Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 3, Article 61, 2003
SOME NEW HARDY TYPE INEQUALITIES AND THEIR LIMITING
INEQUALITIES
ANNA WEDESTIG
D EPARTMENT OF M ATHEMATICS
L ULEÅ U NIVERSITY
SE- 971 87 L ULEÅ
SWEDEN.
annaw@sm.luth.se
Received 12 December, 2002; accepted 08 September, 2003
Communicated by B. Opić
A BSTRACT. 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.
Key words and phrases: Inequalities, Weights, Hardy’s inequality, Pólya-Knopp’s inequality, Multidimensional inequalities,
Sawyer’s two-dimensional operator.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
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 0
p−1
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
Z x
1
Gf (x) = exp
ln f (t)dt .
x 0
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
I thank Professor Lars-Erik Persson and the referee for some comments and good advice which has improved the final version of this paper.
142-02
2
A NNA W EDESTIG
Let 1 < p ≤ q < ∞. In particular, in [3] the authors tried to find a new condition for the
weighted inequality
Z x
Z ∞ p1
Z ∞
q
1q
1
p
(1.1)
exp
ln f (t)dt
u(x)dx
≤C
f (x)v(x)dx
x 0
0
0
by using the weighted Hardy inequality
Z ∞
p1
Z ∞ Z x
q
1q
p
≤C
(1.2)
f (t)dt u(x)dx
f (x)v(x)dx
0
0
0
α
−q
and replacing u(x) and f (x) by u(x)x and f (x) by f (x) respectively in (1.2). Then, by
replacing q with αq and p with αp so that (1.2) becomes
! 1q
Z ∞
p1
αq
Z ∞ Z x
1
α
p
u(x)dx
≤C
f (x)v(x)dx
f (t)dt
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
,
AM = sup
u(t)dt
v(t)
dt
< ∞, p0 =
p−1
x>0
x
0
which, with the same substitutions, will be
Z ∞
1q Z x
p−α
αp
α
q
−α
AM (α) = sup
u(t)t dt
v α−p (t)dt
< ∞.
x>0
x
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
u(t)V (t)q dt
AP.S = sup V (x)
< ∞,
x>0
Rx
0
1−p0
where V (x) = 0 v(t)
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
AP.S ≤ C ≤ p0 AP.S .
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 < ∞ :
Z x
1q
− p1
w(t)dt
< ∞,
DP.S = sup x
x>0
0
where
(1.3)
w(t) =
Z t
pq
1
1
exp
ln
dy
u(t)
t 0
v(y)
and
1
DP.S ≤ C ≤ e p 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
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
s−1
p
χ(t,∞) (x),
s > 1, t > 0.
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H ARDY T YPE I NEQUALITIES
3
By omitting different intervals the following two lower bounds were pointed out:
DP.S ≤ C
and
1q
p1
Z ∞
s−1
w(x)
(s − 1)es )
≤ C.
sup t p
sup
sq dx
1 + (s − 1)es
t>0
s>1
t
xp
Moreover, in [5] the following theorem was proved:
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
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 1 p1
(s − 1)es )
s−1 p
s−1 p
ses
(1.5)
−
=
−1 >0
ses + (1 − es )
1 + (s − 1)es
s
s
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
(1.6)
sups>1
(s − 1)es )
1 + (s − 1)es
p1
DO.G (s, q, p) ≤ Cinf s>1 e
s−1
p
DO.G (s, q, p).
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
p1
1
(s − 1)es )
p
(1.7)
sup
DO.G (s, q, p) ≤ C ≤ e q DO.G (1 + , q, p).
s
1 + (s − 1)e
q
s>1
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
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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4
A NNA W EDESTIG
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).
2. A N EW W EIGHT C HARACTERIZATION OF H ARDY ’ S I NEQUALITY
Our main theorem in this section reads:
Theorem 2.1. Let 1 < p ≤ q < ∞, and s ∈ (1, p) then the inequality
Z ∞
p1
Z ∞ Z x
q
1q
p
≤C
f (x)v(x)dx
(2.1)
f (t)dt u(x)dx
0
0
0
holds for all f ≥ 0 iff
AW (s, q, p) = sup V (t)
(2.2)
q ( p−s
p )
u(x)V (x)
t>0
Rt
∞
Z
s−1
p
1q
< ∞,
dx
t
0
v(x)1−p dx. Moreover if C is the best possible constant in (2.1 ) then
 p  p1
p
1
p−s
p − 1 p0


