On a Hilbert-Type Operator with a Symmetric Homogeneous Kernel of 1-Order and Applications
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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.
2
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
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