Good Lambda Inequalities and Riesz Potentials for Non Doubling Measures in R n

Transcription

1 Good Lambda Inequalities and Riesz Potentials for Non Doubling Measures in R n Mukta Bhandari mukta@math.ksu.edu Advisor Professor Charles Moore Department of Mathematics Kansas State University Prairie Analysis Seminar October 2-3, 2009

2 Some Useful Definitions

3 Some Useful Definitions We need the following definitions for our results:

4 Some Useful Definitions We need the following definitions for our results: (Fractional) Maximal functions. Doubling measure growth condition Riesz Potentials

5 Some Useful Definitions

6 Some Useful Definitions Definition (Doubling Measure) A measure µ in a metric space (X, d) is called doubling if balls in X have finite and positive measure and there is a constant C = C(µ) 1 such that for every ball B in X µ(2b) Cµ(B)

7 Some Useful Definitions Definition (Doubling Measure) A measure µ in a metric space (X, d) is called doubling if balls in X have finite and positive measure and there is a constant C = C(µ) 1 such that for every ball B in X Definition µ(2b) Cµ(B) A measure µ in a metric space (X, d) is said to satisfy growth condition if for every r > 0 there exists a constant C independent of x and r such that µ(b(x, r)) Cr N. This, in particular, allows non-doubling measure. Sometimes we shall refer to this condition by saying that µ is N-dimensional.

8 Some Useful Definitions

9 Some Useful Definitions Definition Let (X, d) be a metric space and µ be any measure on X. Let f be a locally integrable function on X. The maximal function of f, M(f ), is defined by 1 M(f )(x) = sup f (y) dµ(y) r>0 µ (B(x, r)) B(x,r) The operator M is called Hardy-Littlewood maximal operator.

10 Some Useful Definitions Definition Let (X, d) be a metric space and µ be any measure on X. Let f be a locally integrable function on X. The maximal function of f, M(f ), is defined by 1 M(f )(x) = sup f (y) dµ(y) r>0 µ (B(x, r)) B(x,r) The operator M is called Hardy-Littlewood maximal operator. The fractional maximal function of f is defined for 0 < α < N by M α (f )(x) = sup r>0 1 µ(b(x, r)) N α N B(x,r) f (y) dµ(y).

11 Some Useful definitions

12 Some Useful definitions Definition Let 0 < α < N. The Riesz potential operator of order α on X is given by f (y)dµ(y) I α f (x) = d(x, y) N α, x X X

13 Some Useful definitions Definition Let 0 < α < N. The Riesz potential operator of order α on X is given by f (y)dµ(y) I α f (x) = d(x, y) N α, x X X

14 Some Remarks

15 Some Remarks Remark: (X.Tolsa, BMO,H 1 and Calderón-Zygmund operators for non-doubling measures, Math. Ann. 319, No.1, , 2001) If µ satisfies the growth condition then there are lots of big doubling cubes. Precisely speaking, given any point x supp(µ) and c > 0, there exists some (α, β)-doubling cube Q (that is, µ(αq) βµ(q) where α > 1 and β > α n ) centered at x with l(q) c.

16 Some Remarks Remark: (X.Tolsa, BMO,H 1 and Calderón-Zygmund operators for non-doubling measures, Math. Ann. 319, No.1, , 2001) If µ satisfies the growth condition then there are lots of big doubling cubes. Precisely speaking, given any point x supp(µ) and c > 0, there exists some (α, β)-doubling cube Q (that is, µ(αq) βµ(q) where α > 1 and β > α n ) centered at x with l(q) c. There are always small doubling cubes for a Radon measure µ.that is for µ-a.e. x rn there is a sequence of (α, β)-doubling cubes {Q j } j centered at x such that l(q j ) 0 as j. If α, β are not specified then by a doubling cube we will mean a (2, β)-doubling cube where β > 2 n. l(q) denotes the side length of the cube Q whose sides are parallel to the axes.