AW (s, q, p) ≤ C ≤ inf
sup p
AW (s, q, p).
1<s<p
p
p−s
1<s<p
+ 1
where V (t) =
(2.3)
0
p−s
s−1
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
q
1q
Z ∞
p1
1
1
−
(2.4)
g(t) p v(t) p dt u(x)dx
≤C
g(x)dx .
0
0
0
Assume that (2.2) holds. We have, by applying Hölder’s inequality, the fact that DV (t) =
p0
0
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
0
∞
Z
x
Z
1
p
g(t) V (t)
=
0
Z
∞
Z
0
=
≤
≤
− s−1
p
V (t)
− p1
v(t)
q
1q
dt u(x)dx
0
x
≤
s−1
p
pq Z
s−1
g(t)V (t) dt
0
10
p−1
p−s
10
p
(s−1)p0
− p
V (t)
0
− pp
v(t)
! 1q
q0
p
dt
u(x)dx
0
10
p
p − (s − 1)p0
p−1
p−s
x
p
∞
Z
0
Z
x
Z
0
∞
g(t)V (t)s−1
0
Z
∞
V (x)q(
p−s
p
) u(x)dx
pq
! p1
dt
t
Z
p
! 1q
pq
p−(s−1)p0 q
g(t)V (t)s−1 dt V (x) p p0 u(x)dx
AW (s, q, p)
∞
p1
g(t)dt .
0
Hence (2.4) and thus (2.1) holds with a constant satisfying the right hand side inequality in
(2.3).
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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H ARDY T YPE I NEQUALITIES
5
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 p
p1
Z ∞
p
0
0
V (t)−s v(x)1−p dx +
V (x)−s v(x)1−p dx
p
−
s
t
0
p
p1
1
p
1−s
1−s
V (t) −
≤
V (t)
p−s
1−s
Moreover, the left hand side is greater than
Z
t
∞
Z
0
t
s
p
0
V (t)− p v(y)1−p dy +
p−s
Z
x
− ps
V (y)
1−p0
v(y)
1q
q
dy
u(x)dx
t
Z
=
t
∞
p
p−s
q
(1− ps )q
V (x)
1q
u(x)dx
.
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
p
p
1
1− ps )q
(
p
+
V (t)
V (x)
u(x)dx
≤C
p−s
p−s
s−1
t
or, equivalently, that
 p  p1
p
Z ∞
1q
s−1
s
p−s
1−
q
(
)
 p
 V (t) p
V (x) p u(x)dx
≤ C.
p
1
t
+ s−1
p−s
We conclude that (2.2) and the left hand side of the estimate of (2.3) hold. The proof is complete.
Remark 2.2. 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
Z b
1q
p−s
s−1
u(x)V (x)q( p ) dx
< ∞,
AW (s, q, p) = sup V (t) p
t>0
t
Rt
0
where V (t) = a v(x)1−p dx.
3. A W EIGHT C HARACTERIZATION OF P ÓLYA -K NOPP ’ S I NEQUALITY
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 ∞ Z x
q
1q
Z ∞
p1
1
p
(3.1)
exp
ln f (t) dt u(x)dx
≤C
f (x)v(x)dx
x 0
0
0
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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6
A NNA W EDESTIG
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
Remark 3.2. 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].
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 x
Z ∞
p1
1
p
exp
ln f (t)dt
w(x)dx
≤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
(3.3)
0
q
1q
Z x
Z ∞
p1
1
p
f (t)dt w(x)dx
≤C
f (x)dx
x 0
0
holds for all f ≥ 0 if and only if
(3.4)
AW (s, q, p) = sup t
s−1
p
t>0
Z
∞
w(x)
sq
1q
dx
= DO.G (s, q, p) < ∞.
xp
t
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
 DO.G (s, q, p) ≤ C
≤ inf
1<s<p
p−1
p−s
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
q
α
in (3.3) – (3.5), we find that for 1 < s <
Z
(3.6)
0
∞
p
α
and q with
p
α
! 1q
Z ∞
p1
Z x
αq
1
p
α
w(x)dx
≤ Cα
f (x)dx
f (t)dt
x 0
0
holds for all f ≥ 0 if and only if
DO.G
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
q p
α
s, ,
= DO.G
(s, q, p) < ∞.
α α
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H ARDY T YPE I NEQUALITIES
7
Moreover, if Cα is the best possible constant in (3.6), then
 p1
 p

