An Extension of a Boundedness Result for Singular Integral Operators
aa r X i v : . [ m a t h . P R ] J a n An Extension of a Boundedness Result for SingularIntegral Operators
Deniz KarlıDepartment of MathematicsAMF233, I¸sık University34980 S¸ile, Istanbul, [email protected] & [email protected]
Boundedness of singular integral operators has been studied for a long time. Thereare some well-known results which were proved first by using classic analytical tech-niques. In these techniques, there are some important operators providing intermedi-ate steps for the proof. Three often used operators are Lusin Area functional ( A f ),non-tangential maximal function ( N fα ) and G-star functional ( G ∗ f ). They played animportant role in the development of Harmonic Analysis. (See Stein [17] and [18].)With the introduction of probabilistic techniques, alternative proofs have come tothe surface in addition to these analytical tools. In these classical techniques, Brownianmotion plays a central role. One such approach is to consider a ( d + 1)-dimensionalBrownian motion on the upper half-space and provide a probabilistic definition ofharmonic functions in terms of martingales. By means of martingales, one can de-fine Littlewood-Paley functions and hence provide probabilistic proofs of boundednessof some operators. (See, for example, Varopoulos [20], Burkholder and Gundy [7],Burkholder, Gundy and Silverstein [8], Durrett [9] and Bass [2]. For a more detailedliterature study on square functions and these operators, see Ba˜nuelos and Davis [5].)In the paper [12], we studied a more general process in ( d + 1)-dimensional half-space R d × R + . We would like to obtain generalisations of some theorems using thepower of probabilistic techniques and the weaker conditions imposed by the processwhich we start with. This paper can be considered as a continuation of the discussionwhich originates from [12]. This research project is supported by the BAP grant numbered 14B103 at the I¸sık University,Istanbul, Turkey. G ∗ functional, and (ii.) an extension of a classicalmultiplier theorem on singular integrals. This classical version of the multiplier theo-rem, which we will discuss here, focuses on singular integrals with kernels κ : R d → R satisfying the cancelation property Z r< | x |
1, thesame argument applies easily with a slight modification. See also [1, Theorem 1.1].)
Theorem 1.1.
Suppose κ is the kernel of a convolution operator T . If κ ∈ C , itsatisfies the cancelation condition (1.1), | κ ( x ) | ≤ c | x | − d and |∇ κ ( x ) | ≤ c | x | − d − , x = 0 (1.2) then for any < p < ∞ there is a finite constant c p depending only on p such that k T k L p ( R d ) → L p ( R d ) < c p . Our goal is to weaken the condition (1.2) by replacing d in the exponent with d − α/ α ∈ (1 ,
2) when | x | > α = 2,we obtain the condition (1.2).First we introduce our notation and some preliminary results in this section.Throughout the paper c will denote a positive constant. Its value may change fromline to line.We consider a d -dimensional right continuous rotationally symmetric α -stableprocess ( Y t ) t ≥ for α ∈ (0 , Y t ) t ≥ is a right continuous Markov pro-cess with independent and stationary increments whose characteristic function is E ( e iξY s ) = e − s | ξ | α , ξ ∈ R d , s >
0. By p ( s, x, y ), we will denote its (symmetric) tran-sition density such that P x ( Y s ∈ A ) = Z A p ( s, x, y ) dy, and by P s we will denote the corresponding semi-group P s ( f )( x ) = E x ( f ( Y s )). Here P x is the probability measure for the process started at x ∈ R d , and E x is the expectationtaken with respect to P x . The transition density p ( s, x,
0) satisfies the scaling property p ( s, x,
