Lecture 24. Calderon-Zygmund Decomposition Technique: Cube decomposition

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1 Lecure 24 May 3 h, 2004 Our moivaion for his as ecure in he course is o show a resu using our reguariy heory which is oherwise unprovabe using cassica echniques. This is he previous Theorem, and in paricuar he case p 2 (which we haven ye done namey N is a coninuous map L p (Ω o W 2,p (Ω. Cassica mehods give a bes W,p (Ω. For ha we inroduce he Caderon-Zygmund Decomposiion Technique: Cube decomposiion Le K 0 be an n dimensiona cube in R n, f 0 inegrabe and finay fix > 0 such ha K 0 f K 0 Vo(K 0, ha is f. K 0 Nex bisec K 0 ino 2 n equa (in voume subcubes. Le S be he coecion of hose subcubes K for which f >. I.e he subcubes where f is highy concenraed. On each of he remaining K subcubes (hose no in S repea he same procedure, i.e bisec each one ino 2 n sub-subcubes and add hose where f is highy concenraed o S, bisec he res e ceerà... Now for any K S denoe by K is immediae predecessor. Since K S whie K S < f < Vo(K K Vo( K f = Vo(K K Vo(K Vo( K K f 2 n. In summary K S < f 2 n. Denoe F := K S K, G := K 0 \ F F C = K S KC. K We see each poin in G ies in infiniey many nesed cubes wih bounded concenraion of f wih diameers converging o 0: f wih Vo(K i 0. By Lebesgue s Theorem on differeniaion K i he hs f λ a.e (λ denoes Lebesgue s measure on R n, i.e f a.e ( λ a.e on G. In summary, on F we have an average esimae, on G we have a poinwise esimae.

2 The promised proof We resae he desired resu whose proof we gave for p = 2. Theorem. Le f L p (Ω for some < p < and e ω = Nf be he Newonian Poenia of f. Then ω W 2,p (Ω and w = f a.e. and D 2 w Lp (Ω c(n,p,ω f Lp (Ω. For p = 2 we have even R D2 ω 2 = n Ω f 2. Proof. Define an operaor T : L 2 (Ω L 2 (Ω, Tf = D ij Nf. Las ime we showed D ij Nf L2 (Ω = Tf L2 (Ω = f L2 (Ω. In oher words T is srong (2,2 and herefore auomaicay weak (2, 2 i.e ( f L2 (Ω 2 µ Tf ( by he Proposiion in he previous ecure. If we wi now be abe o bound i wih f L (O, he Inerpoaion Theorem wi hen provide he desired bound on D 2 ω for a < p < 2. By duaiy 2 < p < wi hen be aken care of as we (o be made precise. So we Caim. T is weak (, i.e f L 2 (Ω L (Ω µ Tf ( C f L (Ω, > 0. Proof. Exend f riviay ouside Ω (i.e so he exension vanishes on R n \ Ω, and given any fixed > 0 ake a arge enough cube K 0 conaining Ω such ha 2

3 f = K 0 f. Vo(K 0 K 0 The Cube Decomposiion furnishes a counabe number of cubes {K } such ha on each < f 2 n and in addiion f a.e on G := K 0 \ K. Spi f = b + g ino bad, good K f(x on G pars by eing g(x := f on K i.e f coud be osciaing on K, insead we jus repace K i here by is average vaue herein. Then e b := f g, he bad or highy osciaing par. Noe: g 2 n a.e, b(x = 0 on G and K b = 0. We have now Tf = Tb + Tg. And as in he Inerpoaion Theorem of he previous ecure µ Tf ( µ Tb (/2 + µ Tg (/2. We woud ike o bound his wih he L (Ω norm of f. We divide he compuaion ino 3 pars. L (Ω esimae for µ Tg (/2. Using ( on he good par we have µ Tg (/2 ( g L2 (Ω 2 /2 K 0 g 2 (/2 2 and since g/(2 n, ( g/(2 n 2 g /(2 n or ( g/ 2 2 n g / from which 2n+2 g K 0 = 2n+2 + g G K = 2n+2 ( f + f G K K 3

4 = 2n+2 Ω f = 2n+2 f L (Ω. We have no used so far any properies of T. On he bad par we wi, and we wi work wih he kerne of he Newonian Poenia, in jus a momen. L (K 0 \ B esimae for Tb. Le ȳ be he cener of he subcube K. Le B := B(ȳ,δ which n sricy conains K. The diameer of K is δ := diam(ω 2 r if i beongs o he r h subdivision. We conen ourseves wih bounding ony he L norm of Tb on K 0 \ B since by par I of he Proposiion of Lecure 23 (wih p = ha wi bound he disribuion funcion µ Tb isef. Wrie b := I K, he characerisic funcion defined in Lecure 8. b = b. The advanage = of his spiing is ha each erm is compacy suppored unike b isef. Fix some N and approximae b by smooh funcions {b (m } m=0 C0 (K. By varying each wih a consan one can make sure for each n N b (m = b = 0. K K If x K, T(b (m (x = D ij Γ(x yb (m (ydy K [ = Dij Γ(x y D ij Γ(x ȳ ] b (m (ydy K by he zero average b (m. Compuaion. (m Tb (x c δ [dis(x,k ] n+ b (m (y dy. K Proof. Using he above equaion in conjuncion wih he Mean Vaue Theorem of Cacuus here exiss y 0 K (and y y 0 δ y K such ha 4

