On a variant of the Hardy inequality between weighted Orlicz spaces

Transkrypt

STUDIA MATHEMATICA 193 (1) (2009)
On a variant of the Hardy inequality between
weighted Orlicz spaces
by
Agnieszka Kałamajska and
Katarzyna Pietruska-Pałuba (Warszawa)
Abstract. Let M be an N -function satisfying the ∆2 -condition, and let ω, ϕ be two
other functions, with ω ≥ 0. We study Hardy-type inequalities
M (ω(x)|u(x)|) exp(−ϕ(x)) dx ≤ C
M (|u0 (x)|) exp(−ϕ(x)) dx,
R+
R+
where u belongs to some set R of locally absolutely continuous functions containing
C0∞ (R+ ). We give sufficient conditions on the triple (ω, ϕ, M ) for such inequalities to
be valid for all u from a given set R. The set R may be smaller than the set of Hardy
transforms. Bounds for constants are also given, yielding classical Hardy inequalities with
best constants.
1. INTRODUCTION
General framework and classical approach. Hardy type inequalities have
been the subject of intensive research, going back to Hardy, who in the early
1920’s ([25, 26]) obtained inequalities of the form
(1.1)
|u(t)|p tα−p dt ≤ C |u0 (t)|p tα dt.
R+
R+
This inequality was generalized to
q
1/p
t
1/q
≤C
|f (t)|p dν(t)
,
f (τ ) dτ dµ(t)
R+ 0
R+
with two nonnegative Radon measures µ, ν, and further to inequalities in
the Orlicz-space setting:
2000 Mathematics Subject Classification: Primary 26D10; Secondary 46E35.
Key words and phrases: Hardy inequalities, weighted Orlicz spaces.
DOI: 10.4064/sm193-1-1
[1]
c Instytut Matematyczny PAN, 2009
2
(1.2)
A. Kałamajska and K. Pietruska-Pałuba
Q−1
Q(ω(x)|T f (x)|)r(x) dx
R+
≤ P −1
P (C%(x)|f (x)|)v(x) dx .
R+
Lp
Inequalities in
were studied by Muckenhoupt [46], Maz’ya and Rozin
[44], Bradley [8], Kokilashvili [33], Sinnamon [53], Sawyer [51], Bloom and
Kerman [6], Stepanov [54], and many others (we refer to the monographs
[38, 36, 37, 34, 45] and references therein). As to the Orlicz-space result
(1.2), several authors contributed to a complete characterization of admissible weights ω,r, %, v and nondecreasing functions P, Q which allow for (1.2)
x
with T f (x) = 0 K(x, y)f (y) dy being the generalized Hardy operator with
kernel K. To name just a few, we refer to the papers of Bloom and Kerman
[5, 6], Lai [40, 41, 42, 43], Heinig and Maligranda [28], Heinig and Lai [27]
and their references. For a more detailed account of such results, we refer to
Sections 3.1 and 3.2.
Hardy-type inequalities are widely applicable in PDE theory and functional analysis. For example, one can derive various Sobolev embedding theorems in the Lp setting, which can then be used to prove the existence of
solutions of the Cauchy problem for elliptic and parabolic PDEs (see e.g.
[9, 22, 36, 15, 44, 55, 39]), to study the asymptotic behaviour of solutions
[2, 57], as well as their stability [10, 11]. They are present in probability
theory (see e.g. [17, 50, 19]). Hardy inequalities are also interesting in their
own right (see e.g. [37, 38, 56]). For latest results, see the very recent papers
[9, 13, 23, 47] and their references.
Investigation of weighted or nonweighted Orlicz–Sobolev spaces defined
by an N -function different from λp is suggested by physical models (see
e.g. [1, 3, 14, 48, 15]). Therefore it is worthwhile to examine Hardy-type
inequalities in general Orlicz spaces as well.
One of the central problems around Hardy inequalities can be expressed
as follows. Consider the sets H of Hardy transforms and H∗ of conjugate
Hardy transforms:
a
n
o
t
H = u(t) = f (s) ds : |f (τ )| dτ < ∞ for every a > 0
(1.3)
0
1,1
Wloc (R+ )
∞
= {u ∈
n
H∗ = u(t) =
0
: lim u(r) = 0},
r→0
∞
f (s) ds :
t
o
|f (τ )| dτ < ∞ for every a > 0
a
1,1
= {u ∈ Wloc
(R+ ) : lim u(r) = 0}.
r→∞
and let T f be either the Hardy transform of f , or the conjugate Hardy
transform of f . Then the following problem arises.
Hardy inequality between weighted Orlicz spaces
3
Problem 1 (classical). Given two N -functions Q, P , describe all possible weights (ω, r, %, v) for which inequality (1.2) holds with some constant
C independent of f (with T the Hardy transform or the conjugate Hardy
transform).
This problem has been completely solved for the Hardy transform by
Bloom and Kerman in [6] for modular functions P and Q such that Q dominates P in some special sense (for the details see Section 3.2). A further
generalization of (3.3), without the domination restriction, can be found in
Lai’s paper [43]. Therefore Problem 1 can be considered as settled.
Another approach and its motivation. We are concerned with another
problem, which is expressed as follows.
Problem 2 (general). Given two N -functions Q, P and weights (ω, r, %, v),
find a possibly large set R contained in the set of locally absolutely continuous functions for which the inequality (1.2) holds with some constant C
independent of u ∈ R.
Let us make some comments here.
It can happen that the given weights (ω, r, %, v) obey the known requirements for the validity of (1.2), say for T f being the Hardy transform of f,
as described in Problem 1. In that case the set R contains the full set H
of Hardy transforms. On the other hand, if this requirement is not satisfied,
we cannot expect (1.2) to hold for every u ∈ H. In such a case R ∩ H will
be a proper subset of H. The solution to Problem 2 would therefore lead to
Hardy-type inequalities in a (possibly) narrower class of functions.
If we substitute u = T f in (1.2), where T is either the Hardy transform
or the conjugate Hardy transform, then u0 = f , so that (1.2) reduces to the
special case of (1.2).
The reduced problem and partial answers. In this paper we deal with a
special variant of Problem 2, which reads as follows.
Problem 3 (reduced). Given an N -function M satisfying the ∆2 -condition, and a pair of functions (ω, ϕ) where ω ≥ 0, describe a possibly large set
R contained in the set of locally absolutely continuous functions for which
the inequality
(1.4)
M (ω(x)|u(x)|) exp(−ϕ(x)) dx ≤ C M (|u0 (x)|) exp(−ϕ(x)) dx
R+
R+
holds with some constant C independent of u ∈ R.
Variants of (1.4) with M (λ) = λp and R determined by constraints
concerning M, ω, φ have been studied e.g. in [4, 16, 20, 21] (see also their
references). To the best of our knowledge their extension to the Orlicz space
setting has not been considered so far.
4
Our main result (Theorem 2.1) states that if the triple (ω, ϕ, M ) satisfies
certain simple compatibility conditions, then we can indicate a set R such
that (1.4) holds for every u ∈ R. Moreover, we give some bounds on the
constant C, which can be expressed in terms of the Simonenko lower and
upper indices (see [52] and [24], [18] for interesting related results).
The condition u ∈ R depends on the behaviour of u near zero and near
