IJMMS 28:4 (2001) 211–221
PII. S0161171201006937
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
AN IMPROVEMENT OF SOME INEQUALITIES SIMILAR
TO HILBERT’S INEQUALITY
YOUNG-HO KIM
(Received 7 March 2001)
Abstract. We give an improvement of some inequalities similar to Hilbert’s inequality
involving series of nonnegative terms. The integral analogies of the main results are also
given.
2000 Mathematics Subject Classification. 26D15.
1. Introduction. It is well known that the following Hilbert’s double series inequality (see [3, page 226]) plays an important role in many branches of mathematics.
p
p
Theorem 1.1. If p > 1, p = p/(p − 1) and am ≤ A, bn ≤ B, the summations
running 1 to ∞, then
am bn
m+n
<
π
A1/p B 1/p ,
sin(π /p)
(1.1)
unless all the sequence {am } or {bn } is null.
The integral analogue of Hilbert’s inequality can be stated as follows (see [3, page
226]).
Theorem 1.2. If p > 1, p = p/(p − 1) and
∞
0
then
∞∞
0
0
f p (x)dx ≤ F ,
∞
0
g p (y)dy ≤ G,
π
f (x)g(y)
dx dy <
F 1/p G1/p ,
x +y
sin(π /p)
(1.2)
(1.3)
unless f ≡ 0 or g ≡ 0.
These two theorems were studied extensively and numerous variants, generalizations, and extensions appeared in the literature, see [1, 2, 3, 5, 6, 9] and the references
therein.
Recently, Pachpatte [9] gave new inequalities similar to Hilbert’s inequalities given
in the above theorems, involving a series of nonnegative terms as follows.
Theorem 1.3. Let p ≥ 1, q ≥ 1, and let {am } and {bn } be two nonnegative sequences
of real numbers defined for m = 1, 2, . . . , k and n = 1, 2, . . . , r , where k, r are the natural
212
YOUNG-HO KIM
numbers and define Am =
m
s=1 as ,
Bn =
n
t=1 bt .
Then

1/2
r
k k
p q
p−1
2
Am Bn
≤ C(p, q, k, r )
(k − m + 1) Am am 
m+n
m=1 n=1
m=1

1/2
r
q−1 2
×
(r − n + 1) Bn bn  ,
(1.4)
n=1
unless {am } or {bn } is null, where
C(p, q, k, r ) =
1
pq kr .
2
(1.5)
An integral analogue of Theorem 1.3 is given in the following theorem.
Theorem 1.4. Let p ≥ 1, q ≥ 1 and f (σ ) ≥ 0, g(τ) ≥ 0 for σ ∈ (0, x), τ ∈ (0, y),
t
s
where x, y are positive real numbers and define F (s) = 0 f (σ )dσ and G(t) = 0 g(τ)dτ,
for s ∈ (0, x), t ∈ (0, y). Then
xy
0
0
1/2
x
2
F p (s)Gq (t)
ds dt ≤ D(p, q, x, y)
(x − s) F p−1 (s)f (s) ds
s +t
0
1/2
y
q−1
2
×
(y − t) G
(t)g(t) dt
,
(1.6)
0
unless f ≡ 0 or g ≡ 0, where
D(p, q, x, y) =
1
pq xy.
2
(1.7)
In this paper, we give an improvement of the inequalities given in Theorems 1.3 and
1.4 similar to Hilbert’s double series inequality and its integral analogue, involving a
series of nonnegative terms. In addition, we obtain some new Hilbert type inequalities.
These inequalities improve the results obtained by Pachpatte [9].
2. Main results. Our main results are given in the following theorems.
Theorem 2.1. Let p ≥ 1, q ≥ 1, 0 < α, and let {am } and {bn } be two nonnegative
sequences of real numbers defined for m = 1, 2, . . . , k and n = 1, 2, . . . , r , where k, r are
m
n
the natural numbers and define Am = s=1 as , Bn = t=1 bt . Then

1/2
k
p q
p−1
2
A m Bn

≤ C(p, q, k, r ; α)
(k − m + 1) Am am 
α
α 1/α
m=1 n=1 m + n
m=1

1/2
r
q−1 2

×
(r − n + 1) Bn bn  ,
r
k (2.1)
n=1
unless {am } or {bn } is null, where
C(p, q, k, r ; α) =
1/α 1
pq kr .
