A new characterization of Sobolev spaces on R n

Transcription

1 A new characterization of Sobolev spaces on R n Roc Alabern, Joan Mateu and Joan Verdera Abstract In this paper we present a new characterization of Sobolev spaces on R n. Our characterizing condition is obtained via a quadratic multiscale expression which exploits the particular symmetry properties of Euclidean space. An interesting feature of our condition is that depends only on the metric of R n and the Lebesgue measure, so that one can define Sobolev spaces of any order of smoothness on any metric measure space. Introduction In this paper we present a new characterization of the Sobolev spaces W α,p on R n, where the smoothness index α is any positive real number and < p <. Thus W α,p consists of those functions f L p = L p R n ) such that ) α/2 f L p. Here is the Laplacian and ) α/2 f is defined on the Fourier transform side by ξ α f ˆξ). If < α < n this means that f is a function in L p which is the Riesz potential of order α of some other function g in L p, namely f = c n / x n α g. If α is integer, then W α,p is the usual space of those functions in L p such that all distributional derivatives up to order α are in L p. To convey a feeling about the nature of our condition we first discuss the case α =. Consider the square function S f ) 2 x) = f f x) t 2 t, x Rn. Here f is a locally integrable function on R n and f denotes the mean of f on the open ball with center x and radius t. One should think of f f x) as a quotient of increments t of f at the point x. Our characterization of W,p reads as follows. Theorem. If < p <, then the following are equivalent. ) f W,p 2) f L p and S f ) L p. If any of the above conditions holds then S f ) p f p.

2 The symbol A B means, as usual, that for some constant C independent of the relevant parameters attached to the quantities A and B we have C B A C B. Notice that condition 2) in Theorem above is of a metric measure space character, because only involves integrals over balls. It can be used to define in any metric measure space X a notion of Sobolev space W,p X). It is not clear to the authors what are the relations of this space with other known notions of Sobolev space in a metric measure space, in particular with those of Hajlasz [H] or Shanmugalingam [S] see also [HK]). The proof of Theorem follows a classical route see [Str]). The relevant issue is the necessary condition. First, via a Fourier transform estimate we show that S f ) 2 = c f 2, for good functions f. In a second step, we set up a singular integral operator T with values in L 2 /t) such that T f ) L 2 R n, L 2 /t)) = S f ) 2. The kernel of T turns out to satisfy Hörmander s condition, so that we can appeal to a well known result of Benedek, Calderón and Panzone [GR, Theorem 3.4, p. 492] on vector valued Calderón-Zygmund Theory to conclude the proof. The major technical difficulty occurs in checking Hörmander s condition. The proof extends without pain to cover orders of smoothness α with < α < 2. The square function S f ) has to replaced by S α f ) 2 x) = f Bx,t) f x) t α 2 t, x Rn. The result is then that, for < α < 2, f W α,p is equivalent to f L p and S α f ) L p. Notice that S α f ) 2 x) = f y) f x) dy 2 t α t, x Rn, ) where the barred integral on a set stands for the mean over that set. Strichartz [Str]) used long ago the above square function for < α < to characterize W α,p. However the emphasis in [Str] was on a larger variant of S α f ) in which the absolute value is inside the integral in y in ). In the interval α < 2 putting the absolute value inside the integral destroys the characterization, because then one gives up the symmetry properties of R n. For instance, S α f ) vanishes if f is a first degree polynomial. There are in the literature square functions very close to ) which characterize W α,p, for < α < 2. For example, first differences of f may be replaced by second differences and the absolute value may be placed inside the integral [Str] and [St, Chapter V]). The drawback with second differences is that they do not make sense in the setting of metric measure spaces. See also the paper by Dorronsoro [D]. We now proceed to explain the idea for the characterization of W 2,p. Take a smooth function f and consider its Taylor expansion up to order 2 around x f y) = f x) + f x) y x) + β f x)y x) β + R, 2) 2 β =2

3 where R is the remainder and β a multi-index of length 2. Our goal is to devise a square function which plays the role of S f ) see ) for α = ) with respect to second order derivatives. The first remark is that the mean on of the homogeneous polynomial of degree in 2) is zero. Now, the homogeneous Taylor polinomial of degree 2 can be written as β f x) y x) β = Hy x) + β! 2n f x) y x 2, 3) β =2 for a harmonic homogeneous polynomial H of degree 2. Hence the mean on of the homogeneous Taylor polinomial of degree 2 is 2n f x) y x 2 dy. This suggests defining S 2 f )x) 2 = f y) f x) 2n f ) y x 2) t 2 2 dy t, x Rn. 4) We cannot replace f ) by f x) in the preceding definition, because the mean guarantees a little extra smoothness which one needs in a certain Fourier transform computation. Notice that, according to the remarks made before on the mean on the ball of the homogeneous Taylor polynomials of degrees and 2, in the expression above for S 2 f )x) one may add the missing terms to get the full Taylor polynomial of degree 2, except for the fact that f x) should be replaced by f ). Were f smooth enough, one could even add the homogeneous Taylor polynomial of degree 3, because it is odd taking x as the origin) and thus its mean on vanishes. This explains why whatever we can prove for α = 2 will also extend to the range 2 < α < 4 by considering an appropriate square function, which turns out to be S α f )x) 2 = f y) f x) f x) y x 2) 2n t α 2 dy t, x Rn. One should remark that in the range 2 < α < 4 the mean f ) can safely be replaced by f x). Here is our second order theorem. Theorem 2. If < p <, then the following are equivalent. ) f W 2,p 2) f L p and there exists a function g L p such that S 2 f, g) L p, where the square function S 2 f, g) is defined by S 2 f, g)x) 2 = f y) f x) g y x 2) 2 dy t 2 t, x Rn. 3