5 Tb (m (x = DD ij Γ(x y 0 (y y 0 b (m K (ydy DD ij Γ(x y 0 y y 0 b (m (y dy K cδ K x y 0 b(m n+ (y dy cδ [dis(x,k ] n+ b (m (y dy. K This now heps us evauae he L norm Tb (m K 0 \B c δ x ȳ δ ( [dis(x,k ] n+ dx b (m K (y dy. Noe here is some ỹ wih dis(x,k = x ỹ and hen x ȳ x ỹ + ỹ y 0 2dis(x,K 2 n+ c δ x ȳ δ = c b (m K (y dy. ( dx x ȳ n+ b (m K (y dy Le m in he above o ge Tb c b (y dy K 0 \B K i.e we have aken care of hings (have L (Ω esimaes here on K 0 \ B, as can be seen by summing (he b s have disjoin suppors so b = b 5

6 Tb L (K 0 \ B = Tb K 0 \ B = K 0 \B Tb c K b = L ( B esimaes for µ Tb (/2. c b = c f = c f L (Ω. B B µ Tb (/2 = {x Ω : Tb(x > /2} {α K0 \ B : Tb > /2} + B. The firs erm is aken care of (by appying par I of he Proposiion in Lecure 23 wih p = o he esimae above for Tb L (K 0 \ B. For he second, here exiss some consan c such ha B c hence K by he geomery of cubes and bas. Now he K were chosen wih < f, K Vo(K < f L (K. Aogeher µ Tb (/2 c f L (Ω. Combining he above 3 pars f L 2 (Ω µ Tf ( µ Tb (/2 + µ Tg (/2. c f L (Ω. + 2n+ f L (Ω, > 0. Tha is T is weak (, proving he Caim. 6

7 Thus by he Marcinkiewicz Inerpoaion Theorem (MIT exiss c depending on he above consans, i.e on n,p,diam(ω, saisfying f L 2 (Ω Tf Lp (Ω c f Lp (Ω, < p < 2! (2 From he proof of he MIT c bows up as p approaches eiher of he endpoins. We menion wihou proof ha as sronger version of he MIT saes ha if T is srong (r,r and/or srong (q,q hen he consan does no bow-up a r and/or q. Therefore we have in fac < p 2 in (2. As a maer of fac we do no even need o invoke his sronger Theorem since we have done he case p = 2 independeny (wih consan =! in he previous ecure. Ye anoher idea woud be o prove (2 for some one vaue p greaer han 2, and appy he MIT o ge (2 for p = 2 as an inermediae vaue in he inerva (,p! This wi aso concude he proof of our Theorem as p wi be arbirary. To ha end we use he so caed Duaiy Mehod. Le p > 2 be arbirary. (L p (Ω = L q (Ω wih = q +. By he definiion of he dua space (p. 3 of Lecure 7 p Tf Lp (Ω = sup Tf g = g L q (Ω Ω = sup D ij ω g g L q (Ω Ω = = sup ω D ij g g L q (Ω Ω = ( = sup Γ(x yf(ydy D ij g(xdx g L q (Ω Ω Ω = ( = sup Γ(x yd ij g(xf(ydx dy g L q (Ω Ω Ω = ( = sup D ij Γ(x yg(xdx f(ydy g L q (Ω Ω Ω = = sup Tg f g L q (Ω Ω = 7

8 Höder s Ineq. sup f Lp (Ω Tg Lq (Ω g L q (Ω = C sup f Lp (Ω g Lq (Ω g L q (Ω = = C f Lp (Ω. As we wished: T is srong (p,p. In he as inequaiy we simpy used he fac ha < q < 2 is in he range we have aready aken care of. In summary we have shown: If f C0 (Ω, ω := Nf hen ω = f and D 2 ω Lp (Ω c f Lp (Ω for any < p <. Now idenicay o how we finished he proof of he as Theorem in he previous ecure we exend his o a funcions in L p (Ω by approximaing and subsequeny aking imis and making use of Young s Inequaiy (Lecure 23. Our work can be rephrased Coroary. Le Ω R n be a bounded domain and assume u W 2,p 0 (Ω for some < p <. Then D 2 u Lp (Ω c(n,p,ω u Lp (Ω. For p = 2 D 2 u L2 (Ω = u Lp (Ω. Proof. u N( u saisfies Lapace s equaion (u N( u = 0 a.e. In oher words u N( u is a harmonic funcion wih compac suppor in R n hence vanishes idenicay. Hence u = N u and renaming f := u, ω := u gives he inequaiy from our above Theorem. This is quie remarkabe as i es us ha he whoe Hessian, ( n funcions, can be bounded 2 simpy in erms of he sum of is n diagona erms is race. 8

9 Theorem. Le L := a ij D ij +b i D i +c and e Ω R n be a bounded domain. Assume u W 2,p (Ω for some < p < saisfies Lu = f a.e. Assume L sricy eipic wih (a ij > γ I, γ > 0 a ij C 0 (Ω b i,c L (Ω f L p (Ω. Then Ω Ω hods u W 2,p (Ω c ( u Lp (Ω + f Lp (Ω. Proof. We know his for L =, and herefore for any consan coefficiens operaor saisfying he above by Lecure 2. Then perurbing he coefficiens and proceeding jus ike in he Schauder case works, as in Lecure 3, works. Assuming C, boundary, hese esimaes can be exended o hod gobay on a of Ω, as in done in Lecure 4. 9