1011.0667v1

ar X iv :1 01 1. 06 67 v1 [ ma th. CA ] 2 N ov 20 10 A new characterization of Sobolev spaces on Rn Roc Alabern, Joan Mateu and Joan Verdera Abstract In this paper we present a new characterization of Sobolev spaces on Rn. Our characterizing condition is obtained via a quadratic multiscale expression which ex- ploits the particular symmetry properties of Euclidean space. An interesting feature of our condition is that depends only on the metric of Rn and the Lebesgue measure, so that one can define Sobolev spaces of any order of smoothness on any metric measure space. 1 Introduction In this paper we present a new characterization of the Sobolev spaces Wα,p on Rn, where the smoothness index α is any positive real number and 1 < p < ∞. Thus Wα,p consists of those functions f ∈ Lp = Lp(Rn) such that (−∆)α/2 f ∈ Lp. Here ∆ is the Laplacean and (−∆)α/2 f is defined on the Fourier transform side by |ξ|α ˆf (ξ). If 0 < α < n this means that f is a function in Lp which is the Riesz potential of order α of some other function g in Lp, namely f = cn1/|x|n−α ∗ g. If α is integer, then Wα,p is the usual space of those functions in Lp such that all distributional derivatives up to order α are in Lp. To convey a feeling about the nature of our condition we first discuss the case α = 1. Consider the square function S ( f )2(x) = ∫ ∞ 0 ∣∣∣∣∣ fB(x, t) − f (x)t ∣∣∣∣∣ 2 dt t , x ∈ Rn . (1) Here f is a locally integrable function on Rn and fB(x, t) denotes the mean of f on the open ball with center x and radius t. One should think of fB(x, t)− f (x)t as a quotient of increments of f at the point x. Our characterization of W1,p reads as follows. Theorem 1. If 1 < p < ∞, then the following are equivalent. (1) f ∈ W1,p (2) f ∈ Lp and S ( f ) ∈ Lp. If any of the above conditions holds then ‖S ( f )‖p ≃ ‖∇ f ‖p . 1 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−1 B ≤ A ≤ C B. Notice that condition (2) in Theorem 1 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 W1,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 1 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 L2(dt/t) such that ‖T ( f )‖L2(Rn , L2(dt/t)) = ‖S ( f )‖2 . The kernel of T turns out to satisfy Hormander’s condition, so that we can appeal to a well known result of Benedek, Caldero´n and Panzone [GR, Theorem 3.4, p. 492] on vector valued Caldero´n-Zygmund Theory to conclude the proof. The major technical difficulty occurs in checking Hormander’s condition. The proof extends without pain to cover orders of smoothness α with 0 < α < 2. The square function S ( f ) has to replaced by S α( f )2(x) = ∫ ∞ 0 ∣∣∣∣∣ fB(x,t) − f (x)tα ∣∣∣∣∣ 2 dt t , x ∈ Rn . The result is then that, for 0 < α < 2, f ∈ Wα,p is equivalent to f ∈ Lp and S α( f ) ∈ Lp. Notice that S α( f )2(x) = ∫ ∞ 0 ∣∣∣∣∣∣ ? B(x, t) f (y) − f (x) tα dy ∣∣∣∣∣∣ 2 dt t , x ∈ Rn , (2) where the barred integral on a set stands