4 If f W 2,p then one can take g = f /2n and if 2) holds then necessarily g = f /2n, a. e. If any of the above conditions holds then S f, g) p f p. Notice that condition 2) in Theorem 2 only involves the Euclidean distance on R n and integrals with respect to Lebesgue measure. Thus one may define a notion of W 2,p X) on any metric measure space X. For more comments on that see section 5. Again the special symmetry properties of R n play a key role. For instance, S 2 annihilates second order polynomials. Theorem 2 has a natural counterpart for smoothness indexes α satisfying 2 < α < 4. The result states that a function f W α,p if and only if f L p and there exists a function g L p such that S α f, g) L p, where ) S α f, g)x) 2 f y) f x) gx) y x 2 2 = dy t α t, x Rn. Notice that in the range 2 < α < 4 we do not need to replace gx) by the mean g. Before stating our main result, which covers all orders of smoothness and all p with < p <, we need to discuss a couple of preliminary issues. The first is the analogue of 3) for homogeneous polynomials of any even degree. Let P be a homogeneous polynomial of degree 2 j. Then P can be written as Px) = Hx) + j P x 2 j, where = j x 2 j ) and H satisfies j H =. This follows readily from [St, 3..2, p. 69]. Considering the spherical harmonics expansion of Px) we see that Hx) dσ =, x = σ being the surface measure on the unit sphere, and thus that Hx) dx =, t >. The x t precise value of, which can be computed easily, will not be needed. Our main result involves a square function associated with a positive smoothness index α. Let N be the unique integer such that 2N α < 2N + 2. Given locally integrable functions f, g,..., g N we set where S α f, g, g 2,..., g N )x) 2 = R N y, x) t α dy 2 t, x Rn, R N y, x) = f y) f x) g x) y x 2 g N x) y x 2N ) g N ) y x 2N if α = 2N, and R N y, x) = f y) f x) g x) y x 2 g N x) y x 2N ) g N x) y x 2N if 2N < α < 2N

5 Theorem 3. Given α > let N be the unique integer such that 2N α < 2N + 2. If < p <, then the following are equivalent. ) f W α,p 2) f L p and there exist functions g j L p, j N, such that S α f, g, g 2,..., g N ) L p. If f W α,p then one can take g j = j f / and if 2) holds then necessarily g j = j f / a. e. If any of the above conditions holds then S α f, g,..., g N ) p ) α/2 f p. Remark. In the statement above one should understand that for N = or, equivalently, for < α < 2) no function g j appears. The square function involved in this case is S α f ) as defined in ). Again condition 2) in Theorem 3 only involves the Euclidean distance on R n and integrals with respect to Lebesgue measure. Thus one may define a notion of W α,p X) for any positive α and any < p < on any metric measure space X. For previous notions of higher order Sobolev spaces on metric measure spaces see [LLW]. See section 5 for more on that. The proof of Theorem 3 proceeds along the lines sketched before for α =. First we use a Fourier transform computation to obtain the relation S α f, f /L,..., N f /L N ) 2 = c ) α/2 f 2. Then we introduce a singular integral operator with values in L 2 /t 2α+ ) and we check that its kernel satisfies Hörmander s condition. The paper is organized as follows. In sections 2, 3 and 4 we prove respectively Theorems, 2 and 3. In this way readers interested only in first order Sobolev spaces may concentrate in section 2. Those readers interested in the main idea about jumping to orders of smoothness 2 and higher may read section 3. Section 4 is reserved to those interested in the full result. In any case the technical details for the proof of Theorem are somehow different of those for orders of smoothness 2 and higher. The reason is that Hörmander s condition involves essentially taking one derivative of the kernel and is precisely the kernel associated to the first order of smoothness that has minimal differentiability. Our notation and terminology are standard. For instance, we shall adopt the usual convention of denoting by C a constant independent of the relevant variables under consideration and not necessarily the same at each occurrence. If f has derivatives of order M for some non-negative integer M, then M f = β f ) β =M is the vector with components the partial derivatives of order M of f and M f its Euclidean norm. The Zygmund class on R n consists of those continuous functions f such that, for some constant C, f x + h) + f x h) 2 f x) C h, x, h R n. The basic example of a function in the Zygmund class which is not Lipschitz is f x) = x log x, x R n. 5

6 The Schwartz class consists of those infinitely differentiable functions on R n whose partial derivatives of any order decrease faster than any polynomial at. After the first draft of the paper was made public Professor Wheeden brought to our attention his articles [W] and [W2]. In [W] a general result is proven which, in particular, contains Theorem 3 for α not an even integer. In [W2] Sobolev spaces with respect to special homogeneous spaces are considered for < α <. 2 Proof of Theorem The difficult part is the necessity of condition 2) and we start with this. As a first step we show that for a dimensional constant c. Set S f ) 2 = c f 2 5) χx) = B, ) χ B,)x) and χ t x) = t n χ x t ), so that, by Plancherel, R n S f )x) 2 dx = = c f χ t )x) f x) 2 dx R t n 3 ˆχtξ) 2 f ˆξ) 2 dξ R t. n 3 Since ˆχ is radial, ˆχξ) = F ξ ) for a certain function F defined on [, ). Exchange the integration in dξ and in the last integral above and make the change of variables τ = t ξ. Then S f )x) 2 dx = c R n and 5) is reduced to showing that = c R n Fτ) 2 dτ τ 3 ˆ f ξ) 2 ξ 2 dξ Fτ) 2 dτ τ 3 f 2 2 Fτ) 2 dτ τ 3 <. 6) 6

7 Set B = B, ) and e =,,..., ) R n. Then Ft) = ˆχte ) = exp ix t) dx which yields = B = 2 B ix t ) 2 x2 t2 + B x 2 dx t2 +, Ft) = Ot 2 ), as t and shows the convergence of 6) at. Since F ξ ) = ˆχξ) is the Fourier transform of an integrable function, Fτ) is a bounded function and so the integral 6) is clearly convergent at. We are left with the case of a general p between and. If f W,p, then f = g / x n for some g L p with / x n replaced by log x for n = ). Set Ix) = / x n. Then f f x) = f χ t )x) f x) = g K t )x), where K t x) = I χ t )x) Ix) = If we let Tg)x) = g K t )x), x R n, then one can rewrite S f )x) as Then 5) translates into S f )x) = dx g K t )x) 2 ) 2 = Tgx) L2/t3). t 3 R n Tgx) 2 L 2 /t 3 ) dx = c f 2 2 = C g 2 2, Iy) dy Ix). 7) and we conclude that T is an operator mapping isometrically modulo a constant) L 2 R n ) into L 2 R n, L 2 /t 3 )). If the kernel K t x) of T satisfies Hörmander s condition K t x y) K t x) L 2 /t 3 ) dx C, y R n x 2 y then a well known result of Benedek, Calderón and Panzone on vector valued singular integrals see [GR, Theorem 3.4, p. 492]) yields the L p estimate R n Tgx) p L 2 /t 3 ) dx C p g p p, which can be rewritten as S f ) p C p f p. 7

