Volume 9 (2008), Issue 1, Article 9, 13 pp.
HILBERT-PACHPATTE TYPE INEQUALITIES FROM BONSALL’S FORM OF
HILBERT’S INEQUALITY
JOSIP PEČARIĆ, IVAN PERIĆ, AND PREDRAG VUKOVIĆ
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA
pecaric@hazu.hr
FACULTY OF F OOD T ECHNOLOGY AND B IOTECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA
iperic@pbf.hr
FACULTY OF T EACHER E DUCATION
U NIVERSITY OF Z AGREB
A NTE S TAR ČEVI ĆA 55
40000 Č AKOVEC , C ROATIA
predrag.vukovic@vus-ck.hr
Received 19 June, 2007; accepted 09 October, 2007
Communicated by B. Yang
A BSTRACT. The main objective of this paper is to deduce Hilbert-Pachpatte type inequalities
using Bonsall’s form of Hilbert’s and Hardy-Hilbert’s inequalities, both in discrete and continuous case.
Key words and phrases: Inequalities, Hilbert’s inequality, sequences and functions, homogeneous kernels, conjugate and nonconjugate exponents, the Beta function.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
An interesting feature of one of the forms of Hilbert-Pachpatte type inequalities, is that it
controls the size (in the sense of Lp or lp spaces ) of the modified Hilbert transform of a function
or of a series with the size of its derivate or its backward differences, respectively. We start with
the following results of Zhongxue Lü from [9], for both continuous and discrete cases. For a
sequence a : N0 → R, the sequence ∇a : N → R is defined by ∇a(n) = a(n) − a(n − 1). For
a function u : (0, ∞) → R, u0 denotes the usual derivative of u.
205-07
2
J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
Theorem A. Let p > 1, p1 + 1q = 1, s > 2−min{p, q}, and f (x), g(y) be real-valued continuous
functions defined on [0, ∞), respectively, and let f (0) = g(0) = 0, and
Z ∞Z x
Z ∞Z y
1−s 0
p
0<
x |f (τ )| dτ dx < ∞, 0 <
y 1−s |g 0 (δ)|q dδdy < ∞,
0
0
0
0
then
(1.1)
∞
∞
|f (x)||g(y)|
dxdy
+ py q−1 )(x + y)s
0
0
p+s−2
Z ∞ Z x
p1 Z ∞ Z y
1q
B q+s−2
,
q
p
1−s 0
p
1−s 0
q
≤
·
x |f (τ )| dτ dx
y |g (δ)| dδdy .
pq
0
0
0
0
Z
Z
(qxp−1
Theorem B. Let p > 1, p1 + 1q = 1, s > 2 − min{p, q}, and {a(m)} and {b(n)} be two
sequences of real numbers where m, n ∈ N0 , and a(0) = b(0) = 0, and
0<
∞ X
m
X
m
1−s
p
|∇a(τ )| < ∞,
m=1 τ =1
0<
∞ X
n
X
n1−s |∇b(δ)|q < ∞,
n=1 δ=1
then
(1.2)
∞ X
∞
X
|am ||bn |
+ pnq−1 )(m + n)s
m=1 n=1
! p1
p+s−2
∞ X
m
B q+s−2
,
X
q
p
≤
·
m1−s |∇a(τ )|p
pq
m=1 τ =1
(qmp−1
∞ X
n
X
! 1q
n1−s |∇b(δ)|q
.
n=1 δ=1
Note that the condition s > 2 − min{p, q} from Theorem B is not sufficient. Namely, the
author of the proof of Theorem B used the following result
∞
m 2−s
X
1
q+s−2 p+s−2
q
(1.3)
,
m1−s ,
<B
s
(m
+
n)
n
q
p
n=1
for m ∈ {1, 2, . . .} and s > 2 − min{p, q}. For p = q = 2, s = 18 and m = 1, the left-hand
side of (1.3) is greater than the right-hand side of (1.3). Therefore, we refer to a paper of Krnić
and Pečarić, [4], where the next inequality is given:
(1.4)
∞
X
n=1
1
m α1
< m1−s+α1 −α2 B(1 − α2 , s + α2 − 1),
(m + n)s nα2
where 0 < s ≤ 14, 1 − s < α2 < 1 for s ≤ 2 and −1 ≤ α2 < 1 for s > 2. By using this result,
here we shall obtain a generalization of Theorem B but with the condition 2 − min{p, q} < s ≤
2 + min{p, q}. Also, the following result is given in [9]:
∞
m 2−s
X
1
1
q+s−2 p+s−2
q
(1.5)
< B
,
m1−s ,
s + ns
m
n
s
sq
sp
n=1
for m ∈ {1, 2, . . .} and s > 2 − min{p, q}. Similarly as before, for p = q = 2, s = 6 and
m = 1, the left-hand side of (1.5) is greater than the right-hand side of (1.5). The case of
nontrivial weights is essential in Theorem A and Theorem B, since for s = 1 only the trivial
functions and sequences satisfy the assumptions.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
3
In 1951, Bonsall established the following conditions for non-conjugate exponents (see [1]).