for the mean over that set. Stricharzt ([Str]) used long ago the above square function for 0 < α < 1 to characterize Wα,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 (2). In the interval 1 ≤ α < 2 putting the absolute value inside the integral destroys the characterization, because then one gives up the symmetry properties of Rn. For instance, S α( f ) vanishes if f is a first degree polynomial. There are in the literature square functions very close to (2) which characterize Wα,p, for 0 < 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 W2,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|=2 ∂β f (x)(y − x)β + R , (3) 2 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 1( f ) (see (2) for α = 1) with respect to second order derivatives. The first remark is that the mean on B(x, t) of the homogeneous polynomial of degree 1 in (3) is zero. Now, the homogeneous Taylor polinomial of degree 2 can be written as ∑ |β|=2 ∂β f (x) β! (y − x)β = H(y − x) + 1 2n ∆ f (x) |y − x|2 , (4) for a harmonic homogeneous polynomial H of degree 2. Hence the mean on B(x, t) of the homogeneous Taylor polinomial of degree 2 is ? B(x, t) 1 2n ∆ f (x) |y − x|2 dy . This suggests defining S 2( f )(x)2 = ∫ ∞ 0 ∣∣∣∣∣∣∣∣ ? B(x, t) ( f (y) − f (x) − 12n(∆ f )B(x, t) |y − x|2 ) t2 dy ∣∣∣∣∣∣∣∣ 2 dt t , x ∈ Rn . (5) We cannot replace (∆ f )B(x, t) by ∆ f (x) in the preceding definition, because the mean guar- antees a little extra smoothness which one needs in a certain Fourier transform computa- tion. Notice that, according to the remarks made before on the mean on the ball B(x, t) of the homogeneous Taylor polynomials of degrees 1 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 )B(x, t). 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 B(x, t) vanishes. This explains why whatever we can prove for α = 2 will also extend to the range 2 < α < 4 by defining S α( f )(x)2 = ∫ ∞ 0 ∣∣∣∣∣∣∣∣ ? B(x, t) ( f (y) − f (x) − 12n(∆ f )B(x, t) |y − x|2 ) tα dy ∣∣∣∣∣∣∣∣ 2 dt t , x ∈ Rn . (6) Here is our second order theorem. Theorem 2. If 1 < p < ∞, then the following are equivalent. (1) f ∈ W2,p (2) f ∈ Lp and there exists a function g ∈ Lp such that S 2( f , g) ∈ Lp, where the square function S 2( f , g) is defined by S 2( f , g)(x)2 = ∫ ∞ 0 ∣∣∣∣∣∣∣∣ ? B(x, t) ( f (y) − f (x) − gB(x, t) |y − x|2 ) t2 dy ∣∣∣∣∣∣∣∣ 2 dt t , x ∈ Rn . If f ∈ W2,p then one can take g = ∆ f /2n and if (2) holds then necessarily g = ∆ f /2n, a. e. 3 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 Rn and integrals with respect to Lebesgue measure. Thus one may define a notion of W2,p(X) on any metric measure space X. For more comments on that see section 4. Again the special symmetry properties of Rn play a key role. For instance, S 2 anni- hilates second order polynomials. Theorem 2 has a natural counterpart for smoothness indexes α satisfying 2 ≤ α < 4. The result states that a