Orlicz-type integral inequalities for operators

ORLICZ-TYPE INTEGRAL INEQUALITIES FOR OPERATORS

C. J. Neugebauer

Abstract. We examine Orlicz-type integral inequalities for operators and obtain as a corollary a characterization of such inequalities for the Hardy-Littlewood maximal operator extending the well-known L^p- norm inequalities.

1. Introduction

Let f → T f be an operator and for j = 1, 2, · · · , let T

f = T | {z } · · · T

j times

f

be the j-times iterated operator T . The problem which we will address in this note is to nd conditions on T and conditions on , : R

→ R

such that

(1)

∫

Rⁿ

( |T

f |) ≤ c

∫

Rⁿ

( c

|f|)

with the constant c

independent of f . Conversely, we will also examine what the inequality (1) implies about and .

In particular, if T = M , the Hardy-Littlewood maximal operator M f (x) = sup

x∈Q

1

|Q|

∫

Q

|f(t)|dt,

our results lead to a characterization of (1) with T = M . The case T = M and j = 1 has been studied extensively when = is non-decreasing [1].

In fact (2)

∫

Rⁿ

( M f ) ≤ c

∫

Rⁿ

( c |f|)

Received July 26, 2000.

2000 Mathematics Subject Classication: 42B25.

Key words and phrases: Orlicz, integral inequality, Hardy-Littlewood maximal operator.

if and only if ∫

d( t)

t ≤ c( cs)

s , 0 ≤ s < ∞.

In the general case, ̸= , the latest attempt to characterize (2) with in the right side is in [4]. However in [4] the well-known L

-inequalities for M are excluded. Our results will restore the L

-inequalities, and in doing this we were led to study general integral inequalities (1).

In section 2 we examine conditions on and which imply (1) and in section 3 we study when these conditions on and are implied by (1).

All of this will give us a characterization of (1) with T = M . Section 4 is devoted to L log L and exp(L) inequalities of T , and in section 5 we outline similar results for the operator T

f = [T ( |f|

)]

, r > 0.

2. Integral inequalities

We will assume that and can be written as ( t) =

∫

a(s)ds, ( t) =

∫

b(s) ds,

where a, b : R

→ R

. Our results, as in [4], will be expressed in terms of a and b.

Our rst theorem deals with an operator T satisfying the following dis- tributional inequality:

(3) |{|T f| > λ}| ≤ c

λ

∫

λ/c0

|{|f| > s}|ds,

where c

> 1 is independent of λ > 0 and f .

Remark. If T is sublinear, then (3) holds if and only if T is weak-type (1, 1) and (∞, ∞) [3, p.91]. In particular, M satises (3). In Theorem 2 below we will see another characterization of (3).

We need a version of (3) for T

and this is the content of Lemma 1.

Lemma 1. If T satises (3), then (4) |{|T

f | > λ}| ≤ c

(j − 1)!λ

∫

λ/c^j₀

|{|f| > s}| log

(

c

s λ

)

ds

for j = 1, 2, · · · .

Proof. The proof is by induction. The case j = 1 is (3). Assume (4) holds up to j − 1, and let c = c

. Then

|{|T

f | > λ}| ≤ c λ

∫

λ/c

|{|T

f | > s}|ds

≤ c

(j − 2)!λ

∫

λ/c

1 s

∫

s/c^j−1

|{|f| > t}| log

( c

t s

) dtds

= c

(j − 2)!λ

∫

λ/c^j

∫

|{|f| > t}| 1 s log

( c

t s

) dsdt.

The integral in s is

log

(

λ

)

. 

The next theorem gives a characterization of condition (3) in terms of the integral inequality (1).

Theorem 2. The following are equivalent for an operator f → T f . (A) The integral inequality (1) holds for every a, b : R

→ R

satisfying (5)

∫

a(t) t log

( s t )

dt ≤ c

b(c

s), 0 ≤ s < ∞.

(B) T satises condition (3).

Remark. It is easy to verify that the left side of (5) is a j-times iterated integral:

(j − 1)!