Let p and q be real parameters, such that
p > 1, q > 1,
(1.6)
1 1
+ ≥ 1,
p q
and let p0 and q 0 respectively be their conjugate exponents, that is,
Further, define
λ=
(1.7)
1
p
+
1
p0
= 1 and
1
q
+
1
q0
= 1.
1
1
+
p0 q 0
and note that 0 < λ ≤ 1 for all p and q as in (1.6). In particular, λ = 1 holds if and only if
q = p0 , that is, only when p and q are mutually conjugate. Otherwise, we have 0 < λ < 1,
and in such cases p and q will be referred to as non-conjugate exponents. Also, in this paper
we shall obtain some generalizations of (1.1). It will be done in simplier way than in [9]. Our
results will be based on the following results of Pečarić et al., [2], for the non-conjugate and
conjugate exponents.
Theorem C. Let p, p0 , q, q 0 and λ be as in (1.6) and (1.7). If K, ϕ, ψ, f and g are non-negative
measurable functions, then the following inequalities hold and are equivalent
Z
p1 Z
1q
Z
λ
p
q
(ϕF f ) (x)dx
(1.8)
K (x, y)f (x)g(y)dxdy ≤
(ψGg) (y)dy
Ω2
Ω
Ω
and
Z (1.9)
Ω
1
ψG(y)
Z
! 10
q 0
λ
K (x, y)f (x)dx
q
Z
p
≤
dy
p1
,
(ϕF f ) (x)dx
Ω
Ω
where the functions F, G are defined by
Z
10
q
K(x, y)
dy
F (x) =
0
q
Ω ψ (y)
Z
and
K(x, y)
dx
ϕp0 (x)
G(y) =
Ω
10
p
.
The next inequalities from [5] can be seen as a special case of (1.8) and (1.9) respectively for
the conjugate exponents:
Z
p1 Z
1q
Z
(1.10)
K(x, y)f (x)g(y)dxdy ≤
ϕp (x)F (x)f p (x)dx
ψ q (y)G(y)g q (y)dy
Ω2
Ω
Ω
and
Z
G
(1.11)
1−p
p
Z
−p
(y)ψ (y)
Ω
K(x, y)f (x)dx
Z
dy ≤
Ω
ϕp (x)F (x)f p (x)dx,
Ω
where
Z
(1.12)
F (x) =
Ω
K(x, y)
dy
ψ p (y)
Z
and G(y) =
Ω
K(x, y)
dx.
ϕq (x)
In particular, inequalities (1.10) and (1.11) are equivalent.
On the other hand, here we also refer to a paper of Brnetić et al., [8], where a general
Hilbert-type inequality wasP
obtained for n ≥ 2 conjugate exponents, that is, real parameters
p1 , . . . , pn > 1, such that ni=1 p1i = 1. Namely, we let K : Ωn → R and φij : Ω → R,
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
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J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
Q
i, j = 1, . . . , n, be non-negative measurable functions. If ni,j=1 φij (xj ) = 1, then the inequality
p1
Z
n
n Z
Y
Y
i
pi
(1.13)
K(x1 , . . . , xn )
fi (xi ) dx1 . . . dxn ≤
Fi (xi )(φii fi ) (xi ) dxi
,
Ωn
i=1
Ω
i=1
holds for all non-negative measurable functions f1 , . . . , fn : Ω → R, where
Z
n
Y
(1.14)
Fi (xi ) =
K(x1 , . . . , xn )
φpiji (xj )dx1 . . . dxi−1 dxi+1 . . . dxn ,
Ωn−1
j=1,j6=i
for i = 1, . . . , n.
