On sufficient conditions for the boundedness of the Hardy-Littlewood maximal operator between weighted Lp-spaces with different weights

P oc. Lond. Ma h. Socie y, (3) 71 (1995), 135–157 On su icien condi ions o he boundedness o he Ha dy-Li lewood maximal ope a o be ween weigh ed Lp–spaces wi h di e en weigh s Ca los P´e ez 1 In oduc ion and main esul s Le Mbe he Ha dy–Li lewood maximal ope a o de ined o locally in eg able unc ions by M (x) = sup x∈Q 1 |Q|ZQ | (y)|dy, whe e he sup emum is aken o e all he cubes con aing x, and le 1 < p < ∞. B. Muckenhoup [10] cha ac e ized he weigh s wsa is ying he weigh ed no m in- equali y ZRn (w(y)M (y))pdy ≤cZRn (w(y) (y))pdy (1) o all nonnega i e unc ions , as hose weigh s sa is ying he Apcondi ion 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ w(y)−p0dy1/p0 ≤c(2) o all cubes Q. I is na u al o conside a simila p oblem o a couple o weigh s (w, ). Howe e , simple examples show (c . [6] p. 395) ha he analogous necessa y condi ion o (w, ) 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ (y)−p0dy1/p0 ≤c, (3) o all cubes Qis no su icien o he boundedness o M om Lp( p) o Lp(wp). E. Sawye has shown in [13], ha he co ec necessa y and su icien condi ion is gi en by 1 ZQ (w(y)M( −p0χQ)(y))pdy ≤cZQ (y)−p0dy, (4) o all cubes Q. E. Sawye ’s condi ion in ol es he ope a o Mi sel , and i is in e es ing o ob ain su icien condi ions close in o m o he necessa y and simple one (3). The i s esul in ha di ec ion was ob ained by C. Neugebaue in [11]. He no iced ha i (w, ) is a couple o weigh s such ha o some > 1 1 |Q|ZQ w(y)p dy1/p 1 |Q|ZQ (y)−p0 dy1/p0 ≤c(5) o all cubes Q, hen ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy (6) o all nonnega i e unc ions . In his pape we ake up his p oblem and show wi h a di e en app oach ha (6) holds assuming e y weak condi ions on he weigh s. We shall see ha i is enough o eplace he a e age no m associa ed o he weigh −1in (3) by a s onge no m de ined in e ms o any Banach unc ion space whose associa ed space sa is ies ce ain mapping p ope y. To be p ecise we le Xbe a Banach unc ion space o e Rnwi h espec o he Lebesgue measu e dx (c . nex sec ion). Gi en a measu able unc ion and any cube Qwe de ine he X–a e age o o e Qby k kX,Q =  τ`(Q)( χQ)  X ,(7) whe e τδ,δ > 0, is he dila ion ope a o τδ (x) = (δx), χEis he cha ac e is ic unc ion o Eand `(Q) is he sideleng h o he cube Q. We de ine a na u al maximal ope a o associa ed o he space X. De ini ion 1.1 Fo each locally in eg able unc ion he maximal ope a o MXis de ined by MX (x) = sup x∈Q k kX,Q , whe e he sup emum is aken o e all he cubes con aining x. 2 Le X=LBbe he O licz space de ined by he Young unc ion B(c . sec ion 2 o [7] [8]). Then he maximal ope a o MX=MBis de ined in e ms o he a e age k kX,Q =k kB,Q = in {λ > 0 : 1 |Q|ZQ B| (y)| λdy ≤1} (c . [1]). I Xis he Lo en z space X=Ls,q, hen he maximal ope a o is MX (x) = Ms,q (x) = sup x∈Q 1 |Q|1/s   χQ s,q (c . [12], [14] and [3]). Gi en a Banach unc ion space