Publ. Ma . 45 (2001), 175–198
THE LEVEL FUNCTION IN REARRANGEMENT
INVARIANT SPACES
Go d Sinnamon
Abs ac
An exac exp ession o he down no m is gi en in e ms o he
le el unc ion on all ea angemen in a ian spaces and a use-
ul app oxima e exp ession is gi en o he down no m on all e-
a angemen in a ian spaces whose uppe Boyd index is no one.
1. In oduc ion
Le λbe a measu e on Rand ake X o be a ea angemen in a i-
an space o λ-measu able unc ions. We define he down no m o a
λ-measu able unc ion o be
 X↓= sup R
| |gdλ:g≥0,gnon-inc easing,gX≤1.(1.1)
Had we aken he sup emum o e all gin he uni ball o X, he associa e
space o he Banach unc ion space X, we would ha e eco e ed he no m
o in Xso i is immedia e ha
 X↓≤ X.
The significance o he down no m is ha he inequali y
R
gdλ ≤ X↓gX
(1.2)
holds o all and all non-nega i e, non-inc easing unc ions g. Since
he down no m o is smalle han he no m o in X his es ima e
uses he mono onici y hypo hesis on g o imp o e he usual es ima e
R
gdλ ≤ XgX.
2000 Ma hema ics Subjec Classifica ion. P ima y: 46E30; Seconda y: 26D15.
Key wo ds. Le el unc ion, ea angemen in a ian , down no m.
Suppo om he Na u al Sciences and Enginee ing Resea ch Council o Canada is
g a e ully acknowledged.
176 G. Sinnamon
To use inequali y (1.2) effec i ely i is necessa y o unde s and he
down no m. In he case ha Xis Lebesgue space his has been done in
wo ways. Halpe in [6] and Lo en z [9] ga e an exac exp ession o he
down no m when λis a non-nega i e weigh unc ion imes Lebesgue
measu e. Gi en a unc ion hey cons uc ed a ela ed unc ion ◦,
called he le el unc ion o , and showed ha he down no m is p e-
cisely he no m o ◦in X.In[13], he le el unc ion cons uc ion was
ex ended o gene al ( egula ) measu es on R, he down no m was shown
o define a Banach space and he dual space was also cons uc ed.
The second app oach o unde s anding he down no m, gi en in [12]
o weigh ed Lebesgue spaces, was o gi e an equi alen no m in a mo e
ac able o m. The no m in Xo a ce ain a e aging ope a o applied
o was shown o be equi alen o he down no m o . The loss o
exac ness is mo e han made up o because he a e aging ope a o is
linea . This app oach was ex ended o O licz spaces wi h weigh s in [8]
whe e he down no m in sequence spaces was also conside ed.
Ou objec is o look a bo h o hese app oaches in he mo e gen-
e al se ing o ea angemen in a ian spaces. As o en happens when
heo ems a e examined in hei na u al gene ali y, he p oo s educe o
hei essen ial ea u es and g ea e unde s anding is gained. We will see
how he a e aging ope a o in ol ed in he esul s o [12] and [8] a ises
na u ally om he le el unc ion cons uc ion and how he fini eness o
λ(R) affec s ha ope a o . We will also see why he le el unc ion ap-
p oach o he down no m emains alid in all ea angemen in a ian
spaces while a es ic ion is equi ed o he o he app oach o be alid.
In [12] he es ic ion was ha he Lebesgue index be g ea e han one
and in [8]a∆
2condi ion was imposed on he N- unc ions defining he
O licz spaces.
Fo defini ions and no a ion in ol ing Banach unc ion spaces and
ea angemen in a ian spaces we e e o [1]. We adop he con en ion
ha 0·∞ =∞/∞=0. I Aand Ba e exp essions in ol ing , we w i e
A≈B o mean ha he e exis s a posi i e cons an C, no depending on
, such ha C−1A≤B≤CA. The ange o in eg a ion o an in eg al
gi en wi h limi s is aken o be he closed in e al so ha
b
a
dλ=[a,b]
dλ bu b
−∞
dλ=(−∞,b]
dλ.
2. The down no m
Le Xbe a ea angemen in a ian space o e he measu e spa-
ce (R,λ). Fo he down no m o be in e es ing some es ic ions on
The Le el Func ion in R.I. Spaces 177
λa e in o de . Since we wan non-nega i e, non-inc easing unc ions o
be λ-measu able we assume ha se s o he o m (−∞,x] and (−∞,x)
a e λ-measu able which means ha all Bo el se s a e λ-measu able. To
ensu e ha he space Xac ually con ains non- i ial non-nega i e, non-
inc easing unc ions we assume ha o each x,λ(−∞,x]<∞.Fo
echnical easons he measu e λin [13] was assumed o be egula and
since we wish o apply hose esul s we make he same assump ion he e.