infinity, which is very natural in problems arising from PDEs: when analyzing
a particular equation one can often say that its solution (i.e. our function u)
has some “good” properties near the boundary, expressed in terms of its rate
of decay near the boundary.
As an illustration we derive the classical Hardy inequalities (1.1) with
best constants (see Section 3.1.1 for discussion).
We also obtain sufficient conditions for (1.4) to hold for every u ∈ H
(see Proposition 5.2 in Section 5.2.1). These conditions can be easily implemented in practice, and since the verification of the classical Bloom–Kerman
conditions (5.6) from [6] seems rather hard, by using our approach one can
avoid that verification and quickly deduce that (1.4) is satisfied for every
u ∈ H.
Perhaps it is even more interesting to deal with the case when the Bloom–
Kerman conditions (5.6) are not satisfied, so that inequality (1.4) is not valid
for all u ∈ H. Then we find a set R such that R ∩ H is smaller than H and
(1.4) holds for all its elements (see Section 5.2.2).
As a particular type of inequalities like (1.4), we analyze those with
ω = |ϕ0 | (see Section 3.3), and in the class of admissible ϕ’s we obtain the
inequality
(1.5)
M (|ϕ0 (x)|u(x)|) exp(−ϕ(x)) dx ≤ C M (|u0 (x)|) exp(−ϕ(x)) dx.
R+
R+
For M (λ) = λp and ψ(x) = exp{−ϕ(x)/p} we get
(|ψ 0 (x)u(x)|)p dx ≤ C (|u0 (x)ψ(x)|)p dx,
R+
R+
which is nothing other than a particular case of the Caccioppoli inequality
on R+ (see e.g. [12, 29]). Caccioppoli inequalities are commonly used in
regularity theory, and so we believe that our variant (1.5) can be used in
regularity theory as well.
2. PRELIMINARIES AND STATEMENTS OF MAIN RESULTS
2.1. Preliminaries
Orlicz spaces. Let us recall some preliminary facts about Orlicz spaces,
referring e.g. to [49] for details. Here we deal with Orlicz spaces of functions
5
defined on R+ . Suppose that µ is a positive Radon measure on R+ and let
M : [0, ∞) → [0, ∞) be an N -function, i.e. a continuous convex function
satisfying limλ→0 M (λ)/λ = 0 and limλ→∞ M (λ)/λ = ∞.
The weighted Orlicz space LM
µ we deal with is
n
LM
µ := f : R+ → R measurable :
o
M (|f (x)|/K) dµ(x) ≤ 1 for some K > 0 ,
R+
equipped with the Luxemburg norm
o
n
M
(|f
(x)|/K)
dµ(x)
≤
1
.
kf kLM
=
inf
K
>
0
:
µ
R+
This norm is complete and turns LM
µ into a Banach space. When µ is the
Lebesgue measure, it is dropped from the notation. For M (λ) = λp with
p
p > 1, the space LM
µ coincides with the usual Lµ space (defined on R+ ).
The symbol M ∗ denotes the complementary function of an N -function
M , i.e. its Legendre transform: M ∗ (y) = supx>0 [xy − M (x)] for y ≥ 0. It is
again an N -function and from its definition we get the Young inequality
xy ≤ M (x) + M ∗ (y)
for x, y ≥ 0.
M is said to satisfy the ∆2 -condition if, for some constant c > 0 and
every λ > 0, we have
M (2λ) ≤ cM (λ).
(2.1)
In the class of differentiable convex functions the ∆2 -condition is equivalent to
λM 0 (λ) ≤ e
cM (λ) for all λ > 0,
with the constant e
c independent of λ (see e.g. [35, Theorem 4.1]).
We will need the following property of modulars (see [35, formula (9.21)]):
f (x)
(2.2)
M
dµ(x) ≤ 1.
kf kLM
µ
R+
If M satisfies the ∆2 -condition, then (2.2) becomes an equality.
The function M1 is said to dominate M2 if there exist two positive constants K1 , K2 such that M2 (λ) ≤ K1 M1 (K2 λ) for every λ > 0. In that case
we have
(2.3)
k · kLM2 ≤ Kk · kLM1
µ
µ
with K = K2 (K1 + 1).
Functions M1 and M2 are called equivalent when M2 dominates M1 and M1
dominates M2 . In particular, equivalent N -functions give rise to equivalent
Luxemburg norms.
6
On the set LM
µ = {u measurable : M (|u|) dµ < ∞}, one introduces the
so-called dual norm:
n (2.4)
kukL(M ) = sup
u(x)v(x) dµ(x) :
µ
R+
o
∗
v ∈ LM
M ∗ (|v(x)|) dµ(x) ≤ 1 .
µ ,
R+
The advantage of this norm is the Hölder-type inequality:
M∗
(2.5)
f · g dµ ≤ kf kL(M ) kgkL(M ∗ ) for f ∈ LM
µ , g ∈ Lµ .
µ
µ
R+
M
If M satisfies the ∆2 -condition, then LM
µ = Lµ , and in general, the
M
M
Orlicz space Lµ is the completion of Lµ in the dual norm. The Luxemburg
norm and the dual norm are equivalent:
≤ 2kukL(M ) .
kukL(M ) ≤ kukLM
µ
µ
µ
Assumptions. Throughout the paper we assume:
(M) M : [0, ∞) → [0, ∞) is a differentiable N -function, i.e. M is convex,
M (0) = M+0 (0) = 0, M (λ)/λ → ∞ as λ → ∞, and moreover
(2.6)
dM
M (λ)
M (λ)
≤ M 0 (λ) ≤ DM
λ
λ
for every λ > 0,
where DM ≥ dM ≥ 1.
(µ) µ is a Radon measure on R+ , absolutely continuous with respect to
the Lebesgue measure, and µ(dr) = e−ϕ(r) , where ϕ ∈ C 2 (R+ ) and
ϕ0 never vanishes,
(ω) ω : (0, ∞) → [0, ∞) is a C 1 -function.
Remark 2.1. The last inequality in (2.6) implies that M satisfies the ∆2 condition (see (2.1)). The condition dM > 1 is equivalent to the ∆2 -condition
for M ∗ (see e.g. [35, Theorem 4.3] or [30, Proposition 4.1]). Moreover, for
any N -function M, the left-hand inequality in (2.6) holds with dM = 1.
If dM and DM are the best possible constants in (2.6) they obey the
definition of the Simonenko lower and upper indices of M and are related
to Boyd indices of LM (Rn , µ) (see [7], [52] for definitions, and [18], [24], [58]
for discussion on those and other indices of Orlicz spaces).
Notation. Before we proceed, we need to introduce the following quantities. We set
Ω := {r ∈ R+ : ω(r)u(r) 6= 0},
F := {r ∈ R+ : ω(r) 6= 0, ω 0 (r)ϕ0 (r) > 0},
(2.7)
G := {r ∈ R+ : ω(r) 6= 0, ω 0 (r)ϕ0 (r) < 0}.
7
Then we define
ω 0 (r)
ϕ00 (r)
−
b1 (r, ω, ϕ, M ) := 1 + 0
[dM χG (r) + DM χF (r)] ,
(ϕ (r))2 ω(r)ϕ0 (r)
(2.8)
b1 := b1 (ω, ϕ, M ) := inf{b1 (r, ω, ϕ, M ) : r ∈ R+ },
ϕ00 (r)
ω 0 (r)
b2 (r, ω, ϕ, M ) := −1 − 0
+
[dM χF (r) + DM χG (r)] ,
(ϕ (r))2 ω(r)ϕ0 (r)
b2 := b2 (ω, ϕ, M ) := inf{b2 (r, ω, ϕ, M ) : r ∈ R+ },
ω(r)
0
: r ∈ (0, ∞), ϕ (r) 6= 0 .
(2.9)
L = L(ω, ϕ) := sup
|ϕ0 (r)|
We use the conventions sup ∅ = −∞, inf ∅ = +∞, c/∞ = 0; and f χA is the
function f extended by 0 outside A.
2.2. Main results. Our area of interest will be those triples (M, ϕ, ω)
for which either
(B1) b1 > 0, L < ∞, or
(B2) b2 > 0, L < ∞.
We will deal with the following function:
1
M (ω(r)|u(r)|),
(2.10)
hu (r) = h(u,ω,ϕ,M ) (r) := 0
ϕ (r)
which is well defined since ϕ0 (r) is never zero.
Let us introduce the following two classes of functions:
(2.11)
1,1
R+
(ω,ϕ,M ) := {u ∈ Wloc (R+ ) :
lim (hu (Rn ) e−ϕ(Rn ) − hu (sn )e−ϕ(sn ) ) ≥ 0
n→∞
for some sequences sn → 0, Rn → ∞},
(2.12)
R−
(ω,ϕ,M )
:= {u ∈
1,1
Wloc
(R+ )
:
lim (hu (Rn ) e−ϕ(Rn ) − hu (sn )e−ϕ(sn ) ) ≤ 0
n→∞
for some sequences sn → 0, Rn → ∞},
where the limits may be infinite. For simplicity we will usually omit (ω, ϕ, M )
from the notation.
Note that both sets R+ and R− contain the set of compactly supported
1,1
1,1
W
functions and that they sum up to the whole Wloc
(R+ ).
Our main result reads as follows.