8 The reverse inequality follows from polarization from 5) by a well known duality argument [GR, p. 57]) and so the proof of the necessary condition is complete. We are going to prove the following stronger version of Hörmander s condition K t x y) K t x) L 2 /t 3 ) C y x n+, y Rn, 8) for almost all x satisfying x 2 y. To prove 8) we deal separately with three intervals in the variable t. Interval : t < x 3. From the definition of K t in 7) we obtain K t x) = I χ t )x) Ix). Notice that, in the distributions sense, the gradient of I is a constant times the vector valued Riesz transform, namely x I = n )p.v. x. n+ If x 2 y, then the segment [x y, x] does not intersect the ball B, x /2) and thus K t x y) K t x) y sup K t z). 9) z [x y,y] If t < x /3 and z [x y, y], then Bz, t) R n \ B, x /6), and hence K t z) = Iw) Iz)) dw. ) Bz, t) Taylor s formula up to order 2 for Iw) around z yields Iw) = Iz) + 2 w z 2 Iz)w z) + O x ), n+2 where 2 Iz)w z) is the result of applying the matrix 2 Iz) to the vector w z. The mean value of 2 Iz)w z) on Bz, t) is zero, by antisymmetry, and thus, by ), and so, by 9), Integrating in t we finally get x /3 K t z) C t2 x n+2 K t x y) K t x) C y x. n+2 K t x y) K t x) 2 ) 2 t 3 C 8 t 2 y x /3 ) 2 t x n+2 = C y x n+.

9 Interval 2: x /3 < t < 2 x. The function I χ t is continuously differentiable on R n \ S t, S t = {x : x = t}, because its distributional gradient is given by I χ t and each component of χ t is a Radon measure supported on S t. The gradient of I χ t is given at each point x R n \ S t by the principal value integral p.v. I χ t )x) = n )p.v. y dy, y n+ which exists for all such x. The difficulty in the interval under consideration is that it may happen that x = t and then the gradient of I χ t has a singularity at such an x. We need the following estimate. Lemma. p.v. y y n+ dy C log x + t x t, x Rn. Proof. Assume without loss of generality that x = x,,..., ). The coordinates y j, j, change sign under reflection around the y axes. Hence y j p.v. dy =, < j n. y n+ Now, if x < t, If x > t, Bx, p.v. y t) y Bx, p.v. y t) y n+ dy n+ dy = Bx, p.v. y t)\b, t x ) y C = C t+ x t x t y y dy n+ x +t x t t n+ dy = C log t + x t x. = C log x + t x t. Assume without loss of generality that y = y,,..., ). The distributional gradient of I χ t is y n )p.v. y χ n+ t, which is in L 2. Then I χ t W,2 and consequently is absolutely continuous on almost all lines parallel to the first axes. Therefore K t x y) K t x) = 9 K t x τy) y dτ

10 for almost all x and Hence 2 x x /3 K t x y) K t x) C y x n K t x y) K t x) 2 ) 2 t 3 C y x n+ = C y x n+ D, ) x τy + t + log dτ. x τy t 2 x x /3 ) ) 2 x τy + t + log dτ x τy t t where the last identity is a definition of D. Applying Schwarz to the inner integral in D and then changing the order of integration we get D 2 2 x x /3 For each τ make the change of variables + log s = x τy + t x τy t t x τy ) 2 dτ. t 2 to conclude that D 2 4 2/9 + log + s ) 2 ds s s. Interval 3: 2 x t. For each z in the segment [x y, y] we have B, t/4) Bz, t). Then, by ), w K t z) = n ) p.v. dw z ) Bz, t) Bz, t) w n+ z n+ ) = n ) and so Hence, owing to 9), and thus 2 x Bz, t) Bz, t)\b, t/4) w dw z w n+ K t z) C, z [x y, y]. x n K t x y) K t x) C y x n K t x y) K t x) 2 ) 2 y C t 3 x n 2 x z n+ ) 2 y = C t 3 x, n+

11 which completes the proof of the strengthened form of Hörmander s condition 8). We turn now to prove that condition 2) in Theorem is sufficient for f W,p. Let f L p satisfy S f ) L p. Take an infinitely differentiable function φ with compact support in B, ), φ = and set φ ɛ x) = φ x ), ɛ >. Consider the regularized ɛ n ɛ functions f ɛ = f φ ɛ. Then f ɛ is infinitely differentiable and f ɛ p f p φ ɛ, so that f ɛ W,p. Thus, as we have shown before, f ɛ p S f ɛ ) p. We want now to estimate S f ɛ ) p independently of ɛ. Since Minkowsky s integral inequality gives f ɛ ) f ɛ x) = f χ t f ) φ ɛ ) x), S f ɛ )x) = f ɛ ) Bx,t) f ɛ x) L 2 /t 3 ) S f ) φ ɛ )x), and so f ɛ p C S f ) p, ɛ >. For an appropriate sequence ɛ j the sequences k f ɛ j tend in the weak topology of L p to some function g k L p, k n. On the other hand, f ɛ f in L p as ɛ and thus k f ɛ k f, k n in the weak topology of distributions. Therefore k f = g k for all k and so f W,p. 3 Proof of Theorem 2 The difficult direction is ) implies 2) and this is the first we tackle. We start by showing that if f W 2,2 then S 2 f ) 2 = c f 2 ) where the square function S 2 f ) is defined in 4). To apply Plancherel in the x variable it is convenient to write the innermost integrand in 4) as ) ) f z) f y) f x) 2n dz y x 2 dy = B, t) f x + h) f x) B, t) f x + k) 2n ) ) dk h 2 dh. Applying Plancherel we get, for some dimensional constant c, ) ) h c S 2 f ) = ξ 2 exp iξh) + exp iξk) dk R n B, t) B, t) 2n dh ˆ f ξ) 2 dξ t 5. Make appropriate dilations in the integrals with respect to the variables h and k to bring the integrals on B, ). Then use that the Fourier transform of B,) χ B,) is a radial

12 function, and thus of the form F ξ ) for a certain function F defined on [, ). The result is c S 2 f ) 2 2 = Ft ξ ) + t2 ξ 2 Ft ξ ) h 2 dh 2 2n t f ˆξ) 2 dξ. 5 The change of variables τ = t ξ yields R n B,) where I is the integral I = c S 2 f ) 2 2 = I f 2 2 Fτ) + τ2 Fτ) 2n B,) h 2 dh 2 dτ τ. 2) 5 The only task left is to prove that the above integral is finite. Now, as τ, Fτ) = exp ih τ) dh Hence Fτ) + τ 2 Fτ) 2n B,) = B,) B,) = 2 = h 2 dh + ih τ ) 2 h2 τ2 + dh ) h 2 dh τ 2 + Oτ 4 ). B,) h 2 2 dh + ) h 2 dh τ 2 + Oτ 4 ) = Oτ 4 ), B,) 2n B,) because clearly B,) h 2 dh = n B,) h2 dh. Therefore the integral 2) is convergent at τ =. To deal with the case τ we recall that F can be expressed in terms of Bessel functions. Concretely, one has [G, Appendix B.5, p. 429]) B, ) Fτ) = J n/2τ) τ n/2. The asymptotic behavior of J n/2 τ) gives the inequality Fτ) C τ n+ 2 C τ, which shows that the integral 2) is convergent at. 2

