Journal of Inequalities in Pure and
Applied Mathematics
SOME NEW HARDY TYPE INEQUALITIES AND THEIR
LIMITING INEQUALITIES
volume 4, issue 3, article 61,
2003.
ANNA WEDESTIG
Department of Mathematics
Luleå University
SE- 971 87 Luleå
SWEDEN.
EMail: annaw@sm.luth.se
Received 12 December, 2002;
accepted 08 September, 2003.
Communicated by: B. Opić
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
142-02
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Abstract
A new necessary and sufficient condition for the weighted Hardy inequality is
proved for the case 1 < p ≤ q < ∞. The corresponding limiting Pólya-Knopp
inequality is also proved for 0 < p ≤ q < ∞. Moreover, a corresponding limiting
result in two dimensions is proved. This result may be regarded as an endpoint
inequality of Sawyer’s two-dimensional Hardy inequality. But here we need only
one condition to characterize the inequality whereas in Sawyer’s case three
conditions are necessary.
Some New Hardy Type
Inequalities and their Limiting
Inequalities
2000 Mathematics Subject Classification: 26D15.
Key words: Inequalities, Weights, Hardy’s inequality, Pólya-Knopp’s inequality, Multidimensional inequalities, Sawyer’s two-dimensional operator.
Anna Wedestig
I thank Professor Lars-Erik Persson and the referee for some comments and good
advice which has improved the final version of this paper.
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Contents
Contents
1
2
3
4
5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A New Weight Characterization of Hardy’s Inequality . . . . . .
A Weight Characterization of Pólya-Knopp’s Inequality . . . .
Weighted Two-Dimensional Exponential Inequalities . . . . . .
Final Remarks and Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
On Minkowski’s integral inequalities . . . . . . . . . . . . .
5.2
On the conditions and the best constant in the Hardy
inequality (2.1): . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
8
12
15
25
25
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1.
Introduction
We are inspired by the clever Hardy-Pólya observation to the Hardy inequality,
p
p Z ∞
Z ∞ Z x
1
p
f (t)dt dx ≤
f p (x)dx, f ≥ 0, p > 1,
x
p
−
1
0
0
0
1
that by changing f to f p and tending p → ∞ we obtain the Pólya-Knopp
inequality
Z ∞
Z ∞
Gf (x)dx ≤ e
f (x)dx
0
0
with the geometric mean operator
Anna Wedestig
Z x
1
Gf (x) = exp
ln f (t)dt .
x 0
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Let 1 < p ≤ q < ∞. In particular, in [3] the authors tried to find a new
condition for the weighted inequality
Z
(1.1)
0
∞
q
1q
Z x
1
ln f (t)dt
u(x)dx
exp
x 0
Z ∞
p1
p
≤C
f (x)v(x)dx
0
by using the weighted Hardy inequality
Z ∞ Z x
q
1q
Z
(1.2)
f (t)dt u(x)dx
≤C
0
0
Some New Hardy Type
Inequalities and their Limiting
Inequalities
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∞
f p (x)v(x)dx
p1
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and replacing u(x) and f (x) by u(x)x−q and f (x) by f α (x) respectively in
(1.2). Then, by replacing q with αq and p with αp so that (1.2) becomes
! 1q
αq
Z ∞
p1
Z ∞ Z x
1
α
p
f (t)dt
u(x)dx
f (x)v(x)dx
≤C
x 0
0
0
and letting α → 0 we obtain (1.1). The natural choice was of course to try to
use the usual “Muckenhoupt” condition (see [4] and [6])
Z ∞
1q Z x
10
p
p
1−p0
< ∞, p0 =
AM = sup
u(t)dt
v(t)
dt
,
p−1
x>0
x
0
which, with the same substitutions, will be
Z ∞
1q Z
q
−α
AM (α) = sup
u(t)t dt
x>0
x
Anna Wedestig
x
v
α
α−p
p−α
αp
< ∞.
(t)dt
0
However, as α → 0 the first term tends to 0. By making a suitable change in the
condition AM (α) the author was able to give a sufficient condition. In [7] this
problem was solved in a satisfactory way by first proving a new necessary and
sufficient condition for the weighted Hardy inequality (1.2), namely
Z x
1q
− p1
q
AP.S = sup V (x)
u(t)V (t) dt
< ∞,
x>0
Some New Hardy Type
Inequalities and their Limiting
Inequalities
0
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Rx
0
where V (x) = 0 v(t)1−p dt, which was also already proved in the case p = q
in [9], and the bounds for the best possible constant C in (1.2) are
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AP.S ≤ C ≤ p0 AP.S .
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Then, by using the limiting procedure described above, they obtained the following necessary and sufficient condition for the inequality (1.1) to hold for
0<p≤q<∞:
DP.S = sup x
− p1
x
Z
x>0
1q
w(t)dt
< ∞,
0
where
w(t) =
(1.3)
Z t
pq
1
1
ln
dy
u(t)
exp
t 0
v(y)
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
and
1
p
DP.S ≤ C ≤ e DP.S .
The lower bound was also proved in a lemma ([7, Lemma 1]) directly by using
the following test function:
1
s
f (x) = t− p χ[0,t] (x) + (xe)− p t
s−1
p
χ(t,∞) (x),
s > 1, t > 0.
By omitting different intervals the following two lower bounds were pointed
out:
DP.S ≤ C
and
sup
(s − 1)es )
1 + (s − 1)es
p1
sup t
s−1
p
Z
∞
w(x)
sq
xp
Moreover, in [5] the following theorem was proved:
s>1
t>0
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1q
dx
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≤ C.
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Theorem 1.1. Let 0 < p ≤ q < ∞ and u, v be weight functions. Then there
exists a positive constant C < ∞ such that the inequality (1.1) holds for all
f > 0 if and only if there is a s > 1 such that
Z ∞
1q
s−1
w(x)
(1.4)
DO.G = DO.G (s, q, p) = sup t p
< ∞,
sq dx
t>0
t
xp
where w(x) is defined by (1.3). Moreover, if C is the least constant for which
(1.1) holds, then
1
s−1
s−1 p
sup
DO.G (s, q, p) ≤ C ≤ inf e p DO.G (s, q, p).
s>1
s
s>1
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
We see that the condition (1.4) and the condition from Lemma 1 in [7] is the
same but the lower bound is different. Since
p1 1
(s − 1)es )
s−1 p
(1.5)
−
1 + (s − 1)es
s
!
1 p1
ses
s−1 p
−1 >0
=
s
ses + (1 − es )
for all s > 1, we note that the lower bound from Lemma 1 in [7] is better than
that from [5]. This suggests that the bounds for the best constant C in (1.1) with
the condition (1.4) should be
p1
s−1
(s − 1)es )
p D
(1.6) sups>1
D
(s,
q,
p)
≤
Cinf
e
O.G
s>1
O.G (s, q, p).
1 + (s − 1)es
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In [5] the authors also proved that for the upper bound s should be s = 1 + pq .
Thus, the best possible bounds for the constant C should be
(1.7)
sup
s>1
(s − 1)es )
1 + (s − 1)es
p1
1
p
DO.G (s, q, p) ≤ C ≤ e q DO.G (1 + , q, p).
q
In Section 2 of this paper we prove a new necessary and sufficient condition
for the weighted Hardy inequality (1.2) to hold (see Theorem 1). In Section 3 we
make the limiting procedure described above in our new Hardy inequality and
obtain condition (1.4) for the inequality ( 1.1) to hold and we also receive the
expected bounds as in (1.6) or (1.7) (see Theorem 2). Finally, in Section 4 we
prove that a two-dimensional version of the inequality (1.1) can be characterized
by a two-dimensional version of the condition (1.4) and, moreover, the bounds
corresponding to (1.6) or (1.7) hold (see Theorem 3). This result fits perfectly
as an end point inequality of the Hardy inequalities proved by E. Sawyer ([8],
Theorem 1) for the (rectangular) Hardy operator
Z xZ x
H(f )(x1 x2 ) =
f (t1 , t2 )dt1 dt2 .
0
0
We note that for the Hardy case E. Sawyer showed that three conditions were
necessary to characterize the inequality but in our endpoint case only one condition is necessary and sufficient. In Section 5 we give some concluding remarks,
shortly discuss the different weight condition for characterizing the Hardy inequality and prove a two-dimensional Minkowski inequality we needed for the
proof of Theorem 3 but which is also of independent interest (see Proposition
1).
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
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2.
A New Weight Characterization of Hardy’s
Inequality
Our main theorem in this section reads:
Theorem 2.1. Let 1 < p ≤ q < ∞, and s ∈ (1, p) then the inequality
Z
∞
Z
Z
q
1q
≤C
f (t)dt u(x)dx
x
(2.1)
0
0
p1
∞
p
f (x)v(x)dx
Some New Hardy Type
Inequalities and their Limiting
Inequalities
0
holds for all f ≥ 0 iff
(2.2)
AW (s, q, p) = sup V (t)
t>0
where V (t) =
(2.1) then
Rt
0