Finally, in wo king wi h ea angemen in a ian spaces i is usual o as-
sume ha he unde lying measu e space is esonan [1, Defini ion II.2.3
and Theo em II.2.7] so ha , among o he hings, he associa e space will
also be ea angemen in a ian . Fo hese easons we assume hence o h
ha
λis a egula Bo el measu e on R,
λ(−∞,x]<∞ o all x∈R,and
λis nona omic o else comple ely a omic
wi h all a oms ha ing measu e 1.
(2.1)
Wi h hese assump ions on he measu e λwe can show ha X↓, he
collec ion o unc ions sa is ying  X↓<∞, is a no med ec o space:
I is easy o see ha X↓is a ec o space con aining Xand i is clea
om (1.1) ha ·
X↓is non-nega i e, homogeneous, and sa isfies he
iangle inequali y. I emains o show ha only he ze o unc ion has
ze o no m in X↓. Fo each x∈R,λ(−∞,x]<∞so χ(−∞,x]∈Xand
hence, i  X↓= 0 we ha e x
−∞ dλ= 0 o each x. I ollows ha
=0λ-almos e e ywhe e and we ha e shown ha ·
X↓is a no m.
In ac , X↓is a Banach space as we show in Theo em 5.3.
In mos applica ions, he measu e λis weigh ed Lebesgue measu e on
he hal line, dλ(x)=w(x)dx wi h wa non-nega i e, locally in eg able
unc ion on [0,∞). The case ha λis coun ing measu e on he posi i e
in ege s also a ises. The ea angemen in a ian spaces o he la e
measu e include lpand O licz sequence spaces while hose o he o me
measu e a e weigh ed Lebesgue spaces, O licz spaces, Lo en z spaces
and o he s. We has en o poin ou ha while he weigh ed Lebesgue
space Lp
w[0,∞) is no ea angemen in a ian wi h espec o Lebesgue
measu e unless w≡1, i is ea angemen in a ian wi h espec o he
measu e w(x)dx.
We plan o use he le el unc ion o ela e he no m in he space X↓
o he no m on he o iginal space X. The nex p oposi ion in oduces
he le el unc ion as cons uc ed in [13]. Fo con enience we define B
178 G. Sinnamon
o be he collec ion o λ-measu able unc ions on Rwhich a e bounded
and suppo ed in a se o he o m (−∞,M] o some M∈R.
P oposi ion 2.1. Suppose λsa isfies (2.1) and ∈B. Then he e is a
non-nega i e, non-inc easing unc ion ◦∈B, called he le el unc ion
o wi h espec o λ, and ha ing he ollowing p ope ies.
(a) The e exis s a fini e o coun able collec ion o disjoin in e als Ii
o fini e, non-ze o λmeasu e such ha = ◦λ-almos e e ywhe e
on E=R ∪
iIiand o each i,
◦(x)=(1/λIi)Ii
| |dλ
o λ-almos e e y x∈Ii.
(b) I gis non-nega i e and non-inc easing hen
R]
| |gdλ≤R
◦g dλ.
(c) I 1,
2∈Band | 1|≤| 2| hen ◦
1≤ ◦
2.
P oo : The s uc u e o he le el unc ion o is gi en in [13, Theo-
em 4.4, Defini ion 4.6, Co olla y 4.8 and Theo em 4.9]. The e i is
shown ha ◦is non-nega i e and non-inc easing and ha (a) holds. I
is no assumed in [13] ha is suppo ed on (−∞,M] so he possibil-
i y o a le el in e al o infini e λmeasu e is conside ed he e. An easy
a gumen shows ha i is suppo ed on (−∞,M] hen all he le el
in e als Iia e con ained in (−∞,M] and hence a e o fini e λmeasu e.
Clea ly we may disca d hose o ze o λ-measu e.
Pa (b) is gi en in [13, Theo em 4.11] and (c) is p o ed in [13,
Theo em 5.2].
The main esul o his sec ion is gi en in he nex heo em o ∈B
and in Co olla y 2.4 o gene al .
Theo em 2.2. Suppose λsa isfies (2.1),Xis a ea angemen in a i-
an space o e (R,λ), ∈B, and ◦is he le el unc ion o wi h
espec o λ. Then  X↓= ◦X.
P oo : We use he le el in e als o o define he ope a o A .
A h=hχE+
i1
λIiIi
hdλ
χIi.
No e ha A is sel -adjoin , ha is R(A g)hdλ =Rg(A h)dλ o
app op ia e gand h. Also no e ha by [1, Theo em II.4.8] A is a con-
ac ion on any ea angemen in a ian space, in pa icula A hX≤
The Le el Func ion in R.I. Spaces 179
hX o h∈X. I is clea om he defini ion ha A | |= ◦and
since he se s Iia e in e als, A his non-nega i e and non-inc easing
whene e his.