13 We turn our attention to the case < p <. Let I 2 x) stand for the kernel defined on the Fourier transform side by Iˆ 2 ξ) = ξ. 2 In other words, I 2 is minus the standard fundamental solution of the Laplacian. Thus I 2 x) = c n / x n 2 if n 3, I 2 x) = log x if n = 2 and I 2π 2x) = x if n =. Given 2 any f W 2,p there exists g L p such that f = I 2 g indeed, g = f ). We claim that there exists a singular integral operator Tg) taking values in L 2 /t 5 ) such that S 2 f )x) = Tg)x) L 2 /t 5 ). 3) Set and χx) = B, ) χ B,)x) χ t x) = t n χ x t ). Then, letting M = B, ) z 2 dz, f y) f x) 2n f ) y x 2 ) dy = I 2 χ t I 2 M 2n t2 χ t ) g)x) = K t g)x), where K t x) = I 2 χ t )x) I 2 x) M 2n t2 χ t x). Setting Tg)x) = K t g)x) we get 3) from the definition of S 2 f ) in 4). Then ) translates into R n Tgx) 2 L 2 /t 5 ) dx = c f 2 2 = C g 2 2, and we conclude that T is an operator mapping isometrically L 2 R n ) into L 2 R n, L 2 /t 5 )), modulo the constant C. If the kernel K t x) of T satisfies Hörmander s condition K t x y) K t x) L 2 /t 5 ) dx C, y R n, x 2 y then a well known result of Benedek, Calderón and Panzone on vector valued singular integrals see [GR, Theorem 3.4, p. 492]) yields the L p estimate R n Tgx) p L 2 /t 5 ) dx C p g p p, which can be rewritten as S 2 f ) p C p f p. 3

14 The reverse inequality follows from polarization from ) by a well known duality argument [GR, p. 57]) and so the proof of the necessary condition is complete. We are going to prove the following stronger version of Hörmander s condition K t x y) K t x) L 2 /t 5 ) C y /2, x 2 y. x n+/2 For this we deal separately with the kernels H t x) = I 2 χ t )x) I 2 x) and t 2 χ t x). For t 2 χ t x) we first remark that the quantity χ t x y) χ t x) is non-zero only if x y < t < x or x < t < x y, in which cases takes the value /c n t n, c n = B, ). On the other hand, if x 2 y then each z in the segment joining x and x y satisfies z x /2. Assume that x y < x the case x < x y is similar). Then t 2 χ t x y) χ t x))) 2 t 5 ) 2 = c x = c x y ) 2 t 2n+ x y 2n x 2n ) 2 C y /2 x n+/2. We check now that H t satisfies the stronger form of Hörmander s condition. If t < x /2, then the origin does not belong to the ball Bx y, t) nor to the ball. Since I 2 is harmonic off the origin, the mean of I 2 on these balls is the value of I 2 at the center. Therefore H t x y) H t x) = in this case. If t x /2, then H t x y) H t x) y sup H t z) C y z [x y,x] x. n The last inequality follows from H t z) = Bz, t) I 2 z) C / z n C / x n and I 2 w) dw Therefore Bz, t) I 2 w) dw I 2 z), Bz, t) H t x y) H t x) 2 ) 2 y C t 5 x n w C n z. n x /2 ) 2 y = C t 5 x. n+ We turn now to prove that condition 2) in Theorem 2 is sufficient for f W 2,p. Let f and g in L p satisfy S 2 f, g) L p. Take an infinitely differentiable function φ with compact support in B, ), φ = and set φ ɛ x) = φ x ), ɛ >. Consider the ɛ n ɛ 4

15 regularized functions f ɛ = f φ ɛ and g ɛ = g φ ɛ. Then f ɛ is infinitely differentiable and f ɛ p f p φ ɛ, so that f ɛ W 2,p. Recalling that M = B,) z 2 dz, we get, by Minkowsky s integral inequality, S 2 f ɛ, g ɛ )x) = f ɛ χ t )x) f ɛ x) g ɛ χ t )x) M 2 t 2 L 2 /t 5 ) = f χ t ) f g χ t ) M 2 t 2 φ ɛ ) x) L 2 /t 5 ) S 2 f, g) φ ɛ ) x). Now we want to compare /2n) f ɛ and g ɛ. Define Then D ɛ x) = M 2 2n f ɛ χ t )x) g ɛ χ t )x) 2 ) /2. t D ɛ x) S 2 f ɛ )x) + S 2 f ɛ, g ɛ )x) S 2 f ɛ )x) + S 2 f, g)x) φ ɛ ) x), and thus D ɛ x) is an L p function. In particular D ɛ x) <, for almost all x R n. Hence /2n) f ɛ x) g ɛ x) = lim t /2n) f ɛ χ t )x) g ɛ χ t )x) =, for almost all x R n, and so /2n) f ɛ g in L p as ɛ. Since f ɛ f in L p as ɛ, then f ɛ f in the weak topology of distributions. Therefore /2n) f = g and the proof is complete. 4 Proof of Theorem 3 The difficult direction in Theorem 3 is to show that condition 2) is necessary for f W α,p. The proof follows the pattern already described in the preceding sections. One introduces an operator T taking values in L 2 /t 2α+ ) and shows via a Fourier transform estimate that T sends L 2 R n ) into L 2 R n, L 2 /t 2α+ )) isometrically modulo a multiplicative constant). The second step consists in showing that its kernel satisfies Hörmander s condition, after which one appeals to a well known result of Benedek, Calderón and Panzone on vector valued singular integrals to finish the proof. 4. The fundamental solution of ) α/2 Let I α be the fundamental solution of ) α/2, that is, I α is a function such that ˆ I α ξ) = ξ α and is normalized prescribing some behavior at. It is crucial for our proof to have an explicit expression for I α. The result is as follows see [ACL] or [MOPV, p. 3699]). If α is not integer then I α x) = c α,n x α n, x R n, 4) 5