∫

1 s

∫

0

· · · 1 s

∫

0

a(t)

t dtds

· · · ds

. Proof. (B) ⇒(A). From Lemma 1

∫

Rⁿ

( |T

f |) =

∫

0

|{|T

f | > t}|a(t)dt

≤ c

∫

0

a(t) t

∫

t/c^j₀

|{|f| > s}| log

(

c

s t

) dsdt

= c

∫

0

|{|f| > s}|

∫

a(t) t log

( c

s

t )

dtds

c

c

∫

0

|{|f| > s}|b(c

c

s)ds = C

∫

Rⁿ

( c

c

|f|).

Since c

= c

/(j − 1)!, the constant C

= c

/(c

(j − 1)!).

(A)⇒(B). First we show that (5) implies

∫

a(t)

t dt ≤ c

b(c

es), 0 ≤ s < ∞.

Simply note that the integral on the left is

≤

∫

a(t) t log

( es t

) dt ≤

∫

a(t) t log

( es t

)

dt ≤ c

b(c

es).

Thus (A) implies (1) with j = 1, i.e., L ≡

∫

0

|{|T f| > λ}|a(λ)dλ ≤ c

∫

0

|{c|f| > λ}|b(λ)dλ ≡ R,

whenever a, b satisfy (5) with j = 1. For 0 < λ

< ∞ and h > 0, let a(λ) =

χ

(λ) and dene

b(s) =

∫

a(t) t dt =

 



 

0, 0 ≤ s ≤ λ

1

h

log(s/λ

), λ

< s < λ

+ h

1

h

log

0

, s ≥ λ

+ h.

Then

L = 1 h

∫

|{|T f| > λ}|dλ → |{|T f| > λ

}|, as h → 0. The right side

R = c 1 h

∫

|{c|f| > λ}| log(λ/λ

)dλ

+c 1 h log

( λ

+ h λ

) ∫

λ0

|{c|f| > λ}|dλ.

When h → 0, the rst term goes to 0 and the second term goes to

∫

λ0

|{c|f| > λ}| dλ. Hence

|{|T f| > λ}| ≤ c

λ

∫

λ/c

|{|f| > s}|ds.

This is (3) with constant c

. 

Remarks. (i) In (A)⇒(B) we have shown that condition (5) implies the same condition with j = 1. The converse is not true as a(t) = log

t and b(s) = ∫

0 a(t)

t

dt shows. However, if a = b and (5) holds for j = 1, then it also holds for j = 2, 3, · · · . This can be seen from the Remark after the statement of Theorem 2. (ii) This is the special case j = 1 of Theorem 2.

If T satises condition (3) with constant c

, and if

∫

a(t)

t dt ≤ c

b(c

s), 0 ≤ s < ∞,

then ∫

Rⁿ

( |T f|) ≤ c

c

∫

Rⁿ

( c

c

|f|).

Before we examine to what extent (1) implies (5) it will be useful to make an observation. Let us assume that T satises the following reverse weak-type inequality:

(6) c

λ

∫

{|f|>λ}

|f| ≤ |{|T f| > λ}|, with c independent of λ > 0 and f .

Again we need a version of (6) for T

and the next two lemmas deal with this.

Lemma 3. Assume T satises (6) and 0 ≤ k < ∞. Then there exists 0 < c < ∞ such that

c

∫

{|f|>λ}

|f| log

( |f|

λ )

≤

∫

{|T f|>λ}

|T f| log

( |T f|

λ )

.

Proof. Let ( t) = 0, 0 ≤ t ≤ λ, and = t log

(t/λ), t > λ. Then ( t) = ∫

0

a(s)ds, where a(s) = 0, 0 ≤ s ≤ λ, and = log

(s/λ)+k log

(s/λ), s > λ. Let now b(s) = ∫

0 a(t)

t

dt. Then b(s) = 0, 0 ≤ s ≤ λ, and is

≥ c log

(s/λ), s > λ. Replace in (6) λ by s, multiply by a(s) and inte- grate in s from λ to ∞. Interchange the order of integration to obtain the

inequality. 