X,X0will deno e i s associa e space, which is ano he Banach unc ion space (c . nex sec ion). Theo em 1.2 Le 1< p < ∞, and le Xbe a Banach unc ion space such ha MX0:Lp(Rn)→Lp(Rn). Suppose ha (w, )is a couple o weigh s such ha he e is a posi i e cons an K o which 1 |Q|ZQ w(y)pdy1/p k −1kX,Q ≤K, (8) o all cubes Q. Then ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy (9) o all nonnega i e unc ions . A pa icula example is when X=Lp0 , wi h > 1. In his case he associa e space is X0=L(p0 )0whose co esponding maximal ope a o is gi en by MX0 (x) = M(p0 )0 (x) = sup x∈Q1 |Q|ZQ | (y)|(p0 )0dy1/(p0 )0 , which is bounded on Lp(Rn). On he o he hand, obse e ha when = 1 X0=Lp, whose co esponding maximal unc ion Mpis no bounded on Lp(Rn) since Mi sel ails o be bounded on L1(Rn). 3 Co olla y 1.3 Le 1< p < ∞, and suppose ha (w, )is a couple o weigh s such ha o some > 1, he e is a posi i e cons an K o which 1 |Q|ZQ w(y)pdy1/p 1 |Q|ZQ (y)−p0 dy1/p0 ≤K, (10) o all cubes Q. Then ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy (11) o all nonnega i e unc ions . We can deduce a be e esul using he scale o Lo en z spaces: i X=Lp0 ,∞, hen X0=L(p0 )0,1and MX0is bounded on Lp(Rn) (c . sec ion 6). Hence Co olla y 1.4 Le 1< p < ∞, and 1< < ∞. Suppose ha (w, )is a couple o weigh s such ha he e is a posi i e cons an K o which 1 |Q|ZQ w(y)pdy1/p 1 |Q|1/ p0 χQ −1 L p0,∞≤K, (12) o all cubes Q. Then ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy (13) o all nonnega i e unc ions . Mo e in e es ing examples a e p o ided by he heo y o O licz spaces. Theo em 1.5 Le 1< p < ∞, and le Bbe a doubling Young unc ion such ha Z∞ c p0 B( )p−1d <∞,(14) o some posi i e cons an c. i)Le (w, )be a couple o weigh s such ha he e is a posi i e cons an K o which 1 |Q|ZQ w(y)pdy1/p k −1kB,Q ≤K, (15) 4 o all cubes Q. Then ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy (16) o all nonnega i e unc ions . ii) Condi ion (14) is also a necessa y condi ion. Tha is, suppose ha Bhas he p ope y ha ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy o all nonnega i e unc ions , whene e he couple o weigh s (w, )sa is ies 1 |Q|ZQ w(y)pdy1/p k −1kB,Q ≤K, o all cubes Q. Then Bsa is ies (14). Pa icula examples a e gi en by B( )≈ p0logp0−1+δ(1 + ), o he weake one B( )≈ p0logp0−1(1 + )[log log(1 + )]p0−1+δ, wi h δ > 0. The key ac is he boundedness o M¯ Bon Lp(Rn), and he ele an class o Young unc ions is he ollowing. De ini ion 1.6 Le 1< p < ∞. We say ha a doubling Young unc ion Bsa is ies he Bpcondi ion i he e is a posi i e cons an csuch ha Z∞ c B( ) p d ≈Z∞ c p0 ¯ B( )p−1d <∞. Then we ha e he ollowing cha ac e iza ion. Theo em 1.7 Le 1<p<∞. Suppose ha Bis a Young unc ion. Then he ollowing a e equi alen . i) B∈Bp; (17) 5 ii) he e is a cons an csuch ha ZRn MB (y)pdy ≤cZRn (y)pdy (18) o all nonnega i e unc ions ; iii) he e is a cons an csuch ha ZRn MB (y)pw(y)dy ≤cZRn (y)pMw(y)dy (19) o all nonnega i e unc ions and w; i ) he e is a cons an csuch ha ZRn M (y)pw(y) [M¯ B(u1/p)(y)]pdy ≤cZRn (y)pMw(y) u(y)dy, (20) o all nonnega i e unc ions ,wand u. A consequence is he ollowing inequali y: Co olla y 1.8 Le 1< p < ∞. Suppose ha wis a weigh . Then ZRn M (y)pM[p0]+1w(y)1−pdy ≤cZRn (y)pw(y)1−pdy (21) o all nonnega i e unc ions . As usual [ ] deno es he in ege pa o . Acnowledgemen . We wish o hank he e e ee o he ca e ul eading o his pape . 