Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 47812, 9 pages
doi:10.1155/2007/47812
Research Article
On a Hilbert-Type Operator with a Symmetric Homogeneous
Kernel of −1-Order and Applications
Bicheng Yang
Received 21 March 2007; Accepted 12 July 2007
Recommended by Shusen Ding
Some character of the symmetric homogenous kernel of −1-order in Hilbert-type operator T : lr → lr (r > 1) is obtained. Two equivalent inequalities with the symmetric homogenous kernel of −λ-order are given. As applications, some new Hilbert-type inequalities with the best constant factors and the equivalent forms as the particular cases are
established.
Copyright © 2007 Bicheng Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
If the real function k(x, y) is measurable in (0, ∞) × (0, ∞), satisfying k(y,x) = k(x, y),
for x, y ∈ (0, ∞), then one calls k(x, y) the symmetric function. Suppose that p > 1, 1/ p +
1/q = 1, lr (r = p, q) are two real normal spaces, and k(x, y) is a nonnegative symmetric
p
function in (0, ∞) × (0, ∞). Define the operator T as follows: for a = {am }∞
m=1 ∈ l ,
(Ta)(n) :=
∞
k(m,n)am ,
n ∈ N;
(1.1)
m ∈ N.
(1.2)
m=1
q
or for b = {bn }∞
n =1 ∈ l ,
(Tb)(m) :=
∞
k(m,n)bn ,
n =1
The function k(x, y) is said to be the symmetric kernel of T.
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Journal of Inequalities and Applications
If k(x, y) is a symmetric function, for ε(≥ 0) small enough and x > 0, set kr (ε,x) as
kr (ε,x) :=
∞
(1+ε)/r
k(x,t)
0
x
t
dt
(r = p, q).
(1.3)
In 2007, Yang [1] gave three theorems as follows.
Theorem 1.1. (i) If for fixed x > 0, and r = p, q, the functions k(x,t)(x/t)1/r are decreasing
in t ∈ (0, ∞), and
kr (0,x) :=
∞
1/r
x
t
k(x,t)
0
dt = k p
(r = p, q),
(1.4)
where k p is a positive constant independent of x, then T ∈ B(lr →lr ), T is called the Hilberttype operator and T r ≤ k p (r = p, q);
(ii) if for fixed x > 0, ε ≥ 0 and r = p, q, the functions k(x,t)(x/t)(1+ε)/r are decreasing
in t ∈ (0, ∞); kr (ε,x) = k p (ε) (r = p, q; ε ≥ 0) is independent of x, satisfying k p (ε) = k p +
o(1) (ε→0+ ), and
∞
1
1+ε
m
m=1
1
0
m
t
k(m,t)
(1+ε)/r
dt = O(1)
ε→0+ ; r = p, q ,
(1.5)
then T r = k p (r = p, q).
Theorem 1.2. Suppose that p > 1, 1/ p + 1/q = 1, and kr (0,x) (r = p, q; x > 0) in (1.3)
∞
p
q
satisfy condition (i) in Theorem 1.1. If am ,bn ≥ 0 and a = {am }∞
m=1 ∈ l , b = {bn }n=1 ∈ l ,
then one has the following two equivalent inequalities:
∞ ∞
k(m,n)am bn ≤ k p a p bq ;
n=1 m=1
∞
p 1/ p
∞
k(m,n)am
n=1 m=1
where the positive constant factor k p (=
∞
0
(1.6)
≤ k p a p ,
k(x,t)(x/t)1/q dt) is independent of x > 0.
Theorem 1.3. Suppose that p > 1, 1/ p + 1/q = 1, and kr (ε,x) (r = p, q; x > 0, ε ≥ 0)
p
in (1.3) satisfy condition (ii) in Theorem 1.1. If am ,bn ≥ 0 and a = {am }∞
m=1 ∈ l , b =
∞
q
{bn }n=1 ∈ l , and a p , bq > 0, T is defined by (1.1), and the formal inner product of
Ta and b is defined by
(Ta,b) :=
∞ ∞
k(m,n)am bn = (a,Tb),
n=1 m=1
(1.7)
Bicheng Yang 3
then one has the following two equivalent inequalities:
(Ta,b) < T p a p bq ;
(1.8)
Ta p < T p a p ,
where the constant factor T p =
∞
0
k(x,t)(x/t)1/q dt(> 0) is the best possible.