Lemma 4. Assume T satises (6). Then there exists 0 < c

< ∞ such

that c

λ

∫

{|f|>λ}

|f| log

( |f|

λ )

≤ |{|T

f | > λ}|

for j = 1, 2, · · · . Proof. (6) gives us

|{|T

f | > λ}| ≥ c λ

∫

{|T^(j−1)f|>λ}

|T

f | and by Lemma 3 the last integral is

≥ c

∫

{|T^(j−2)f|>λ}

|T

f | log

( |T

f | λ

)

≥ · · · c

∫

{|f|>λ}

|f| log

( |f|

λ )

. 

Theorem 5. Assume T satises (6). If a : R

→ R

and b(s) =

∫

a(t)

t

log

(

t

) dt, then

c

∫

Rⁿ

|f|b(|f|) ≤

∫

Rⁿ

( |T

f |), where as before ( t) = ∫

0

a(s)ds.

Proof. Multiply the inequality in Lemma 4 by a(λ) and integrate in λ from 0 to ∞. The right side is what we want, and the left side is obtained by an interchange of the order of integration. 

Corollary 6. Assume T satises both condition (3) and (6). If a, b are as in Theorem 5 and ( t) = ∫

0

a(s)ds, ( t) = ∫

0

b(s)ds, then c

∫

Rⁿ

|f|b(|f|) ≤

∫

Rⁿ

( |T

f |) ≤ C

∫

Rⁿ

( C |f|).

Proof. The left side is Theorem 5 and the right side is Theorem 2. 

Remark. (i) The Corollary is especially interesting in the case where

the left and right sides are essentially the same. An example is a(t) =

t

, α > 0. Other examples will be examined in section 4. (ii) An example

of an operator satisfying both (3) and (6) is M as we shall point out in the

next section.

3. (1) implies (5)

For this implication we need to assume that the function b : R

→ R

satises (i) b(t) > 0, t > 0, and (ii) b is quasi-increasing , i.e., there is a constant 0 < c

< ∞ such that b(t

) ≤ c

b(c

t

) for 0 ≤ t

≤ t

.

Remark. In [4] b(t) was assumed to be positive for t ≥ 0 thus not allowing b(t) = t

, α > 0. (ii) is the same as in [4].

Theorem 7. Assume a, b : R

→ R

with b satisfying (i) and (ii). If T satises the condition (6), then the integral inequality (1) implies (5).

Proof. We deny (5) and have then 0 < s

< ∞ such that

∫

0

a(t) t log

( s

t )

dt > 2

b(2

ks

), k = 1, 2, · · · . Let {Q

} be a disjoint collection of cubes with

|Q

| = 1 2

(2

s

) .

Let 0 < c

< ∞ be given. We claim that there is f : R

→ R

such that

∫

Rⁿ

( c

f ) < ∞ and

∫

Rⁿ

( |T

f |) = ∞.

Dene

f (x) = 1 c

∑ 2

s

χ

(x).

Then ∫

Rⁿ

( c

f ) = ∑ ∫

Qk

( c

f ) = ∑

(2

s

) |Q

| = ∑ 1 2

< ∞.

Next, by Lemma 4,

∫

Rⁿ

( |T

f |) =

∫

0

|{|T

f | > λ}|a(λ)dλ ≥

c

∫

0

(∫

{|f|>λ}

log

( f (x)

λ )

dx )

a(λ) λ dλ = c

∫

Rⁿ

f (x)

∫

a(λ) λ log

( f (x) λ

)

dλdx = c

c

∑ 2

s

(∫

0

a(λ) λ log

( 2

s

c

λ )

dλ )

|Q

| ≥ c

c

∑

k≥k0

2

s

2

b(2

ks

) 1

2

(2

s

) ≡ L,

where k

> c

is chosen so large that 2

≥ c

. Since b is quasi-increasing we see that

(2

s

) =

∫

0

b(s)ds ≤ c

b(c

2

s

)2

s

. Hence

L ≥ c c

c

∑

k≥k0

b(2

ks

) b(c

2

s

) = ∞,

since for k ≥ c

, c

2

s

≤ (k2

s

)/c

, from which we get b(c

2

s

) ≤

c

b(k2

s

). 

