Some intersections of Lorentz spaces

0.13 MB·English

by  F. Abtahi

SOME INTERSECTIONS OF LORENTZ SPACES
5 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
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0
2

n
Abstract. Let (X,µ) be a measure space. For p,q ∈ (0,∞] and arbitrary
a
J subsets P,Q of (0,∞], we introduce and characterize some intersections of
Lorentzspaces,denoted byILp,Q(X,µ),ILJ,q(X,µ)andILJ,Q(X,µ).
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3
0. Introduction

]
A Let (X,µ) be a measure space. For 0<p≤∞, the space Lp(X,µ) is the usual
Lebesgue space as defined in [3] and [6]. Let us remark that for 1≤p<∞
F
. 1/p
h
kfk := |f(x)|pdµ(x)
t p
a (cid:18)ZX (cid:19)
m defines a norm on Lp(X,µ) such that (Lp(X,µ),k.k ) is a Banach space. Also for
p
[ 0<p<1,

1 kfk := |f(x)|pdµ(x)
p
v ZX
9 defines a quasi norm on Lp(X,µ) such that (Lp(X,µ),k.k ) is a complete metric
p
5 space. Moreover for p=∞,
1
0 kfk∞ =inf{B >0:µ({x∈X :|f(x)|>B})=0}
0 defines a norm on L∞(X,µ) such that (L∞(X,µ),k.k ) is a Banach space. In [1],
. ∞
2 we considered an arbitrary intersection of the Lp−spaces denoted by Lp(G),
p∈J
0
where G is a locally compact group with a left Haar measure λ and J ⊆ [1,∞].
5 T
Then we introduced the subspace IL (G) of Lp(G) as
1 J p∈J
:
v IL (G)={f ∈ Lp(G):kfTk =supkfk <∞},
J J p
i p∈J
X p∈J
\
r and studied ILJ(G) as a Banach algebra under convolution product, for the case
a
where 1 ∈ J. Also in [2], we generalized the results of [1] to the weighted case.
In fact for an arbitrary family Ω of the weight functions on G and 1 ≤ p < ∞,
we introduced the subspace IL (G,Ω) of the locally convex space Lp(G,Ω) =
p
Lp(G,ω). Moreover,we providedsome sufficient conditions on G and also Ω
ω∈Ω
to construct a norm on IL (G,Ω). The fourth section of [2] has been assigned to
p
T
someintersectionsofLorentzspaces. Indeedforthecasewherepisfixedandqruns
through J ⊆(0,∞), we introduced IL (G) as a subspace of ∩ L (G), where
p,J q∈J p,q
L (G) is the Lorentz space with indices p and q. As the main result, we proved
p,q
that IL (G)=L (G), in the case where m =inf{q:q ∈J} is positive.
p,J p,mJ J
In the present work, we continue our study concerning the intersections of
Lorentz spaces on the measure space (X,µ), to complete our results in this di-
rection. Precisely,we verifymost the results givenin the secondand thirdsections
of [1], for Lorentz spaces.
2000 Mathematics Subject Classification. Primary43A15,Secondary43A20.
Key words and phrases. Lp−spaces,Lorentzspaces.
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2 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
1. Preliminaries
In this section, we give some preliminaries and definitions which will be used
throughout the paper. We refer to [3], as a good introductory book.
Let (X,µ) be a measure space and f be a complex valued measurable function
on X. For each α>0, let
d (α)=µ({x∈X :|f(x)|>α}).
f
The decreasing rearrangementof f is the function f∗ :[0,∞)→[0,∞] defined by
f∗(t)=inf{s>0:d (s)≤t}.
f
We adopt the convention inf∅ = ∞, thus having f∗(t) = ∞ whenever d (α) > t
f
for all α≥0. For 0<p≤∞ and 0<q <∞, define
(1.1) kfkLp,q = ∞ tp1f∗(t) q dtt 1/q,
(cid:18)Z0 (cid:16) (cid:17) (cid:19)
where dt is the Lebesgue measure. In the case where q =∞, define
(1.2) kfkLp,∞ =suptp1f∗(t).
t>0
The setofall f with kfk <∞is denotedby L (X,µ)andis calledthe Lorentz
p,q p,q
space with indices p and q. As in Lp−spaces, two functions in L (X,µ) are
p,q
considered equal if they are equal µ−almost everywhere on X. It is worth noting
that by [3, Proposition 1.4.5] for each 0<p<∞ we have
∞
(1.3) |f(x)|pdµ(x)= f∗(t)pdt.
