈ R n Z R n e | x | −| y | p ′ X N c ( x, y ) (cid:13)(cid:13) t∂ t P A−I t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) dx < ∞ . It follows that the operator g ,X ∗ { P A−I t } t> ; glob is bounded from L p ′ X ∗ ( R n , γ − ) into itself.On the other hand, by proceeding as above and by taking into account that estimation (3.6)holds for q = 1 and A − I (because n ≥
2) we obtain (cid:13)(cid:13) t∂ t (cid:2) P A−I t ( x, y ) − P t ( x − y ) (cid:3)(cid:13)(cid:13) L ((0 , ∞ ) dtt ) ≤ C (cid:13)(cid:13)(cid:13) v∂ v [ T A− Iv ( x, y ) − W v ( x − y )] (cid:13)(cid:13)(cid:13) L ((0 , ∞ ) , dvv ) ≤ C p | x || x − y | n − , ( x, y ) ∈ N . We deduce that (cid:13)(cid:13) g ,X ∗ { P A−I t } t> ; loc ( f ( · , w ))( x ) − g ,X ∗ { P t } t> ; loc ( f ( · , w ))( x ) (cid:13)(cid:13) L p ′ X ∗ ( R n ,γ − ) ≤ C k f k L p ′ X ∗ ( R n ,γ − ) , for every f ∈ L p ′ X ∗ ( R n , γ − ).Finally, since X ∗ has the UMD property, [19, Theorem 1.6] leads to g ,X ∗ { P t } t> is a boundedoperator from L p ′ X ∗ ( R n , γ − ) into itself. We can also write (cid:13)(cid:13) t∂ t P t ( x − y ) (cid:13)(cid:13) L ((0 , ∞ ) dtt ) ≤ C Z ∞ | ∂ v W v ( x − y ) | dv ≤ C Z ∞ e − c | x − y | v v n +1 dv ≤ C | x − y | n ≤ C p | x || x − y | n − , ( x, y ) ∈ N c . Thus, g ,X ∗ { P t } t> ; glob , and as a consequence g ,X ∗ { P t } t> ; loc , are bounded operators from L p ′ X ∗ ( R n , γ − )into itself. Thus (4.3) is established.In order to prove (4.4) we proceed in the same way. In this case we write (cid:13)(cid:13) t∂ x i P A t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) = (cid:13)(cid:13)(cid:13) t √ π Z ∞ e − t u u ∂ x i T A u ( x, y ) du (cid:13)(cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C Z ∞ (cid:13)(cid:13) t e − t u (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) | ∂ x i T A u ( x, y ) | duu ≤ C Z ∞ | ∂ x i T A u ( x, y ) | du √ u , x, y ∈ R n . Since ∂ x i T A u ( x, y ) = 2 e − nu ( x i − e − u y i )(1 − e − u ) n +1 e − | x − e − uy | − e − u , x, y ∈ R n , u > , it follows that (cid:13)(cid:13) t∂ x i P A t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C Z ∞ e − nu e − η | x − e − uy | − e − u (1 − e − u ) n +12 du √ u ≤ Ce − η ( | x | −| y | ) Z ∞ e − nu e − η | y − e − ux | − e − u (1 − e − u ) n +12 du √ u , x, y ∈ R n , with 0 < η < R A m,k in (2.5) we get, for very ( x, y ) / ∈ N , (cid:13)(cid:13) t∂ x i P A t ( x, y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C e − η | x | , if h x, y i ≤ , | x + y | n exp (cid:16) − η (cid:0) | x + y || x − y | − | y | − | x | (cid:1)(cid:17) , if h x, y i ≥ . By choosing p < η < g Xi, { P A t } t> ; glob is bounded from L pX ( R n , γ − )into itself.On the other hand, as above, we have that (cid:13)(cid:13) t∂ x i [ P A t ( x, y ) − P t ( x − y )] (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C Z ∞ | ∂ x i [ T A u ( x, y ) − W u ( x − y )] | du √ u , x, y ∈ R n . We can write ∂ x i ( T A u ( x, y ) − W u ( x − y )) = − x i − y i e − u )1 − e − u T A u ( x, y ) + x i − y i u W u ( x − y )= x i − y i u ( W u ( x − y ) − T A u ( x, y )) + (cid:18) x i − y i u − x i − y i e − u )1 − e − u (cid:19) T A u ( x, y ) , x, y ∈ R n , u > . Then, we get | ∂ x i ( T A u ( x, y ) − W u ( x − y )) | ≤ C e − nu (1 − e − u ) n +12 + 1 u n +12 ! ≤ Cu n +12 , x, y ∈ R n , u > . Also, by using that (cid:12)(cid:12)(cid:12)(cid:12) x i − y i u − x i − y i e − u )1 − e − u (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) | x − y | (cid:12)(cid:12)(cid:12)(cid:12) u − − e − u (cid:12)(cid:12)(cid:12)(cid:12) + 11 − e − u | x i − y i − ( x i − e − u y i ) | (cid:19) ≤ C ( | x − y | + | y | ) , x, y ∈ R n , u > , the estimation e − | x − e − u | − e − u ≤ Ce − c | x − y | u , ( x, y ) ∈ N , u ∈ (0 , , and (3.3), we obtain that, when ( x, y ) ∈ N and u ∈ (0 , | ∂ x i ( T A u ( x, y ) − W u ( x − y )) | ≤ Ce − c | x − y | u (cid:18) | x − y | u n (cid:16) | y | + | y |√ u (cid:17) + e − nu ( | x − y | + | y | )(1 − e − u ) n (cid:19) ≤ C e − c | x − y | u u n − (cid:16) | y | + | y |√ u (cid:17) . Taking into account (3.1), (3.4) and (3.5) for q = 1, we get (cid:13)(cid:13) t∂ x i [ P A t ( x, y ) − P t ( x − y )] (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C Z m ( x )0 e − c | x − y | u u n (cid:16) | y | + | y |√ u (cid:17) du + Z ∞ m ( x ) duu n +1 ! ≤ C p | x || x − y | n − , ( x, y ) ∈ N , and deduce that (cid:13)(cid:13) g Xi, { P A t } t> ; loc ( f ( · , w ))( x ) − g Xi, { P t } t> ; loc ( f ( · , w ))( x ) (cid:13)(cid:13) L pX ( R n ,γ − ) ≤ C k f k L pX ( R n ,γ − ) , for every f ∈ L pX ( R n γ − ).We now study g Xi, { P t } t> ; loc by using the vector-valued Calder´on-Zygmund theory. We considerthe operator G i defined by G i ( f )( t, x, w ) = t∂ x i P t ( f ( · , w ))( x ) , x ∈ R n , t > , w ∈ Ω . This is a convolution operator defined by the kernel K it ( x ) = c n x i t ( t + 2 | x | ) n +32 , x ∈ R n , t > , where c n = − n +1 ( n + 1) π − n +12 Γ( n +12 ). The Fourier transforms c K it of K it is given by c K it ( y ) = Ciy i e − ct | y | , y ∈ R n , t > . Then, (cid:13)(cid:13) c K it ( y ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) = C (cid:18)Z ∞ y i t e − ct | y | dtt (cid:19) ≤ C, y ∈ R n . We also have that (cid:13)(cid:13) K it ( x ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) = c n (cid:18)Z ∞ x i t ( t + 2 | x | ) n +3 dt (cid:19) ≤ C (cid:18)Z ∞ dt ( t + | x | ) n +1 (cid:19) ≤ C | x | n , x ∈ R n \ { } . In a similar way we can see that, for j = 1 , ..., n , (cid:13)(cid:13) ∂ x j K it ( x ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) ≤ C | x | n +1 , x ∈ R n \ { } . By using Bourgain’s results (see [8] and [9]) we conclude that the operator G i is bounded from L pX ( R n , dx ) into L pX ( L ((0 , ∞ ) , dtt )) ( R n , dx ).Here X ( L ((0 , ∞ ) , dtt )) represents the K¨othe Bochner function space consisting of all the mea-surable functions g : Ω −→ L ((0 , ∞ ) , dtt ) such that the function h ( w ) = (cid:13)(cid:13) g ( w ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) , w ∈ Ω , belongs to X . X ( L ((0 , ∞ ) , dtt )) is endowed with the natural norm k · k X ( L ((0 , ∞ ) , dtt )) defined by k g k X ( L ((0 , ∞ ) , dtt )) = (cid:13)(cid:13)(cid:13)(cid:13) g ( · , w ) (cid:13)(cid:13) L ((0 , ∞ ) , dtt ) (cid:13)(cid:13) X , g ∈ X ( L ((0 , ∞ ) , dtt )) . Hence, the operator g Xi, { P t } t> is bounded from L pX ( R n , dx ) into itself.According to [28, Propositions 3.2.5 and 3.2.7], g Xi, { P t } t> ; loc is bounded from L pX ( R n , γ − ) intoitself.We conclude that g Xi, { P A t } t> is bounded from L pX ( R n , γ − ) into itself. Thus the property ( ii ) isestablished. References [1]
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V´ıctor Almeida, Jorge J. Betancor, Juan C. Fari˜na, Lourdes Rodr´ıguez-MesaDepartamento de An´alisis Matem´atico, Universidad de La Laguna,Campus de Anchieta, Avda. Astrof´ısico S´anchez, s/n,38721 La Laguna (Sta. Cruz de Tenerife), Spain
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