(2.3)
sup  1<s<p
1−p0
v(x)
p
p−s
p
p−s
p
Z
∞
q ( p−s
p )
u(x)V (x)
Anna Wedestig
1q
dx
< ∞,
t
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dx. Moreover if C is the best possible constant in
1
s−1
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 p1
p
+
s−1
p
 AW (s, q, p)
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≤ C ≤ inf
1<s<p
p−1
p−s
10
p
AW (s, q, p).
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Proof of Theorem 2.1. Let f p (x)v(x) = g(x) in (2.1). Then (2.1) is equivalent
to
Z ∞ Z x
1q
q
p1
Z ∞
1
1
−
(2.4)
g(t) p v(t) p dt u(x)dx
g(x)dx .
≤C
0
0
0
Assume that (2.2) holds. We have, by applying Hölder’s inequality, the fact
p0
0
that DV (t) = v(t)1−p = v(t)− p and Minkowski’s inequality,
Z ∞ Z x
q
1q
1
1
−
g(t) p v(t) p dt u(x)dx
0
∞
Z
x
Z
1
p
g(t) V (t)
=
0
Z
∞
Z
0
=
≤
s−1
p
− s−1
p
V (t)
− p1
v(t)
q
1q
dt u(x)dx
Anna Wedestig
0
x
≤
≤
Some New Hardy Type
Inequalities and their Limiting
Inequalities
0
pq Z
s−1
g(t)V (t) dt
10
p
p − (s − 1)p0
10
p−1
p−s
10
p
(s−1)p0
− p
V (t)
0
− pp
v(t)
! 1q
q0
p
u(x)dx
dt
0
0
p−1
p−s
x
p
∞ Z
Z
0
Z
g(t)V (t)s−1
0
Z
p
Contents
! 1q
pq
x
p−(s−1)p0 q
g(t)V (t)s−1 dt V (x) p p0 u(x)dx
0
∞
AW (s, q, p)
Z
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∞
V (x)q(
p−s
p
) u(x)dx
pq
! p1
dt
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Hence (2.4) and thus (2.1) holds with a constant satisfying the right hand side
inequality in (2.3).
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Now we assume that (2.1) and thus (2.4) holds and choose the test function
p
p
0
0
g(x) =
V (t)−s v(x)1−p χ(0,t) (x) + V (x)−s v(x)1−p χ(t,∞) (x),
p−s
where t is a fixed number > 0. Then the right hand side is equal to
Z t 0
p
p−s
p
−s
1−p0
V (t) v(x)
Z
∞
dx +
1−p0
−s
V (x) v(x)
p1
dx
t
≤
p
p−s
p
1−s
V (t)
1
−
V (t)1−s
1−s
p1
Moreover, the left hand side is greater than
Z
t
∞
Z
0
t
q
1q
Z x
p
− ps
− ps
1−p0
1−p0
dy +
V (y) v(y)
dy u(x)dx
V (t) v(y)
p−s
t
Z ∞ q
1q
s
p
1−
q
=
V (x)( p ) u(x)dx .
p−s
t
Hence, (2.4) implies that
Z ∞
1q
p
p1
1−s
p
p
1
1− ps )q
(
V (x)
u(x)dx
≤C
+
V (t) p
p−s
p−s
s−1
t
i.e., that
p
− p1
Z ∞
1q
s−1
s
p
p
1
1−
q
(
)
+
V (t) p
V (x) p u(x)dx
≤C
p−s
p−s
s−1
t
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
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or, equivalently, that