2. I NTEGRAL C ASE
In this section we shall state our main results. We suppose that all integrals converge and
shall omit these types of conditions. Thus, we have the following.
Theorem 2.1. Let p1 + 1q = 1 with p > 1. If K(x, y), ϕ(x), ψ(y) are non-negative functions and
f (x), g(y) are absolutely continuous functions such that f (0) = g(0) = 0, then the following
inequalities hold
Z ∞Z ∞
K(x, y)|f (x)| |g(y)|
(2.1)
dxdy
qxp−1 + py q−1
0
Z ∞0 Z ∞
1 1
≤
K(x, y)|f (x)| |g(y)|d x p d y q
0
1
≤
pq
0
Z
∞
Z
0
x
p1 Z
ϕ (x)F (x)|f (τ )| dτ dx
0
p
∞
p
0
y
Z
0
q
q
1q
ψ (y)G(y)|g (δ)| dδdy
0
,
0
and
Z
∞
G
(2.2)
1−p
−p
Z
(y)ψ (y)
0
0
∞
1
p
p
dy
K(x, y)|f (x)|d x
Z Z
1 ∞ x p
≤ p
ϕ (x)F (x)|f 0 (τ )|p dτ dx,
p 0
0
where F (x) and G(y) are defined as in (1.12).
Proof. By using Hölder’s inequality, (see also [9]), we have
Z x
p1 Z
1
1
0
p
|f (τ )| dτ
(2.3)
|f (x)| |g(y)| ≤ x q y p
0
1q
y
0
q
|g (δ)| dδ
.
0
From (2.3) and using the elementary inequality
xy ≤
(2.4)
xp y q
1 1
+ , x ≥ 0, y ≥ 0,
+ = 1, p > 1,
p
q
p q
we observe that
(2.5)
pq|f (x)| |g(y)|
|f (x)| |g(y)|
≤
≤
1
1
p−1
q−1
qx
+ py
xq y p
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
Z
x
0
p
p1 Z
|f (τ )| dτ
0
y
0
q
1q
|g (δ)| dδ
0
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
5
and therefore
∞
Z
∞
Z
pq
(2.6)
0
0
K(x, y)|f (x)| |g(y)|
dxdy
qxp−1 + py q−1
Z ∞Z ∞
K(x, y)
≤
1
1 |f (x)||g(y)|dxdy
0
0
xq y p
Z x
p1 Z
Z ∞Z ∞
0
p
≤
K(x, y)
|f (τ )| dτ
0
0
0
1q
y
0
q
|g (δ)| dδ
dxdy.
0
Applying the substitutions
Z
p1
x
0
p
|f (τ )| dτ
f1 (x) =
,
1q
y
Z
0
q
|g (δ)| dδ
g1 (y) =
0
0
and (1.10), we obtain
Z ∞Z ∞
(2.7)
K(x, y)f1 (x)g1 (y)dxdy
0
0
Z
∞
≤
ϕ
p
(x)F (x)f1p (x)dx
p1 Z
ψ
0
Z
q
(y)G(y)g1q (y)dy
1q
0
∞
Z
=
0
∞
x
p1 Z
ϕ (x)F (x)|f (τ )| dτ dx
0
p
p
0
0
∞
Z
y
q
0
q
ψ (y)G(y)|g (δ)| dδdy
1q
.
0
By using (2.6) and (2.7) we obtain (2.1). The second inequality (2.2) can be proved by applying
(1.11) and the inequality
Z x
p1
1
|f (x)| ≤ x q
|f 0 (t)|p dt .
0
Now we can apply our main result to non-negative homogeneous functions. Recall that for a
homogeneous function of degree −s, s > 0, the equality K(tx, ty) = t−s K(x, y) is satisfied.
Further, we define
Z ∞
k(α) :=
K(1, u)u−α du
0
and suppose that k(α) < ∞ for 1 − s < α < 1. To prove first application of our main results
we need the following lemma.
