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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 inequal- ity (see [3, page 226]) plays an important role in many branches of mathematics.
Theorem1.1. Ifp >1,p=p/(p−1)and
apm≤A,
bpn≤B, the summations running1to∞, then
ambn
m+n< π
sin(π /p)A1/pB1/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]).
Theorem1.2. Ifp >1,p=p/(p−1)and ∞
0
fp(x)dx≤F , ∞
0
gp(y)dy≤G, (1.2)
then ∞
0
∞
0
f (x)g(y)
x+y dx dy < π
sin(π /p)F1/pG1/p, (1.3) unlessf≡0org≡0.
These two theorems were studied extensively and numerous variants, generaliza- tions, 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.
Theorem1.3. Letp≥1, q≥1, and let{am}and{bn}be two nonnegative sequences of real numbers defined form=1,2, . . . , kandn=1,2, . . . , r ,wherek,rare the natural
numbers and defineAm=m
s=1as,Bn=n
t=1bt. Then k
m=1
r n=1
ApmBnq
m+n≤C(p, q, k, r )
k
m=1
(k−m+1) Ap−1m am
2
1/2
×
r
n=1
(r−n+1) Bnq−1bn
2
1/2
,
(1.4)
unless{am}or{bn}is null, where
C(p, q, k, r )=1 2pq
kr . (1.5)
An integral analogue ofTheorem 1.3is given in the following theorem.
Theorem1.4. Letp≥1,q≥1andf (σ )≥0, g(τ)≥0forσ ∈(0, x),τ∈(0, y), wherex,yare positive real numbers and defineF (s)=s
0f (σ )dσandG(t)=t
0g(τ)dτ, fors∈(0, x),t∈(0, y). Then
x 0
y 0
Fp(s)Gq(t)
s+t ds dt≤D(p, q, x, y) x
0(x−s)
Fp−1(s)f (s)2
ds 1/2
× y
0
(y−t)
Gq−1(t)g(t)2
dt 1/2
,
(1.6)
unlessf≡0org≡0, where
D(p, q, x, y)=1 2pq
xy. (1.7)
In this paper, we give an improvement of the inequalities given in Theorems1.3and 1.4similar 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.
Theorem2.1. Letp≥1,q≥1,0< α, and let{am}and{bn}be two nonnegative sequences of real numbers defined form=1,2, . . . , kandn=1,2, . . . , r ,wherek,rare the natural numbers and defineAm=m
s=1as,Bn=n
t=1bt. Then k
m=1
r n=1
ApmBnq
mα+nα1/α ≤C(p, q, k, r;α)
k
m=1
(k−m+1)
Ap−1m am2
1/2
×
r
n=1
(r−n+1) Bnq−1bn
2
1/2
,
(2.1)
unless{am}or{bn}is null, where
C(p, q, k, r;α)= 1
2 1/α
pq
kr . (2.2)
Proof. By using the following inequality (see [1,7]), n
m=1
zm
β
≤β n m=1
zm
m
k=1
zk
β−1
, (2.3)
whereβ≥1 is a constant andzm≥0,(m=1,2, . . .), it is easy to observe that Apm≤p
m s=1
asAp−1s , m=1,2, . . . , k, Bnq≤q n t=1
atBtq−1, n=1,2, . . . , r . (2.4) From (2.4) and using the Schwarz inequality and the elementary inequality
n
i=1
ai
1/n
≤ n
i=1
aαi n
1/α
, 0< α,(see [4]), (2.5) (forai,i=1,2, . . . , n, nonnegative real numbers) we observe that
ApmBnq≤pq m
s=1
asAp−1s
n
t=1
atBtq−1
≤pq(m)1/2 m
s=1
asAp−1s 2
1/2
(n)1/2 n
t=1
atBq−1t 2
1/2
≤pq
mα+nα 2
1/αm
s=1
asAp−1s 2
1/2n
t=1
atBtq−12
1/2
.