0) = s − d/α p (1 , x/s /α , , x ∈ R d , s > . (1.3)Similarly, we denote a one-dimensional Brownian motion (independent from Y s ) by Z s and the probability measure for the process started at t > P t . The process ofinterest is the product X s = ( Y s , Z s ) started at ( x, t ) ∈ R d × R + , the corresponding2robability measure and the expectation are P ( x,t ) and E ( x,t ) , respectively. Define thestopping time T = inf { s ≥ Z s = 0 } which is the first time X t hits the boundary of R d × R + . It is clear that T and the process Y are independent since T is expressedin terms of Z only.To provide a connection between probabilistic and deterministic integrals, we willuse two tools; a new measure P m a and the vertical Green function. Denoting theLebesgue measure on R d by m ( · ), we define the measure P m a by P m a = Z R d P ( x,a ) m ( dx ) , a > . Let E m a denote the expectation with respect to this measure. We note that the lawof X T under this measure is m ( · ). Moreover, the semi-group P t is invariant under theLebesgue measure, that is, Z R d P t f ( x ) m ( dx ) = Z R d f ( x ) m ( dx ) . (1.4)This follows from the symmetry of the kernel and the conservativeness of Y .Second, for a positive Borel function f , the vertical Green function, which is theGreen function for one-dimensional Brownian motion, is given by E a (cid:20)Z T f ( Z s ) ds (cid:21) = Z ∞ ( s ∧ a ) f ( s ) ds. (1.5)Harmonic functions play a key role in showing boundedness of Littlewood-Paleyoperators. Here we adapt the probabilistic interpretation of a harmonic function (withrespect to the process X ). A continuous function u : R d × R + → R is said to beharmonic (or α -harmonic) if u ( X s ∧ T ) is a martingale with respect to the filtration F s = σ ( X r ∧ T : r ≤ s ) and the probability measure P ( x,t ) for any starting point( x, t ) ∈ R d × R + . One way to obtain such a harmonic function is to start with abounded Borel function f : R d → R and define its extension u by u ( x, t ) := E ( x,t ) f ( Y T ) = Z ∞ E x f ( Y s ) P t ( T ∈ ds ) , where P t ( T ∈ ds ) is the exit distribution of one-dimensional Brownian motion from(0 , ∞ ) which is given by µ t ( ds ) := P t ( T ∈ ds ) = t √ π e − t / s s − / ds (see [14]). By a slight abuse of notation, we will denote both the function on R d andits extension to the upper-half space by the same letter, that is, f t ( x ) := f ( x, t ) = E ( x,t ) f ( Y T ). Next, we define the semi-group Q t = R ∞ P s µ t ( ds ). This semi-groupprovides us a representation of the extension f t ( x ) = f ( x, t ) = Q t f ( x ) = Z R d f ( y ) Z ∞ p ( s, x, y ) µ t ( ds ) dy.
3e note that this is a convolution with the probability kernel q t ( x ) = Z ∞ p ( s, x, µ t ( ds ) , whose Fourier transform is e − t |·| α/ . So q t ( x ) can be identified with the density of asymmetric α/ q t ( x ) is radially decreasing in x . To see this, it is enough to write therepresentation p (1 , x,
0) = Z R d πs ) d/ e −| x | / (4 s ) g α/ (1 , s ) ds, (1.6)where g α/ is the density of an α/ R ∞ e − λv g ( s, v ) dv = e − sλ α/ . (See [16, p. 261] for details.)One of the key tools in proving certain inequalities is the density estimates on p ( s, x, c ( s − d/α ∧ s | x − y | d + α ) ≤ p ( s, x, y ) ≤ c ( s − d/α ∧ s | x − y | d + α ) , (1.7)( s, x, y ) ∈ R + × R d × R d , which allows us to control the tail of the transition density. (See[6, Theorem 2.1].) This estimate leads to an estimate on q t ( x ) due to the observationthat it coincides with the density of a symmetric α/ c ( t − d/α ∧ t | x | d + α ) ≤ q t ( x ) ≤ c ( t − d/α ∧ t | x | d + α ) . (1.8)In addition, we will need to control the derivative of p ( s, x, ∂ kx j denote the k th partial derivative in the direction of j th coordinate. Lemma 1.2.