16 for some constant c α,n depending only on α and n. The same formula works if α is an even integer and the dimension is odd or if α is an odd integer and the dimension is even. The remaining cases, that is, α and n are even integers or α and n are odd integers are special in some cases. If α < n formula 4) still holds, but if α is of the form n + 2N, for some non-negative integer N, then I α x) = c α,n x α n A + B log x ), x R n, where c α,n, A and B are constants depending on α and n, and B. Thus in this cases and only in this cases) a logarithmic factor is present. For instance, if α = n, then I α x) = B log x. If n = and α = 2, then I 2 x) = /2) x and there is no logarithmic factor. 4.2 The case p = 2 Given a positive real number α let N be the unique integer satisfying 2N α < 2N + 2. Define the square function associated with α by S α f ) 2 x) = ρ N y, x) dy 2 t α t, x Rn, 5) where ρ N y, x) is f y) f x) 2n f x) y x 2 L N N f x) y x 2N ) L N N f ) y x 2N if α = 2N, and f y) f x) 2n f x) y x 2 L N N f x) y x 2N ) L N N f x) y x 2N if 2N < α < 2N + 2. Recall that = j x 2 j ) and that the role which the play in Taylor expansions was discussed just before the statement of Theorem 3 in the introduction. In this subsection we prove that S α f ) 2 = c ) α/2 f ) 2. 6) We first consider the case α = 2N and then we indicate how to proceed in the simpler) case 2N < α < 2N + 2. Our plan is to integrate in x in 5), interchange the integration in x and t and then apply Plancherel in x. Before we remark that by making the change of variables y = x + th we transform integrals on in integrals on B, ) and we get ρ N y, x) dy = B, ) B, ) f x + th) dh N j= j f x) N f x + th) dh t 2N 6 t 2 j B, ) B, ) h 2N dh. h 2 j dh

17 Now apply Plancherel in x, as explained before, and make the change of variables τ = t ξ, where ξ is the variable in the frequency side. We obtain where I = S α f ) 2 2 = c I )α/2 f 2 2, 7) N Fτ) ) j τ 2 j M j j= ) N τ 2N M N L N Fτ) 2 dτ τ 2α+. Here F is a function defined on [, ) such that F ξ ) gives the Fourier transform of the radial function χ B,) B,) at the point ξ, and we have introduced the notation M j = B, h 2 j dh. We have to show that the integral I is finite. Using the series expansion of the exponential we see that, as τ, Fτ) = exp ih τ) dh B,) = + + ) N τ 2N 2N)! B,) h 2N dh + Oτ 2N+2 ). We need to compare B,) h2n dh with B,) h 2N dh. The linear functionals P 2 j P) and P P, defined on the space H B,) 2 j of homogeneous polynomials of degree 2 j, have the same kernel. This follows from the discussion before the statement of Theorem 3 in the introduction. Therefore, for some constant c, 2 j P) = c P, P H 2 j. B,) Taking Px) = x 2 j we get = c B,) x 2 j dx, and taking Px) = x 2 j 2 j)! = c B,) x2 j dx. Hence 2 j)! B,) x 2 j dx = B,) x 2 j dx = M j, we get and thus, owing to the definition of I and the fact that Fτ) = + Oτ 2 ), as τ, I = Oτ 22N+2) ) dτ, as τ. τ2α+ Then I is convergent at because α < 2N + 2 indeed, now α = 2N, but this part of the argument works for the full range 2N α < 2N + 2.) We turn to the case τ. Notice that the only difficulty is the last term in the integrand of I, because τ 4 j dτ <, j N, τ2α+ 7

18 provided 2N α. To deal with the term τ 2N Fτ) 2 dτ τ 2α+ 8) we only need to recall that F can be expressed in terms of Bessel functions. Concretely, one has [G, Appendix B.5, p. 429]) B, ) Fτ) = J n/2τ) τ n/2. The asymptotic behaviour of J n/2 τ) gives the inequality, as τ, Fτ) C τ n+ 2 C τ, which shows that the integral 8) is finite provided 2N < α, which is the case because α = 2N. Observe that the argument works for 2N < α < 2N + 2 provided ρ N y, x) is defined replacing N f x) by N f ). In the case 2N < α < 2N + 2 we argue similarly. After applying Plancherel we obtain 7) where now I is I = N Fτ) ) j τ 2 j M j j= 2 dτ τ 2α+. The proof that I is finite as τ is exactly as before. As τ the situation now is simpler because the worst term in I is the last one, namely, which is finite because 2N < α. τ 4N dτ τ, 2α+ 4.3 A vector valued operator and its kernel Given f W α,p, there exists a function g L p such that f = I α g. Indeed, g = ) α/2 f ). Then ρ N y, x) dy = K t g)x), where the kernel K t x) is K t x) = I αy) if α = 2N, and N j= K t x) = j I α x) y x 2 j I αy) j= N I α ) y x 2N L N dy 9) N j I α x) y x 2 j dy 2) 8

19 if 2N < α < 2N + 2. Hence the square function associated with the smoothness index α is S α f ) 2 x) = K t g)x) 2 Define an operator T acting on functions f L 2 R n ) by Tgx) = K t g)x), x R n. t 2α+, x Rn. The identity 6) in subsection 4.2 says that T takes values in L 2 R n, L 2 /t 2α+ )) and, more precisely, that R n Tgx) 2 L 2 /t 2α+ ) dx = S α f ) 2 2 = c g 2 2. Therefore T is an operator mapping isometrically modulo a multiplicative constant) L 2 R n ) into L 2 R n, L 2 /t 2α+ )) and we have an explicit expression for its kernel. If we can prove that K t x) satisfies Hörmander s condition K t x y) K t x) L 2 /t 2α+ ) dx C, y R n, x 2 y then the proof is finished by appealing to a well known result of Benedek, Calderón and Panzone [GR, Theorem 3.4, p. 492]; see also [GR, p. 57]). In fact, we will show the following stronger version of Hörmander s condition K t x y) K t x) L 2 /t 2α+ ) C y γ, x 2 y, 2) x n+γ for some γ > depending on α and n. The proof of 2) is lengthy. In the next subsection we will consider the case of small increments in t, namely t < x / Hörmander s condition: t < x /3 We distinguish two cases: 2N < α < 2N + 2 and α = 2N. Assume first that 2N < α < 2N + 2. Set χx) = B, ) χ B,)x) and χ t x) = t χ x n t ). To compute the gradient of K t we remark that K t x) = I α χ t )x) N j= M j t 2 j j I α x), 9

