Volume 10 (2009), Issue 4, Article 115, 19 pp.
HARDY-HILBERT TYPE INEQUALITIES WITH FRACTIONAL KERNEL IN Rn
MARIO KRNIĆ, JOSIP PEČARIĆ, IVAN PERIĆ, AND PREDRAG VUKOVIĆ
FACULTY OF E LECTRICAL E NGINEERING AND C OMPUTING
U NIVERSITY OF Z AGREB
U NSKA 3, Z AGREB , CROATIA
mario.krnic@fer.hr
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6
10000 Z AGREB , CROATIA
pecaric@hazu.hr
iperic@pbf.hr
T EACHER T RAINING C OLLEGE Č AKOVEC
A NTE S TAR ČEVI ĆA 55
40000 Č AKOVEC , CROATIA
predrag.vukovic@vus-ck.hr
Received 01 July, 2009; accepted 18 November, 2009
Communicated by G. Sinnamon
A BSTRACT. 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.
Key words and phrases: 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.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In order to obtain our general results, we need to present the definitions of non-conjugate
parameters. Let pi , i = 1, 2, . . . , k, be the real parameters which satisfy
(1.1)
180-09
k
X
1
≥ 1 and
p
i=1 i
pi > 1,
i = 1, 2, . . . , k.
2
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
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
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)
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
1
1
and
+ 1 − λ = , i = 1, 2, . . . , k.
q
qi
pi
i=1 i
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)
p1
p2
(1.6)
dxdy ≤ K
f (x)dx
g (y)dy
.
s
(x + y)
0
0
0
0
λ=
Hardy, Littlewood and Pólya did not give a specific value for the constantK 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 multi-dimensional 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 Hilbert-type
and Hardy-Hilbert-type inequalities were obtained for non-conjugate parameters, where k ≥ 2,
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
3
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 satisfying
P
(1.1) – (1.5) and ki=1 Aij = 0, j = 1, 2, . . . , k. Then the following inequalities hold and are
equivalent:
p1
Z ∞
Z ∞ Qk
k Z ∞
Y
pi
i
f
(x
)
i
i
(k−1−s)+p
α
p
i
i
i=1
(1.7)
···
xi qi
fi i (xi )dxi
P
λs dx1 . . . dxk < K
k
0
0
0
i=1
j=1 xj
and

Z



(1.8)
p0k

∞
(1−λp0k )(k−1−s)−p0k αk
xk
0
Z


∞
∞
Z
···
0
0
 10
p
k
Qk−1


i=1 fi (xi )
P
λs dx1 . . . dxk−1  dxk 

k
j=1 xj
<K
k−1
Y Z ∞
i=1
xi
pi
(k−1−s)+pi αi
qi
fipi (xi )dxi
p1
i
,
0
where
k
1
1 Y
qi
Γ(s
−
k
+
1
−
q
α
+
q
A
)
K=
i
i
i
ii
Γ(s)λ i=1
αi =
Pk
j=1
Aij , Aij > − q1i , i 6= j and Aii − αi >
k
Y
1
Γ(qi Aij + 1) qi ,
i,j=1,i6=j
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 non-negative and
measurable, so that all integrals converge. Further, the Euclidean norm of the vector x ∈ Rn
will be denoted by |x|.
2. P RELIMINARIES
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 )
dx1 dx2 . . . dxk =
|y| i=0 αi −n ,
Pk
Γn
i=0 αi
Pk
for arbitrary k, n ∈ N and 0 < αi < n such that 0 <
i=0 αi < n. The constant Γn (α)
introduces the n−dimensional gamma function and is defined by the formula
n
π 2 2α Γ α2
,
(2.2)
Γn (α) =
0 < α < n,
Γ n2 − α2
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
4
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
where Γ is the well known gamma function. Further, from the definition of the n−dimensional
gamma function it easily follows that
(2.3)
Γn (n − α) =
(2π)n
,
Γn (α)
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].
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
βi < s < kn −
i=1
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
P
Pk−1
where 0 < βi < n, 0 < s < n and (k − 1)n − k−1
i < s < kn −
i=1 βP
i=1 βi . Obviously, if
k−1
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.
3. BASIC R ESULT
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
(3.1)
k
X
Aij = 0,
j = 1, 2, . . . , k.
i=1
We also define
(3.2)
αi =
k
X
Aij ,
i = 1, 2, . . . , k.
j=1
The main result of this paper is as follows:
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
5
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:
Qk
p1
Z
k Z
Y
pi (k−1)n−pi s
i
+p
α
p
i
i
i=1 fi (xi )
i
qi
fi (xi )dxi
(3.3)
|xi |
P
λs dx1 dx2 . . . dxk ≤ K
k
(Rn )k Rn
i=1
i=1 xi 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 ∈
formula
K=
− qni , 0
, αi − Aii <
|xi |
Rn
s−(k−1)n
,
qi
dxk
 10
pk