For k = 1 , and j = 1 , ..., d , we havei. (cid:12)(cid:12)(cid:12) ∂ kx j p (1 , x, (cid:12)(cid:12)(cid:12) ≤ c (cid:18) ∧ | x | k (cid:19) p (1 , x, andii. (cid:12)(cid:12)(cid:12) ∂ kx j p ( t, x, (cid:12)(cid:12)(cid:12) ≤ c (cid:18) t − k/α ∧ | x | k (cid:19) p ( t, x, whenever t > . This Lemma is a direct consequence of Proposition 3.3 of [3] and the inequality(1.7) above.For the rest of the paper, we will need some results and definitions from [12]. Tokeep this paper as much self-contained as possible, we provide some of these theoremsand definitions here. For details, we refer to [12]. One of the main results of [12] isthat harmonic functions, defined above, satisfy the Harnack inequality. We will use4his result to show boundedness of some operators in the next section. Let us denoteby D r the open rectangular box with center ( y, s ) ∈ R d × R + D r = { ( x, t ) ∈ R d × R + : | x i − y i | < r /α , i = 1 , ..., d, x = ( x , ..., x d ) , | s − t | < r } . When using these rectangular boxes, we will consider nested boxes with the same cen-ter. That is why we don’t include the center point in the notation for simplicity, andjust write D r for these rectangular boxes. Theorem 1.3 (K. ’11) . There exists c > such that if u is non-negative and boundedon R d × R + , harmonic in D and D ⊂ R d × R + , then u ( x, t ) ≤ c u ( x ′ , t ′ ) , ( x, t ) , ( x ′ , t ′ ) ∈ D . Using this inequality, we proved a Littlewood-Paley Theorem. We defined a newoperator with respect to our product process X s = ( Y s , Z s ). The horizontal componentof the classical operator is replaced by the one corresponding to the symmetric stableprocess. The two components are defined as −→ G f ( x ) = (cid:20)Z ∞ t Z R d [ f t ( x + h ) − f t ( x )] | h | d + α dh dt (cid:21) / , and G ↑ f ( x ) = "Z ∞ t (cid:20) ∂∂t f ( x, t ) (cid:21) dt / , and hence the Littlewood-Paley operator G f is defined as G f = h ( −→ G f ) + ( G ↑ f ) i / . Unlike the Brownian motion case, the Littlewood-Paley Theorem (Theorem 1.4 part(i.)) cannot be extended to p ∈ (1 , −→ G f,α , −→ G f,α ( x ) = (cid:20)Z ∞ t Γ α ( f t , f t )( x ) dt (cid:21) / , where Γ α ( f t , f t )( x ) = Z | h |
2. One of our main results in the paper [12] (part iii. ofTheorem 1.4) allows us to generalize this inequality first by considering the harmonicextension Q t f and then writing the integrand as the singular integral (1.9) instead ofthe differential ∂ αx on a restricted domain to provide some control over the large jumpterms. Without this restriction, it is not possible to extend this result to p ∈ (1 , p > Lemma 1.5. If p > and f ∈ L p ( R d ) ∩ L ( R d ) then k f k p ≤ c k G ↑ f k p .Proof. First note that by the Plancherel identity, k G ↑ f k = c Z ∞ t Z R d (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | ξ | α e − t | ξ | α/ dξ dt = c k f k , (1.10)since ( Q t f ) b ( · ) = e − t |·| α/ b f ( · ).Second, if h ∈ L q ( R d ) ∩ L ( R d ), where 1 /p + 1 /q = 1, then using polarization6dentity and equality (1.10), Z R d f ( x ) h ( x ) dx = 14 (cid:0) k f + h k − k f − h k (cid:1) = c (cid:16) k G ↑ f + h k − k G ↑ f − h k (cid:17) = c Z R d Z ∞ t ∂f∂t ( x, t ) ∂h∂t ( x, t ) dtdx. Using the Cauchy-Schwartz inequality and then the H¨older inequality, we obtain Z R d f ( x ) h ( x ) dx ≤ c Z R d G ↑ f ( x ) G ↑ h ( x ) dx ≤ c k G ↑ f k p k G ↑ h k q ≤ c k G ↑ f k p k h k q where the last inequality follows from Theorem 1.4.Finally, the result follows if we take supremum over all such h with k h k q ≤ L p ( R d ). Among these operators, twoimportant ones are the Area functional and G ∗ functional. The Area functional in oursetup is given by A f ( x ) = (cid:20)Z ∞ Z | y |
