HARDY-HILBERT TYPE INEQUALITIES WITH
FRACTIONAL KERNEL IN Rn
MARIO KRNIĆ
JOSIP PEČARIĆ
Faculty of Electrical Engineering and Computing
University of Zagreb, Unska 3, Zagreb, Croatia
EMail: mario.krnic@fer.hr
Faculty of Textile Technology, University of Zagreb
Pierottijeva 6, 10000 Zagreb, Croatia
EMail: pecaric@hazu.hr
IVAN PERIĆ
PREDRAG VUKOVIĆ
Faculty of Textile Technology, University of Zagreb
Pierottijeva 6, 10000 Zagreb, Croatia
EMail: iperic@pbf.hr
Teacher Training College Čakovec
Ante Starčevića 55, 40000 Čakovec, Croatia
EMail: predrag.vukovic@vus-ck.hr
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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Contents
Received:
01 July, 2009
Accepted:
18 November, 2009
Communicated by:
JJ
II
G. Sinnamon
J
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2000 AMS Sub. Class.:
26D15.
Page 1 of 39
Key words:
Inequalities, multiple Hilbert’s inequality, multiple Hardy-Hilbert’s inequality,
equivalent inequalities, non-conjugate parameters, gamma function, Selberg’s
integral, the best possible constant, symmetric-decreasing function, general rearrangement inequality, hypergeometric function.
Abstract:
The main objective of this paper is some new special Hilbert-type and HardyHilbert-type inequalities in (Rn )k with k ≥ 2 non-conjugate parameters which
are obtained by using the well known Selberg’s integral formula for fractional
integrals in an appropriate form. In such a way we obtain extensions over the
whole set of real numbers, of some earlier results, previously known from the literature, where the integrals were taken only over the set of positive real numbers.
Also, we obtain the best possible constants in the conjugate case.
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Contents
1
Introduction
3
2
Preliminaries
7
3
Basic Result
9
4
The Best Possible Constants in the Conjugate Case
15
5
Some Applications
25
6
Trilinear Version of a Standard Beta Integral
28
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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1.
Introduction
In order to obtain our general results, we need to present the definitions of nonconjugate parameters. Let pi , i = 1, 2, . . . , k, be the real parameters which satisfy
(1.1)
k
X
1
≥ 1 and pi > 1,
p
i
i=1
i = 1, 2, . . . , k.
Further, the parameters pi 0 , i = 1, 2, . . . , k are defined by the equations
1
1
(1.2)
+ 0 = 1, i = 1, 2, . . . , k.
pi pi
Since pi > 1, i = 1, 2, . . . , k, it is obvious that p0i > 1, i = 1, 2, . . . , k. We define
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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k
1 X 1
λ :=
.
k − 1 i=1 p0i
(1.3)
It is easy to deduce that 0 < λ ≤ 1. Also, we introduce parameters qi , i =
1, 2, . . . , k, defined by the relations
1
1
(1.4)
= λ − 0 , i = 1, 2, . . . , k.
qi
pi
In order to obtain our results we require
qi > 0 i = 1, 2, . . . , k.
(1.5)
and
1
1
+1−λ= ,
qi
pi
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It is easy to see that the above conditions do not automatically imply (1.5). The
above conditions were also given by Bonsall (see [2]). It is easy to see that
k
X
1
λ=
q
i=1 i
JJ
i = 1, 2, . . . , k.
P
Of course, if λ = 1, then ki=1 p1i = 1, so the conditions (1.1) – (1.4) reduce to the
case of conjugate parameters.
Considering the two-dimensional case of non-conjugate parameters (k = 2),
Hardy, Littlewood and Pólya, (see [7]), proved that there exists a constant K, dependent only on the parameters p1 and p2 such that the following Hilbert-type inequality holds for all non-negative measurable functions f ∈ Lp1 (h0, ∞i) and
g ∈ Lp2 (h0, ∞i) :
p1
Z ∞
p1 Z ∞
Z ∞Z ∞
1
2
f (x)g(y)
p2
p1
(1.6)
g
(y)dy
.
dxdy
≤
K
f
(x)dx
(x + y)s
0
0
0
0
Hardy, Littlewood and Pólya did not give a specific value for the constant K
in the previous
inequality.
An alternative proof by Levin (see [9]) established that
K = B s sp11 0 , sp12 0 , where B is the beta function, but the paper did not determine
whether this was the best possible constant. This question still remains open. The
inequality (1.6) was also generalized by F.F. Bonsall (see [2]).
Hilbert and Hardy-Hilbert type inequalities (see [2]) are very significant weight
inequalities which play an important role in many fields of mathematics. Similar
inequalities, in operator form, appear in harmonic analysis where one investigates the
boundedness properties of such operators. This is the reason why Hilbert’s inequality
is so popular and is of great interest to numerous mathematicians.
In the last century Hilbert-type inequalities have been generalized in many different directions and numerous mathematicians have reproved them using various techniques. Some possibilities of generalizing such inequalities are, for example, various
choices of non-negative measures, kernels, sets of integration, extension to the multidimensional case, etc. Several generalizations involve very important notions such
as Hilbert’s transform, Laplace transform, singular integrals, Weyl operators.
In this paper we refer to a recent paper of Brnetić et al, [4], where a general
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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Hilbert-type and Hardy-Hilbert-type inequalities were obtained for non-conjugate
parameters, where k ≥ 2, with positive σ−finite measures on Ω. However, we shall
keep our attention on a result with Lebesgue measures and a special homogeneous
function of degree −s. This is contained in:
Theorem 1.1. Let k ≥ 2 be an integer, pi , p0i , qi , i = 1, 2, . . . , k, be real numbers
P
satisfying (1.1) – (1.5) and ki=1 Aij = 0, j = 1, 2, . . . , k. Then the following
inequalities hold and are equivalent:
Z ∞
Z ∞ Qk
i=1 fi (xi )
(1.7)
···
P
λs dx1 . . . dxk
k
0
0
j=1 xj
p1
k Z ∞
Y
pi
i
(k−1−s)+pi αi pi
qi
<K
xi
fi (xi )dxi
and
(1.8)
0

Z
∞
(1−λp0k )(k−1−s)−p0k αk 
xk

 10
p0k
∞
Z
···
0
<K
∞
0
p
Contents
xi
pi
(k−1−s)+pi αi
qi
fipi (xi )dxi
0
k
1
1 Y
qi
Γ(s
−
k
+
1
−
q
α
+
q
A
)
i
i
i
ii
Γ(s)λ i=1
k
Y
i,j=1,i6=j
1
Γ(qi Aij + 1) qi ,
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p1
i
JJ
Page 5 of 39

fi (xi )

λs dx1 ...dxk−1 dxk 

k
j=1 xj
i=1
P
where
K=
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k
Qk−1
k−1
Y Z ∞
i=1
vol. 10, iss. 4, art. 115, 2009
0
i=1