pi (k−1)n−pi s
+pi αi
qi
fipi (xi )dxi
p1
i
,
where the constant K is given by the
k
k
Y
1 Y
1
1
qi
Γ
(n
+
q
A
)
Γn (s − (k − 1)n − qi αi + qi Aii ) qi .
n
i ij
λ
Γn (s) i,j=1,i6=j
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
Z
(Rn )k
Qk
fi (xi )
Pi=1 λs dx1 dx2 . . . dxk
k
i=1 xi 
 q1
Qk
i
Z
k
pi Aii
qi Aij
Y
|xi |
|xj |
j=1,j6
=
i
p
p
−q
i
i
i

P
s
=
Fi
(xi )fi (xi )
k
(Rn )k i=1
x
j=1 j " k
#1−λ
Y
·
|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 Now by using Selberg’s integral formula (2.4) it follows easily that
" Qk
# q1
i
(k−1)n−s
+αi −Aii
j=1,j6=i Γn (n + qi Aij )Γn (n − s)
(3.5)
Fi (xi ) =
|xi | qi
.
Γn (kn + qi αi − qi Aii − s)
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
6
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
P
Further, since ki=1 q1i + 1 − λ = 1, qi > 0 and 0 < λ ≤ 1, we can apply Hölder’s inequality
1
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
Rn
i=1
=
k Z
Y
i=1
i=1
pi Aii
|xi |
Rn
p1
pi
(Fi fi ) (xi )dxi
i
,
Rn
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

Z
p 0
− qk [(k−1)n−s]−pk 0 αk 
fk (xk ) = |xk | k

(Rn )k−1
in the inequality (3.3), we obtain
k−1
Y Z
p0k
I(xk ) ≤ K
i=1
|xi |
 ppk 0

i=1 fi (xi )
P
λs dx1 dx2 . . . dxk−1 
k
i=1 xi pi (k−1)n−pi s
+pi αi
qi
Rn
k
Qk−1
fipi (xi )dxi
p1
i
p0k
pk
I(xk ) ,
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:
Qk
Z
Z
(k−1)n−s
+αk
i=1 fi (xi )
|xk | qk
fk (xk )
P
λs dx1 dx2 . . . dxk =
k
Rn
(Rn )k i=1 xi 

Qk−1
Z
(k−1)n−s


−
−αk
i=1 fi (xi )
qk
· |xk |
P
λs dx1 dx2 . . . dxk−1  dxk .
k
(Rn )k−1 i=1 xi Applying Hölder’s inequality with conjugate parameters pk and p0k to the above transformation,
we have
Qk
Z
p1
Z
pk (k−1)n−pk s
k
f
(x
)
+p
α
i
i
p
k k
qk
· I(xk ),
|xk |
fk k (xk )dxk
Pi=1 λs dx1 dx2 . . . dxk ≤
k
(Rn )k Rn
x
i=1 i 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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
7
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).
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
k
k
X −s
Y
|xj |qi Aij i = 1, 2, . . . , k
|xi |pi Aii
xj Fi pi −qi (xi )fipi (xi ),
j=1
j=1,j6=i
and
k
Y
|xi |pi Aii (Fi fi )pi (xi )
i=1
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
x
=
C
|xi |(k−1)n−s+qi (αi −Aii ) ,
i
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.
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
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).
4. T HE B EST P OSSIBLE C ONSTANTS IN THE C ONJUGATE C ASE
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
Y
1 Y
1
1
pi
K=
Γn (n + pi Aij )
Γn (s − (k − 1)n − pi αi + pi Aii ) pi .
Γn (s) i,j=1,i6=j
i=1
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
(4.1)
s − (k − 1)n + pi Aii − pi αi = n + pj Aji ,
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
j 6= i,
i, j ∈ {1, 2, . . . , k}.
http://jipam.vu.edu.au/
8
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
In this case, the above constant K takes the form
k
1 Y
fi ),
K =
Γn (n + A
Γn (s) i=1
∗
(4.2)
where
j 6= i
fi = pj Aji ,
A
(4.3)
fi < 0.
−n<A
and
fi satisfy the relation
It is easy to see that the parameters A
k
X
(4.4)
fi = s − kn.
A
i=1
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