2
(2.2)
213
SOME INEQUALITIES SIMILAR TO HILBERT’S INEQUALITY
Proof. By using the following inequality (see [1, 7]),
β
n
n
≤β
zm
m=1
m
zm
m=1
β−1
zk
,
(2.3)
k=1
where β ≥ 1 is a constant and zm ≥ 0, (m = 1, 2, . . .), it is easy to observe that
p
Am ≤ p
m
p−1
as As
q
m = 1, 2, . . . , k,
,
Bn ≤ q
s=1
n
q−1
at Bt
n = 1, 2, . . . , r .
,
(2.4)
t=1
From (2.4) and using the Schwarz inequality and the elementary inequality
n
1/n
≤
ai
i=1
1/α
n
aα
i
n
i=1
,
0 < α, (see [4]),
(2.5)
(for ai , i = 1, 2, . . . , n, nonnegative real numbers) we observe that
m
n
p q
p−1
q−1
Am Bn ≤ pq
as As
at Bt
s=1
≤ pq(m)1/2
t=1
m
p−1 2
1/2
s=1
≤ pq
m α + nα
2
1/α m
n
(n)1/2
as A s
q−1 2
1/2
at Bt
t=1
p−1 2
1/2 as As
n
q−1 2
(2.6)
1/2
at Bt
s=1
.
t=1
Dividing both sides of (2.6) by (mα + nα )1/α , and then taking the sum over n from 1
to r first and then the sum over m from 1 to k and using the Schwarz inequality and
then interchanging the order of the summations (see [7, 8]) we observe that
r
k p
q
Am Bn
α
α 1/α
m=1 n=1 m + n
m
1/2 r n
1/2 1/α k
1
p−1 2
q−1 2
≤ pq
as As
bt Bt
2
m=1 s=1
n=1 t=1
k m
1/2
r n
1/2
1/α
1
p−1 2
q−1 2
1/2
1/2
(k)
(r )
as A s
bt Bt
≤ pq
2
m=1 s=1
n=1 t=1
k 1/2 r
r 1/2
1/α k
1
p−1 2
q−1 2
= pq kr
1
1
as A s
b t Bt
2
m=s
s=1
t=1
n=t
= C(p, q, k, r ; α)
1/2 k
p−1 2
(k − s + 1)
as A s
s=1
= C(p, q, k, r ; α)
k
r
q−1 2
(r
b t Bt
1/2
− t + 1)
t=1
(k − m + 1)
m=1
p−1 2
am A m
1/2 r
q−1 2
(r − n + 1) bn Bn
1/2
.
n=1
(2.7)
This completes the proof.
214
YOUNG-HO KIM
Remark 2.2. In Theorem 2.1, setting α ≡ 1, we have Theorem 1.3. If we take p =
q = 1 in Theorem 2.1, then the inequality of the result of Theorem 2.1 reduces to the
following inequality:
1/2
k
2
Am Bn
≤ C(1, 1, k, r ; α)
(k − m + 1) am
α
α 1/α
m=1 n=1 m + n
m=1
1/2
r
2
(r − n + 1) bn
,
×
r
k (2.8)
n=1
where C(1, 1, k, r ; α) is obtained by taking p = q = 1 in (2.2).
Our next result deals with further generalization of the inequality obtained in
Remark 2.2.