Theorem 2.1. Suppose that M, ϕ, ω satisfy (M), (µ), (ω) and (B1) (respectively (B2)). Then
(2.13)
M (ω(r))|u(r)|) µ(dr) ≤ C M (|u0 (r)|) µ(dr)
R+
R+
8
−
2
for every u ∈ R+
(ω,ϕ,M ) (resp. u ∈ R(ω,ϕ,M ) ), where C = c(LDM /b1 dM )
2 /b d )) with c(x) = max(xdM , xDM ).
(resp. C = c(LDM
2 M
As a direct consequence we also obtain the following theorem.
Theorem 2.2. Suppose that M, ϕ, ω satisfy (M), (µ), (ω) and (B1) (respectively (B2)). Then
e 0 kLM
kωukLM
≤ Cku
µ
µ
−
for every u ∈ R+
(ω,ϕ,M ) (resp. u ∈ R(ω,ϕ,M ) ) such that for every λ > 0, λu ∈
e = c(LD2 /b1 dM ) + 1
R+
(resp. λu ∈ R−
) as well , and where C
(ω,ϕ,M )
M
(ω,ϕ,M )
e = c(LD2 /b2 dM ) + 1) with c(x) = max(xdM , xDM ).
(resp. C
M
3. PARTICULAR CASES
The main goal of this section is to illustrate Theorem 2.1 in various
contexts. First we discuss the case when M (λ) = λp (Subsection 3.1). Then
we turn to general M (Subsection 3.2). Finally, in Subsection 3.3, we restrict
ourselves to weights ω = |ϕ0 |.
3.1. Inequalities in the Lp setting. When M (λ) = λp with p > 1,
our conditions get simpler. In particular, dM = DM = p, and since ϕ0 is
assumed to be nonzero everywhere, we have
ω 0 (r)
ϕ00 (r)
−
p
χ
(r) = −b2 (r, ϕ, ω).
b1 (r, ϕ, ω) = 1 + 0
(ϕ (r))2
ω(r)ϕ0 (r) {ω(r)6=0}
It follows that our theorem yields results when L < ∞ and either
inf r>0 b1 (r, ϕ, ω) > 0, or supr>0 b1 (r, ϕ, ω) < 0.
3.1.1. Classical Hardy inequalities. As the first example illustrating our
methods, we get the classical Hardy inequality (see e.g. [26, Theorem 330] for
the classical source, Theorem 5.2 in [36], or [38] for the statement, historical
framework and discussion).
Theorem 3.1. Let 1 < p < ∞ and α 6= p − 1. Suppose
∞ 0thatp uα= u(t) is
an absolutely continuous function in (0, ∞) such that 0 |u (t)| t dt < ∞,
and let
u+ (0) := lim u(t) = 0
for α < p − 1,
u(∞) := lim u(t) = 0
for α > p − 1.
t→0
t→∞
Then
∞
(3.1)
p α−p
|u(t)| t
0
where C = (p/|α − p +
dt ≤ C
∞
0
1|)p .
|u0 (t)|p tα dt,
9
We consider the case α 6= 0. Let us explain how this theorem follows from
our results. Setting
M (r) = rp ,
µ(dr) = rα dr = exp(α ln r),
ω(r) = 1/r,
we have
ϕ(r) = −α ln r,
ω 0 (r) = −1/r2 ,
ϕ0 (r) = −α/r,
ϕ00 (r) = α/r2 ,
ω 0 (r)ϕ0 (r) = α/r2 > 0,
dM = DM = p.
By a direct check we see that
b1 =
α − (p − 1)
,
α
b2 =
(p − 1) − α
,
α
L=
1
.
|α|
Therefore for α > p−1 and for α < 0 we have b1 > 0, while for 0 < α < p−1
we have b2 > 0. In either case ϕ0 never vanishes. The constant C in (2.13) is
equal to (p/|α − (p − 1)|)p , which coincides with the classical statement.
The only thing that remains to be checked is that any function u as in
the statement of Theorem 3.1 for which the right hand side in (3.1) is finite
belongs to R+ when α > p − 1 or α < 0, and to R− when 0 < α < p − 1.
We will use standard arguments (see e.g. [36, proof of Theorem 5.2]).
First suppose that α > p − 1, and let u be as in the assumptions of
∞
Theorem 3.1. Then for any t > 0 we define U (t) := t |u(τ )| dτ. One has
∞
U (t) =
|u0 (τ )| dτ ≤
t
t
≤
∞
∞
|u0 (τ )|p τ α dτ
t
=
p−1
α − (p − 1)
|u0 (τ )|τ α/p τ −α/p dτ
1/p ∞
τ −α/(p−1) dτ
(p−1)/p
t
(p−1)/p ∞
|u0 (τ )|p τ α dτ
1/p
t−(α−(p−1))/p < ∞.
t
∞
From this chain of inequalities and the condition 0 |u0 (τ )|p τ α dτ < ∞
1,1
we infer not only that u ∈ Wloc
(0, ∞) and that U is well defined, but
p
α−(p−1)
also that limR→∞ U (R) R
= 0. Indeed, taking into account that
limt→∞ u(t) = 0, we have
∞
∞
0
|u(t)| = u (τ ) dτ ≤ |u0 (τ )| dτ = U (t),
t
t
and therefore limR→∞ |u(R)|p Rα−(p−1) = 0 as well.
Recall now the formulas defining the class R+ . The function h(r)e−ϕ(r)
appearing there is now equal to −α−1 |u(r)|p rα−(p−1) , vanishing for r tending
to infinity. Therefore u ∈ R+ .
10
t
When α < p−1, we proceed similarly, but now we take U (t) := 0 |u0 (τ )| dτ.
Again, U is well defined and limr→0 U (r)p rα−(p−1) = 0, and since now
|u(r)| ≤ U (r) as well, this yields u ∈ R− . We are done.
3.1.2. General approach within Lp -spaces. The following result is considered classical (see e.g. [44, Theorem 1 of Section 1.3.1]).
Theorem 3.2. Let µ, ν be nonnegative Borel measures on (0, ∞), let ν ∗
be the absolutely continuous part of ν, and 1 ≤ p ≤ q ≤ ∞. Then
(3.2)
q
t
1/q
1/p
≤C
|f (t)|p dν(t)
f (τ ) dτ dµ(t)
R+ 0
R+
for all locally integrable functions f if and only if
r ∗ −1/(p−1) (p−1)/p
dν
1/q
B := sup (µ[r, ∞))
dτ
< ∞.
dτ
r>0
0
The case p = q ≥ 1 is due to Muckenhoupt [46]. Extensions to general p, q
were proven by Mazya and Rozin ([44, Theorem 1 of Section 1.3.1]), Bradley
[8] and Kokilashvili [33]. Some other generalizations (admitting also p, q below 1) were obtained by Sinnamon [53], Sawyer [51], Bloom and Kerman [6],
Stepanov [54] and others.
Observe that inequality (2.13) corresponding to ω(r) = r is a particular
case of (3.2) when one takes ν(dr) = e−ϕ(r) dr, p = q, µ(dr) = rp e−ϕ(r) dr,
t
but only for the representative u(t) = 0 u0 (τ ) dτ . In particular, u+ (0) = 0,
which we have not required (Theorem 3.1 shows that in general the condition
u+ (0) = 0 may not hold). In this case (3.2) reads
|u(τ )ω(τ )|p dν(τ ) ≤ C |u0 (τ )|p dν(τ ),
t
R+
R+
and u(t) = 0 u0 (s) ds, u is an absolutely continuous function. The condition
B < ∞ is equivalent to
p−1
∞
r
ϕ(x)
dx
< ∞.
sup
xp exp(−ϕ(x)) dx
exp
p
−
1
r>0
r
0
It is of different nature than our conditions (B1) and (B2) and is usually not
easy to handle. But since B < ∞ is equivalent to the inequality (3.2) holding
for all u in the set H of Hardy transforms (see (1.3)), our assumptions can
serve as a tool towards verifying B < ∞.
We may as well deal with the set H∗ of
t conjugate Hardy transforms
∞
u = − t f (τ ) dτ (see (1.3)) instead of u = 0 f (τ ) dτ ∈ H as illustrated in
Theorem 3.1, the case α > p − 1.
11
It may happen that inequalities (3.2) do not hold in general on the whole
set H, but they do hold on some smaller sets. These are the sets R+ , R−
defined by (2.11) and (2.12).
3.2. Results in Orlicz spaces. The papers of Maz’ya, Bloom–Kerman
and Lai [44, 6, 40, 41, 42, 43, 27] are concerned with inequalities of the
form
P (C%(x)|f (x)|)v(x) dx ,
Q(ω(x)|T f (x)|)r(x) dx ≤ P −1
(3.3) Q−1
R+
R+
where T f is the Hardy-type operator
x
T f (x) = K(x, y)f (y) dy,
x > 0,
0
with a suitable kernel K. The case K = 1, corresponding to the classical
Hardy operator, is also covered. P and Q are assumed to be nondecreasing
functions on R+ satisfying
lim P (t) = lim Q(t) = 0,
t→0+
t→0+
lim P (t) = lim Q(t) = ∞.
t→∞
t→∞
Bloom and Kerman proved in [6] that within the class of modular functions
P and Q satisfying the following domination property:
P
P
• there exists a constant η > 0 for which
QP −1 (aj ) ≤ QP −1 (η aj )
whenever {aj } is any nonnegative sequence,
(3.3) is equivalent to the conditions
y 
∗ G(ε, y)K(y, x)


