IJMMS 30:9 (2002) 515–520
PII. S0161171202111276
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
ON NEW GENERALIZATIONS OF HARDY’S INTEGRAL
INEQUALITIES
LÜ ZHONGXUE and XIE HONGZHENG
Received 1 November 2001
We give some new generalizations of Hardy’s integral inequalities.
2000 Mathematics Subject Classification: 26D15.
1. Introduction. The classical Hardy inequality [3] states that: for f (x) ≥ 0, p > 1,
∞
1/p + 1/q = 1, and 0 < 0 f p (x)dx < ∞,
∞
0
1
x
x
p
f (t)dt
dx < qp
∞
0
f p (t)dt,
(1.1)
0
where q = p/(p − 1) is the best possible constant.
∞
The dual form of (1.1) is as follows: if 0 < 0 (xf (x))p dx < ∞, then
∞ ∞
p
f (t)dt
0
dx < p p
x
∞
p
tf (t) dt,
(1.2)
0
where the constant p p in (1.2) is still best possible.
Bicheng et al. [2] gave some new generalizations of (1.1) which can be stated as
follows:
b x
p
1/q p b
1
a
f (t)dt dx < qp 1 −
f p (t)dt;
(1.3)
x a
b
a
a
∞
∞ x
p
1
1 − θp (t) f p (t)dt
f (t)dt dx < qp
0 < θp (t) < 1 ;
(1.4)
x a
a
a
b
b x
p
1/q 1
t
f (t)dt dx < qp
(1.5)
1−
f p (t)dt,
x
b
0
0
0
∞ p where θp (t) = (1/p) k=1 k+1
(−1)k−1 (a/t)k/q > 0 for t > a > 0, and θp (a) = 1/q.
Recently, Becheng and Debnath [1] gave improvement of (1.3) and some generalizations of (1.2):
b
p
1 x
f (t)dt dx < qp ηp (a, b)
f p (t)dt;
x a
a
a
∞ ∞
∞
p
1/p p
a
f (t)dt dx < p p
1−
tf (t) dt;
t
a
x
a
b
b b
p
p
p
f (t)dt dx < p
µp (t) tf (t) dt,
b
0
x
0
(1.6)
516
L. ZHONGXUE AND X. HONGZHENG
b
where the constants ηp (a, b) = max a≤t≤b {(1/q)t 1/q t x −1−1/q [1 − (a/x)1/q ]p−1 dx},
µp (t) = (1/p){1 − (t/b)1/p }p (b/t)1/p .
In this paper, we show some new improvements and generalizations of the inequalities (1.1) and (1.2).
2. Main results
∞
Lemma 2.1. Let a ≥ 0, p > 1, 1/p +1/q = 1−1/r , f ≥ 0, r > 1, and 0 < a f p (t)dt <
∞. Then, there exists a real number x0 ∈ (a, ∞) such that, for any x > x0 , the following
inequality is true:
x
p
f (t)dt
p−1 pq(p − 1)
1 p−1
1−
(p + q)(p − 1) − p
r
p−1 x
× x 1−1/(1−1/r )q(p−1) − a1−1/(1−1/r )q(p−1)
t 1/(1−1/r )q f p (t)dt.
<
a
a
(2.1)
Proof. By Hölder’s inequality, we have
x
x
p
f (t)dt
a
=
x
≤
t 1/(1−1/r )pq f (t)t −1/(1−1/r )pq dt
a
t
1/(1−1/r )q
p
f (t)dt
a
t −1/(1−1/r )pq
p/(p−1)
p−1
dt
a
p−1 pq(p − 1)
1 p−1
1−
(p + q)(p − 1) − p
r
p−1 x
× x 1−1/(1−1/r )q(p−1) − a1−1/(1−1/r )q(p−1)
t 1/(1−1/r )q f p (t)dt.
=
x p
a
(2.2)
We need to show that there exists a real number x0 ∈ (a, ∞), such that (2.2) does
not assume equality for any x > x0 . Otherwise, there exists x = xn ∈ (a, ∞), where
n = 1, 2, 3, . . . , xn ↑ ∞, such that (2.2) becomes an equality. By the same argument, there
exists a real number c > 0, and an integer N, such that for n > N,
p/(p−1)
p
t 1/(1−1/r )pq f (t) = c t −1/(1−1/r )pq
a.e. in a, xn .