Z



Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
,
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αi =
Pk
j=1
Aij , Aij > − q1i , i 6= j and Aii − αi >
k−s−1
.
qi
Our main objective is to obtain inequalities similar to the inequalities in Theorem
1.1, which will include the integrals taken over the whole set of real numbers.
The techniques that will be used in the proofs are mainly based on classical real
analysis, especially on the well known Hölder inequality and on Fubini’s theorem.
Conventions. Throughout this paper we suppose that all the functions are nonnegative and measurable, so that all integrals converge. Further, the Euclidean norm
of the vector x ∈ Rn will be denoted by |x|.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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2.
Preliminaries
The main results in this paper will be based on the well-known Selberg formula for
the fractional integral
Z
(2.1)
|xk |αk −n |xk − xk−1 |αk−1 −n |xk−1 − xk−2 |αk−2 −n · · · |x2 − x1 |α1 −n
(Rn )k
α0 −n
· |x1 − y|
Qk
Pk
i=0 Γn (αi )
i=0 αi −n ,
dx1 dx2 . . . dxk =
|y|
Pk
Γn
i=0 αi
P
for arbitrary k, n ∈ N and 0 < αi < n such that 0 < ki=0 αi < n. The constant
Γn (α) introduces the n−dimensional gamma function and is defined by the formula
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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n
2
(2.2)
π 2α Γ α2
,
Γn (α) =
Γ n2 − α2
0 < α < n,
where Γ is the well known gamma function. Further, from the definition of the
n−dimensional gamma function it easily follows that
(2.3)
(2π)n
Γn (n − α) =
,
Γn (α)
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0 < α < n.
In the book [13], Stein derived the formula (2.1) with two parameters using the Riesz
potential.
Multiple integrals similar to the one in (2.1) are known as Selberg’s integrals and
their exact values are useful in representation theory and in mathematical physics.
These integrals have only been computed for special cases. For a treatment of Selberg’s integral, the reader can consult Section 17.11 of [11].
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Now, by using the integral equality (2.1), we can easily compute the integral
Qk−1
Z
−βi
i=1 |xi | P
dx1 dx2 . . . dxk−1 ,
k
s
(Rn )k−1 x
i=1 i
where 0 < βi < n, 0 < s < n and
(k − 1)n −
k−1
X
i=1
βi < s < kn −
k−1
X
βi .
i=1
Such an integral will be more suitable for our computations. Namely, by using the
substitution x1 = t1 − xk and xi = ti − ti−1 , i = 2, 3, . . . , k − 1 (see also [5]), one
obtains the formula
Qk−1
Z
−βi
i=1 |xi | P
(2.4)
dx1 dx2 . . . dxk−1
k
s
(Rn )k−1 i=1 xi Q
Pk−1
Γn (n − s) k−1
βi )
i=1 Γn (n − |xk |(k−1)n−s− i=1 βi ,
xk 6= 0
=
Pk−1
Γn kn − s − i=1 βi
Pk−1
P
where 0 < βi < n, 0 < s < n and (k − 1)n − k−1
i=1 βi < s < kn −
i=1 βi .
Pk−1
Obviously, if 0 < βi < n and 0 < s < n then the condition s < kn − i=1 βi is
trivially satisfied.
We shall use the relation (2.4) in the next section, to obtain generalizations of the
multiple Hilbert inequality, over the set of real numbers.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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3.
Basic Result
As we have already mentioned, we shall obtain some extensions of the multiple
Hilbert inequality on the whole set of real numbers. We also obtain the equivalent
inequality, usually called the Hardy-Hilbert inequality. For more details about equivalent inequalities the reader can consult [7]. To obtain our results we introduce the
real parameters Aij , i, j = 1, 2, . . . , k satisfying
k
X
(3.1)
Aij = 0,
j = 1, 2, . . . , k.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
i=1
We also define
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αi =
(3.2)
k
X
Aij ,
Contents
i = 1, 2, . . . , k.
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The main result of this paper is as follows:
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Theorem 3.1. Let k ≥ 2 be an integer and pi , p0i , qi , i = 1, 2, . . . , k, be real numbers
satisfying (1.1) – (1.5). Further, let Aij , i, j = 1, 2, . . . , k be real parameters defined
by (3.1) and (3.2). Then the following inequalities hold and are equivalent:
Page 9 of 39
j=1
Z
(3.3)
(Rn )k
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Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
i=1 xi ≤K
Close
k Z
Y
i=1
Rn
|xi |
pi (k−1)n−pi s
+pi αi
qi
fipi (xi )dxi
p1
i
and
Z
−
|xk |
(3.4)
pk 0
[(k−1)n−s]−pk 0 αk
qk
Rn
pk 0

Z

·
Qk−1
(Rn )k−1

i=1 fi (xi )
P
λs dx1 dx2 . . . dxk−1 
k
i=1 xi ≤K
k−1
Y Z
i=1
for any 0 < s < n, Aij ∈
given by the formula
K=
1
k
Y
Γλn (s) i,j=1,i6=j
− qni , 0
|xi |
dxk
 10
pk



pi (k−1)n−pi s
+pi αi
qi
Rn
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković



fipi (xi )dxi
p1
i
vol. 10, iss. 4, art. 115, 2009
,
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, αi − Aii <
Γn (n + qi Aij )
1
qi
k
Y
s−(k−1)n
,
qi
where the constant K is
1
qi
Γn (s − (k − 1)n − qi αi + qi Aii ) .
i=1
Proof. We start with the inequality (3.3). The left-hand side of the inequality (3.3)
can easily be transformed in the following way
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Z
(Rn )k
Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
i=1 xi 
 q1
Qk
i
Z
k
p
A
q
A
i
ii
i
ij
Y |xi |
|xj |
j=1,j6
=
i
p
p
−q
i

P
s
=
Fi i i (xi )fi (xi )
k
(Rn )k i=1
j=1 xj Close
"
·
k
Y
#1−λ
|xi |pi Aii (Fi fi )pi (xi )
dx1 dx2 . . . dxk ,
i=1
where