sup  p
(3.7)
1<s< α
p
p−αs
p
p−αs
αp
α
+
1
s−1

 DO.G (s, q, p) ≤ C
≤ inf p
1<s< α
p−α
p − αs
p−α
αp
DO.G (s, q, p).
We also note that
α1
Z x
Z
1 x
1
α
↓ exp
ln f (t)dt, as α → 0+ .
f (t)dt
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.
4. W EIGHTED T WO -D IMENSIONAL E XPONENTIAL I NEQUALITIES
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


q
 1q
Z ∞ Z ∞ Z x Zx2


f (t1, t2 ) dt1 dt2  w(x1, x2 )dx1 dx2 
0
0
0
0
∞
Z
Z
∞
f p (x1 , x2 ) v (x2 , x2 ) dx1 dx2
≤C
0
p1
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
Z
b1
Z
(4.1)
0
0
b2
q
1q
Z x1 Z x2
1
exp
log f (y1 , y2 )dy1 dy2
u(x1 , x2 )dx1 dx2
x1 x2 0
0
Z b1 Z b2
p1
≤C
f p (x1 , x2 )v(x1 , x2 )dx1 dx2
0
0
if and only if
(4.2) DW (s1, s2 , p, q) :=
sup y
s1 −1
p
1
y1 ∈(0,b1 )
y2 ∈(0,b2 )
y
s2 −1
p
2
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
log
dt1 dt2
u(x1 , x2 )
w(x1 , x2 ) = exp
x1 x2 0
v(t1 , t2 )
0
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8
A NNA W EDESTIG
and the best possible constant C in (4.1) can be estimated in the following way:
s1
p1 s2
p1
e (s1 − 1)
e (s2 − 1)
sup
(4.3)
DW (s1 , s2 , p, q)
es1 (s1 − 1) + 1
es2 (s2 − 1) + 1
s1 ,s2 >1
≤C
≤ inf e
s1 +s2 −2
p
s1 ,s2 >1
DW (s1 , s2 , p, q).
Remark 4.2. 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.
We will need a two-dimensional version of the following well-known Minkowski integral
inequality:
Z b
Z x
r r1 Z b
Z b
r1
(4.4)
Φ(x)
Ψ(y)dy dx
≤
Ψ(y)
Φ(x)dx dy.
a
a
a
y
The following proposition will be required in the proof of Theorem 4.1. Proposition 4.3 will be
proved in Section 5.
Proposition 4.3. 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
b1
Z
b2
Z
Z
x1
Φ(x1 , x2 )
(4.5)
a1
a2
Ψ(y1 , y2 )dy1 dy2
a1
r1
r
x2
Z
dx1 dx2
a2
Z
b1
Z
b2
≤
Z
b1
Z
Ψ(y1 , y2 )
a1
a2
r1
b2
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):
pq
Z b1 Z b2 Z x1 Z x2
1
log g(y1 , y2 )dy1 dy2
exp
x1 x2 0
0
0
0
! 1q
pq
Z x1 Z x2
1
1
× exp
log
dt1 dt2
u(x1 , x2 )dx1 dx2
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
dt1 dt2
w(x1 , x2 ) = exp
log
u(x1 , x2 ),
x1 x2 0
v(t1 , t2 )
0
then we can equivalently write (4.1) as
Z
b1
Z
(4.6)
0
0
b2
! 1q
pq
Z x1 Z x2
1
w(x1 , x2 )dx1 dx2
exp
log g(y1 , y2 )dy1 dy2
x1 x2 0
0
Z b1 Z b2
p1
.
≤C
g(x1 , x2 )dx1 dx2
0
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H ARDY T YPE I NEQUALITIES
9
Let y1 = x1 t1 and y2 = x2 t2 , then (4.6) becomes
b1
Z
Z
b2
(4.7)
0
1
Z
exp
log g(x1 t1 , x2 t2 )dt1 dt2
0
0
! 1q
pq
1
Z
w(x1 , x2 )dx1 dx2
0
b1
Z
Z
p1
b2
≤C
g(x1 , x2 )dx1 dx2
0
.
0
By using the result
1
Z
1
Z
log ts11 −1 ts22 −1 dt1 dt2