2 P elimina ies In his sec ion we p o ide he necessa y backg ound om he heo y o unc ion spaces ha will be used la e . We begin by ecalling some basic ac s abou he heo y o Banach unc ion spaces in oduced by W.A.J. Luxembu g in [9], and we shall e e he eade o [2] o a comple e accoun . Le (R, µ) be a measu e space, and le M+(R) be he cone o µ–measu able unc ions on Rwhose alues lie in [0,∞]. A mapping ρ:M+(R)→[0,∞] is called a Banach unc ion no m i , o all , g, n,(n= 1,2,3, . . .) in M+(R), o all cons an s a≥0, and o all µ–measu able subse s Eo R, he ollowing p ope ies hold: 6 i) ρ( ) = 0 i = 0 µ–a.e.; ρ(a ) = aρ( ); ρ( +g)≤ρ( ) + ρ(g) ii) 0 ≤g≤ µ–a.e. implies ρ(g)≤ρ( ) iii) 0 ≤ n↑ µ–a.e. implies ρ( n)↑ρ( ) i ) µ(E)<∞implies ρ(χE)<∞ ) µ(E)<∞implies RE dµ ≤CEρ( ), o some cons an CE, 0 < CE<∞, depending on Eand ρbu independen o . Le M(R) deno e he collec ion o all µ–measu able unc ions on R. The collec ion X=X(ρ) o all unc ions ∈M(R) o which ρ(| |) = k kX<∞is called a Banach unc ion space. The mos impo an p ope y o he Banach unc ion spaces is he gene alized H¨olde inequali y ZR | (y)g(y)|dµ(y)≤ k kXkgkX0,(22) whe e X0is he associa e space o X. A Banach unc ion space Xis said o be ea angemen –in a ian i whene e , g ∈Xa e equimeasu able, hen k kX=kgkX. Recall ha wo unc ions a e equimeasu able i µ ( ) = µg( ), > 0, whe e µ ( ) = µ{x∈R:| (x)|> },is he dis ibu ion o . Mos o he p ope ies o he ea angemen –in a ian spaces can be o mula ed in e ms o he undamen al unc ion o X,ϕX, gi en by ϕX( ) = kχEkX, whe e µ(E) = . Obse e ha he pa icula choice o he se Ewi h µ(E) = is imma e ial by he ea angemen –in a iance o X.ϕXis quasiconca e and con inuous, excep pe haps a he o igin. Fu he mo e, i X0is he associa ed space o X he ollowing iden i y holds ϕX( )ϕX0( ) = , > 0.(23) Examples o ea angemen –in a ian spaces include he Lebesgue Lpspaces, he minimal and maximal Lo en z spaces Λ, M, (c . [2]). Also, he O licz and Ls,q spaces ha we a e going o desc ibe b ie ly nex . A unc ion B: [0,∞)→[0,∞) is a Young unc ion i i is con inuous, con ex and inc easing sa is ying B(0) = 0 and B( )→ ∞ as → ∞. We shall assume ha Bis no malized so ha B(1) = 1. Also, we shall equi e ha Bsa is ies he doubling condi ion 7 B(2 )≤C B( ), > k (24) o some cons an s C > 0, k≥0. We shall make use o he ollowing p ope y B( )≈ B0( ), > 0,(25) and ha →B( ) is inc easing. Each Young unc ion Bhas associa ed a complemen a y Young unc ion ¯ B ha sa is ies ≤B−1( )¯ B−1( )≤2 , > 0.