Z

Fi (xi ) =
(Rn )k−1
 q1
Qk
i
qi Aij
j=1,j6=i |xj |
P
s dx1 dx2 . . . dxi−1 dxi+1 . . . dxn  .
k
j=1 xj vol. 10, iss. 4, art. 115, 2009
Now by using Selberg’s integral formula (2.4) it follows easily that
" Qk
j=1,j6=i Γn (n + qi Aij )Γn (n − s)
Γn (kn + qi αi − qi Aii − s)
Fi (xi ) =
(3.5)
# q1
i
|xi |
(k−1)n−s
+αi −Aii
qi
.
P
Further, since ki=1 q1i + 1 − λ = 1, qi > 0 and 0 < λ ≤ 1, we can apply Hölder’s
1
inequality with conjugate parameters q1 , q2 , . . . , qk and 1−λ
, on the above transformation. In such a way, we obtain the inequality
Qk
Z
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
(Rn )k i=1 xi q1 Y
1−λ
k Z
k Z
Y
i
pi Aii
pi
pi Aii
pi
≤
|xi |
(Fi fi ) (xi )dxi
|xi |
(Fi fi ) (xi )dxi
=
R
i=1
Z
k
Y
i=1
n
Rn
i=1
pi Aii
|xi |
pi
(Fi fi ) (xi )dxi
p1
i
,
Rn
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
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since q1i + 1 − λ = p1i . Finally, by using definition (3.5) of the functions Fi , i =
1, 2, . . . , k, one obtains the inequality (3.3).
Let us show that the inequalities (3.3) and (3.4) are equivalent. Suppose that the
inequality (3.3) is valid. If we put the function fn : Rn 7→ R, defined by
 ppk 0

Z
0
p
− qk [(k−1)n−s]−pk 0 αk 
k
fk (xk ) = |xk |

k
Qk−1
fi (xi )

P
λs dx1 dx2 . . . dxk−1 
k
i=1 xi i=1
(Rn )k−1
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
in the inequality (3.3), we obtain
p0k
I(xk )
≤K
k−1
Y Z
i=1
|xi |
pi (k−1)n−pi s
+pi αi
qi
Rn
fipi (xi )dxi
p1
i
p0k
pk
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I(xk ) ,
Contents
where I(xk ) denotes the left-hand side of the inequality (3.4). This gives the inequality (3.4).
It remains to prove that the inequality (3.3) is a consequence of the inequality
(3.4). For this purpose, let us suppose that the inequality (3.4) is valid. Then the
left-hand side of the inequality (3.3) can be transformed in the following way:
Z
(Rn )k
Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk =
k
i=1 xi 
Z
(k−1)n−s

−
−αk
qk
· |xk |
Z
|xk |
(k−1)n−s
+αk
qk
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fk (xk )
Rn
(Rn )k−1
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
Qk−1

i=1 fi (xi )
P
λs dx1 dx2 . . . dxk−1  dxk .
k
i=1 xi Applying Hölder’s inequality with conjugate parameters pk and p0k to the above transformation, we have
Z
(Rn )k
Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
i=1 xi Z
p1
pk (k−1)n−pk s
k
+pk αk pk
qk
|xk |
≤
fk (xk )dxk
· I(xk ),
Rn
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
and the result follows from (3.4). Hence, we have shown that the inequalities (3.3)
and (3.4) are equivalent. Since the first inequality is valid, the second one is also
valid. This completes the proof.
vol. 10, iss. 4, art. 115, 2009
Clearly, by putting n = 1 in Theorem 3.1, we obtain inequalities which are similar to the inequalities in Theorem 1.1. The integrals are taken over the whole set of
real numbers, the weight functions are the same and the constant is of the same form
as in Theorem 1.1, where the ordinary gamma function is replaced with Γ1 (α).
Contents
Remark 1. Observe that equality in the inequality (3.3) holds if and only if it holds
in Hölder’s inequality. By using the notation from Theorem 3.1, it means that the
functions
−s
k
k
X
Y
|xi |pi Aii
|xj |qi Aij xj Fi pi −qi (xi )fipi (xi ),
i = 1, 2, . . . , k
j=1,j6=i
and
j=1
k
Y
i=1
|xi |pi Aii (Fi fi )pi (xi )
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are effectively proportional. So, if we suppose that the functions fi , i = 1, 2, . . . , k
are not equal to zero, straightforward computation (see also [4, Remark 1]) leads to
the condition
−s
k
k
X
Y
xi = C
|xi |(k−1)n−s+qi (αi −Aii ) ,
i=1
i=1
where C is an appropriate constant, and that is a contradiction. So equality in Theorem 3.1 holds if and only if at least one of the functions fi is identically equal to
zero. Otherwise, for non-negative and non-zero functions, the inequalities (3.3) and
(3.4) are strict.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Remark 2. If the parameters pi , i = 1, 2, . . . , k are chosen in such a way that
(3.6)
qj > 0, for some j ∈ {1, 2, . . . n},
qi < 0, i 6= j
and
λ<1
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or
(3.7)
qi < 0,
i = 1, 2, . . . , n
then the exponents from the proof of Theorem 3.1 fulfill the conditions for the reverse Hölder inequality (for details see e.g. [12, Chapter V]), which gives the reverse
of the inequalities (3.3) and (3.4).
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4.
The Best Possible Constants in the Conjugate Case
In this section we shall focus on the case of the conjugate exponent, to obtain the
best possible constants in Theorem 3.1, for some general cases. It seems to be a
difficult problem to obtain the best possible constant in the case of non-conjugate
parameters.
It follows easily that the constant K from the previous theorem, in the conjugate
case (λ = 1, pi = qi ), takes the form
K=
k
k
Y
1
1 Y
1
Γn (n + pi Aij ) pi
Γn (s − (k − 1)n − pi αi + pi Aii ) pi .
Γn (s) i,j=1,i6=j
i=1
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
However, we shall deal with an appropriate form of the inequalities obtained in
the previous section in the conjugate case. The main idea is to simplify the above
constant K, i.e. to obtain the constant without exponents. For this sake, it is natural
to consider real parameters Aij satisfying the following constraint
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(4.1) s − (k − 1)n + pi Aii − pi αi = n + pj Aji ,
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j 6= i,
i, j ∈ {1, 2, . . . , k}.
In this case, the above constant K takes the form
(4.2)
K∗ =
1
Γn (s)
k
Y
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Page 15 of 39
fi ),
Γn (n + A
i=1
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where
(4.3)
fi = pj Aji ,
A
j 6= i
and
fi < 0.
−n<A
fi satisfy the relation
It is easy to see that the parameters A
(4.4)
k
X
i=1
fi = s − kn.
A
Close
Further, the inequalities (3.3) and (3.4) with the parameters Aij , satisfying (4.1),
become
Qk
p1
Z
k Z
Y
i
fi (xi )
fi pi
∗
−n−p
A
i=1
i
P
s dx1 dx2 . . . dxk ≤ K
(4.5)
|xi |
fi (xi )dxi
k
(Rn )k Rn
i=1
i=1 xi and
(4.6)


Z
pk 0  p1k 0

Qk−1

fi (xi )
i=1
P
s dx1 dx2 ...dxk−1  dxk
k

(Rn )k−1 
i=1 xi p1
k−1
Y Z
i
fi pi
∗
−n−pi A
.
≤K
|xi |
fi (xi )dxi

Z
fk )
(1−p0k )(−n−pk A
|xk |
·

 Rn
i=1
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Title Page
Rn
Contents
We shall see that the constant K ∗ in (4.5) and (4.6) is the best possible in the
sense that we cannot replace the constant K ∗ in inequalities (4.5) and (4.6) with the
smaller constant, so that inequalities are fulfilled for all non-negative measurable
functions. Before we prove the facts we have to establish the following two lemmas:
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n
Lemma 4.1. Let k ≥ 2 be an integer, xk ∈ R , and xk 6= 0. We define