(2.3)
Hence
xn
a
f p (t)dt =
=
xn
c
a
xn
t −1/(1−1/r )q(p−1)
dt
t 1/(1−1/r )q
ct
−p/(1−1/r )q(p−1)
a
This is a contradiction to the fact that 0 <
the proof is complete.
∞
a
(2.4)
dt → ∞ as n → ∞.
f p (t)dt < ∞. Hence, (2.1) holds true and
517
ON NEW GENERALIZATIONS OF HARDY’S INTEGRAL INEQUALITIES
Lemma 2.2. Let b > 0, p > 1, 1/p + 1/q = 1 − 1/r , f ≥ 0, r > 1, and let 0 <
b p−1+1/(1−1/r ) p
f (t)dt < ∞. Then, there exists a real number x0 ∈ (0, b) such that,
0 t
for any x ∈ (0, x0 ), the following inequality is true:
b
p
f (t)dt
x
<
p−1 p−1
1
1−
p
x −1/(1−1/r )p − b−1/(1−1/r )p
r
b
×
t p−1+(p−1)/(1−1/r )p f p (t)dt.
(2.5)
x
Proof. For any x ∈ (0, b), by Hölder’s inequality, we have
b
b
p
f (t)dt
x
=
2
2
t (1+(1−1/r )p)(p−1)/(1−1/r )p f (t)t −(1+(1−1/r )p)(p−1)/(1−1/r )p dt
p
x
b
b
t (1+(1−1/r )p)(p−1)/(1−1/r )p f p (t)dt
≤
x
t −(1+(1−1/r )p)/(1−1/r )p dt
p−1
x
p−1 p−1
1
p
=
1−
x −1/(1−1/r )p − b−1/(1−1/r )p
r
b
×
t p−1+(p−1)/(1−1/r )p f p (t)dt.
x
(2.6)
We need to show that there exists a real number x0 ∈ (0, b), such that (2.6) does not
assume equality for any x ∈ (0, x0 ). Otherwise, there exists x = xn ∈ (0, b), where
n = 1, 2, 3, . . . , xn ↓ 0, such that (2.6) becomes an equality. Then there exist cn and dn
which are not always zero, such that (see [4, page 29])
p
2
cn t (1+(1−1/r )p)(p−1)/(1−1/r )p f (t)
2 p/(p−1)
= dn t −(1+(1−1/r )p)(p−1)/(1−1/r )p
a.e. in xn , b .
(2.7)
Since f (t) ≠ 0 a.e. in (0, b), there exists an integer N such that, for n > N, f (t) ≠ 0
a.e. in (0, xn ). Thus, for both cn = c ≠ 0 and dn = d ≠ 0 for n > N,
b
0
b
t p−1+1/(1−1/r ) f p (t)dt = lim
n→∞ xn
d
t −(1+1/(1−1/r )p)
dt = lim
t 1−(1+1/(1−1/r )p)
c n→∞
b
xn
dt
= ∞.
t
(2.8)
b
This contradicts the fact that 0 < 0 t p−1+1/(1−1/r ) f p (t)dt < ∞. Hence, (2.5) is valid
and this completes the proof of the lemma.
Lemma 2.3. Let a > 0, p > 1, 1/p + 1/q = 1 − 1/r , f ≥ 0, r > 1, and 0 <
∞ p−1+1/(1−1/r ) p
f (t)dt < ∞. Then, there exists a real number x0 ∈ (a, ∞) such that,
a t
for any x ∈ (a, x0 ), the following inequality is true:
∞
p
f (t)dt
x
<
∞
p−1
1
1−
x −(p−1)/(1−1/r )p
t p−1+(p−1)/(1−1/r )p f p (t)dt.
p
r
x
(2.9)
518
L. ZHONGXUE AND X. HONGZHENG
Proof. For any x ∈ (a, ∞), by Hölder’s inequality, we have
∞
p
f (t)dt
x
≤
∞
p−1
1
1−
x −(p−1)/(1−1/r )p
t p−1+(p−1)/(1−1/r )p f p (t)dt.