(26) Le (X, µ) be a measu e space and le Bbe a Young unc ion. The O licz space LB(µ) consis s o all µ–measu able unc ions such ha ZX B| (y)| λdµ(y)<∞, o some λ > 0. LB(µ) can be no med by he Luxembu g no m de ined by k kB,µ = in {λ > 0 : ZX B| (y)| λdµ(y)≤1}. LB(µ) is a ea angemen –in a ian space wi h undamen al unc ion gi en by ϕB( ) = ϕLB(µ)( ) = 1 B−1(1 ).(27) In pa icula i Eis a measu able subse o X, hen kχEkB,µ =1 B−1(1 µ(E)).(28) Finally, he associa ed space o LB(µ) is L¯ B(µ). A unc ion belongs o he Lo en z space Ls,q, 0 < s, q ≤ ∞, i k kLs,q(µ)=qZ∞ 0 µ{x∈Rn:| (x)|> }1/sqd 1/q <∞, whene e q < ∞, and sup 0< <∞ µ{x∈Rn:| (x)|> }1/s <∞, 8 i q=∞. Fo each 1 < s, q ≤ ∞ Ls,q is a ea angemen –in a ian Banach unc ion space wi h undamen al unc ion ϕ( ) = 1/s, and associa ed space Ls0,q0. 3 The gene al case Le Xbe a Banach unc ion space o e Rnwi h espec o he Lebesgue measu e. Recall ha o any measu able unc ion and a bi a y cube Qwe de ined he X–a e age o o e Qby k kX,Q =  τ`(Q)( χQ)  X ,(29) whe e τδ,δ > 0, is he dila ion ope a o τδ (x) = (δx), χEis cha ac e is ic unc ion o Eand `(Q) is he sideleng h o he cube Q. No e ha H¨olde ’s inequali y o Banach unc ion spaces (22) yields a e he change o a iable y=`(Q)z 1 |Q|ZQ (y)g(y)dy ≤ k kX,Q kgkX0,Q .(30) We also in oduce he ollowing maximal ope a o associa ed o he space X. Fo each locally in eg able unc ion we ha e also de ined MXby MX (x) = sup x∈Q k kX,Q , whe e he sup emum is aken o e all he cubes con aining x. P oo o Theo em 1.2: Since he se o bounded unc ions wi h compac suppo is dense in Lp( p) i is enough o show ha he e is a cons an csuch ha ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy, (31) o each nonnega i e bounded unc ion wi h compac suppo . Fo each in ege k, and o any cons an a > 2nwe le Ωkand Dkbe he se s Ωk={x∈Rn:ak< M (x)}, 9 =CZRn (y)pdy Z∞ 1/2 B( ) p d =CZRn (y)pdy, since B∈Bp. This p o es ha i) implies ii). Fo he p oo ha ii) implies iii) we disc e ize as in Theo em 1.2. We ix a cons an a > 2n, and o each in ege kwe le Ωk, and Dkbe he se s Ωk={x∈Rn:MB (x)> ak}, Dk={x∈Rn:Md B (x)>ak 4n}. He e Md Bdeno es he dyadic e sion o MB. Hence, by Lemma 4.1 wi h =ak he e is a amily o maximal nono e lapping dyadic cubes {Qk,j} o which Ωk⊂ ∪j3Qk,j, Dk=∪jQk,j, and ak 4n<k kB,Qk,j ≤ak 2n.(47) We shall need he ollowing lemma. Lemma 4.2 Suppose a > 2n. Fo all in ege s k, j we le Ek,j =Qk,j −Qk,j ∩Dk+1. Then {Ek,j}is a disjoin amily o se s which sa is y |Qk,j ∩Dk+1|<2n a|Qk,j|,(48) and |Qk,j|<1 1−2n a |Ek,j|.(49) We pos pone he p oo o his also un il he end o he p oo o he heo em. Now, using (47), and (49) we es ima e he le side o (19) as in he p oo o Theo em 1.2 by ZRn MB (y)pw(y)dy =X kZΩk−Ωk+1 MB (y)pw(y)dy ≤(50) ≤apX k akpw(Ωk)≤CX k,j akpw(3Qk,j)≤ ≤CX k,j k kp B,Qk,j w(3Qk,j) = CX k,j k kp B,Qk,j w(3Qk,j) |3Qk,j||Qk,j| ≤ 16 ≤CX k,j      w(3Qk,j ) |3Qk,j|1/p     p B,Qk,j |Ek,j| ≤ ≤CX k,j ZEk,j MB( (Mw)1/p)(y)pdy ≤CZRn MB( (Mw)1/p)(y)pdy ≤ ≤CZRn (y)pMw(y)dy, since we a e assuming ii). This p o es iii). Le us assume ha iii) holds. Obse ing ha (20) is equi alen wi h ZRn M( g)(y)pw(y) [M¯ B(g)(y)]pdy ≤cZRn (y)pMw(y)dy, o