Recently, Yang [2] also considered some frondose character of the symmetric kernel for
p = q = 2; Yang et al. [3–6] considered the character of the norm in Hilbert-type integral
operator and some applications.
Definition 1.4. If k(x, y) is a nonnegative function in (0, ∞) × (0, ∞), and there exists
λ > 0, satisfying k(xu,xv) = x−λ k(u,v), for any x,u,v ∈ (0, ∞), then k(x, y) is said to be
the homogeneous function of −λ-order.
In this paper, for keeping on research of the thesis in [1, 2], some frondose character of
the symmetric homogeneous kernel of −1-order satisfying condition (ii) of Theorem 1.1
is considered. One also considers two equivalent inequalities with the symmetric homogeneous kernel of −λ-order. As applications, some new Hilbert-type inequalities with the
best constant factors and the equivalent forms as the particular cases of the kernel are
established.
For this, one needs the formula of the Beta function B(u,v) as (see [7])
B(u,v) =
∞
0
1
t −u+1 du = B(v,u)
(1 + t)u+v
(u,v > 0).
(1.9)
2. A lemma and a theorem
Suppose that the symmetric kernel k(x, y) is homogeneous function of −1-order. Setting
∞
u = t/x in (1.3), one finds kr (ε,x) is independent of x > 0 and kr (ε) := 0 k(1,u)u−(1+ε)/r du
= kr (ε,x) (r = p, q). If k p := kr (0,x) is a positive constant, then setting v = 1/u, one
∞
∞
obtains kq = 0 k(1,u)u−1/q du = 0 k(v,1)v−1/ p dv = k p > 0, and kr (0,x) = k p (r = p, q).
Hence based on the above conditions, if for fixed x > 0 and r = p, q, the functions
k(x,t)(x/t)1/r are decreasing in t ∈ (0, ∞), then the kernel k(x, y) satisfies condition (i)
of Theorem 1.1 and suits using Theorem 1.2.
Lemma 2.1. Let p > 1, 1/ p + 1/q = 1, let the symmetric kernel k(x, y) be homogeneous
function of −1-order, and for fixed x > 0, r = p, q, the functions k(x,t)(x/t)1/r be decreasing in t ∈ (0, ∞). If k(1,u) is positive and continuous in (0,1], and there exist constant η <
that lim u→0+ uη k(1,u) = C, then for ε ∈ [0,min { p, q}(1 −
min {1/ p,1/q} and
∞C ≥ 0, such
−(1+ε)/r
η) − 1), kr (ε) := 0 k(1,u)u
du are positive constants satisfying k p (ε) = k p + o(1)
(ε→0+ ; r = p, q), and expression (1.5) is valid. Hence k(x, y) satisfies condition (ii) of
Theorem 1.1 and suits using Theorem 1.3.
Proof. For fixed x > 0, ε ≥ 0, and r = p, q, the functions k(x,t)(x/t)(1+ε)/r = k(x,t)(x/
t)1/r (x/t)ε/r are still decreasing in t ∈ (0, ∞). Since lim u→0+ uη k(1,u) = C and uη k(1,u)
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Journal of Inequalities and Applications
is positive and continuous in (0,1], there exists a constant L > 0, such that uη k(1,u) ≤
L (u ∈ [0,1]). Setting u = 1/v in the following second integral, since k(1,1/v) = vk(v,1),
one finds
1
0 < k p (ε) =
k(1,u)u−(1+ε)/ p du +
0
1
0
1
=
=
0
≤L
k(1,u)u−(1+ε)/ p du +
1
k(v,1)v(1+ε)/ p−1 dv
0
u−(1+ε)/ p−η + u(1+ε)/ p−η−1 du = L
∞
0
1
0
1 ε
− −η
q p
−1
+
1+ε
−η
p
−1 .
