VII BMO and Carleson Measures 7.1. The sharp maximal functions and BMO spaces .......... 237 7.2. John-Nirenberg theorem ....................... 243 7.3. Non-tangential maximal functions and Carleson measures ... 248 7.4. BMO functions and Carleson measures ............... 253 References ................................... 255 § 7.1 The sharp maximal functions and BMO spaces Functions of bounded mean oscillation were introduced by F. John and L. Niren- berg [JN61], in connection with differential equations. Definition 7.1.1. The mean oscillation of a locally integrable function f (i.e. a function be- 1 Rn Rn longing to Lloc( )) over a cube QZ in is defined as the following integral: ~ 1 fQ = jf(x) − Avg fjdx; jQj Q Q where Avg f is the average value of f on the cube Q, i.e. Q Z 1 Avg f = f(x)dx: Q jQj Q Definition 7.1.2. 1 Rn A BMO function is any function f belonging to Lloc( ) whose mean oscil- lation has a finite supremum over the set of all cubesa Q contained in Rn. For 2 1 Rn f Lloc( ), we define the maximal BMO function or the sharp maximal function Z 1 M #f(x) = sup jf(t) − Avg fjdt; Q3x jQj Q Q where the supremum is take over all cubes Q in Rn that contains the given point x, and M # is called the sharp maximal operator. Then we denote the # norm of f in this space by kfkBMO = kM fk1. The set Rn f 2 1 Rn k k 1g BMO( ) = f Lloc( ): f BMO < is called the function space of bounded mean oscillation or the BMO space. aThe use of cubes Q in Rn as the integration domains on which the mean oscillation is calculated, is not mandatory: Wiegerinck (2001) uses balls instead and, as remarked by Stein ([Ste93], p. 140), in doing so a perfectly equivalent of definition of functions of bounded mean Lecture Notes on Harmonic Analysis Updated: April 28, 2020 © Chengchun HAO Institute of Mathematics AMSS CAS - 237 - - 238 - Chengchun HAO oscillation arises. Several remarks are in order. First, it is a simple fact that BMO(Rn) is a linear space, that is, if f; g 2 BMO(Rn) and λ 2 C, then f + g and λf are also in BMO(Rn) and kf + gkBMO 6kfkBMO + kgkBMO; kλfkBMO =jλjkfkBMO: But k · kBMO is not a norm. The problem is that if kfkBMO = 0, this does not imply that f = 0 but that f is a constant. Moreover, every constant function c satisfies kckBMO = 0. Consequently, functions f and f + c have the same BMO norms whenever c is a constant. In the sequel, we keep in mind that elements of BMO whose difference is a constant are identified. Although k · kBMO is only a semi-norm, we occasionally refer to it as a norm when there is no possibility of confusion. We give a list of basic properties of BMO. Proposition 7.1.3. The following properties of the space BMO(Rn) are valid: (1) If kfkBMO = 0, then f is a.e. equal to a constant. 1 n n (2) L (R ) ,! BMO(R ) and kfkBMO 6 2kfk1. (3) Suppose that there exists an A > 0 such that for all cubes Q in Rn there exists a constant cQ such thatZ 1 sup jf(x) − cQjdx 6 A: (7.1.1) Q jQj Q n Then f 2 BMO(R ) and kfkBMO 6 2A. f 2 L1 (Rn) (4) For all loc , we have Z 1 1 kfkBMO 6 sup inf jf(x) − cQjdx 6 kfkBMO: 2 Q jQj cQ Q (5) If f 2 BMO(Rn), h 2 Rn and τ h(f) is given by τ h(f)(x) = f(x − h), then τ h(f) is also in BMO(Rn) and h kτ (f)kBMO = kfkBMO: (6) If f 2 BMO(Rn) and λ > 0, then the function δλ(f) defined by δλ(f)(x) = f(λx) is also in BMO(Rn) and λ kδ (f)kBMO = kfkBMO: (7) If f 2 BMO(Rn), then so is jfj. Similarly, if f, g are real-valued BMO functions, then so are max(f; g) and min(f; g). In other words, BMO is a