P
v(x) dx ≤ G(ε, y) < ∞,


Bεv(x)%(x)
0
(3.4)
y

H(ε, y)

∗


 P Bεv(x)%(x) v(x) dx ≤ H(ε, y) < ∞,
0
holding for all y > 0 and ε > 0, where P ∗ is the Legendre transform
of P ,
∞
−1
G(ε, y) = P Q
Q(εω(x))r(x) dx ,
y
H(ε, y) = P Q−1
∞
Q(εω(x)K(x, y))r(x) dx ,
y
and B > 0 is a constant.
In our particular case: P = Q (= M ) with M satisfying the ∆2 -condition,
r = v = exp(−ϕ(x)), K ≡ 1, % ≡ 1, inequality (3.3) reduces to
12
(3.5)
M (ω(x)T f (x)) exp(−ϕ(x)) dx ≤ C
R+
M (f (x)) exp(−ϕ(x)) dx ,
R+
which is an inequality of the type we are dealing with.
In this case conditions (3.4) simplify to
y
G(ε, y)
(3.6)
M∗
exp(−ϕ(x)) dx ≤ G(ε, y) < ∞
Bε exp(−ϕ(x))
0
for all y > 0 and ε > 0, where
∞
G(ε, y) = M (εω(x)) exp(−ϕ(x)) dx,
y
and B > 0 is a constant. A further generalization of (3.3), without the
restriction P Q, can be found in Lai’s paper [43].
Condition (3.6) as well as Lai’s condition are not easily implemented:
in practice, for given ω, ϕ, M, it is usually hard to see whether (3.6) holds
or not. Conditions (B1) and (B2) are much simpler. When they are satisfied, and when we know
∞ that the Hardy operator T f (x) (or the dual Hardy
∗
operator T f (x) = x f (τ ) dτ , see [43, last remark on page 671]) belongs
to the set R− or R+ , then inequality (3.5) is just the statement of Theorem 2.1.
3.3. Special choice of weights. The case of ω = |ϕ0 |. Another case
that substantially simplifies the approach is that of ω = |ϕ0 | (in fact this
was used in the proof of the classical Hardy inequality). Since we require
ϕ to be C 1 with nonzero derivative, ϕ0 is either always positive, or always
negative.
This time, we have

ϕ00 (r)


 1 + (1 − dM ) 0 2 if ϕ00 (r) ≤ 0,
ϕ (r)
b1 (r, |ϕ0 |, ϕ, M ) =
ϕ00 (r)


 1 + (1 − DM ) 0 2 if ϕ00 (r) ≥ 0,
ϕ (r)

ϕ00 (r)


if ϕ00 (r) ≤ 0,
 −1 + (DM − 1) 0
(ϕ (r))2
0
b2 (r, |ϕ |, ϕ, M ) =
ϕ00 (r)