p
p−s
p
p−s
p
 p1
p
+
1
s−1
 V (t)
s−1
p
∞
Z
s
V (x)(1− p )q u(x)dx
1q
≤ C.
t
We conclude that (2.2) and the left hand side of the estimate of (2.3) hold. The
proof is complete.
Remark 2.1. If we replace the interval (0, ∞) in (2.1) with the interval (a, b),
then, by modifying the proof above, we see that Theorem 2.1 is still valid with
the same bounds and the condition
AW (s, q, p) = sup V (t)
t>0
s−1
p
Z
b
q ( p−s
p )
u(x)V (x)
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
1q
dx
< ∞,
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t
Contents
where V (t) =
Rt
0
v(x)1−p dx.
a
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3.
A Weight Characterization of Pólya-Knopp’s
Inequality
In this section we prove that a slightly improved version of Theorem 1.1 can be
obtained just as a natural limit of our Theorem 2.1.
Theorem 3.1. Let 0 < p ≤ q < ∞ and s > 1. Then the inequality
Z
(3.1)
0
∞
Z x
q
1q
1
exp
ln f (t) dt u(x)dx
x 0
Z ∞
p1
p
≤C
f (x)v(x)dx
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
0
holds for all f > 0 if and only if DO.G (s, q, p) < ∞, where DO.G (s, q, p) is
defined by (1.4). Moreover, if C is the best possible constant in (3.1), then
(3.2)
sup
s>1
(s − 1)es
1 + (s − 1)es
p1
1
q
DO.G (s, q, p) ≤ C ≤ e DO.G
p
1 + , q, p .
q
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Remark 3.1. Theorem 3.1 is due to B. Opic and P. Gurka, but our lower bound
in (3.2) is strictly better (see (1.5)). As mentioned before, other weight characterizations of (3.1) have been proved by L.E. Persson and V. Stepanov [7] and
H.P. Heinig, R. Kerman and M. Krbec [1].
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Proof. If we in the inequality (3.1) replace f p (x)v(x) with f p (x) and let w(x)
be defined as in (1.3), then we see that (3.1) is equivalent to
∞
Z
0
q
1q
Z ∞
p1
Z x
1
p
w(x)dx
exp
ln f (t)dt
≤C
f (x)dx .
x 0
0
Further, by using Theorem 2.1 with u(x) = w(x)x−q and v(x) = 1, we have
that
Z ∞
p1
Z ∞ Z x
q
1q
1
p
≤C
f (x)dx
f (t)dt w(x)dx
(3.3)
x 0
0
0
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
holds for all f ≥ 0 if and only if
(3.4)
AW (s, q, p) = sup t
s−1
p
t>0
Z
t
∞
w(x)
x
sq
p
1q
dx
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= DO.G (s, q, p) < ∞.
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J
Moreover, if C is the best possible constant in (3.3), then