(2.1)
k(1,u)u−(1+ε)/ p du is a positive constant. Since by (2.1), one
k(1,u) u−(1+ε)/ p − u−1/ p + u(1+ε)/ p−1 − u−1/q du
uη k(1,u) u−(1+ε)/ p−η − u−1/ p−η + u(1+ε)/ p−1−η − u−1/q−η du
0
≤L
0
k(1,u)u−(1+ε)/ p du
0 ≤ k p (ε) − k p = ≤
1
1
uη k(1,u) u−(1+ε)/ p−η + u(1+ε)/ p−η−1 du
Hence the integral k p (ε) =
obtains
1
∞
1
0
−(1+ε)/ p−η
u
− u−1/ p−η + u−1/q−η − u(1+ε)/ p−1−η du
(2.2)
1
−(1+ε)/ p−η
1 −1/q−η
=L u
− u−1/ p−η du
u
− u(1+ε)/ p−1−η du
+
0
0
−1 −1 −1 −1 1
1
ε
1
1+ε
+
.
− −η
−
−η
−η
−
−η
=L q
p
q
p
p
Then |k p (ε) − k p |→0 (ε→0+ ) and k p (ε) = k p + o(1) (ε→0+ ). Similarly, kq (ε) is also a positive constant and kq (ε) = kq + o(1) = k p + o(1) (ε→0+ ). Hence kr (ε) is a positive constant with kr (ε) = k p + o(1) (ε→0+ ; r = p, q). Since for ε ∈ [0,min { p, q}(1 − η) − 1) and
r = p, q, one obtains
0<
=
∞
1
m1+ε
m=1
∞
1
m2+ε
m=1
≤L
1
0
k(m,t)
1
0
t
m
m
t
(1+ε)/r
η k 1,
dt =
t
m
t
m
m=1
m
0
m
m
1
m2+ε
m=1
1 0
k 1,
t
m
m
t
(1+ε)/r
dt
−(1+ε)/r −η
∞
1 1 t −(1+ε)/r −η
t
d
∞
dt
(2.3)
∞
=
L
1
< ∞,
1 − (1 + ε)/r − η m=1 m2−(1+ε)/r −η
and then (1.5) is valid. The lemma is proved.
Note. In applying Lemma 2.1, if k(1,u) is continuous in [0,1], then one can set η = 0 and
does not consider the limit.
Bicheng Yang 5
If kλ (x, y) is the homogeneous function of −λ-order (λ > 0), then k(x, y) = kλ (x,
y)(xy)(1/2)(λ−1) is obviously homogeneous function of −1-order. Suppose that k(x, y) satisfies the conditions of Lemma 2.1, setting ωr (x) = x(r/2)(1−λ) (r = p, q), since
∞ ∞
kλ (m,n)am bn =
n=
1 m=1
∞
1− p
ωq (n)
n =1
∞
∞ ∞
n=1 m
=1
p
kλ (m,n)am
1/ p
1/q
k(m,n) ω p (m)am ωq (n)bn ;
=
m=1
∞
∞
k(m,n)
1/ p
ω p (m)am
p
(2.4)
,
n=1 m=1
by (1.8), one has the following theorem.
Theorem 2.2. Let p > 1, 1/ p + 1/q = 1, let the symmetric kernel kλ (x, y) be homogeneous function of −λ-order (λ > 0), and let the functions k(x, y) = kλ (x, y)(xy)(1/2)(λ−1) satisfy the conditions of Lemma 2.1. If ωr (x) = x(r/2)(1−λ) (r = p, q), am ,bn ≥ 0, a = {am }∞
m=1 ∈
∞ (p/2)(1−λ) p 1/ p
p
q
∞
lω p , b = {bn }n=1 ∈ lωq , such that a p,ω p = { n=1 n
an } > 0, bq,ωq =
{
∞
n =1 n
(q/2)(1−λ) bq }1/q
n
> 0, then one has the following two equivalent inequalities:
∞ ∞
∞
kλ (m,n)am bn < k p a p,ω p bq,ωq ;
n=1 m=1
1− p
ωq (n)
n =1
∞
p 1/ p
< k p a p,ω p ,
kλ (m,n)am
m=1
where the constant factor k p =
∞
0
(2.5)
k(1,u)u−1/ p dt is the best possible.
