International Journal of Mathematics and Mathematical Sciences Volume 2008, Article ID 165089,12pages
doi:10.1155/2008/165089
Research Article
On Hilbert’s Inequality for Double Series and
Its Applications
Zhou Yu and Gao Mingzhe
Department of Mathematics and Computer Science, Normal College of Jishou University, Jishou, Hunan 416000, China
Correspondence should be addressed to Gao Mingzhe,[email protected]
Received 2 September 2007; Accepted 26 March 2008
Recommended by Feng Qi
This study shows that a refinement of the Hilbert inequality for double series can be established by introducing a real functionuxand a parameterλ. In particular, some sharp results of the classical Hilbert inequality are obtained by means of a sharpening of the Cauchy inequality. As applications, some refinements of both the Fejer-Riesz inequality and Hardy inequality inHpfunction are given.
Copyrightq2008 Z. Yu and G. Mingzhe. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let{an}and{bn}be two sequences of complex numbers. Ifλ0,1, then
∞
m1−λ ∞
n1−λ
ambn
mnλ ≤ π
∞
n1−λ
|an|2
1/2 ∞
n1−λ
|bn|2
1/2
, 1.1
∞
m1−λ
∞
n1−λ
m /n
ambn
m−n ≤ π
∞
m1−λ
|an|2
1/2 ∞
n1−λ
|bn|2
1/2
, 1.2
that is,
∞
m1−λ ∞
n1−λ
ambn
mnλ 2
∞
m1−λ ∞
n1−λ m /n
ambn
m−n 2
≤ π2
∞
n1−λ
|an|2 ∞
n1−λ
|bn|2. 1.3
Recently, the various extensions on 1.1 appeared in some papers e.g., 3, 4, etc..They
focalize on changing the denominator of the function of the left-hand side of1.1. Such as the
denominator mnλ is replaced byαmβnμ in3, and the denominatormnλ
is replaced by mum nvnμ in 4. Some new results in these papers were yielded.
The inequality 1.3 is obviously a significant refinement of1.1and 1.2. However, both
extensions and refinements on1.2and1.3do not commonly appear in previous papers. The main purpose of the present paper is to establish both an extension and a significant refinement on1.3. Explicitly, letux>0x∈0, ∞be a real function and let limx→∞ux ∞. If the
denominatormnλof the first term of the left-hand side of1.3is replaced byumun, and the denominatorm−nof the second term of the left-hand side of1.3is replaced by um−un, then a new inequality established is significant in theory and applications; and as applications, we will give both extensions and refinements on Fejer-Riesz’s inequality and Hardy’s inequality. For convenience, we introduce some notations and functions as follows:
Va, b
∞
m0
∞
n0
um/vn
ambn
um− un,
Uka, b ∞
m0
∞
n0
ambn
um unk, k1,2,
x2∞
n0
|xn|2,
Tkx ∞
n0
|xn|
ukn, k1,2,
f, g 2π
0
ftgtdt,
α2 2π
0
|α|2dt, where αf, g, h.
1.4
In particular, whenba, the notationsUka, aandVa, aare denoted, respectively, byUka
andVa. Throughout this paper, we will frequently use these notations, and we stipulate that Zdenotes integer set and that un Znλ/2, whereZn ∈ Z,n ∈ N0, λis an integer or
0< λ <1.
2. Lemmas
Lemma 2.1. If both∞n0anand∞n0bnare absolute convergent, then
i∞m0 ∞
n0ambnis absolute convergent,
iia2andb2are convergent.
The proof of it has been given in the paper2. Hence, it is omitted here.
Lemma 2.2. Letf, g, h∈ L20, 2π, and lethbe a variable unit-vector. Then,
f, g2 ≤ f2g2
1−
|f, h|
f −
|g, h| g
2
. 2.1
In particular, whenhis orthogonal tof, we have
f, g2 ≤ f2g21−|g, h|
g
2
; 2.2
and the equality in2.2holds if and only iff, gandhare linearly dependent.
The proof of these results has been given in5,6.
Define a function by
rx πU1xsin 2λπ−U2xsin2λπ. 2.3
Lemma 2.3. Let{an}and{bn}be two sequences of complex numbers. Ifλis an integer or 1/2 ≤ λ <1,
then
rarb≥0. 2.4
Proof. Whenλis an integer, it is clear thatrarb 0. So we consider only the case for 1/2 ≤ λ <1. It is easy to deduce that
U1x 1
0 ∞
m0
xmtum−1/2
2dt >0,
U2x 1
0
ds s
s
0 ∞
n0
xntun−1/2
2dt >0.
2.5
When 1/2≤λ <1, it follows from2.3thatrx<0 for anyx∈C. Hence, we haverarb>0.