I gis non-nega i e and non-inc easing and gX≤1 hen by P opo-
si ion 2.1(b),
R
| |gdλ≤R
◦gdλ≤ ◦X
so we ha e  X↓≤ ◦X. Now we p o e he e e se inequali y. A
simple limi ing a gumen shows ha
 ◦X= sup R
◦hdλ
whe e he sup emum is aken o e non-nega i e unc ions h∈Bsa -
is ying hX≤1. Fo such an h, since ◦is non-nega i e and non-
inc easing, we ha e
R
◦hdλ≤R
◦h◦dλ =R
(A | |)(Ahh)dλ =R
| |(A (Ahh)) dλ.
Se g=A (Ahh). Since Ahh=h◦is non-nega i e and non-inc easing
we see ha gis non-nega i e and non-inc easing. Mo eo e , gX≤
hX≤1 since bo h A and Aha e con ac ions on X. We conclude
ha
 ◦X≤sup R
| |gdλ:g≥0,gnon-inc easing,gX≤1= X↓.
This comple es he p oo .
To ex end he defini ion o he le el unc ion om ∈B o all λ-mea-
su able unc ions we use P oposi ion 2.1(c).
Defini ion 2.3. I is a λ-measu able unc ion le
n= min(| |,n)χ(−∞,n]and se ◦= limn→∞ ◦
n.
Clea ly n∈B o each nso ◦
nis defined. By P oposi ion 2.1(c),
{ ◦
n}is a non-dec easing sequence so he limi in Defini ion 2.3 always
exis s as a unc ion which akes alues in [0,∞]. Mo eo e , i ∈B
hen n= o sufficien ly la ge nso he new defini ion o ◦ag ees
wi h he o iginal one.
I is immedia e ha , wi h his defini ion o he le el unc ion, P opo-
si ion 2.1(c) emains alid o a bi a y unc ions. An applica ion o
he Mono one Con e gence Theo em shows ha Pa (b) also ex ends.
Pa (a) does no hold o a bi a y unc ions because he igh mos le el
in e al may ha e infini e λmeasu e. To wha ex en he s uc u e o

180 G. Sinnamon
◦ o an a bi a y unc ion can be desc ibed in e ms o le el in e als
is no clea .
The sequence n= min(| |,n)χ(−∞,n]in Defini ion 2.3 is chosen o
con enience, in Sec ion 5 we show ha he defini ion o ◦is independen
o he app oxima ing sequence.
Co olla y 2.4. Suppose λsa isfies (2.1) and Xis a ea angemen in-
a ian space o e (R,λ).I ∈X↓ hen ◦is fini e λ-almos e e y-
whe e, belongs o X, and  X↓= ◦X.
P oo : Since n= min(| |,n)χ(−∞,n]is a non-dec easing sequence, P o-
posi ion 2.1(c) shows ha ◦
nis also. The Fa ou p ope y o he Banach
unc ion space X, Theo em 2.2, and he obse a ion ha n≤| |show
ha
 ◦X= lim
n→∞  ◦
nX= lim
n→∞  nX↓≤ X↓.
Now [1, Lemma I.1.5(i) and Theo em I.1.4] show ha ◦∈Xand is
he e o e fini e λ-almos e e ywhe e.
Fo each non-nega i e, non-inc easing unc ion gwi h gX≤1we
ha e, by he Mono one Con e gence Theo em and P oposi ion 2.1(b),
R
| |gdλ= lim
n→∞ R
ngdλ≤lim
n→∞ R
◦
ngdλ≤lim
n→∞  ◦
nX= ◦X.
Taking he sup emum o e all such gyields  X↓≤ ◦Xand com-
ple es he p oo .
3. An equi alen no m
Exp essing he down no m o a unc ion in e ms o he le el unc-
ion o , al hough exac , has a majo d awback. The map → ◦
is no linea , in ac , i is no e en sublinea . In he Lebesgue space
case i was shown in [12] ha he space X↓has an equi alen no m
which can be exp essed in e ms o a linea a e aging ope a o applied
o . The same a e aging ope a o was shown o wo k in O licz spaces
in [8]. The linea i y o his a e aging ope a o leads o a duali y p inci-
ple which educes weigh ed inequali ies o a gene al ope a o conside ed
o e mono one unc ions o weigh ed inequali ies o a modified ope a o
conside ed o e all unc ions. In his sec ion we show ha he equi a-
len no m and he duali y p inciple emain alid o a wide ange o
ea angemen in a ian spaces. Since he echniques in ol ed a e qui e
diffe en his also p o ides new p oo s o some esul s o [12] and [8].
The Le el Func ion in R.I. Spaces 181
Since λsa isfies (2.1), i s cumula i e dis ibu ion unc ion is fini e
on R. Le Λ(x)=x
−∞ dλ o x∈[−∞,∞] and define he a e aging
ope a o Pby
P (x)=Λ(x)−1x
−∞
dλ+Λ(∞)−1∞
−∞
dλ.
By ou con en ion he second e m is absen , ega dless o , when
Λ(∞)=∞.