Qk−1
Z
Z
fi
A
f
i=2 |xi |
P
|x1 |A1 
I1ε (xk ) =
dx . . . dxk−1  dx1 ,
k
s 2
Kn (ε)
(Rn )k−2 x
i=1 i where ε > 0, Kn (ε) is the closed n−dimensional ball of radius ε and parameters
fi , i = 1, 2, . . . , k are defined by (4.3). Then there exists a positive constant Ck such
A
that
(4.7)
I1ε (xk ) ≤ Ck εn+A1 |xk |−2n−A1 −Ak ,
f
f
f
when
ε → 0.
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Proof. We treat two cases. If k = 2 we have
Z
f
|x1 |A1
ε
I1 (x2 ) =
dx1 .
s
Kn (ε) |x1 + x2 |
By letting ε → 0, we easily conclude that there exists a positive constant c2 such that
Z
f
ε
−s
I1 (x2 ) ≤ c2 |x2 |
|x1 |A1 dx1 .
Kn (ε)
The previous integral can be calculated by using n−dimensional spherical coordinates. More precisely, we have
Z
f
(4.8)
|x1 |A1 dx1
Kn (ε)
Z π
Z πZ
···
=
0
Z
=
0
2πZ
0
ε
r
f1 −1
n+A
0
ε
rn+A1 −1 sinn−2 θn−1 sinn−3 θn−2 · · · sin θ2 drdθ1 . . . dθn−1
f
0
|Sn |εn+A1
dr
dS =
,
f1
n+A
Sn
Z
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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f
n
where |Sn | = 2π 2 Γ−1 ( n2 ) is the Lebesgue measure of the unit sphere in Rn . Consequently,
f
c2 |Sn |εn+A1
ε
I1 (x2 ) ≤
|x2 |−s ,
f
n + A1
f1 − A
f2 = −s holds for k = 2.
so the inequality holds when ε → 0, since −2n − A
Further, if k > 2, then by letting ε → 0, since |x1 | → 0, we easily conclude that
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there exists a positive constant ck such that

Z
Z
f
|x1 |A1 dx1 
(4.9) I1ε (xk ) ≤ ck
(Rn )k−2
Kn (ε)

Qk−1
fi
A
i=2 |xi |
P
dx . . . dxk−1  .
k
s 2
i=2 xi We have already calculated the first integral in the inequality (4.9), and the second
one is the Selberg integral. Namely, by using the formulas (2.3), (2.4) and (4.4) we
have
Qk−1
Z
f
|xi |Ai
i=2
P
s dx2 . . . dxk−1
(4.10)
k
(Rn )k−2 x
i=2 i f1 + A
fk ) Qk−1 Γn (n + A
fi )
Γn (2n + A
f f
i=2
=
|xk |−2n−A1 −Ak .
Γn (s)
Finally, by using (4.8), (4.9) and (4.10), we obtain the inequality (4.7) and the proof
is completed.
Similarly, we have
vol. 10, iss. 4, art. 115, 2009
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Page 18 of 39
Lemma 4.2. Let k ≥ 2 be an integer and xk ∈ Rn . We define


Qk−1
Z
Z
fi
A
−1
f
i=2 |xi |

P
I1ε (xk ) =
|x1 |A1 
s dx2 . . . dxk−1 dx1 ,
k
k−2
n
n
−1
n
(R )
R \K (ε )
i=1 xi fi , i = 1, 2, . . . , k are defined by (4.3). Then there
where ε > 0 and parameters A
exists a positive constant Dk such that
(4.11)
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
−1
I1ε (xk ) ≤ Dk εn+Ak ,
f
when ε → 0.
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Proof. We treat again two cases. If k = 2 we have
Z
f
|x1 |A1
ε−1
I1 (x2 ) =
dx1 .
s
Rn \Kn (ε−1 ) |x1 + x2 |
If ε → 0, then |x1 | → ∞, so we easily conclude that there exists a positive constant
d2 such that
Z
−1
I1ε (x2 ) ≤ d2
|x1 |A1 −s dx1 ,
f
Rn \Kn (ε−1 )
and by using spherical coordinates for calculating the integral on the right-hand side
of the previous inequality, we obtain
−1
I1ε (x2 ) ≤
d2 |Sn | n+Af2
ε
.
f2
n+A
Further, if k > 2, then by using (2.3), (2.4) and (4.4), we have
Qk−1
Z
fi
A
i=2 |xi |
(4.12)
dx . . . dxk−1
Pk
s 2
(Rn )k−2 x
i
i=1
f1 + A
fk ) Qk−1 Γn (n + A
fi )
Γn (2n + A
f f
i=2
=
|x1 + xk |−2n−A1 −Ak .
Γn (s)
So, we get
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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f1 + A
fk ) Qk−1 Γn (n + A
fi )
Γn (2n + A
ε−1
i=2
(4.13) I1 (xk ) =
Γn (s)
Z
f
f f
·
|x1 |A1 |x1 + xk |−2n−A1 −Ak dx1 .
Rn \Kn (ε−1 )
By letting ε → 0, then |x1 | → ∞, so there exists a positive constant dk such that
Z
f
ε−1
I1 (xk ) ≤ dk
|x1 |−2n−Ak dx1 .
Rn \Kn (ε−1 )
Since,
Z
fk
−2n−A
|x1 |
Rn \Kn (ε−1 )
|Sn |εn+Ak
,
dx1 =
fk
n+A
f
the inequality (4.11) holds.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Now, we are able to obtain the main result, i.e. the best possible constants in
the inequalities (4.5) and (4.6). Clearly, inequalities (4.5) and (4.6) do not contain
ei , i =
parameters Aij , i, j = 1, 2, . . . , k, so we can regard these inequalities with A
1, 2, . . . , k, as primitive parameters. More precisely, we have
ei , i = 1, 2, . . . , k, are real parameters fulfilling constraint
Theorem 4.3. Suppose A
fi < 0, i = 1, 2, . . . , k. Then, the constant K ∗ is the best possible
(4.4) and −n < A
in both inequalities (4.5) and (4.6).
Proof. Let us denote by Kn (ε) the closed n−dimensional ball of radius ε with the
center in 0.
Let 0 < ε < 1. We define the functions fei : Rn 7→ R, i = 1, 2, . . . , k in the
following way
(
f
|xi |Ai , xi ∈ Kn (ε−1 ) \ Kn (ε),
e
fi (xi ) =
0,
otherwise.
If we put defined functions in the inequality (4.5), then the right-hand side of the
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inequality (4.5) becomes
K
∗
k Z
Y
i=1
−n
|xi |
p1
dxi
i
=K
∗
Z
|xi |−n dxi .
Kn (ε−1 )\Kn (ε)
Kn (ε−1 )\Kn (ε)
By using n−dimensional spherical coordinates we obtain for the above integral
Z ε−1
Z
Z
1
−n
−1
|xi | dxi =
r dr
dS = |Sn | ln 2 ,
ε
Kn (ε−1 )\Kn (ε)
ε
Sn
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
n
2
Γ−1 ( n2 )
n
where |Sn | = 2π
is the Lebesgue measure of the unit sphere in R . So for
the above choice of functions fi the right-hand side of the inequality (4.5) becomes
(4.14)
K ∗ |Sn | ln
Title Page
1
.
ε2
Now let J denote the left-hand side of the inequality (4.5). By using Fubini’s theorem, for the above choice of functions fi , we have
Qk
Z
fi
A
i=1 |xi |
P
J=
dx dx . . . dxk
k
s 1 2
(Kn (ε−1 )\Kn (ε))k i=1 xi Z
f
=
|xk |Ak
Kn (ε−1 )\Kn (ε)