ZX Z0
ItfollowsthatL (X,µ)=Lp(X,µ). Furthermorebythe definitiongiveninequa-
p,p
tion (1.2), one can observe that L (X,µ) = L∞(X,µ). Note that in the case
∞,∞
where p = ∞, one can conclude that the only simple function with finite k.k
L∞,q
normis the zerofunction. For this reason,L (X,µ)={0},forevery0<q <∞;
∞,q
see [3, page 49].
In [2], for locally compact group G and 0<p<∞ and also an arbitrary subset
Q of (0,∞) with
m =inf{q :q ∈Q}>0,
Q
we introduced IL (G) as a subset of ∩ L (G) by
p,Q q∈Q p,q
(1.4) IL (G)={f ∈ L (G):kfk = supkfk <∞}.
p,Q p,q Lp,Q Lp,q
q∈Q
q∈Q
\
As the main result of the third section in [2], we proved the following theorem.
Theorem 1.1. [2, Theorem 12] Let G be a locally compact group, 0<p<∞ and
Q be an arbitrary subset of (0,∞) such that m >0. Then IL (G)=L (G).
Q p,Q p,mQ
Moreover, for each f ∈L (G),
p,mQ
m 1/mQ
Q
(1.5) kfk ≤kfk ≤max 1, kfk .
Lp,mQ Lp,Q ( (cid:18) p (cid:19) ) Lp,mQ
Note that in the definition of IL (G) given in (1.4), one can replace G by an
p,Q
arbitrary measure space (X,µ). Precisely if let
(1.6) IL (X,µ)={f ∈ L (X,µ):kfk = supkfk <∞},
p,Q p,q Lp,Q Lp,q
q∈Q
q∈Q
\
3
then IL (X,µ)=L (X,µ). Moreover for each f ∈L (X,µ), (1.5) is sat-
p,Q p,mQ p,mQ
isfied. Furthermore, Theorem [2, Theorem 12] is also valid for IL (X,µ). In the
p,Q
presentwork,inasimilarway,weintroduceandcharacterizethespacesIL (X,µ)
J,q
andalsoIL (X,µ), as otherintersections ofLorentzspaces. Moreoverwe obtain
J,Q
some results about Lorentz space related to a Banach spaces E, introduced in [4].
2. Main Results
At the beginning of the present section we recall [3, Exercise 1.4.2], which will
be used several times in our further arguments. A simple proof is given here.
Proposition 2.1. Let (X,µ) be a measure space and 0<p <p ≤∞. Then
1 2
L (X,µ) L (X,µ)⊆ L (X,µ).
p1,∞ p2,∞ p,s
\ p1<p<p\2,0<s≤∞
Proof. Let f ∈ L (X,µ) ∩ L (X,µ). If kfk = 0, one can readily
p1,∞ p2,∞ Lp1,∞
obtained that f ∈ L (X,µ), for all p < p < p and 0 < s ≤ ∞. Now let
p,s 1 2
kfk 6=0 and first suppose that p <∞. We show that f ∈L (X,µ), for all
Lp1,∞ 2 p,s
p <p<p and 0<s<∞. It is clear that for each α>0
1 2
kfkp1 kfkp2
(2.1) d (α)≤min Lp1,∞, Lp2,∞ .
f αp1 αp2 !
Set
1
kfkp2 p2−p1
B = Lp2,∞ .
kfkp1
Lp1,∞!
Thus
kfkLp,s = p ∞(df(α)1p α)sdαα s1 = p ∞df(α)sp αs−1dα s1
(cid:18) Z0 (cid:19) (cid:18) Z0 (cid:19)
1 1
B kfkp1 ps s ∞ kfkp2 ps s
≤ p αs−1 Lp1,∞ dα + p αs−1 Lp2,∞ dα
 Z0 αp1 !   ZB αp2 ! 