Lemma 2.2. If s > 0, 1 − s < α < 1 and K(x, y) is a non-negative homogeneous function of
degree −s, then
α
Z ∞
x
(2.8)
K(x, y)
dy = x1−s k(α).
y
0
Proof. By using the substitution u =
equation (2.8) follows easily.
y
x
and the fact that K(x, y) is homogeneous function, the
Corollary 2.3. Let s > 0, p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely continuous functions such that f (0) = g(0) = 0, and K(x, y) is a non-negative symmetrical and homogeneous
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
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6
J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
function of degree −s, then the following inequalities hold
Z ∞Z ∞
K(x, y)|f (x)| |g(y)|
(2.9)
dxdy
qxp−1 + py q−1
0
0
Z ∞Z ∞
1 1
≤
K(x, y)|f (x)| |g(y)|d x p d y q
0
L
≤
pq
0
∞
Z
Z
x
x
0
1−s+p(A1 −A2 )
p1
|f (τ )| dτ dx
0
p
0
Z
∞
Z
×
y
0
1q
y
1−s+q(A2 −A1 )
0
q
|g (δ)| dδdy
0
and
Z
∞
y
(2.10)
(p−1)(s−1)+p(A1 −A2 )
∞
Z
0
0
p
K(x, y)|f (x)|d(x ) dy
p Z ∞ Z x
L
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
1
p
1
1
where A1 ∈ ( 1−s
, 1q ), A2 ∈ ( 1−s
, p1 ) and L = k(pA2 ) p k(qA1 ) q .
q
p
Proof. Let F (x), G(y) be the functions defined as in (1.12). Setting ϕ(x) = xA1 and ψ(y) =
y A2 , by Lemma 2.2 we obtain
Z ∞Z x
(2.11)
ϕp (x)F (x)|f 0p dτ dx
0
0
pA2 !
Z ∞Z x
Z ∞
x
0
p
=
|f (τ )|
K(x, y)
dy xp(A1 −A2 ) dτ dx
y
0
0
0
Z ∞Z x
= k(pA2 )
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
0
and similarly
Z ∞Z
(2.12)
0
0
y
q
0
q
Z
∞
Z
ψ (y)G(y)|g (δ)| dδdy = k(qA1 )
0
0
y
y 1−s+q(A2 −A1 ) |g 0 (δ)|q dδdy.
0
From (2.1), (2.11) and (2.12), we get (2.9). Similarly, the inequality (2.10) follows from (2.2).
We proceed with some special homogeneous functions. First, by putting K(x, y) =
Corollary 2.3, we get the following.
1
(x+y)s
in
Corollary 2.4. Let s > 0, p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely continuous
functions such that f (0) = g(0) = 0, then the following inequalities hold
Z ∞Z ∞
|f (x)| |g(y)|
dxdy
p−1 + py q−1 )(x + y)s
0
0 (qx
Z ∞Z ∞
|f (x)| |g(y)| p1 1q ≤
d x d y
(x + y)s
0
0
Z ∞ Z x
p1 Z ∞ Z y
1q
L1
1−s+p(A1 −A2 ) 0
p
1−s+q(A2 −A1 ) 0
q
≤
x
|f (τ )| dτ dx
y
|g (δ)| dδdy
pq
0
0
0
0
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
7
and
Z
∞
y
(p−1)(s−1)+p(A1 −A2 )
Z
0
0
∞
p
|f (x)| p1 d x
dy
(x + y)s
p Z ∞ Z x
L1
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
where A1 ∈ ( 1−s
, 1q ), A2 ∈ ( 1−s
, p1 ) and
q
p
1
1
L1 = [B(1 − pA2 , pA2 + s − 1)] p [B(1 − qA1 , qA1 + s − 1)] q .
Remark 2.5. By putting A1 = A2 = 2−s
in Corollary 2.4, with the condition s > 2−min{p, q},
pq
we obtain Theorem A from the introduction.
Since the function K(x, y) =
Corollary 2.3 we obtain:
y
ln x
y−x
is symmetrical and homogeneous of degree −1, by using
Corollary 2.6. Let p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely continuous functions
such that f (0) = g(0) = 0, then the following inequalities hold
Z ∞Z ∞
ln xy |f (x)| |g(y)|
dxdy
p−1 + py q−1 )(y − x)
0
0 (qx
Z ∞Z ∞ y
ln x |f (x)| |g(y)| 1 1 ≤
d xp d y q
y−x
0
0
Z ∞ Z x
p1 Z ∞ Z y
1q
L2
p(A1 −A2 ) 0
p
q(A2 −A1 ) 0
q
≤
x
|f (τ )| dτ dx
y
|g (δ)| dδdy
pq
0
0
0
0
and
Z ∞
p Z ∞ Z x
Z ∞
|f (x)| ln xy 1 p
L2
p(A1 −A2 )
p
y
d x
xp(A1 −A2 ) |f 0 (τ )|p dτ dx,
dy ≤
y−x
p
0
0
0
0
where A1 ∈ (0, 1q ), A2 ∈ (0, p1 ) and
2
2
L2 = π 2 (sin pA2 π)− p (sin qA1 π)− q .