1. Note that k K λt k = k K λ k = c d . Hence the normalized function c − d K λt is a bounded approximate identity. Using this kernel we define two componentsby −→ G ∗ λ,f ( x ) = (cid:20)Z ∞ t · K λt ∗ Γ α ( f t , f t )( x ) dt (cid:21) / ,G ∗ , ↑ λ,f ( x ) = (cid:20)Z ∞ t · K λt ∗ ( ∂∂t f t ( · )) ( x ) dt (cid:21) / G ∗ functional is G ∗ λ,f ( x ) = (cid:20)h −→ G ∗ λ,f ( x ) i + h G ∗ , ↑ λ,f ( x ) i (cid:21) / . As we can see in definitions of the operators, we mostly restrict our domain of integra-tion to a parabolic-like domain in the upper half-space. By taking the scaling factorinto account, we focus on the set { ( y, t ) ∈ R d × R + : | y − x | < t /α } with vertexat x ∈ R d . Our first observation is that the growth of an extension function is con-trolled by the Hardy-Littlewood maximal function M ( · ), where the Hardy-Littlewoodmaximal function is given by M ( f )( x ) = sup r> | B (0 , | · r d Z | y |
Let p > and f ∈ L p ( R d ) . Theni. N fα ( x ) ≤ c M ( f )( x ) , x ∈ R d , ii. N fα ∈ L p ( R d ) and k N fα k p ≤ c k f k p . Proof.
It is enough to consider positive functions to prove the first statement. If f isnot positive, then we can consider the decomposition f = f + − f − , where f + , f − ≥ Q t , the inequalities N fα ≤ N f + α + N f − α and M ( f + ) + M ( f − ) ≤ M ( f )and the fact that both Q t f + and Q t f − are positive harmonic to prove the result for f . Hence we can reduce our problem to positive functions. So suppose f >
0. Thenfor a fixed t > y ∈ B ( x, t /α ), Theorem 1.3 applied several times implies that8 t ( y ) ≤ c f t ( x ). Here we should emphasize that the constant c does not depend on thevariable t , since these balls scale as t varies and so the same number of application ofthe Harnack inequality suffices at each t for fixed x .Moreover, f t ( x ) = f ∗ q t ( x ) where q t is radially decreasing and its L -norm equalsone. To see this we note that the transition density p ( s, x,
0) is obtained from thecharacteristic function e − s | x | α by the inverse Fourier transform. Hence we can write p ( s, x,
0) as in equation (1.6). Thus p ( s, x,
0) is radially decreasing in the variable x and so is q t ( x ). Then f t ( x ) ≤ c M ( f )( x ) for any t > N fα ( x ) ≤ c M ( f )( x ). Finally, using the fact kM ( f ) k p ≤ c k f k p , p > , one can obtain the result.Before we study the Area functional, we define an auxiliary operator L ∗ f . Thisoperator is in a close relation with −→ G ∗ λ,f for a particular value of λ and hence it providesan intermediate step to prove boundedness of the Area functional. Moreover, the classicversion L ∗ f is used to give a probabilistic proof of boundedness of Littlewood-Paleyfunction.For a given f ∈ L p ( R d ), we define this operator as L ∗ f ( x ) = (cid:20)Z ∞ t · Q t Γ α ( f t , f t )( x ) dt (cid:21) / , where Γ α is as in (1.9). This operator is bounded on L p ( R d ) whenever p > Theorem 2.2.
Let p > and f ∈ L p ( R d ) . Then we have k L ∗ f k p ≤ c k f k p . Proof.