exp
0
pq
q
= e−(s1 +s2 −2) p
0
and Jensen’s inequality, the left hand side of (4.7) becomes
e
s1 +s2 −2
p
Z
b1
Z
0
≤e
s1 +s2 −2
p
b2
0
Z
b1
Z
=e
Z
0
b1
Z
b2
0
0
Z
log ts11 −1 ts22 −1 g(x1 t1 , x2 t2 ) dt1 dt2
! 1q
pq
w(x1 , x2 )dx1 dx2
0
1
Z
0
b2
Z
1
1
Z
0
0
s1 +s2 −2
p
1
Z
exp
ts11 −1 ts22 −1 g(x1 t1 , x2 t2 )dt1 dt2
! 1q
pq
w(x1 , x2 )dx1 dx2
0
x1
x2
Z
0
y1s1 −1 y2s2 −1 g(y1 , y2 )dy1 dy2
pq
−s1 q −s2 q
x1 p x2 p w(x1 , x2 )dx1 dx2
! 1q
.
0
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:
≤e
s1 +s2 −2
p
Z
0
≤e
s1 +s2 −2
p
b1
Z
b2
y1s1 −1 y2s2 −1 g(y1 , y2 )
Z
0
b1
y1
Z
b1
Z
DW (s1 , s2 , q, p) ·
Z
b2
−s1 q −s2 q
x1 p x2 p w(x1 , x2 )dx1 dx2
dy1 dy2
y2
p1
b2
g(y1 , y2 )dy1 dy2
0
! p1
pq
.
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 , we choose the
test function
−1
g (x1 , x2 ) = g0 (x1 , x2 ) = t−1
1 t2 χ(0,t1 ) (x1 ) χ(0,t2 ) (x2 )
s −1
+ t−1
1 χ(0,t1 ) (x1 )
+
e−s2 t22
xs22
χ(t2 ,∞) (x2 )
e−s1 ts11 −1
χ(t1 ,∞) (x1 ) t−1
2 χ(0,t2 ) (x2 )
xs11
s −1
e−(s1 +s2 ) ts11 −1 t22
+
xs11 xs22
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
χ(t1 ,∞) (x1 ) χ(t2 ,∞) (x2 ) .
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10
A NNA W EDESTIG
Then the right side of (4.6) yields
Z b1 Z b2
p1
g0 (y1 , y2 ) dy1 dy2
0
0
Z t1 Z t2
Z
−1 −1
=
t1 t2 dy1 dy2 +
e−s2 ts22 −1
dy1 dy2
y2s2
0
0
0
t2
p1
Z b1 Z b2 −(s1 +s2 ) s1 −1 s2 −1
Z b1 Z t2
−s1 s1 −1
e
t
t
e
t
1
2
1
dy1 dy2 +
dy1 dy2
+
t−1
2
s1
s1 s2
y
y
y
t1
t2
t1
0
1
1 2
s2 −1 !
s1 −1 !
t2
e−s1
t1
e−s2
= 1+
1−
+
1−
s2 − 1
b2
s1 − 1
b1
s1 −1 !
s2 −1 !! p1
e−s1 e−s2
t1
t2
+
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)
t1
Z
b2
t−1
1
i.e.,
b1
Z
p1
b2
Z
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
Z
b1
b2
Z
(4.9)
0
0
Z
! 1q
pq
Z x1 Z x2
1
w(x1 , x2 ) exp
dx1 dx2
log g(y1 , y2 )dy1 dy2
x1 x2 0
0
b1
Z
b2
≥
t1
t2
! 1q
pq
Z x1 Z x2
1
w(x1 , x2 ) exp
log g(y1 , y2 )dy1 dy2
dx1 dx2
.
x1 x2 0
0
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
where
1
I1 =
x1 x2
Z
t1
Z
t2
t1 t2
t1 t2
−1
log t−1
dy1 dy2 = −
log t1 −
log t2 ,
1 t2
x1 x2
x1 x2
0
0
Z t1 Z x2
−s2 s2 −1
1
t2
−1 e
I2 =
log t1
dy1 dy2
x1 x2 0 t2
y2s2
t1
t1 t2
t1
t1 t2
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
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H ARDY T YPE I NEQUALITIES
11
and
e−(s1 +s2 ) ts11 −1 ts22 −1
log
dy1 dy2
y1s1 y2s2
t1
t2
t2
t1 t2
= (s1 − 1) log t1 − (s1 − 1) log t1 −
log t1
x2
x1 x2
t1 t2
t1
+ (s2 − 1) log t2 − (s2 − 1) log t2 −
log t2
x1
x1 x2
t2
t1
− s1 log x1 +
log t1 + s1 log x1
x1
x2
t1
t2
− s2 log x2 +
log t2 + s2 log x2 .