Qk−1
Z
fi
A
i=1 |xi |

P
·
s dx1 dx2 . . . dxk−1 dxk .
k
k−1
n
−1
n
(K (ε )\K (ε))
i=1 xi Note that the integral J can be transformed in the following way: J = J1 − J2 − J3 ,
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where
Z
J1 =
Kn (ε−1 )\Kn (ε)
Z

Z
fk
A
|xk | 
(Rn )k−1
fk
A
|xk |
J2 =
Kn (ε−1 )\Kn (ε)
Z
Kn (ε−1 )\Kn (ε)
Ijε (xk )dxk ,
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
j=1
fk
A
|xk |
J3 =
k−1
X

Qk−1
fi
A
|x
|
i
i=1
P
s dx1 dx2 . . . dxk−1  dxk ,
k
i=1 xi k−1
X
−1
Ijε (xk )dxk .
vol. 10, iss. 4, art. 115, 2009
j=1
−1
Here, for j = 1, 2, . . . , k − 1, the integrals Ijε (xk ) and Ijε (xk ) are defined by
Z Qk−1
fi
A
ε
i=1 |xi |
P
Ij (xk ) =
dx dx . . . dxk−1 ,
k
s 1 2
Pj x
i=1 i satisfying Pj = {(U1 , U2 , . . . , Uk−1 ) ; Uj = Kn (ε), Ul = Rn , l 6= j}, and
Z Qk−1
f
|xi |Ai
ε−1
i=1
P
s dx1 dx2 . . . dxk−1 ,
Ij (xk ) =
k
Qj x
i=1 i satisfying Qj = {(U1 , U2 , . . . , Uk−1 ) ; Uj = Rn \ Kn (ε−1 ), Ul = Rn , l 6= j}.
Now, the main idea is to find the lower bound for J. The first part J1 can easily be
computed. Namely by using Selberg’s integral formula (2.4) and since the relation
fi , it easily follows that
(4.4) holds for parameters A
Qk−1
Z
fi
A
fk −n
∗
−A
i=1 |xi |
P
,
s dx1 dx2 . . . dxk−1 = K |xk |
k
k−1
(Rn )
i=1 xi Title Page
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and consequently, by using n−dimensional spherical coordinates, as we did for computing the right-hand side of the inequality (4.5), we obtain that
1
(4.15)
J1 = K ∗ |Sn | ln 2 .
ε
Now we shall show that the parts J2 and J3 converge when ε → 0. For that sake,
without loss of generality, it is enough to estimate the integrals
Z
Z
fk ε
f −1
A
|xk | I1 (xk )dxk and
|xk |Ak I1ε (xk )dxk .
Kn (ε−1 )\Kn (ε)
Kn (ε−1 )\Kn (ε)
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
By using Lemma 4.1 and n−dimensional spherical coordinates we obtain
Z
Z
fk ε
f1
f
A
n+A
|xk | I1 (xk )dxk ≤ Ck ε
|xk |−2n−A1 dxk
Kn (ε−1 )\Kn (ε)
= Ck |Sn |ε
f1
n+A
Z
Contents
ε−1
r
f1 −1
−n−A
dr
ε
Ck |Sn | f
=
1 − ε2(n+A1 ) .
f1
n+A
Further, we use Lemma 4.2 to estimate the second integral
Z
f −1
|xk |Ak I1ε (xk )dxk .
Kn (ε−1 )\Kn (ε)
Similarly to before, by using spherical coordinates we obtain the inequality
Z
Z
fk
fk ε−1
f
A
n+A
|xk | I1 (xk )dxk ≤ Dk ε
|xk |Ak dxk
Kn (ε−1 )\Kn (ε)
Title Page
Kn (ε−1 )\Kn (ε)
Kn (ε−1 )\Kn (ε)
Dk |Sn | f
=
1 − ε2(n+Ak ) .
fk
n+A
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fi > 0, i = 1, 2, . . . , k, the above computation shows that J2 + J3 ≤
Now, since n + A
O(1) when ε → 0. Hence, for the right-hand side of the inequality (4.5), by using
(4.15), we obtain
(4.16)
J ≥ K ∗ |Sn | ln
1
− O(1),
ε2
when
ε → 0.
Now, let us suppose that the constant K ∗ is not the best possible. That means that
there exists a smaller positive constant L∗ , 0 < L∗ < K ∗ , such that the inequality (4.5) holds, if we replace K ∗ with L∗ . In that case, for the above choice of
functions fei , the right hand-side of the inequality (4.5) becomes L∗ |Sn | ln ε12 . Since
L∗ |Sn | ln ε12 ≥ J, by using (4.16), we obtain the inequality
(4.17)
1
(K − L ) |Sn | ln 2 ≤ O(1),
ε
∗
∗
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Title Page
when
ε → 0.
Now, by letting ε → 0, we obtain from (4.17) a contradiction, since the left hand
side of the inequality goes to infinity. This contradiction shows that the constant K ∗
is the best possible in the inequality (4.5).
Finally, the equivalence of the inequalities (4.5) and (4.6) means that the constant
K ∗ is also the best possible in the inequality (4.6). That completes the proof.
Remark 3. In the papers [3] and [8] we have also obtained the best possible constants,
but only for n = 1 and for the inequalities which involve the integrals taken over the
set of non-negative real numbers.
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5.
Some Applications
In this section we shall consider some special choices of real parameters Aij , i, j =
1, 2, . . . , k, in Theorem 3.1. In such a way, we shall obtain some extensions (on
the set of real numbers) of the numerous versions of multiple Hilbert’s and HardyHilbert’s inequality, previously known from the literature. Further, in the conjugate
case we shall obtain the best possible constants in some cases.
To begin with, let us define real parameters Aij , i, j = 1, 2, . . . , k, by Aii =
(nk − s) λqqi2−1 and Aij = (s − nk) qi1qj , i 6= j, i, j = 1, 2, . . . , k. Then we have
i
k
X
Aij =
i=1
X s − nk
i6=j
qi qj
+ (nk − s)
λqj − 1
qj 2
=
s − nk
qj
k
X
1
−λ
q
i
i=1
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
!
= 0,
for j = 1, P
2, . . . , k. Clearly, the parameters Aij are symmetric and it directly follows
that αi = nj=1 Aij = 0, for j = 1, 2, . . . , k. In such a way we obtain the following
result:
pi , p0i , qi ,
Corollary 5.1. Let k ≥ 2 be an integer and
i = 1, 2, . . . , k, be real numbers
satisfying (1.1) – (1.5). Then the following inequalities hold and are equivalent:
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Page 25 of 39
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Z
(5.1)
(Rn )k
Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
i=1 xi ≤L
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k Z
Y
i=1
Rn
|xi |
pi (k−1)n−pi s
qi
fipi (xi )dxi
p1
i
and



