Cho, Yonggeun

Koh, Eunhee

Lee, Sanghyuk

Institutional Repositories DataBase (IRDB)

A maximal inequality for filtration on some function spaces

Osaka City University Repository

Cho, Y., Koh, E. and Lee, S.
Osaka J. Math.
41 (2004), 267–276
A MAXIMAL INEQUALITY FOR FILTRATION
ON SOME FUNCTION SPACES
YONGGEUN CHO, EUNHEE KOH and SANGHYUK LEE
(Received October 1, 2002)
1. Introduction
In this paper we consider a maximal inequality associated with filtration on
Lorentz spaces and Orlicz spaces. Let ( µ), ( ν) be arbitrary measure spaces and
let be a bounded linear operator from a function space defined on ( ν) to a func-
tion space on ( µ). Let be a sequence of measurable subsets of which are
nested: ⊂ +1 for all . Such a sequence is called a filtration of . Denote by
χ the characteristic function of . M. Christ and A. Kiselev in [3] considered the
maximal operator
∗ ( ) = sup | ( χ )( )|
which was studied to obtain the a.e. convergence of an integral operator [4]. They ob-
tained the following result.
Theorem 1.1. Let 1 ≤ , < ∞, and suppose that : ( ) → ( ) is a
bounded linear operator. Then for any nested sequence of measurable subsets { } ⊂
, the maximal operator ∗ is a bounded operator from ( ) to ( ) provided
< . Moreover,
‖ ∗‖ ≤ (1− 2−{(1/ )−(1/ )})−1‖ ‖
where ‖ ‖ denotes the operator norm of from ( ) to ( ).
It should be noted that the phenomena for the maximal inequality occur because
of the strict difference of convexity between two functions ( ) generating the func-
tion spaces ( and ). Based on this fact, we extend the theorem above to some
different function spaces which naturally contain the Lebesgue spaces. Especially, we
thus show a version of Theorem 1.1 still holds on Lorentz spaces and Orlicz spaces
reflecting the difference of convexity. For another reference concerning the Lorentz
space, see the paper [5].
268 Y. CHO, E. KOH AND S. LEE
Let ( ) = ( µ) denote the space of all measurable functions satisfying
‖ ‖ =
( ∫ ∞
0
[ 1/ ∗( )] )1/ <∞
where ∗ is the decreasing rearrangement of (see [6]). Then we first have the fol-
lowing result:
Theorem 1.2. Let 1 ≤ ≤ < ≤ <∞, and suppose : ( ) → ( )
is a bounded linear operator. Then ∗ is bounded from ( ) to ( ). Moreover,
(1.1) ‖ ∗‖ → ≤
(
1− 2−{(1/ )−(1/ )})−1‖ ‖ →
where ‖ ‖ → denotes the operator norm of from to .
Now we consider a generalization to Orlicz spaces. The Young function is
given by ( ) = ∫0 φ( ) for an increasing left continuous function φ with φ(0) = 0.
For the Young function, the Luxemburg norm is defined by
ρ ( ) = inf
{
:
∫ ( | ( )|)
ν( ) ≤ 1
}
Then the Orlicz space ( ) = ( ν) is the function space with the norm ‖·‖ =
ρ (·). For further details, see pp. 265–280 in [2].
Next, we consider a pair of Young functions and . We impose several as-
sumptions on , . For any , ≥ 0, let us assume
(1.2) ( ) ∼ ( ) ( )
Here ∼ means that there is a constant > 0 such that
−1 ≤ ≤
For the function , we assume that there is a strictly convex function ˜ such that for
any α ≥ 1,
(α ) ≤ ˜ (α) ( ) and ˜ (α) ∼ ˜ (1/α)−1(1.3)
Then the second result is the following:
Theorem 1.3. Let be a bounded linear operator from ( ) to ( ). As-
sume and satisfy (1 3) and (1 2), respectively, and further assume
(1.4)
∫ 1
0
−1( ) −1( −1) <∞
A MAXIMAL INEQUALITY FOR FILTRATION 269
Then there is a constant such that ‖ ∗ ‖ ≤ ‖ ‖
Compared with the result in [3] where ( ) = and ( ) = , the result above