function f ∈ Wα,p if and only if f ∈ Lp and there exists a function g ∈ Lp such that S α( f , g) ∈ Lp, where S α( f , g)(x)2 = ∫ ∞ 0 ∣∣∣∣∣∣∣∣ ? B(x, t) ( f (y) − f (x) − gB(x, t) |y − x|2 ) tα dy ∣∣∣∣∣∣∣∣ 2 dt t , x ∈ Rn . We proceed now to state our main result, which covers all orders of smoothness and all p with 1 < p < ∞. Before it is convenient to discuss the analogue of (4) for homoge- neous polynomials of any even degree. Let P be a homogeneous polynomial of degree 2 j. Then P can be written as P(x) = H(x) + ∆ jP 1 L j |x |2 j , where L j = ∆ j(|x |2 j) and H satisfies ∆ jH = 0. This follows readily from [St, 3.1.2, p. 69]. Considering the spherical harmonics expansion of P(x) we see that ∫ |x|=1 H(x) dσ = 0, σ being the surface measure on the unit sphere, and thus that ∫ |x|≤t H(x) dx = 0, t > 0. The precise value of L j, which can be computed easily, will not be needed. Theorem 3. Given α > 0 choose an integer N such that 2N ≤ α < 2N + 2. If 1 < p < ∞, then the following are equivalent. (1) f ∈ Wα,p (2) f ∈ Lp and there exist functions g j ∈ Lp, 1 ≤ j ≤ N such that S α( f , g1, g2, . . . , gN)∈ Lp, where the square function S α( f , g1, g2, . . . , gN) is defined by S α( f , g1, g2, . . . , gN)(x)2 = ∫ ∞ 0 ∣∣∣∣∣∣ ? B(x, t) RN(y, x) tα dy ∣∣∣∣∣∣ 2 dt t , x ∈ Rn , and RN(y, x) is RN(y, x) = f (y) − f (x) − g1(x) |y − x|2 + · · · + gN−1(x) |y − x|2(N−1) + (gN)B(x, t)|y − x|2N . If f ∈ Wα,p then one can take g j = ∆ j f /L j and if (2) holds then necessarily g j = ∆ j f /L j, a. e. If any of the above conditions holds then ‖S α( f , g1, . . . , gN)‖p ≃ ‖(−∆)α/2 f ‖p . 4 Again condition (2) in Theorem 2 only involves the Euclidean distance on Rn and integrals with respect to Lebesgue measure. Thus one may define a notion of Wα,p(X) for any positive α and any 1 < p < ∞ on any metric measure space X. For previous notions of higher order Sobolev spaces on metric measure spaces see [LLW]. See section 4 for more on that. The proof of Theorem 3 proceeds along the lines sketched before for α = 1. First we use a Fourier transform computation to obtain the relation ‖S α( f ,∆ f /L1, . . . ,∆N f /LN)‖2 = c ‖(−∆)α/2 f ‖2 . Then we introduce a singular integral operator with values in L2(dt/t2α+1) and we check that its kernel satisfies Hormander’s condition. The paper is organized as follows. In sections 1, 2 and 3 we prove respectively The- orems 1, 2 and 3. In this way readers interested only in first order Sobolev spaces may concentrate in section 1. Those readers interested in the main idea about jumping to orders of smoothness 2 and higher may read section 2. Section 3 is reserved to those interested in the full result. In any case the technical details for the proof of Theorem 1 are somehow different of those for orders of smoothness 2 and higher. The reason is that Horman- der’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 con- sideration 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 Eu- clidean norm. The Zygmund class on Rn 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 ∈ Rn . The basic example of a function in the Zygmund class which is not Lipschitz is f (x) = |x| log |x|, x ∈ Rn. The Scharwtz class consists of those infinitely differentiable functions on Rn whose partial derivatives of any order decrease faster than any polynomial at ∞. 