Z



Rn

Z
p 0
− qk [(k−1)n−s] 
|xk | k

pk 0
Qk−1
(Rn )k−1

i=1 fi (xi )
P
λs dx1 dx2 . . . dxk−1 
k
i=1 xi ≤L
k−1
Y Z
i=1
|xi |
pi (k−1)n−pi s
qi
Rn
dxk
 10
pk






fipi (xi )dxi
p1
i
,
vol. 10, iss. 4, art. 115, 2009
where 0 < nk − s < n min{pi , qj , i, j = 1, 2, . . . , k} and the constant L is defined
by the formula
L=
1
k
Y
Γλn (s) i=1
Γn n −
1 k
nk − s p0i Y
qi
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
Γn n −
i=1
nk − s
pi
Title Page
q1
i
Contents
.
The equality in both inequalities holds if and only if at least one of the functions fi ,
i = 1, 2, . . . , k, is equal to zero.
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Remark 4. Straightforward computation shows that parameters Aij from Corollary
5.1, in the conjugate case, satisfy equation (4.1). Hence, the constant L from the
previous corollary becomes
k
1 Y
nk − s
L=
Γn n −
,
Γn (s) i=1
pi
and that is the best possible constant in the conjugate case.
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Remark 5. Similar to the previous corollary, if we define the parameters Aij by
i −1)
and Aij = − λqni qj , i 6= j, i, j ∈ {1, 2, . . . , k}, then we have
Aii = n(λq
λq 2
i
k
X
i=1
Aij =
X
−
i6=j
n
n(λqj − 1)
n
+
=−
2
λqi qj
λqj
λqj
k
X
1
−λ
q
i=1 i
!
= 0,
for
Pn j = 1, 2, . . . , k. Since the parameters Aij are symmetric one obtains αi =
j=1 Aij = 0, for j = 1, 2, . . . , k. So, by putting these parameters in Theorem
3.1 we obtain the same inequalities as those in Corollary 5.1, with the constant L
replaced by
k
1 Y
Γn
L0 = λ
Γn (s) i=1
where (k − 1)n − s <
n
λp0i
n
λp0i
λ− q1 Y
k
i
Γn
i=1
n
s + 0 − (k − 1)n
λpi
q1
i
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Title Page
,
< n, i = 1, 2, . . . , k.
It is important to mention that the results in this section, as well as Theorem
3.1, are extensions of our papers [3] and [4], obtained by using Selberg’s integral
formula.
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6.
Trilinear Version of a Standard Beta Integral
As we know, Selberg’s integral formula is the k−fold generalization of a standard
beta integral on Rn . A few years ago, by using a Fourier transform (see [6]), the
following trilinear version of a standard beta integral was obtained:
Z
|t|α+β−2n
|x − y|n−α−β
(6.1)
dt = B(α, β, n) n−β n−α ,
α
β
|x| |y|
Rn |x − t| |y − t|
where x, y ∈ Rn , x 6= y 6= 0, 0 < α, β < n, α + β > n and
n−α
Γ n−β
Γ α+β−n
n Γ
2
2
2
.
B(α, β, n) = π 2
Γ α2 Γ β2 Γ n − α+β
2
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Title Page
By using the definition (2.2) of the n−dimensional gamma function we easily obtain
that
Γn (n − α)Γn (n − β)
(6.2)
B(α, β, n) =
.
Γn (2n − α − β)
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We also define
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(6.3)
B ∗ (α, β, n) =
Γn (α)Γn (β)
.
Γn (α + β)
It is still unclear whether or not there is a corresponding k−fold analogue of (6.1).
In spite of that, we shall use the trilinear formula (6.1) to obtain a 2−fold inequality
of Hilbert type for the kernel K(x, y) = |x − y|α−n |x + y|β−n , where 0 < α, β < n,
α + β < n.
In the 2−dimensional case we denote non-conjugate exponents in the following
way: p1 = p, p2 = q, p01 = p0 and p02 = q 0 . So, with the above notation, we have the
following result:
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Theorem 6.1. Let α and β be real parameters satisfying 0 < α, β < n and α + β <
n. Then, the following inequalities hold and are equivalent
Z
f (x)g(y)
(6.4)
dxdy
λ(n−α) |x + y|λ(n−β)
(Rn )2 |x − y|
p1
Z
(p−1)(α+β+n)−pnλ p
≤N
|x|
f (x)dx
Rn
Z
(q−1)(α+β+n)−qnλ q
|y|
×
1q
g (y)dy
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
Rn
and
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(Z
n(λq 0 −1)−α−β
Z
|y|
(6.5)
Rn
(Rn )
) 10
q0
q
f (x)dx
dy
|x − y|λ(n−α) |x + y|λ(n−β)
Z
p1
(p−1)(α+β+n)−pnλ p
≤N
|x|
f (x)dx ,
Rn
where the constant N is defined by N = 2λ(α+β−n) B ∗ (α, β, n)λ .
Proof. The main idea is the same as in Theorem 3.1, i.e. to reduce the case of nonconjugate exponents to the case of conjugate exponents. Note that the right-hand
side of the first inequality (6.4) can be transformed in the following way:
Z
Z
1
1
f (x)g(y)
dxdy =
P1 q0 P2 p0 P3 1−λ dxdy,
λ(n−α)
λ(n−β)
|x + y|
(Rn )2 |x − y|
(Rn )2
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where
|x|(p−1)(α+β) |y|−α−β p
f (x),
|x − y|n−α |x + y|n−β
|y|(q−1)(α+β) |x|−α−β q
P2 =
g (y),
|x − y|n−α |x + y|n−β
P3 = |x|(p−1)(α+β) |y|(q−1)(α+β) f p (x)g q (y).
P1 =
Therefore, respectively, by applying Hölder’s inequality with conjugate exponents
1
q 0 , p0 , 1−λ
and Fubini’s theorem, we obtain the inequality (6.4).
Let us show that the inequalities (6.4) and (6.5) are equivalent. To this aim,
suppose that the inequality (6.4) is valid. If we put the function
n(λq 0 −1)−α−β
Z
g(y) = |y|
Rn
qq0
f (x)
dx
|x − y|λ(n−α) |x + y|λ(n−β)
in the inequality (6.4), then the left-hand side of (6.4) becomes J, where J is the
left-hand side of the inequality (6.5). Also, the second factor on the right-hand side
1
inequality (6.4) becomes J q , so (6.5) follows easily.
It remains to prove that (6.4) is a consequence of (6.5). For this purpose, let’s
suppose that the inequality (6.5) is valid. Then the left-hand side of the inequality
(6.4) can be transformed in the following way:
Z
f (x)g(y)
dxdy
λ(n−α)
|x + y|λ(n−β)
(Rn )2 |x − y|
Z
Z
α+β+n
α+β+n
f
(x)
−nλ
−
+nλ
0
0
=
|y| q
g(y) |y| q
dx dy.
λ(n−α) |x + y|λ(n−β)
Rn
Rn |x − y|
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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Finally, by applying Holder’s inequality with conjugate exponents q and q 0 on the
previous transformation, and by using the inequality (6.5) one easily obtains (6.4).
Hence, the inequalities are equivalent and the proof is completed.
Real parameters α and β in (6.4) and (6.5) satisfy the condition α+β < n. In what
follows we shall obtain similar inequalities which are, in some way, complementary
to the inequalities (6.4) and (6.5). The first step is to consider the case when the