Similarly, for the symmetrical homogeneous function of degree −s, K(x, y) =
have:
1
,
max{x,y}s
we
Corollary 2.7. Let s > 0, p1 + 1q = 1 with p > 1. If f (x), g(y) are absolutely continuous
functions such that f (0) = g(0) = 0, then the following inequalities hold
Z ∞Z ∞
|f (x)| |g(y)|
dxdy
p−1
+ py q−1 ) max{x, y}s
0
0 (qx
Z ∞Z ∞
|f (x)| |g(y)| p1 1q ≤
d x d y
max{x, y}s
0
0
Z ∞ Z x
p1 Z ∞ Z y
1q
L3
1−s+p(A1 −A2 ) 0
p
1−s+q(A2 −A1 ) 0
q
≤
x
|f (τ )| dτ dx
y
|g (δ)| dδdy
pq
0
0
0
0
and
Z ∞
Z ∞
1 p
|f (x)|
(p−1)(s−1)+p(A1 −A2 )
y
d xp
dy
max{x, y}s
0
0
p Z ∞ Z x
L3
≤
x1−s+p(A1 −A2 ) |f 0 (τ )|p dτ dx,
p
0
0
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
http://jipam.vu.edu.au/
8
J OSIP P E ČARI Ć , I VAN P ERI Ć ,
where A1 ∈
1−s 1
,q
q
, A2 ∈
1−s 1
,p
p
AND
P REDRAG V UKOVI Ć
1
1
and L3 = k(pA2 ) p k(qA1 ) q , where k(α) =
s
.
(1−α)(s+α−1)
At the end of this section we give a generalization of the inequality (2.1) from Theorem 2.1.
In the proof we used a general Hilbert-type inequality (1.13) of Brnetić et al., [8].
P
Theorem 2.8. Let n ∈ N, n ≥ 2, ni=1 p1i = 1 with pi > 1, i = 1, . . . , n. Let qi , αi , i =
Q
1, . . . , n, are defined with q1i = 1 − p1i and αi = nj=1,j6=i pj . If K(x1 , . . . , xn ), φij (xj ), i, j =
Q
1, . . . , n, are non-negative functions such that ni,j=1 φij (xj ) = 1, and fi (xi ), i = 1, . . . , n, are
absolutely continuous functions such that fi (0) = 0, i = 1, . . . , n, then the following inequality
holds
Q
Z ∞
Z ∞
K(x1 , . . . , xn ) ni=1 |fi (xi )|
dx1 . . . dxn
...
Pn
pi −1
0
0
i=1 αi xi
1
1
Z ∞
Z ∞
n
Y
p1
≤
...
K(x1 , . . . , xn )
|fi (xi )|d x1 . . . d xnpn
0
0
1
≤
p1 . . . pn
i=1
n Z
Y
∞
0
i=1
xi
Z
0
φpiii (xi )Fi (xi )|fi (τi )|pi dτi dxi
p1
i
,
0
where Fi (xi ) are defined as in (1.14) for i = 1, . . . , n.
3. D ISCRETE C ASE
We also give results for the discrete case. For that, we apply the following result from [5].
Theorem 3.1. If {a(m)} and {b(n)} are non-negative real sequences, K(x, y) is non-negative
homogeneous function of degree −s strictly decreasing in both parameters x and y, p1 + 1q = 1,
p > 1, A, B, α, β > 0, then the following inequalities hold and are equivalent
(3.1)
∞ X
∞
X
K(Amα , Bnβ )am bn
m=1 n=1
<N
∞
X
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) apm
m=1
·
∞
X
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) bqn
,
n=1
and
(3.2)
∞
X
∞
X
nβ(s−1)(p−1)+pβ(A1 −A2 )+β−1
n=1
!p
K(Amα , Bnβ )am
m=1
<N
p
∞
X
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) apm ,
m=1
, α−1
}, 1q ), A2 ∈ max
where A1 ∈ (max{ 1−s
q
αq
(3.3)
1
1
N = α− q β − p A
2−s
+A1 −A2 −1
p
B
n
1−s β−1
, βp
p
2−s
+A2 −A1 −1
q
o
, p1 and
1
1
k(pA2 ) p k(2 − s − qA1 ) q .