Theo em 3.1. Suppose λsa isfies (2.1) and Xis a ea angemen in-
a ian space o e (R,λ). Then  X↓≈P X o all ≥0i and
only i P:X→Xis bounded.
P oo : Suppose fi s ha  X↓≈P X o all ≥0. Then he e
exis s a cons an Csuch ha o any ∈X,
P X≤C X↓≤C X
so P:X→Xis bounded.
Con e sely, suppose ha P:X→Xis bounded and hence con in-
uous. Then he e exis s a cons an Csuch ha P X≤C X o
all ∈X.I ≥0 hen se n= min( ,n)χ(−∞,n]so ha ◦is he
poin wise limi o he inc easing sequence { ◦
n}. By P oposi ion 2.1(b)
wi h g=χ(−∞,x]we ha e x
−∞ ndλ ≤x
−∞ ◦
ndλ o each nand each
x∈R.Thus
P nX≤P( ◦
n)X≤C ◦
nX
and so, using he Mono one Con e gence Theo em and he Fa ou p op-
e y o X, we ha e
P X≤C ◦X=C X↓.
On he o he hand, since ◦
nis non-nega i e and non-inc easing, ◦
n≤
P( ◦
n) and hence
 X↓= ◦X= lim
n→∞  ◦
nX≤lim
n→∞ P( ◦
n)X.
To comple e he p oo i will suffice o p o e he ollowing lemma since
hen we will ha e
 X↓≤3 lim
n→∞ P nX=3P X.
Lemma 3.2. Suppose λsa isfies (2.1) and Xis a ea angemen in-
a ian space o e (R,λ). Then o any non-nega i e ∈Bwe ha e
P( ◦)X≤3P X.
182 G. Sinnamon
P oo : We show ha P( ◦)−P X≤2P X, om which he esul
is immedia e. Fo his a gumen we need a ew de ails om [13, Defini-
ion 4.6] in addi ion o hose p esen ed in P oposi ion 2.1. Wi h Iiand
Eas in P oposi ion 2.1, i we define aiand biby (ai,b
i)⊂Ii⊂[ai,b
i]
hen we ha e he ollowing: The poin x∈Ei and only i
(−∞,x)
◦dλ =(−∞,x)
dλ and (−∞,x]
◦dλ =(−∞,x]
dλ.
The poin xis in e io o one o he in e als Iii and only i
(−∞,x)
◦dλ > (−∞,x)
dλ and (−∞,x]
◦dλ > (−∞,x]
dλ.
The le endpoin , ai∈Iii and only i
(−∞,ai]
◦dλ > (−∞,ai]
dλ.
The igh endpoin , bi∈Iii and only i
(−∞,bi)
◦dλ > (−∞,bi)
dλ.
I ollows ha
P( ◦)(x)−P (x)
=
i
Λ(x)−1x
−∞
◦dλ −x
−∞
dλ
χIi(x)
=
i
Λ(x)−1Ii∩(−∞,x]
◦dλ −Ii∩(−∞,x]
dλ
χIi(x)
≤
i
Λ(x)−1Ii∩(−∞,x]
◦dλχIi(x).
The second equali y abo e is easy o p o e in wo cases depending on
whe he ai∈Iio no .
We use P oposi ion 2.1(a) o con inue he calcula ion.
P( ◦)(x)−P (x)
≤
i
Λ(x)−1Ii∩(−∞,x]
dλ Ii
dλ−1Ii
dλχIi(x)
=
i
λ(−∞,x]−1λ(Ii∩(−∞,x])λ(Ii)−1Ii
dλχIi(x).
The Le el Func ion in R.I. Spaces 183
Now we use he ob ious inequali y
λ(Ii∩(−∞,x])λ(Ii∪(−∞,x]) ≤λ(−∞,x]λ(Ii)
o ge
P( ◦)(x)−P (x)≤
i
λ(Ii∪(−∞,x])−1Ii
dλχIi(x).(3.1)
No e ha o x∈Ii,Ii∪(−∞,x] does no depend on x. I is ei he
(−∞,b
i) o (−∞,b
i] depending on whe he o no biis in Ii. Se Bi=
λ(Ii∪(−∞,x]).
Define I0and I1by
I0={i:2Bi<Λ(∞)}and I1={i:2Bi≥Λ(∞)}.
Fo each i∈I
0choose ci∈Rsuch ha Λ(ci)=2Bi. This is possible i
λis non-a omic because Λ is con inuous in ha case. I is also possible
i λconsis s o equal a oms because he condi ion λ(−∞,x]<∞ensu es
ha he a oms do no clus e . The eason o choosing such a ciis so
ha he se I
i=(−∞,c
i] (Ii∪(−∞,x]) has λ-measu e Bi o each
x∈Ii.