function g ∈ Lq (Rn ) is symmetric-decreasing, that is, g(x) ≥ g(y) whenever |x| ≤
|y|. Since q > 1, for such a function and y ∈ Rn , y 6= 0, we have
Z
1
q
g (y) ≤
(6.6)
g q (x) dx
|B(|y|)| B(|y|)
Z
1
≤
g q (x) dx
|B(|y|)| Rn
n
=
|y|−n kgkqq ,
|Sn |
where B(|y|) denotes the ball of radius |y| in Rn , centered at the origin, and |B(|y|)| =
|y|n |Snn | is its volume.
Theorem 6.2. Letα and β be real parameters satisfying 0 < α < n, 0 < β < n,
α+β = n p1 + 1q > n. If f and g are nonnegative functions such that f ∈ Lp (Rn ),
g ∈ Lq (Rn ), then the following inequalities hold and are equivalent
1−λ
Z
f (x)g(y)
n
(6.7)
dxdy ≤
C(p, q; α, β; n)kf kp kgkq ,
n−α |x + y|n−β
|Sn |
(Rn )2 |x − y|
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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and
(Z
Z
(6.8)
Rn
Rn
f (x)dx
|x − y|n−α |x + y|n−β
) 10
q0
q
dy
≤
n
|Sn |
1−λ
C(p, q; α, β; n)kf kp ,
with the constant
Z
(6.9)
C(p, q; α, β; n) =
Rn
n
|x|− q dx
,
|e1 − x|n−α |e1 + x|n−β
n
where e1 = (1, 0, . . . , 0) ∈ R and |Sn | is the Lebesgue measure of the unit sphere
in Rn .
Proof. Since we shall use a general rearrangement inequality (see e.g. [10]) it is
enough to prove the inequality for symmetric-decreasing functions f and g. First,
1
using Hölder’s inequality with parameters q 0 , p0 and 1−λ
, we have
Z
1
1
f (x)g(y)
q0
p0 1−λ
(6.10)
dxdy
≤
I
I
I3 ,
1
2
n−α |x + y|n−β
R2n |x − y|
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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where
Z
I1 =
R2n
Z
I2 =
R2n
Z
I3 =
R2n
n
n
|x| p0 |y|− q
f p (x)dxdy,
|x − y|n−α |x + y|n−β
−n
p
n
q0
|x| |y|
g q (y)dxdy,
|x − y|n−α |x + y|n−β
n
n
|x| p0 |y| q0
f p (x)g q (y)dxdy.
n−α
n−β
|x − y| |x + y|
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Further, using the substitution y = |x|u (so dy = |x|n du) and rotational invariance
of the Lebesgue integral in Rn we easily get:
Z
Z
−n
q
n
|y|
p
0
|x| p f (x)
I1 =
dydx
n−α |x + y|n−β
Rn
Rn |x − y|
n
Z
Z
n
n
|u|− q
0 − q +α+β−n p
p
|x|
=
f (x)
n−α n−β dudx
x
Rn
Rn x
|x| − u
|x| + u
n
Z
|u|− q du
=
kf kpp .
n−α
n−β
|e1 + u|
Rn |e1 − u|
vol. 10, iss. 4, art. 115, 2009
Title Page
Analogously,
Z
I2 =
Rn
−n
p
|u|
n−α
|e1 − u|
du
n−β
|e1 + u|
I3 ≤
n
|Sn |
It remains to prove that
Z
Z
Rn
−n
q
|u|
n−α
|e1 − u|
du
n−β
|e1 + u|
Contents
kgkqq ,
and, by (6.6),
Rn
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
kf kpp kgkqq .
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−n
p
|x| dx
=
|e1 − x|n−α |e1 + x|n−β
Z
Rn
−n
q
|x| dx
.
|e1 − x|n−α |e1 + x|n−β
We transform the left integral in polar coordinates using x = tθ, t ≥ 0, θ ∈ Sn and
Close
the substitution t = u1 to obtain:
n
Z
|x|− p dx
n−α |e + x|n−β
1
Rn |e1 − x|
n
Z
Z ∞
t− p tn−1 dt
=
dθ
|e1 − tθ|n−α |e1 + tθ|n−β
Sn
0
n
Z
Z ∞
t− p tn−1 dt
=
dθ
n−β
n−α
Sn
0
(1 + t2 − 2the1 , θi) 2 (1 + t2 + 2the1 , θi) 2
n
Z
Z ∞
u p −α−β un−1 du
=
dθ
n−β
n−α
Sn
0
(1 + u2 − 2uhe1 , θi) 2 (1 + u2 + 2uhe1 , θi) 2
n
Z
|x|− q dx
.
=
n−α |e + x|n−β
1
Rn |e1 − x|
To complete the proof, we need to consider the general case, that is, for arbitrary
nonnegative functions f and g. Since x 7→ |x|n−α , x 7→ |x|n−β are symmetricdecreasing functions vanishing at infinity, the general rearrangement inequality implies that
Z
Z
f (x)g(y)
f ∗ (x)g ∗ (y)
dxdy
≤
dxdy.
(6.11)
n−α |x + y|n−β
n−α |x + y|n−β
R2n |x − y|
R2n |x − y|
Clearly, by (6.10), the right-hand side of (6.11) is not greater than
1−λ
n
(6.12)
C(p, q; α, β; n)kf ∗ kp kg ∗ kq
|Sn |
1−λ
n
=
C(p, q; α, β; n)kf kp kgkq ,
|Sn |
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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where C(p, q; α, β; n) is the constant from the right-hand side of (6.7). To achieve
equality in (6.12), we used the fact that the symmetric-decreasing rearrangement is
norm preserving.
On the other hand, by putting the function
Z
g(y) =
Rn
qq0
f (x)
dx
|x − y|n−α |x + y|n−β
in the inequality (6.7) we obtain (6.8). The equivalence of the inequalities (6.7) and
(6.8) can be shown in the same way as in Theorem 6.1.
The case n = 1 of the previous theorem is interesting as for that case the constant
C(p, q; α, β; n) can be expressed in terms of the hypergeometric function. More
precisely, using the definition of hypergeometric functions (for more details see [1])
it is easy to see that the following identity holds for 0 < d1 , d2 , d3 < 1, d1 +d2 +d3 >
1:
Z
|t|−d2 |1 − t|−d3 |1 + t|−d1 dt
R
= B(1 − d2 , 1 − d3 )F (d1 , 1 − d2 ; 2 − d2 − d3 ; −1)
+ B(1 − d2 , 1 − d1 )F (d3 , 1 − d2 ; 2 − d2 − d1 ; −1)
+ B(d1 + d2 + d3 − 1, 1 − d3 )F (d1 , d1 + d2 + d3 − 1; d1 + d2 ; −1)
+ B(d1 + d2 + d3 − 1, 1 − d1 )F (d3 , d1 + d2 + d3 − 1; d3 + d2 ; −1).
Hence, for n = 1 we have
Corollary 6.3. Let α and β be real parameters satisfying 0 < α < 1, 0 < β < 1,
α + β = p1 + 1q > 1. If f and g are nonnegative functions such that f ∈ Lp (R),
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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g ∈ Lq (R), then the following inequalities hold and are equivalent
Z
f (x)g(y)
(6.13)
dxdy ≤ 2λ−1 C(p, q; α, β)kf kp kgkq ,
1−α
1−β
|x
−
y|
|x
+
y|
R2
and
(Z Z
(6.14)
R
R
f (x)dx
|x − y|1−α |x + y|1−β
) 10
q0
q
dy
≤ 2λ−1 C(p, q; α, β)kf kp ,
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
where
1 1
1
(6.15) C(p, q; α, β) = B
, α F 1 − β, 0 ; 0 + α; −1
q0
q q
1
1 1
+B
, β F 1 − α, 0 ; 0 + β; −1
q0
q q
1
1 1
+B
, α F 1 − β, 0 ; 0 + α; −1
p0
p p
1
1 1
, β F 1 − α, 0 ; 0 + β; −1 ,
+B
p0
p p
B(·, ·) is the usual (one-dimensional) beta function and F (d1 , d2 ; d3 ; z) is the hypergeometric function.
The following corollary should be compared with Theorem 6.1.
Corollary 6.4. If f and g are nonnegative functions such that f ∈ Lp (R) and g ∈
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Lq (R), then the following inequalities hold and are equivalent
Z
f (x)g(y)dxdy
(6.16)
λ
R2
|x2 − y 2 | 2
λ 1
λ 1
λ−1
≤2
B 1− , 0 +B 1− , 0
kf kp kgkq .
2 2p
2 2q
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
and
(6.17)
#q 0
 "
Z Z
f (x)dx
R
|x2 − y 2 | 2
λ