Lemma 2.4. Letfz ∞
n0anzun−1/2.Iffzis analytic in the unit-disc|z|<1, then
Proof.
−ππ tf−eit2dt
π
−πt f
−eitf−eitdt
π
−πt ∞
m0
∞
n0
amancosπt isinπtum−1/2
cosπ−t isinπ−tun−1/2dt
2π
⎛ ⎜ ⎜ ⎝
∞
m0
∞
n0
um/un
aman
um−un ⎞ ⎟ ⎟
⎠sinλπi ⎛ ⎜ ⎜ ⎝
∞
m0
∞
n0
um/un
aman
um−un ⎞ ⎟ ⎟ ⎠cosλπ
2πVa.
2.7
Thereby, the relation2.6holds.
3. Theorems and their corollaries
In order to prove our assertions, we need also to introduce the following functions:
s1x
√
2|T1xcosλπ−1/2πT2xsinλπ|
π2x2− rx1/2 ,
s2x
√
2|T1xsinλπ−1/πT2xsin2λπ/2|
π2x2 rx1/2 .
3.1
Theorem 3.1. Letrxbe a function defined by2.3, let{an}and{bn}be two nonzero sequences of
complex numbers, and let both∞n0anand∞n0bnbe absolute convergent. Then,
iifλis an integer, then
U1a, b2Va, b2≤π2a2b21−R2, 3.2
where
R2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
1 π2
T1a
a −
T2b
πb
2
T1b
b −
T2a
πa
2
ifλis odd,
1 π2
T1a
a
2
T1b
b
2
ifλis even;
3.3
iiif 0< λ <1, then
U1a, bcos 2λπ−
1
2πU2a, bsin 2λπ
2Va, b2≤
π2a2b2− 1
π2rarb
where R2 min{s
1a−s2b2, s1b−s2a2}, six i 1,2 is defined by 3.1. In
particular, when 1/2≤λ <1, we haverarb>0.
Proof. Define two functions by
fa, t ∞
m0
am
√
tsinumt,
gb, t ∞
n0
bn
√
tcosunt, t ∈ 0, 2π.
3.5
Since both ∞n0an and ∞n0bn are absolute convergent by Lemma 2.1, the double series
∞ m0
∞
n0ambn is absolute convergent. Accordingly, fa, tgb, t is uniformly convergent
in the interval0, 2π. Thereby, the interchange in order of summation and integration can be made. In what follows, we stipulate that the interchanges in order of summation and integration are justified. It is easy to deduce that
f2π2a2−ra,
g2π2b2rb, f, g
2π
0
fa, tgb, tdt
πVa, b
U1a, bcos 2λπ−
1
2πU2a, bsin 2λπ ,
3.6
whererxis a function defined by2.3. ByLemma 2.2, we have
Va, b
U1a, bcos 2λπ−
1
2πU2a, bsin 2λπ
2
1
π2f, g 2≤ 1
π2f
2g21−r
1 π2
π2a2−raπ2b2rb1−r,
3.7
wherer |f, h|/f − |g, h|/g2,his a variable unit-vector, it can be properly chosen in accordance with our requirement.
iWhenλis an integer, it is known from2.3thatrx 0. We selecth1
√
2t/2πit is easy to deduce thath11,and
f, h1√ 2
∞
m0
am
um . 3.8
Since the series∞n0anis absolute convergent, it is justified that the complex numberan
is replaced by|an|in3.8. Hence, we have
f, h1
√
Similarly
g, h1 ⎧ ⎪ ⎨ ⎪ ⎩
√ 2
πT2b ifλis odd, 0 ifλis even.
3.10
We therefore obtain that
r1
f, h1
f −
g, h1
g
2
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
2 π2
T 1a
a −
T2b
πb
2
ifλ is odd,
2 π2
T1a
a
2
ifλis even.
3.11
Hence, the inequality3.7can be reduced to
Va, b U1a, b2≤ {π2a2b2}1−r1. 3.12
Notice thatU1b, a U1a, bandVb, a −Va, b. If we still select the unit-vector
h2
√
2t/2π, then, interchangingaandbin3.12, we have
−Va, b U1a, b 2
≤π2a2b21−r2, 3.13
wherer2is defined by
r2
f, h2
f −
g, h2
g 2
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
2 π2
T1b
b −
T2a
πa 2
ifλis odd,
2 π2
T1b
b
2
ifλis even.
3.14
Adding3.12and3.13, then the inequality3.2follows after simplifications. iiWhen 0< λ <1, we firstly considerhin3.7. We still select unit-vectorh1
√ 2t/2π. It is easy to deduce that
f, h1√ 2
∞
m0
am
um
cosλπ− 1 2π
∞
m0
am
um2
sinλπ,
g, h1
√ 2
∞
n0
bn
un
sinλπ− 1 π
∞
n0
bn
un2
sin2λπ 2
.