Rn

Z
fk )
(1−p0k )(−n−pk A

|xk |
·
(Rn )k−1
 10
pk 0
pk

Qk−1

i=1 fi (xi )
P
dx1 dx2 . . . dxk−1  dxk
k
s


i=1 xi ≤ K∗
k−1
Y Z
i=1
fi
−n−pi A
|xi |
Rn
fipi (xi )dxi
p1
i
.
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:
Lemma 4.1. Let k ≥ 2 be an integer, xk ∈ Rn , and xk 6= 0. We define


Qk−1
Z
Z
fi
A
f
i=2 |xi |

P
I1ε (xk ) =
|x1 |A1 
s dx2 . . . dxk−1 dx1 ,
k
k−2
n
n
K (ε)
(R )
i=1 xi fi , i =
where ε > 0, Kn (ε) is the closed n−dimensional ball of radius ε and parameters A
1, 2, . . . , k are defined by (4.3). Then there exists a positive constant Ck such that
(4.7)
I1ε (xk ) ≤ Ck εn+A1 |xk |−2n−A1 −Ak ,
f
f
Proof. We treat two cases. If k = 2 we have
Z
ε
I1 (x2 ) =
Kn (ε)
f
when ε → 0.
|x1 |A1
dx1 .
|x1 + x2 |s
f
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)
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
9
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
π
Z
2π
Z
···
=
0
Z
0
0
0
rn+A1 −1 dr
Z
Sn
|Sn |εn+A1
,
f1
n+A
f
dS =
f
0
rn+A1 −1 sinn−2 θn−1 sinn−3 θn−2 · · · sin θ2 drdθ1 . . . dθn−1
f
ε
=
ε
n
2
where |Sn | = 2π Γ−1 ( n2 ) is the Lebesgue measure of the unit sphere in Rn . Consequently,
I1ε (x2 ) ≤
c2 |Sn |εn+A1
|x2 |−s ,
f
n + A1
f
f1 − A
f2 = −s holds for k = 2. Further, if
so the inequality holds when ε → 0, since −2n − A
k > 2, then by letting ε → 0, since |x1 | → 0, we easily conclude that there exists a positive
constant ck such that