Theorem 2.3. Let {am }, {bn }, Am , and Bn be as defined in Theorem 2.1. Let {pm }
and {qn } be two nonnegative sequences for m = 1, 2, . . . , k and n = 1, 2, . . . , r , and define
m
n
Pm = s=1 ps , Qn = t=1 qt . Let φ and ψ be two real-valued, nonnegative, convex, and
submultiplicative functions defined on R+ = [0, ∞). Then
k
2 1/2
r
k φ Am ψ Bn
am
≤
M(k,
r
;
α)
(k
−
m
+
1)
p
φ
m
α
α 1/α
pm
m=1 n=1 m + n
m=1
r
2 1/2
bn
×
(r − n + 1) qn φ
,
qn
n=1
where
M(k, r ; α) =
2 1/2 r 1/α k
ψ Qn 2 1/2
φ Pm
1
.
2
Pm
Qn
m=1
n=1
(2.9)
(2.10)
Proof. From the hypotheses of φ and ψ and by using Jensen’s inequality and the
Schwarz inequality (see [5]), it is easy to observe that
φ Am = φ
Pm
m
m
s=1 ps as /ps
s=1 ps as /ps
m
m
≤ φ Pm φ
s=1 ps
s=1 ps
2 1/2
m m
φ Pm
φ Pm as
as
ps φ
(m)1/2
,
≤
ps φ
≤
Pm s=1
ps
Pm
p
s
s=1
and similarly,
ψ Bn

2 1/2
n
ψ Qn
bt
 .
(n)1/2 
≤
qt ψ
Qn
qt
t=1
(2.11)
(2.12)
From (2.11) and (2.12) and using the elementary inequality
n
i=1
1/n
ai
≤
n
aα
i
n
i=1
1/α
,
0 < α,
(2.13)
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SOME INEQUALITIES SIMILAR TO HILBERT’S INEQUALITY
(for ai , i = 1, 2, . . . , n, nonnegative real numbers) we observe that
φ Am ψ Bn ≤
m α + nα
2
1/α 2 1/2 m φ Pm
as
ps φ
Pm
p
s
s=1
(2.14)
2 1/2 n ψ Qn
bt
qt ψ
.
Qn
qt
t=1
×
Dividing both sides of the above inequality by (mα +nα )1/α , and then taking the sum
over n from 1 to r first and then the sum over m from 1 to k and using the Schwarz
inequality and then interchanging the order of the summations we observe that
r
k φ Am ψ Bn
α
α 1/α
m=1 n=1 m + n
2 1/2 1/α k m φ Pm
as
1
≤
ps φ
2
P
p
m
s
m=1
s=1
×
r
n=1
2 1/2 n ψ Qn
bt
qt ψ
Qn
qt
t=1
2 1/2
1/α k 2 1/2 k m φ Pm
as
1
ps φ
≤
2
P
p
m
s
m=1
m=1 s=1
×
r
n=1
2 1/2
2 1/2 r n ψ Qn
bt
qt ψ
Qn
qt
n=1 t=1
= M(k, r ; α)
k
ps φ
s=1
= M(k, r ; α)
k
k
as
ps φ
ps
s=1
= M(k, r ; α)
as
ps
2 1/2 k
1
m=s
r
(k − s + 1)
r
t=1
m=1
qt ψ
bt
qt
t=1
1/2 2
(k − m + 1) pm φ
am
pm
2 1/2
2 bt
qt ψ
qt
r
r
1/2
1
n=t
1/2
2
(r − t + 1)
(r − n + 1) qn ψ
n=1
bn
qn
2 1/2
.
(2.15)
The proof is complete.
Remark 2.4. By applying the elementary inequality (see [4])
n
i=1
1/n
ai
≤
γ
n
ai
n
i=1
1/γ
,
0 < γ,
(2.16)
(for ai , i = 1, 2, . . . , n, nonnegative real numbers) on the right sides of the inequalities
216
YOUNG-HO KIM
in Theorems 2.1 and 2.3, we get, respectively, the following inequalities:
r
k q
p
Am Bn
α + nα 1/α
m
m=1 n=1
k
γ r
γ 1/γ
p−1
q−1
2
2
≤ C1
(k − m + 1) Am am
+
(r − n + 1) Bn bm
,
m=1
(2.17)
n=1
where C1 = (1/2)1/γ C(p, q, k, r ; α), and
r
k φ Am ψ Bn
α
α 1/α
m=1 n=1 m + n
k
2 γ r
2 γ 1/γ
am
bn
≤ M1
(k − m + 1) pm φ
+
(r − n + 1) qn ψ
,
pm
qn
m=1
n=1
(2.18)
where M1 = (1/2)1/γ M(k, r ; α), which we believe are new to the literature.