(2.6)
Dividing both sides of (2.6) by(mα+nα)1/α, and then taking the sum overnfrom 1 torfirst and then the sum overmfrom 1 tokand using the Schwarz inequality and then interchanging the order of the summations (see [7,8]) we observe that
k m=1
r n=1
ApmBqn
mα+nα1/α
≤pq 1
2
1/αk
m=1
m
s=1
asAp−1s 2
1/2r
n=1
n
t=1
btBq−1t 2
1/2
≤pq 1
2 1/α
(k)1/2 k
m=1
m
s=1
asAp−1s 2
1/2
(r )1/2 r
n=1
n
t=1
btBtq−12
1/2
=pq kr
1 2
1/αk
s=1
asAp−1s 2
k
m=s
1
1/2r
t=1
btBq−1t 2
r
n=t
1 1/2
=C(p, q, k, r;α) k
s=1
asAps−12
(k−s+1)
1/2r
t=1
btBtq−12
(r−t+1) 1/2
=C(p, q, k, r;α) k
m=1
(k−m+1)
amAp−1m 2
1/2r
n=1
(r−n+1)
bnBnq−12
1/2
. (2.7) This completes the proof.
Remark2.2. InTheorem 2.1, settingα≡1, we haveTheorem 1.3. If we takep= q=1 inTheorem 2.1, then the inequality of the result ofTheorem 2.1reduces to the following inequality:
k m=1
r n=1
AmBn
mα+nα1/α≤C(1,1, k, r;α) k
m=1
(k−m+1) am
2
1/2
× r
n=1
(r−n+1) bn
2
1/2
,
(2.8)
whereC(1,1, k, r;α)is obtained by takingp=q=1 in (2.2).
Our next result deals with further generalization of the inequality obtained in Remark 2.2.
Theorem2.3. Let{am},{bn},Am, andBnbe as defined inTheorem 2.1. Let{pm} and{qn}be two nonnegative sequences form=1,2, . . . , kandn=1,2, . . . , r, and define Pm=m
s=1ps,Qn=n
t=1qt. Letφandψbe two real-valued, nonnegative, convex, and submultiplicative functions defined onR+=[0,∞). Then
k m=1
r n=1
φ Am
ψ Bn
mα+nα1/α ≤M(k, r;α) k
m=1
(k−m+1)
pmφ am
pm
21/2
× r
n=1
(r−n+1)
qnφ bn
qn
21/2
,
(2.9)
where
M(k, r;α)=1 2
1/α k
m=1
φ Pm Pm
21/2r
n=1
ψ Qn Qn
21/2
. (2.10)
Proof. From the hypotheses ofφandψand by using Jensen’s inequality and the Schwarz inequality (see [5]), it is easy to observe that
φ Am
=φ Pm
m
s=1psas/ps
m s=1ps
≤φ Pm
φ
ms=1mpsas/ps s=1ps
≤φ Pm
Pm
m s=1
psφ as
ps
≤φ Pm
Pm
(m)1/2 m
s=1
psφ
as
ps
21/2
,
(2.11)
and similarly,
ψ Bn
≤ψ Qn Qn
(n)1/2
n
t=1
qtψ
bt
qt
2
1/2
. (2.12)
From (2.11) and (2.12) and using the elementary inequality n
i=1
ai
1/n
≤ n
i=1
aαi n
1/α
, 0< α, (2.13)
(forai,i=1,2, . . . , n, nonnegative real numbers) we observe that
φ Am
ψ Bn
≤
mα+nα 2
1/α φ
Pm Pm
m
s=1
psφ
as
ps
21/2
× ψ
Qn
Qn
n
t=1
qtψ
bt
qt
21/2 .
(2.14)
Dividing both sides of the above inequality by(mα+nα)1/α, and then taking the sum overnfrom 1 tor first and then the sum overmfrom 1 tokand using the Schwarz inequality and then interchanging the order of the summations we observe that
k m=1
r n=1
φ Am
ψ Bn
mα+nα1/α
≤ 1
2
1/α k
m=1
φ Pm
Pm
m
s=1
psφ
as
ps
21/2
× r
n=1
ψ Qn
Qn
n
t=1
qtψ
bt
qt
21/2
≤ 1
2
1/α k
m=1
φ Pm
Pm
21/2 k
m=1
m
s=1
psφ
as
ps
21/2
× r
n=1
ψ Qn Qn
21/2r
n=1
n
t=1
qtψ
bt
qt
21/2
=M(k, r;α) k
s=1
psφ
as
ps
2k
m=s
1
1/2r
t=1
qtψ
bt
qt
2r
n=t
1 1/2
=M(k, r;α) k
s=1
psφ
as
ps
2
(k−s+1)
1/2r
t=1
qtψ
bt
qt
2
(r−t+1) 1/2
=M(k, r;α) k
m=1
(k−m+1)
pmφ am
pm
21/2r
n=1
(r−n+1)
qnψ bn
qn
21/2
.
(2.15) The proof is complete.