Qk−1
Z
Z
fi
A
f
i=2 |xi |
P
(4.9)
I1ε (xk ) ≤ ck
|x1 |A1 dx1 
dx . . . dxk−1  .
k
s 2
Kn (ε)
(Rn )k−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
Z
(4.10)
(Rn )k−2
Qk−1
fi
A
i=2 |xi |
P
dx . . . dxk−1
k
s 2
i=2 xi 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
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 
dx . . . dxk−1  dx1 ,
k
s 2
Rn \Kn (ε−1 )
(Rn )k−2 x
i=1 i fi , i = 1, 2, . . . , k are defined by (4.3). Then there exists a positive
where ε > 0 and parameters A
constant Dk such that
(4.11)
−1
I1ε (xk ) ≤ Dk εn+Ak ,
f
Proof. We treat again two cases. If k = 2 we have
Z
ε−1
I1 (x2 ) =
Rn \Kn (ε−1 )
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
when ε → 0.
|x1 |A1
dx1 .
|x1 + x2 |s
f
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M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
If ε → 0, then |x1 | → ∞, so we easily conclude that there exists a positive constant d2 such
that
Z
f
ε−1
I1 (x2 ) ≤ d2
|x1 |A1 −s dx1 ,
Rn \Kn (ε−1 )
and by using spherical coordinates for calculating the integral on the right-hand side of the
previous inequality, we obtain
d2 |Sn | n+Af2
−1
I1ε (x2 ) ≤
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
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.
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 parameters Aij , i, j =
ei , i = 1, 2, . . . , k, as primitive parameters.
1, 2, . . . , k, so we can regard these inequalities with A
More precisely, we have
ei , i = 1, 2, . . . , k, are real parameters fulfilling constraint (4.4) and
Theorem 4.3. Suppose A
fi < 0, i = 1, 2, . . . , k. Then, the constant K ∗ is the best possible in both inequalities
−n < A
(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
fi
A
n −1
n
fei (xi ) = |xi | , xi ∈ K (ε ) \ K (ε),
0,
otherwise.
If we put defined functions in the inequality (4.5), then the right-hand side of the inequality
(4.5) becomes
p1
Z
k Z
Y
i
∗
−n
∗
=K
|xi |−n dxi .
K
|xi | dxi
i=1
Kn (ε−1 )\Kn (ε)
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
Kn (ε−1 )\Kn (ε)
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H ARDY-H ILBERT T YPE I NEQUALITIES
11
By using n−dimensional spherical coordinates we obtain for the above integral
Z
Z ε−1
Z
1
−n
−1
|xi | dxi =
r dr
dS = |Sn | ln 2 ,
ε
Kn (ε−1 )\Kn (ε)
ε
Sn
n
where |Sn | = 2π 2 Γ−1 ( n2 ) is the Lebesgue measure of the unit sphere in Rn . So for the above
choice of functions fi the right-hand side of the inequality (4.5) becomes
1
(4.14)
K ∗ |Sn | ln 2 .
ε
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 |
J=
dx dx . . . dxk
Pk
s 1 2
(Kn (ε−1 )\Kn (ε))k x
i
i=1
Z
f
=
|xk |Ak
Kn (ε−1 )\Kn (ε)


Qk−1
Z
fi
A
i=1 |xi |
P
·
dx dx . . . dxk−1  dxk .
k
s 1 2
(Kn (ε−1 )\Kn (ε))k−1 i=1 xi Note that the integral J can be transformed in the following way: J = J1 − J2 − J3 , where


Qk−1
Z
Z
fi
A
f
i=1 |xi |

P
J1 =
|xk |Ak 
s dx1 dx2 . . . dxk−1 dxk ,
k
k−1 n
−1
n
n
K (ε )\K (ε)
(R )
i=1 xi Z
k−1
X
fk
A
J2 =
|xk |
Ijε (xk )dxk ,
Kn (ε−1 )\Kn (ε)
Z
J3 =
j=1
|xk |Ak
f
Kn (ε−1 )\Kn (ε)
k−1
X
−1
Ijε (xk )dxk .
j=1
−1
Here, for j = 1, 2, . . . , k − 1, the integrals Ijε (xk ) and Ijε (xk ) are defined by
Z Qk−1
f
|xi |Ai
ε
i=1
P
s dx1 dx2 . . . dxk−1 ,
Ij (xk ) =
k
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 (4.4) holds for
fi , it easily follows that
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 J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
12
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
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)
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)
Kn (ε−1 )\Kn (ε)
= Ck |Sn |ε
f1
n+A
Z
ε−1
r−n−A1 −1 dr
f
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)
Kn (ε−1 )\Kn (ε)
Dk |Sn | fk )
2(n+A
=
1−ε
.
fk
n+A
fi > 0, i = 1, 2, . . . , k, the above computation shows that J2 + J3 ≤ O(1)
Now, since n + A
when ε → 0. Hence, for the right-hand side of the inequality (4.5), by using (4.15), we obtain
1
(4.16)
J ≥ K ∗ |Sn | ln 2 − O(1), when ε → 0.
ε
∗
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
1
(4.17)
(K ∗ − L∗ ) |Sn | ln 2 ≤ O(1), when ε → 0.
ε
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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
13
5. S OME A PPLICATIONS
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 Hardy-Hilbert’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
i
and Aij = (s − nk) qi1qj , i 6= j, i, j = 1, 2, . . . , k. Then we have
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
!
= 0,
for j P
= 1, 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:
Corollary 5.1. Let k ≥ 2 be an integer and pi , p0i , qi , i = 1, 2, . . . , k, be real numbers satisfying
(1.1) – (1.5). Then the following inequalities hold and are equivalent:
Z
(5.1)
(Rn )k
1
p
k Z
Y
pi (k−1)n−pi s
i
fi (xi )
p
i
qi
|xi |
fi (xi )dxi
P
λs dx1 dx2 . . . dxk ≤ L
k
Rn
i=1
i=1 xi Qk
i=1
and