The following theorems deal with slight variants of (2.9) given in Theorem 2.3.
Theorem 2.5. Let {am } and {bn } be as defined in Theorem 2.1, and define Am =
m
n
1/m s=1 as and Bn = 1/n t=1 bt , for m = 1, 2, . . . , k and n = 1, 2, . . . , r , where k, r
are the natural numbers. Let φ and ψ be two real-valued, nonnegative, and convex
functions defined on R+ = [0, ∞). Then
r
k m=1 n=1
mn
m α + nα
1/α φ Am ψ Bn ≤ C(1, 1, k, r ; α)
k
2
(k − m + 1) φ am
1/2
m=1
×
r
2
(r − n + 1) ψ bn
1/2
,
n=1
(2.19)
where C(1, 1, k, r ) is defined by taking p = q = 1 in (2.2).
Proof. From the hypotheses and by using Jensen’s inequality and the Schwarz
inequality, it is easy to observe that
φ Am = φ
m
1 as
m s=1
≤
m
1
1 (m)1/2
φ as ≤
m s=1
m
m
2
φ as
1/2
s=1
n
n
n
2 1/2
1 1 1
1/2
ψ Bm = ψ
bt ≤
ψ bt ≤ (n)
.
ψ bt
n t=1
n t=1
n
t=1
,
(2.20)
The rest of the proof can be completed by following the same steps as in the proofs
of Theorems 2.1 and 2.3 with suitable changes and hence we omit the details.
Theorem 2.6. Let {am }, {bn }, {pm }, {qn }, Pm , and Qn be as in Theorem 2.3,
m
n
and define Am = 1/Pm s=1 ps as and Bn = 1/Qn t=1 qt bt , for m = 1, 2, . . . , k and
SOME INEQUALITIES SIMILAR TO HILBERT’S INEQUALITY
217
n = 1, 2, . . . , r , where k, r are the natural numbers. Let φ and ψ be as defined in
Theorem 2.5. Then
k
1/2
r
k 2
Pm Qn φ Am ψ Bn
≤ C(1, 1, k, r ; α)
(k − m + 1) ps φ as
1/α
m α + nα
m=1 n=1
m=1
r
1/2
2
×
(r − n + 1) qt ψ bt
,
(2.21)
n=1
where C(1, 1, k, r ; α) is defined by taking p = q = 1 in (2.2).
Proof. From the hypotheses and by using Jensen’s inequality and the Schwarz
inequality, it is easy to observe that
m
m
m
2 1/2
1 1 1
ps as ≤
ps φ as ≤
(m)1/2
ps φ as
,
Pm s=1
Pm s=1
Pm
s=1
n
1/2
n
n
2
1 1
1 qt bt ≤
qt ψ bt ≤
(n)1/2
.
qt ψ bt
φ Bn = ψ
Qn t=1
Qn t=1
Qn
t=1
(2.22)
φ Am = φ
The rest of the proof can be completed by following the same steps as in the proofs
of Theorems 2.1 and 2.3 with suitable changes and hence we omit the details.
3. Integral analogues. In this section, we present the integral analogues of the inequalities given in Theorems 2.1, 2.3, 2.5, and 2.6, which in fact are motivated by the
integral analogue of Hilbert’s inequality given in Theorem 1.2.
An integral analogue of Theorem 2.1 is given in the following theorem.