20 where M j = B,) z 2 j dz. Thus K t x) = I αy) N j I α )x) y x 2 j dy. j= Let P m F, x) stand for the Taylor polynomial of degree m of the function F around the point x. Therefore K t x) = I α y) P 2N+ I α, x)y)) dy, because the terms which have been added have zero integral on the ball, either because they are Taylor homogeneous polynomials of I α of odd degree or because they are the zero integral part of a Taylor homogeneous polynomial of I α of even degree see the discussion before the statement of Theorem 3 in the introduction). Given x and y such that x 2 y, apply the formula above to estimate K t z) for z in the segment from x y to y. The standard estimate for the Taylor remainder gives K t z) t 2N+2 sup 2N+3 I α w). w Bz, t) Notice that if z [x y, y], w Bz, t) and t x /3, then w x /6. Now, one has to observe that 2N+3 I α w) C w α n 2N 3, owing to the fact that logarithmic factors do not appear because the exponent α n 2N 3 < n is negative. By the mean value Theorem we then get Since we obtain K t x y) K t x) y sup K t z) C y t 2N+2 x α n 2N 3. z [x y,y] x /3 x /3 t 22N+2) ) /2 = C x 2N+2 α, t 2α+ ) /2 K t x y) K t x) 2 C y t 2α+ x, n+ which is the stronger form of Hörmander s condition 2) with γ= in the domain t < x /3. Let us consider now the case α = 2N. Since N I 2N is the Dirac delta at, N I 2N x) =. Hence K t x) = K ) t x) K 2) t x), where K ) t is given by 2) with α replaced by 2N and K 2) t x) = N I 2N ) y x 2N dy = M N t 2N N I 2N ). L N L N The kernel K ) t K 2) t is estimated exactly as in the first case by just setting α = 2N. The kernel requires a different argument. 2

21 Since N I 2N is the Dirac delta at the origin, K 2) t is a constant multiple of t 2N χ t. We show now that this kernel satisfies the strong form of Hörmander s condition. The quantity χ t x y) χ t x) is non-zero only if x y < t < x or x < t < x y, in which cases takes the value /c n t n, c n = B, ). On the other hand, if x 2 y then each z in the segment joining x and x y satisfies z x /2. Assume that x y < x the case x < x y is similar). Then ) t 2N χ t x y) χ t x))) 2 2 x = C t 4N+ which is 2) with γ = /2. = C 4.5 Hörmander s condition: t x /3 x y ) 2 t 2n+ x y ) 2 y /2 C 2n x 2n x, n+/2 We distinguish three cases: α < n +, α = n + and α > n +. If α < n +, then all terms in the expressions 9) and 2) defining K t satisfy Hörmander s condition in the domain t x /3. If α = 2N the last term in 9) is of the form L N N I α ) y x 2N dy = M N L N t 2N N I α ) = C t 2N χ t x), which has been dealt with in the previous subsection. Let us consider the terms of the form t 2 j j I α x), j. One has the gradient estimate t 2 j j I α x) C t 2 j x α n 2 j, 22) because no logarithmic factors appear, the reason being that the exponent α n 2 j α n + ) is negative. Since ) /2 22 j) = C x 2 j α, t x /3 t 2α+ we get Hörmander s condition with γ = in the domain t x /3. It remains to take care of the first term I αy) dy in 9). We have the following obvious estimate for its gradient I α y) dy Bx, C y α n dy. t) Notice that there are no logarithmic factors precisely because α < n +. The integrand in the last integral is locally integrable if and only if α >. Assume for the moment that < α < n +. Then I α y) dy C t n+α. 2

22 Since t 2α n ) x /3 t 2α+ ) /2 = C x n, we get Hörmander s condition with γ = in the domain t x /3. The case α = has been treated in section, so we can assume that < α <. We need the following well known and easily proved inequality Lemma. Let E be a measurable subset of R n and < β < n. Then dz z C n β E β/n, where E is the Lebesgue measure of E. E Denoting by D the symmetric difference between and Bx y, t), we obtain, by the Lemma, I α y) dy I α y) dy C t n dy y n α Since Bx y, t) t 2α n α/n) x /3 t 2α+ D C t n t n y ) α/n = C t α n α/n y α/n. ) /2 = C x n α/n, we get Hörmander s condition with γ = α/n in the domain t x /3. We tackle now the case α = n +. Since α and n are integers with different parity no logarithmic factor will appear in I α. Thus I α x) = C x. The proof above shows that the terms t 2 j j I α x) appearing in the expression 9) of the kernel K t still satisfy Hörmander s condition for j. The remaining term is I α y) I α x)) dy and its gradient is estimated by remarking that the function x satisfies a Lipschitz condition. We obtain I α y) I α x)) dy C. But clearly x /3 ) /2 = C x α = C x n, t 2α+ which completes the argument. We turn our attention to the case α > n +. If α = 2N the part of K t which has to be estimated is N H t x) = I αy) j I α x) y x 2 j dy. j= 22

23 Since 2N I α is the Dirac delta at the origin and x we have 2N I α x) = and so H t x) coincides with the expression 2) for K t x) in the case 2N < α < 2N + 2. We are then going to deal with this kernel in the full range 2N α < 2N + 2. Let M be the unique positive integer M such that < α n 2M. We split K t into two terms according to M, that is, K t = K ) t K 2) t, where and K ) t x) = K 2) t x) = I αy) j=m M j= j I α x) y x 2 j dy N j I α x) y x 2 j dy. The estimate of each of the terms in K 2) t is performed as we did for the case α < n + if α = 2N, then j N ). The gradient estimate is exactly 22). Now no logarithmic factors appear because the exponent satisfies α n 2 j α n 2M. The rest is as before. To estimate K ) t we distinguish three cases: < α n 2M <, < α n 2M and α n 2M =. In the first case we write the gradient of K ) as K ) t x) = = I αy) M j= j I α )x) y x 2 j dy I α y) P 2M 2 I α, x)y) ) dy, where P 2M 2 is the Taylor polynomial of degree 2M 2 of I α around the point x. As before, the added terms have zero integral on either because they are homogeneous Taylor polynomials of odd degree or the zero integral part of homogeneous Taylor polynomials of even degree. Now fix y in but not in the half line issuing from x and passing through the origin. Define a function g on the interval [, ] by gτ) = I α x + τy x)) P 2M 2 I α, x)x + τy x)), τ. Then g C [, ] because the segment with endpoints x and y omits the origin. Since g j) ) =, j 2M 2, where g j) stands for the derivative of g of order j, we have I α y) P 2M 2 I α, x)y) = g) = 2M 2 j= g j) ) j! τ) 2M 2 2M 2)! g2m ) τ) dτ, by the integral form of Taylor s remainder. The obvious estimate for the derivative of g of order 2M is g 2M ) τ) 2M I α x + τy x)) y x 2M C 23 t t 2M x + τy x) n α 2M).