R
Proof. Set α = β = 1 −
dy
 10
q

λ 1
λ 1
λ−1
≤2
B 1− , 0 +B 1− , 0
kf kp ,
2 2p
2 2q
λ
2
in the previous corollary.
Note that inequalities (6.16) and (6.17) could not be obtained from Theorem 6.1.
In other words there are no such α and β for which the kernel in inequalities (6.4)
λ
and (6.5) reduces to |x2 − y 2 |− 2 .
vol. 10, iss. 4, art. 115, 2009
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References
[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions with Formulae, Graphs and Mathematical Tables, National Bureau of
Standards, Applied Math. Series 55, 4th printing, Washington, 1965.
[2] F.F. BONSALL, Inequalities with non-conjugate parameters, Quart. Journ.
Math. Oxford, (2), 2 (1951), 135–150
[3] Y. BICHENG, I. BRNETIĆ, M. KRNIĆ AND J. PEČARIĆ, Generalization
of Hilbert and Hardy-Hilbert integral inequality, Math. Inequal. Appl., 8(2)
(2005), 259–272.
[4] I. BRNETIĆ, M. KRNIĆ AND J. PEČARIĆ, Multiple Hilbert and HardyHilbert inequalities with non-conjugate parameters, Bull. Austral. Math. Soc.,
71 (2005), 447–457.
[5] A. ČIŽMEŠIJA, I. PERIĆ AND P. VUKOVIĆ, Inequalities of the Hilbert type
in Rn with non-conjugate exponents, Proc. Edinb. Math. Soc. (2), 51(1) (2008),
11–26.
[6] L. GRAFAKOS AND C. MORPURGO, A Selberg integral formula and applications, Pacific Journ. Math., 191(1) (1999), 85–94.
[7] G.H. HARDY, J.E. LITTLEWOOD
Univ. Press, Cambridge, 1952.
AND
G. PÓLYA, Inequalities, Cambridge
[8] M. KRNIĆ, G. MINGHZE, J. PEČARIĆ AND G. XUEMEI, On the best constant in Hilbert’s inequality, Math. Inequal. Appl., 8(2) (2005), 317–329
[9] V. LEVIN, On the two parameter extension and analogue of Hilbert’s inequality, J. London Math. Soc., 11 (1936), 119–124
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
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vol. 10, iss. 4, art. 115, 2009
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[10] E.H. LIEB AND M. LOSS, Analysis, 2nd edition, Graduate Studies in Mathematics, Vol. 14, American Mathematical Society, Providence, RI, 2001.
[11] M.L. MEHTA, Random Matrices, 2nd ed, Academic Press, Boston, MA, 1990.
[12] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and
New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993
[13] E.M. STEIN, Singular Integrals and Differentiability Properties of Functions,
Princeton Univ. Press, Princeton, NJ, 1970.
Hardy-Hilbert Type Inequalities
M. Krnić, J. Pečarić,
I. Perić and P. Vuković
vol. 10, iss. 4, art. 115, 2009
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