if ϕ00 (r) ≥ 0,
 −1 + (dM − 1) 0
(ϕ (r))2
L = 1.
In particular (as DM ≥ dM ≥ 1), we get
b1 > 0
if and only if
sup
r>0
ϕ00 (r)
1
eM ,
<
=: D
0
2
(ϕ (r))
DM − 1
b2 > 0 if and only if
13
ϕ00 (r)
1
>
=: deM .
0
2
r>0 (ϕ (r))
dM − 1
inf
This leads to the following conclusion, which is of independent interest.
Corollary 3.1. Assume that conditions (M) and (µ) are satisfied , and
either
sup
r>0
Then
ϕ00 (r)
eM
<D
(ϕ0 (r))2
M (|ϕ0 (r)| |u(r)|) exp(−ϕ(r)) dr ≤ C
R+
inf
or
r>0
ϕ00 (r)
> deM .
(ϕ0 (r))2
M (|u0 (r)|) exp(−ϕ(r)) dr
R+
and
e 0 kLM
kϕ0 ukLM
≤ Cku
µ
µ
2
2
e
for every u ∈ R+
(|ϕ0 |,ϕ,M ) with C = c(DM /b1 dM ), C = c(DM /b1 dM ) + 1
2
in the first case, and for every u ∈ R−
(|ϕ0 |,ϕ,M ) with C = c(DM /b2 dM ),
e = c(D2 /b2 dM ) + 1 in the other case, where c(r) = max(rdM , rDM ).
C
M
Example 3.1 (classical inequalities). To illustrate this corollary we consider again M (λ) = λp , p > 1, ϕ(λ) = −α ln r as in Section 3.1.1. In this
case
ϕ00
1
dM = DM = p,
≡ .
0
2
(ϕ )
α
We have
ϕ00 (r)
1
1
eM
≡ sup 0
<
=D
2
α
p−1
r>0 (ϕ (r))
1
ϕ00 (r)
1
≡ inf 0
>
= deM
α r>0 (ϕ (r))2
p−1
for α ∈ (−∞, 0) ∪ (p − 1, ∞),
for α ∈ (0, p − 1).
Therefore the classical Hardy inequality follows from Corollary 3.1 as well.
Example 3.2 (Hardy inequalities with logarithmic-type weights). As another illustration we show what inequality can be obtained for the measures
µ(dr) = rα (ln(1 + r))β dr with α, β > 0.
In this case we have
ϕ(r) =: ϕα,β (r) = −α ln(r) − β ln ln(1 + r),
β
1
α
ϕ0 (r) = − −
,
r
ln(1 + r) 1 + r
α
β
1
1
00
ϕ (r) = 2 +
1+
.
r
(1 + r)2 ln(1 + r)
ln(1 + r)
14
Choose ω(r) = |ϕ0 (r)|; then
b1 (r) = 1 + (1 − DM )
ϕ00 (r)
,
(ϕ0 (r))2
and so b1 = 1 − (DM − 1) supr>0 ϕ00 (r)/(ϕ0 (r))2 . As ϕ0 (r) is of order 1/r and
ϕ00 (r) is of order 1/r2 on R+ , the supremum involved is finite. Set
ϕ00 (r)
.
(3.7)
sα,β = sup 0
2
r>0 (ϕ (r))
Then
1
b1 > 0 ⇔ DM < 1 +
.
sα,β
As 1/r ∼ ω(r), we arrive at the following.
Theorem 3.3. Suppose α, β > 0 and let M be an N -function satisfying
(M) such that DM < 1 + 1/sα,β , where sα,β is given by (3.7). Then there
exists a constant C > 0 such that
u(r) α
r (ln(1 + r))β dr ≤ C M (|u0 (r)|)rα (ln(1 + r))β dr
M
r
R+
R+
for all u ∈ R+ (|ϕ0 |, ϕ, M ) ⊃ C01 (R+ ).
A similar analysis can be performed for negative α or β.
How to easily verify whether u ∈ R+ or u ∈ R− will be shown in Subsection 5.1.
4. PROOFS OF THEOREMS 2.1 AND 2.2
Before we pass to the actual proofs, let us formulate two easy lemmas concerning Young functions. Although the lemmas may be known to specialists
(see e.g. [24], [52] and [18] for related results), for the reader’s convenience
we supply the proofs.
Lemma 4.1. Let M be a differentiable N -function.
(i) Suppose that there exists a constant DM ≥ 1 such that
M (r)
for every r > 0.
(4.1)
M 0 (r) ≤ DM
r
Then for all r > 0 and λ ≥ 1,
M (λr) ≤ λDM M (r).
(ii) Suppose that there exists a constant dM ≥ 1 such that
M (r)
dM
≤ M 0 (r) for every r > 0.
r
Then for all r > 0 and λ ≤ 1,
M (λr) ≤ λdM M (r).
15
(iii) Suppose that there exist constants 1 ≤ dM ≤ DM such that
dM
M (r)
M (r)
≤ M 0 (r) ≤ DM
r
r
for every r > 0.
Then for all r, λ > 0,
(4.2)
M (λr) ≤ max(λdM , λDM )M (r) =: c(λ)M (r).
We recall Remark 2.1 for interpretation of the constants dM , DM .
Proof. We only prove (i); (ii) is proven analogously, while (iii) is their
direct consequence.
From (4.1) we get M 0 (r)/M (r) ≤ DM /r, and further, for any r > 0 and
λ > 1,
λr
λr
D
M 0 (t)
M
dt ≤
dt,
M (t)
t
r
r
DM ]λr , and further M (λr) ≤
which after integrating gives [ln M (t)]λr
r ≤ [ln t
r
D
λ M M (r).
Lemma 4.2. Let M be a differentiable N -function, and suppose that 1 ≤
dM ≤ DM are two constants such that
M (r)
M (r)
≤ M 0 (r) ≤ DM
r
r
Then for all r, s > 0,
(4.3)
(4.4)
dM
for every r > 0.
M (r)
DM − 1
1
s≤
M (r) +
M (s).
r
dM
dM
Proof. Using the Young inequality rs ≤ M ∗ (r) + M (s) together with
(4.3) we have
M (r)
1
M ∗ (M 0 (r)) M (s)
s≤
M 0 (r)s ≤
+
.
r
dM
dM
dM
From the very definition of the conjugate function M ∗ we have
(4.5)
M (r) = rM 0 (r) − M ∗ (M 0 (r)),
and so
M ∗ (M 0 (r)) ≤ DM M (r) − M (r) = (DM − 1)M (r).
Inserting this into (4.5) we get (4.4).
−
Proof of Theorem 2.1. Suppose u ∈ R+
(ω,ϕ,M ) (resp. u ∈ R(ω,ϕ,M ) ).
Choose sn → 0 and Rn → ∞ as in (2.11) (resp. (2.12)). To abbreviate,
16
we write
∞
J :=
−ϕ(r)
M (ω(r))|u(r)|)e
Rn
dr,
Jn :=
0
sn
∞
Rn
H :=
M (|u0 (r)|)e−ϕ(r) dr,
Hn :=
M (ω(r))|u(r)|)e−ϕ(r) dr,
M (|u0 (r)|)e−ϕ(r) dr.
sn
0
hu
Let
be given by (2.10). Under our assumptions, it is well defined for
1,1
(Rn ) and M is locally Lipschitz, we infer that
every r > 0. Since u ∈ Wloc
1,1
hu ∈ Wloc
(R+ ) and
d
1
(4.6) (hu )0 (r) =
M (ω(r)|u(r)|)
dr ϕ0 (r)
+
1
ϕ0 (r)
M 0 (ω(r)|u(r)|)(ω 0 (r)|u(r)| + ω(r)u0 (r) sgn u(r)),
in the sense of distributions and almost everywhere, and h is absolutely
