ON THE HILBERT INEQUALITY
ZHOU YU and GAO MINGZHE
Abstract. In this paper it is shown that the√Hilbert
“ √ inequality
” for double series can be improved
n
n−1
ln n
√
−
by introducing a weight function of the form n+1
, where n ∈ N . A similar result for
π
n+1
the Hilbert integral inequality is also given. As applications, some sharp results of Hardy-Littlewood’s
theorem and Widder’s theorem are obtained.
1. Introduction
Let {an } and {bn } be two sequences of complex numbers. It is all-round known that the inequality
(1.1)
∞ ∞
∞
∞
X X a b̄ 2
X
X
m n 2
2
|an |
|bn |
≤ π2
m
+
n
n=1
n=1 m=1
n=1
is called the Hilbert inequality for double series, where
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∞
P
n=1
2
|an | < +∞ and
∞
P
2
|bn | < +∞, and
n=1
that the constant factor π 2 in (1.1) is the best possible. The equality in (1.1) holds if and only if
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Received August 3, 2007; revised February 21, 2008.
2000 Mathematics Subject Classification. Primary 26D15.
Key words and phrases. Hilbert’s inequality; weight function; double series; monotonic function; HardyLittlewood’s theorem; Widder’s theorem.
A Project Supported by Scientific Research Fund of Hunan Provincial Education Department (06C657).
{an }, or {bn } is a zero-sequence (see [?]). The corresponding integral form of (1.1) is that
∞∞
2
∞
 ∞

Z Z
Z
Z
f
(x)g(y)
2
2
dxdy ≤ π 2  |f (x)| dx  |g (x)| dx
(1.2)
x+y
0
where
R∞
0
0
2
|f (x)| dx < +∞ and
0
R∞
0
2
|g (x)| dx < +∞, and that the constant factor π 2 in (1.2) is
0
also the best possible. The equality in (1.2) holds if and only if f (x) = 0, or g (x) = 0. Recently,
various improvements and extensions of (1.1) and (1.2) appeared in a great deal of papers (see [?]).
The purpose of the present paper √
is to build the Hilbert inequality with the weights by means of a
x
monotonic function of the form 1+√
, thereby new refinements of (1.1) and (1.2) are established,
x
and then to give some of their important applications.
For convenience, we need the following lemmas.
Lemma 1.1. Let n ∈ N. Then
Z∞
dx
1
π
1
√ + ln n
(1.3)
=
(n + x2 )(1 + x)
n+1 2 n 2
0
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Proof. Let a, e and f be real numbers. Then
Z
dx
2
2
(a + x )(e + f x)
1
1
e
x
2
2
= 2
f ln |e + f x| − ln(a + x ) + arctan
+C
e + a2 f 2
2
a
a
where C is an arbitrary constant. This result has been given in the papers (see [3]–[4]). Based on
this indefinite integral it is easy to deduce that the equality (1.3) is true.
Lemma 1.2. Let n ∈ N, x ∈ (0, +∞). Define two functions by
√
√
1 n 12
x
n
√
√
f (x) =
1−
−
x+n x
1+ x 1+ n
√
√
x
n
1 n 12
√ −
√
,
g (x) =
1+
x+n x
1+ x 1+ n
then f (x) and g (x) are monotonously decreasing in (0, + ∞), and
Z∞
(1.4)
f (x) dx = π − πω (n)
0
Z∞
(1.5)
g (x) dx = π + πω (n)
0
where the weight function ω is defined by
√ √
n
n − 1 ln n
√
ω (n) =
(1.6)
−
n+1
π
n+1
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√
x
Proof. At first, notice that 1 − 1+√
= 1+1√x , hence we can write f (x) in form f (x) = f1 (x) +
x
f2 (x), where
√
1
n
n
√
√
√ √ .
f1 (x) =
,
f2 (x) =
(x + n) x
1+ n
(x + n) (1 + x) x
It is obvious that f1 (x) and f2 (x) are monotonously decreasing
in (0, + ∞). Hence f (x) is
√
n
1√
monotonously decreasing in (0, +∞ ). Next, notice that 1 − 1+√
=
, we can write g (x) in
n
1+ n
form g (x) = g1 (x) + g2 (x), where
√
√
n
√
√ ,
g1 (x) =
(1 + n) (x + n) x
g2 (x) =
n
√ .
(x + n) (1 + x)
It is obvious that g1 (x) and g2 (x) are monotonously decreasing in (0, + ∞). Hence g (x) is also
monotonously decreasing in (0, + ∞). Further we need only to compute two integrals.
Z∞
Z∞ f (x) dx =
0
1 n 12
x+n x
1+
√
n
x
√ −
√
dx
1+ n 1+ x
√
0
= 1+
√
1+
n
√
Z∞
n
1 n 12
x+n x
Z∞
dx−
0
√
=
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1+
n
√
√
1+
x
√
x
0
Z∞ π−
1 n 12
x+n x
√
x
√
dx
1+ x
1+ n
0