all nonnega i e unc ions ,g, and w, i ) ollows immedia ely om (19) a e an applica ion o he inequali y M( g)(y)≤MB (y)M¯ Bg(y), y ∈Rn, which is a consequence o he gene alized H¨olde ’s inequali y (30). To p o e ha i ) implies i) we le w= 1 in (20) ob aining ZRn M (y)p1 [M¯ B(u1/p)(y)]pdy ≤cZRn (y)p1 u(y)dy, o all nonnega i e unc ions , and u. Since his is (36) in Theo em 3.1 wi h X=L¯ B, we can apply ha p oposi ion o ge a cons an c > 0 o which Zc 0 ϕB( )p d <∞.(51) He e ϕB=ϕLBis he undamen al unc ion o LB. We claim ha (51) is equi alen wi h B∈Bp. Indeed, by (27) and (25) i eadily ollows ha Zc 0 ϕB( )p d =Zc 0 1 B−1(1 )p 1 d ≈(52) ≈Z∞ c B( ) p d , 17 om which we ob ain he claim. This concludes he p oo o he Theo em sa e o he p oo s o Lemmas 4.1 and 4.2. P oo o Lemma 4.1: The p oo is a simple adap a ion o a gumen s in [6] Ch. 2. Since is bounded wi h compac suppo , say supp ⊂K, k kB,Q ≤ k kL∞kχKkB,Q = =k kL∞ 1 B−1|Q| |Q∩K|, and i ollows ha k kB,Q →0 as Q↑Rn. Hence, i he e a e any dyadic cubes Qwi h k kB,Q > , hey a e con ained in cubes o his ype which a e maximal wi h espec o inclusion. We le C ={Pj}be he amily o he dyadic maximal nono e lapping cubes sa is ying < k kB,Pj. Le P0 jbe he only dyadic cube con aining Pjwi h sideleng h wice ha o Pj. Then < k kB,Pj≤2nk kB,P 0 j . The las inequali y can easily be deduced om he de ini ion o he Luxembu g no m using he ac ha →B( ) is inc easing. Hence by he maximali y o he cubes {Pj}we ge < k kB,Pj≤2n . (53) Obse e ha om his discussion i is clea ha {y∈Rn:Md B (y)> }=∪jPj.(54) Le x∈Ω . By de ini ion, he e is a cube Rcon aining xsuch ha < k kB,R .(55) Le kbe he unique in ege such ha 2−(k+1)n<|R| ≤ 2−kn. The e is some dyadic cube wi h side leng h 2−k, and a mos 2no hem, {Ji:i= 1, . . . , n}, mee ing he in e io o R. I is easy o see ha o one o hese cubes, say J1, 2n< χJ1  B,R .(56) 18 This can be seen as ollows. I o each i= 1,...,2nwe had   χJi   B,R ≤ 2n, we would ge since R⊂ ∪2n i=1Ji ha k kB,R =  χ∪2n i=1Ji   B,R ≤ ≤ 2n X i=1   χJi   B,R ≤2n 2n= , con adic ing (55). Using ha |R| ≤ |J1|<2n|R|one can also show 4n<k kB,J1.(57) By le ing C /(4)n={Qj}, we ha e by (53) ha 4n<k kB,Qj≤ 2n,(58) o each j, yielding (44). (46) also ollows since {y∈Rn:Md B (y)> 4n}=∪jQj. Also, we see om (57) ha J1⊂Qk, o some k, and hen R⊂3J1⊂3Qk. This gi es Ω ⊂ ∪j3Qj, which is (43). Now, by he le side o he inequali y (58), and he de ini ion o k kB,Q we ge |Ω | ≤ CX j |Qj| ≤ ≤CX jZQj B4n (y) dy ≤CZRn B (y) dy. (59) To ob ain (45) we jus use he s anda d idea o w i ing as = 1+ 2, whe e 1(x) = (x) i (x)> 2, and 1(x) = 0 o he wise. Then MB (x)≤MB 1(x) + MB 2(x)≤MB 1(x) + 2. Finally, since (59) holds o each ≥0, > 0 we ha e |Ω | ≤ {y∈Rn:MB 1(y)> 2}≤CZRn B 1(y) dy = 19 =CZ{y∈Rn: (y)> /2} B (y) dy, concluding he p oo o Lemma 4.1. 