continuous on each interval [s, R] ⊆ (0, ∞) (see e.g. [44, Theorems 1 and 2,
Sec. 1.1.3]). Moreover, for 0 < s < R < ∞,
R
u 0
−ϕ(r)
u
(h ) (r)e
dr = h
(r)e−ϕ(r) |R
s
s
R
+ M (ω(r)|u(r)|)e−ϕ(r) dr
s
R
=: θ(R, s) + M (ω(r)|u(r)|)e−ϕ(r) dr,
s
and so we have
Rn
Jn =
(4.7)
(hu )0 (r)e−ϕ(r) dr − θn ,
sn
where θn := θ(Rn , sn ) → α ∈ [0, ∞] (resp. [−∞, 0]).
Inserting (4.6) in (4.7) and using (2.7) yields, after some rearrangement,
Rn Jn =
(4.8)
sn
−
ϕ00 (r)
M (ω(r)|u(r)|)e−ϕ(r) dr
− 0
(ϕ (r))2
Rn
sn
Rn
+
sn
1
|ϕ0 (r)|
1
|ϕ0 (r)|
M 0 (ω(r)|u(r)|)|ω 0 (r)| |u(r)|χG (r)e−ϕ(r) dr
M 0 (ω(r)|u(r)|)(|ω 0 (r)| |u(r)|χF (r)+ω(r)u0 (r)sgn u(r))e−ϕ(r) dr−θn
=: In − IIn + IIIn − θn .
17
From now on, the proofs for the two cases: (B1) with u ∈ R+ , and (B2)
with u ∈ R− , differ slightly.
Case 1. Assume that (B1) is satisfied and u ∈ R+ . Since we have
−M 0 (λ) ≤ −dM M (λ)/λ, it follows that
(4.9) −IIn ≤ −
Rn
dM M (ω(r)|u(r)|)
sn
Rn
=
|ω 0 (r)|
χG (r)e−ϕ(r) dr
ω(r)|ϕ0 (r)|
ω 0 (r)
χG (r)e−ϕ(r) dr =: IVn .
ω(r)ϕ0 (r)
dM M (ω(r)|u(r)|)
sn
For IIIn , we first estimate u0 (r) sgn u(r) by |u0 (r)|, and then use the inequality M 0 (λ) ≤ DM M (λ)/λ. For every r ∈ Ω (see (2.7)) we have
(4.10)
M 0 (ω(r)|u(r)|)(|ω 0 (r)| |u(r)|)χF (r) + ω(r)u0 (r) sgn u0 (r))
|u0 (r)|
|ω 0 (r)|
χF (r) + M (ω(r)|u(r)|)
≤ DM M (ω(r)|u(r)|)
ω(r)
|u(r)|
=: DM [A1 (r) + A2 (r)].
This implies
(4.11) IIIn ≤ DM
1
Ω∩[sn ,Rn ]
where
Vn =
|ω 0 (r)|
χF (r)M (ω(r)|u(r)|)e−ϕ(r) dr
ω(r)|ϕ0 (r)|
DM
ω 0 (r)
χF (r)M (ω(r)|u(r)|)e−ϕ(r) dr,
ω(r)ϕ0 (r)
DM
|u0 (r)|
M (ω(r)|u(r)|)e−ϕ(r) dr.
|u(r)||ϕ0 (r)|
Ω∩[sn ,Rn ]
V In =
[A1 (r) + A2 (r)]e−ϕ(r) dr =: Vn + V In ,
DM
Ω∩[sn ,Rn ]
=
|ϕ0 (r)|
Ω∩[sn ,Rn ]
The definition (2.8) of the constant b1 yields
b1 Jn ≤ Jn − In − IVn − Vn ,
and (4.8), (4.9), (4.11) give
Jn = In − IIn + IIIn − θn ≤ In + IVn + Vn + V In − θn .
Combining the two we get
b1 Jn ≤ V In − θn .
Consequently, since b1 is assumed to be positive,
DM
A2 (r) −ϕ(r)
θn
(4.12)
Jn ≤
e
dr − .
0
b1
|ϕ (r)|
b1
Ω∩[sn ,Rn ]
18
Now we use the estimates from Lemmas 4.1 and 4.2. For any 0 < δ ≤ 1
(or any δ > 0 when M (λ) = λp ) and r ∈ Ω,
M (ω(r)|u(r)|) |u0 (r)|
ω(r)|u(r)|
δ
DM − 1
≤ δω(r)
M (ω(r)|u(r)|) +
dM
DM − 1
M (ω(r)|u(r)|) +
≤ δω(r)
dM
A2 (r) = δω(r)
0
1
|u (r)|
M
dM
δ
1
1
0
c
M (|u (r)|)
dM
δ
(Lemma 4.1 was used in the very last line).
Using now this estimate on A2 (r), we obtain from (4.12)
R
DM δ DM − 1 n ω(r)
Jn ≤
0 (r)|
b1
dM
|ϕ
s
n
Rn
1
1
ω(r)
θn
0
−ϕ(r)
+
c
M (|u (r)|)e
dr −
0
dM
δ s |ϕ (r)|
b1
n
1
1
θn
DM Lδ DM − 1
Jn +
c
Hn − ,
≤
b1
dM
dM
δ
b1
or, after rearranging,
DM Lδ
1
θn
DM Lδ DM − 1
Jn ≤
Hn .
c
+ 1−
b1
b1
dM
b1 dM
δ
Choose now δ0 =
b1 dM
2
LDM
to obtain
DM θ n
+ Jn ≤ c
b1
1
δ0
Hn .
Only now do we let n → ∞. As the limits limn→∞ Jn = J, limn→∞ Hn
= H and limn→∞ θn are all well defined and nonnegative (finite or not), this
implies
2 LDM
J ≤c
H
b1 dM
and finishes the proof.
Case 2. We now prove the statement under assumption (B2), for u ∈ R− .
We start with an expression similar to (4.8), but the integrals are rearranged
somewhat differently.
19
This time we write
Rn ϕ00 (r)
− 0
(4.13) Jn =
2
(ϕ
(r))
s
n
Rn
+
Rn
−
sn
sn
1
|ϕ0 (r)|
1
|ϕ0 (r)|
M 0 (ω(r)|u(r)|)|ω 0 (r)| |u(r)|χF (r) e−ϕ(r) dr
M 0 (ω(r)|u(r)|)(|ω 0 (r)| |u(r)|χG (r)+ω(r)u0 (r)sgn u(r))e−ϕ(r) dr−θn
=: In + IIn0 − IIIn0 − θn .
As before, since M 0 (λ) ≥ dM M (λ)/λ, we get
IIn0 ≥
Rn
dM M (ω(r)|u(r)|)
sn
ω 0 (r)
χF (r)e−ϕ(r) dr =: IVn0
ω(r)ϕ0 (r)
(one can omit the absolute values because r ∈ F ). Furthermore, since
−M 0 (λ) ≥ −DM M (λ)/λ, for every r ∈ Ω we have
−M 0 (ω(r)|u(r)|) |ω 0 (r)| |u(r)|χG (r) + ω(r)|u0 (r)|
|u0 (r)|
|ω 0 (r)|
χG (r) + M (ω(r)|u(r)|)
≥ −DM M (ω(r)|u(r)|)
ω(r)
|u(r)|
=: −DM [B(r) + A2 (r)],
where A2 (r) is as in (4.10). This last estimate combined with (4.13) implies
1
−IIIn0 ≥ −DM
[B(r) + A2 (r)]e−ϕ(r) dr =: Vn0 + V In0 .
0
|ϕ (r)|
Ω∩[sn ,Rn ]
On the other hand,
b2 Jn ≤ In + IVn0 + Vn0 − Jn ,
Jn ≥ In + IIn0 − IIIn0 − θn ≥ In + IVn0 + Vn0 + V In0 − θn .
We get b2 Jn ≤ −V In0 + θn , which leads to
DM
A2 (r) −ϕ(r)
θn
Jn ≤
e
dr + .
0
b2
|ϕ (r)|
b2
Ω∩[sn ,Rn ]
Now the proof follows the lines of the first case, starting from (4.12), with
b2 replacing b1 and θn replacing −θn .
Remark 4.1. Observe that for M (λ) = λp we have dM = DM = p and
c(x) = xp . Therefore under the assumptions of Theorem 2.1 the constant C
equals either (Lp/b1 )p if b1 > 0, or (Lp/b2 )p if b2 > 0.
20
Proof of Theorem 2.2. Without loss of generality we can assume that
is finite. Let us substitute uA := u/A in (2.13). Then we get
A = ku0 kLM
µ
M ω|u| exp(−ϕ) ≤ 1,
C
A
R+
which implies kωukLM/C ≤ A. As the N -functions M/C and M are equivaµ
lent, we see by (2.3) that
kωµkLM
≤ (C + 1)kωukLM/C ≤ (C + 1)A = (C + 1)ku0 kLM
.
µ
µ
µ
Therefore the result follows.
Remark 4.2. One can compare our results with those recently proven