= p1s kfkpp1 Bαs−1−spp1dαs1 +p1s kfkpp2 ∞αs−1−spp2dαs1
Lp1,∞ Z0 ! Lp2,∞ (cid:18)ZB (cid:19)
1 1
1 p1 Bs−spp1 s 1 p2 Bs−spp2 s
= ps kfkLpp1,∞ s− spp1 ! +ps kfkLpp2,∞ spp2 −s!
1
= p s kfkpp1.(pp22−−pp1) kfkpp2 (pp2−−pp11)
(s− spp1)! Lp1,∞ Lp2,∞
1
+ p s kfkpp1.(pp22−−pp1) kfkpp2 (pp2−−pp11)
(spp2 −s)! Lp1,∞ Lp2,∞
1 1
= p s + p s kfkpp1(pp22−−pp1) kfkpp2(pp2−−pp11)
 (s− spp1)! (spp2 −s)!  Lp1,∞ Lp2,∞
< ∞. 
4 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
Thus f ∈ L (X,µ). For p = ∞, since d (α) = 0 for each α > kfk , inequality
p,s 2 f ∞
(2.1) implies that
kfksLp,s ≤ s−psp1kfkps1pp,1∞kfks∞−spp1,
p
which implies f ∈L (X,µ). In the case where s=∞, by [3, Proposition 1.1.14],
p,s
for p <r <p we have
1 2
L (X,µ)∩L (X,µ)⊆Lr(X,µ)⊆L (X,µ).
p1,∞ p2,∞ r,∞
This gives the proposition. (cid:3)
Proposition2.2. Let(X,µ)beameasurespace, 0<q ≤∞and0<p <p ≤∞.
1 2
Then
L (X,µ)=L (X,µ)∩L (X,µ).
r,q p1,q p2,q
p1≤\r≤p2
Moreover for each f ∈L (X,µ)∩L (X,µ) and p <r <p ,
p1,q p2,q 1 2
kfk ≤21/qmax{kfk ,kfk }.
Lr,q Lp1,q Lp2,q
Proof. By [3, Proposition 1.4.10] and Proposition 2.1 we have
L (X,µ)∩L (X,µ) ⊆ L (X,µ)∩L (X,µ)
p1,q p2,q p1,∞ p2,∞
⊆ L (X,µ).
r,s
p1<r<p\2,0<s≤∞
It follows that
L (X,µ)∩L (X,µ)⊆ L (X,µ).
p1,q p2,q r,q
p1≤\r≤p2
The converse of the inclusion is clearly valid. Thus
L (X,µ)∩L (X,µ)= L (X,µ).
p1,q p2,q r,q
p1≤\r≤p2
Now let q <∞. For each f ∈L (X,µ)∩L (X,µ), we have
p1,q p2,q
kfkq = ∞ t1rf∗(t) q dt
Lr,q t
Z0 (cid:16) (cid:17)
≤ 1 tp12f∗(t) q dt + ∞ tp11f∗(t) q dt
t t
≤ Zkf0kq(cid:16) +kf(cid:17)kq Z1 (cid:16) (cid:17)
Lp2,q Lp1,q
≤ 2 max{kfkq ,kfkq }
Lp2,q Lp1,q
Also for q =∞ we have
kfkLr,∞ = suptr1f∗(t)
t>0
≤ max{ sup tp12f∗(t),suptp11f∗(t)}
0<t<1 t≥1
≤ max{kfk ,kfk }.
Lp2,∞ Lp1,∞
This completes the proof. (cid:3)
Weareinapositiontoprove[1,Proposition2.3]forLorentzspaces. Itisobtained
in the following proposition. Recall from [1] that for a subset J of (0,∞),
M =sup{p:p∈J}.
J
Proposition 2.3. Let (X,µ) be a measure space, 0<q ≤∞ and J be a subset of
(0,∞) such that 0<m . Then the following assertions hold.
J
5
(i) If m ,M ∈J, then
J J
L (X,µ)= L (X,µ)=L (X,µ)∩L (X,µ).
p,q p,q mJ,q MJ,q
p∈[m\J,MJ] p\∈J
(ii) If m ∈J and M ∈/ J, then L (X,µ)= L (X,µ).
