Tamsui Oxford Journal of Information and Mathematical Sciences 28(3) (2012) 281-292
Aletheia University
A New Multiple Inequality Similar to
Hardy-Hilbert’s Integral Inequality ∗
Vandanjav Adiyasuren†
Department of Mathematical Analysis,
National University of Mongolia, Ulaanbaatar, Mongolia
and
Tserendorj Batbold‡
Institute of Mathematics, National University of Mongolia,
P.O. Box 46A/104, Ulaanbaatar 14201, Mongolia
Received February 9, 2011, Accepted October 24, 2011.
Abstract
In this paper, by introducing some parameters we establish a new
multiple inequality similar to Hardy-Hilbert’s integral inequality with
the best constant factor which involves the Γ function. Some particular
results are obtained.
Keywords and Phrases: Hardy-Hilbert’s inequality, Hölder’s inequality,
Hardy’s inequality.
∗
1991 Mathematics Subject Classification. Primary 26D15.
E-mail: V [email protected]
‡
Corresponding author.
E-mail: [email protected]
†
282
Vandanjav Adiyasuren and Tserendorj Batbold
1. Introduction
If p > 1, p1 + 1q = 1, f, g ≥ 0, satisfy 0 <
R∞ q
g (x)dx < ∞, then
0
∞
Z
∞
Z
0
0
Z
∞
∞
Z
0
f (x)g(y)
π
dxdy <
x+y
sin( πp )
0
f (x)g(y)
dxdy < pq
max{x, y}
R∞
0
∞
Z
f p (x)dx
f p (x)dx < ∞ and 0 <
p1 Z
0
g q (x)dx
1q
,
(1.1)
0
p1 Z
∞
Z
∞
p
∞
q
1q
g (x)dx
f (x)dx
(1.2)
0
0
where the constant factors π/(sin π/p) and pq are the best possible. Inequalities (1.1) and (1.2) are called Hardy-Hilbert’s inequalities (see [1]) and are
important in analysis and their applications (cf. Mitrinović et al. [2]). At
present, because of the requirement of higher-dimensional harmonic analysis
and higher-dimensional operator theory, multiple Hardy-Hilbert’s integral inequalities are researched (see [3, 4, 5]).
In 2005, Yang ([3]) gave the following inequality.
P
Theorem 1.1. If n ∈ N\{1}, pi > 1, ni=1 p1i = 1, λ > 0, fi ≥ 0, satisfy
Z
∞
0<
xpi −1−λ fipi (x)dx < ∞
(i = 1, 2, . . . , n),
0
then
Z
∞
Z
···
0
0
∞
n
Y
1
P
n
j=1
xj
λ
fi (xi )dx1 . . . dxn
i=1
n
1 Y
<
Γ
Γ(λ) i=1
where the constant factor
1
Γ(λ)
Qn
i=1
Γ
Z ∞
p1
i
λ
pi −1−λ pi
x
fi (x)dx
pi
0
λ
pi
(1.3)
is the best possible.
In the recent years, many new inequalities similar to (1.1) and (1.2) have
been established; see ([6, 7, 9]). Recently Das and Sahoo ([6]) have given a
new inequality similar to Hardy-Hilbert’s inequality (1.1) as follows:
283
Multiple Inequality Similar to Hardy-Hilbert’s Inequality
Theorem 1.2. Let p > 1, p1 + 1q = 1, λ, s, r > 0, r + s = λ, f, g ≥ 0 and F (x) =
Rx
Rx
R∞
R∞
f (t)dt, G(x) = 0 g(t)dt. If 0 < 0 f p (x)dx < ∞ and 0 < 0 g q (x)dx <
0
∞, then
Z
∞
0
∞
Z
0
1
1
xr− q −1 y s− p −1
F (x)G(y)dxdy
(x + y)λ
Z ∞
p1 Z
p
< pqB(r, s)
f (x)dx
1q
∞
q
g (x)dx
(1.4)
0
0
where the constant factor pqB(r, s) is the best possible.