2 Proof of Theorem 1 The difficult part is the necessity of condition (2) and we start with this. As a first step we show that ‖S 1( f )‖2 = c ‖∇ f ‖2 (7) for a dimensional constant c. Set χ(x) = 1 |B(0, 1)| χB(0,1)(x) 5 and χt(x) = 1tnχ( x t ) , so that, by Plancherel,∫ Rn S 1( f )(x)2 dx = ∫ ∞ 0 ∫ Rn |( f ∗ χt)(x) − f (x)|2 dx dtt3 = c ∫ ∞ 0 ∫ Rn |χˆ(tξ) − 1|2 ∣∣∣ ˆf (ξ)∣∣∣2 dξ dt t3 . Since χˆ is radial, χˆ(ξ) = F(|ξ|) for a certain function F defined on [0,∞). Exchange the integration in dξ and dt in the last integral above and make the change of variables τ = t |ξ|. Then ∫ Rn S 1( f )(x)2 dx = c ∫ Rn ∫ ∞ 0 |(F(τ) − 1|2 dτ τ3 | ˆf (ξ)|2|ξ|2 dξ = c ∫ ∞ 0 |(F(τ) − 1|2 dτ τ3 ‖∇ f ‖22 and (7) is reduced to showing that∫ ∞ 0 |(F(τ) − 1|2 dτ τ3 < ∞ . (8) Set B = B(0, 1) and e1 = (1, 0, . . . , 0) ∈ Rn. Then F(t) = χˆ(te1) = ? B exp (−ıx1t) dx = ? B ( 1 − ıx1t − 1 2 x21t 2 + · · · ) dx = 1 − 1 2 ? B x21 dx t2 + · · · , which yields F(t) − 1 = O(t2), as t → 0 and shows the convergence of (8) at 0. Since F(|ξ|) = χˆ(ξ) is the Fourier transform of an integrable function, F(τ) is a bounded function and so the integral (8) is clearly convergent at ∞. We are left with the case of a general p between 1 and ∞. If f ∈ W1,p, then f = g∗1/|x|n−1 for some g ∈ Lp (with 1/|x|n−1 replaced by log |x| for n = 1). Set I(x) = 1/|x|n−1. Then fB(x, t) − f (x) = ( f ∗ χt)(x) − f (x) = (g ∗ Kt)(x) , where Kt(x) = (I ∗ χt)(x) − I(x) = ? B(x, t) I(y) dy − I(x) . (9) 6 If we let T (g)(x) = (g ∗ Kt)(x), x ∈ Rn, then one can rewrite S 1( f )(x) as S 1( f )(x) = (∫ ∞ 0 |(g ∗ Kt)(x)|2 dtt3 ) 1 2 = ‖Tg(x)‖L2(dt/t3) . Then (7) translates into ∫ Rn ‖Tg(x)‖2L2(dt/t3) dx = c ‖g‖22 , and we conclude that T is an operator mapping isometrically L2(Rn) into L2(Rn, L2(dt/t3)). If the kernel Kt(x) of T satisfies Hormander’s condition∫ |x|≥2|y| ‖Kt(x − y) − Kt(x)‖L2(dt/t3) ≤ C, y ∈ Rn then a well known result of Benedek, Caldero´n and Panzone on vector valued singular integrals (see [GR, Theorem 3.4, p. 492]) yields the Lp estimate ∫ Rn ‖Tg(x)‖pL2(dt/t3) dx ≤ Cp ‖g‖pp , which can be rewritten as ‖S 1( f )‖p ≤ Cp ‖∇ f ‖p . The reverse inequality follows from polarization from (7) by a well known argument ([GR, p. 507]) and so the proof of the necessary condition is complete. We are going to prove the following stronger version of Hormander’s condition ‖Kt(x − y) − Kt(x)‖L2(dt/t3) ≤ C |y| |x|n+1 , y ∈ Rn , (10) for almost all x satisfying |x| ≥ 2|y|. To prove (10) we deal separately