J J p∈J p,q p∈[mJ,MJ) p,q
(iii) If m ∈/ J and M ∈J, then L (X,µ)= L (X,µ).
J J Tp∈J p,q Tp∈(mJ,MJ] p,q
(iv) If m ,M ∈/ J, then L (X,µ)= L (X,µ).
J J p∈J p,Tq p∈(mJ,MTJ) p,q
Proof. (i). It is clearly obtainTby Proposition 2.2T.
(ii). Letf ∈ L (X,µ)andtakem <t<M . Thenthereexistt ,t ∈J
p∈J p,q J J 1 2
such that t <t<t . So by Proposition 2.2
1 2
T
f ∈L (X,µ)∩L (X,µ)= L (X,µ)
t1,q t2,q p,q
t1≤\p≤t2
and thus f ∈L (X,µ). It follows that
t,q
L (X,µ)⊆L (X,µ),
p,q t,q
p∈J
\
for each t∈[m ,M ). Consequently
J J
L (X,µ)⊆ L (X,µ).
p,q p,q
p\∈J p∈[m\J,MJ)
The converse of the inclusion is clear.
(iii) and (iv) are proved in a similar way. (cid:3)
Similar to the definition of IL (X,µ) given in (1.6), for J,Q⊆(0,∞) let
p,Q
IL (X,µ)={f ∈ L (X,µ): kfk =sup kfk <∞}
J,q p∈J p,q LJ,q p∈J Lp,q
and
T
IL (X,µ)={f ∈ L (X,µ): kfk =sup kfk <∞}.
J,Q p∈J,q∈Q p,q LJ,Q p∈J,q∈Q Lp,q
Proposition 2.4. LeTt (X,µ) be a measure space, 0<q ≤∞ and J ⊆(0,∞) such
that m >0. Then
J
IL (X,µ)⊆L (X,µ)∩L (X,µ).
J,q mJ,q MJ,q
Proof. First, let q <∞. We follow a proof similar to the proof of [2, Theorem 12].
Suppose that M < ∞ and (x ) is a sequences in J such that lim x =M . For
J n n n J
f ∈IL (X,µ) by Fatou’s lemma, we have
J,q
kfkqLMJ,q == Z0∞∞l(cid:16)imtM1iJn.ff∗t(xt1n)(cid:17).fq∗d(ttt) q dt
n t
≤ Zlim0 inf ∞(cid:16)tx1n.f∗(t)(cid:17)q dt
n t
= liminfkZf0kq(cid:16) (cid:17)
n Lxn,q
≤ kfkq
J,q
< ∞.
If M =∞ and f ∈IL (X,µ), then
J J,q
∞f∗(t)qdt 1/q = ∞liminf tx1nf∗(t) q dt 1/q
t n t
(cid:18)Z0 (cid:19) (cid:18)Z0 (cid:16) (cid:17) (cid:19)
≤ liminfkfk ≤kfk <∞.
Lxn,q J,q
n
6 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
On the other hand, as we mentioned in section 1, since q < ∞ then L (X,µ)=
∞,q
{0} and since ∞f∗(t)qdt < ∞, so we have f = 0, µ−almost every where on X.
0 t
Thus IL (X,µ)=L (X,µ)={0}. It follows that IL (X,µ)⊆L (X,µ).
J,q R ∞,q J,q MJ,q
Now suppose that q =∞ and f ∈IL (X,µ). Then
J,q
kfkLMJ,∞ = st>up0tM1J f∗(t)=stu>p0 linmtx1n.f∗(t)
(cid:16) (cid:17)
≤ sup limkfk =kfk <∞,
t>0 n Lxn,∞ LJ,∞
(cid:16) (cid:17)
and so f ∈ L (X,µ). Thus we proved that IL (X,µ) ⊆ L (X,µ), for
MJ,∞ J,q MJ,q
each0<q ≤∞. Usingsomesimilararguments,onecanobtainthatIL (X,µ)⊆
J,q
L (X,µ). Consequently the proof is complete. (cid:3)
mJ,q
ThefollowingpropositionisobtainedimmediatelyfromPropositions2.2,2.3and
2.4.