Applying Theorem 3.1, we obtain the following.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
9
Corollary 3.2. Let s > 0, p1 + 1q = 1 with p > 1. Let {a(m)} and {b(n)} be two sequences of real
numbers where m, n ∈ N0 , and a(0) = b(0) = 0. If K(x, y) is a non-negative homogeneous
function of degree −s strictly decreasing in both parameters x and y, A, B, α, β > 0, then the
following inequalities hold
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
qmp−1 + pnq−1
m=1 n=1
∞ ∞
1 X X K(Amα , Bnβ )|am | |bn |
≤
1
1
pq m=1 n=1
mq np
(3.4)
N
<
pq
∞ X
m
X
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) |∇a(τ )|p
m=1 τ =1
∞ X
n
X
·
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) |∇b(δ)|q
,
n=1 δ=1
and
(3.5)
∞
X
∞
X
nβ(s−1)(p−1)+pβ(A1 −A2 )+β−1
n=1
K(Amα , Bnβ )
<N
!p
1
mq
m=1
p
|am |
∞ X
m
X
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) |∇a(τ )|p ,
m=1 τ =1
where A1 ∈ max
as in (3.3).
n
1−s α−1
, αq
q
o
,
1
q
, A2 ∈ max
n
1−s β−1
, βp
p
o
,
1
p
and the constant N is defined
Proof. By using Hölder’s inequality, (see also [9]), we have
1
q
|am | |bn | ≤ m n
(3.6)
1
p
m
X
! p1
|∇a(τ )|p
τ =1
n
X
! 1q
|∇b(δ)|q
.
δ=1
From (2.4) and (3.6), we get
pq|am | |bn |
|am | |bn |
≤
≤
1
1
p−1
q−1
qm
+ pn
mq np
(3.7)
m
X
! p1
|∇a(τ )|p
τ =1
n
X
! 1q
|∇b(δ)|q
,
δ=1
and therefore
(3.8)
pq
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
qmp−1 + pnq−1
m=1 n=1
≤
∞ X
∞
X
K(Amα , Bnβ )|am | |bn |
1
≤
∞ X
∞
X
1
mq np
m=1 n=1
K(Amα , Bnβ )
m=1 n=1
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
m
X
τ =1
! p1
|∇a(τ )|p
n
X
! 1q
|∇b(δ)|q
.
δ=1
http://jipam.vu.edu.au/
10
J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
1
1
P
Pn
p p e
q q
Applying the substitutions e
am = ( m
τ =1 |∇a(τ )| ) , bn = (
δ=1 |∇b(δ)| ) and (3.1), we
have
∞ X
∞
X
(3.9)
K(Amα , Bnβ )e
amebn
m=1 n=1
∞
X
<N
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α)e
apm
m=1
·
∞
X
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β)ebqn
,
n=1
=
∞ X
m
X
! p1
mα(1−s)+αp(A1 −A2 )+(p−1)(1−α) |∇a(τ )|p
m=1 τ =1
·
∞ X
n
X
! 1q
nβ(1−s)+βq(A2 −A1 )+(q−1)(1−β) |∇b(δ)|q
,
n=1 δ=1
where A1 ∈
A2 ∈ (max{ 1−s
, β−1
}, p1 ) and the constant N is defined as
p
βp
in (3.3). Now, by applying (3.8) and (3.9) we obtain (3.4). The second inequality (3.5) can be
proved by using (3.2) and the inequality
! p1
m
X
1
|am | ≤ m q
|∇a(τ )|p
.
(max{ 1−s
, α−1
}, 1q ),
q
αq
τ =1
Remark 3.3. If the function K(x, y) from the previous corollary is symmetrical, then k(2 −
1
2−s
s − qA1 ) = k(qA1 ). So, if K(x, y) = (x+y)
,A = B =
s , then we can put A1 = A2 =
pq
α = β = 1 in Corollary 3.2 and obtain Theorem B from the introduction but with the condition
2 − min{p, q} < s < 2.