Sulaiman ([8, 9]) derived three new integral inequalities similar to (1.1) as
follows:
Theorem 1.3 ([8]). Let f, g, h ≥ 0, p, q, r > 2,
x
Z
+ 1q +
1
r
= 1,
y
Z
f (t)dt,
F (x) =
1
p
g(t)dt,
G(y) =
h(t)dt.
H(z) =
0
0
0
z
Z
Then
∞
∞
∞
p
x1/p y 1/q z 1/r xyzF (x)G(y)H(z)
dxdydz
(x + y + z)6
0
0
0
Z ∞
1/p Z ∞
1/q Z
p/2
q/2
< Kp Kq Kr
f (x)dx
g (y)dy
Z
Z
Z
0
0
1/r
∞
r/2
h
(z)dz
0
(1.5)
where
s
Kp =
p/2
B 1/p
p/2 − 1
1 1
,
p p
B 1/p (p, p).
Theorem 1.4 ([9]). Let f, g ≥ 0, p > 1, p1 + 1q = 1, λ > 1, αp = p(λ − 1) + 1.
Define
Z x
Z y
F (x) =
f (t)dt, G(y) =
g(t)dt.
0
0
284
Vandanjav Adiyasuren and Tserendorj Batbold
Then
Z ∞Z
0
∞
F
0
αq
αpp αqq
(x)G q (y)
αq
αp
dxdy <B(λ, λ)
(x + y)2λ
αp − 1
αq − 1
p1 Z ∞
1q
Z ∞
αp
αq
f (x)dx
g (y)dy .
×
αp
p
0
0
(1.6)
Theorem 1.5 ([9]). Let f, g, h ≥ 0, p, q, r > 1, p1 + 1q +
αs,t = λ(1 + s/t) + (λ − 1)(s − 1),
1
r
= 1, s, t ∈ {p, q, r},
λ > 1.
Define
Z
x
Z
f (t)dt,
F (x) =
h(t)dt.
H(z) =
0
0
Suppose also that
Z ∞
f αp,q (t)dt < ∞,
0<
∞
Z
g
0<
αq,r
Z
(t)dt < ∞,
∞
hαr,p (t)dt < ∞.
0
∞
F αp,q /p (x)Gαq,r /q (y)H αr,p /r (z)
dxdydz
(x + y + z)4λ
0
0
0
1/q Z
1/p Z ∞
Z ∞
αq,r
αp,q
g (y)dy
f (x)dx
<K
Z
∞
0<
0
0
z
Z
g(t)dt,
G(y) =
0
Then
Z ∞Z
y
αr,p
h
(z)dz
(1.7)
0
0
0
1/r
∞
where
K = B(λ, λ)B(2λ, 2λ)
αp,q
αp,q − 1
αp,q /p αq,r
αq,r − 1
αq,r /q αr,p
αr,p − 1
αr,p /r
.
But he cannot show that the constant factors in the new inequalities are
the best possible.
In this paper, by introducing some parameters we establish a new multiple
generalization of the above four inequalities similar to Hardy-Hilbert’s integral
inequality (1.3) with the best constant factor. Some particular results are
obtained.
Multiple Inequality Similar to Hardy-Hilbert’s Inequality
285
2. Preliminary Lemmas
In order to prove main result, we need the following lemmas.
Pk+1
Lemma 2.1. If k ∈ N, ri > 1(i = 1, 2, . . . , k + 1), and
Z
∞
Z
∞
···
0
0
k
Y
1
1+
Pk
j=1
uj
λ(k)
i=1
ri = λ(k), then
Qk+1
r −1
uj j du1
. . . duk =
j=1
Γ(ri )
.
Γ(λ(k))
i=1
(2.1)
For the proof of Lemma 2.1 the reader is referred to ([3, 4]).
Lemma
2.2 (Hardy’s inequality, cf. [1]). If p > 1, f ≥ 0 and F (x) =
Rx
f (t)dt, then
0
Z
0
∞
F (x)
x
p
dx <
p
p−1
p Z
∞
f p (x)dx
(2.2)
0
unless f ≡ 0. The constant is the best possible.
Lemma 2.3. Let p > β1 , 0 < β ≤ 1, 0 < ε < βp − 1 for x ≥ 1, then
(x
βp−(1+ε)
βp
− 1)β ≥ x
βp−(1+ε)
p
− 1.