in [9]. Namely, in Theorem 3.1, p. 416 there, the authors obtain the following
inequality for the Gaussian measure, for M (λ) = λp :
2
2
p x
x
p
p
0 p
|ωu| exp −
|u | exp −
dx ≤
dx,
2
p−1
2
R+
R+
where
exp(x2 /2(p − 1))
,
2
0 exp(σ /2(p − 1)) dσ
ω(x) = x
holding for every u ∈ W01,p (R+ , dµ) (the completion of C0∞ (R+ ) in the
weighted Sobolev space W 1,p (R+ , dµ), where µ(x) = exp(−x2 /2) dx is the
Gaussian measure). In this case we have ϕ(x) = x2 /2 and the quantity
ω(x)/ϕ0 (x) is not bounded as required by our Theorem 3.1. Instead, the
weight ω obeys a different requirement, namely the ODE xω − (p − 1)ω 0 =
(p − 1)ω 2 .
5. ANALYSIS OF SETS R
5.1. Verification of the condition u ∈ R. The purpose of this subsection is two-fold. First we give an easy practical method to verify whether
u ∈ R (here R stands for R+ or R− ). Further, we discuss when the condition
(5.1)
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞
R+
together with u ∈ H (respectively u ∈ H∗ ) implies that u ∈ R.
We start with the following result.
Proposition 5.1. Suppose M, ϕ satisfy the conditions (M), (µ), (ω),
ϕ ∈ C 2 (R+ ), ϕ0 never vanishes and L < ∞ (see (2.9)). Then the following
statements hold true.
21
1,1
(i) Assume that ϕ0 (R) → 0 as R → ∞, and u ∈ Wloc
(R+ ) is such that
−ϕ(R)
u(R) and u(R)e
are bounded for R → ∞. Then for ϕ0 < 0 we
+
have u ∈ R(ω,ϕ,M ) , while for ϕ0 > 0 we have u ∈ R−
(ω,ϕ,M ) .
1,1
(R+ ) is such that
(ii) Assume that ϕ0 (r) → 0 as r → 0, and u ∈ Wloc
−ϕ(r)
u(r) and u(r)e
are bounded for r → 0. Then for ϕ0 > 0 we have
+
u ∈ R(ω,ϕ,M ) , while for ϕ0 < 0 we have u ∈ R−
(ω,ϕ,M ) .
Proof. (i) We have (hu was defined in (2.10))
M (ω(R)|u(R)|)
exp(−ϕ(R))
|ϕ0 (R)|
M |ϕ0 (R)| |u(R)| |ϕω(R)
0 (R)|
exp(−ϕ(R))
=
|ϕ0 (R)|
M (|ϕ0 (R)| |u(R)|)
≤ c(L)
exp(−ϕ(R))χ{u(R)6=0} (R),
|ϕ0 (R)|
|hu (R)| exp(−ϕ(R)) =
where c(·) is defined in (4.2). The property M (r)/r → 0 as r → 0 and our
assumptions imply
M (|ϕ0 (R)| |u(R)|)
χ{u(R)6=0} (R) → 0
|ϕ0 (R)| |u(R)|
as R → ∞
and so c(L)|u(R)|exp(−ϕ(R)) is bounded for R → ∞. Therefore |hu (R)|
× exp(−ϕ(R)) → 0 as R → ∞ and the statement now follows from the
definition of R+ and R− .
(ii) As in the proof of (i), we check that |hu (r)| exp(−ϕ(r)) → 0 as
r → 0.
In the remaining part of this subsection we examine the property (5.1).
Let us set some additional notation. First, recalling c(·) from (4.2), define,
for r > 0,
(5.2)
fϕ (r) = c−1 (e−ϕ(r) ),
Aϕ (r) = k1/fϕ kL(M ∗ ) (0,r) ,
Bϕ (r) = k1/fϕ kL(M ∗ ) (r,∞) .
The norms here are the dual norms defined by (2.4).
We will distinguish the cases when Aϕ is well defined, and when Bϕ is
well defined.
Case 1. Assume that Aϕ (r) is well defined for small r’s. We now analyze
when u ∈ H which satisfies (5.1) belongs to R.
We start with the following lemma.
Lemma 5.1. Assume that M, ϕ, ω satisfy (M), (ϕ), (ω), and the function
K(r) =
M (ω(r)Aϕ (r)) −ϕ(r)
e
|ϕ0 (r)|
22
1,1
is bounded for r → 0. Then for every u ∈ Wloc
(R+ ) such that
∞
M (|u0 (r)|)e−ϕ(r) dr < ∞ and lim u(r) = 0
the function
0
u
h
r→0
defined by (2.10) satisfies
lim hu (r)e−ϕ(r) = 0.
(5.3)
r→0
Proof. Set
r
U (r) = |u0 (%)| d%.
0
From inequality (2.5) we have
r
r
(5.4)
U (r) = |u0 (%)| d% = |u0 (%)fϕ (%)| ·
0
0
1
d%
|fϕ (%)|
0
≤ ku · fϕ kL(M ) (0,r) · Aϕ (r).
We have
ku0 fϕ kLM (0,r) ≤ ku0 kLM
.
µ (0,r)
(5.5)
Indeed, if A = ku0 fϕ kLM (0,r) then the definition (5.2) of fϕ and the
property (4.2) yield
0 0 0 |u |fϕ
|u |
|u |
1=
M
dx ≤
c(fϕ )M
dx =
M
exp(−ϕ) dx.
A
A
A
(0,r)
(0,r)
(0,r)
ku0 kLM
,
µ (0,r)
Therefore A ≤
which is (5.5).
As the dual and Luxemburg norms are equivalent, we get
ku0 fϕ kL(M ) (0,r) ≤ Aku0 kL(M ) (0,r) ≤ Aku0 kL(M ) (0,∞) < ∞,
µ
µ
where A is some universal constant. Therefore U is well defined.
From the assumption limr→0 u(r) = 0 we get |u(r)| ≤ U (r), and the
estimate (5.4) holds true for u in place of U as well.
Now,
0
A (r))
)
M (ω(r)|u(r)|) −ϕ(r) M (ω(r)ku kL(M
(0,r) ϕ
µ
u
−ϕ(r)
|h (r)e
|=
e
≤
e−ϕ(r)
|ϕ0 (r)|
|ϕ0 (r)|
≤
M (ω(r)Aϕ (r)) −ϕ(r)
e
c(ku0 kL(M ) (0,r) ) = K(r)c(ku0 kL(M ) (0,r) ).
µ
µ
|ϕ0 (r)|
But since c(x) → 0 as x → 0, and ku0 kL(M ) (0,r) → 0 as r → 0, the assertion
µ
(5.3) follows from the boundedness of K(r) for small r’s.
As a corollary, we obtain straight from the definition of R+ , R− :
23
Corollary 5.1. Suppose that the assumptions of Lemma 5.1 are satisfied.
(i) If ϕ0 > 0, then
n
o
u∈H:
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞ ⊂ R+ .
R+
(ii) If ϕ0 < 0, then
n
o
u∈H:
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞ ⊂ R− .
R+
To illustrate the statements above we now discuss the following example.
Example 5.1. Let ϕ(r) = −α ln r, α < p−1, M (λ) = λp and ω(r) = 1/r
as in Theorem 3.1. Then M ∗ (λ) = cp λp/(p−1) and, in this range of α’s,
fϕ (r) = rα/p ,
M ∗ (fϕ (r)) = cp rα/(p−1) ,
Aϕ (r) = ap r(p−1−α)/p .
∞