∞
√
Z
Z∞
 √
1
1
nπ 
√
dt −
dt −
= π − 2 n
2
2

(n + t )
(n + t ) (1 + t)
1 + n
0
0


∞
√
Z

√
1
nπ 
√
=π− π−2 n
dt −
2

(n + t ) (1 + t)
1 + n
0
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1 n 12
x+n x
dx
By Lemma 1.1, we obtain
Z∞
(1.7)
√
√
π
n ln n
nπ
√
+
−
f (x) dx = π − π −
n+1
n+1
1+ n
0
The equality (1.4) follows from (1.7) at once after some simple computations and simplifications.
Similarly, the equality (1.5) can be obtained.
2. Main Results
First, we establish a new refinement of (1.1).
Theorem 2.1. Let {an } and {bn } be two sequences of complex numbers. If
∞
P
and
n=1
2
|bn | < +∞, then
n=1
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(2.1)

∞ ∞
!2
!2 
∞
∞
X X a b̄ 4
 X

X
m n 2
2
|an |
−
ω (n) |an |
≤ π4


m + n
m=1 n=1
n=1
n=1

!2
!2 
∞
∞
 X

X
2
2
×
|bn |
−
ω (n) |bn |


n=1
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where the weight function ω (n) is defined by (1.6).
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∞
P
n=1
2
|an | < +∞
Proof. Let c(x) be a real function and satisfy the condition 1 − c (n) + c (m) ≥ 0, (n, m ∈ N ).
Firstly we suppose that bn = an . Applying Cauchy’s inequality we have
2 2
∞ X
∞
∞ X
∞
X
X
am ān am ān
(1 − c (n) + c (m))
=
m
+
n
m
+
n
m=1 n=1
m=1 n=1
∞ ∞
!
X X a (1 − c(n) + c(m))1/2 m 1/4
m
=
n
(m + n)1/2
m=1 n=1
!2
1/2 an (1 − c(n) + c(m))
n 1/4 ×
m
(m + n)1/2
≤ J1 J2
(2.2)
where
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∞
∞ P
P
J1 =
m=1 n=1
∞ P
∞
P
|am |2
m+n
J2 =
m=1 n=1
|ān |2
m+n
m
n
12
n
m
1
2
(1 − c(n) + c(m))
(1 − c(n) + c(m))
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We can write the double series J1 in the following form:
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∞
X
n=1
∞
X
!
1 n 12
2
(1 − c(m) + c(n)) |an | .
m
+
n
m
m=1
√
√
√
x
x
n
Let c (x) = 1+√
. It is obvious that 1 − 1+√
+ 1+√
≥ 0. It is known from Lemma 1.2 that the
x
x
n
function f (x) is monotonously decreasing. Hence we have
!
√
√
1 n 21
m
n
2
√ +
√
J1 =
1−
|an |
m
+
n
m
1
+
m
1
+
n
n=1 m=1