Fo each i∈I
1se I
i=Ii. We claim ha





i
B−1
iIi
dλχIi



X
≤




i
B−1
iIi
dλχI
i



X
.(3.2)
This is a amilia calcula ion in ea angemen in a ian spaces which
ollows om Lemma 3.3 below. Now, i i∈I
0and x∈I
i hen x≤ciso
2Bi=Λ(ci)≥Λ(x). I ollows ha

i∈I0
B−1
iIi
dλχI
i(x)≤2Λ(x)−1
i∈I0Ii
dλχ
I
i(x)≤2Λ(x)−1x
−∞
dλ
since he in e als Iia e disjoin . I i∈I
1 hen I
i=Iiand 2Bi≥Λ(∞)
so, once again using disjoin ness, we ha e

i∈I1
B−1
iIi
dλχ
I
i(x)≤2Λ(∞)−1∞
−∞
dλ.
Combining hese las wo es ima es wi h (3.1) and (3.2) yields he desi ed
inequali y
P( ◦)−P X≤2P X
and comple es he p oo .
190 G. Sinnamon
P oo : Since ¯αX<1, Theo em 4.4 shows ha he e exis s a posi i e
cons an Csuch ha
C−1PϕX≤ϕX↓≤CPϕX
o all ϕ∈X↓.
I a C1exis s sa is ying (4.5) hen o any ∈Yand any non-nega i e,
non-inc easing g∈Xwe ha e
M
(T∗g)dµ =R
(T )gdλ≤T X↓gX
≤CPT XgX≤CC1 YgX.
Taking he sup emum o e all wi h  Y≤1 yields (4.6) wi h C2=
CC1.
Con e sely, i he e exis s a C2sa is ying (4.6) hen o any ∈Y
and any non-nega i e, non-inc easing g∈Xwe ha e
R
(T )gdλ=M
(T∗g)dµ ≤ YT∗gY≤C2 YgX.
Taking he sup emum o e all non-nega i e, non-inc easing g∈Xwi h
gX≤1 we ha e T X↓≤C2 Yand hence
PT X≤CT X↓≤CC2 Y
and so (4.5) holds wi h C1=CC2.
The Boyd indices a e known o many classes o ea angemen in a i-
an spaces. The simples is he class o Lebesgue spaces. Fo 1 ≤p≤
∞le Lp
λdeno e he collec ion o λ-measu able unc ions such ha
 Lp
λ<∞whe e
 Lp
λ≡R
| |pdλ1/p
o p<∞and  L∞
λ≡ess supλ
x∈R
| (x)|.
I is well known ha he uppe Boyd index o Lp
λis 1/p. Theo em 4.4
educes o he ollowing.
P oposi ion 4.6. Suppose λsa isfies (2.1) and 1≤p≤∞. Then
P:Lp
λ→Lp
λi and only i 1<p≤∞i and only i
sup R
gdλ :g≥0,gnon-inc easing,gLp
λ
≤1
≡ Lp
λ↓≈P Lp
λ.

The Le el Func ion in R.I. Spaces 191
No e ha since λmay be coun ing measu e on he se o posi i e
in ege s his includes he case Lp
λ=lp.
I 1 <p<∞and is a non-nega i e weigh defined on (0,∞) hen
we may define λby dλ(x)=χ(0,∞)(x) (x)dx and eplace by / o
ob ain
sup ∞
0
g :g≥0,gnon-inc easing,gLp
≤1
≈∞
0x
0
px
0
−p
(x)dx1/p
+∞
0
∞
0
−1/p
which was p o ed in [12, Theo em 1].
Conside able p og ess has been made on de e mining he Boyd in-
dices o O licz spaces in [3], [4], [5], [10] and o he s bu only a small
po ion o his heo y is equi ed o ou pu poses. We e e o [11] o
he defini ions o a Young’s unc ion Φ, i s complemen a y Young’s unc-
ion Ψ, and he O licz space LΦ
λ. We say a Young’s unc ion sa isfies he
∆2condi ion and w i e Φ ∈∆2p o ided he e exis s a cons an C>1
such ha Φ(2x)≤CΦ(x) o all x>0. We say ha Φ sa isfies he ∆∞
2
condi ion and w i e Φ ∈∆∞
2p o ided he e exis cons an s N>0 and
C>1 such ha Φ(2x)≤CΦ(x) o all x>N.
P oposi ion 4.7. Suppose λsa isfies (2.1),Φis a Young’s unc ion,
and Ψis i s complemen a y Young’s unc ion. Then P:LΦ
λ→LΦ
λi
and only i Ψ∈∆∞
2i and only i
(4.7) sup R
gdλ :g≥0,gnon-inc easing,gLΨ
λ≤1
≡ LΦ
λ↓≈P LΦ
λ.
P oo : The associa e space o LΦ
λis LΨ
λwi h equi alen no ms so all
ha is needed o deduce his esul om Theo em 4.4 is o e i y ha
he uppe Boyd index o LΦ
λis less han one i and only i Ψ ∈∆∞
2.