Thus K(r) is just a constant. Moreover, every u ∈ H with 0 |u0 |p xα dx < ∞
+
belongs to R−
(1/x,xα ,λp ) for α < 0, and to R(1/x,xα ,λp ) when α > 0.
Case 2. We now assume that Bϕ (R) is well defined for large R’s. In
this case we have the following dual statements dealing with the question
whether u ∈ H∗ which satisfies (5.1) belongs to R.
Lemma 5.2. Assume that M, ϕ, ω satisfy (M), (ϕ), (ω), and the function
M (ω(R)Bϕ (R)) −ϕ(R)
L(R) =
e
|ϕ0 (R)|
∞
is bounded for R → ∞. Then for every u ∈ H∗ with 0 M (|u0 (r)|)e−ϕ(r) dr
< ∞ the function hu defined by (2.10) satisfies
lim hu (R)e−ϕ(R) = 0.
R→∞
It is almost identical
r Proof.
∞ 0 to that of Lemma 5.1: just replace U (r) =
0
∗
0 |u (%)| d% with U (R) = R |u (%)| d% and proceed as before.
As a counterpart of Corollary 5.1 we have the following.
Corollary 5.2. Suppose that the assumptions of Lemma 5.2 are satisfied.
(i) If ϕ0 > 0, then
o
n
u ∈ H∗ :
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞ ⊂ R− .
R+
(ii) If ϕ0 < 0, then
n
o
u ∈ H∗ :
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞ ⊂ R+ .
R+
24
This result is illustrated by the following example.
Example 5.2. We again consider the case ϕ(r) = −α ln r, M (λ) = λp
and ω(r) = 1/r as in Theorem 3.1, but now with α > p − 1. We have
Bϕ (r) = bp r(−α+p−1)/p
Therefore every u ∈ H∗
∞ 0 p αand L(r) is a constant.
+
which satisfies 0 |u | x dx < ∞ belongs to R(1/x,xα ,λp ) .
Remark 5.1. Suppose that the assumptions of Theorem 2.1 are satisfied
and put R = R+ in the case of (B1) and R = R− in the case of (B2). One
could ask whether the spaces
o
n
M (|u0 (r)|) exp(−ϕ(r)) dr < ∞ ,
u∈R:
R+
playing the crucial role in the inequality (2.13), can possibly be nonlinear.
We do not know the answer to this question.
5.2. Relation to Bloom–Kerman results. We now compare our results with those of Bloom and Kerman [6].
5.2.1. When our conditions imply Bloom and Kerman’s. Here we give
an example where our assumptions yield inequality (3.5) for all functions
t
T f = 0 f (τ ) dτ ∈ H, so the Bloom–Kerman condition is satisfied.
Corollary 5.1 yields the following proposition.
Proposition 5.2. Assume that (M, ϕ, ω) satisfy the assumptions of
Lemma 5.1 and either (ϕ0 > 0, b1 > 0, L < ∞) or (ϕ0 < 0, b2 > 0, L < ∞).
Then:
(i) There exists a constant C > 0 such that
∞
∞
M (ω(x)|u(x)|) exp(−ϕ(x)) dx ≤ C M (|u0 (x)|) exp(−ϕ(x)) dx
0
0
for every u ∈ H.
(ii) The triple (ω, ϕ, M ) satisfies the Bloom–Kerman condition:
y
G(ε, y)
∗
(5.6)
M
exp(−ϕ(x)) dx ≤ G(ε, y) < ∞,
Bε exp(−ϕ(x))
0
for all y > 0 and ε > 0, where
∞
G(ε, y) = M (εω(x))exp(−ϕ(x)) dx
y
and B > 0 is a constant.
25
Proof. (i) This is
x just a combination of Corollary 5.1 and Theorem 2.1.
(ii) Let u(x) = 0 f (τ ) dτ = (T f )(x) ∈ H be the Hardy transform of f .
Then (i) implies
∞
∞
M (ω(x)T f (x)) exp(−ϕ(x)) dx ≤ C M (f (x)) exp(−ϕ(x)) dx,
0
0
which is equivalent to (5.6) (see Theorem 1.7 in [6] and our comments in
Subsection 3.2).
5.2.2. When Bloom and Kerman conditions are not satisfied. It may
happen that our conditions are satisfied and the Bloom–Kerman conditions
are not. In that case inequalities (2.13) cannot hold for every Hardy transform u ∈ H (see (1.3)) but they hold on proper subsets in the set of Hardy
transforms. This is illustrated by the following result.
Proposition 5.3. There exists a triple (ω, ϕ, M ) such that conditions
(M), (µ), (ω) are satisfied and :
(i) (ω, ϕ, M ) satisfies (B1), in particular
M (ω(r))|u(r)|) µ(dr) ≤ C M (|u0 (r)|) µ(dr)
(5.7)
R+
R+
R+
(ω,ϕ,M ) .
for all u ∈
(ii) (ω, ϕ, M ) does not satisfy the Bloom–Kerman condition (5.6).
(iii) The set
(0,+)
R(ω,ϕ,M ) := R+
(ω,ϕ,M ) ∩ H ⊆ H
is a proper subset of H. Moreover , there is no constant C such that
inequality (5.7) is satisfied for every u ∈ H.
Proof. Let p > 1 and
M (λ) = λp , ϕ(x) = − 12 x2 , ω(x) = x.
In particular, conditions (M), (µ), (ω) are satisfied, b1 = 1 + x12 (p − 1) > 0,
L = 1 and condition (B1) is also satisfied. Therefore (i) follows by Theorem 2.1.
(ii) The Bloom–Kerman condition does not hold: one has
∞
2
G(ε, y) = (εx)p ex /2 dx = ∞,
y
and so (5.6) is violated.
(iii) The Laplace function
r
u(r) = exp(−τ 2 ) dτ
0
belongs to H \
R0,+
(ω,ϕ,M )
and does not satisfy (5.7).
26
Acknowledgements. The authors would like to thank Miroslav Krbec
for helpful advice and discussions.
The work of both authors was supported by a KBN grant no. 1-PO3A008-29. It was also partially supported by EC FP6 Marie Curie programmes
SPADE2 and CODY.
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Institute of Mathematics
University of Warsaw
Banacha 2
02-097 Warszawa, Poland
E-mail: [email protected]
[email protected]
Received June 26, 2007
Revised version March 11, 2009
(6180)

On a variant of the Hardy inequality between weighted Orlicz spaces

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