2 We now conclude he p oo o he Theo em by p o ing Lemma 4.2. P oo o Lemma 4.2: The amily Ek,j is clea ly disjoin . We no e ha (47) and he de ini ion o he Luxembu g no m implies ha 1<1 |Qk,j|ZQk,j B4n ak (y)dy, and 1 |Qk,j|ZQk,j B2n ak (y)dy ≤1. Hence by s anda d p ope ies o he dyadic cubes we can es ima e wha po ion o Qk,j is co e ed by Dk+1 as in [4] (c . [6] p. 398) |Qk,j ∩Dk+1| |Qk,j|=X i |Qk,j ∩Qk+1,i| |Qk,j|= =X i:Qk+1,i⊂Qk,j |Qk+1,i| |Qk,j|< <X i:Qk+1,i⊂Qk,j 1 |Qk,j|ZQk+1,i B4n ak+1 (y)dy ≤ ≤2n a 1 |Qk,j|ZQk,j ∩∪iQk+1,i B2n ak (y)dy ≤ ≤2n a. He e we ha e used ha B(2n a )≤2n aB( ), > 0, since 2n a<1, and because →B( ) is inc easing. This gi es (48). Finally |Ek,j| |Qk,j|>1−2n a>0, comple ing he p oo o he Lemma and hence ha o Theo em 1.7. 2 20 4.2 P oo o Co olla y 1.8 I we le w= 1 and uis eplaced by wp−1in (20) we ha e he weigh ed inequali y ZRn M (y)p1 [M¯ B(w(p−1)/p)(y)]pdy ≤cZRn (y)p1 w(y)p−1dy, when B∈Bp. Le δ= [p0]−p0+ 1 >0, and ake B( )≈ p log1+δ(1+ ). Then ¯ B( )≈ p0log[p0](1+ ) and [M¯ B(w(p−1)/p)(y)]p= [MA(w)(y)]p−1, whe e A( )≈ log[p0](1+ ). Then Co olla y 1.8 will ollow i we p o e he poin wise inequali y MAw(x)≤C M[p0]+1w(x). I is enough o p o e ha he e is a cons an Csuch ha o each cube Q k kA,Q ≤C |Q|ZQ M[p0] (y)dy. By homogenei y we can assume ha he igh hand side is equal o C. Then, by he de ini ion o he Luxembu g no m we need o p o e 1 |Q|ZQ A(w(y)) dy =1 |Q|ZQ w(y) log[p0](1 + w(y)) dy ≤C. Bu his is a consequence o i e a ing he ollowing inequali y o E.M. S ein [15] ZQ w(y) logk(1 + w(y)) dy ≤CZQ Mw(y) logk−1(1 + Mw(y)) dy, (60) wi h k= 1,2,3,· · · . 2 4.3 Some u he conside a ions abou he class Bp We obse e ha 1 < p < q < ∞implies ha Bp⊂Bq. A ypical Young unc ion ha belongs o he class Bpis B( ) = swi h 1 ≤s<p. Ano he mo e in e es ing example is he unc ion Bgi en by B( )≈ p log1+δ(1 + ), 21 o B( )≈ p log(1 + ) [log log(1 + )]1+δ, wi h δ > 0. Since he unc ion B( ) = sbelongs o Bpwe ha e 1 < s < p and so implies ha B∈Bp−wi h 0 <  < p −s, one could hink ha he same p ope y would hold o any Young unc ion in Bp. Howe e , his is alse as he ollowing example shows. Fo δ > 0, conside he example men ioned abo e B( )≈ p log1+δ(1 + ). Then, B∈Bp, bu i can be easily shown ha he e is no  > 0 o which B∈Bp−. We can emedy his si ua ion i we es ic a en ion o hose Young unc ions ha a e submul iplica i e. We say ha he Young unc ion Bis submul iplica i e i B( s)≤B( )B(s) o each , s > 0. Lemma 4.3 Le 1< p < ∞. Assume ha Bis a submul iplica i e Young unc ion such ha B∈Bp. Then he e exis s  > 0 o which B∈Bp−. P oo : This is a simple consequence o he ac ha B∈Bpi and only ¯α(B)< p. (C . o ins ance [2] Ch. 5.) He e ¯α(B) deno es ¯α(B) = lim →∞ logB( ) log =in >1 logB( ) log , and i can be shown ha he limi exis s, is ini e, and s ic ly posi i e. 2 Le us make he ollowing obse a ion conce ning a pa icula case o (20). Tak- ing he weigh w= 1, inequali y (20) becomes ZRn M (y)p1 [M¯ B(u1/p)(y)]pdy ≤cZRn (y)p1 u(y)dy, (61) 22 o all nonnega i e unc ions , and u. Le 1 < < ∞, and conside B( ) = (p0 )0. Then B∈Bp, and (61) is ZRn M (y)p1 [M(u (p0−1))(y)](p−1)/ dy ≤cZRn (y)p1 u(y)dy. Howe e , his es ima e ollows om well–known esul