Remark. The proof of Theorem 4 is along the lines of [4](see also [1, p.14]).

Theorems 2 and 7 imply the following characterization of (1) for T = M , the Hardy-Littlewood maximal operator.

Theorem 8. Assume a, b : R

→ R

with b satisfying (i) and (ii).

Then ∫

Rⁿ

( M

f ) ≤ c

∫

Rⁿ

( c

|f|) if and only if

∫

a(t) t log

( s t )

dt ≤ c

b(c

s), 0 ≤ s < ∞.

Proof. Since M is sublinear, weak-type (1, 1), and ( ∞, ∞), M satises the condition (3)[3, p.91]. Hence Theorem 2 gives the suciency. For the necessity, simply observe that M also satises the condition (6). For f ∈ L

( R

), this follows immediately from a Calderon-Zygmund decomposition at height λ [2, p.23]. For general f ∈ L

(R

), approximate by f

(x) =

f (x)χ

(x). 

4. LlogL and exp(L) inequalities

In this section we will present some more examples for which the left and right sides in Corollary 6 are essentially the same. This leads then to a characterization for ∫

Rⁿ

( |T

f |) to be nite.

For the next lemma it is useful to use the notation f (s) ∼ g(s) on E

to mean: there are constants 0 < c < C < ∞ such that c ≤ f(s)/g(s) ≤

C, s ∈ E.

Lemma 9. For 0 < k < ∞ and j a positive integer, let ϕ(s) =

∫

log

(1 + t) t log

( s t )

dt.

Then (i) ϕ(s) ∼ s

on [0, 1] and (ii) ϕ(s) ∼ log

(1 + s) on [1, ∞).

Proof. (i) This follows from the fact that ϕ(s) can be written as a j- times iterated integral of (log

(1 + t))/t (Remark after Theorem 2) and log(1 + t) ∼ t on [0, 1]. For (ii) simply note that ϕ(s) > 0 and continuous for s ≥ 1. Thus all we have to verify is that ϕ(s)/ log

(1 + s) → L as s → ∞ and 0 < L < ∞. This follows from a j-times repeated application

of L’Hˆ opital’s rule. 

Let for 0 < k < ∞ and j = 1, 2, · · · , L

(f ) =

∫

{|f|≤1}

|f|

+

∫

{|f|>1}

|f| log

(1 + |f|), K

(f ) =

∫

Rⁿ

|T

f | log

(1 + |T

f |).

Theorem 10. Let 0 < k < ∞. If T satises the conditions (3) and (6), then there are constants 0 < c

< C

< ∞ such that

c

L

(f ) ≤ K

(f ) ≤ C

L

(f ).

Proof. Let ( t) = t log

(1 + t). Then ( t) = ∫

0

a(s)ds, where a(s) = log

(1 + s) + ks

1 + s log

(1 + s).

Let b(s) = ∫

a(t)

t

log

(

t

) dt. By Corollary 6,

c

∫

Rⁿ

|f(x)|b(|f(x)|) ≤

∫

Rⁿ

|T

f | log

(1 + |T

f |) ≤ C

∫

Rⁿ

( C |f|), where ( t) = ∫

0

b(s)ds and where we may take C > 1. By Lemma 9 the left side is

≥ c (∫

{|f|≤1}

|f|

+

∫

{|f|>1}

|f| log

(1 + |f|) )

.

Again from Lemma 9 we see that ( t) ∼ t

on [0, 1] and ( t) ∼ t log

(1+

t) on [1, ∞). Since for C > 1, log(1 + Ct) ≤ C log(1 + t), the right side is

≤ C

∫

Rⁿ

( |f|) ≤ C

L

(f ). 

The k = 0 case reads as follows:

Theorem 11. Assume T satises the conditions (3) and (6). Then there are constants 0 < c

< C

< ∞ and there is a constant 0 < c

< 1 such that

c

∫

{|f|≥1}

|f| log

|f| log(1 + |f|) ≤

∫

{|T^(j)f|≥1}

|T

f |

≤ C

∫

{|f|>c0}

|f| log

( |f|

c

) log

( 1 + |f|

c

) .