Let f ∈ L p ( R d ), r = 2 p and q be the conjugate of r , that is, 1 /r + 1 /q = 1. Let h be a continuously differentiable function with compact support. Then E ( x,a ) (cid:20)Z T Γ α ( f Z s , f Z s )( Y s ) ds · h ( X T ) (cid:21) = Z ∞ E ( x,a ) (cid:2) E ( x,a ) (cid:2) { s 1, then (cid:20)Z R d (cid:0) L ∗ f ( x ) (cid:1) r dx (cid:21) /r ≤ c k f k r , which gives the result if we replace r with p/ λ = (2 d + α ) / (2 d ) then we can see the relation between twofunctionals L ∗ f and −→ G ∗ λ ,f . Hence we can show boundedness of the area functional A f . Theorem 2.3. Suppose p > and f ∈ L p ( R d ) . Then we havei. for λ > , A f ≤ c λ −→ G ∗ λ,f .ii. If λ = (2 d + α ) / (2 d ) , then k−→ G ∗ λ ,f k p ≤ c k f k p . iii. k A f k p ≤ c k f k p .Proof. Part (i.) is easy when we observe (cid:20) t /α t /α + | y | (cid:21) λd ≥ − λd for | y | < t /α . Part (iii.) is a corollary of (i.) and (ii.). So it is enough to prove (ii.).First we recall that K λ t ( x ) = t ( t /α + | x | ) d + α = t − d/α 11 + | x | t /α ! d + α . We also know that q t ( x ) is comparable to t − d/α ∧ t | x | d + α = t − d/α ∧ (cid:16) | x | t /α (cid:17) d + α , by (1.8). Hence q t is comparable to K λ t and we have K λ t ≤ c q t ( x ) . This leads to −→ G ∗ λ ,f ( x ) ≤ cL ∗ f ( x ) . Then the result follows from Theorem 2.2.11he result of the previous theorem is not restricted to the horizontal componentwith parameter λ . We can generalise this result to the case including the verticalcomponent and any parameter λ > Theorem 2.4. If λ > , p ≥ and f ∈ L p ( R d ) then k G ∗ λ,f k p ≤ c k f k p . Proof. Let us denote by g α ( y, t ) the functionΓ α ( f t , f t )( y ) + (cid:18) ∂∂t f ( y, t ) (cid:19) . Assume h ∈ C K ( R d ). Then by the symmetry of K λt ( x ) in x , Z R d h ( x ) ( G ∗ λ,f ( x )) dx = Z ∞ t Z R d h ( x ) Z R d K λt ( x − y ) g α ( y, t ) dy dx dt = Z ∞ t Z R d g α ( y, t ) · h ∗ K λt ( y ) dy dt. Since K λt is radially decreasing and integrable, h ∗ K λt ( y ) ≤ c M ( h )( y ). Hence Z R d h ( x ) ( G ∗ λ,f ( x )) dx ≤ c Z R d M ( h )( x ) ( G f,α ( x )) dx. (2.2)For p = 2, it is enough to consider h = 1. Then by parts (ii.) and (iii.) of Theorem1.4, k G ∗ λ,f k ≤ c k G f,α k ≤ c k f k . Now suppose p > 2. We take r = p/ q > /r + 1 /q = 1. UsingH¨older’s inequality in (2.2), Z R d h ( x ) ( G ∗ λ,f ( x )) dx ≤ c (cid:20)Z R d ( M ( h )( x )) q dx (cid:21) /q · (cid:20)Z R d ( G f,α ( x )) r dx (cid:21) /r ≤ c k h k q k G f,α k p . If we take supremum over all such h with k h k q ≤ 1, then we obtain k G ∗ λ,f k p = k ( G ∗ λ,f ) k r ≤ c k G f,α k p . Finally, using the boundedness of the operator G f,α when p > Q t . 