17 Useful Theorems

18 Useful Theorems José García-Cuerva and A. Eduardo Gatto proved the following theorem in 2003

19 Useful Theorems José García-Cuerva and A. Eduardo Gatto proved the following theorem in 2003 Theorem Let (X, d, µ) be a metric measure space where the measure µ is Borel which satisfies the growth condition µ(b(x, r)) Cr n. Then, for 1 p < n α and 1 q = 1 p α n, we have ( ) C f q L µ ({x X : I α f (x) > λ}) p (µ), λ that is, I α is a bounded operator from L p (µ) into the Lorentz space L q, (µ).

20 Theorems on Good λ inequalities:

21 Theorems on Good λ inequalities: B.Muckenhoupt and R. L. Wheeden proved the following good λ- inequality in 1972 in Weighted Norm Inequalities for Fractional Integrals, Trans. Amer. Math. Soc., 192, 1974 Theorem Let µ be positive Radon measure in R n. Then there exists a > 1, b > 0 such that for every λ > 0 and for every ɛ, 0 < ɛ 1 {x : I α µ(x) > aλ} bɛ n/(n α) {x : I α µ(x) > λ} + {x : M α µ(x) > ɛλ} where I α µ(x) = 1 γ(α) R n dµ(y) x y n α.

22 Theorems on Good λ inequalities:

23 Theorems on Good λ inequalities: As a special case we have the following theorem:

24 Theorems on Good λ inequalities: As a special case we have the following theorem: Let f L 1 loc (Rn ), 0 < α < N, I α f (x) = R n f (y) x y N α dy Theorem There exists constants k, c such that for every λ > 0 and 0 < ɛ < 1, {x : I α f (x) > kλ, M α f < ɛλ} Cɛ N N α {x : Iα f (x) > λ}.

25 Main Results:

26 Main Results: Theorem Let 0 < α < N. The Riesz potential operator of order α on R n is given by f (y)dµ(y) I α f (x) =, x Rn d(x, y) N α R n for measurable function f on R n. We consider a positive Radon measure µ on R n satisfying the growth condition µ (B(x, r)) Cr N. Then there exists constants k 1, C such that for every λ > 0 and ɛ, 0 < ɛ 1, µ ({x : I α f (x) > kλ, M α f (x) ɛλ}) Cɛ N N α µ ({x : Iα f (x) > λ}).

27 Sketch of the Proof: Let E λ = {x R n : I α f (x) > λ}. Then E λ is open. So it has the following Whitney decomposition: There exists a countable family of dyadic cubes {Q j } j with disjoint interiors such that (i) E λ = Q j, (ii) diam(q j ) dist(q j, E c λ) 4diam(Q j ) for every j. (iii) If the boundaries of any two cubes Q l and Q m touch, then 1 4 l(q l) l(q m ) 4. (iv) For every 0 < δ < 1 4, (1 + δ)q E λ, (v) Q χ (1+δ)Q(x) 12 N χ Eλ (x) for every x E λ. This implies that µ ((1 + δ)q) (x) 12 N µ(e λ ). Q j

28 Sketch of the Proof: Reference: Appendix J on Classical and Modern Fourier Analysis,page A-34 by Loukas Grafakos. Fix x E λ and a Q in {Q j } j with x Q Let k > 1 and consider the set {x Q : I α f (x) > kλ}. Suppose Q {M α f (x) ɛλ} φ. Otherwise there is nothing to prove. Let Q x be the cube center at x and l(q x ) = δl(q). We may fix δ = 1 8. Then Q x (1 + δ)q and 16Q x 2Q. Consider 1 Q 2 j x, j = 1, 2, 3,.... Take the first which is doubling and denote it by Q x := 1 Q 2 m+1 x where m depends on x and Q x 1 2 Q x.

29 Sketch of the Proof This means that (a) For every x E λ {Mα f (x) ɛλ} there exists a doubling cube Q x such that µ(2 Q x ) βµ( Q x ) where β > 2 n. We may assume β = 2 n+ɛ.