p
r
x
(2.10)
We show that there exists a real number x0 ∈ (a, ∞), such that (2.10) does not assume equality for any x ∈ (a, x0 ). Otherwise, there exists x = xn ∈ (a, ∞), where
n = 1, 2, 3, . . . , xn ↓ a, such that (2.10) becomes an equality. By the same argument
there exist a real number c > 0, and an integer N, such that for n > N,
p
2
t (1+(1−1/r )p)(p−1)/(1−1/r )p f (t)
2 p/(p−1)
= c t −(1+(1−1/r )p)(p−1)/(1−1/r )p
a.e. in xn , ∞ ,
(2.11)
∞
∞
and hence a t p−1+1/(1−1/r ) f p (t)dt = c limn→∞ xn (dt/t) = ∞. This contradicts the fact
∞ p−1+1/(1−1/r ) p
f (t)dt < ∞. Hence (2.9) is valid and this completes the proof
that 0 < a t
of the lemma.
Theorem 2.4. Let 0 < a < b, p > 1, 1/p + 1/q = 1 − 1/r , f ≥ 0, r > 1, and 0 <
∞ p
a f (t)dt < ∞. Then
b
a
1
x
x
p
f (t)dt
dx <
a
pq(p − 1)
(p + q)(p − 1) − p
b
p 1 p
η(a, b)
f p (t)dt,
1−
r
a
(2.12)
where the constant
(p + q)(p − 1) − p 1/(1−1/r )q
t
pq(p − 1)(1 − 1/r )
b
1−1/(1−1/r )q(p−1) p−1
a
×
x −1−1/(1−1/r )q 1 −
dx ,
x
t
1−1/(1−1/r )q(p−1) p
a
(p + q)(p − 1) − p
1−
η(a, b) <
.
p(p − 1)
b
η(a, b) = max
a≤t≤b
(2.13)
Proof. In view of the proof of Lemma 2.1, we obtain
b
a
1
x
x
p
f (t)dt
dx
a
p−1 pq(p − 1)
1 p−1
1−
(p + q)(p − 1) − p
r
b b
1−1/(1−1/r )q(p−1) p−1
a
×
x −1−1/(1−1/r )q 1 −
dx t 1/(1−1/r )q f p (t)dt
x
a
t
p p b
1
pq(p − 1)
g(t)f p (t)dt,
1−
=
(p + q)(p − 1) − p
r
a
(2.14)
<
ON NEW GENERALIZATIONS OF HARDY’S INTEGRAL INEQUALITIES
519
where the weight function g(t) is defined by
(p + q)(p − 1) − p 1/(1−1/r )q
t
pq(p − 1)(1 − 1/r )
b
1−1/(1−1/r )q(p−1) p−1
a
×
x −1−1/(1−1/r )q 1 −
dx,
x
t
g(t) :=
(2.15)
t ∈ [a, b].
Setting η(a, b) := max a≤t≤b g(t), since g(t) is a nonconstant continuous function,
then by (2.14) we have (2.12). Since g(b) = 0, and for any t ∈ [a, b),
(p + q)(p − 1) − p 1/(1−1/r )q
t
g(t) <
pq(p − 1)(1 − 1/r )
=
b
x
t
−1−1/(1−1/r )q
1−1/(1−1/r )q(p−1) p−1
a
dx
1−
b
1/(1−1/r )q 1−1/(1−1/r )q(p−1) p−1 t
(p + q)(p − 1) − p
a
1−
1−
pq(p − 1)
b
b
1−1/(1−1/r )q(p−1) p−1 1/(1−1/r )q a
a
(p + q)(p − 1) − p
1−
1−
≤
pq(p − 1)
b
b
1−1/(1−1/r )q(p−1) p
a
(p + q)(p − 1) − p
.
1−
<
pq(p − 1)
b
(2.16)
This completes the proof.