(2.3)
Proof. For x ≥ 1, we set
f (x) = (x
βp−(1+ε)
βp
− 1)β − x
βp−(1+ε)
p
+1
Simple computations yield for x > 1
f 0 (x) =
1+ε−βp
βp − (1 + ε) (β−1)p−(1+ε)
p
x
(1 − x βp )β−1 − 1 > 0.
p
f is increasing function on (1, ∞) and continuous on [1, ∞). In particular, we
have f (x) ≥ f (1) = 0, which gives the desired inequality.
286
Vandanjav Adiyasuren and Tserendorj Batbold
3. Main Results
P
Theorem 3.1. Let n ∈ N\{1}, pi > α1i , 0 < αi ≤ 1, ni=1 p1i = 1, λ, si > 1, and
Rx
Pn
Assume Fi (x) = 0 fi (t)dt (i = 1, 2, . . . , n). If fi ≥ 0 satisfy
i=1
R s∞i =αi pλ.
0 < 0 fi i (x)dx < ∞ (i = 1, 2, . . . , n) then the following inequality holds:
Z
∞
∞
Z
···
0
n
Y
1
P
n
j=1
0
xj
λ
si + p1 −1−αi
xi
i
i=1
n
1 Y
Γ(si )
<
Γ(λ) i=1
1
Γ(λ)
where the constant factor
Fiαi (xi )dx1 . . . dxn
Qn
i=1
α i pi
αi pi − 1
Γ(si )
αi Z
∞
fiαi pi (x)dx
p1
i
(3.1)
0
αi p i
αi pi −1
αi
is the best possible.
Proof. By Hölder’s inequality and Lemma 2.1, we have
Z
∞
Z
∞
···
J =
P
0
0
n
Y
1
n
j=1
λ
xj
si + p1 −1−αi
xi
i
Fiαi (xi )dx1 . . . dxn
i=1

Z
∞
Z
···
=
0
0
≤

Z
n 

Y
i=1



Qn
=
0
∞
∞
n
j=1
xj
i=1
∞
Z
···
0
n
Γ(sj ) Y
Γ(λ)
i=1
j=1
n
n
sj −1
Y
 psi −αi Y

pi
i

x
Fiαi (xi )
x
λ
j
 i
 dx1 . . . dxn
1
P
1
P
n
j=1
Z
0
∞

xj
j=1
(j6=i)
s
λ xi i
Fi (xi )
xi
n
Y
s −1
xj j
j=1
(j6=i)
Fi (xi )
xi
αi pi
dx1 . . . dxn
1
pi






p1
αi pi
dxi
i
.
(3.2)
Then by Hardy inequality (2.2), (3.1) is valid. For the best constant factor,
let for sufficiently small ε > 0,
(
0,
for x ∈ (0, 1)
fei (x) =
(i = 1, 2, . . . , n)
− α1+ε
p
x i i , for x ∈ [1, ∞)
Multiple Inequality Similar to Hardy-Hilbert’s Inequality
then we find
n Z
Y
∞
feiαi pi (x)dx
p1
i
=
0
i=1
1
ε
287
(3.3)
and
(
for xi ∈ (0, 1)
0,
Fei (xi ) =
(i = 1, 2, . . . , n).
− 1), for xi ∈ [1, ∞)
αi
αi
Q Q i pi
i
Denote φ(ε) = ni=1 αi piα−(1+ε)
. Then φ(ε) → ni=1 ααi piip−1
, as ε → 0+
and for xi ≥ 1, by Lemma 2.3, we have
n
n
n
αi pi −(1+ε)
αi pi −(1+ε)
Y
Y
Y
αi pi
pi
αi
αi
e
− 1) > φ(ε) (xi
− 1).