(3.5)
sup  1<s<p
p
p−s
p
p−s
p
 p1
p
+
1
s−1
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≤ C ≤ inf
1<s<p
p−1
p−s
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10
p
DO.G (s, q, p).
Now, we replace f in (3.3) with f α , 0 < α < p, and after that we replace p with
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p
α
and q with
q
α
Z
in (3.3) – (3.5), we find that for 1 < s <
∞
(3.6)
0
p
α
! 1q
Z x
Z ∞
p1
αq
1
α
p
f (t)dt
w(x)dx
≤ Cα
f (x)dx
x 0
0
holds for all f ≥ 0 if and only if
q p
α
DO.G s, ,
= DO.G
(s, q, p) < ∞.
α α
Moreover, if Cα is the best possible constant in (3.6), then

(3.7)

sup  p
1<s< α
p
p−αs
p
p−αs
αp
 p1
αp
+
Some New Hardy Type
Inequalities and their Limiting
Inequalities
1
s−1
Anna Wedestig

 DO.G (s, q, p)
Title Page
≤ C ≤ inf p
1<s< α
p−α
p − αs
p−α
αp
DO.G (s, q, p).
We also note that
Z x
α1
Z
1
1 x
α
f (t)dt
↓ exp
ln f (t)dt, as α → 0+ .
x 0
x 0
We conclude that (3.1) holds exactly when limα→0+ Cα < ∞ and this holds,
according to (3.7), exactly when (3.4) holds. Moreover, when α → 0+ (3.7)
implies that (3.2) holds, where we have inserted the optimal value s = 1 + pq on
the right hand side as pointed out in [5]. The proof is complete.
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4.
Weighted Two-Dimensional Exponential
Inequalities
In [8], E. Sawyer proved a two-dimensional weighted Hardy inequality for the
case 1 < p ≤ q < ∞. More exactly, he showed that for the inequality

