An. Şt. Univ. Ovidius Constanţa
Vol. 16(1), 2008, 67–72
On the inverse inequalities of two new type
Hilbert integral inequalities
Zhao Changjian and Mihály Bencze
Abstract
The Hilbert integral inequality is an important inequality in applications. In 1998, two new inequalities similar to Hilbert integral inequality
were given by B. G. Pachpatte. The main purpose of the present article is to give two new inverse inequalities similar to these two new
inequalities by Jensen inequality and Holder integral inequality.
Hilbert integral inequality plays an important role in mathematical analysis
and its applications. In 1998, B. G. Pachpatte gave two new integral inequalities similar to Hilbert integral inequality (see [2], p. 226) in [1]. In this paper
we give two new inverse inequalities similar to these new inequalities.
Our main results are given in the following theorems:
Theorem 1. Let h ≥ 1, l ≥ 1 and f (σ) > 0, g (τ ) > 0 for σ ∈ (0, x) , τ ∈
s
(0, y) , where x, y are positive real numbers and define F (s) = f (σ) dσ and
G (t) =
Then
(1)
t
0
0
g (τ ) dτ for s ∈ (0, x) , t ∈ (0, y) and
x y
0 0
F h (s)Gl (t)
C(s,t,p) dsdt
1
p
+
1
q
= 1, p < 0 or 0 < p < 1.
≥
Key Words: Hilbert integral inequality; Hölder integral inequality; Jensen inequality.
Mathematics Subject Classification: 26D15.
Received: 31 October, 2007
Accepted: January, 2008
67
68
Zhao Changjian and Mihály Bencze
≥ hl (xy)
1/p
x
0
where C (s, t, p) =
y
1/q
h−1 q
l−1 q
(x − s) F
(s) ds 1/q ×
(y − t) G
(t) dt
s
f p (σ) dσ
1/p t
0
0
1/p
g p (τ ) dτ
.
0
Proof. From the hypotheses, it is easy to observe that
s
F h−1 (σ) f (σ) dσ, s ∈ (0, x)
F h (s) = h
0
t
Gl−1 (τ ) g (τ ) dτ, t ∈ (0, y)
l
G (t) = l
0
Therefore
(2) F h (s) Gl (t) = hl
s
F h−1 (σ) f (σ) dσ
t
0
Gl−1 (τ ) g (τ ) dτ
0
On the other hand, according to Holder integral inequality (see [2] P. 154)
we have
s
1/q s
1/p
q
s
h−1
p
(3) F h−1 (σ) f (σ) dσ ≥
F
(σ) dσ
·
f (σ) dσ
0
(4)
t
0
0
Gl−1 (τ ) g (τ ) dτ ≥
t
0
Gl−1 (τ )
q
0
1/q t
1/p
p
dτ
·
g (τ ) dτ
0
By (2), (3) and (4) yield that
F h (s) Gl (t) ≥ hl
×
s
f p (σ) dσ
0
1/p t
1/p
×
g p (τ ) dτ
0
1/q t
1/q
s
h−1
l−1
q
q
F
G
(σ) dσ
(τ ) dτ
.
0
Thus
0
F h (s)Gl (t)
C(s,t,p) dsdt
s
1/q t
1/q
h−1
l−1
q
q
F
G (τ ) dτ
≥ hl
(σ) dσ
.
0
0
Integrating over t from 0 to y first and then integrating the resulting inequality over s from 0 to x and using special case of Holder integral inequality (we take f (x) = 1 in to Holder integral inequality), we observe that:
x y F h (s)Gl (t)
C(s,t,p) dsdt ≥
0 0 1/q y t
1/q x s h−1
l−1
q
q
F
G
(σ) dσ
ds ×
(τ ) dτ
dt ≥
≥ hl
0
0
0
0
1/q
1/q
x s
y t
q
q
h−1
l−1
1/p
1/p
≥ hlx
(σ) dσ ds
·xy
(τ ) dτ dt
=
F
G
0
0
0
0
69
On the inverse inequalities
1/p
= hl (xy)
x
0
q
(x − s) F h−1 (s) ds
1/q
×
y
0
q
(y − t) Gl−1 (t) dt
1/q
.
The proof is complete.
This is just a inverse inequality similar to following which was given by B.
G. Pachpatte in [1].
x y
0
0
⎛ x
⎞1/2
h−1 2
1
F h (s) Gl (t)
1/2 ⎝
dsdt ≤ h (xy)
(x − s) F
(s) ds⎠ ×
s+t
2
0
⎛ y
⎞1/2
l−1 2
× ⎝ (y − t) G (t) dt⎠ .
