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