Z



Rn

Z
0
p
− qk [(k−1)n−s] 
k
|xk |

(Rn )k−1
pk 0
Qk−1

i=1 fi (xi )
P
λs dx1 dx2 . . . dxk−1 
k
i=1 xi ≤L
k−1
Y Z
i=1
Rn
|xi |
dxk
 10
pk






pi (k−1)n−pi s
qi
fipi (xi )dxi
p1
i
,
where 0 < nk − s < n min{pi , qj , i, j = 1, 2, . . . , k} and the constant L is defined by the
formula
1 k
1
k
nk − s p0i Y
nk − s qi
1 Y
L= λ
Γn n −
Γn n −
.
Γn (s) i=1
qi
pi
i=1
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.
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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
14
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
i −1)
Remark 5. Similar to the previous corollary, if we define the parameters Aij by Aii = n(λq
λqi2
and Aij = − λqni qj , i 6= j, i, j ∈ {1, 2, . . . , k}, then we have
!
k
k
X
X
n
n(λqj − 1)
n X1
Aij =
−
+
=−
− λ = 0,
2
λq
q
λq
λq
q
i
j
j
j
i
i=1
i=1
i6=j
P
for j = 1, 2, . . . , k. Since the parameters Aij are symmetric one obtains αi = nj=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
λ− q1 Y
q1
k
k
i
i
1 Y
n
n
0
L = λ
Γn
Γn s + 0 − (k − 1)n
,
0
λp
Γn (s) i=1
λpi
i
i=1
where (k − 1)n − s <
n
λp0i
< 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.
6. T RILINEAR V ERSION
OF A
S TANDARD B ETA I NTEGRAL
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)
,
α
β
|x|n−β |y|n−α
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
By using the definition (2.2) of the n−dimensional gamma function we easily obtain that
B(α, β, n) =
(6.2)
Γn (n − α)Γn (n − β)
.
Γn (2n − α − β)
We also define
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:
B ∗ (α, β, n) =
(6.3)
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|
Z
p1 Z
1q
(p−1)(α+β+n)−pnλ p
(q−1)(α+β+n)−qnλ q
|y|
g (y)dy
≤N
|x|
f (x)dx
Rn
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
Rn
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
15
and
(Z
(6.5)
0
|y|n(λq −1)−α−β
Z
Rn
(Rn )
q0 ) q10
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 non-conjugate
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
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
P1 =
P3 = |x|(p−1)(α+β) |y|(q−1)(α+β) f p (x)g q (y).
1
Therefore, respectively, by applying Hölder’s inequality with conjugate exponents 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 inequality (6.4) becomes
1
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
f (x)
−nλ
− α+β+n
+nλ
0
0
q
q
g(y) |y|
dx dy.
=
|y|
λ(n−α) |x + y|λ(n−β)
Rn |x − y|
Rn
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)
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
16
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
and (6.5). The first step is to consider the case when the function g ∈ Lq (Rn ) is symmetricdecreasing, 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
n
≤
g q (x) dx =
|y|−n kgkqq ,
|B(|y|)| Rn
|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. Let α and β be real parameters satisfying 0 < α < n, 0 < β < n, α + β =
1
1
n p + q > n. If f and g are nonnegative functions such that f ∈ Lp (Rn ), g ∈ Lq (Rn ), then
the following inequalities hold and are equivalent
Z
f (x)g(y)
dxdy ≤
|x − y|n−α |x + y|n−β
(6.7)