By using (1.4), see [4], we will obtain a larger interval for the parameter s. More precisely,
we have:
Corollary 3.4. Let p1 + 1q = 1 with p > 1 and 2 − min{p, q} < s ≤ 2 + min{p, q}. Let {a(m)}
and {b(n)} be two sequences of real numbers where m, n ∈ N0 , and a(0) = b(0) = 0. Then
the following inequalities hold
∞ X
∞
X
|am | |bn |
(3.10)
p−1
(qm
+ pnq−1 )(m + n)s
m=1 n=1
∞ ∞
1 XX
|am | |bn |
≤
1
pq m=1 n=1 m q n p1 (m + n)s
N1
<
pq
∞ X
m
X
! p1
m1−s |∇a(τ )|p
·
m=1 τ =1
∞ X
n
X
! 1q
n1−s |∇b(δ)|q
,
n=1 δ=1
and
(3.11)
∞
X
n=1
n
(s−1)(p−1)
∞
X
m=1
!p
|am |
1
q
m (m +
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
<
n)s
N1p
∞ X
m
X
m1−s |∇a(τ )|p ,
m=1 τ =1
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
11
where N1 = B( s+q−2
, s+p−2
).
q
p
Proof. As in the proof of Corollary 3.2, by using Hölder’s inequality we obtain
∞ X
∞
X
|am | |bn |
(qmp−1
m=1 n=1
∞ X
∞
X
≤
1
pq
m=1 n=1
+ pnq−1 )(m + n)s
|am | |bn |
1
q
1
m n p (m + n)s
∞ ∞
1 XX
1
≤
pq m=1 n=1 (m + n)s
≤
1
pq
∞ X
m
∞
X
X
m=1 τ =1
·
n=1
∞ X
n
X
n=1 δ=1
m
X
! p1
p
|∇a(τ )|
τ =1
1
(m + n)s
n
X
! 1q
q
|∇b(δ)|
δ=1
m 2−s
q
n
∞
X
n 2−s
m 2−s
pq
pq
m
n
! p1
!
|∇a(τ )|p
n 2−s
1
p
s
(m + n) m
m=1
! 1q
!
|∇b(δ)|q
.
Now, the inequality (3.10) follows from (1.4). Let us show that the inequality (3.11) is valid.
For this purpose we use the following inequality from [4]
!p
∞
∞
∞
X
X
X
a
m
(s−1)(p−1)
< L1
m1−s apm ,
(3.12)
n
s
(m
+
n)
m=1
n=1
m=1
where 2 − min{p, q} < s ≤ 2 + min{p, q} and L1 = B( s+p−2
, s+q−2
). Setting
p
q
! p1
m
X
am =
|∇a(τ )|p
τ =1
in (3.12) and using
|am | ≤ m
1
q
m
X
! p1
|∇a(τ )|p
,
τ =1
the inequality (3.11) follows easily.
4. N ON - CONJUGATE E XPONENTS
Let p, p0 , q, q 0 and λ be as in (1.6) and (1.7). To obtain an analogous result for the case
of non-conjugate exponents, we introduce real parameters r0 , r such that p ≤ r0 ≤ q 0 and
1
+ 1r = 1. For example, we can define r10 = q10 + 1−λ
or r0 = (2 − λ)p.
r0
2
It is easy to see that
0
1
1
r
r
1
0 q0
0
0
0
p
q
p
(4.1)
x y ≤ 0 rx + r y
,
x ≥ 0, y ≥ 0,
rr
and
Z x
p1 Z y
1q
1
1
0
p
0
q
0
0
|f (τ )| dτ
|g (δ)| dδ ,
(4.2)
|f (x)| |g(y)| ≤ x p y q
0
0
hold, where f (x), g(y) are absolutely continuous functions on (0, ∞).
Applying Theorem C, (4.1) and (4.2) in the same way as in the proof of Theorem 2.1, we
obtain the following result for non-conjugate exponents.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
http://jipam.vu.edu.au/
12
J OSIP P E ČARI Ć , I VAN P ERI Ć ,
AND
P REDRAG V UKOVI Ć
Theorem 4.1. Let p, p0 , q, q 0 and λ be as in (1.6) and (1.7). Let r0 , r be real parameters
such that p ≤ r0 ≤ q 0 and r10 + 1r = 1. If K(x, y), ϕ(x), ψ(y) are non-negative functions and
f (x), g(y) are absolutely continuous functions such that f (0) = g(0) = 0, then the following
inequalities hold
Z ∞Z ∞ λ
K (x, y)|f (x)| |g(y)|
(4.3)
dxdy
r
r0
0
0
rx p0 + r0 y q0
Z Z
1 1
pq ∞ ∞ λ
≤ 0
K (x, y)|f (x)| |g(y)|d x p d y q
rr 0
0
Z ∞ Z x
p1 Z ∞ Z y
1q
1
p
0
p
q
0
q
≤ 0
(ϕF ) (x)|f (τ )| dτ dx
(ψG) (y)|g (δ)| dδdy ,
rr
0
0
0
0
and
Z
∞
(4.4)
0
1
ψG(y)
Z
∞
q0 ! q10
K λ (x, y)|f (x)|d(x )
dy
1
p
0
1
≤
p
∞
Z
x
Z
0
p1
(ϕF ) (x)|f (τ )| dτ dx ,
0
p
p
0
where F (x) and G(y) are defined as in Theorem C.