x2
x1
1
I4 =
x1 x2
Z
x1
Z
x2
Now we see that
(s −1) (s −1)
t1 1 t2 2
xs11 xs22
I1 + I2 + I3 + I4 = log
!
so that, by (4.9),
Z
b1
Z
0
b2
0
! 1q
pq
Z x1 Z x2
1
log g0 (y1 , y2 )dy1 dy2
dx1 dx2
w(x1 , x2 ) exp
x1 x2 0
0
b b
 1q
"
#q
Z1Z2
(s1 −1) (s2 −1) p
t
t
w(x1 , x2 ) 1 s1 2s2
≥
dx1 dx2  .
x1 x2
t1 t2
Hence, by (4.6) and (4.8),
s1 −1
p
t1
s2 −1
p
t2
b b
 1q
Z1Z2 q
q
− s1 − 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.
p1
p1 s2
e (s2 − 1)
es1 (s1 − 1)
DW (s1 , s2 , q, p) ≤ C.
es2 (s2 − 1) + 1
es1 (s1 − 1) + 1
We conclude that (4.2) holds and that the left hand inequality in (4.3) holds. The proof is
complete.
Corollary 4.4. Let 0 < p ≤ q < ∞, and let f be a positive function on R2+ . Then
Z
∞
Z
(4.10)
0
0
∞
q
1q
Z x1 Z x2
1
exp
log f (y1 , y2 )dy1 dy2
xα1 1 xα2 2 )dx1 dx2
x1 x2 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
α1 + 1
β1 + 1
=
q
p
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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12
A NNA W EDESTIG
and
α2 + 1
β2 + 1
=
q
p
and the best constant C has the following estimate:
s1
p1 1q
2q
β1 +β2
e (s1 − 1)
p
es2 (s2 − 1)
1
1
p
sup
·
·
e
s
s
e 1 (s1 − 1) + 1 e 2 (s2 − 1) + 1
s1 − 1 s2 − 1
q
s1 ,s2 >1
≤ C ≤ inf e
s1 +s2
+ 2q
p
s1 ,s2 >1
Proof. Apply Theorem 3.1 with the weights u(x1 , x2 ) = xα1 1 xα2 2 and v(x1 , x2 ) = xβ1 1 xβ2 2 .
Remark 4.5. If p = q, then the inequality (4.10) is sharp with the constant C = e
Theorem 2.2 in [2]
β1 +β2 +2
p
.
, see
5. F INAL R EMARKS AND P ROOF
5.1. On Minkowski’s integral inequalities. In order to prove the two-dimensional Minkowski
integral inequality in Proposition 4.3 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.
b) If K(x, y) is a measurable function on [a, b] × [c, d] , then
r ! r1 Z d Z b
r1
Z b Z d
(5.1)
K(x, y)dy dx
≤
K r (x, y)dx dy.
a
c
a
c
For the reader’s convenience we include here a simple proof.
Proof.
0
r
. By using the sharpness in Hölder’s inequality, Fubini’s theorem and an
b) Let r = r−1
obvious estimate we have
Z d
r ! r1
Z b Z d
Z b
=
sup
ϕ(x)
K(x, y)dy dx
K(x, y)dy dx
a
kϕ(x)kL
c
r 0 (a,b)
≤1
a
Z
=
c
d Z
b
sup
kϕ(x)kL
Z
K(x, y)ϕ(x)dx dy
≤1
r 0 (a,b)
c
d
a
b
Z
≤
sup
kϕ(x)kL
c
Z
d
Z
=
≤1
r 0 (a,b)
K(x, y)ϕ(x)dx dy
a
b
r
r1
K (x, y)dx
c
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.
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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H ARDY T YPE I NEQUALITIES
13
Proof of Proposition 4.3. 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 and dν(x) =Φ(x)dx.
Then the inequality (4.5) reads:

 r1
r

Z

Z
Z
dµ(y) dν(x) ≤

a≤y≤x
a≤x≤b
 r1

Z
dν(x) dµ(y).

a≤y≤b
y≤x≤b
We use the Hölder inequality and the Fubini theorem and get

 r1
r

Z

Z

Z
dµ(y) dν(x) =

a≤x≤b
sup
kgk
a≤y≤x

Z
g(x) 
0
Lr (dν)≤1
dµ(y) dν(x)
a≤y≤x
a≤x≤b

Z
=
sup
kgk
L
r0

Z
g(x)dν(x) dµ(y)

(dν)≤1
a≤y≤b
y≤x≤b
 r1

Z
Z
≤

a≤y≤b
dν(x) dµ(y).
y≤x≤b
The proof is complete.
Remark 5.2. By using the technique in the proof of Proposition 4.3, we find that the following
n-dimensional version of (4.5) holds:
Let n ∈ Z+ and r > 1. Then
Z
b1
Z
bn
···
(5.2)
a1
Φ(x1 , x2 , . . . , xn )
an
Z
x1
×
Z
···
Ψ(y1 , y2 , . . . , yn )dy1 . . . dyn
a1
Z
an
b1
≤
r1
r
xn
Z
dx1 . . . dxn
bn
···
Ψ(y1 , . . . , yn )
a1
an
Z
b1
Z
···
×
y1
r1
bn
Φ(x1 , x2 , . . . , xn )dx1 . . . dxn
dy1 . . . dyn .
yn
Remark 5.3. In view of our proof of Theorem 4.1 and Remark 5.2, 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 5.4. For r = 1 the inequalities in (4.4 ), (4.5), (5.1), and (5.2) are reduced to equalities
according to the Fubini theorem.
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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14
A NNA W EDESTIG
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) 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
AM
AP.S
1q 10
1q p
1
p−1
p
,
=
p − (b + 1)
p − (b + 1)
q
1q 1q 10
p
1
p
1
p−1
=
(p − 1) q
q
p − (b + 1)
p − (b + 1)
1
= AM (p − 1) q ,
1q 10 1
1q p
1
p−1
p−1 q
p
AW (s, q, p) =
q
p − (b + 1)
p − (b + 1)
s−1
1q
p−1
= AM
.
s−1
Moreover, if C is the best constant in (2.1), then C has the following estimates:
1 1
q q
p 0 p0
AM ≤ C ≤ 1 + 0
1+
AM ,
p
q
AP.S ≤ C ≤ p0 AP.S ,
and

sup  1<s<p
p
p−s
p
p−s
p
 p1
p
+
Note in this case, with s =
1
s−1
 AW (s, q, p) ≤ C ≤ inf
1<s<p
p−1
p−s
10
p
AW (s, q, p).
p2 +qp−q
,
p+qp−q
we have from the upper bound, that
1
p − 1 p0
C ≤ inf AW (s, q, p)
1<s<p
p−s
1q 1
p−1
p − 1 p0
= AM inf
1<s<p
s−1
p−s
1
1q q
p 0 p0
= 1+ 0
1+
AM .
p
q
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,
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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H ARDY T YPE I NEQUALITIES
15
with the condition AP.S we have the following bounds:
AM · 1.189207115 ≤ C ≤ AM · 1.783810673,
and with the condition AW (s) we have the following bounds:
AM · 1.396254480 ≤ C ≤ AM · 1.711077405.
R EFERENCES
[1] H.P. HEINIG, R. KERMAN
J., 8(1) (2001), 69–86.
AND
M. KRBEC, Weighted exponential inequalities, Georgian Math.
[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.
[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.
[9] G. TOMASELLI, A class of inequalities, Boll. Un. Mat. Ital., 2 (1969), 622–631.
J. Inequal. Pure and Appl. Math., 4(3) Art. 61, 2003
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