Z
∞
Z

∞
Z

x
f (t1, t2 ) dt1 dt2  w(x1, x2 )dx1 dx2 

0
0
 1q
q
Zx2
0
0
Z
∞
Z
∞
p
≤C
p1
f (x1 , x2 ) v (x2 , x2 ) dx1 dx2
0
Some New Hardy Type
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Inequalities
Anna Wedestig
0
to hold, three different weight conditions must be satisfied. Here we will show
that when we consider the endpoint inequality of this Hardy inequality, we only
need one weight condition to characterize the inequality. Our main result in this
section reads:
Theorem 4.1. Let 0 < p ≤ q < ∞, and let u, v and f be positive functions
on R2+ . If 0 < b1 , b2 ≤ ∞, then
q
1q
Z x1Z x2
b2
1
exp
log f (y1 , y2 )dy1 dy2
u(x1 , x2 )dx1 dx2
x1 x2 0 0
0
Z b1 Z b2
p1
p
≤C
f (x1 , x2 )v(x1 , x2 )dx1 dx2
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(4.1)
0
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if and only if
(4.2) DW (s1, s2 , p, q)
:=
sup y
s1 −1
p
1
y
s2 −1
p
2
y1 ∈(0,b1 )
y2 ∈(0,b2 )
Z
b1
y1
Z
b2
−
x1
s1 q
p
−
x2
s2 q
p
1q
w(x1 , x2 )dx1 dx2
< ∞,
y2
where s1 , s2 > 1 and
pq
Z x1 Z x2
1
1
dt1 dt2
w(x1 , x2 ) = exp
u(x1 , x2 )
log
x1 x2 0
v(t1 , t2 )
0
and the best possible constant C in (4.1) can be estimated in the following way:
p1 s2
p1
s1
e (s2 − 1)
e (s1 − 1)
(4.3)
DW (s1 , s2 , p, q)
sup
es2 (s2 − 1) + 1
es1 (s1 − 1) + 1
s1 ,s2 >1
≤C
≤ inf e
s1 +s2 −2
p
s1 ,s2 >1
We will need a two-dimensional version of the following well-known Minkowski
integral inequality:
Z b
Z b
r1
Z x
r r1 Z b
(4.4)
Φ(x)
Ψ(y)dy dx
≤
Ψ(y)
Φ(x)dx dy.
a
a
Anna Wedestig
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DW (s1 , s2 , p, q).
Remark 4.1. For the case p = q = 1, b1 = b2 = ∞ a similar result was
recently proved by H. P. Heinig, R. Kerman and M. Krbec [1] but without the
estimates of the operator norm (= the best constant C in (4.1)) pointed out in
(4.3) here.
a
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The following proposition will be required in the proof of Theorem 4.1. Proposition 4.2 will be proved in Section 5.
Proposition 4.2. Let r > 1, a1 , a2 , b1 , b2 ∈ R, a1 < b1, a2 < b2 and let Φ and
Ψ be measurable functions on [a1 , b1 ] × [a2 , b2 ] . Then
Z
b1
Z
b2
Z
x1
Z
Φ(x1 , x2 )
(4.5)
a1
a2
Ψ(y1 , y2 )dy1 dy2
a1
Z
b1
Z
Z
b1
Z
a2
r1
b2
Ψ(y1 , y2 )
a1
dx1 dx2
a2
b2
≤
r1
r
x2
Φ(x1 , x2 )dx1 d2
y1
dy1 dy2 .
y2
Proof of Theorem 4.1. Assume that (4.2) holds. Let g(x1 , x2 ) = f p (x1 , x2 )v(x1 , x2 )
in (4.1):
Z
b1
0
Z
0
b2
pq
Z x1 Z x2
1
exp
log g(y1 , y2 )dy1 dy2
x1 x2 0
0
! 1q
pq
Z x1 Z x2
1
1
log
dt1 dt2
u(x1 , x2 )dx1 dx2
× exp
x1 x2 0
v(t1 , t2 )
0
Z ∞ Z ∞
p1
≤C
g(x1 , x2 )dx1 dx2
.
0
0
If we let
pq
Z x1 Z x2
1
1
w(x1 , x2 ) = exp
log
dt1 dt2
u(x1 , x2 ),
x1 x2 0
v(t1 , t2 )
0
Some New Hardy Type
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Inequalities
Anna Wedestig
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then we can equivalently write (4.1) as
Z b1Z
(4.6)
0
0
! 1q
pq
Z x1Z x2
1
log g(y1 , y2 )dy1 dy2
w(x1 , x2 )dx1 dx2
exp
x1 x2 0 0
Z b1 Z b2
p1
≤C
g(x1 , x2 )dx1 dx2
.
b2
0
0
Let y1 = x1 t1 and y2 = x2 t2 , then (4.6) becomes
Z b1Z
!
Z 1Z 1
pq
w(x1 , x2 )dx1 dx2
exp
log g(x1 t1 , x2 t2 )dt1 dt2
b2 (4.7)
0
0
0
b1
Z
≤C
0
Z
1
log ts11 −1 ts22 −1 dt1 dt2
p1
b2
g(x1 , x2 )dx1 dx2
0
1
pq
0
0
q
= e−(s1 +s2 −2) p
0
0
0
.
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Contents
and Jensen’s inequality, the left hand side of (4.7) becomes
pq
Z b1 Z b2 Z 1Z 1
s1 +s2 −2
s1 −1 s2 −1
e p
exp
log t1 t2 g(x1 t1 , x2 t2 ) dt1 dt2
0
Anna Wedestig
0
Z
By using the result
Z
exp
Some New Hardy Type
Inequalities and their Limiting
Inequalities
1
q
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1q
× w(x1 , x2 )dx1 dx2
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≤e
s1 +s2 −2
p
b1
Z
b2
Z
0
1
Z
0
1
Z
0
ts11 −1 ts22 −1 g(x1 t1 , x2 t2 )dt1 dt2
pq
