HARMONIC ANALYSIS PIOTR HAJLASZ 1. Maximal function 1 n For a locally integrable function f 2 Lloc(R ) the Hardy-Littlewood max- imal function is defined by Z n Mf(x) = sup jf(y)j dy; x 2 R : r>0 B(x;r) The operator M is not linear but it is subadditive. We say that an operator T from a space of measurable functions into a space of measurable functions is subadditive if jT (f1 + f2)(x)j ≤ jT f1(x)j + jT f2(x)j a.e. and jT (kf)(x)j = jkjjT f(x)j for k 2 C. The following integrability result, known also as the maximal theorem, plays a fundamental role in many areas of mathematical analysis. p n Theorem 1.1 (Hardy-Littlewood-Wiener). If f 2 L (R ), 1 ≤ p ≤ 1, then Mf < 1 a.e. Moreover 1 n (a) For f 2 L (R ) 5n Z (1.1) jfx : Mf(x) > tgj ≤ jfj for all t > 0. t Rn p n p n (b) If f 2 L (R ), 1 < p ≤ 1, then Mf 2 L (R ) and p 1=p kMfk ≤ 2 · 5n=p kfk for 1 < p < 1, p p − 1 p kMfk1 ≤ kfk1 : Date: March 28, 2012. 1 2 PIOTR HAJLASZ The estimate (1.1) is called weak type estimate. 1 n 1 n Note that if f 2 L (R ) is a nonzero function, then Mf 62 L (R ). Indeed, R if λ = B(0;R) jfj > 0, then for jxj > R we have Z λ Mf(x) ≥ jfj ≥ n ; B(x;R+jxj) !n(R + jxj) n and the function on the right hand side is not integrable on R . Thus the statement (b) of the theorem is not true for p = 1. 1 n If g 2 L (R ), then the Chebyschev inequality 1 Z jfx : jg(x)j > tgj ≤ jgj for t > 0 t Rn is easy to prove. Hence the inequality at (a) would follow from boundedness of of Mf in L1. Unfortunately Mf is not integrable and (a) is the best what we can get for p = 1. Before we prove the theorem we will show that it implies the Lebesgue differentiation theorem. 1 n Theorem 1.2 (Lebesgue differentiation theorem). If f 2 Lloc(R ), then Z lim f(y) dy = f(x) a.e. r!0 B(x;r) 1 n Proof. Since the theorem is local in nature we can assume that f 2 L (R ). R Let fr(x) = B(x;r)f(y) dy and define Ωf(x) = lim sup fr(x) − lim inf fr(x) : r!0 r!0 1 It suffices to prove that Ωf = 0 a.e. and that fr ! f in L . Indeed, the first property means that fr converges a.e. to a measurable function g while the second one implies that for a subsequence fri ! f a.e. and hence g = f a.e. Observe that Ωf ≤ 2Mf and hence for any " > 0 Theorem 1.1(a) yields C Z jfx :Ωf(x) > "gj ≤ jfj : " Rn 2 Let h be a continuous function such that kf − hk1 < " . Continuity of h implies Ωh = 0 everywhere and hence Ωf ≤ Ω(f − h) + Ωh = Ω(f − h) ; so C Z jfΩf > "gj ≤ jfΩ(f − h) > "gj ≤ jf − hj ≤ C": " Rn HARMONIC ANALYSIS 3 Since " > 0 can be arbitrarily small we conclude Ωf = 0 a.e. We are left 1 with the proof that fr ! f in L . We have Z Z Z jfr(x) − f(x)j dx ≤ jf(y) − f(x)j dy dx Rn Rn B(x;r) Z Z = jf(x + y) − f(x)j dy dx Rn B(0;r) Z (1.2) = kfy − fk1 dy ; B(0;r) 1 where fy(x) = f(x + y). Since fy ! f in L as y ! 0 the right hand side of (1.2) converges to 0 as r ! 0. 2 1 n If f 2 Lloc(R ), then we can define f at every point by the formula Z (1.3) f(x) := lim sup f(y) dy : r!0 B(x;r) According to the Lebesgue differentiation theorem this is a representative of f in the class of functions that coincide with f a.e. 1 n n Definition. Let f 2 Lloc(R ). We say that x 2 R is a Lebesgue point of f if Z lim jf(y) − f(x)j dy = 0 ; r!0 B(x;r) where f(x) is defined by (1.3). 1 n Theorem 1.3. If f 2 Lloc(R ), then the set of points that are not Lebesgue points of f has measure zero. Proof. For c 2 Q let Ec be the set of points for which Z (1.4) lim jf(y) − cj dy = jf(x) − cj r!0 B(x;r) does not hold. Clearly jE j = 0 and hence the set E = S E has measure c c2Q c n n zero. Thus for x 2 R n E and all c 2 Q, (1.4) is satisfied. If x 2 R n E and f(x) 2 R, approximating f(x) by rational numbers one can easily check that Z lim jf(y) − f(x)j dy = jf(x) − f(x)j = 0 : r!0 B(x;r) The proof is complete. 2 n We can generalize the above result as follows. We say that x 2 R is a p p-Lebesgue point of f 2 Lloc, 1 ≤ p < 1 if Z lim jf(y) − f(x)jp dy = 0 : r!0 B(x;r) The same method as the one used above leads to the following result that we leave an an exercise. 4 PIOTR HAJLASZ p n Theorem 1.4. If f 2 Lloc(R ), 1 ≤ p < 1, then the set of points that are not p-Lebesgue points of f has measure zero. n n Definition. Let E ⊂ R be a measurable set. We say that x 2 R is a density point of E if jB(x; r) \ Ej lim = 1 : r!0 jB(x; r)j Applying the Lebesgue theorem to f = χE we obtain n Theorem 1.5. Almost every point of a measurable set E ⊂ R is its density n point and a.e. point of R n E is not a density point of E. In the Lebesgue theorem we have seen that the averages of f over balls converge to f(x) and it is natural to inquire if we can replace balls by other sets like cubes or even balls, but not centered at x. n Definition. We say that a family F of measurable subsets of R is regular n at x 2 R if (a) The sets are bounded and have positive measure; (b) There is a sequence Sn 2 F with jSnj ! 0; (c) There is a constant C > 0 such that every S 2 F is contained in a ball B ⊃ S centered at x such that jSj ≥ CjBj. n Example. The family of all cubes Q in R such that the distance of Q to x is no more than Cdiam Q is regular. 1 n Theorem 1.6. If f 2 Lloc(R ), x is a Lebesgue point of f, and F is regular at x, then Z lim f(y) dy = f(x) : S2F jSj!0 S Proof. For S 2 F denote by rS the radius of a ball BS = B(x; rS) such that S ⊂ BS and jSj ≥ CjBSj. Clearly if jSj ! 0, then rS ! 0. We have Z Z f(y) dy − f(x) ≤ jf(y) − f(x)j dy S S Z ≤ jSj−1 jf(y) − f(x)j dy BS Z ≤ C−1 jf(y) − f(x)j dy ! 0 BS as jSj ! 0. 2 HARMONIC ANALYSIS 5 Note that if F is a regular family at 0 and we define the maximal function associated with F by Z MF f(x) = sup jf(x − y)j dy ; S2F S then it is a routine calculation to show that (1.5) MF f(x) ≤ CMf(x) ; so MF satisfies the claim of Theorem 1.1 (with different constants). In p particular MF is a bounded operator in L , 1 < p ≤ 1. In the proof of Theorem 1.1 we will need the following two results. Theorem 1.7 (Cavalieri's principle). If µ is a σ-finite measure on X and Φ : [0; 1) ! [0; 1) is increasing, absolutely continuous and Φ(0) = 0, then Z Z 1 Φ(jfj) dµ = Φ0(t)µ(fjfj > tg) dt : X 0 Proof. The result follows immediately from the equality Z Z Z jf(x)j Φ(jf(x)j) dµ(x) = Φ0(t) dt dµ(x) X X 0 and the Fubini theorem. 2 Corollary 1.8. If µ is a σ-finite measure on X and 0 < p < 1, then Z Z 1 jfjp dµ = p tp−1µ(fjfj > tg) dt : X 0 The next result has many applications that go beyond the maximal the- orem. Theorem 1.9 (5r-covering lemma). Let B be a family of balls in a metric space such that supfdiam B : B 2 Bg < 1. Then there is a subfamily of pairwise disjoint balls B0 ⊂ B such that [ [ B ⊂ 5B: B2B B2B0 If the metric space is separable, then the family B0 is countable and we can 0 1 arrange it as a sequence B = fBigi=1, so 1 [ [ B ⊂ 5Bi : B2B i=1 Remark. Here B can be either a family of open balls or closed balls.