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III the Maximal Function and Calderón-Zygmund Decomposition III The Maximal Function and Calderón-Zygmund Decomposition 3.1. Two covering lemmas ......................... 77 3.2. Hardy-Littlewood maximal function ................ 79 3.2.1. Hardy-Littlewood maximal operator ............ 80 3.2.2. Control of other maximal operators ............ 86 3.2.3. Applications to differentiation theory ............ 87 3.2.4. An application to Sobolev’s inequality ........... 90 3.3. Calderón-Zygmund decomposition ................. 91 References .................................. 97 § 3.1 Two covering lemmas Lemma 3.1.1: Finite version of Vitali covering lemma n Suppose B = fB1;B2;··· ;BNg is a finite collection of open balls in R . Then, there exists a disjoint sub­collection B j , B j , ···, B j of B such that 1 2! k [N k 6 n m m B` 3 ∑ (B ji ): `=1 i=1 Proof. The argument we give is constructive and relies on the following simple observation: Suppose B and B0 are a pair of balls that intersect, with the radius of B0 being not greater than that of B. Then B0 is con­ tained in the ball B˜ that is concentric with B but with 3 times 0 its radius. (See Fig 3.1.) B B As a first step, we pick a ball B j1 in with maximal (i.e., B largest) radius, and then delete from the ball B j1 as well as any B balls that intersect B j1 . Thus, all the balls that are deleted are ˜ contained in the ball B j1 concentric with B j1 , but with 3 times B~ its radius. The remaining balls yield a new collection B0, for which Figure 3.1: The balls B Figure 1:˜ The balls B and B~ we repeat the procedure. We pick B j2 and any ball that inter­ and B sects B j2 . Continuing this way, we find, after at most N steps, ··· a collection of disjoint balls B j1 , B j2 , , B jk . Finally, to prove that this disjoint collection of balls satisfies the inequality in the lemma, ˜ we use the observation made at the beginning of the proof. Let B ji denote the ball concentric Lecture Notes on Harmonic Analysis (subject to change anytime!) Typeset: March 17, 2020 © Chengchun HAO Institute of Mathematics AMSS CAS - 77 - - 78 - 3. The Maximal Function and Calderón-Zygmund Decomposition B with B ji , but with 3 times its radius. Since any ball B in must intersect a ball B ji and have [ ⊂ ˜ equal or smaller radius than B ji , we must have B\B j =6 ?B B ji , thus ! ! i [N [k k k 6 ˜ 6 m ˜ n m m B` m B ji ∑ (B ji ) = 3 ∑ (B ji ): `=1 i=1 i=1 i=1 In the last step, we have used the fact that in Rn a dilation of a set by d > 0 results in the multiplication by d n of the Lebesgue measure of this set. For the infinite version of Vitali covering lemma, one can see the textbook [Ste70, the lemma on p.9]. The decomposition of a given set into a disjoint union of cubes (or balls) is a fundamental tool in the theory described in this chapter. By cubes, we mean closed cubes; by disjoint we mean that their interiors are disjoint. We have in mind the idea first introduced by Whitney and formulated as follows. Theorem 3.1.2: Whitney covering lemma Let F be a non­empty closed set in Rn and W be its complement. Then there exists a F f g¥ countable collection of cubes = Qk k=1 whose sides are parallel to the axes, such that S ¥ W c (i) k=1 Qk = = F ; (ii) Q˚ j \ Q˚k = ? if j =6 k, where Q˚ denotes the interior of Q; (iii) there exist two constants c1;c2 > 0 independent of F (In fact we may take c1 = 1 and c2 = 4.), such that c1 diam(Qk) 6 dist(Qk;F) 6 c2 diam(Qk): Proof. Consider the lattice of points62 Chapter 4 Calderon-Zygmund´ Decomposition in Rn whose coordinates are integers. This lattice deter­ M mines a mesh 0, which is M 1 − a collection of cubes: namely M0 Mk Mk all cubes of unit length, whose M1 Q vertices are points of the above O 1 2 3 lattice. The mesh M0 leads F to a two­way infinite chain of ¥ Ωk+1 such meshes fMkg−¥, with −k Ωk Mk = 2 M0. Thus, each cube in theFig. 