Proof. Let a(t) = χ

(t). Then ( t) = ∫

0

a(s)ds = 0, 0 ≤ t ≤ 1, and

= t − 1, t > 1. If b(s) = ∫

a(t)

t

log

(

t

) dt, then b(s) = 0, 0 ≤ s ≤ 1, and

= (log

s)/j, s > 1. Hence from Corollary 6

c

∫

Rⁿ

|f|b(|f|) ≤

∫

{|T^(j)f|>1}

( |T

f | − 1) ≤ C

∫

Rⁿ

( C |f|), where again ( t) = ∫

0

b(s)ds. Add |{|T

f | > 1}| to the inequality and let L be the left side and R the right side. By Lemma 4,

L ≥ c (∫

{|f|≥1}

|f| log

|f| +

∫

{|f|≥1}

|f| log

|f|

)

≥ c

∫

{|f|≥1}

|f| log

|f|(1 + log |f|)

≥ c

∫

{|f|≥1}

|f| log

|f| log(1 + |f|).

By Lemma 1,

R ≤ C (∫

Rⁿ

( C |f|) +

∫

c^′

|{|f| > s}| log

( s c

) ds

)

for some 0 < c

< 1. Note that ( s) = 0, 0 ≤ s ≤ 1 and = j

∫

1

log

tdt ≤ cs log

s, s > 1. Hence

∫

Rⁿ

( C |f|) ≤ C

∫

{|f|>1/C}

|f| log

(C |f|).

If we set

(s) = ∫

c^′

log

(

c^′

) dt for s ≥ c

and = 0 for 0 ≤ s ≤ c

, then the second integral in R is

∫

c^′

|{|f| > s}|d

(s) =

∫

0

|{

( |f|) > t}|dt =

∫

{|f|>c^′} j

( |f|).

Since

(s) ≤ s log

(s/c

), s ≥ c

, we see that

∫

c^′

|{|f| > s}|d

(s) ≤

∫

{|f|>c^′}

|f| log

( |f|

c

)

. Let now c

= min(1/C, c

). Then

R ≤ C

∫

{|f|>c0}

|f| log

( |f|

c

) ( 1 + log

( |f|

c

)) .

Since 1 + log s ≤ 2 log(1 + s), s ≥ 0, the proof is complete. 

Remarks. (i) If T only satises condition (6), then in Theorems 10 and 11 we only get the left inequality, whereas if T only satises condition (3) we get only the right inequality. (ii) Since M satises both (3) and (6), Theorems 10 and 11 are valid for T = M , and for j = 1 are the R

- versions of the well-known fact: if B is a nite ball and k ≥ 0, then M f log

(1 + M f ) ∈ L

(B) if and only if |f| log

(1 + |f|) ∈ L

(B) [2, p.23].

The set-up for the exp(L) inequalities is the following: a : [1, ∞) → R

, b : R

→ R

, ( t) = ∫

1

a(s)ds, and ( t) = ∫

0

b(s)ds.

Theorem 12. If ( ∗)

∫

a(t)

log t dt ≤ c

b(c

s), 0 ≤ s < ∞, and if T satises condition (3) with constant c

, then

∫

Rⁿ

( e

) ≤ c

c

∫

Rⁿ

( c

c

|f|).

Proof. The substitution t = e

in ( ∗) gives

∫

a(e

)e

u du ≤ c

b(c

s).

Hence from Remark (ii) after Theorem 2,

∫

Rⁿ ∗

( |T f|) ≤ c

c

∫

Rⁿ

( c

c

|f|),

where

(t) = ∫

0

a(e

)e

du = ∫

1

a(s)ds = ( e

). 

We will now briey discuss two examples illustrating Theorem 12. In both examples we assume that T satises condition (3) with constant c

.

(I) If 1 < p < ∞, then

∫

Rⁿ

(

e

− 1 )

≤ C

{∫

{|f|≤1/c0]}

|f|

+

∫

{|f|>1/c0}

(

e

− 1 )

} .