12 heorem 2.5. Let p > . Suppose T is a convolution operator on L p ( R d ) with kernel κ , that is, T f ( x ) = f ∗ κ ( x ) . Suppose further that there exists λ > such that | ∂ t Q t κ ( x ) | ≤ c t − − d/α (cid:18) t /α t /α + | x | (cid:19) λd = c t − K λt ( x ) . (2.3) Then for f ∈ C K (, that is, f ∈ C with compact support) k T f k p ≤ c k f k p . The condition (2.3) above may not seem very useful in terms of applications whenwe consider its current form. Hence we will provide a sufficient and a more usefulcondition later in Theorem 2.8 below. Proof. First suppose p > 2. We note that by the semi-group property, we have Q t = Q t/ Q t/ and q t = q t/ ∗ q t/ , which leads to ∂ t q t = 2 q t/ ∗ ∂ t q t/ . Next, we observe that ∂ t Q t T f ( x ) = 2 Q t/ T ( ∂ t Q t/ f )( x ), since their Fourier transforms are equal , that is, \ (cid:2) Q t/ T ( ∂ t Q t/ f ) (cid:3) = 2 c q t/ b κ \ ( ∂ t q t/ ) b f = \ (2 q t/ ∗ ∂ t q t/ ) b κ b f = d ∂ t q t b κ b f = \ ∂ t Q t T f . Then (cid:16) G ↑ T f ( x ) (cid:17) = Z ∞ t | ∂ t Q t T f ( x ) | dt = 4 Z ∞ t (cid:12)(cid:12) Q t/ T ( ∂ t Q t/ f )( x ) (cid:12)(cid:12) dt. Using our assumption (2.3), we can see that Q t/ T ( ∂ t Q t/ f ) = (1 / ∂ t Q t T f ( x ) → t → ∞ . Hence the last line above equals4 Z ∞ t (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ t ss ∂ s Q s/ T ( ∂ s Q s/ f )( x ) ds (cid:12)(cid:12)(cid:12)(cid:12) dt. If we apply the Cauchy-Schwartz inequality first, and then change the order of theintegrals, we get (cid:16) G ↑ T f ( x ) (cid:17) ≤ c Z ∞ t (cid:20)Z ∞ t s − ds (cid:21) · (cid:20)Z ∞ t s ( ∂ s Q s/ T ( ∂ s Q s/ f )( x )) ds (cid:21) dt = c Z ∞ Z ∞ t s ( ∂ s Q s/ T ( ∂ s Q s/ f )( x )) dsdt = c Z ∞ s ( ∂ s Q s/ T ( ∂ s Q s/ f )( x )) ds. Using the bound in (2.3) and Jensen’s inequality, (cid:16) G ↑ T f ( x ) (cid:17) ≤ c Z ∞ s (cid:2) ( s − K λs/ ) ∗ ( ∂ s Q s/ f )( x )) (cid:3) ds ≤ c Z ∞ s K λs/ ∗ ( ∂ s Q s/ f ) ( x ) ds ≤ c (cid:0) G ∗ λ,f ( x ) (cid:1) . p > k T f k p ≤ c k G ↑ T f k p ≤ c k G ∗ λ,f k p ≤ c k f k p , by Lemma 1.5 and Theorem 2.4.For p ∈ (1 , 2) we use a duality argument. For this purpose let q be such that1 /p +1 /q = 1. First we observe that if κ ∗ ( x ) = κ ( − x ) and T ∗ is the convolution operatorcorresponding to κ ∗ , then the condition (2.3) holds for κ ∗ . Then for h ∈ L q ( R d ) (cid:12)(cid:12)(cid:12)(cid:12)Z R d h ( x ) T f ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R d T ∗ h ( x ) f ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ c k T ∗ h k q k f k p ≤ k h k q k f k p by the first part of the proof. Finally, if we take supremum over all such h with k h k q ≤ 1, the result follows.Before any further discussion, we recall the definition of the measure µ t ( ds ) = t √ π e − t / s s − / ds and show the following estimates. Lemma 2.6. For M > , we havei. Z M | s − / | µ ( ds ) ≤ √ π M / ,ii. Z ∞ M | − / (2 s ) | µ ( ds ) ≤ √ π M − / .Proof. (i.) We note that | s − / | s − / e − / (4 s ) ≤ s > 0. Hence the result follows.