3.15
Since the series∞n0an and∞n0bn are absolute convergent, it is justified that the complex
numbers an and bn are replaced, respectively, by |an| and |bn| in the above relations.
LetR2
1 s1a−s2b
2. We obtain from3.7
Va, b
U1a, bcos 2λπ− 1
2πU2a, bsin 2λπ
2
≤
π2a2b2−b2ra− a2rb− 1
π2rarb
1−R21.
3.16
Notice thatU1b, a U1a, b,U2b, a U2a, b,andVb, a −Va, b. If we still
select the unit-vectorh2
√
2t/2π, then, interchangingaandbin3.16, we have
−Va, b
U1a, bcos 2λπ− 1
2πU2a, bsin 2λπ
2
≤π2a2b2−a2rb− b2ra− 1
π2rbra
1−R2 2
,
3.17
whereR2
2 s1b−s2a2. LetR2 min{R21, R22}. Adding3.16and3.17, the inequality
3.4can be gotten after simplifications. In particular, when 1/2 ≤ λ < 1, by Lemma 2.3, we haverarb≥0. The proof ofTheorem 3.1is completed.
Corollary 3.2. Let∞
n0anbe absolute convergent. Then,
iifλis an integer, then
U1a2|Va|2≤π2a41−R2, 3.18
where
R2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
2 π2a2
T1a−T2a
π
2
ifλis odd,
2
π2a2T1a
2 ifλis even;
3.19
iiif 0< λ <1, then
U1acos 2λπ−
1
2πU2asin 2λπ
2|Va|2≤
π2a4− 1
π2r
2a1−R2,
3.20
whereR2 s
1a−s2a2, six i1,2is defined by3.1.
In particular, whenun nλ/2, according to3.2, one obtains a refinement of1.3immediately.
Corollary 3.3. Ifλ0,1, then
∞
m1−λ ∞
n1−λ
ambn
mnλ 2
∞
m1−λ ∞
n1−λ m /n
ambn
m−n 2
≤ π2 ∞
n1−λ
|an|2 ∞
n1−λ
|bn|2 1−R2
where
R2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
1 π2
T1a
a − T2b
πb
2
T1b
b − T2a
πa
2
ifλ1,
1 π2
T1a
a
2
T1b
b
2
ifλ0,
3.22
where
Tkx ∞
n1−λ
|xn|
nλ/2k, k1,2. 3.23
Corollary 3.4. Ifun n1/4, then
∞
m0
∞
n0
ambn
mn1/2
2
∞
m0
∞
n0,m /n
ambn
m−n
2≤π2a2b2− 1
π2rarb
1−R2,
3.24
where
rarb>0, R2min{s1a−s2b2,s1b−s2a2},
s1x
−1/√
2πT2x
π2x2− rx1/2
,
s2x
√
2|T1x −1/2πT2x|
π2x2 rx1/2 ,
Tkx ∞
n0
|xn|
n 1/4k, k1,2.
3.25
Sinceλ1/2, it is known fromLemma 2.3thatrarb>0.
IfrarbandRin3.24are replaced by zero, then the inequality3.24can be reduced to
∞
m0
∞
n0
ambn
mn1/2
2
∞
m0
∞
n0,m /n
ambn
m−n
2< π2a2b2. 3.26
The inequalities3.24and3.26are refinements of the Hilbert-Ingham inequality
∞
m0
∞
n0
ambn
mn1/2
≤πab. 3.27
Theorem 3.5. With the assumptions asTheorem 3.1, ifλ1/4, then
2π1 U2a, b
2Va, b2≤
π2a2b2− 1
π2rarb
1−R2, 3.28
where
R2min{s1a−s2b2,s1b−s2a2},
s1x |
T1x −1/2πT2x|
π2x2− rx1/2 ,
s2x |
T1x −1/2πT2x
√ 2−1| π2x2 rx1/2 ,
Tkx ∞
n0
|xn|
n1/8k, k1,2.
3.29
4. Applications toHP function
Letfzbe analytic in the unit-disc|z|<1. Iffz ∞n0anzn∈Hpwithp >0,then
1
0 ftp
dt≤1 2
2π
0 f
eitpdt, 4.1
where the coefficient 1/2 is the best possible. It is called the Fejer-Riesz inequality in Hp
function7.
We will give both an extension and a refinement of4.1in what follows.
Theorem 4.1. Letfz ∞n0anzun−1/2∈Hp,wherep >0 andun Zn λ/2Zn∈Z, λ∈
N0. Iffzis analytic in the unit-disc|z|<1, then
1
0
ftpdt
2
1 2π
π
−πt
f−eitpdt
2
≤
1 2
2π
0
feitpdt
2
1−R2, 4.2
whereR2>0.