0
1q
× w(x1 , x2 )dx1 dx2
=e
s1 +s2 −2
p
b1
Z
b2
Z
0
0
×
x1
Z
x2
Z
0
y1s1 −1 y2s2 −1 g(y1 , y2 )dy1 dy2
pq
−s1 pq
x1
0
−s2 q
x2 p w(x1 , x2 )dx1 dx2
1q
Some New Hardy Type
Inequalities and their Limiting
Inequalities
.
Therefore, by also using Minkowski’s integral inequality (4.5) for p < q and
Fubini’s theorem for p = q, we find that the left hand side in (4.6) can be
estimated as follows:
Z b1 Z b2
s1 +s2 −2
p
≤e
y1s1 −1 y2s2 −1 g(y1 , y2 )
0
0
Z
b1 Z
b2
×
y1
≤e
s1 +s2 −2
p
−s1 pq
x1
−s2 pq
x2
! p1
pq
w(x1 , x2 )dx1 dx2
dy1 dy2
y2
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Z
b1
Z
DW (s1 , s2 , q, p) ·
p1
b2
g(y1 , y2 )dy1 dy2
0
.
0
Hence, (4.6) and, thus, (4.1) holds with a constant C satisfying the right hand
side estimate in (4.3).
Now, assume that (4.1) holds. For fixed t1 and t2 , 0 < t1 < b1 , 0 < t2 < b2 ,
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we choose the test function
g (x1 , x2 ) = g0 (x1 , x2 )
−1
= t−1
1 t2 χ(0,t1 ) (x1 ) χ(0,t2 ) (x2 )
s −1
e−s2 t22
(x1 )
xs22
+
t−1
1 χ(0,t1 )
+
e−s1 ts11 −1
χ(t1 ,∞) (x1 ) t−1
2 χ(0,t2 ) (x2 )
xs11
+
χ(t2 ,∞) (x2 )
s −1
e−(s1 +s2 ) ts11 −1 t22
χ(t1 ,∞)
xs11 xs22
(x1 ) χ(t2 ,∞) (x2 ) .
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
Then the right side of (4.6) yields
Z
b1
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p1
b2
Z
g0 (y1 , y2 ) dy1 dy2
Z t1 Z t2
Z
−1 −1
=
t1 t2 dy1 dy2 +
0
0
Z
0
0
t1
1+
t1
Z
s2 − 1
1−
b2
b2
t2
t2
−s1 s1 −1
t1
−1 e
dy1 dy2 +
t2
y1s1
0
s2 −1 !
e−s2
t2
b1 Z
+
=
Contents
0
e−(s1 +s2 ) ts11 −1 ts22 −1
dy1 dy2
y1s1 y2s2
t1
t2
s1 −1 !
e−s1
t1
+
1−
s1 − 1
b1
Z
JJ
J
e−s2 ts22 −1
t−1
dy1 dy2
1
y2s2
b1
Z
b2
p1
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s1 −1 !
s2 −1 !! p1
t2
e−s1 e−s2
t1
1−
+
1−
(s1 − 1) (s2 − 1)
b1
b2
1
e−s2
e−s1
e−s1 e−s2
p
,
≤ 1+
+
+
s2 − 1 s1 − 1 (s1 − 1) (s2 − 1)
i.e.,
Z
b1
Z
p1
b2
Some New Hardy Type
Inequalities and their Limiting
Inequalities
g0 (y1 , y2 ) dy1 dy2
(4.8)
0
0
≤
es1 (s1 − 1) + 1
es1 (s1 − 1)
p1 es2 (s2 − 1) + 1
es2 (s2 − 1)
p1
.
Moreover, for the left hand side in (4.6) we have
! 1q
pq
Z x1Z x2
1
(4.9)
w(x1 , x2 ) exp
log g(y1 , y2 )dy1 dy2 dx1 dx2
x1 x2 0 0
0
0
! 1q
pq
Z x1Z x2
Z b1Z b2
1
log g(y1 , y2 )dy1 dy2
dx1 dx2 .
≥
w(x1 , x2 ) exp
x1 x2 0 0
t1 t2
Z b1Z
b2
With the function g0 (y1 , y2 ) we get that
Z x1 Z x2
1
exp
log g0 (y1 , y2 )dy1 dy2 = exp (I1 + I2 + I3 + I4 ) ,
x1 x2 0
0
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where
1
I1 =
x1 x2
Z
t1
t2
Z
0
0
t1 t2
t1 t2
−1
log t1 −
log t2 ,
log t−1
t
dy1 dy2 = −
1 2
x1 x2
x1 x2
Z t1 Z x2
−s2 s2 −1
1
t2
−1 e
I2 =
log t1
dy1 dy2
x1 x2 0 t2
y2s2
t1 t2
t1
t1 t2
t1
t1
= − log t1 +
log t1 + (s2 − 1) log t2 +
log t2 − s2 log x2 ,
x1
x1 x2
x1
x1 x2
x1
Z x1 Z t2
−s1 s1 −1
1
t1
−1 e
I3 =
log t2
dy1 dy2
x1 x2 t1 0
y1s1
t2
t1 t2
t2
t1 t2
t2
= − log t2 +
log t2 + (s1 − 1) log t1 +
log t1 − s1 log x1 ,
x2
x1 x2
x2
x1 x2
x2
and
Some New Hardy Type
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Inequalities
Anna Wedestig
Title Page
Contents
I4 =
1
x1 x2
Z
x1
Z
x2
log
t1
t2
e−(s1 +s2 ) ts11 −1 ts22 −1
dy1 dy2
y1s1 y2s2
t2
t1 t2
log t1 −
log t1
x2
x1 x2
t1
t1 t2
+ (s2 − 1) log t2 − (s2 − 1) log t2 −
log t2
x1
x1 x2
t1
t2
− s1 log x1 +
log t1 + s1 log x1
x1
x2
t1
t2
log t2 + s2 log x2 .
− s2 log x2 +
x2
x1
= (s1 − 1) log t1 − (s1 − 1)
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Now we see that
(s −1) (s −1)
I1 + I2 + I3 + I4 = log
t1 1 t2 2
xs11 xs22
!
so that, by (4.9),
Z
b1
0
Z
b2
0
! 1q
pq
Z x1 Z x2
1
w(x1 , x2 ) exp
log g0 (y1 , y2 )dy1 dy2
dx1 dx2
x1 x2 0
0
 1q
b b
"
#q
Z1Z2
(s1 −1) (s2 −1) p
t
t
≥
w(x1 , x2 ) 1 s1 2s2
dx1 dx2  .
x1 x2
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
t1 t2
Hence, by (4.6) and (4.8),
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s1 −1
s2 −1