s. Indeed, i is enough o show ha [M¯ B(u1/p)(y)]−pis an Apweigh by he heo em o Muckenhoup and he Lebesgue di e en ia ion heo em. Now, ecall ha a weigh wbelongs o Api and only i w=w1w1−p 2whe e w1and w2a e A1weigh s, and ha (Mg)δ∈A1 0< δ < 1 (see [6] p. 436). Then i is clea ha [M¯ B(u1/p)(y)]−p= [M(u (p0−1))(y)1/ ](1−p) is an Apweigh . This a gumen may sugges ha [M¯ B(u1/p)(y)]−psa is ies he Apcondi ion o each B∈Bp. Howe e , he ollowing example indica es ha his is no ue in gene al, and hus abo e a gumen is no sha p enough o ge (61). Le 1 < p < ∞,δ > 0, and le Bbe he Young unc ion such ha ¯ B( )≈ p0logp0−1+δ(1 + ). Then B∈Bpbu w=M¯ B(χQ(0,1) )−p6∈ Ap. O he wise he e would exis  > 0 such ha w∈Ap−(c . [6] p. 399.) Hence o each M > 0 we would ha e Z|y|>M w(y) |y|n(p−)dy < ∞.(62) (c . [6] p. 412.) Howe e (39), and (27) yield w(y)≈¯ B−1(b|y|n)p,|y|> a, o some posi i e dimensional cons an a, b. Thus using pola coo dina es and (25) Z|y|>a w(y) |y|n(p−)dy ≈Z|y|>a ¯ B−1(b|y|n)p |y|n(p−)dy ≈Z∞ a1 ¯ B−1( )p p−1− d ≈Z∞ a2 p ¯ B( )p−1− d ≈Z∞ a2 p0 log1+δ(p−1)−(p0−1+δ)( ) d =∞, con adic ing (62). 23 5 O licz spaces and wo-weigh inequali ies P oo o Theo em 1.5 Pa i) ollows immedia ely om Theo em 1.2 oge he wi h Theo em 1.7. Indeed, le X=LBwi h associa e space X0=L¯ B. Then he hypo hesis MX0=M¯ B:Lp(Rn)→Lp(Rn) in Theo em 1.2 is equi alen wi h ¯ B∈Bp by Theo em 1.7. Pa ii) ollows om P oposi ion 3.2 and he compu a ion in (52). 2 As an easy consequence o his heo em we can ob ain su icien condi ions much in he spi i o [5]. Co olla y 5.1 Le 1<p<∞. Suppose ha ϕ: (0,∞)→(0,∞)is inc easing, ha ϕ(2 )≤Cϕ( ), > 0, and ha o some posi i e cons an c Z∞ c 1 ϕ( )p−1 d <∞.(63) Assume ha (w, )is a couple o weigh s which sa is ies o some posi i e cons an K 1 |Q|ZQ w(y)pdy1/p "1 |Q|ZQ (y)−p0ϕ wp(Q) |Q|1/p (y)−1!dy#1/p0 ≤K, (64) o all cubes Q. Then ZRn (w(y)M (y))pdy ≤cZRn ( (y) (y))pdy, (65) o all nonnega i e unc ions . P oo : Conside he Young unc ion de ined by B( )≈ p0ϕ( ) Kp0which sa is ies (14) (i.e. ha ¯ B∈Bp). Now, (64) implies 1 |Q|ZQ B wp(Q) |Q|1/p (y)−1!dy ≤1, o equi alen ly     wp(Q) |Q|1/p −1    B,Q ≤1. 24 F om his and by homogenei y we ge (15). Then Theo em 1.5 applies. 2 6 Lo en z spaces and wo-weigh inequali ies In his sec ion we s udy wo weigh ed no m inequali ies o he Ha dy–Li lewood maximal ope a o whene e he weigh s sa is y (8) wi h Xbeing a Lo en z space. We begin by ecalling ha he maximal ope a o associa ed o he Lo en z space Ls,q is gi en by Ms,q (x) = sup x∈Q 1 |Q|1/s   χQ s,q. We now s a e a esul simila o Theo em 1.7. Theo em 6.1 Le 1< p, s < ∞, and 1≤q < ∞. Then he ollowing a e equi a- len . i) s < p ; (66) ii) he e is a cons an csuch ha ZRn Ms,q (y)pdy ≤cZRn (y)pdy (67) o all nonnega i e unc ions ; iii) he e is a cons an csuch ha ZRn Ms,q (y)pw(y)dy ≤cZRn (y)pMw(y)dy (68) o all nonnega i e unc ions , and w; i ) he e is a cons an csuch ha ZRn M (y)pw(y) [Ms0,q0(u1/p)(y)]pdy ≤cZRn (y)pMw(y) u(y)dy, (69) o all nonnega i e unc ions ,wand u. 25