Fi (xi ) = φ(ε) (xi
αi pi −(1+ε)
αi p i
αi pi
(xi
αi pi −(1+ε)
i=1
i=1
i=1
Hence, we have
Z ∞
Z
J(ε) =
···
0
∞
P
n
j=1
0
Z
∞
Z
∞
n
j=1
···
=φ(ε)
n
j=1
i
Feiαi (xi )dx1 . . . dxn
i=1
n
Y
xj
λ
si + p1 −1−αi
xi
n
Y
xj
i
αi pi −(1+ε)
pi
(xi
− 1)dx1 . . . dxn
i=1
1
P
1
1
si + p1 −1−αi
xi
1
∞
Z
λ
P
1
1
xj
∞
Z
···
>φ(ε)
n
Y
1
λ
si − pε −1
xi
i
i=1
!
n
k
n
X
XY
sj − pε −1
si + p1 −1−αi Y
j
k
(−1)
+
xi i
xj
dx1 . . . dxn
cyc i=1
k=1
Z
∞
1
X
j=k+1
∞
···
>φ(ε)
−
Z
1
k
XY
n
Y
1
P
n
j=1
xj
λ
si + p1 −1−αi
xi
1≤k≤n cyc i=1
k−odd
i
1≤k≤n cyc
k−odd
i
i=1
n
Y
j=k+1
X X
=φ(ε) M −
N (k) .
si − pε −1
xi

sj − pε
j
xj
−1 
 dx1 . . . dxn
288
Vandanjav Adiyasuren and Tserendorj Batbold
Taking ui =
xi+1
(i
x1
Z
= 1, 2, . . . , n − 1) and using (2.1), we have
∞
∞
Z
···
M =
1
1
si − pε −1
Qn
i=1 xi
P
n
j=1
Z
∞
x−ε−1
dx1
1
=
i
xj
∞
Z
Qn−1
i=1
∞
x1−ε−1 dx1
1+
0
0
−
∞
Z
···
1
Z
λ dx1 . . . dxn
1/x1
Z
1/x1
Z
i=1
Qn−1
i=1
···
1+
0
0
1
ε
−1
i+1
si+1 − p
ui
Pn−1
ui
λ du1 . . . dun−1
si+1 − p
ui
Pn−1
i=1
ε
−1
i+1
ui
λ du1 . . . dun−1
Pn−1 ε Qn−1
ε
1 Γ(s1 − i=1 pi+1 ) i=1 Γ(si+1 − pi+1 )
>
ε
Γ(λ)
Z ∞
Z 1/x1
Z 1/x1 n−1
Y si+1 − p ε −1
i+1
−ε−1
du1 . . . dun−1
−
x1 dx1
···
ui
1
0
1 Γ(s1 −
=
ε
−
ε
p1
+
Again taking ui =
Z
∞
Z
Pn−1
Pn−1
i=1
Qk
···
N (k) =
1
Z
=
ε
i=1 pi+1 )
xi+1
(i
x1
∞
si+1
x−1−γ
dx1
1
i=1
ε
)
pi+1
Γ(si+1 −
Γ(λ)
1
Q
n−1
i=1 (si+1 −
si + p1 −1−αi
i=1
xi
i
P
Qn
n
j=1
Z
i=1
ε
.
)
pi+1
= 1, 2, . . . , n − 1) and using (2.1), we have
1
∞
0
Qn−1
∞
Z
j=k+1
λ
sj − pε −1
xj
j
dx1 · · · dxn
xj
∞
···
g(u1 , . . . , un−1 )du1 . . . dun−1
Z ∞
Z 1/x1
Z 1/x1
−1−γ
−
x1
dx1
···
g(u1 , . . . , un−1 )du1 . . . dun−1
1
0
0
Z ∞
Z ∞
Z ∞
−1−γ
<
x1
dx1
···
g(u1 , . . . , un−1 )du1 . . . dun−1
1
0
0
Qk−1
Qn−1
1
1
1 Γ(s1 + γ − α1 − p1 ) i=1 Γ(si+1 + pi+1 − αi+1 ) i=k Γ(si+1 −
= ·
γ
Γ(λ)
1
0
0
ε
)
pi+1
289
Multiple Inequality Similar to Hardy-Hilbert’s Inequality
where γ =
Pk
i=1
αi −
Pk
1
i=1 pi
+
Pn
Qk−1
i=1
g(u1 , . . . , un−1 ) =
ε
i=k+1 pi ,
and
1
−1−αi+1
i+1
si+1 + p
ui
1+
Qn−1
i=k
λ
i=1 ui
si+1 − p
ui
Pn−1
ε
−1
i+1
.