with three intervals in the variable t. Interval 1: t < |x|3 . From the definition of Kt in (9) we obtain ∇Kt(x) = (∇I ∗ χt)(x) − ∇I(x) . (11) Notice that, in the distributions sense, the gradient of I is a constant times the vector valued Riesz transform, namely ∇I = −(n − 1)p.v. x |x|n+1 . If |x| ≥ 2|y|, then the segment [x − y, x] does not intersect the ball B(0, |x |/2) and thus |Kt(x − y) − Kt(x)| ≤ |y| sup z∈[x−y,y] |∇Kt(z)| . (12) 7 If t < |x|/3 and z ∈ [x − y, y], then B(z, t) ⊂ Rn \ B(0, |x |/6), and hence ∇Kt(z) = ? B(z, t) (∇I(w) − ∇I(z)) dw . (13) Taylor’s formula up to order 2 for ∇I(w) around z yields ∇I(w) = ∇I(z) + ∇2I(z)(w − z) + O( |w − z| 2 |x|n+2 ) , where ∇2I(z)(w − z) is the result of applying the matrix ∇2I(z) to the vector w − z. The mean value of ∇2I(z)(w − z) on B(z, t) is zero, by antisymmetry, and thus, by (13), |∇Kt(z)| ≤ C t 2 |x|n+2 and so, by (12) |Kt(x − y) − Kt(x)| ≤ C |y| t 2 |x|n+2 . Integrating in t we finally get (∫ |x|/3 0 |Kt(x − y) − Kt(x)|2 dtt3 ) 12 ≤ C |y| |x|n+2 (∫ |x|/3 0 t dt ) 12 = C |y| |x|n+1 . (14) Interval 2: |x |/3 < t < 2|x |. The function I ∗ χt is continuously differentiable on R n \ S t, S t = {x : |x| = t}, because its distributional gradient is given by I ∗ ∇χt and each component of ∇χt is a Radon measure supported on S t. The gradient of I ∗ χt is given at each point x ∈ Rn \ S t by the principal value integral p.v.(∇I ∗ χt)(x) = −(n − 1)p.v. ? B(x, t) y |y |n+1 , 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 ∗ χt has a singularity at such an x. We need the following estimate. Lemma 1. ∣∣∣∣∣∣p.v. ∫ B(x, t) y |y|n+1 dy ∣∣∣∣∣∣ ≤ C log |x | + t ||x | − t| , x ∈ Rn . Proof. Assume without loss of generality that x = (x1, 0, . . . , 0). The coordinates y j, j , 1, change sign under reflection around the y1 axes. Hence p.v. ∫ B(x, t) y j |y|n+1 dy = 0, 1 < j ≤ n . 8 Now, if |x | < t, ∣∣∣∣∣∣p.v. ∫ B(x, t) y1 |y|n+1 dy ∣∣∣∣∣∣ = ∣∣∣∣∣∣p.v. ∫ B(x, t)\B(0, t−|x|) y1 |y|n+1 dy ∣∣∣∣∣∣ ≤ C ∫ t+|x| t−|x| dt t = C log t + |x| t − |x| . If |x | > t, ∣∣∣∣∣∣p.v. ∫ B(x, t) y1 |y|n+1 dy ∣∣∣∣∣∣ = ∣∣∣∣∣∣ ∫ B(x, t) y1 |y|n+1 dy ∣∣∣∣∣∣ ≤ C ∫ |x|+t |x|−t dt t = C log |x| + t |x| − t . � Assume without loss of generality that y = (y1, 0, . . . , 0). The distributional gradient of I ∗ χt is −(n − 1)p.v. y |y|n+1 ∗ χt , which is in L2. Then I ∗χt ∈ W1,2 and consequently is absolutely continuous on almost all lines parallel to the first axes. Therefore Kt(x − y) − Kt(x) = − ∫ 1 0 ∇Kt(x − τy) · y dτ for almost all x and |Kt(x − y) − Kt(x)| ≤ C |y| |x|n ∫ 1 0 ( 1 + log |x − τy| + t ||x − τy| − t| ) dτ . Hence (∫ 2|x| |x|/3 |Kt(x − y) − Kt(x)|2 dtt3 ) 1 2 ≤ C |y| |x|n+1  ∫ 2|x| |x|/3 (∫ 1 0 ( 1 + log |x − τy| + t ||x − τy| − t| ) dτ )2 dt t  1 2 = C |y| |x|n+1 D , 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 D2 ≤ ∫ 1 0  ∫ 2|x| |x|/3 ( 1 + log |x − τy| + t ||x − τy| − t| )2 dt t  dτ . 