t1 p t2 p 
Zb1 Zb2
 1q
Contents
− q s1 − q s2
x1 p x2 p w(x1 , x2 )dx1 dx2 
t1 t2
≤C
es1 (s1 − 1) + 1
es1 (s1 − 1)
p1 es2 (s2 − 1) + 1
es2 (s2 − 1)
p1
i.e.
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es1 (s1 − 1)
es1 (s1 − 1) + 1
p1 es2 (s2 − 1)
es2 (s2 − 1) + 1
p1
DW (s1 , s2 , q, p) ≤ C.
We conclude that (4.2) holds and that the left hand inequality in (4.3) holds.
The proof is complete.
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Corollary 4.3. Let 0 < p ≤ q < ∞, and let f be a positive function on R2+ .
Then
Z ∞Z ∞ q
1q
Z x1Z x2
1
α1 α2
(4.10)
exp
log f (y1 , y2 )dy1 dy2 x1 x2 )dx1 dx2
x1 x2 0 0
0
0
Z ∞ Z ∞
p1
β1 β2
p
≤C
f (x1 , x2 )x1 x2 dx1 dx2
0
0
holds with a finite constant C if and only if
Some New Hardy Type
Inequalities and their Limiting
Inequalities
α1 + 1
β1 + 1
=
q
p
Anna Wedestig
and
α2 + 1
β2 + 1
=
q
p
and the best constant C has the following estimate:
sup
s1 ,s2 >1
es1 (s1 − 1)
es2 (s2 − 1)
·
es1 (s1 − 1) + 1 es2 (s2 − 1) + 1
p1 Title Page
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1
1
·
s1 − 1 s2 − 1
1q
≤ C ≤ inf e
s1 ,s2 >1
e
β1 +β2
p
2q
p
q
s1 +s2
+ 2q
p
.
Proof. Apply Theorem 3.1 with the weights u(x1 , x2 ) = xα1 1 xα2 2 and v(x1 , x2 ) =
xβ1 1 xβ2 2 .
Remark 4.2. If p = q, then the inequality (4.10) is sharp with the constant
β1 +β2 +2
C = e p , see Theorem 2.2 in [2]
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5.
Final Remarks and Proof
5.1.
On Minkowski’s integral inequalities
In order to prove the two-dimensional Minkowski integral inequality in Proposition 4.2 we in fact need the following forms of Minkowski’s integral inequality
in one dimension:
Lemma 5.1.
a) Let r > 1, a, b ∈ R, a < b and c < d. If Φ and Ψ are positive measurable
functions on [a, b] , then (4.4) holds.
Anna Wedestig
b) If K(x, y) is a measurable function on [a, b] × [c, d] , then
Z b Z
K(x, y)dy
(5.1)
a
c
! r1
r
d
dx
Z
d
Z
≤
c
b
K r (x, y)dx
Some New Hardy Type
Inequalities and their Limiting
Inequalities
r1
Title Page
dy.
a
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For the reader’s convenience we include here a simple proof.
Proof.
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0
b) Let r =
r
.
r−1
By using the sharpness in Hölder’s inequality, Fubini’s
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theorem and an obvious estimate we have
r ! r1
Z Z
b
d
dx
K(x, y)dy
a
c
Z
=
b
sup
kϕ(x)kL
≤1
Z
c
d
b
Z
≤1
c
d
a
b
Z
sup
kϕ(x)kL
c
Z
d
Z
=
r 0 (a,b)
Anna Wedestig
a
b
r
Some New Hardy Type
Inequalities and their Limiting
Inequalities
K(x, y)ϕ(x)dx dy
≤1
K (x, y)dx
c
K(x, y)ϕ(x)dx dy
r 0 (a,b)
≤
K(x, y)dy dx
a
sup
kϕ(x)kL
d
ϕ(x)
r 0 (a,b)
Z
=
Z
r1
dy.
a
a) The proof follows by applying (5.1) with c = a, d = b and
 1
 Φ r (x)Ψ(y), a ≤ y ≤ x,
K(x, y) =
 0,
x < y ≤ b.
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Proof of Proposition 4.2. Put x = (x1 , x2 ), y = (y1 , y2 ), a = (a1 , a2 ), b =
(b1 , b2 ); by the symbol x ≤ y we mean x1 ≤ y1 and x2 ≤ y2 , etc.; dµ(y) =Ψ(y)dy
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and dν(x) =Φ(x)dx. Then the inequality (4.5) reads:

 r1
r

Z

a≤y≤x
a≤x≤b
Z
dµ(y) dν(x) ≤

 r1

Z
Z
dν(x) dµ(y).

a≤y≤b
y≤x≤b
We use the Hölder inequality and the Fubini theorem and get


a≤x≤b
 r1
r

Z
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Z
dµ(y) dν(x)

a≤y≤x


Z
=
sup
kgk
g(x) 
0
Lr (dν)≤1
dµ(y) dν(x)
a≤y≤x
a≤x≤b

Z
=
a≤y≤b
Z
y≤x≤b
 r1

Z

a≤y≤b
g(x)dν(x) dµ(y)

0
Lr (dν)≤1
≤

Z
sup
kgk
Anna Wedestig
Z
dν(x) dµ(y).
y≤x≤b
The proof is complete.
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Remark 5.1. By using the technique in the proof of Proposition 4.2, we find
that the following n-dimensional version of (4.5) holds:
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Let n ∈ Z+ and r > 1. Then
Z b1
Z bn
(5.2)
···
Φ(x1 , x2 , . . . , xn )
a1
an
Z
x1
×
···
Ψ(y1 , y2 , . . . , yn )dy1 . . . dyn
a1
r1
r
xn
Z
dx1 . . . dxn
an
Z
b1
≤
Z
bn
···
a1
Ψ(y1 , . . . , yn )
an
Z
b1
Z
bn
···
×
y1
r1
Φ(x1 , x2 , . . . , xn )dx1 . . . dxn
dy1 . . . dyn .
yn
Remark 5.2. In view of our proof of Theorem 4.1 and Remark 5.1, we find that
there exists also a n-dimensional variant of Theorem 4.1 (n ∈ Z+ ), where the
actual endpoint Hardy type inequality can be characterized by only one weight
condition.
Remark 5.3. For r = 1 the inequalities in (4.4 ), (4.5), (5.1), and (5.2) are
reduced to equalities according to the Fubini theorem.
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
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5.2.
On the conditions and the best constant in the Hardy
inequality (2.1):
As we have seen, there are at least three different conditions to characterize the
Hardy inequality (2.1) for 1 < p ≤ q < ∞, namely the classical (Muckenhoupt)
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condition (see [4], [6]), the condition by L.E. Persson and V. Stepanov (see
[7]) and the new condition derived in Theorem 2.1. It is difficult to make a
comparison in the general case and here we only consider the power weight
case, i.e. when u(x) = xa−q , v(x) = xb . In this case the inequality ( 2.1) holds
for all f ≥ 0 iff
a+1
b+1
=
q
p
and
1q 1q 10
p
p
1
p−1
AM =
,
q
p − (b + 1)
p − (b + 1)
AP.S
1q 1q 10
p
1
p
1
p−1
=
(p − 1) q
q
p − (b + 1)
p − (b + 1)
1
= AM (p − 1) q ,
Some New Hardy Type
Inequalities and their Limiting
Inequalities
Anna Wedestig
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1
1q 1q 10 p
p
1
p−1
p−1 q
AW (s, q, p) =
s−1
q
p − (b + 1)
p − (b + 1)
1q
p−1
= AM
.
s−1
Moreover, if C is the best constant in (2.1), then C has the following estimates:
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AM ≤ C ≤
1+
q
p0
1q 0
1+
p
q
10
p
AM ,
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AP.S ≤ C ≤ p0 AP.S ,
and

sup  1<s<p
p
p−s
p
p−s
p
 p1
p
+
1
s−1
 AW (s, q, p) ≤ C ≤ inf
Note in this case, with s =
1<s<p
p2 +qp−q
,
p+qp−q
p−1
p−s
10
p
AW (s, q, p).
we have from the upper bound, that
10
p−1 p
1<s<p
p−s
1q 1
p − 1 p0
p−1
= AM inf
1<s<p
p−s
s−1
1
1q q
p 0 p0
= 1+ 0
1+
AM .
p
q
C ≤ inf AW (s, q, p)
We finish this paper by giving a numerical example.
Example 5.1. Let p = 3, q = 4, and s = 1.15 for the lower bound of AW (s, q, p).
Then with the condition AM we have the following bounds:
AM ≤ C ≤ AM · 1.711077405,
with the condition AP.S we have the following bounds:
AM · 1.189207115 ≤ C ≤ AM · 1.783810673,
Some New Hardy Type
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and with the condition AW (s) we have the following bounds:
AM · 1.396254480 ≤ C ≤ AM · 1.711077405.
Some New Hardy Type
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Inequalities
Anna Wedestig
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References
[1] H.P. HEINIG, R. KERMAN AND M. KRBEC, Weighted exponential inequalities, Georgian Math. J., 8(1) (2001), 69–86.
[2] P. JAIN AND R. HASSIJA, Some remarks on two dimensional Knopp type
inequalities, to appear.
[3] P. JAIN, L.E. PERSSON AND A. WEDESTIG, From Hardy to Carleman
and general mean-type inequalities, Function Spaces and Applications,
CRC Press (New York)/Narosa Publishing House (New Delhi)/Alpha Science (Pangbourne) (2000), 117–130.
[4] A. KUFNER AND L.E. PERSSON, Weighted Inequalities of Hardy Type,
World Scientific, to appear, 2003.
[5] B. OPIĆ AND P. GURKA, Weighted inequalities for geometric means, Proc.
Amer. Math. Soc., 3 (1994), 771–779.
[6] B. OPIĆ AND A. KUFNER, Hardy-Type Inequalities, Pitman Research
Notes in Mathematics Series, Vol 211, Longman Scientific and Technical
Harlow, 1990.B.
Some New Hardy Type
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Inequalities
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[7] L.E. PERSSON AND V.D. STEPANOV, Weighted integral inequalities with
the geometric mean operator, J. Inequal. Appl., 7(5) (2002), 727–746. (An
abbreviated version can also be found in Russian Akad. Sci. Dokl. Math., 63
(2001), 201–202).
[8] E. SAWYER, Weighted inequalities for the two-dimensional Hardy operator, Studia Math., 82(1) (1985), 1–16.
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[9] G. TOMASELLI, A class of inequalities, Boll. Un. Mat. Ital., 2 (1969),
622–631.
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