Proposition 2.5. Let (X,µ) be a measure space, 0<q ≤∞ and J ⊆(0,∞) such
that m >0 and M <∞. Then
J J
IL (X,µ) = IL (X,µ)=IL (X,µ)
J,q (mJ,MJ),q [mJ,MJ),q
= IL (X,µ)=IL (X,µ)
(mJ,MJ],q [mJ,MJ],q
= L (X,µ)∩L (X,µ).
mJ,q MJ,q
Furthermore, for each f ∈IL (X,µ) and p∈J,
J,q
kfk ≤21/qmax{kfk ,kfk }
Lp,q LmJ,q LMJ,q
Theorem2.6. Let(X,µ)beameasurespaceandJ,Q⊆(0,∞)suchthatm ,m >
J Q
0. Then
(2.2) IL (X,µ)=L (X,µ)∩L (X,µ).
J,Q mJ,mQ MJ,mQ
Moreover for each f ∈IL (X,µ)
J,Q
max{kfk ,kfk } ≤ sup kfk
LmJ,mQ LMJ,mQ p∈J,q∈Q Lp,q
≤ Kmax{kfk ,kfk },
LmJ,mQ LMJ,mQ
for some positive constant K >0.
Proof. Let f ∈IL (X,µ). Then by proposition 2.5, we have
J,Q
f ∈L (X,µ)∩L (X,µ),
mJ,q MJ,q
for each q ∈Q, and so
f ∈(∩ L (X,µ))∩(∩ L (X,µ)).
q∈Q mJ,q q∈Q MJ,q
Thus [2, Theorem 12] implies that f ∈ L (X,µ)∩L (X,µ). Also by
mJ,mQ MJ,mQ
Fatou’s lemma, one can readily obtain that
kfk ≤ sup kfk <∞
LmJ,mQ p∈J,q∈Q Lp,q
and also
kfk ≤ sup kfk <∞.
LMJ,mQ p∈J,q∈Q Lp,q
For the converse,note that by Proposition2.2 and [3, Proposition1.4.10],we have
L (X,µ)∩L (X,µ) = L (X,µ)
mJ,mQ MJ,mQ r,mQ
mJ≤\r≤MJ
⊆ L (X,µ).
r,t
mJ≤r≤MJ\,mQ≤t≤MQ
7
It follows that
L (X,µ)∩L (X,µ)⊆ L (X,µ).
mJ,mQ MJ,mQ p,q
p∈J,q∈Q
\
ByProposition2.5and[2,Theorem12],foreachf ∈L (X,µ)∩L (X,µ)
mJ,mQ MJ,mQ
we have
max{kfk ,kfk } ≤ sup kfk
LmJ,mQ LMJ,mQ mJ≤p≤MJ,mQ≤q≤MQ Lp,q
m 1
≤ mJ≤sup≤pMJ(cid:20)max{1,( pQ)mQ}kfkLp,mQ(cid:21)
m 1
≤ 21/mQmax{1,(mQJ)mQ} max{kfkLmJ,mQ,kfkLMJ,mQ}
and so the inequality is provided by choosing
m 1
K =21/mQmax{1,( Q)mQ}.
m
J
Moreoverf ∈IL (X,µ) and the equality (2.2) is satisfied. (cid:3)
J,Q
Proposition 2.7. Let (X,µ) be a measure space and 0 < p ≤ ∞. Then fg ∈
L (X,µ), for each f ∈L∞(X,µ) and g ∈L (X,µ).
p,∞ p,∞
Proof. By [3, Proposition 1.4.5] parts (7) and (15), we have
kf gkp,∞ = sup tp1 (f g)∗(t) ≤sup tp1f∗(t)g∗(t)
2 2
t>0 t>0(cid:18) (cid:19)
(cid:16) (cid:17)
= sup (2t)p1 f∗(t)g∗(t) ≤2p1 kgkp,∞ kfk∞
t>0
(cid:16) (cid:17)
< ∞.