30 Sketch of the Proof This means that (a) For every x E λ {Mα f (x) ɛλ} there exists a doubling cube Q x such that µ(2 Q x ) βµ( Q x ) where β > 2 n. We may assume β = 2 n+ɛ. (b) µ ( j+1 Q x ) βµ ( 1 2 j+1 Q x ) for every j < m.

31 Sketch of the Proof This means that (a) For every x E λ {Mα f (x) ɛλ} there exists a doubling cube Q x such that µ(2 Q x ) βµ( Q x ) where β > 2 n. We may assume β = 2 n+ɛ. (b) µ ( j+1 Q x ) βµ ( 1 2 j+1 Q x ) for every j < m. (c) Therefore, {x Q : I α f (x) > kλ, M α f (x) < ɛλ} x Q Q x.

32 Sketch of the Proof: The Besicovitch covering lemma implies that there exists Q xj, j = 1, 2, 3,... such that {x Q : I α f (x) > kλ, M α f (x) < ɛλ} j Q xj where χ Qxj 4 n χ (1+δ)Q. ( That is, µ Q ) j xj µ ((1 + δ)q). j

33 Sketch of the Proof: The Besicovitch covering lemma implies that there exists Q xj, j = 1, 2, 3,... such that {x Q : I α f (x) > kλ, M α f (x) < ɛλ} j Q xj where χ Qxj 4 n χ (1+δ)Q. ( That is, µ Q ) j xj µ ((1 + δ)q). j Let f = f 1 + f 2 where f 1 = χ 2Q f and f 2 = f f 1.

34 Sketch of the Proof: The Besicovitch covering lemma implies that there exists Q xj, j = 1, 2, 3,... such that {x Q : I α f (x) > kλ, M α f (x) < ɛλ} j Q xj where χ Qxj 4 n χ (1+δ)Q. ( That is, µ Q ) j xj µ ((1 + δ)q). j Let f = f 1 + f 2 where f 1 = χ 2Q f and f 2 = f f 1. Then, {x Q : I α f 1 (x) > kλ, M α f (x) < ɛλ} j {x Q xj : I α f 1 (x) > kλ, M α f (x) < ɛλ}.

35 Sketch of the Proof Fix a doubling cube Q xj. Write f 1 = χ 2 Qxj f 1 + χ (2 Qxj ) c f 1 = f 11 + f 12.

36 Sketch of the Proof Fix a doubling cube Q xj. Write f 1 = χ 2 Qxj f 1 + χ (2 Qxj ) c f 1 = f 11 + f 12. Then ( µ {x 2 Q ) ( ) 1 N/N α xj : I α f 11 > kλ} kλ f 11 1 Cɛ N/N α µ( Q xj ).

37 Sketch of the Proof Fix a doubling cube Q xj. Write f 1 = χ 2 Qxj f 1 + χ (2 Qxj ) c f 1 = f 11 + f 12. Then ( µ {x 2 Q ) ( ) 1 N/N α xj : I α f 11 > kλ} kλ f 11 1 Cɛ N/N α µ( Q xj ). Note that µ(q) Cl(Q) N. That is ( ) (N α)/n C µ(q) 1. l(q) N α

38 Sketch of the Proof Fix a doubling cube Q xj. Write f 1 = χ 2 Qxj f 1 + χ (2 Qxj ) c f 1 = f 11 + f 12. Then ( µ {x 2 Q ) ( ) 1 N/N α xj : I α f 11 > kλ} kλ f 11 1 Cɛ N/N α µ( Q xj ). Note that µ(q) Cl(Q) N. That is ( ) (N α)/n C µ(q) 1. l(q) N α We also have µ(q xj ) βµ for every i < m. ( ) ( ) Q x j... β i µ 2 i Q x j

39 Sketch of the Proof: For x Q xj, f 1 (y) I α f 12 (x) = dµ(y) (2 Q xj ) c d(x, y) N α m 1 + k=0 cλ. 16Q xj \2Q 1 2 k Qx j \ 1 2 k+1 Qx j f 1 (y) dµ(y) d(x, y) N α f 1 (y) dµ(y). d(x, y) N α Therefore I α f 12 (x) Cλ for every x Q xj.

