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The Space of Functions of Bounded Mean Oscillation (BMO) Plays an Impor- Tant Role in Harmonic Analysis

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The Space of Functions of Bounded Mean Oscillation (BMO) Plays an Impor- Tant Role in Harmonic Analysis 8BMO The space of functions of Bounded Mean Oscillation (BMO) plays an impor- tant role in Harmonic Analysis. d A function f,inL1(loc)inR is said to be a BMO function if Z 1 sup inf |f(y) − a|dy = kukBMO < ∞ (8.1) a | | x;r Bx;r y∈Bx;r where Bx;r is the ball of radius r centered at x,and|Bx;r| is its volume. Remark. The infimum over a can be replaced by the choice of Z 1 a =¯a = | | f(y)dy Bx;r y∈Bx;r giving us an equivalent definition. We note that for any a, Z | − |≤ 1 | − | a a¯ | | f(y) a dy Bx;r y∈Bx;r and therefore if a∗ is the optimal a, ∗ |a¯ − a |≤kfkBMO Remark. Any bounded function is in the class BMO and kfkBMO ≤kfk∞. Theorem 8.1 (John-Nirenberg).R Let f beaBMOfunctiononacubeQ | | k k ≤ of volume Q =1satisfying Q f(x)dx =0and f BMO 1. Then there are finite positive constants c1;c2, independent of f, such that, for any `>0 ` |{x : |f(x)|≥`}| ≤ c1 exp[− ] (8.2) c2 Proof. Let us define F (`)=sup|{x : |f(x)|≥`} f k k ≤ Rwhere the supremum isR taken over all functions with f BMO 1and k k1 ≤kk ≤ Q f(x)dx =0.Since Q f(x)dx = 0 implies that f f BMO 1, ≤ 1 d F (`) ` . Let us subdivide the cube into 2 subcubes with sides one half theR original cube. We pick a number a>1 and keep theR cubes Qi with 1 |f(x)|dx ≥ a. We subdivide again those with 1 |f(x)|dx < a |Qi| Qi |Qi| Qi and keep going. In this manner we get an atmost countable collection of dis- joint cubes that we enumerate as {Qj}, that have the following properties: 49 R 1. 1 |f(x)|dx ≥ a. |Qj| Qj 0 2. Each Qj is contained inR a bigger cube Qj with sides double the size of 1 the sides of Qj and 0 0 |f(x)|dx < a. |Qj| Qj c 3. By the Lebesgue theorem |f(x)|≤a on Q ∩ (∪jQj) . R If we denote by a = 1 f(x)dx,wehave j |Qj| Qj Z Z d | |≤ 1 | | ≤ 2 | | ≤ d aj f(x) dx 0 f(x) dx 2 a | | | | 0 Qj Qj Qj Qj by property 2). On the other hand f − aj has mean 0 on Qj and BMO norm at most 1. Therefore (scaling up the cube to standard size) d |Qj ∩{x : |f(x)|≥2 a + `}| ≤ |Qj ∩{x : |f(x) − aj|≥`}| ≤|Qj|F (`) Summing over j, because of property 3), X d |{x : |f(x)|≥2 a + `}| ≤ F (`) |Qj| j P | |≤ 1 On the other hand property 1) implies that j Qj a giving us 1 F (2da + `) ≤ F (`) a which is enough to prove the theorem. Corollary 8.1. For any p>1 there is a constant Cd;p depending only on the dimension d and p such that Z Z 1 1 sup |f(x) − f(x)dx|pdx ≤ C kfkp | | | | d;p BMO Q Q Q Q Q The importance of BMO, lies partly in the fact that it is dual to H1. Theorem 8.2. There are constants 0 <c≤ C<∞ such that Z ckfkBMO ≤ sup | f(x)g(x)dx|≤CkfkBMO (8.3) :k k ≤1 g g H1 and every bounded linear functional on H1 is of the above type. 50 The proof of the theorem depends on some lemmas. Lemma 8.1. The Riesz transforms Ri map L∞ → BMO boundedly. In fact Ω(x) convolution by any kernel of the form K(x)= |x|d where Ω(x) is homoge- neous of degree zero, has mean 0 on Sd−1 and satisfies a H¨older condition on d−1 S maps L∞ →BMO boundedly. Proof. Let us suppose that Q is the unit cube centered around the origin and denote by 2Q the doubled cube. We write f = f1 + f2 where f1 = f12Q and f2 = f − f1 = f1(2Q)c . g(x)=g1(x)+g2(x) where Z gi(x)= K(x − y)fi(y)dy Rd Z b d b |g1(x)|dx ≤kg1k2 ≤ sup |K(ξ)|kf1k2 ≤ 2 2 sup |K(ξ)|kfk∞ Q ξ ξ R − 2 On the other hand with aQ = Q K( y)f (y)dy Z |g2(x) − aQ|dx Q Z Z ≤ dx |K(x − y) − K(−y)|f2(y)dy Q ZZRd ≤kfk∞ |K(x − y) − K(−y)|dxdy z∈Q Zy=∈2Q ≤kfk∞ sup |K(x − y) − K(−y)|dy x |y|≥2|x| ≤ Bkfk∞ The proof for arbitrary cube is just a matter of translation and scaling. The H¨older continuity is used to prove the boundedness of Kb(ξ). Lemma 8.2. Any bounded linear function Λ on H1 is given by Xd Z Z Xd Λ(f)= (Rif)(x)gi(x)dx = − f(x) (Rigi)(x)dx i=0 i=0 where R0 = I and Ri for 1 ≤ i ≤ d are the Riesz transforms. 