Singular Integrals. 1 Calderon-Zygmund decomposition

Transcription

1 Singular Integrals Analysis III Calderon-Zygmund decomposition Let f be an integrable function f dx <. We will decompose the R n function at height α > 0, f = g + b with g Cα almost everywhere, with b dx C f R n L (R n ), additional properties. We denote by A the Lebesgue measure of A. Theorem Let f L (R n ), let α > 0. There exists a countable collection of cubes with sides parallel to the axes, Q j with disjoint interiors, such that, for each j, α < f dx 2 n α Q j Q j Consider Ω = j Q j F = R n \ Ω. Then, Moreover, Ω α f L (R n ). f(x) α holds almost everywhere for x F. There exists a decomposition f(x) = g(x) + b(x) such that g(x) 2 n α almost everywhere, moreover, for every p, b(x) = 0 for x F, for each Q j, g L p (R n ) α p p ( + 2 np ) p f p L (R n ) Q j b(x)dx = 0 b L (R n ) ( + 2 n ) f L (R n ).

2 Proof. We take cubes Q with sides parallel to the axes. We start by using all such cubes with equal side size large enough so that f L (R n ) α Q. We take each such cube divide it into 2 n daughters, Q with sides parallel to the axes. We ask whether f dx α? Q Q If the answer is yes, we continue subdividing. If the answer is no, we retain the cube as one of the Q j -s. Clearly, by construction, the fathers Q of the retained daughters had f dx α therefore Q Q α < f dx 2 n α Q j Q j holds because Q j Q Q = 2 n Q j. The interiors of cubes Q j are mutually disjoint, by induction. We consider the set Ω = j Q j. We have Q j < α Q j f dx therefore we obtain Ω α f L (R n ) just by summing. The Lebesgue differentiation theorem can be stated as follows: f(x) = lim f(y)dy Q x Q holds x-almost everywhere, where Q is the family of cubes that contain x Q x means that we take their diameters to converge to zero. Note that the cubes are not required to be centered at x, but a simple argument shows that M cubes (f)(x) = sup f dx x Q Q obeys M cubes (f)(x) cm(f)(x), with c depening on dimension only, M(f) the usual maximal function. If x F is not in the exceptional set of measure zero, then the limit of averages of integrals on cubes exists equals f(x), there exists a subsequence of cubes Q containing x whose diameters converge to zero which were not retained as Q j -s, so f(x) lim sup Q x f dx α. We define Q Q { g(x) = Q f(x), if x F Q j Q j f(x)dx, if x Int(Q j ). 2 Q

3 This defines g almost everywhere. Also, by construction almost everywhere. Now Ω F g(x) 2 n α g p dx α p f L (R n ) g p dx α p 2 np Ω α p 2 np f L (R n ). Clearly b = f g is defined almost everywhere vanishes on F satisfies Q j b(x)dx = 0. Finally, because b f + g, we have b dx f L (R n ) + 2 n α Ω ( + 2 n ) f L (R n ) Ω this concludes the proof. 2 Marcinkiewicz Interpolation Theorem We recall that a sublinear operator is said to be weak type (p,q) if ( ) q C f L {x T f(x) > α} p (R n ) α holds for all α > 0 with C independent of α f. This definition applies only to q <. For q = the notion is replaced by the strong notion of boundedness. Note that if T f L q (R n ) C f L p (R n ) the inequality above follows by Chebyshev s inequality. We recall that if p < p < p 2 then L p (R n ) L p (R n ) + L p 2 (R n ) Indeed, for any f L p (R n ) any γ > 0 we can write f = f γ + f γ with f γ (x) = { f(x), if f(x) > γ 0 if f(x) γ 3

4 Then clearly { 0, if f(x) > γ f γ (x) = f(x) if f(x) γ f γ p L p (R n ) γp p f p L p (R n ) f γ p 2 L p 2 (R n ) γp 2 p f p L p (R n ) Theorem 2 Let < r. Assume that T is subadditive weak type (,) weak type (r,r). Then for every < p < r there exists a constant C p such that T f L p (R n ) C p f L p (R n ) Proof. We take first r <. We take α > 0 denote λ(α) = {x T f(x) > α} We decompose f = f γ + f γ as above but with γ = α. Then, because T f(x) > α implies at least one one of the inequalities T f α (x) > α or 2 T f α (x) > α, we deduce from the assumptions 2 λ(α) C fα dx + C r α 2α r f α r dx C α f dx + f >α Cr 2α r f α f r dx We multiply by pα p integrate. For the first term we use: ( ) α p α f dx dα = 0 f >α ( ) f(x) f(x) α p 2 dα dx = f(x) p dx R n 0 p R n For the second term we use ( ) α p α r 0 f α f(x) r dx dα = ( ) f(x) r dx R n = r p f(x) αp r dα R n f(x) p dx This concludes the proof for r <. If r = we know that T f α is essentially bounded, T f α (x) C 2 f α L (R n ) C 2 α holds almost everywhere. Therefore the measure of the set where T f(x) > 2C 2 α is not larger than the measure of the set where T f α (x) > C 2 α we continue by estimating the distribution function of T f as above. 4