9 For each τ make the change of variables s = t |x − τy| to conclude that D2 ≤ ∫ 4 2/9 ( 1 + log 1 + s |1 − s| )2 ds s . Interval 3: 2|x | ≤ t. For each z in the segment [x − y, y] we have B(0, t/4) ⊂ B(z, t). Then, by (13), ∇Kt(z) = −(n − 1) ( p.v. 1 |B(z, t)| ∫ B(z, t) w |w|n+1 dw − z |z|n+1 ) = −(n − 1) ( 1 |B(z, t)| ∫ B(z, t)\B(0, t/4) w |w|n+1 dw − z |z|n+1 ) and so |∇Kt(z)| ≤ C 1 |x|n , z ∈ [x − y, y] . Hence, owing to (12), |Kt(x − y) − Kt(x)| ≤ C |y| |x|n and thus (∫ ∞ 2|x| |Kt(x − y) − Kt(x)|2 dtt3 ) 1 2 ≤ C |y| |x|n (∫ ∞ 2|x| dt t3 ) 1 2 = C |y| |x|n+1 , which completes the proof of the strengthened form of Hormander’s condition (10). We turn now to prove that condition (2) in Theorem 1 is sufficient for f ∈ W1,p. Let f ∈ Lp satisfy S 1( f ) ∈ Lp. Take an infinitely differentiable function φ ≥ 0 with compact support in B(0, 1), ∫ φ = 1 and set φǫ(x) = 1ǫnφ( xǫ ), ǫ > 0. Consider the regularized functions fǫ = f ∗ φǫ . Then fǫ is infinitely differentiable and ‖∇ fǫ‖p ≤ ‖ f ‖p‖∇φǫ‖1, so that fǫ ∈ W1,p. Thus, as we have shown before, ‖∇ fǫ‖ ≃ ‖S 1( fǫ)‖p . We want now to estimate ‖S 1( fǫ)‖p independently of ǫ. Since ( fǫ)B(x, t) − fǫ(x) = (( f ∗ χt − f ) ∗ φǫ) (x) , Minkowsky’s integral inequality gives S 1( fǫ)(x) = ‖( fǫ)B(x,t) − fǫ(x)‖L2(dt/t3) ≤ (S 1( f ) ∗ φǫ)(x) , and so ‖∇ fǫ‖ ≤ C ‖S 1( f )‖p, ǫ > 0. For an appropriate sequence ǫ j → 0 the sequences ∂k fǫ j tend in the weak ⋆ topology of Lp to some function gk ∈ Lp, 1 ≤ k ≤ n. On the other hand, fǫ → f in Lp as ǫ → 0 and thus ∂k fǫ → ∂k f , 1 ≤ k ≤ n in the weak topology of distributions. Therefore ∂k f = gk for all k and so f ∈ W1,p. 10 3 Proof of Theorem 2 The difficult direction is (1) implies (2) and this is the first we tackle. We start by showing that if f ∈ W2,2 then ‖S 2( f )‖2 = c ‖∆ f ‖2 (15) where the square function S 2( f ) is defined in (5). To apply Plancherel in the x variable it is convenient to write the innermost integrand in (5) as ? B(x, t) ( f (y) − f (x) − (? B(x, t) ∆ f (z) 2n dz ) |y − x|2 ) dy = ? B(0, t) ( f (x + h) − f (x) − (? B(0, t) ∆ f (x + k) 2n dk ) |h|2 ) dh . Applying Plancherel we get, for some dimensional constant c, c ‖S 2( f )‖22= ∫ ∞ 0 ∫ Rn ? B(0, t) ( exp (ıξh) − 1 + (? B(0, t) exp (ıξk) dk ) |h|2|ξ|2 2n ) dh | ˆf (ξ)|2 dξ dt t5 . Make appropriate dilations in the integrals with respect to the variables h and k to bring the integrals on B(0, 1). Then use that the Fourier transform of 1 |B(0,1)|χB(0,1) is a radial function, and thus of the form F(|ξ|) for a certain function F defined on [0,∞). The result is c ‖S 2( f )‖22 = ∫ Rn ∫ ∞ 0 ∣∣∣∣∣∣F(t |ξ|) − 1 + t2|ξ|2F(t |ξ|) 1 2n ? B(0,1) |h|2 dh ∣∣∣∣∣∣ 2 dt t5 | ˆf (ξ)|2 dξ . The change of variables τ = t |ξ| yields c ‖S 2( f )‖22 = I ‖∆ f ‖22 where I is the integral I = ∫ ∞ 0 ∣∣∣∣∣∣F(τ) − 1 + τ2F(τ) 1 2n ? B(0,1) |h|2 dh ∣∣∣∣∣∣ 2 dτ τ5 . (16) The only task left is to prove that the above integral is finite. Now, as τ →