Obviously, Theorem 4.1 is the generalization of Theorem 2.1. Namely, if λ = 1, r0 = p and
r = q, then the inequalities (4.3) and (4.4) become respectively the inequalities (2.1) and (2.2).
If K(x, y) is a non-negative symmetrical and homogeneous function of degree −s, s > 0, then
we obtain:
Corollary 4.2. Let s > 0, p, p0 , q, q 0 and λ be as in (1.6) and (1.7). If f (x), g(y) are absolutely
continuous functions such that f (0) = g(0) = 0, and K(x, y) is a non-negative symmetrical
and homogeneous function of degree −s, then the following inequalities hold
Z ∞Z ∞
K λ (x, y)|f (x)| |g(y)|
(4.5)
dxdy
(p−1)(2−λ) + py (q−1)(2−λ)
0
0 qx
Z ∞Z ∞
1 1
1
≤
K λ (x, y)|f (x)| |g(y)|d x p d y q
2−λ 0
0
Z ∞ Z x
p1
p
M
0
p
0 (1−s)+p(A1 −A2 )
q
≤
x
|f (τ )| dτ dx
pq(2 − λ)
0
0
Z ∞ Z y
1q
q
(1−s)+q(A
−A
)
0
q
2
1
0
|g (δ)| dδdy
·
yp
0
0
and
Z
∞
y
(4.6)
q0
(s−1)+q 0 (A1 −A2 )
p0
Z
∞
K λ (x, y)|f (x)|d x
q 0
! 10
q
dy
0
0
M
≤
p
where A1 ∈
1
p
1−s 1
, p0
p0
, A2 ∈
1−s 1
, q0
q0
Z
∞
Z
x
x
0
p
(1−s)+p(A1 −A2 )
q0
p1
|f (τ )| dτ dx ,
0
p
0
1
1
and M = k(p0 A1 ) p0 k(q 0 A2 ) q0 .
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 9, 13 pp.
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H ILBERT-PACHPATTE T YPE I NEQUALITIES F ROM B ONSALL’ S F ORM ...
13
Proof. The proof follows directly from Theorem 4.1 setting r0 = (2 − λ)p, r = (2 − λ)q,
ϕ(x) = xA1 and ψ(y) = y A2 in the inequalities (4.3) and (4.4). Namely, if F (x) and G(y) are
the functions defined by
10
Z ∞
10
Z ∞
q
p
K(x, y)
K(x, y)
F (x) =
dy
and
G(y)
=
dx
,
0
0
ψ q (y)
ϕp (x)
0
0
then applying Lemma 2.2 we have
Z ∞
p0
q
p
pA1
−q 0 A2
(4.7)
(ϕF ) (x) = x
K(x, y)y
dy
0
= xpA1 −pA2
Z
0
=x
∞
q0 A2 ! qp0
x
K(x, y)
dy
y
p
(1−s)+p(A1 −A2 )
q0
p
k q0 (q 0 A2 ),
and similarly
q
(4.8)
(ψG)q (y) = y p0
(1−s)+q(A2 −A1 )
q
k p0 (p0 A1 ).
Now, by using (4.3), (4.7) and (4.8) we obtain (4.5).
The second inequality (4.6) follows directly from (4.4).
Remark 4.3. Setting K(x, y) =
1
(x+y)s
in Corollary 4.2 we obtain that the constant M is equal
1
p0
1
to M = B(1 − p0 A1 , p0 A1 + s − 1) B(1 − q 0 A2 , q 0 A2 + s − 1) q0 .
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http://jipam.vu.edu.au/