4.1 Meshes and layers: M0 with dashed (green) lines; M1Mwith dotted lines; M 1 with solid (blue)M lines Figure 3.2: Meshes and layers: 0 with dashed− lines; 1 mesh M gives rise to 2n k with dotted lines; M−1 with solid lines M diam (Q) 6 dist (Q; F) 6 4 diam (Q); Q F : (4.1) cubes in the mesh k+1 by bi­ 2 0 k secting the sides. The cubesLet in us prove (4.1) first. Suppose Q Mk; then diam (Q) = pn2− . Since Q F0, there exists an 2 k+1 p 2 x Q Ωk. Thus− distk (Q; F) 6 dist (x; F) 6 c2− , and dist (−Q;kF) > dist (x; F) diam (Q) > the mesh M each have sides2 of length\ 2 and are thus of diameter n2 . − k c2 k pn2 k. If we choose c = 2 pn we get (4.1). − − − MThen by (4.1) the cubes Q F are disjointW from F and clearly cover Ω. Therefore, (i) is also In addition to the meshes k, we consider the2 layers0 k, defined by proved.n o Notice that− thek collection F has all our− requiredk+1 properties, except that the cubes in it are not Wk = x : c2 < dist(x;0 F) 6 c2 ; necessarily disjoint. To finish the proof of the theorem, we need to refine ourS choice leading to F0, eliminating those cubes which were really unnecessary. W ¥ W where c is a positive constant which we shall fix momentarily. Obviously, = k=−¥ k. We require the following simple observation. Suppose Q1 and Q2 are two cubes (taken respectively from the mesh Mk1 and Mk2 ). Then if Q1 and Q2 are not disjoint, one of the two must be contained in the other. (In particular, Q Q , if k > k .) 1 ⊂ 2 1 2 Start now with any cube Q F , and consider the maximal cube in F which contains it. In This draft is subject to change anytime! 2 0 Typeset: March0 17, 2020 view of the inequality (4.1), for any cube Q0 F which contains Q F , we have diam (Q0) 6 2 0 2 0 dist (Q0; F) 6 dist(Q; F) 6 4 diam (Q). Moreover, any two cubes Q0 and Q00 which contain Q have obviously a non-trivial intersection. Thus by the observation made above each cube Q F has 2 0 a unique maximal cube in F0 which contains it. By the same taken these maximal cubes are also disjoint. We let F denote the collection of maximal cubes of F0. Then obviously (i) Q F Q = Ω, 2 (ii) The cubes of F are disjoint, S (iii) diam (Q) 6 dist (Q; F) 6 4 diam (Q), Q F . 2 Therefore, we complete the proof. ut 4.2 Calderon-Zygmund´ Fundamental Lemma Now, we give an important theorem in harmonic analysis. §3.2. Hardy-Littlewood maximal function - 79 - Now we make an initial choice of cubes, and denote the resulting collection by F0. Our choice is made as follows. We consider the cubes of the mesh Mk, (each such cube is of −k size approximately 2 ), and include a cube of this mesh in F0 if it intersects Wk, (the points of the latter are all approximately at a distance 2−k from F). Namely, [ F0 = fQ 2 Mk : Q \ Wk =6 ?g: k For appropriate choice of c, we claim that diam(Q) 6 dist(Q;F) 6 4diam(Q); Q 2 F0: (3.1.1) p −k Let us prove (3.1.1) first. Suppose Q 2 Mk; then diam(Q) = n2 . Since Q 2 F0, there 2 \W 6 6 −k+1 > − exists an x Q kp. Thus, dist(Q;F) dist(xp;F) c2 , and dist(Q;F) dist(x;F) diam(Q) > c2−k − n2−k. If we choose c = 2 n, we get (3.1.1). Then by (3.1.1) the cubes Q 2 F0 are disjoint from F and clearly cover W. Therefore, (i) is also proved. Notice that the collection F0 has all our required properties, except that the cubes in it are not necessarily disjoint. To finish the proof of the theorem, we need to refine our choice leading to F0, eliminating those cubes which were really unnecessary. We require the following simple observation. Suppose Q1 and Q2 are two cubes (taken M M respectively from the mesh k1 and k2 ). Then if Q1 and Q2 are not disjoint, one of the two must be contained in the other. (In particular, Q1 ⊂ Q2, if k1 > k2.) Start now with any cube Q 2 F0, and consider the maximal cube in F0 which contains 0 it. In view of the inequality (3.1.1), for any cube Q 2 F0 which contains Q 2 F0, we have diam(Q0) 6 dist(Q0;F) 6 dist(Q;F) 6 4diam(Q). Moreover, any two cubes Q0 and Q00 which contain Q have obviously a non­trivial intersection. Thus, by the observation made above each cube Q 2 F0 has a unique maximal cube in F0 which contains it. By the same taken these maximal cubes are also disjoint.
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