24 Since we are in the first case, α is not integer and thus no logarithmic factor exists. Moreover < n α 2M) <, which implies that and that / z n α 2M) is locally integrable in any dimension. Therefore ) K ) t x) C t 2M n dy dτ, x + τy x) n α 2M) Since = C t 2M n C t 2M n t τ) = t α n Bx,t) Bx,t τ ) α 2M dτ τ n dz ) dτ z n α 2M) τ n dτ τ n α 2M) = C tα n. t 2α n ) x /3 t 2α+ ) /2 = C x n, we get Hörmander s condition with γ = in the domain t x /3. Let us consider the second case: < α n 2M. This time we express the gradient of K ) t by means of a Taylor polynomial of degree 2M : M K ) t x) = I αy) j I α )x) y x 2 j dy = j= I α y) P 2M I α, x)y) ) dy. Using again the integral form of the Taylor remainder of the function g, with P 2M 2 replaced by P 2M, we obtain ) K ) t x) C t 2M n dy dτ, Bx,t) x + τy x) n α 2M ) = C t 2M n dz ) dτ z n α 2M ) τ n C t 2M n t τ) = t α n Bx,t τ ) α 2M dτ τ n dτ τ n α 2M ) = C tα n, from which we get the desired estimate as before. We turn now to the last case left, α = n + 2M, with M a positive integer. In this case I α x) = C x 2M A + B log x ), x R n, B, 24

25 where A, B and C are constants depending on n and M. We also have and M I α x) = C x 2 A + B log x ), x R n M I α x) = C x A 2 + B 2 log x ), x R n. In particular M I α is in the Zygmund class on R n. We have M 2 K ) t x) = I αy) j I α )x) y x 2 j M I α )x) y x 2M 2 L M dy = j= I α y) P 2M 3 I α, x)y) L M M I α )x) y x 2M 2 Introduce the function g as above, with P 2M 2 replaced by P 2M 3, so that Now I α y) P 2M 3 I α, x)y) = g) g 2M 2) τ) 2M 2)! = = β =2M 2 β =2M 2 + β =2M 2 = 2M 3 j= g j) ) j! 2M 2) τ) 2M 3 g2m 2) τ) 2M 2)! dτ. ) β I α x + τy x)) y x) β β! β I α x + τy x)) β I α x) ) y x) β ) β I α x) y x) β. β! β! ) dy. The last term in the preceding equation is the homogeneous Taylor polynomial of degree 2M 2 of the vector I α around the point x. It is then equal to a homogeneous polynomial of the same degree with zero integral on plus L M M I α )x) y x 2M 2 by the discussion before the statement of Theorem 3 in the introduction). Hence ) β I α x) y x) β M I α )x) y x 2M 2 β! L M dy =, Bx,t ) β =2M 2 and therefore, remarking that 2M 2) τ)2m 3 dτ =, K ) t x) = 2M 2) τ) 2M 3 β =2M 2 β I α x+τy x)) β I α x) ) y x) β β! dτ dy. 25

26 Thus K ) t x) C β =2M 2 β I α x + τy x)) β I α x) ) y x) β dy β! dτ. Making the change of variables h = τy x) the integral in y above becomes J = τ β β I α x + h) β I α x) ) h β β! dh, B, t τ) which is invariant under the change of variables h = h, because β is even. Hence 2J = τ β β I α x + h) + β I α x h) 2 β I α x) ) h β β! dh. B, t τ) Now we claim that β I α is in the Zygmund class for β = 2M 2. This follows from the fact that the Zygmund class in invariant under homogeneous smooth Calderón -Zygmund operators, M is an elliptic operator and M I α is in the Zygmund class. Hence J C τ β h + β dh C t 2M τ. Thus Since B, t τ) K ) t x) C t 2M. t 22M ) x /3 t 2α+ ) /2 = C x n, we get Hörmander s condition with γ = in the domain t x / The sufficient condition In this section we prove that condition 2) in Theorem 3 is sufficient for f W α,p. Let f, g,..., g N L p satisfy S α f, g,..., g N ) L p. Take an infinitely differentiable function φ with compact support in B, ), φ = and set φ ɛ x) = φ x ), ɛ >. Consider the regularized functions f ɛ = f φ ɛ, g j,ɛ = g j φ ɛ, j N. We want to show first ɛ n ɛ that the infinitely differentiable function f ɛ is in W α,p. We have ) α/2 f ɛ = f ) α/2 φ ɛ. We need a lemma. Lemma 2. i) If ϕ is a function in the Schwartz class and α any positive number, then ) α/2 ϕ belongs to all L q spaces, q. ii) If f L p, p, then ) α/2 f is a tempered distribution. 26

27 Proof. Set ψ = ) α/2 ϕ. If α = 2m with m a positive integer, then ψ = ) m ϕ is in the Schwartz class and so the conclusion in i) follows. If α = 2m +, then ψ = ) /2 ) m ϕ = i n R j j ) m ϕ ), where R j are the Riesz transforms, that is, the Calderón-Zygmund operators whose Fourier multiplier is ξ j / ξ. It is clear from the formula above that ψ is infinitely differentiable on R n and so the integrability issue is only at. Since j ) m ϕ ), has zero integral, one has, as x, ψx) C x n, and so the conclusion follows. Assume now that m < α < m, for some positive integer m. Thus j= ˆψξ) = ξ α ˆϕξ) = ξ m ˆϕξ) ξ. m α If m is even, of the form m = 2M for some positive integer M, then ψ = M ϕ I m α, where I m α x) = C x m α n. Hence ψ is infinitely differentiable on R n. Since M ϕ has zero integral, ψx) C x m α n, as x. But α m + > and thus ψ is in all L q spaces. If m is odd, of the form m = 2M + for some non-negative integer M, then ψ = i n R j j M ϕ) I m α. j= Again ψ is infinitely differentiable on R n and, since R j j M ϕ) has zero integral just look at the Fourier transform and remark that it vanishes at the origin), we get ψx) C x m α n, as x, which completes the proof of i). To prove ii) take a function ϕ in the Schwartz class. Let q be the exponent conjugate to p. Define the action of ) α/2 f on the Schwartz function ϕ as f, ) α/2 ϕ. By part i) and Hölder s inequality one has ) α/2 f, ϕ = f, ) α/2 ϕ C f p ) α/2 ϕ q, which completes the proof of ii). Let us continue the proof of the sufficiency of condition 2). By the lemma ) α/2 φ ɛ is in L and so Hence f ɛ W α,p. Next, we claim that ) α/2 f ɛ p = f ) α/2 φ ɛ p f p ) α/2 φ ɛ. S α f ɛ, g,ɛ,..., g N,ɛ )x) S α f, g,..., g N ) φ ɛ )x), x R n. 23) 27