51 d Proof. The space H1 is a closed subspace of the direct sum ⊕L1(R )ofd +1 d copies of L1(R ). Hahn-Banach theorem allows us to extend Λ boundedly ⊕ d { } to PL1(R ) and the Riesz representation theorem gives us gi . Finally d g0 + i=1 Rigi is in BMO. Lemma 8.3. If g ∈BMO then Z |g(y)| dy < ∞ (8.4) | |d+1 Rd 1+ y and Z G(t; x)= g(y)p(t; x − y)dy exists where p(· ; ·) is the Poisson kernel for the half space t>0. Moreover g(t; x) satisfies Z sup t|∇G(t; y)|2dtdy ≤ Akgk2 hd (8.5) |y−x|<h BMO x 0<t<h for some constant independent of g.Here∇G is the full gradient in t and x. R |g(x)| Proof. First let us estimate Rd 1+|x|d+1 dx.IfwedenotebyQn the cube of side 2n around the origin Z Z Z | | | | X | | g(x) ≤ g(x) g(x) d+1 dx d+1 dx + d+1 dx d 1+|x| 1+|x| ∩ c 1+|x| R ZQ0 n QnZ+1 Qn X 1 ≤ |g(x)|dx + |g(x)|dx 2n(d+1) ZQ0 n ZQn+1 X 1 ≤ |g(x)|dx + |g(x) − a +1|dx 2n(d+1) n Q0 n Qn+1 X |a +1| + n 2n Z n X (n+1)d X 2 |a +1| ≤ |g(x)|dx + kgk + n BMO 2n(d+1) 2n Q0 n n 52 where Z 1 +1 ≤ an (n+1)d g(x)dx 2 Qn+1 Moreover Z 1 | − | | − | ≤ dk k a2Q aQ = | | g(x) a2Q dx 2 g BMO Q Q and this provides a bound of the form Z | |≤ k k | | aQn Cn g BMO + g(x) dx Q0 establishing (8.4). We now turn to proving (8.5). Again because of the homogeneity under translations and rescaling, we can assume that x =0 and h = 1. So we only need to control Z t|∇G(t; y)|2dtdy ≤ Akgk2 |y|<1 BMO 0<t<1 We denote by Q4 the cube |x|≤2andwriteg as − g = aQ4 +(g1 aQ4 )+g2 − c where g1 = g1Q4 , g2 = g g1 = g1Q4 . Since constants do notR contribute to (8.5), we can assume that a = 0, and therefore the integral |g(x)|dx can Q4 Q4 be estimated in terms of kgkBMO. An easy calculation, writing G = G1 + G2 yields Z | | |∇ |≤ g(x) ≤ k k G2(t; y) +1 dx A g BMO c | |d Q4 1+ x As for the G1 contribution in terms of the Fourier transform we can control it by Z ∞ Z Z ∞ Z Z 2 2 −2t|ξ| 2 2 t|∇G| dtdy = t|ξ| e |gb1(ξ)| dξdt = |gb1(ξ)| dξ 0 Rd 0 Rd Rd which is controlled by kgkBMO because of the John-Nirenberg theorem. 53 Lemma 8.4. Any function g whose Poisson intgeral G satisfies (8.5) defines a bounded linear functional on H1. Proof. The idea of the proof is to write Z ∞ Z Z ∞ Z 2 t∇G(t; x)∇F (t; x)dtdx =4 te−2t|ξ||ξ|2fb(ξ)gb¯(ξ)dξdt 0 Rd Z 0 Rd = fb(ξ)gb¯(ξ)dξ ZRd = f(x)g(x)dx Rd and concentrate on Z ∞ Z t|∇xG(t; x)||∇xF (t; x)|dtdx 0 Rd We need the auxiliary function ZZ 1− 2 1 d 2 (Shu)(x)=[ t |∇u| dydt] |x−y|<t<h Clearly (Shu)(x)isincreasinginh and we show in the next lemma that kS∞F k1 ≤ CkfkH1 Let us assume it and complete the proof. Define h(x)=sup{h :(ShF )(x) ≤ MC} then (Sh(x)F )(x) ≤ MC In addition it follows from (8.5) that Z 2 d sup |(ShF )(x)| dx ≤ Ch y;h |y−x|≤h Now h(x) <hmeans (ShF )(x) >MCand therefore Chd |{x : |x − y| <h;h(x) <h}| ≤ M 2 54 By the proper choice of M, we can be sure that |{x : |x − y| <h;h(x) ≥ h}| ≥ chd Now we complete the proof. Z ∞ Z t|∇xG(t; x)||∇xF (t; x)|dtdx 0 Rd Z ∞ Z Z 1−d ≤ C t |∇xG(t; x)||∇xF (t; x)|dtdxdy 0 Rd |y−x|<t≤h(y) Z Z ∞ Z 1− 2 1 d 2 ≤ dy( t |∇xG(t; x)| dxdt) Rd 0 |y−x|<t≤h(y) Z ∞ Z 1− 2 1 d 2 × ( t |∇xF (t; x)| dxdt) Z 0 |y−x|<t≤h(y) ≤ M (S∞F )(y)dy ≤ MkfkH1 Rd Lemma 8.5.
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