It follows that fg ∈L (X,µ). (cid:3)
p,∞
Proposition 2.8. Let (X,µ) be a measure space, 0 < p ≤ ∞ and J,Q ⊆ (0,∞)
such that m >0, m >0 and m ∈Q. Then IL (X,µ)=A∩B, where
J Q Q J,Q
A={f ∈ L (X,µ), M =supkfk <∞, ∀q ∈Q}
p,q q Lp,q
p∈J
p∈J,q∈Q
\
and
B ={f ∈ L (X,µ), M = supkfk <∞, ∀p∈J}.
p,q p Lp,q
q∈Q
p∈J,q∈Q
\
Proof. ItisclearthatIL (X,µ)⊆A∩B. Fortheconverseassumethatf ∈A∩B.
J,Q
By [2, Theorem 12] implies that for each p∈J
m 1
qs∈uQpkfkLp,q ≤max{1,( pQ)mQ}kfkLp,mQ
and so
m 1
p∈Js,upq∈QkfkLp,q ≤ max{1,(mQJ )mQ}spu∈pJkfkLp,mQ
m 1
= max{1,(mQ)mQ}MmQ
J
< ∞.
It follows that f ∈IL (X,µ). (cid:3)
J,Q
8 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
In the sequel, we investigatesomepreviousresults, forthe specialLorentzspace
ℓ {E},introducedin [4]. Inthe further discussions, E stands for aBanachspace.
p,q
Also K is the real or complex field and I is the set of positive integers. We first
provide the required preliminaries, which follow from [4].
Definition 2.9. For 1 ≤p ≤ ∞, 1≤ q < ∞ or 1 ≤ p< ∞, q = ∞ let ℓ {E} be
p,q
the space of all E-valued zero sequences {x } such that
i
1
∞ iq/p−1kx kq q for 1≤p≤∞, 1≤q <∞
k{x }k = i=1 φ(i)
i p,q 1
( (cid:0)P supiipkxφ(i)k (cid:1) for 1≤p<∞, q =∞.
is finite, where {kx k} is the non-increasingrearrangementof {kx k}. If E =K,
φ(i) i
then ℓ {K} is denoted by ℓ .
p,q p,q
In particular, ℓ {E} coincides with ℓ {E} and k.k =k.k ; see [5].
p,p p p,p p
The following result will be used in the final result of this paper. It is in fact [4,
Proposition 2].
Proposition 2.10. Let E be a Banach space.
(i) If 1 ≤ p < ∞,1 ≤ q < q ≤ ∞, then ℓ {E} ⊆ ℓ {E} and for every
1 p,q p,q1
{x }∈ℓ {E}
i p,q
1− 1
q q q1
k{x }k ≤ k{x }k ,
i p,q1 p i p,q
(cid:18) (cid:19)
for p<q and
k{x }k ≤k{x }k ,
i p,q1 i p,q
for p≥q. In fact
q
k{x }k ≤max{1, }k{x }k .
i p,q1 p i p,q
(ii) Let eather 1≤p<p ≤∞, 1≤q <∞ or 1≤p<p <∞, q =∞. Then
1 1
ℓ {E}⊆ℓ {E}
p,q p1,q
and for every {x }∈ℓ {E}
i p,q
k{x }k ≤k{x }k
i p1,q i p,q
Now for J,Q⊆[1,∞) let
IL ={{x }∈ ℓ : k{x }k =sup k{x }k <∞}.
J,Q i p∈J,q∈Q p,q i J,Q p∈J,q∈Q i p,q
We finish this work with the following result, which determines the structure of
T
IL .
J,Q
Proposition 2.11. Let J,Q⊆[1,∞). Then IL =ℓ .
J,Q mJ,mQ
Proof. Some similar arguments to [2, Theorem 12] implies that IL = ℓ .
p,Q p,mQ
Indeed, by Proposition 2.10 for each q ∈ Q, ℓ ⊆ ℓ . Also for each {x } ∈
p,mQ p,q i
ℓ ,
p,mQ
m
Q
k{x }k ≤max{1, }k{x }k .
i p,q p i p,mQ
It follows that {x }∈IL and
i p,Q
m
Q
k{x }k ≤max{1, }k{x }k .