3. Applications to some Hilbert-type inequalities
In the following, suppose that p > 1, 1/ p + 1/q = 1, ωr (n) = n(r/2)(1−λ) (r = p, q), am ,bn ≥
∞
p
q
p 1/ p
∞
0, a = {am }∞
> 0,
m=1 ∈ lω p , b = {bn }n=1 ∈ lωq , such that a p,ω p = { n=1 ω p (n)an }
bq,ωq = {
∞
q 1/q
n=1 ωq (n)bn }
> 0, and one omits the words that the constant factors are
the best possible.
(a) Let kλ (x, y) = (1/(xα + y α )λ/α )(α > 0,0 ≤ 1 − 2min {1/ p,1/q} < λ ≤ 1 + 2min {1/ p,
1/q}), and k(x, y) = (xy)(λ−1)/2 /(xα + y α )λ/α . Then for fixed x > 0 and r = p, q,
((xt)(λ−1)/2 /(xα + t α )λ/α )(x/t)1/r = (x(1/2)(λ−1)+1/r /(xα + t α )λ/α )(1/t)1/r+(1/2)(1−λ) are decreasing in t ∈ (0, ∞). Since k(1,u) = u(λ−1)/2 /(1 + y α )λ/α is continuous in (0,1], there exists
η = (1/2)(1 − λ) < min {1/ p,1/q}, such that lim u→0+ uη k(1,u) = 1; setting t = uα in the
following, one obtains
kp =
∞
0
1
λ/α u
1 + uα
(λ−1)/2−1/ p
1
du =
α
∞
1 1 λ+1 1 1 λ+1 1
= B
−
,
−
α α
2
p α
2
q
0
1
t (1/α)[(λ+1)/2−1/ p]−1 dt
(1 + t)λ/α
=: k p α,λ .
(3.1)
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Journal of Inequalities and Applications
Then by (2.5), one has the following corollary.
Corollary 3.1. The following inequalities are equivalent:
∞ ∞
n=1 m=1
∞
am bn
λ/α
α
m + nα
< k p (α,λ)a p,ω p bq,ωq ;
n(p/2)(λ−1)
n =1
∞
m=1
p 1/ p
am
mα + n α
(3.2)
< k p (α,λ)a p,ω p .
λ/α
In particular, (i) for α = 1, one has k p (1,λ) = B((λ + 1)/2 − 1/ p,(λ + 1)/2 − 1/q) and
∞ ∞
am bn
λ+1 1 λ+1 1
− ,
−
a p,ω p bq,ωq ;
λ <B
2
p 2
q
n=1 m=1 (m + n)
∞
n(p/2)(λ−1)
n =1
p 1/ p
∞
am
(m
+ n)λ
m=1
<B
(3.3)
λ+1 1 λ+1 1
− ,
−
a p,ω p ;
2
p 2
q
(3.4)
(ii) for α = λ, one has k p (λ,λ) = (1/λ)B((1/λ)((λ + 1)/2 − 1/ p),(1/λ)((λ + 1)/2 − 1/q))
and
∞ ∞
am bn
1 1 λ+1 1 1 λ+1 1
< B
−
,
−
λ + nλ
m
λ
λ
2
p λ
2
q
n=1 m=1
∞
n(p/2)(λ−1)
n =1
∞
am
λ + nλ
m
m=1
p 1/ p
a p,ω p bq,ωq ;
1 1 λ+1 1 1 λ+1 1
< B
−
,
−
λ λ
2
p λ
2
q
(3.5)
a p,ω p .