Since he uppe Boyd index o LΦ
λis one minus he lowe Boyd index
o LΨ
λwe wish o show ha he lowe Boyd index o Ψ is g ea e han
ze o i and only i Ψ ∈∆∞
2. This ollows om [10, Theo em 3.2b and
Theo em 4.2].
When λis weigh ed Lebesgue measu e on he hal line, o λis coun ing
measu e on he posi i e in ege s (4.7) was es ablished in [8, Theo em 2.2
and Theo em 3.2] unde he assump ion ha bo h Φ and Ψ sa is y he
∆2condi ion. Fo sequence spaces Heinig and Ku ne gi e somewha
192 G. Sinnamon
mo e. Thei Theo em 3.2 includes a weigh ed e sion o he down no m
which sugges s he ollowing p oblem.
P oblem 4.8. Suppose λsa isfies (2.1), Xis a ea angemen in a i-
an space o e (R,λ), and is a non-nega i e, λ-measu able unc ion.
Cha ac e ize he no m
 X↓
= sup R
| |gdλ:g≥0,gnon-inc easing,g X≤1.
5. Comple eness and duali y
We ha e seen ha X↓is a no med ec o space. In his sec ion we
show ha X↓is a Banach space o unc ions which is no , in gene al, a
Banach unc ion space. We also cha ac e ize he dual space o X↓.To
begin we show ha he map → ◦p ese es inc easing limi s.
P oposi ion 5.1. Suppose ha λsa isfies (2.1) and ∈B.I 0≤ n↑
| | hen ◦
n↑ ◦.
P oo : Since ∈B, n∈B o all nand hence , n∈L2
λ⊂L2
λ↓ o all
n.By[13, Theo em 5.4] ◦is he unique 2-le el unc ion o and ◦
n
is he unique 2-le el unc ion o n.Now[13, Lemma 5.3] wi h hn= ◦
n
shows ha limn→∞ ◦
nis also a 2-le el unc ion o . We conclude ha
limn→∞ ◦
n= ◦as equi ed.
Theo em 5.2. Suppose ha λsa isfies (2.1) and Xis a ea angemen
in a ian space o e (R,λ).I 0≤ n↑| | hen ◦
n↑ ◦and  nX↓↑
 X↓.
P oo : Fi s no e ha P oposi ion 2.1(c) easily ex ends o a bi a y
unc ions and he e o e n≤| |implies ◦
n≤ ◦and we ha e limn→∞ ◦
n≤
◦.
To p o e he o he inequali y le h=| |, se hn= min(h, n)χ(−∞,n]
and define
mn,k = min( n,h
k).
Since n↑h≥hk o all k, we ha e limn→∞ mn,k =hk o all k. Since
hk∈B, P oposi ion 5.1 shows ha limn→∞ m◦
n,k =h◦
k o all k.Now
by Defini ion 2.3
◦= lim
k→∞ h◦
k= lim
k→∞ lim
n→∞ m◦
n,k ≤lim
k→∞ lim
n→∞ ◦
n= lim
n→∞ ◦
n.
The Le el Func ion in R.I. Spaces 193
Thus we ha e ◦
n↑ ◦. Now we apply Co olla y 2.4 and he Fa ou
p ope y in X o ge
lim
n→∞  nX↓= lim
n→∞  ◦
nX= ◦X= X↓.
This comple es he p oo .
Theo em 5.3. I λsa isfies (2.1) and Xis a ea angemen in a ian
space o e (R,λ) hen X↓is a Banach space.
P oo : We ha e al eady shown ha X↓is a no med linea space, i
emains o p o e comple eness. To do his we show ha e e y ab-
solu ely summable sequence in X↓is summable in X↓. Suppose ha
n∈X↓ o all nand ∞
n=1  nX↓<∞. Then | n|∈X↓and so
SN≡N
n=1 | n|∈X↓ o each N. Le Sbe he poin wise limi o he
non-dec easing sequence SN, ha is, S=∞
n=1 | n|. Since SN↑Sand
lim
N→∞ SNX↓≤
N

n=1
 nX↓≤
∞

n=1
 nX↓<∞
we ha e SX↓<∞by Theo em 5.2 and hence S∈X↓. In pa icula
his implies ha Sis fini e λ-almos e e ywhe e because o any M∈R,
χ(−∞,M]is non-inc easing so
M
−∞
Sdλ≤SX↓χ(−∞,M]X<∞.
(Since λ(−∞,M] is fini e, χ(−∞,M]∈X.) Thus, Sis fini e λ-almos
e e ywhe e on (−∞,M] bu since Mwas a bi a y, Sis fini e λ-almos
e e ywhe e on R.