5 3 Singular Integrals We consider kernels K L loc (Rn \ {0}) satisfying the following properties: There exists a constant B such that holds for all 0 x R n, x >2 y K(x) B x n () K(x y) K(x) dx B (2) holds for any y 0, r < x <r 2 K(x)dx = 0 (3) holds for any 0 < r < r 2 <. The aim is to prove that convolution integrals with these kernels are bounded in L p, < p <. We start by making a few general observations. For > 0 let { K(x) if x, (C (K))(x) = 0 if x let for f L p (R n ) ( x ) (τ K)(x) = n K (δ f)(x) = f(x). We denote by T K the convolution operator T K (f)(x) = K(x y)f(y)dy. R n This operator is well defined if, for instance, f L (R n ) K L p (R n ). This latter property is not true in general if all we know about K is (-3). Part of the theorem will be to make sense of the operators. Let us observe that Proposition i. If K has properties (-3) then τ K has the same properties with the same constant B, uniformly, for all > 0. 5

6 ii. For any > 0 C (K) = τ (C (τ (K))) iii. If K has properties (-3) then C (K) has the same properties (-3), with a constant B > 0 depending on B dimension of space only. iv. If K L p (R n ) f L (R n ) then ( ) T K δ f = T τkf δ v. If T : L p (R n ) L p (R n ) is a bounded linear operator then the family T δ is uniformly bounded, i.e. δ sup >0 δ T δ L(L p (R n ),L p (R n )) C vi. If f(ξ) = Ff(ξ) = e ix ξ f(x)dx R n denotes the Fourier transform, then. F(C (K))(ξ) = F(C (τ K))(ξ) Proof. The proofs of i, ii, iv, v vi are direct consequences of the definitions are left as an exercise. The proof of iii uses the following important observation: if K is a function obeying () if λ >, ρ > 0 then the integral K(x) ω n B log λ (4) ρ x λρ is bounded uniformly, with ω n the area of the unit sphere S n in R n, independently of ρ. This is used to bound K(x y) dx ω n B log 3 2 x >2 y, x y >, x < x >2 y, x y <, x > K(x) dx ω n B log 2 It follows that if K obeys (-3), then C (K) obeys the same with B = ( + ω n log 3)B instead of B. 6

7 Lemma Let K obey (-3). There exists a constant γ depending on dimension of space only so that sup >0 sup FC K(ξ) γb ξ R n Proof. Clearly, by points vi i in Proposition above, it is enough to prove that K (ξ) γb holds for all ξ, where K = C (K) K satisfies (-3). Now clearly, K L 2 (R n ) (by Proposition above, C (K) L 2 (R n )) K (ξ) = lim e ix ξ K (x)dx. R x R We fix ξ 0 study separately the integrals for x 2π for larger x. I (ξ) = e ix ξ K x 2π (x)dx = ( x e ix ξ ) K 2π (x)dx x B x n dx = 2πω x 2π n B. We consider now the integrals I 2 (ξ) = e ix ξ K (x)dx 2π x R for R > 3π show they are uniformly bounded. In order to do so we choose η = π ξ satisfying e iξ η =. Note that η = π. We translate by 2 η: I 2 (ξ) = e ix ξ 2π (K x R (x) K (x η) + K (x η)) dx = e ix ξ 2π (K x R (x) K (x η)) dx + e ix ξ 2π K x R (x η)dx The last piece is e i((x η) ξ+η ξ) 2π K x R (x η)dx = e ix ξ 2π K x +η R (x )dx = I 2 (ξ) + E(ξ) 7

8 where with A = B = E(ξ) = A e ix ξ K (x)dx { x 2π } { x R \ x 2π { x 2π B e ix ξ K (x)dx } x + η R } { x + η R \ x 2π } x R We have thus 2I 2 (ξ) = e ix ξ (K (x) K (x η)) dx + E(ξ) 2π x R Now A B { x 2π { x π because R > 3π. Therefore } { 3π x x R π } x R } { 2π x x R x R + π } A B K dx cω n B with c > an absolute constant follows from () via (4). Thus E(ξ) cω n B. On the other h, because x 2π = 2 η, K (x) K (x η) dx ( + ω n log 3)B 2π x R thus I 2 (ξ) 2 [cω nb + ( + ω n log 3)B] concludes the proof of the lemma. 8