(3.6)
(b) Let kλ (x, y) = (ln (x/ y)/(xλ − y λ )) (0 ≤ 1 − 2min {1/ p,1/q} < λ ≤ 1 + 2min {1/ p,
1/q}), k(x, y) = (ln(x/ y)/(xλ − y λ ))(xy)(1/2)(λ−1) . Since ln(t/x)/((t/x)λ − 1) is decreasing
in t ∈ (0, ∞) (see [8]), then for fixed x > 0 and r = p, q,
1/r
x
ln(x/t)
(xt)(1/2)(λ−1)
xλ − tλ
t
= x−(1/2)(λ+1)+1/r
1/r+(1/2)(1−λ)
ln(t/x) 1
(t/x)λ − 1 t
(3.7)
are decreasing in t ∈ (0, ∞). Since k(1,u) = (lnu)u(λ−1)/2 /(uλ − 1) is continuous in
(0,1](k(1,1) = lim u→1 k(1,u)), and (1 − λ)/2 < min {1/ p,1/q}, there exists ε > 0, such that
η = (1/2)(1 − λ) + ε < min {1/ p,1/q}, and lim u→0+ uη k(1,u) = 0, then setting t = uλ in the
following, and using the formula as (see [9])
∞
0
ln t a−1
π
t du =
t−1
sin aπ
2
2
= B(a,1 − a)
(0 < a < 1),
(3.8)
Bicheng Yang 7
one obtains
kp =
∞
0
ln u (λ−1)/2−1/ p
1
du = 2
u
uλ − 1
λ
∞
0
lnt 1/2+(1/λ)(1/q−1/2)−1
dt
t
t−1
1 1 1 1 1 1 1 1 1
=
−
, +
−
B +
λ 2 λ q 2 2 λ p 2
2
(3.9)
.
Then by (2.5), one has the following corollary.
Corollary 3.2. The following inequalities are equivalent:
2
∞ ∞
ln(m/n)am bn
1 1 1 1 1 1 1 1 1
<
−
,
−
a p,ω p bq,ωq ;
B
+
+
mλ − n λ
λ 2 λ q 2 2 λ p 2
n=1 m=1
(3.10)
∞
∞
ln(m/n)am
mλ − n λ
m=1
p 1/ p
n(p/2)(λ−1)
n =1
1 1 1 1 1 1 1 1 1
<
−
, +
−
B
+
λ 2 λ q 2 2 λ p 2
(3.11)
2
a p,ω p .
(c) Let kλ (x, y) = 1/ max {xλ , y λ } (0 ≤ 1 − 2min {1/ p,1/q} < λ ≤ 1 + 2min {1/ p,1/q}),
and k(x, y) = (1/ max {xλ , y λ })(xy)(1/2)(λ−1) . Then for fixed x > 0 and r = p, q,
1/r
1
x
(xt)(1/2)(λ−1)
λ
λ
t
max x ,t
= x(1/2)(λ−1)+1/r
1/r+(1/2)(1−λ)
1
1
λ
λ
t
max x ,t
(3.12)
are decreasing in t ∈ (0, ∞). Since k(1,u) = (u(λ−1)/2 / max {1,uλ } ) (u ∈ (0,1]) is continuous in (0,1], there exists η = (1/2)(1 − λ) < min {1/ p,1/q}, and lim u→0+ uη k(1,u) = 1, one
finds
kp =
∞
0
=
1
u(λ−1)/2−1/ p du =
max 1,uλ
λ−1 1
+
2
q
−1
λ−1 1
+
+
2
p
1
0
u
(λ−1)/2−1/ p
∞
du +
−1 1
u(λ−1)/2−λ−1/ p du
(3.13)
.
Then by (2.5), one has the following corollary.