Remark2.4. By applying the elementary inequality (see [4]) n
i=1
ai
1/n
≤ n
i=1
aγi n
1/γ
, 0< γ, (2.16)
(forai,i=1,2, . . . , n, nonnegative real numbers) on the right sides of the inequalities
in Theorems2.1and2.3, we get, respectively, the following inequalities:
k m=1
r n=1
ApmBqn
mα+nα1/α
≤C1
k
m=1
(k−m+1) Apm−1am
2
γ
+ r
n=1
(r−n+1) Bqn−1bm
2
γ1/γ
,
(2.17)
whereC1=(1/2)1/γC(p, q, k, r;α), and k
m=1
r n=1
φ Am
ψ Bn
mα+nα1/α
≤M1
k
m=1
(k−m+1)
pmφ am
pm
2γ
+ r
n=1
(r−n+1)
qnψ bn
qn
2γ1/γ
,
(2.18) whereM1=(1/2)1/γM(k, r;α), which we believe are new to the literature.
The following theorems deal with slight variants of (2.9) given inTheorem 2.3.
Theorem2.5. Let{am}and{bn}be as defined inTheorem 2.1, and defineAm= 1/mm
s=1as andBn=1/nn
t=1bt, form=1,2, . . . , kandn=1,2, . . . , r, wherek, r are the natural numbers. Let φandψ be two real-valued, nonnegative, and convex functions defined onR+=[0,∞). Then
k m=1
r n=1
mn
mα+nα1/αφ Am
ψ Bn
≤C(1,1, k, r;α) k
m=1
(k−m+1) φ
am2
1/2
× r
n=1
(r−n+1) ψ
bn
2
1/2
,
(2.19) whereC(1,1, k, r )is defined by takingp=q=1in (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 1
m m s=1
as
≤ 1 m
m s=1
φ as
≤ 1 m(m)1/2
m
s=1
φ as
2
1/2
,
ψ Bm
=ψ 1
n n t=1
bt
≤ 1 n
n t=1
ψ bt
≤ 1 n(n)1/2
n
t=1
ψ bt
2
1/2
.
(2.20)
The rest of the proof can be completed by following the same steps as in the proofs of Theorems2.1and2.3with 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, and define Am =1/Pmm
s=1psas and Bn = 1/Qnn
t=1qtbt, for m= 1,2, . . . , k and
n=1,2, . . . , r, where k, r are the natural numbers. Let φ and ψ be as defined in Theorem 2.5. Then
k m=1
r n=1
PmQnφ Am
ψ Bn
mα+nα1/α ≤C(1,1, k, r;α) k
m=1
(k−m+1) psφ
as
2
1/2
× r
n=1
(r−n+1) qtψ
bt
2
1/2
,
(2.21)
whereC(1,1, k, r;α)is defined by takingp=q=1in (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 1
Pm
m s=1
psas
≤ 1 Pm
m s=1
psφ as
≤ 1 Pm
(m)1/2 m
s=1
ps φ
as2
1/2
,
φ Bn
=ψ 1
Qn
n t=1
qtbt
≤ 1 Qn
n t=1
qtψ bt
≤ 1 Qn
(n)1/2 n
t=1
qtψ bt
2
1/2
.
(2.22) The rest of the proof can be completed by following the same steps as in the proofs of Theorems2.1and2.3with suitable changes and hence we omit the details.
3. Integral analogues. In this section, we present the integral analogues of the in- equalities given in Theorems2.1,2.3,2.5, and2.6, which in fact are motivated by the integral analogue of Hilbert’s inequality given inTheorem 1.2.
An integral analogue ofTheorem 2.1is given in the following theorem.
Theorem3.1. Letp≥1,q≥1,0< α≤1andf (σ )≥0,g(τ)≥0forσ ∈(0, x), τ∈(0, y), wherex,yare positive real numbers, defineF (s)=s
0f (σ )dσ andG(t)= t
0g(τ)dτ, fors∈(0, x),t∈(0, y). Then x
0
y 0
Fp(s)Gq(t)
sα+tα1/αds dt≤D(p, q, x, y;α) x
0(x−s) Ff(s)2
ds 1/2
× y
0
(y−t) Gg(t)2
dt 1/2
,
(3.1)
unlessf≡0org≡0, whereFf(s)=Fp−1(s)f (s),Gg(t)=Gq−1(t)g(t), and D(p, q, x, y;α)=
1 2
1/α
pq
xy. (3.2)
Proof. From the hypotheses ofF (s)andG(t), it is easy to observe that Fp(s)=p
s
0Fp−1(σ )f (σ )dσ , s∈(0, x), Gq(t)=q
t
0Gq−1(τ)g(τ)dτ, t∈(0, y).