√

√
∞ Z∞ 1
X
1
x
n
n 2
2
√ −
√
≤
1−
dx |an |


x
+
n
x
1
+
x
1
+
n
n=1
∞
X
∞
X
0
=π
∞
X
2
|an | − π
n=1
∞
X
2
ω (n) |an |
n=1
where the weight function ω (n) is defined by (1.6).
Similarly,
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J2 ≤
n=1
=π
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
∞
X
1 n 21
x+n x
Whence J1 J2 ≤ π
2
(
1+
0
2
|an | + π
n=1
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
∞ Z∞
X
∞
P
n=1
∞
X
√


√
x
n
2
√ −
√
dx |ān |

1 + x 1 + n
2
ω (n) |an | .
n=1
2
|an |
2
−
∞
P
n=1
2 )
ω (n) |an |
.
2
Consequently, we have
(2.3)

2
!2
∞ X
∞
∞
X
 X
am ān 2
|an |
−
≤ π2

m
+
n
m=1 n=1
n=1
∞
X
2
ω (n) |an |
!2 


n=1
where the weight function ω (n) is defined by (1.6).
If bn 6= an , then we can apply Schwarz’s inequality to estimate the right-hand side of (2.1) as
follows:

4 Z1
∞ X
∞
X

am b̄n = 
m + n

m=1 n=1
0
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(2.4)
∞
X
!
1
am tm− 2
m=1
∞
X
n=1
!
1
b̄n tn− 2
2 2


dt


1
!2
!2 2
Z
Z1 X
∞
∞
X
1
1
≤ |am | tm− 2
dt
|bn | tn− 2
dt
n=1
m=1
0
0
2 2
∞ ∞
∞ X
∞
X
am ān X X bm b̄n =
m + n m=1 n=1 m + n m=1 n=1
And then by using the relation (2.3), from (2.4) and the inequality (2.1), we obtain at once.
Similarly, we can establish a new refinement of (1.2).
Theorem 2.2. Let f (x) and g (x) be two functions in complex number field. If
+∞,
R∞
R∞
2
|f (x)| dx <
0
2
|g(x)| dx < +∞, then
0

∞∞
4
2  ∞
2 

Z Z

Z∞
Z


f (x)g (x)
2
2
4




|f
(x)|
dx
−
ω
(x)
|f
(x)|
dx
≤
π
dxdy


x+y


0
0
0 0


2
2  ∞


Z

 Z∞
2
2
×  |g (x)| dx −  ω (x) |g (x)| dx




(2.5)
0
0
where the weight function ω is defined by

 0√ √
x
x − 1 ln x
ω (x) =
(2.6)
√
−

x+1
π
x+1
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x>0
Its proof is similar to that of Theorem 2.1, it is omitted here.
For the convenience of the applications, we list the following result.
R∞
2
Corollary 2.3. Let f (x) be a function in complex number field. If 0 |f (x)| dx < + ∞, then

∞∞
2  ∞
2 
Z Z
2


Z∞
Z


f (x)f (y)
2
2
2




(2.7) dxdy ≤ π
|f (x)| dx
−
ω(x) |f (x)| dx


x+y


0
0
0
where the weight function ω is defined by (2.6).
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x=0
0
3. Applications
As applications, we shall give some new refinements of Hardy-Littlewood’s theorem and Widder’s
theorem.
R1
Let f (x) ∈ L2 (0, 1) and f (x) 6= 0 for all x. Define a sequence {an } by an = 0 xn f (x)dx,
n = 0, 1, 2, . . .. Hardy-Littlewood ([1]) proved that
∞
X
(3.1)
a2n < π
n=0
Z1
f 2 (x)dx,
0
where π is the best constant that the inequality (3.1) keeps valid.
Theorem 3.1. Let f (x) ∈ L2 (0, 1) and f (x) 6= 0 for all x. Define a sequence {an } by
R1
an = 0 xn−1/2 f (x)dx n = 1, 2, . . . . Then
(3.2)
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∞
X
!2
a2n
≤π
n=1

∞
 X

!2
a2n
−
n=1
∞
X
ω (n) a2n
!2  21 Z1

n=1
where ω(n) is defined by (1.6).
Proof. By our assumptions, we may write a2n in the form
a2n =
Z1
0
an xn−1/2 f (x)dx.