0
Theorem 2. Let f, g, F, G be as in Theorem 1. Let p (σ) and q (τ ) be
two positive functions defined for σ ∈ (0, x) , τ ∈ (0, y) and define P (s) =
s
t
p (σ) dσ, Q (t) = q (τ ) dτ for s ∈ (0, x) , t ∈ (0, y) where x, y are positive
0
0
real numbers and p, q are real numbers and 1p + 1q = 1, p < 0 or 0 < p < 1. Let
φ and ψ be two real-valued nonnegative, concave, and supermultiplicative functions (f is said to be supermultiplicative function if f (xy) ≥ f (x) f (y) , x, y ∈
R+ ) defined on R+ = [0, +∞) . Then
x y F h (s)Gl (t)
(5)
D(s,t,p) dsdt ≥
0 0
x
q 1/q y
q 1/q
(s)
(x − s) φ fp(s)
ds
×
(y − t) ψ g(t)
dt
,
≥ L (x, y, p)
q(t)
0
where
0
⎛
L (x, y, p) = ⎝
x 0
φ (P (s))
P (s)
p
⎞1/p ⎛
ds⎠
⎝
y 0
ψ (g (t))
q (t)
p
⎞1/p
dt⎠
and
⎛ s
⎞1/p ⎛ t
⎞1/p
D (s, t, p) = ⎝ pp (σ) dσ ⎠ ⎝ q p (τ ) dτ ⎠ .
0
0
Proof. From the hypotheses and by using Jensen inequality and Holder
integral inequality, it is easy to observe that
70
Zhao Changjian and Mihály Bencze
⎛
(6) φ (F (s)) = φ ⎝
⎛ Ês
≥ φ (P (s)) φ ⎝ 0
≥
φ(P (s))
P (s)
and similarly,
p(σ)dσ
0
s
p (σ) dσ
0
0
Ês
(σ)
p(σ) fp(σ)
dσ
Ês
Ês
P (s)
f (σ)
p(σ) p(σ) dσ
p(σ)dσ
⎞
⎠≥
0
⎞
⎠≥
φ(P (s))
P (s)
s
p (σ) φ
0
f (σ)
p(σ)
dσ ≥
1/p s q 1/q
(σ)
φ fp(σ)
dσ
0
(7) ψ (G (t)) ≥
ψ(Q(t))
Q(t)
t
0
1/p t q 1/q
g(τ )
ψ g(τ )
q (τ ) dτ
dτ
.
p
0
By (6) and (7), we get that:
≥
(8)
φ(F (s))ψ(G(t))
D(s,t,p)
φ(P (s))
P (s)
≥
s q 1/q t q 1/q
ψ(Q(t))
f (σ)
g(τ )
φ p(σ)
ψ g(τ )
dσ
×
dτ
.
Q(t)
0
0
Integrating two sides of (8) over t from 0 to y first and then integrating the
resulting inequality over s from 0 to x and using Holder integral inequality, we
observe that
1/q x y φ(F (s))ψ(G(t))
x φ(P (s)) s f (σ) q
φ p(σ)
dsdt ≥
dσ
ds ×
D(s,t,p)
P (s)
0 0
0
0
1/q y ψ(Q(t)) t g(τ ) q
×
ψ g(τ )
dτ
dt ≥
Q(t)
0
≥
×
0
x
φ(P (s)) p
P (s)
0
y
ψ(Q(t)) p
Q(t)
0
= L (x, y, p)
x
0
ds
1/p x s q 1/q
(σ)
φ fp(σ)
dσ ds
×
0
0
y t q 1/q
)
ψ g(τ
dt
dτ dt
=
g(τ )
0
0
q 1/q y
q 1/q
(σ)
)
(x − s) φ fp(σ)
ds
×
(y − t) ψ g(τ
dt
.
g(τ )
0
This is just another inverse inequality similar to following inequality which
was given by B. G. Pachpatte in [1].
x
2 1/2
x y φ(F (s))ψ(G(t))
f (σ)
(9)
dsdt ≤ L (x, y)
(x − s) p (s) φ p(σ)
ds
×
s+t
0 0
×
y
0
0
2 1/2
g(τ )
(y − t) g (t) ψ g(τ )
dt
, where
71
On the inverse inequalities
⎛
L (x, y) =
1⎝
2
x 0
φ (P (s))
P (s)
2
⎞1/2 ⎛ y
⎞1/2
2
ψ (Q (t))
ds⎠ ⎝
dt⎠ .
Q (t)
0
Acknowledgment. We wish to express my gratitude to professor Li Wenrong for his valuable help in writing this paper.
References
[1] B. G. Pachpatte, On some new inequalities similar to Hilbert‘s inequality , J. Math.
Anal. Appl.,226(1)(1998), 166-179.
[2] G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge Univ. Press, Cambridge, 1952.
[3] D. S. Mitrinovic, Analytic inequalities, Springer-Verlag, 1970.
[4] Gao Minzhe, On Hilbert inequality and its applications, J. Math. Anal. Appl.,
212(1997), 316-323.
[5] Hu Ke: On Hilbert inequality and its application [J], Advances Mathematics, 1993,
22(2): 160-163.
[6] Yang Bicheng and L. Debnath, Generalizations of Hardy integral inequalities, Internat.
J. Math. and Math. Sci. 22(3)(1999), 535-542.
Department of Mathematics, Binzhou Teachers College, Shandong 256604,
China
e-mail: chjzhao@163.com
Str. Harmanului 6, 505600 Sacele-Négyfalu, Jud. Brasov, Romania
benczemihaly@yahoo.com
72
Zhao Changjian and Mihály Bencze