i p,Q p i p,mQ
9
Thus ℓ ⊆IL . The reverseof this inclusionis clearwhenever m ∈Q. Now
p,mQ p,Q Q
let mQ ∈/ Q. Thus there is a sequence (yn)n∈N in Q, converging to mQ. For each
{x }∈IL , Fatou’s lemma implies that
i p,Q
∞
k{xi}kmp,mQQ = impQ−1kxΦ(i)kmQ
i=1
X
∞
= liminf iypn−1kxΦ(i)kyn
n
Xi=1 (cid:16) (cid:17)
∞
≤ liminf iypn−1kxΦ(i)kyn
n
Xi=1(cid:16) (cid:17)
= liminfk{x }kyn ≤liminfk{x }kyn
n i p,yn n i p,Q
= k{x }kmQ,
i p,Q
which implies {x } ∈ ℓ . Consequently IL = ℓ . In the sequel, we show
i p,mQ p,Q p,mQ
that IL ⊆ ℓ , for each q ∈ Q. Suppose that (y ) be a sequences in J such
J,q mJ,q n
that lim x =m and {x }∈IL Then by Fatou’s lemma, we have
n n J i J,q
∞
k{xi}kqmJ,q = imqJ−1kxΦ(i)kq
Xi=1(cid:16) (cid:17)
∞
= liminf iyqn−1kxΦ(i)kq
n
Xi=1 (cid:16) (cid:17)
∞
≤ liminf iyqn−1kxΦ(i)kq
n
Xi=1(cid:16) (cid:17)
= liminfk{x }kq
n i yn,q
≤ k{x }kq
i J,q
< ∞.
Hence IL ⊆ ℓ . Now suppose that {x } ∈ IL . Then for each q ∈ Q,
J,q mJ,q i J,Q
{x }∈IL and so {x }∈ℓ . Onthe other hand by the aboveinequalities, for
i J,q i mJ,q
each 1 ≤ q < ∞, we have k{x }k ≤ k{x }k . So {x } ∈ IL ⊆ ℓ ,
i mJ,q i J,q i mJ,Q mJ,mQ
which implies IL ⊆ ℓ . Also by Proposition 2.10, for each p ≥ m and
J,Q mJ,mQ J
q ≥m we have ℓ ⊆ℓ ⊆ℓ . Consequently
Q mJ,mQ mJ,q p,q
ℓ ⊆ ℓ .
mJ,mQ p,q
p∈J,q∈Q
\
Moreoverfor each {x }∈ℓ ,
i p,q
m
Q
sup k{x }k ≤ supk{x }k ≤max{1, }k{x }k .
i p,q i mJ,q m i mJ,mQ
p∈J,q∈Q q∈Q J
It follows that
ℓ ⊆IL ⊆ℓ .
mJ,mQ J,Q mJ,mQ
Therefore IL =ℓ , as claimed. (cid:3)
J,Q mJ,mQ
Acknowledgment. This researchwas partially supported by the Banach alge-
bra Center of Excellence for Mathematics, University of Isfahan.
10 F.ABTAHI,H.G.AMINI,H.A.LOTFIANDA.REJALI
References
[1] Abtahi,F.,Amini,H.G.,Lotfi,H.A.andRejali,A.,AnarbitraryintersectionofLp−spaces,
Bull.Aust.Math.Soc.,85,(2012), 433-445.
[2] Abtahi, F., Amini, H. G., Lotfi, H. A. and Rejali, A., Some intersections of the weighted
Lp−spaces,Abstr.Appl.Anal.,ArticleID986857, 12page,2013.
[3] Grafakos, L., Classical Fourier Analysis, 2nd edn., Springer Science+Business Media, LLC,
(2008).
[4] Kato,M.,OnLorentz spaces lp,q(E),HiroshimaMath.J.,6,(1976), 73-93.
[5] Miyazaki,K.,(p−q)-nuclearand(p−q)-integraloperators,HiroshimaMath.J.,4,(1974),
99-132.
[6] Rudin,W.,Realandcomplexanalysis,thirdedn,McGraw-HillBookCo.,NewYork,1987.
F.Abtahi
DepartmentofMathematics,UniversityofIsfahan,Isfahan,Iran
[email protected]
H.G.Amini
DepartmentofMathematics,UniversityofIsfahan,Isfahan,Iran
A.Lotfi
DepartmentofMathematics,UniversityofIsfahan,Isfahan,Iran
hali-lotfi@yahoo.com
A.Rejali
DepartmentofMathematics,UniversityofIsfahan,Isfahan,Iran
[email protected]