0
f 2 (x) dx
Applying Cauchy-Schwarz’s inequality we estimate the right hand side of (3.2) as follows
2
2  1
!2  ∞ Z1
!
∞
∞
Z

X
X
X
a2n
=
an xn−1/2 f (x)dx =
an xn−1/2 f (x)dx


n=1
n=1 0
Z1
≤
=
∞
X
an xn−1/2
0
n=1
Z1
∞ X
∞
X
dx
=
f 2 (x)dx
0
m+n−1
am an x
Z1
dx
0 m=1 n=1
(3.3)
n=1
0
Z1
!2
∞ X
∞
X
f 2 (x)dx
0
am an
m
+n
m= n=1
! Z1
f 2 (x)dx
0
It is known from (2.3) and (3.3) that the inequality (3.2) is valid. Therefore the theorem is
proved.
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Let an ≥ 0 (n = 0, 1, 2, . . . .), A(x) =
∞
P
an xn , A∗ (x) =
n=0
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(3.4)
A2 (x)dx ≤ π
0
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This is Widder’s theorem (see [1]).
∞
P
n=0
Z∞
0
an xn
n! .
2
e−x A∗ (x) dx
Then
Theorem 3.2. With the assumptions as the above-mentioned, it yields

2  ∞
2 
 1
2


Z
Z

 Z∞
2
2
e−x A∗ (x) dx −  ω (x) e−x A∗ (x) dx
(3.5)  A2 (x) dx ≤ π 2 




0
0
0
where ω (x) is defined by (2.6).
Proof. At first we have the following relation:
Z∞
Z∞
∞
n
X
an (xt)
−t ∗
e A (tx)dt = e−t
dt
n!
n=0
0
0
=
Z∞
∞
∞
X
X
an xn
tn e−t dt =
an xn = A(x)
n!
n=0
n=0
0
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Let tx = s. Then we have

2

2
Z1
Z1 Z∞
Zi Z∞

s
1
A2 (x)dx =
e−t A∗ (tx) dt dx =  e− x A∗ (s) ds 2 dx


x
0
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0
0
0
(3.6)
0
0
0

2
Z∞ Z∞
Z∞ Z∞
f (s) f (t)
dsdt
=  e−su f (s) ds du =
s + t
0
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0

2

2
Z∞ Z∞
Z∞ Z∞
=  e−sy A∗ (s)ds dy =  e−s(u+1) A∗ (s) ds du
0
0
0
where f (x) = e−x A∗ (x). By Corollary 2.3, the inequality (3.5) follows from (3.6) at once.
1. Hardy G. H., Littlewood J. E. and Polya G., Inequalities, Cambridge Univ. Press, Cambridge, U.K., 1952.
2. Gao Mingzhe and Hsu Lizhi, A survey of various refinements and generalizations of Hilbert’s inequalities, Journal
of Mathematical Research and Exposition 25(2) (2005), 227–243.
3. Zwillinger D. et al., CRC Standard Mathematical Tables and Formulae, CRC Press, 1988.
4. Gradshteyn I. S. and Ryzhik I. M., Table of Integrals, Series, and Products, Academic Press, 2000.
Zhou Yu, Department of Mathematics and Computer Science Normal College, Jishou University Jishou Hunan
416000, P. R. China, e-mail: hong2990@163.com
Gao Mingzhe, Department of Mathematics and Computer Science Normal College, Jishou University Jishou Hunan
416000, P. R. China, e-mail: mingzhegao@163.com
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