is more general. For this particular example, the conditions (1.3) and (1.2) are satisfied
and ∫ 1
0
−1( ) −1( −1) =
∫ 1
0
(1/ )−(1/ ) <∞
provided < . We obtain another example if we set ( ) = , ( ) = (log(2 + ))β
with β > 0. The condition (1.2) is clearly satisfied. It is easily verified that for any
α ≥ 1, there exists ε > 0 such that (α ) . α ε ( ) with ε = + εβ. So if we
set ˜ ( ) = ε , then (1.3) is satisfied and we can find ε so that ˜ satisfies the condi-
tion (1.4) for < .
The proof of these theorems follows the line of argument in [3]. But some techni-
cal difficulties arising in the consideration of Lorentz and Orlicz spaces will be settled
by introducing several lemmas.
2. Proof of Theorem 1.2
We begin by proving an elementary but crucial lemma concerning Lorentz space.
Lemma 2.1. Let , be disjoint measurable sets in and let , be measur-
able functions on . If ≤ <∞, then
(2.1) ‖ χ + χ ‖ ≤ ‖ χ ‖ + ‖ χ ‖
and if ≤ , then
(2.2) ‖ χ + χ ‖ ≥ ‖ χ ‖ + ‖ χ ‖
Proof. By a limiting argument, we may assume that and are simple func-
tions. Without loss of generality, we may write χ , χ as χ =
∑
=1 χ ,
χ =
∑
=1 χ respectively, where , are measurable sets contained in ,
respectively. We may also assume
| 1| ≥ | 2| ≥ · · · ≥ | | ≥ | +1| ≥ · · ·
Set = ν( ), = ν( ). Also for 1 ≤ ≤ , set = ∑
=1 , =
∑
=1 .
Then the decreasing rearrangements of χ , χ are given by
( χ )∗( ) =
{
| | if −1 ≤ <
0 if ≤
270 Y. CHO, E. KOH AND S. LEE
( χ )∗( ) =
{
| | if −1 ≤ <
0 if ≤
Since the supports of and are disjoint, we have + = ∑ χ ∪ . Thus we
have
( χ + χ )∗( ) =
{
| | if −1 + −1 ≤ < +
0 if + ≤
Now for = 1 . . . , let us set
S = ( + ) / − ( −1 + −1) / − / + /−1 − / + /−1
Then a simple computation shows that
‖ + ‖ − ‖ ‖ − ‖ ‖ =
∑
| | S
Finally, we only need to observe that S ≤ 0 if 0 < / ≤ 1 and S ≥ 0 if
/ ≥ 1. This completes the proof of Lemma 2.1.
Now we prove Theorem 1.2. Fix , , , so that 1 ≤ ≤ < ≤ < ∞.
Without loss of generality, we may assume ‖ ‖ ( ) = 1. Define a function M from
measurable sets of ( ν) to R by
M( ) = ‖ χ ‖ ( )
As mentioned in [3], we may assume that for λ > 0 and for any measurable set
, if λ ≤M( ), then there is a measurable subset such that ⊂ and M( ) = λ.
This can be achieved by replacing by ×[0 1], ν by the product of ν and Lebesgue
measure on [0 1], by ◦ π where π ( ) = ∫ 10 ( ) , and by × [0 1].
Then we see that the boundedness of ∗ is implied by the boundedness of ( ◦ π)∗.
Indeed, assume that ( ◦ π)∗ is bounded from ( × [0 1]) to ( ) and (1.1)
holds for ( ◦ π)∗ instead of ∗. Given ∈ ( ), apply the above assumption to
⊗ χ[0 1]. Since
( ◦ π)∗ ⊗ χ[0 1] = sup
(∫ 1
0
χ ×[0 1]( ⊗ χ[0 1])
)
=
∗
and since ‖ ⊗ χ[0 1]‖ ( ×[0 1]) = ‖ ‖ ( ), (1.1) follows.
We also need the following lemma which is a modification of the one in [3].
Lemma 2.2. Let be a measurable function with ‖ ‖ ( ) = 1. Then there is
a collection { } of measurable subsets of , with ∈ {0 1 2 . . .} and 1 ≤ ≤ 2 ,
A MAXIMAL INEQUALITY FOR FILTRATION 271
satisfying the following conditions.
1. { : 1 ≤ ≤ 2 } is a partition of into disjoint measurable subsets.
2. ‖χ ‖ ( ) ≤ 2− for 1 ≤ ≤ 2 .
3. For each , can be decomposed as an empty, finite or countable union such
that for some sequences , ,
=
⋃
≥1
⋃ 1 < 2 < 3 < · · ·
where is a measurable set for which M( ) = 0.
Proof of Lemma 2.2. Define for 1 ≤ ≤ 2 − 1,
= min
{ ∈ N : M( ) ≥ 2− }
By the divisibility assumption for 1 ≤ ≤ 2 −1, we can choose a subset of in