9 Theorem 3 Let K L 2 (R n ) satisfy (-3) sup K(ξ) γb ξ R n for some γ > 0. Then for each < p <, the operator f T K (f) = K(x y)f(y)dy R n defined for f L (R n ) L p (R n ) has a unique bounded extension T K : L p (R n ) L p (R n ) T K L(L p (R n ),L p (R n )) C holds with C = C(n, p, B) a uniform constant. Proof. Because of the uniform bound on the Fourier transform, we have immediately the result for p = 2. We will prove that T K is weak type (, ), conclude from the Marcienkewicz theorem that T K is bounded in L p (R n ) for < p < 2. Then we will use duality to deduce the theorem for p > 2. Let α > 0. Let f L (R n ) We consider the Calderon-Zygmund decomposition at height α. We have T K (f) = T K (g) + T K (b) Therefore, {x T K (f)(x) > α} {x T K (g)(x) > α 2 } + {x TK (b)(x) > α 2 } We will estimate the two pieces separately. We know from Theorem that g 2 L 2 (R n ) ( + 22n )α f L (R n ) therefore {x TK (g)(x) > α 2 } 4 α 2 T K (g) 2 L 2 (R n ) cb 2 α 2 g 2 L 2 (R n ) cb2 α f L (R n ) It remains to bound the term involving b. In order to do so, we will encase the cubes Q j of the Calderon-Zygmund decompostion in concentric, larger cubes Q j of diameters 2 n times larger. We consider Ω = j Q j 9

10 F = R n \ Ω. We note that Ω j Q j γ j Q j = Ω γα f L (R n ) holds with a constant γ depending only on n. Note also that, if we denote y j the common center of Q j Q j, then if x Q j, then x y j 2 y y j holds for all y Q j, in other words, the distance from x to the center of Q j is larger than twice the radius of the concentric ball circumscribing Q j. We write b(x) = b j (x) where b j (x) = b(x) if x Q j, b j (x) = 0 otherwise. Because the interiors of Q j are mutually disjoint because b(x) = 0 on F, for almost all x the sum reduces to one term only. Note that T K (b j )(x) = K(x y)b(y)dy = Q j (K(x y) K(x y j ))b(y)dy Q j because Q j b(y)dy = 0. We are going now to estimate the L norm of T K (B) in F = j (R n \ Q j). T F K (b)(x) dx j T F K b j (x) dx j dx x Q j y Q j K(x y) K(x y j ) b(y) dy = [ ] j y Q j b(y) K(x y) K(x y x Q j ) dx dy j [ ] j Q j b(y) K(x y x y j 2 y y j j (y y j )) K(x y j ) dx dy B j Q j b(y) dy = B b L (R n ) γb f L (R n ). Thus, by Chebyshev, {x F T K (b)(x) > α 2 } 2α F T K (b)(x) dx 2γα B f L (R n ) The rest of the set where T K (b)(x) > α is a subset of 2 Ω, therefore its measure is bounded above by the measure of Ω, that is bounded by γα f L (R n ). This concludes the proof of the fact that T K is weak type (,). The boundedness in L 2 together with the Marcinkiewicz interpolation theorem then imply the result for < p < 2. On the other h the adjoint 0

11 of T K is computed convolving with K( x), which is a kernel that satisfies the properties in the theorem. Therefore the adjoint (T K ) is bounded in L q (R n ), < q < 2 thus T K is bounded in L p (R n ), 2 < p <. This concludes the proof of the theorem. Theorem 4 Let K satisfy (-3). Let < p < consider the operators T (f)(x) = f(x y)k(y)dy. y There exists a constant C p depending on p, n B only such that T f L p (R n ) C p f L p (R n ) holds uniformly for all > 0. Moreover, for each f L p (R n ) the strong limit T K (f) = lim 0 T (f) exists in the norm of L p (R n ). The operator T K obeys the same norm bound as T. is bounded in L p (R n ) Proof. The operators T are convolution operators with kernels C (K). In view of Proposition Lemma, these kernels satisfy the assumptions of Theorem 2, with uniform constant B. Therefore they are uniformly bounded in each L p (R n ), independently of. It is then enough to check the convergence on a dense subset of L p (R n ). Let f C0 (R n ). Then (T f)(x) = K(y)f(x y)dy + K(y)(f(x y) f(x))dy. y < y The first integral is a fixed function in L p (R n ) because C (K) is in L p (R n ) for < p < f is in L (R n ). The limit as 0 of the second functions is strong in L p (R n ) because they are all supported in a fixed compact the convergence is uniform. This concludes the proof of the theorem.