Theorem 2.5. Let a > 0, p > 1, 1/p + 1/q = 1 − 1/r , f ≥ 0, r > 1, and 0 <
∞
∞ p−1+1/(1+1/r ) p
p
f (t)dt < ∞. Then
a (tf (t)) dt < ∞, 0 < a t
∞ ∞
p
f (t)dt
a
dx <
x
1−
∞
p
(r −p)/(r −1)p p
r
1
a
p
1−
tf (t) dt.
r
r −p a
t
(2.17)
Proof. Applying (2.9), we have
∞ ∞
p
f (t)dt
a
dx
x
∞
p−1 ∞
1
p
x −(p−1)/(1−1/r )p
t p−1+∗(p−1)/(1−1/r )p f p (t)dt dx
r
a
x
p−1 ∞ t
1
p
x −(p−1)/(1−1/r )p dx t p−1+(p−1)/(1−1/r )p f p (t)dt
=
1−
r
a
a
∞
(r −p)/(r −1)p p
p
r
1
a
=
1−
1−
tf (t) dt.
p
r
r −p a
t
<
1−
(2.18)
Hence, (2.17) is valid. This completes the proof of the theorem.
Theorem 2.6. Let b > 0, p > 1, 1/p + 1/q = 1 − 1/r , f ≥ 0, r > 1, and 0 <
b
b p−1+1/(1−1/r ) p
p
f (t)dt < ∞. Then
0 (tf (t)) dt < ∞, 0 < 0 t
b b
p
f (t)dt
0
x
p b
p
1
dx <
1−
µ(t) tf (t) dt,
p
r
0
(2.19)
520
L. ZHONGXUE AND X. HONGZHENG
t
where µ(t) := 1/(1 − 1/r )p{ 0 x −(p−1)/(1−1/r )p [1−(x/b)1/(1−1/r )p ]p−1 dx}t (p−r )/(r −1)p ,
t ∈ (0, b].
Proof. Applying (2.5), we have
b b
p
f (t)dt
0
1−
dx <
x
p−1 b p−1
1
x −1/(1−1/r )p − b−1/(1−1/r )p
p
r
0
b
×
t p−1+(p−1)/(1−1/r )p f p (t)dt dx
x
p−1 b t
1/(1−1/r )p p−1
1
x
p
x −(p−1)(1−1/r )p 1−
dx
r
b
0
0
p
(p−r )/(r −1)p
×t
tf (t) dt
=
1−
=
1−
p b
p
1
p
µ(t) tf (t) dt,
r
0
(2.20)
t
where µ(t) := 1/(1 − 1/r )p{ 0 x −(p−1)/(1−1/r )p [1−(x/b)1/(1−1/r )p ]p−1 dx}t (p−r )/(r −1)p ,
t ∈ (0, b]. This proves (2.19) and the proof of the theorem is complete.
Remark 2.7. Let r → ∞, (2.1) changes into [2, (2.3)]. Hence, (2.1) is a generalization
of [2, (2.3)].
Remark 2.8. Let r → ∞, (2.5) and (2.9) change into [1, (3.1) and (3.5)], respectively.
Hence (2.5) and (2.9) is generalization of [1, (3.1) and (3.5)], respectively.
Remark 2.9. Let r → ∞, (2.12), (2.17), and (2.19) change into [1, (3), (4), and (5)],
respectively. Hence, (2.12), (2.17), and (2.19) is generalization of [1, (3), (4), and (5)],
respectively.
References
[1]
[2]
[3]
[4]
Y. Bicheng and L. Debnath, Generalizations of Hardy’s integral inequalities, Int. J. Math.
Math. Sci. 22 (1999), no. 3, 535–542.
Y. Bicheng, Z. Zhuohua, and L. Debnath, Note on new generalizations of Hardy’s integral
inequality, J. Math. Anal. Appl. 217 (1998), no. 2, 321–327.
G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, 2nd ed., Cambridge University
Press, London, 1952.
J. A. Oguntuase and C. O. Imoru, New generalizations of Hardy’s integral inequality, J.
Math. Anal. Appl. 241 (2000), no. 1, 73–82.
Lü Zhongxue: Department of Basic Science of Technology College, Xuzhou Normal
University, Xuzhou 221011, China
E-mail address: lvzx1@163.net
Xie Hongzheng: Department of Mathematics, Harbin Institute of Technology,
Harbin 150001, China