(ii.) Similarly, we also have | s − / | e − / (4 s ) ≤ , which results in the desired inequality. Lemma 2.7. Suppose ψ ( x ) = ( ∂ t q t ( x )) t =1 = Z ∞ p ( s, x, (cid:18) − s (cid:19) µ ( ds ) . Then forsome positive constants c, c , c we havei. | ψ ( x ) | ≤ c (cid:0) ∧ | x | − d − α (cid:1) ≤ c q ( x ) andii. | ∂ x i ψ ( x ) | ≤ c (cid:0) ∧ | x | − d − − α/ (cid:1) , i = 1 , ..., d . roof. (i.) First note that | ψ ( x ) | ≤ c Z ∞ p ( s, x, | − s | s − / e − / (4 s ) ds ≤ c Z ∞ | − s | s − − dα e − / (4 s ) ds < ∞ by the estimate (1.7) on the density p ( s, x, | ψ ( x ) | ≤ c Z | x | α s | x | d + α | − s | µ ( ds ) + c Z ∞| x | α s − d/α | − s | µ ( ds ) ≤ c | x | d + α Z | x | α | s − | µ ( ds ) + c | x | d Z ∞| x | α | − s | µ ( ds ) . By Lemma 2.6, | ψ ( x ) | ≤ c (cid:0) ∧ | x | − d − α/ (cid:1) . The second inequality follows from the estimate (1.8) on q ( x ).(ii.) Similarly, using the bound on ∂ x i p ( s, x, 0) (Lemma 1.2), we obtain | ∂ x i ψ ( x ) | ≤ c Z ∞ | ∂ x i p ( s, x, || − s | µ ( ds ) ≤ c Z ∞ | − s | s − − d +1 α e − / (4 s ) ds < ∞ and | ∂ x i ψ ( x ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ∂ x i p ( s, x, − s ) µ ( ds ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c Z | x | α s | x | d +1+ α | − s | µ ( ds ) + c Z ∞| x | α s − ( d +1) /α | − s | µ ( ds ) ≤ c | x | d +1+ α Z | x | α | s − | µ ( ds ) + c | x | d +1 Z ∞| x | α | − s | µ ( ds ) . When we use Lemma 2.6, we obtain the desired result.In the previous theorem, we stated a boundedness condition on the kernel of convo-lution operator by means of the action of the semi-group Q t . In order for this conditionto be more useful, we want to state an application in some purely analytic language.We provide two conditions in Theorem 2.8, under which the condition (2.3) of Theorem2.5 holds and hence the result follows. Theorem 2.8. Suppose α ∈ (1 , , κ : R d → R is a function with the cancelationproperty (1.1) such thati. | κ ( x ) | ≤ c | x | d {| x |≤ } + c | x | d − α/ {| x | > } ,ii. |∇ κ ( x ) | ≤ c | x | d +1 {| x |≤ } + c | x | d + α/ {| x | > } . uppose T is a convolution operator with kernel κ . Then for f ∈ C K and p > wehave k T f k p ≤ c k f k p . Proof. First let φ be a smooth function on R such that φ ( r ) = 1 whenever | r | ≤ φ ( r ) = 0 whenever | r | > 2. Now let κ ( x ) = κ ( x ) φ ( | x | ) , κ ( x ) = κ ( x )(1 − φ ( | x | ))and T f = f ∗ κ , T f = f ∗ κ . Then T f = T f + T f . By the classical case (Theorem 1.1), k T f k p ≤ c k f k p . Sowithout loss of generality we may assume T = T and κ = κ and ignore the indices.As before, set ψ ( x ) = ( ∂ t q t ( x )) t =1 . By scaling, we have ∂ t q t ( x ) = t − − d/α ψ ( x/t /α ).So by Theorem 2.5 and scaling, it is enough to show that | ( ∂ t Q t κ ( x )) t =1 | ≤ c (1 + | x | ) λd for some λ > 1. Here we will take λ = 1 + ( α − / (2 d ).First assume | x | ≤ 1. Then( ∂ t Q t κ ( x )) t =1 = Z | y | > κ ( y ) ψ ( x − y ) dy (2.4)and we have by Lemma 2.7 (i), | ( ∂ t Q t κ ( x )) t =1 | ≤ Z | y | > | κ ( y ) | | ψ ( x − y ) | dy ≤ c Z | y | > | κ ( y ) | q ( x − y ) dy. Then the assumption (i) on κ ( · ) in the hypothesis gives us that | κ ( y ) | ≤ c | y | d − α/ ≤ c whenever | y | > . Thus Z | y | > | κ ( y ) | q ( x − y ) dy ≤ c Z | y | > q ( x − y ) dy ≤ c Z R d q ( x − y ) dy = c, since q is a probability kernel. Hence | ( ∂ t Q t κ ( x )) t =1 | ≤ c ≤ c (1 + | x | ) λd . (2.5)16ow assume | x | > 1. Consider three subsets of R d : D = { y ∈ R d : | y | < | x | / } , D = { y ∈ R d : | y − x | < | x | / } and D = R d − ( D ∪ D ). Now split the integral (2.4)into three integrals with respect to these subsets. Z | y | > κ ( y ) ψ ( x − y ) dy = Z D + Z D + Z D := I + I + I . Since κ satisfies the cancelation condition (1.1), | I | = (cid:12)(cid:12)(cid:12)(cid:12)Z D κ ( y ) ( ψ ( x − y ) − ψ ( x )) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ sup | z − x | < | x | / |∇ ψ ( z ) | Z D c | y | d − α/ | y | dy ≤ c | x | − α/ sup | z − x | < | x | / |∇ ψ ( z ) | . By Lemma 2.7, the gradient above is bounded by c | x | − d − − α/ . This gives the inequality | I | ≤ c | x | d − α ≤ c | x | λd . For I , we note that c | y | ≤ | x − y | ≤ c ′ | y | whenever y ∈ D . So using Lemma 2.7 again, | I | ≤ c Z D | x − y | d + α/ | y | d − α/ dy ≤ c Z | y |≥| x | / | y | − d +1 − α dy ≤ c | x | λd . For the last part, namely I , we use our assumption on ∇ κ . By a change of variables,we have | I | = (cid:12)(cid:12)(cid:12)(cid:12)Z | y − x | < | x | / ψ ( x − y ) κ ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z | y | < | x | / ψ ( y ) κ ( x − y ) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z | y | < | x | / | ψ ( y ) | | κ ( x − y ) − κ ( x ) | dy + | κ ( x ) | (cid:12)(cid:12)(cid:12)(cid:12)Z | y | < | x | / ψ ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) . First observe that Z R d ψ ( y ) dy = Z ∞ Z R d p ( s, y, dy (1 − s ) µ ( ds ) = 0 . Hence (cid:12)(cid:12)(cid:12)(cid:12)Z | y | < | x | / ψ ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z | y |≥| x | / ψ ( y ) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ c Z | y |≥| x | / | y | − d − α/ dy ≤ c | x | α/ . We also note that if | y | < | x | / |∇ κ | , we obtain | κ ( x − y ) − κ ( x ) | ≤ c | y || x | d + α/ ≤ c | y | / | x | d +( α − / = c | y | / | x | λd . | I | ≤ c | x | λd Z R d | ψ ( y ) | · | y | / dy + c | x | d − α . (2.6)If we show that the integral in (2.6) is bounded by a constant, then we have | I | ≤ c | x | λd . To show the boundedness of the integral in (2.6), we consider the cases | y | < | y | ≥ 1. Note that Z R d | ψ ( y ) | · | y | / dy ≤ Z | y | < | ψ ( y ) | dy + Z | y |≥ c | y | d + α/ · | y | / dy by Lemma 2.7. The second term is convergent since α > 1. The first term is boundedby Z | y | < Z ∞ p ( s, y, | − s | µ ( ds ) dy ≤ Z ∞ Z R d p ( s, y, dy | − s | µ ( ds ) ≤ Z / s µ ( ds ) + Z ∞ / µ ( ds ) ≤ c. Hence | ( ∂Q t κ ( x )) t =1 | ≤ | I | + | I | + | I | ≤ c | x | λd ≤ c (1 + | x | ) λd (2.7)whenever | x | > 1. Finally, inequalities (2.5) and (2.7) and Theorem 2.5 imply that k T f k p ≤ c k f k p for p > f ∈ C K , which finishes the proof. 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