lattice. Moreover, kjfjkBMO 62kfkBMO; 3 k max(f; g)k 6 (kfk + kgk ) ; BMO 2 BMO BMO Lecture Notes on Harmonic Analysis Updated: April 28, 2020 §7.1. The sharp maximal functions and BMO spaces - 239 - 3 k min(f; g)k 6 (kfk + kgk ) : BMO 2 BMO BMO f 2 L1 (Rn) (8) For loc , define Z 1 kfk = sup jf(x) − Avg fjdx; BMOballs (7.1.2) B jBj B B where the supremum is taken over all balls B in Rn. Then there are positive constants cn, Cn such that c kfk 6 kfk 6 C kfk : n BMO BMOballs n BMO Proof. To prove (1), note that f has to be a.e. equal to its average cN over every cube [−N; N]n. Since [−N; N]n is contained in [−N − 1;N + 1]n, it follows that cN = cN+1 for all N. This implies the required conclusion. To prove (2), observe that ! Avg jf − Avg fj 6 Avg jfj + j Avg fj 6 2 Avg jfj 6 2kfk1: Q Q Q Q Q For (3), note that Z 1 jf − Avg fj 6 jf − cQj + j Avg f − cQj 6 jf − cQj + jf(t) − cQjdt: Q Q jQj Q Averaging over Q and using (7.1.1), we obtain that kfkBMO 6 2A. The lower inequality in (4) follows from (3) while the upper one is trivial. (5) is immediate. For (6), note that Avg δλ(f) = Avg f and thus Z Q λQ Z 1 1 jf(λx) − Avg δλ(f)jdx = jf(x) − Avg fjdx: jQj Q Q jλQj λQ λQ The first inequality in (7) is a consequence of the fact that Z Z 1 1 j j − j j j j − j j j j − j j f(x) Avg f = f(x) f(t) dt = ( f(x) f(t) )dt Q jQj Q jQj Q Z 1 6 (jf(x) − f(t)j)dt jQj Q Z Z 1 1 6 j − j j − j f(x) Avg f + Avg f f(t) dt jQj Q Q jQj Q Q 6jf − Avg fj + Avg jf − Avg fj: Q Q Q The second and the third inequalities in (7) follow from the first one in (7) and the facts that f + g + jf − gj f + g − jf − gj max(f; g) = ; min(f; g) = : 2 2 Rn We now turn to (8). Given anyp cube Q in , let B be the smallest ball that − p contains it. Then jBj/jQj = 2 nV nn due to jQj = (2r)n and jBj = V ( nr)n, and Z n Z p n 1 jBj 1 V nn jf(x) − Avg fjdx 6 jf(x) − Avg fjdx 6 n kfk : n BMOballs jQj Q B jQj jBj B B 2 It follows from (3) that p kfk 6 21−nV nnkfk : BMO n BMOballs Lecture Notes on Harmonic Analysis Updated: April 28, 2020 - 240 - Chengchun HAO To obtain the reverse conclusion, given any ball B find the smallest cube Q that n n contains it, with jBj = Vnr and jQj = (2r) , and argue similarly using a version of (3) for the space BMOballs. 1 Example 7.1.4. It is trivial that any bounded function is in BMO, i.e., L ,! BMO. 1 The converse is false, that is, L (Rn) is a proper subspace of BMO(Rn). A simple example that already typifies some of the essential properties of BMO is given by n the function f(x) = ln jxj. To check that this function is in BMO, for every x0 2 R j j j − j and R > 0, we find a constant Cx0;R such that the average of ln x Cx0;R over n the ball B(0;R) = fx 2 R : jx − x0j 6 Rg is uniformly bounded. The constant j j j j j j 6 Cx0;R = ln x0 if x0 > 2R and Cx0;R = ln R if x0 2R has this property. Indeed, jx j > 2R if 0 , then Z 1 j j j − j n ln x Cx0;R dx V R j − j6 n Z x x0 R 1 jxj = n ln dx V R j − j6 jx j n x x0 R 0 3 1 6 max ln ; ln 2 2 = ln 2; since 1 jx j 6 jxj 6 3 jx j when jx − x j 6 R and jx j > 2R. Also, if jx j 6 2R, then 2 0 2 Z0 0 0 0 1 j j j − j n ln x Cx0;R dx V R j − j6 n Z x x0 R Z j j j j 1 x 6 1 x = n ln dx n ln dx VnR j − j6 R VnR j j6 R Z x x0 R Z x 3R 3 1 ! − = jln jxjj dx = n 1 rn−1j ln rjdr Vn j j6 Vn Z x 3 Z 0 1 dr 3 =n (−1)rn ln r + n rn−1 ln rdr r Z0 Z 1 1 3 6n te−ntdt + n ln 3 rn−1dr 0 1 1 = + 3n ln 3: n Thus, ln jxj is in BMO. It is interesting to observe that an abrupt cutoff of a BMO function may not give a function in the same space. 1 R Example 7.1.5.