Corollary 3.3. The following inequalities are equivalent:
∞ ∞
am bn
<
max
mλ ,nλ
n=1 m=1
∞
n(p/2)(λ−1)
n =1
λ−1 1
+
2
q
∞
ln(m/n)am
m=1 max
mλ ,nλ
−1
p 1/ p
λ−1 1
+
+
2
p
<
λ−1 1
+
2
q
−1 a p,ω p bq,ωq ;
−1
+
λ−1 1
+
2
p
(3.14)
−1 a p,ω p .
(3.15)
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Journal of Inequalities and Applications
Remarks 3.4. (i) For p = q = 2 in (3.3), (3.5), (3.10), and (3.14), setting ω(n) = n1−λ (0 <
λ ≤ 2), one has some Hilbert-type inequalities with a parameter (see [8, 10–12]):
∞ ∞
am bn
λ λ
λ < B 2 , 2 a2,ω b2,ω ;
n=1 m=1 (m + n)
∞ ∞
am bn
mλ + n λ
n=1 m=1
<
π
a2,ω b2,ω ;
λ
2
∞ ∞
ln (m/n)am bn
π
<
a2,ω b2,ω ;
λ − nλ
m
λ
n=1 m=1
∞ ∞
4
am bn
< a2,ω b2,ω .
λ ,nλ
λ
max
m
n=1 m=1
(3.16)
(3.17)
(3.18)
(3.19)
(ii) For λ = 1 in (3.17), (3.18), and (3.19), one has the following base Hilbert-type
inequalities (see [9]):
∞ ∞
am bn
n=1 m=1
m+n
< π a 2 b 2 ;
∞ ∞
ln(m/n)am bn
< π 2 a 2 b 2 ;
m−n
n=1 m=1
(3.20)
∞ ∞
am bn
< 4 a 2 b 2 .
max
{m,n}
n=1 m=1
References
[1] B. Yang, “On the norm of a Hilbert’s type linear operator and applications,” Journal of Mathematical Analysis and Applications, vol. 325, no. 1, pp. 529–541, 2007.
[2] B. Yang, “On the norm of a self-adjoint operator and applications to the Hilbert’s type inequalities,” Bulletin of the Belgian Mathematical Society, vol. 13, no. 4, pp. 577–584, 2006.
[3] B. Yang, “On the norm of an integral operator and applications,” Journal of Mathematical Analysis and Applications, vol. 321, no. 1, pp. 182–192, 2006.
[4] B. Yang, “On the norm of a self-adjoint operator and a new bilinear integral inequality,” Acta
Mathematica Sinica, vol. 23, no. 7, pp. 1311–1316, 2007.
[5] Á. Bényi and C. Oh, “Best constants for certain multilinear integral operators,” Journal of Inequalities and Applications, vol. 2006, Article ID 28582, 12 pages, 2006.
[6] B. Yang, “On the norm of a certain self-adjiont integral operator and applications to bilinear
integral inequalities,” to appear in Taiwanese Journal of Mathematics.
[7] Z. Wang and D. Gua, An Introduction to Special Functions, Science Press, Bejing, China, 1979.
[8] B. Yang, “Generalization of a Hilbert-type inequality with the best constant factor and its applications,” Journal of Mathematical Research and Exposition, vol. 25, no. 2, pp. 341–346, 2005.
[9] B. Yang, “On new generalizations of Hilbert’s inequality,” Journal of Mathematical Analysis and
Applications, vol. 248, no. 1, pp. 29–40, 2000.
[10] B. Yang, “An extension of Hardy-Hilbert’s inequality,” Chinese Annals of Mathematics, vol. 23,
no. 2, pp. 247–254, 2002.
Bicheng Yang 9
[11] G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952.
[12] B. Yang, “On a generalization of a Hilbert’s type inequality and its applications,” Chinese Journal
of Engineering Mathematics, vol. 21, no. 5, pp. 821–824, 2004.
Bicheng Yang: Department of Mathematics, Guangdong Institute of Education, Guangzhou,
Guangdong 510303, China
Email address: bcyang@pub.guangzhou.gd.cn