Proof. At first, we prove the theorem for casep 2.Letfz ∞m0amzum−1/2. It is easy to
deduce that
1
0 ft2
dt
∞
m0
∞
n0
um un/0
aman
um un U1a,
1 2π
2π
0
feit2dta2.
ByLemma 2.4, we have
2π1 π
−πtf
−eit2dt
∞
m0
∞
n0
um/un
aman
um−un
Va. 4.4
Since the series∞n0anis absolutely convergent, it is justified that the complex numberanis
replaced by|an|.
According3.18, we have
1
0ft 2
dt
2
1 2π
π
−πt
f−eit2dt
2
≤
1 2
2π
0
feit2dt
2
1−R2, 4.5
whereR2is defined by3.19. It is easy to deduce that
2 π2a2
4 π
2π
0
feit2dt
−1
,
T1a 1
0
t−1/2ftdt,
T2a 1
0
ds s
s
0
t−1/2ftdt.
4.6
Becauseun Znλ/2/1/π,T1a−T2a/π /0. It shows thatR2>0. Hence, the inequality
4.2is valid whenp 2. Ifp /2, then by the Blaschke decomposition theorem, it holds that fz BzGz, whereBz is Blaschke function andGz/0,|Bz| ≤ 1 in |z| < 1 and |Beit|1.
LetFz Gzp/2∈H2. According to the above result forp2,we have 1
0 ftp
dt
2
1 2π
π
−πt
f−eitpdt
2
1
0 Ft2
dt
2
1 2π
π
−πt
F−eit2dt
2
≤1 2
2π
0 F
eit2dt 2
1−R2F
1 2
2π
0
Geitpdt 2
1−R2G
1 2
2π
0
feitpdt
2
1−R2.
4.7
Based on the case for p 2, we haveR2F > 0. Hence,R2 > 0.The proof ofTheorem 4.1 is
completed. Let
fz ∞
m0
Then,
∞
n0
|cn|
n1 ≤ 1 2
π
−π
f
eitdt. 4.9
It is called the Hardy inequality inHpfunction7.
We will give both an extension and a refinement of4.9as follows.
Theorem 4.2. Letfz ∞m0cmzumbe analytic in the unit-disc|z|<1, whereum Zmλ/2
(withZm∈Zandλ∈N0) andf∈H1. Then, ∞
n0
|cn|
un 2 1 2π π −πt
f−eit2dt
2 ≤ 1 2 π −π
feitdt
2
1−R2, 4.10
where R2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 4 π 1
0t−1ftdt−
1 π 1 0 ds s s 0
t−1ftdt
2 2π
0
feitdt −1 ifλis odd,
4 π
1
0
t−1ftdt
2 2π
0
feitdt
−1
ifλis even.
4.11
Proof. By Blaschke decomposition theorem, we have
fz BzGz BzG1/2zG1/2z f
1zf2z, 4.12
whereBzis Blaschke function,f1, f2 ∈ H2. Letf1z BzG1/2z ∞
m0amzum, f2z
G1/2z ∞
n0bnzun.It is easy to deduce that
a2b2∞
m0
|am|2 ∞
n0
|bn|2 1
2π π
−π|G
1/2eit|2dt
1 2π
π
−π|Be
itG1/2eit|2dt 1
2π 2π
0
feitdt.
4.13
Owing tofz f1zf2z, it holds that 1
0t−1ftdt 1
0t−1f1t 2
dt. It is easy to deduce
that
∞
n0
|cn|
un≤
∞
n0
rsn
|ar| |as|
ur us
∞
m0
∞
n0
|am| |an|
um un. 4.14
ByLemma 2.4, we find that
2π1 −ππ tf−eitdt 1 2π
π
−πt
f1
−eit2dt ∞
m0
∞
n0
um/un
|am| |an|
um −un
It follows from4.13,4.14,4.15, andCorollary 3.2that
∞
n0
|cn|
un
2
1 2π
π
−πt
f−eitdt
2
≤
1 2
2π
0
feitdt
2
1−R2, 4.16
whereR2is defined by3.19. It is easy to deduce that
2 π2a2
4 π
2π
0
feit2dt
−1
,
T1a 1
0
t−1f1t2dt 1
0
t−1ftdt,
T2a 1
0
ds s
s
0
t−1f1t2dt 1
0
ds s
s
0
t−1ftdt.
4.17
These show that the inequality4.10is valid.
Acknowledgment
The research is supported by the Scientific Research Fund of Hunan Provincial Education Departmentno. 06C657.
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