28 One has where S α f, g,..., g N )x) = R α x, t) L 2 /t 2α+ ), R α x, t) = f χ t )x) f x) N j= M j g j x) t 2 j M N g N χ t )x) t 2N if α = 2N and R α x, t) = f χ t )x) f x) N M j g j x) t 2 j if 2N < α < 2N + 2. As before we have set M j = B,) z 2 j dz. Minkowsky s integral inequality now readily yields 23). Set N D ɛ x) = j= M j j f ɛ x) g j,ɛ x)) t 2 j M N j= N f ɛ L N g N,ɛ ) χ t ) x) t 2N L 2 /t 2α+ ) if α = 2N and if 2N < α < 2N + 2. By 23) D ɛ x) = N j= M j j f ɛ x) g j,ɛ x)) t 2 j L 2 /t 2α+ ) j= D ɛ x) S α f ɛ )x) + S α f ɛ, g,ɛ,..., g N,ɛ )x) S α f ɛ )x) + S α f, g,..., g N ) φ ɛ )x), and so D ɛ L p. In particular, D ɛ x) is finite for almost all x R n. Thus N lim inf t M j ) j f ɛ x) g j,ɛ x)) t 2 j M N N f ɛ g N,ɛ ) χ t x) t 2N t L N α =, for almost all x R n, if α = 2N, and N lim inf t M j j f ɛ x) g j,ɛ x)) t 2 j t α =, j= for almost all x R n, if 2N < α < 2N + 2. It is easy to conclude that the only way this may happen is whenever j f ɛ x) = g j,ɛ x), j N, 28

29 for almost all x R n. Hence j f ɛ g j, j N, in L p as ɛ. Since f ɛ f in L p as ɛ, j f ɛ j f, j N, in the weak topology of tempered distributions. Hence j f = g j, j N. We claim now that the functions f ɛ are uniformly bounded in W α,p. Indeed, by the proof of necessity of condition 2) and by 23), ) α/2 f ɛ p S α f ɛ, f ɛ /L,..., N f ɛ /L N ) p S α f, f /L,..., N f /L N ) p <. Hence there exist a function h L p and a sequence ɛ j as j such that ) α/2 f ɛ j h as j in the weak topology of L p. On the other hand, by Lemma 2, ) α/2 f is a tempered distribution and so ) α/2 f ɛ ) α/2 f as ɛ in the weak topology of tempered distributions. Therefore ) α/2 f = h L p and the proof is complete. 5 Final remarks Let X, d, µ) be a metric measure space, that is, X is a metric space with distance d and µ is a Borel measure on X. We assume that the support of µ is X. Then, given α > and < p <, we can define the Sobolev space W α,p X) as follows. Let N be the unique integer such that 2N α < 2N + 2. Given locally integrable functions f, g,..., g N define a square function by S α f, g, g 2,..., g N )x) 2 = D where D is the diameter of X and R N y, x) is R N y, x) t α dµy) 2 t, x Rn, R N y, x) = f y) f x) g x) dy, x) 2 + g N x) dy, x) 2N ) g N ) dy, x) 2N 29

30 if α = 2N and R N y, x) = f y) f x) g x) dy, x) 2 + g N x) dy, x) 2N ) g N x) dy, x) 2N if 2N < α < 2N + 2. Here the barred integral stands for the mean with respect to µ on the indicated set, is the open ball with center x and radius t and g Bx,t) is the mean of the function g on. We say that a function f belongs to the Sobolev space W α,p X) provided f L p µ) and there exist functions g, g 2,..., g N L p µ) such that S α f, g, g 2,..., g N ) L p µ). We have seen in the previous sections that this definition yields the usual Sobolev spaces if X = R n is endowed with the Euclidean distance and µ is Lebesgue measure. One can prove with some effort that the same is true if R n is replaced by a half-space. Very likely this should also work for smoothly bounded domains, but we have not gone that far. There are many interesting questions one may ask about these new Sobolev spaces. For instance, how do they compare, for α =, with the known first order Sobolev spaces, notably those introduced by Hajlasz in [H] or the Newtonian spaces of [S]? For higher orders of smoothness one would like to compare them with those introduced by Liu, Lu and Wheeden in [LLW]. One may also wonder about their intrinsic properties, namely, about versions of the Sobolev imbedding theorem, the Poincaré inequality and so on. For the Sobolev imbedding theorem the following remark might be useful. In R n the L p space can be characterized by means of the following zero smoothness square function: S f ) 2 x) = f f 2 Bx, 2t) t, x Rn. The result is then that a locally integrable function f is in L p if and only if S f ) L p. The proof follows the pattern described several times in this paper. One first deals with the case p = 2 via a Fourier transform computation. Then one introduces a L 2 /t)-valued operator T such that T f ) L 2 R n, L 2 /t)) = c S f ) 2 and one shows that its kernel satisfies Hörmander s condition. Acknowledgements. We are grateful to Piotr Hajlasz, who kindly encouraged us to publish the paper, and to Richard Wheeden for a useful correspondence concerning [W] and [W2]. This work has been partially supported by grants 29SGR42 Generalitat de Catalunya) and MTM Spanish Ministry of Science). The first named author acknowledges the generous support received from a fellowship from the Spanish Ministry of Science and from the Ferran Sunyer Balaguer Foundation. He is thankful to Professor Hajlasz for his hospitality at Pittsburg University and for many enlightening conversations. 3

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32 Roc Alabern Departament de Matemàtiques Universitat Autònoma de Barcelona 893 Bellaterra, Barcelona, Catalonia Joan Mateu Departament de Matemàtiques Universitat Autònoma de Barcelona 893 Bellaterra, Barcelona, Catalonia Joan Verdera Departament de Matemàtiques Universitat Autònoma de Barcelona 893 Bellaterra, Barcelona, Catalonia 32