(3.3)
From (3.3) and using the Schwarz inequality and the elementary inequality n
i=1
ai
1/n
≤ n
i=1
aαi n
1/α
, 0< α, (3.4)
(forai,i=1,2, . . . , n, nonnegative real numbers) we observe that
Fp(s)Gq(t)=pq s
0
Fp−1(σ )f (σ )dσ t
0
Gq−1(τ)g(τ)dτ
≤pq(s)1/2 s
0
Fp−1(σ )f (σ )2
dσ 1/2
(t)1/2 t
0
Gq−1(τ)g(τ)2
dτ 1/2
≤pq
sα+tα 2
1/αs 0
Fp−1(σ )f (σ )2
dσ 1/2t
0
Gq−1(τ)g(τ)2
dτ 1/2
. (3.5) Dividing both sides of the above inequality by(sα+tα)1/α, and then integrating over tfrom 0 toyfirst and then integrating the resulting inequality oversfrom 0 toxand using the Schwarz inequality we observe that
x 0
y 0
Fp(s)Gq(t) sα+tα1/αds dt
≤pq 1
2
1/αx 0
s 0
Ff(σ )2
dσ 1/2
dt y
0
t 0
Gg(τ)2
dτ 1/2
dt
≤pq 1
2 1/α
(xy)1/2 x
0
s 0
Ff(σ )2
dσ
ds
1/2y 0
t 0
Gg(τ)2
dτ
dt 1/2
=D(p, q, x, y;α) x
0
(x−s) Ff(s)2
ds
1/2y 0
(y−t) Gg(t)2
dt 1/2
,
(3.6) whereFf(σ )=Fp−1(σ )f (σ ),Gg(τ)=Gq−1(τ)g(τ). This completes the proof.
Remark3.2. In the special case whenp=q=1, inequality (3.1) inTheorem 3.1 reduces to the following inequality:
x 0
y 0
F (s)G(t) sα+tα1/αds dt
=D(1,1, x, y;α) x
0
(x−s)f2(s)ds
1/2y 0
(y−t)g2(t)dt 1/2
,
(3.7)
whereD(1,1, x, y;α)is obtained by takingp=q=1 in (3.2).
The integral analogues of the inequalities in Theorems2.3,2.5, and2.6are estab- lished in the following theorems.
Theorem3.3. Letf,g,F,Gbe as inTheorem 3.1. Letp(σ )andq(τ)be two positive functions defined forσ ∈(0, x),τ∈(0, y), and defineP (s)=s
0p(σ )dσ andQ(t)= t
0q(τ)dτ, fors∈(0, x),t∈(0, y), wherex,yare positive real numbers. Letφand ψbe as inTheorem 2.3. Then
x 0
y 0
φ F p(s)
ψ G(t)
sα+tα1/α ds dt≤L(x, y;α) x
0
(x−s)
p(s)φ f (s)
p(s) 2
ds 1/2
× y
0(y−t)
q(t)ψ g(t)
q(t) 2
dt 1/2
,
(3.8)
where
L(x, y;α)=1 2
1/αx 0
φ P (s) P (s)
2
ds
1/2x 0
ψ Q(t) Q(t)
2
dt 1/2
. (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
P (s) s
0
P (s)
f (σ )/p(σ ) dσ
s
0
p(σ )dσ
≤φ P (s) P (s)
s 0P (σ )φ
f (σ ) p(σ )
dσ
≤φ P (s) P (s) (s)1/2
s 0
P (σ )φ
f (σ ) p(σ )
2
dσ 1/2
,
(3.10)
and similarly,
ψ G(t)
≤ψ Q(t) Q(t) (t)1/2
t 0
q(τ)ψ
g(τ) q(τ)
2
dτ 1/2
. (3.11)
From (3.10) and (3.11) and using the elementary inequality n
i=1
ai
1/n
≤ n
i=1
aαi n
1/α
, 0< α, (3.12)
(forai,i=1,2, . . . , n, nonnegative real numbers) we observe that
φ F (s)
ψ G(t)
≤
sα+tα 2
1/α φ
P (s) P (s) (s)1/2
s 0
P (σ )φ
f (σ ) p(σ )
2
dσ 1/2
× ψ
Q(t) Q(t) (t)1/2
t 0
q(τ)ψ
g(τ) q(τ)
2
dτ 1/2
.