(Rn )2
n
|Sn |
1−λ
C(p, q; α, β; n)kf kp kgkq ,
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
where e1 = (1, 0, . . . , 0) ∈ Rn 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, using Hölder’s inequality with
1
parameters q 0 , p0 and 1−λ
, we have
Z
1
1
f (x)g(y)
q0
p0 1−λ
dxdy
≤
I
I
,
1 2 I3
|x − y|n−α |x + y|n−β
(6.10)
R2n
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
n
|x|− p |y| q0
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.
|x − y|n−α |x + y|n−β
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
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H ARDY-H ILBERT T YPE I NEQUALITIES
17
Further, using the substitution y = |x|u (so dy = |x|n du) and rotational invariance of the
Lebesgue integral in Rn we easily get:
n
Z
Z
n
|y|− q
p
0
p
I1 =
|x| f (x)
dydx
n−α |x + y|n−β
Rn
Rn |x − y|
Z
Z
−n
q
n
n
|u|
+α+β−n
−
p
0
f (x)
=
|x| p q
n−α n−β dudx
x
Rn
Rn x
|x| + u
|x| − u
n
Z
|u|− q du
=
kf kpp .
n−α
n−β
n
|e1 + u|
R |e1 − u|
Analogously,
n
Z
|u|− p du
I2 =
kgkqq ,
n−α
n−β
n
|e1 + u|
R |e1 − u|
and, by (6.6),
n
Z
n
|u|− q du
I3 ≤
kf kpp kgkqq .
|Sn | Rn |e1 − u|n−α |e1 + u|n−β
It remains to prove that
n
n
Z
Z
|x|− p dx
|x|− q dx
=
.
n−α |e + x|n−β
n−α |e + x|n−β
1
1
Rn |e1 − x|
Rn |e1 − x|
We transform the left integral in polar coordinates using x = tθ, t ≥ 0, θ ∈ Sn and 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 symmetric-decreasing functions
vanishing at infinity, the general rearrangement inequality implies that
Z
Z
f (x)g(y)
f ∗ (x)g ∗ (y)
(6.11)
dxdy
≤
dxdy.
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
1−λ
n
n
∗
∗
(6.12)
C(p, q; α, β; n)kf kp kg kq =
C(p, q; α, β; n)kf kp kgkq ,
|Sn |
|Sn |
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.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
18
M ARIO K RNI Ć , J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
On the other hand, by putting the function
qq0
Z
f (x)
g(y) =
dx
n−α |x + y|n−β
Rn |x − y|
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 =
1
+ 1q > 1. If f and g are nonnegative functions such that f ∈ Lp (R), g ∈ Lq (R), then the
p
following inequalities hold and are equivalent
Z
f (x)g(y)
dxdy ≤ 2λ−1 C(p, q; α, β)kf kp kgkq ,
(6.13)
1−α |x + y|1−β
|x
−
y|
2
R
and
(Z Z
q0 ) q10
f (x)dx
(6.14)
dy
≤ 2λ−1 C(p, q; α, β)kf kp ,
1−α |x + y|1−β
|x
−
y|
R
R
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
+B
, β F 1 − α, 0 ; 0 + β; −1 ,
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 ∈ Lq (R), then
the following inequalities hold and are equivalent
Z
λ 1
λ 1
f (x)g(y)dxdy
λ−1
≤2
B 1− , 0 +B 1− , 0
kf kp kgkq .
(6.16)
λ
2 2p
2 2q
R2
|x2 − y 2 | 2
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/
H ARDY-H ILBERT T YPE I NEQUALITIES
19
and
(6.17)
#q 0
 "
Z Z
f (x)dx
R
|x2 − y 2 | 2
λ

R
Proof. Set α = β = 1 −
λ
2
dy
 10
q
λ−1
≤2

λ 1
λ 1
B 1− , 0 +B 1− , 0
kf kp ,
2 2p
2 2q
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) and (6.5) reduces to
λ
|x2 − y 2 |− 2 .
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J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 115, 19 pp.
http://jipam.vu.edu.au/