We ha e shown ha ∞
n=1 | n|con e ges poin wise λ-almos e e y-
whe e and i ollows ha ∞
n=1 ncon e ges poin wise λ-almos e e y-
whe e. Le FN=N
n=1 nand F=∞
n=1 n. Fix K, se IN=
in n≥N|Fn−FK|≤|FN−FK| o N>Kand no e ha IN∈X↓
wi h INX↓≤N
n=K+1  nX↓≤∞
n=K+1  nX↓. The sequence IN
is non-dec easing and con e ges poin wise o |F−FK|. Thus, applying
Theo em 5.2 again,
F−FKX↓= lim
N→∞ INX↓≤
∞

n=K+1
 nX↓
and so F−FKX↓ ends o ze o as K→∞. Tha is, FK→Fin X↓
as K→∞. This comple es he p oo .
194 G. Sinnamon
Al hough X↓is a Banach space, i is no a Banach unc ion space in
gene al as he ollowing example shows: Take λ o be Lebesgue measu e
on he hal line. We show ha condi ion [1, Defini ion I.1.1(P5)] ails
o he space L2
λ↓. To do his we exhibi a se Eo fini e measu e and a
sequence o unc ions { n}in L2
λ↓such ha E n/ nL2
λ↓is unbounded.
Se
E=
∞

n=1
[n−2−n,n]
and no e ha λ(E)=∞
n=1 2−n<∞.I n=2
nχ[n−2−n,n]we compu e
E n= 1 and ◦
n=(1/n)χ[0,n].Thus nL2
λ↓= ◦
nL2
λ=n−1/2and
so E n/ nL2
λ↓is unbounded o la ge n.
Fo he emainde o his sec ion we in es iga e he dual space o X↓.
Defini ion 5.4. Suppose ha gis a λ-measu able unc ion. Define ¯gby
¯g(x) = ess sup ≥x|g( )|, se gX↓=¯gX, and le X↓be he collec ion
o unc ions g o which gX↓<∞.
No e ha ¯gis non-nega i e and non-inc easing and ha , by a s an-
da d measu e heo y a gumen , ¯g≥|g|λ-almos e e ywhe e. The
space X↓is a subspace o Xsince we ha e gX≤gX↓.I is
easy o see ha ·
X↓is a no m.
Al hough he no a ion X↓sugges s he associa e space o X↓ his is
no asse ed he e. In ac , since X↓is no necessa ily a Banach unc ion
space, i is no clea ha i has a well-defined associa e space. The
space X↓does beha e like an associa e space, howe e , as we see in
Theo ems 5.6 and 5.7 below. Theo em 5.8 shows ha he dual space
o X↓o en coincides wi h X↓. To p epa e o hese h ee heo ems we
need ano he esul om [13].
P oposi ion 5.5. Suppose λsa isfies (2.1),α∈(0,1), and ,g a e
λ-measu able unc ions such ha ◦and ¯ga e fini e λ-almos e e y-
whe e. Then he e exis s a non-nega i e λ-measu able unc ion hsuch
ha
R
h|g|dλ ≥α2R
| |¯gdλ and R
hϕ dλ ≤R
| |ϕdλ
o all non-nega i e, non-inc easing, λ-measu able unc ions ϕ.
P oo : This is p o ed in [13, Lemma 6.5] unde he assump ion ha
◦∈Lp
λand ¯g∈Lp
λ o some p∈(1,∞]. Only he weake assump ion
ha ◦and ¯ga e fini e λ-almos e e ywhe e is used in he p oo . I
emains alid in his mo e gene al si ua ion wi hou al e a ion.
The Le el Func ion in R.I. Spaces 195
Theo em 5.6. Suppose λsa isfies (2.1),Xis a ea angemen in a i-
an space o e (R,λ), and ∈X↓. Then
 X↓= sup R
gdλ
:gX↓≤1.(5.1)
P oo : Since ¯gis non-inc easing and ¯g≥|g|λ-almos e e ywhe e we
ha e
(5.2) R
gdλ
≤R
| ||g|dλ ≤R
| |¯gdλ≤R
◦¯gdλ
≤ ◦X¯gX= X↓gX↓.
This p o es ha he le side o (5.1) is no less han he igh side.
To p o e he o he inequali y no e ha i gis non-nega i e and non-
inc easing wi h gX≤1 hen sgn( )gX↓≤1 and R| |gdλ =
R sgn( )gdλ
so by (1.1)
 X↓= sup R
| |gdλ:g≥0,gnon-inc easing,gX≤1
≤sup R
gdλ
:gX↓≤1.
This comple es he p oo .
Theo em 5.7. Suppose λsa isfies (2.1),Xis a ea angemen in a i-
an space o e (R,λ), and g∈X↓. Then
gX↓= sup R
gdλ
: X↓≤1.(5.3)
P oo : The calcula ion in (5.2) shows ha he le hand side o (5.3)
is no less han he igh hand side. To p o e he o he inequali y we
equi e P oposi ion 5.5. Fix g∈X↓. Then ¯g∈Xso ¯gis fini e λ-almos
e e ywhe e. Fix α∈(0,1) and choose a non-nega i e unc ion wi h
 X≤1 such ha
gX↓=¯gX≤1
αR
¯g dλ.