Let a(t) = (t−1)

, t ≥ 1, and let b(s) = ∫

(t−1)^p−1

log t

dt. The left side is what we want, and for the right side note rst that b(s) ≤ c

s

, 0 ≤ s ≤ 1, and b(s) ≤ c

(e

− 1)

, 1 ≤ s < ∞. This can be seen by showing that

s

lim

b(s)

s

= 1, lim

s→∞

b(s)

(e

− 1)

= 0.

From this we see that ( t) ≤ c

t

/p, 0 < t ≤ 1, and for t > 1,

( t) ≤ c

/p + c

∫

(e

− 1)

ds.

Since ∫

1

(e

− 1)

ds /

(e

− 1)

→ 1/p as t → ∞, the last integral is ≤ c

(e

− 1)

. Consequently, ( t) ≤ c

(e

− 1)

, t > 1 for some constant c

.

(II) There is a version of the inequality in (I) for 0 < p ≤ 1 and it reads as follows:

∫

Rⁿ

(

e

− 1 )

(

e

( |T f| − 1) + 1 )

≤ C

{∫

{|f|≤1/c0}

|f|

+

∫

{|f|>1/c0}

(

e

− 1 )

}

.

To obtain this inequality we let a(t) = (t − 1)

log t, t ≥ 1, and

b(s) =

∫

(t − 1)

dt = 1

p (e

− 1)

.

Since (e

− 1)

≤ c

s

on [0, 1], we see that ( t) ≤ c

t

on [0, 1], and for t > 1, ( t) ≤ c

(e

− 1)

as in (I) above. This gives us the right side of the inequality. For the left side, since p − 1 < 0, ( t) ≥ (t − 1)

∫

1

log s ds

= (t − 1)

(t log t − t + 1).

5. The operator T

For 0 < r < ∞ let T

f (x) = [T (|f|

)(x)]

, and let T

f = [T

(|f|

)]

be the operator T

j-times iterated. Theorem 2 for T

reads as follows:

Theorem 13. Assume that T satises condition (3) and that (7)

∫

a(t) t

log

( s t )

dt ≤ c

b(c

s)

s

, 0 ≤ s < ∞.

If ( t) = ∫

0

a(s)ds and ( t) = ∫

0

b(s)ds, then (8)

∫

Rⁿ

( |T

f |) ≤ c

∫

Rⁿ

( c

|f|).

Proof. The reader will have no diculty to adapt the proof of Theorem

2, (B)⇒(A), to T

. 

Remark. The version of Theorem 2, (A)⇒(B), for T

reads as follows:

If (8) holds whenever a, b : R

→ R

satises (7), then T satises condition (3) for non-negative functions. The proof is the same by rst reducing (7) to j = 1, and then letting a(λ) =

χ

(λ), and b(s) = s

∫

0 a(t)

t^r

dt.

Theorem 14. Assume a, b : R

→ R

with b satisfying (i) and (ii) of

section 3. Let and be as in Theorem 13. If T satises the condition

(6) and if (8) holds, then we have (7).

Proof. With the choice of

|Q

| = 1 2

(2

s

)

the proof is, apart from obvious modications, the same as the proof of

Theorem 7. 

Since the Hardy-Littlewood maximal operator satises the conditions (3) and (6) we have:

Corollary 15. If a, b : R

→ R

with b satisfying (i) and (ii) of section 3. If and are as in Theorem 12, then

∫

Rⁿ

( M

f ) ≤ c

∫

Rⁿ

( c

|f|) if and only if

∫

a(t) t

log

( s t )

dt ≤ c

b(c

s)

s

, 0 ≤ s < ∞.

References

[1] V. Kokilashvili and M. Krbec, Weighted inequalities in Lorentz and Orlicz spaces, World Scientic, 1991.

[2] E. M. Stein, Singular integrals and differentiability properties of functions, Princeton University Press, Princeton, 1970.

[3] A. Torchinsky, Real-Varable methods in harmonic analysis, Academic Press, 1986.

[4] Yoon Jae Yoo, Hardy-Littlewood maximal functions in Orlicz spaces, Bull. Korean Math. Soc. 26 (1999), no. 2, 225–231.

Department of Mathematics Purdue University

Lafayette, IN 47907, USA

E-mail : [email protected]