40 Sketch of the Proof: For x Q xj, f 1 (y) I α f 12 (x) = dµ(y) (2 Q xj ) c d(x, y) N α m 1 + k=0 cλ. 16Q xj \2Q 1 2 k Qx j \ 1 2 k+1 Qx j f 1 (y) dµ(y) d(x, y) N α f 1 (y) dµ(y). d(x, y) N α Therefore I α f 12 (x) Cλ for every x Q xj.

41 Sketch of the Proof: We now estimate I α f 2. Fix a point x Q. Consider the ball B = B(x, 6diam(Q)) 2Q. Let x 0 B E λ c such that Then for any y (2Q) c diam(q) dist(q, x 0 ) 4diam(Q). d(x 0, y) d(x, x 0 ) + d(x, y) Cdiam(Q) + d(x, y) Cdist(Q, Eλ c ) + d(x, y) Cd(x, y) + d(x, y). Thus d(x 0, y) Cd(x, y) for x 0 B E c λ Therefore, I α f 2 (x) = (2Q) c f (y)dµ(y) d(x,y) N α and x Q. CI α f (x 0 ) < Cλ.

42 Sketch of the Proof: Finally summing over all Whitney cubes yields the inequality. µ ({x : I α f (x) > kλ, M α f (x) < ɛλ}) = µ ({x Q : I α f (x) > kλ, M α f (x) < ɛλ}) Q Q = Q < Q Q Q µ ({x Q : I α f 1 > (k C)λ, M α f < ɛλ}) j j ( µ {x Q ) xj : I α f 1 > (k C)λ, M α f < ɛλ} ( µ {x Q ) xj : I α f 11 > (k 2C)λ, M α f < ɛλ} Cɛ N/N α µ( Q xj ) j Cɛ N/N α 4 n µ((1 + δ)q) = 4 N 12 N Cɛ N/N α µ({i α f > λ}).

43 Thus, finally we have µ ({x : I α f > kλ}) µ ({I α f > kλ, M α f ɛλ}) + µ ({M α f > ɛλ}) Cɛ N/N α µ ({I α f > λ}) + µ ({M α f > ɛλ}). Next, we derive the same inequality for the weights w A p (µ) where µ satisfies the growth condition. That is for any measurable set E, w(e) = w(x)dµ(x). Details about this type of weight can be obtained from the paper A P Weights for Nondoubling Measures in R n and Applications, Trans.Amer. Math. Soc, 354, 5, , (by Joan Orobitg and Carlos Pérez) A (µ) is defined as A (µ) = p>1 A p(µ). E

44 Good Lambda inequality for A (µ) weights Theorem Let w A (µ), and let α > 0, and 0 < p <. Then there exists a positive constant a 1 for which, for every η > 0, there is an ɛ, 0 < ɛ 1, such that the inequality w ({x R n : I α f (x) > aλ}) ηw ({x R n : I α f (x) > λ}) + w ({x R n : M α f (x) > ɛλ}) holds for every λ > 0.

45 Proof of the theorem: This theorem follows from the good lambda imequality for nondoubling measure (last theorem) and the following remark: Remark: w A (µ) if and only if there are positive constants c and δ such that, for any cube Q and any measurable set E contained in Q, w(e) w(q) c ( ) µ(e) δ. µ(q)

46 Applications: We give two applications of above two theorems to derive potential inequalities. Here the measure µ under consideration is a positive Radon measure which satisfies the growth condition. They are given by the following two theorems: Theorem Let 1 < p < and 0 < α < N. Then there is a constant A such that for any measurable function f on R n and a measure µ satisfying the growth condition we have Theorem I α f p A M α f p. Let w A (µ), 0 < α < N, and let 0 < p <. Then there exists a positive constant C, that only depends on N, p, and A (µ) constants of w, such that I α f p wdµ C (M α (f )) p wdµ R N R N

47 !!! Thank You!!!