Theorem 3.1. Let p ≥ 1, q ≥ 1, 0 < α ≤ 1 and f (σ ) ≥ 0, g(τ) ≥ 0 for σ ∈ (0, x),
s
τ ∈ (0, y), where x, y are positive real numbers, define F (s) = 0 f (σ )dσ and G(t) =
t
0 g(τ)dτ, for s ∈ (0, x), t ∈ (0, y). Then
xy
0
0
1/2
x
2
F p (s)Gq (t)
(x − s) Ff (s) ds
1/α ds dt ≤ D(p, q, x, y; α)
0
sα + tα
1/2
y
2
×
(y − t) Gg (t) dt
,
(3.1)
0
unless f ≡ 0 or g ≡ 0, where Ff (s) = F p−1 (s)f (s), Gg (t) = Gq−1 (t)g(t), and
D(p, q, x, y; α) =
1/α
1
pq xy.
2
(3.2)
Proof. From the hypotheses of F (s) and G(t), it is easy to observe that
s
F p (s) = p
F p−1 (σ )f (σ )dσ ,
0
t
q
G (t) = q
(3.3)
G
0
s ∈ (0, x),
q−1
(τ)g(τ)dτ,
t ∈ (0, y).
218
YOUNG-HO KIM
From (3.3) and using the Schwarz inequality and the elementary inequality
n
1/n
≤
ai
i=1
n
aα
i
n
i=1
1/α
,
0 < α,
(3.4)
(for ai , i = 1, 2, . . . , n, nonnegative real numbers) we observe that
p
F (s)G (t) = pq
s
q
F
0
p−1
≤ pq(s)1/2
t
(σ )f (σ )dσ
G
q−1
0
s
2
F p−1 (σ )f (σ ) dσ
(τ)g(τ)dτ
1/2
(t)1/2
0
sα + tα
≤ pq
2
1/α 2
Gq−1 (τ)g(τ) dτ
t
0
s
F
p−1
2
(σ )f (σ ) dσ
1/2 t
G
0
q−1
1/2
1/2
2
(τ)g(τ) dτ
.
0
(3.5)
Dividing both sides of the above inequality by (s α + t α )1/α , and then integrating over
t from 0 to y first and then integrating the resulting inequality over s from 0 to x and
using the Schwarz inequality we observe that
xy
0
F p (s)Gq (t)
1/α ds dt
0
sα + tα
1/2 1/2 1/α x
s
y
t
2
2
1
dt
dt
Ff (σ ) dσ
Gg (τ) dτ
≤ pq
2
0
0
0
0
1/α
1/2 1/2
x
s
y
t
2
2
1
≤ pq
(xy)1/2
Ff (σ ) dσ ds
Gg (τ) dτ dt
2
0
0
0
0
1/2 1/2
x
y
2
2
= D(p, q, x, y; α)
(x − s) Ff (s) ds
(y − t) Gg (t) dt
,
0
0
(3.6)
where Ff (σ ) = F p−1 (σ )f (σ ), Gg (τ) = Gq−1 (τ)g(τ). This completes the proof.
Remark 3.2. In the special case when p = q = 1, inequality (3.1) in Theorem 3.1
reduces to the following inequality:
xy
0
0
F (s)G(t)
1/α ds dt
sα + tα
= D(1, 1, x, y; α)
1/2 x
0
(x − s)f (s)ds
1/2
y
2
2
0
(y − t)g (t)dt
(3.7)
,
where D(1, 1, x, y; α) is obtained by taking p = q = 1 in (3.2).
The integral analogues of the inequalities in Theorems 2.3, 2.5, and 2.6 are established in the following theorems.
SOME INEQUALITIES SIMILAR TO HILBERT’S INEQUALITY
219
Theorem 3.3. Let f , g, F , G be as in Theorem 3.1. Let p(σ ) and q(τ) be two positive
s
functions defined for σ ∈ (0, x), τ ∈ (0, y), and define P (s) = 0 p(σ )dσ and Q(t) =
t
0 q(τ)dτ, for s ∈ (0, x), t ∈ (0, y), where x, y are positive real numbers. Let φ and
ψ be as in Theorem 2.3. Then
xy
0
0
2 1/2
x
φ F p(s) ψ G(t)
f (s)
ds dt ≤ L(x, y; α)
(x − s) p(s)φ
ds
1/α
p(s)
0
sα + tα
(3.8)
2 1/2
y
g(t)
×
(y − t) q(t)ψ
dt
,
q(t)
0
where
2 1/2 x 2 1/2
1/α x φ P (s)
ψ Q(t)
1
L(x, y; α) =
ds
dt
.