Since  X≤1, ∈X⊂X↓so ◦is fini e λ-almos e e ywhe e by
Co olla y 2.4. The unc ion ho P oposi ion 5.5 sa isfies
hX↓= sup R
hϕ dλ ≤sup R
| |ϕdλ≤ X≤1

196 G. Sinnamon
whe e he sup ema a e aken o e all non-nega i e, non-inc easing unc-
ions ϕwi h ϕX≤1. The e o e
α3gX↓≤α2R
¯gdλ≤R
h|g|dλ ≤sup R
gdλ
: X↓≤1.
Since his holds o all α∈(0,1) we may le α→1 o ob ain he
emaining inequali y in (5.3).
Defini ion 5.8. Suppose ha Ais a Banach space o unc ions. We
say he space Ahas absolu ely con inuous no m p o ided e e y non-
inc easing sequence o unc ions in Awhich con e ges o ze o poin wise,
con e ges o ze o in A.
In iew o [1, P oposi ion I.3.5] his defini ion ag ees wi h [1, Defini-
ion I.3.1] when Ais a Banach Func ion Space.
Theo em 5.9. Suppose λsa isfies (2.1),Xis a ea angemen in a i-
an space o e (R,λ)and bo h Xand X↓ha e absolu ely con inuous
no m. Then he dual space o X↓is X↓. Mo e p ecisely, each unc-
ion g∈X↓gi es ise o a con inuous linea unc ional Lgon X↓gi en
by Lg( )=R gdλ. The no m o Lgis gX↓and e e y con inuous
linea unc ional on X↓is Lg o some g∈X↓.
P oo : By [1, Co olla y I.4.3] X=X∗.I g∈X↓ hen Lgis a clea ly
linea and Theo em 5.7 shows ha Lgis con inuous on X↓, ha ing
no m gX↓. Suppose now ha Lis a con inuous, linea unc ional
on X↓. We wish o show ha L=Lg o some g∈X↓.
Since Xis a subspace o X↓(wi h ·
X↓≤·
X) we may conside
Las a con inuous linea unc ional on X. The hypo hesis ha X∗=X
shows ha he e is a unc ion g∈Xsuch ha L =R gdλ o all
∈X. To comple e he p oo we show ha L =R gdλ o all
∈X↓and ha g∈X↓.
To do he fi s we fix ∈X↓, se n= min(n, max(−n, ))χ(−∞,n],
and conside he sequence {| ng|}. This inc eases poin wise o | g|. The
Mono one Con e gence Theo em yields
R
| g|dλ = lim
n→∞ R
| ng|dλ = lim
n→∞ L(| n|sgn(g))
≤LX→Rlim
n→∞  nX↓≤LX→R X↓<∞.
The Le el Func ion in R.I. Spaces 197
Thus g ∈L1
λ. Now conside { n}as a sequence in X↓. Since {| − n|}
dec eases o ze o poin wise and X↓has absolu ely con inuous no m we
see ha { n}con e ges o in X↓. Since Lis con inuous,
L = lim
n→∞ L( n) = lim
n→∞ R
ngdλ=R
gdλ
whe e he las inequali y ollows om he Domina ed Con e gence The-
o em using ou obse a ion ha g ∈L1
λ.
The second ask is o show ha g∈X↓. Se
gn(x) = min(n, |g(x)|)χ(∞,n]and no e ha gn∈X↓and {gn}inc eases
poin wise o |g|.Thus{¯gn}inc eases poin wise o ¯g. The Fa ou p ope y
o he Banach unc ion space Ximplies ha
lim
n→∞ gnX↓= lim
n→∞ ¯gnX=¯gX=gX↓.
Bu
gnX↓= sup R
¯gndλ
≤sup R
| ||g|dλ
= sup L(| |sgn(g)) ≤LX↓→R.
He e he sup ema a e aken o e all unc ions wi h  X↓≤1. The
conclusion is ha gX↓≤LX↓→Rso ha g∈X↓as equi ed.
Co olla y 5.10. I λsa isfies (2.1),Xis a ea angemen in a ian
space o e (R,λ)and bo h Xand X↓ha e absolu ely con inuous no m
hen X↓is comple e.
P oo : The dual space o any no med linea space is comple e.
See [13, Example 6.9] o an example o show ha X↓need no be
eflexi e e en when bo h Xand X↓ha e absolu ely con inuous no m.
I may be ha i Xhas absolu ely con inuous no m hen so does X↓
bu we ha e no p oo o coun e example. In e y many cases, howe e ,
i is ue. We lea e he ollowing as a (non- i ial) exe cise: Suppose λ
sa isfies (2.1), Λ(x)=x
−∞ dλ, and hM(x) = min(M,1/Λ(x)) o M>0.
I Xhas absolu ely con inuous no m and hM∈X o all M>0 hen
X↓has absolu ely con inuous no m.
198 G. Sinnamon
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