such a way that M( ) = 2− and 2 = . Since is increasing, we may assume
that ⊂ +1 and −1 = 2 . Now we define by
1 = 1
2 = ( 2 \ 1) . . . = ( \ −1) . . .
2 = ( 2 \ 2 −1)
Since ≤ , by (2.2) in Lemma 2.1 M( 1 ∪ 2) ≥ M( 1) +M( 2) if 1 and 2 are
disjoint. So for all 1 ≤ ≤ 2 , we have
M( ) = M( \ −1) ≤M( )−M( −1) = 2−
Form the construction, it follows that for each , there are sequences { } { }
so that
⊂ ⊂ +1
+1
lim
→∞
M( ) = M( )
and is strictly increasing as increase. Indeed, using binary expansion, we can write
M( ) = ∑∞
=1 2
− where is strictly increasing as increases. By our construction
of the sets { }, we see that for each ∈ N, there is a such that ⊂ and
M( ) = ∑
=1 2
−
. Since ⊂ +1 and −1 = 2 , we have ⊂ +1+1 .
Now observe
(
+1
+1
\
)
=
+1 for some sequence { }. Since ⋃ =⋃
, by the monotone convergence theorem, we have M
(
\⋃ ) = 0. Now
we set = \⋃ . This completes the proof of Lemma 2.2.
272 Y. CHO, E. KOH AND S. LEE
Let : → Z be a measurable function. Define an operator ( ) =
( χ ( ) )( ). To prove Theorem 1.2, it is sufficient to show that
‖ ‖ ( ) ≤ ‖ ‖ ( )
where is independent of . Set = { : ( ) = } and define to be the
index set { : appears in the decomposition of } Define measurable sets
by =
⋃
∈ . Observe ∩ = ∅ if 6= . Suppose not. Then there is an
such that ⊂ ∩ because is pair-wise disjoint. So and appear in
the decomposition of . But scale-2 element is contained at most once in . It is
a contradiction.
Note χ =
∑
( ) : =S
χ ∪ . We write
=
∑
χ ( χ ) =
∑ ∑
( ) : =S
χ ( χ ∪ )
=
∑∑
χ ( χ ∪ )
Since is bounded from ( ) to ( ), we may drop in the above expres-
sion. Since > 1, the Lorentz space is a Banach space (see 1.6 of [1]). Thus we
have
‖ ‖ ≤
∞∑
=0
∥∥∥∥∥∥
∑
χ ( χ )
∥∥∥∥∥∥
Now fix and note that ≥ and { } are disjoint. By (2.1) in Lemma 2.1, we have
the following. ∥∥∥∥∑χ ( χ )∥∥∥∥ ≤∑∥∥∥χ ( χ )∥∥∥
≤
∑(‖ ‖ → ) ∥∥∥ χ ∥∥∥
The second inequality is trivial. By the decomposition in Lemma 2.2, the last in the
above inequality is bounded by∑(‖ ‖ → ) 2−{( / )−1} ‖ χ ‖
Since ≤ and for each , are disjoint, another application of Lemma 2.1 implies
A MAXIMAL INEQUALITY FOR FILTRATION 273∑ ‖ χ ‖ ≤ ‖ ‖ . Putting all things together, we have
‖ ‖ ≤
∞∑
=0
2− (
−1− −1)(‖ ‖ → )‖ ‖
≤ (1− 2−{(1/ )−(1/ )})−1(‖ ‖ → )
since < and ‖ ‖ = 1. This completes the proof of Theorem 1.2.
3. Proof of Theorem 1.3
We begin with making several observations. Since is strictly increasing, its in-
verse −1 satisfies
(3.1) −1( ) −1( ) ≤ −1( ) −1
( )
≤ −1( ) −1( )
Let be an Orlicz space with Young’s function . If ( ) ≥ ( ) ( ) for some
, then by the definition of Orlicz space norm, we have
∫ (| ( )|/‖ ‖ ) = 1.
The condition on implies 1 ≤ ∫ (| ( )|)/ (‖ ‖ ) and hence (‖ ‖ ) ≤∫ (| ( )|) . Conversely if we assume ( ) ≤ ( ) ( ) for some , then we
have (‖ ‖ ) ≥ ∫ (| ( )|) . By the assumptions (1.2) on we have
(‖ ‖ ) ∼
∫
(| ( )|)
In the similar way it is easy to see that for satisfying ‖ ‖ ≤ 1,
˜ (‖ ‖ ) ≤ ∫ ˜ (| ( )|)
As before, it is sufficient to show for all measurable : → Z, the operator
given by
( ) = ( χ ( ) )( )
is bounded from to . Without loss of generality we may assume ‖ ‖ = 1.
Now we introduce a decomposition for functions which is similar to Lemma 2.2.
Lemma 3.1. Let be a measurable function with ‖ ‖ = 1. Then there is a
collection { } of measurable sets in , indexed by ∈ {0 1 2 . . . } and 1 ≤ ≤
2 , satisfying the following conditions:
1. { : 1 ≤ ≤ 2 } is a partition of into disjoint measurable subsets.