Hence, we have
(
J(ε) > φ(ε)
1 Γ(s1 −
ε
1
Γ(λ)
Pn−1
ε
i=1 pi+1 )
Qn−1
i=1
Γ(si+1 −
Γ(λ)
Qn
αi p i
αi pi −1
ε
)
pi+1
)
− (1) .
(3.4)
αi
in (3.1) is not the best possible,
αi
Qn
αi p i
1
Γ(s
)
then there exists a positive constant K such that K < Γ(λ)
i
i=1
αi pi −1
αi
Q
n
αi p i
1
is replaced by K. In
and (3.1) still remains valid if Γ(λ) i=1 Γ(si ) αi pi −1
particular by (3.3) and (3.4), we have
(
)
P
Qn−1
ε
ε
Γ(s1 − n−1
i=1 pi+1 )
i=1 Γ(si+1 − pi+1 )
φ(ε)
− ε (1)
Γ(λ)
Z ∞
Z ∞
n
Y
si + p1 −1−αi α
1
xi i
<ε
···
Fei i (xi )dx1 . . . dxn
λ
P
n
0
0
i=1
j=1 xj
p1
n Z ∞
Y
i
α
fei i (x)dx
< εK
= K.
If the constant factor
i=1
i=1
Γ(si )
0
αi
αi p i
Γ(s
)
≤ K as ε → 0+ . This contradicts the fact K <
i
i=1
αi pi −1
αi
αi
Qn
αi p i
αi p i
1
Γ(s
)
.
Hence
the
constant
factor
Γ(s
)
i
i
i=1
i=1
Γ(λ)
αi pi −1
Γ(λ)
αi pi −1
in (3.1) is the best possible. The theorem is proved.
1
Then Γ(λ)
Qn
1
Qn
Now we discuss some particular cases of (3.1). Taking αi = 1 in Theorem
3.1, we get the following multiple extension of (1.4), with the constant factor
is the best possible.
P
P
Corollary 3.2. Let n ∈ N\{1}, pi > 1, ni=1 p1i = 1, λ, si > 1, and ni=1 si =
Rx
λ. Assume Fi (x) = 0 fi (t)dt (i = 1, 2, . . . , n). If fi ≥ 0 satisfy 0 <
290
Vandanjav Adiyasuren and Tserendorj Batbold
R∞
fipi (x)dx < ∞ (i = 1, 2, . . . , n) then the following inequality holds:
0
Z
∞
∞
Z
···
P
n
j=1
0
0
n
Y
1
xj
λ
si + p1 −2
xi
i
Fi (xi )dx1 . . . dxn
i=1
p1
Z ∞
n
i
1 Y
pi
pi
<
fi (x)dx
Γ(si )
Γ(λ) i=1
pi − 1
0
Qn
pi
1
where the constant factor Γ(λ)
is the best possible.
i=1 Γ(si ) pi −1
(3.5)
Taking n = 3, αi = 1/2 in Theorem 3.1, we get the following generalization
of (1.5), with the constant factor is the best possible.
Corollary 3.3. Let p, q, r > 2, p1 + 1q + 1r = 1, λ, s1 , s2 , s3 > 1, and s1 +s2 +s3 =
Rx
Rx
Rx
λ. Assume F (x) = 0 f (t)dt, G(x) = 0 g(t)dt, and H(x) = 0 h(t)dt.
R∞ q
R∞ p
h ≥ 0 satisfy 0 < 0 f 2 (x)dx < ∞, 0 < 0 g 2 (x)dx < ∞, 0 <
RIf∞f, g,
r
h 2 (x)dx < ∞ then the following inequality holds:
0
Z
0
1
3
1
3
1
3
xs 1 + p − 2 y s 2 + q − 2 z s 3 + r − 2 p
F (x)G(y)H(z)dxdydz
(x + y + z)λ
0
0
1/r
1/q Z ∞
1/p Z ∞
Z ∞
r/2
q/2
p/2
(3.6)
h (z)dz
g (y)dy
< Cλ
f (x)dx
∞
Z
∞
Z
∞
0
where the constant factor Cλ =
q
p/2
.