2
P (s)
Q(t)
0
0
(3.9)
Proof. From the hypotheses and by using Jensen’s inequality and the Schwarz
inequality, it is easy to observe that



φ F (s) = φ 

s
P (s) P (s) f (σ )/p(σ ) dσ
0
s
p(σ )dσ





0
φ P (s) s
f (σ )
≤
dσ
P (σ )φ
P (s)
p(σ )
0
2
1/2
s
φ P (s)
f (σ )
1/2
(s)
dσ
,
P (σ )φ
≤
P (s)
p(σ )
0
(3.10)
and similarly,
2 1/2
t
ψ Q(t)
g(τ)
ψ G(t) ≤
(t)1/2
dτ
.
q(τ)ψ
Q(t)
q(τ)
0
(3.11)
From (3.10) and (3.11) and using the elementary inequality
n
i=1
1/n
ai
≤
n
aα
i
n
i=1
1/α
,
0 < α,
(3.12)
(for ai , i = 1, 2, . . . , n, nonnegative real numbers) we observe that
φ F (s) ψ G(t) ≤
1/α 2
1/2 s
φ P (s)
sα + tα
f (σ )
(s)1/2
dσ
P (σ )φ
2
P (s)
p(σ )
0
2 1/2 t
ψ Q(t)
g(τ)
×
(t)1/2
dτ
q(τ)ψ
.
Q(t)
q(τ)
0
(3.13)
220
YOUNG-HO KIM
The rest of the proof can be completed by following the same steps as in the proof of
Theorem 3.1 and closely looking at the proof of Theorem 2.3, and hence we omit the
details.
s
Theorem 3.4. Let f , g, be as in Theorem 3.1, and define F (s) = 0 f (σ )dσ and
t
G(t) = 0 g(τ)dτ, for s ∈ (0, x), t ∈ (0, y), where x, y are positive real numbers. Let
φ and ψ be as in Theorem 2.5. Then
xy
0
0
st
sα + tα
1/α φ F (s) ψ G(t) ds dt
≤ D(1, 1, x, y; α)
x
0
2
(x − s) φ f (σ ) ds
1/2 y
0
2
(y − t) ψ g(t) dt
1/2
,
(3.14)
where D(1, 1, x, y; α) is obtained by taking p = q = 1 in (3.2).
Theorem 3.5. Let f , g, p, q, P , and Q be as in Theorem 3.3, and define F (s) =
s
t
1/P (s) 0 p(σ )f (σ )dσ and G(t) = 1/Q(t) 0 q(τ)g(τ)dτ, for s ∈ (0, x), t ∈ (0, y),
where x, y are positive real numbers. Let φ and ψ be as in Theorem 2.5. Then
xy
P (s)Q(s)φ F (s) ψ G(t)
ds dt
1/α
0 0
sα + tα
1/2
x
2
≤ D(1, 1, x, y; α)
(x − s) p(s)φ f (s) ds
(3.15)
×
0
y
0
2
(y − t) q(t)ψ g(t)
dt
1/2
,
where D(1, 1, x, y; α) is obtained by taking p = q = 1 in (3.2).
The proofs of Theorems 3.4 and 3.5 can be completed by following the proof of
Theorem 3.3 and by closely looking at the proofs of Theorems 2.5 and 2.6 and by
making use of the integral versions of Jensen’s and the Schwarz inequalities. Here, we
omit the details.
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Young-Ho Kim: Department of Applied Mathematics (or, Brain Korea 21 Project
Corps), Changwon National University, Changwon 641-773, Korea
E-mail address: yhkim@sarim.changwon.ac.kr