2.
∫ (| |χ ) = 2− for all 1 ≤ ≤ 2 .
274 Y. CHO, E. KOH AND S. LEE
3. For each , can be decomposed as an empty, finite or countable union such
that for some sequences , ,
=
⋃
≥1
⋃ 1 < 2 < 3 < · · ·
where M( ) = 0.
The proof of the above lemma can be obtained by following the same line of argument
as in [3]. So we omit the detailed proof. According to Lemma 3.1, we decompose
with the same notations for , , as in the proof of Theorem 1.2. We write
( ) =
∞∑
=1
( χ )( )χ ( )
=
∑
=1
∑
( χ ∪ )( )χ ( ) =
∑
( )( )χ ( )
where = χ . By the condition (1.2) on and the fact that are mutually
disjoint for each fixed , we have∥∥∥∥∥∥
∑
( )χ
∥∥∥∥∥∥
 ≤ ∑∫ (| ( )( )|χ ( ))
On the other hand, using the boundedness of from to , we have
(‖ ‖ ) ≥ (‖ ‖ ) ∼ ∫ (| |)
By the decomposition and the condition (1.3) on , we see that
˜ (‖ ‖ ) ≤ ∫ (| |) ∼ 2−
Hence we have∥∥∥∥∥∥
∑
( )χ
∥∥∥∥∥∥
 ≤ ∑ (‖ ‖ )
≤
∑ (˜ (2− )) ≤ 2 (˜ (2− ))
since the number of is not greater than 2 for each . By the triangle inequality, we
A MAXIMAL INEQUALITY FOR FILTRATION 275
have
‖ ‖ ≤
∑∥∥∥∥∥∥
∑
( )χ
∥∥∥∥∥∥
Summing with respect to we get
‖ ‖ ≤
∑˜ (2− ) (2 )(3.2)
Finally, (1.4) implies the left hand side of the above is finite. This completes the proof
of Theorem 1.3.
REMARK 1. In Theorem 1 3, if we set ( ) = (log(2 + ))β (β > 0) and ( ) =
, then the inequality (3.2) can be expressed as
‖ ‖ ≤
∑
2−(1/ ε−1/ ) =
(
1− 2−(1/ ε−1/ ))−1
Thus we have the similar result as in Theorem 1 1. It is interesting to prove Theo-
rem 1 3 for the case ( ) = (log(2 + ))β and ( ) = where the convexity differ-
ence between and is logarithmic. But the lack of convexity difference causes a
difficulty in controlling the inequality (3 2).
REMARK 2. Theorem 1 1 can be easily extended to the vector valued function
spaces (e.g. where is a Banach space). For example, if is a linear operator
from ( ν) to ( µ) with 1 ≤ , ≤ ∞ and { } is a nested set sequence,
then the maximal operator ∗ defined by
∗
= sup || ( χ )||
satisfies the same inequality as in Theorem 1 1.
References
[1] J. Bergh and J. Lo¨fstro¨m: Interpolation Spaces, Springer-Verlag, Berlin, Heidelberg, New York,
1976.
[2] C. Bennett and R. Sharpley: Interpolation of Operators, Academic Press, New York, 1988.
[3] M. Christ and A. Kiselev: Maximal functions associated to filtrations, J. Func. Anal. 179
(1998), 409–425.
[4] M. Christ and A. Kiselev: Absolutely Continuous Spectrum for One dimensional Schro¨dinger
Operators with Slowly Decaying Potentials: Some Optimal Results, J. Amer. Math. Soc. 11
(1998), 771–797.
276 Y. CHO, E. KOH AND S. LEE
[5] A. Kiselev: An Interpolation Theorem Related to the A.E. Convergence of Integral Operators,
Proc. Amer. Math. Soc. 127 (1999), 1781–1788.
[6] E.M. Stein and G. Weiss: Harmonic analysis on Euclidean spaces, Princeton University Press,
Princeton, 1971.
Yonggeun Cho
Department of Mathematics
Pohang University of Science and Technology
Pohang 790-784
Korea
e-mail: bonobono@postech.ac.kr
Eunhee Koh
Korea Institute for Advanced Study
207-43 Cheongnyangni 2-dong, Dongdaemun-gu
Seoul, 130-722
Korea
e-mail: ehkoh@kias.re.kr
Sanghyuk Lee
Department of Mathematics
Pohang University of Science and Technology
Pohang 790-784
Korea
e-mail: huk@euclid.postech.ac.kr


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A maximal inequality for filtration on some function spaces

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