Kp,s1 = Γ(s1 ) p/2−1
0
0
1
K K K
Γ(λ) p,s1 q,s2 r,s3
is the best possible and
Taking n = 2, α1 = αp /p, α2 = αq /q, s1 = s2 = λ in Theorem 3.1, we get
the following result.
Corollary 3.4. Let p > α11 , q > α12 , p1 + 1q = 1, λ > 1, αp = p(λ − 1) + 1, and
Rx
1 − min{ p1 , 1q } < λ ≤ 2 − max{ p1 , 1q }. Assume F (x) = 0 f (t)dt, and G(y) =
Ry
R∞ α
R∞ α
p
g(t)dt.
If
f,
g
≥
0
satisfy
0
<
f
(x)dx
<
∞,
0
<
g q (x)dx < ∞
0
0
0
then the following inequality holds:
αq
αpp αqq
Z ∞ Z ∞ αpp
αp
αq
F (x)G q (y)
dxdy <B(λ, λ)
(x + y)2λ
αp − 1
αq − 1
0
0
Z ∞
p1 Z ∞
1q
αp
αq
×
f (x)dx
g (y)dy
(3.7)
0
0
291
Multiple Inequality Similar to Hardy-Hilbert’s Inequality
where the constant factor B(λ, λ)
αp
αp −1
αpp αq
αq −1
αqq
is the best possible.
Taking n = 3, α1 = αp,q /p, α2 = αq,r /q, α3 = αr,p /r, s1 = λ 1q + 1 , s2 =
λ 1r + 1 , s3 = λ p1 + 1 in Theorem 3.1, we get the following result.
Corollary 3.5. Let p >
2−max{ p1 , 1q , r1 }
1+max{ p1 , 1q , r1 }
1
,q
α1
>
1
,r
α2
>
1 1
,
α3 p
+
1
q
+
1
r
= 1, s, t ∈ {p, q, r},1 <
λ ≤
, and αs,t = λ(1 + s/t) + (λ − 1)(s − 1). Assume F (x) =
Rx
Ry
Rz
G(y) = 0 g(t)dt,
and H(z) = 0 h(t)dt.
R0∞f (t)dt,
R
R ∞If αf, g, h ≥ 0 satisfy 0 <
∞ αq,r
αp,q
f (t)dt < ∞, 0 < 0 g (t)dt < ∞, 0 < 0 h r,p (t)dt < ∞ then the
0
following inequality holds:
∞
∞
∞
F αp,q /p (x)Gαq,r /q (y)H αr,p /r (z)
dxdydz
(x + y + z)4λ
0
0
0
1/q Z
1/p Z ∞
Z ∞
αq,r
αp,q
g (y)dy
f (x)dx
<K
Z
Z
Z
0
0
1/r
∞
αr,p
h
(z)dz
(3.8)
0
where the constant factor
αp,q /p αq,r /q αr,p /r
Γ(s1 )Γ(s2 )Γ(s3 )
αp,q
αq,r
αr,p
K=
Γ(4λ)
αp,q − 1
αq,r − 1
αr,p − 1
is the best possible.
References
[1] G. H. Hardy, J. E. Littlewood, and G. Polya, Inequalities, Cambridge
University Press Cambridge, 1952.
[2] D. S. Mitrinović, J. E. Pečarić, and A. M. Fink, Inequalities Involving
Function and Their Integrals and Derivatives, Kluwer Acadamic Publishers, Boston, 1991.
[3] B. Yang, On a New Multiple Extension of Hilbert’s integral inequality, J.
Math. Pure and Appl. Math., 6 no. 2 (2005), Art. 39.
292
Vandanjav Adiyasuren and Tserendorj Batbold
[4] B. Yang and L. Debnath, On An Extended Multiple Hardy-Hilbert’s Integral Inequality, Tamsui Oxford J. Math. Sci, 21 no. 2 (2005), 157-169.
[5] Q. Huang and B. Yang, On a Multiple Hilbert-Type Integral Operator
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[6] N. Das and S. Sahoo, New Inequalities Similar to Hardy-Hilbert’s inequality, Turk. J. Math, 34 (2010), 153-165.
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