(3.13)
The rest of the proof can be completed by following the same steps as in the proof of Theorem 3.1and closely looking at the proof ofTheorem 2.3, and hence we omit the details.
Theorem 3.4. Letf, g, be as in Theorem 3.1, and define F (s)=s
0f (σ )dσ and G(t)=t
0g(τ)dτ, fors∈(0, x),t∈(0, y), wherex,yare positive real numbers. Let φandψbe as inTheorem 2.5. Then
x 0
y 0
st
sα+tα1/αφ F (s)
ψ G(t)
ds dt
≤D(1,1, x, y;α) x
0(x−s) φ
f (σ )2
ds
1/2y 0(y−t)
ψ g(t)2
dt 1/2
, (3.14) whereD(1,1, x, y;α)is obtained by takingp=q=1in (3.2).
Theorem 3.5. Letf,g,p, q, P, andQbe as in Theorem 3.3, and defineF (s)= 1/P (s)s
0p(σ )f (σ )dσ andG(t)=1/Q(t)t
0q(τ)g(τ)dτ, for s∈(0, x), t∈(0, y), wherex,yare positive real numbers. Letφandψbe as inTheorem 2.5. Then
x 0
y 0
P (s)Q(s)φ F (s)
ψ G(t) sα+tα1/α ds dt
≤D(1,1, x, y;α) x
0(x−s) p(s)φ
f (s)2
ds 1/2
× y
0
(y−t) q(t)ψ
g(t)2
dt 1/2
,
(3.15)
whereD(1,1, x, y;α)is obtained by takingp=q=1in (3.2).
The proofs of Theorems3.4and3.5can be completed by following the proof of Theorem 3.3and by closely looking at the proofs of Theorems 2.5and 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, Changwon641-773, Korea
E-mail address:[email protected]
Journal of Applied Mathematics and Decision Sciences
Special Issue on
Decision Support for Intermodal Transport
Call for Papers
Intermodal transport refers to the movement of goods in a single loading unit which uses successive various modes of transport (road, rail, water) without handling the goods during mode transfers. Intermodal transport has become an important policy issue, mainly because it is considered to be one of the means to lower the congestion caused by single-mode road transport and to be more environmentally friendly than the single-mode road transport. Both consider- ations have been followed by an increase in attention toward intermodal freight transportation research.
Various intermodal freight transport decision problems are in demand of mathematical models of supporting them.
As the intermodal transport system is more complex than a single-mode system, this fact offers interesting and challeng- ing opportunities to modelers in applied mathematics. This special issue aims to fill in some gaps in the research agenda of decision-making in intermodal transport.
The mathematical models may be of the optimization type or of the evaluation type to gain an insight in intermodal operations. The mathematical models aim to support deci- sions on the strategic, tactical, and operational levels. The decision-makers belong to the various players in the inter- modal transport world, namely, drayage operators, terminal operators, network operators, or intermodal operators.
Topics of relevance to this type of decision-making both in time horizon as in terms of operators are:
•
Intermodal terminal design
•
Infrastructure network configuration
•
Location of terminals
•
Cooperation between drayage companies
•
Allocation of shippers/receivers to a terminal
•
Pricing strategies
•
Capacity levels of equipment and labour
•
Operational routines and lay-out structure
•
Redistribution of load units, railcars, barges, and so forth
•
Scheduling of trips or jobs
•
Allocation of capacity to jobs
•
Loading orders
•
Selection of routing and service
Before submission authors should carefully read over the journal’s Author Guidelines, which are located at
http://www .hindawi.com/journals/jamds/guidelines.html.Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking Sys- tem at
http://mts.hindawi.com/, according to the followingtimetable:
Manuscript Due June 1, 2009 First Round of Reviews September 1, 2009 Publication Date December 1, 2009
Lead Guest Editor
Gerrit K. Janssens,
Transportation Research Institute (IMOB), Hasselt University, Agoralaan, Building D, 3590 Diepenbeek (Hasselt), Belgium;
[email protected]Guest Editor
Cathy Macharis,
Department of Mathematics, Operational Research, Statistics and Information for Systems (MOSI), Transport and Logistics Research Group, Management School, Vrije Universiteit Brussel, Pleinlaan 2, 1050 Brussel, Belgium;
[email protected]Hindawi Publishing Corporation http://www.hindawi.com