On Some Extensions of Hardy-Hilbert ’ s Inequality and Applications

Laith Emil Azar Department of Mathematics, Al Al-Bayt University, P.O. Box. 130095, Mafraq 25113, Jordan Correspondence should be addressed to Laith Emil Azar, azar laith@yahoo.com Received 23 October 2007; Accepted 2 January 2008 Recommended by Shusen Ding By introducing some parameters we establish an extension of Hardy-Hilbert’s integral inequality and the corresponding inequality for series. As an application, the reverses, some particular results and their equivalent forms are considered. Copyright q 2008 Laith Emil Azar. 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.

The corresponding inequalities for series 1.3 and 1.4 are where the sequences {a n } and {b n } are such that 0 ∞, and the constant factor π/ sin π/p and pq are the best possible.By introducing a parameter 0 < λ ≤ 2, some extensions of 1.6 p q 2 were given by Yang 5,6 as Very recently, in 7 the following extensions were given: where the constant factor D A, B see 7, Lemma 2.1 is the best possible in both inequalities.For more information related to this subject see, for example, 8, 9 .
In this paper by introducing some parameters, we generalize 1.8 and we obtain the reverse form for each of them.Some particular results and the equivalent form are also considered.

2.3
Proof.Setting t y/x λ , we get

2.4
i for A, B > 0, we obtain ii for A 0, B > 0, we find then for ε → 0 , Proof.Setting t x/y λ , we find dt dy

2.9
On the other hand,

2.10
Hence, 2.8 is valid.The lemma is proved.
where the constant factor C λ A, B defined in 2.3 is the best possible.In particular, π, and inequality 2.11 reduces to Hardy-Hilbert's inequality

and 2.11 reduces to Hardy-Hilbert's-type inequality
Proof.By the Holder inequality, taking into account 2.1 , we get 2.14 If 2.14 takes the form of equality, then there exist constants M and N which are not all zero such that B max x λ , y λ g q y , Mx p 1−λ/2 f p x Ny q 1−λ/2 g q y , a.e. in 0, ∞ × 0, ∞

2.15
Hence, there exists a constant c such that Mx p 1−λ/2 f p x Ny q 1−λ/2 g q y c a.e. in 0, ∞ .

2.16
We claim that M 0. In fact, if M / 0, then which contradicts the fact that 0 Hence, by 2.14 we get 2.11 .If the constant factor C λ A, B is not the best possible, then there exists a positive constant

2.18
By using 2.8 , we find

2.19
Therefore, we get For ε → 0 , it follows that C λ A, B ≤ K which contradicts the fact that K < C λ A, B .Hence, the constant factor C λ A, B in 2.11 is the best possible.The theorem is proved.
where the constant factor C λ A, B defined in 2.3 is the best possible.In particular, π, and inequality 2.22 reduces to Hardy-Hilbert's inequality ii for A 0, λ B 1, C 1 0, 1 4 and 2.22 reduces to Hardy-Hilbert's-type inequality

2.25
By using 2.8 , we find

2.26
Therefore, we get

2.29
where the constant factor C λ A, B p is the best possible.Inequalities 2.11 and 2.29 are equivalent. Proof.Setting then by 2.11 we have ∞ 0 y q 1−λ/2 −1 g q y dy ∞ 0 Hence, we obtain

2.32
Thus, by 2.11 , both 2.31 and 2.32 keep the form of strict inequalities, then we have 2.29 .By Holder's inequality, we find B max x λ , y λ dx y 1/p−λ/2 g y dy

2.34
where the constant factor C λ A, B p is the best possible.Inequalities 2.22 and 2.34 are equivalent.
The proof of Theorem 2.6 is similar to that of Theorem 2.5, so we omit it.

3.5
On the other hand, we have

3.6
where . Therefore, 3.3 is valid.By the symmetry, 3.4 is still valid.The lemma is proved.

3.10
Proof.Setting t x/n λ in the following, by 3.4 , we have

3.11
Thus, inequality 3.10 holds.The lemma is proved.
where the constant factor C λ A, B defined in 2.3 is the best possible.In particular, i for λ A B 1, C 1 1, 1 π, and inequality 3.12 reduces to Hardy-Hilbert's inequality ii for A 0, λ B 1, C 1 0, 1 4 and 3.12 reduces to Hardy-Hilbert's-type inequality Proof.By the Holder inequality, taking into account 3.1 , we get Then, by 3.3 and 3.4 we obtain 3.12 .It remains to show that the constant factor C λ A, B is the best possible, to do that we set for 0 < ε < pλ/2, a m m λ/2−1−ε/p ; b n n λ/2−1−ε/q , by 3.9 we have If there exists a constant 0 < K ≤ C λ A, B such that 3.12 is still valid if we replace C λ A, B by K, then in particular by 3.10 we find is the best constant factor in 3.12 .The theorem is proved.
where the constant factor C λ A, B defined in 2.4 is the best possible.In particular, Proof.By reverse Holder's inequality, we get

3.21
Then by 3.3 and 3.4 , in view of q < 0, we have 3.18 .For 0 < ε < pλ/2, setting a m m λ/2−1−ε/p , b n n λ/2−1−ε/q m, n ∈ N .If there exists a constant K ≥ C λ A, B such that 3.18 is still valid if we replace C λ A, B by K, then in particular by 3.9 and 3.10 we find

3.23
Hence, if ε → 0 , we get C λ A, B ≥ K. Thus, K C λ A, B is the best constant factor in 3.18 .
a m b n A min m λ , n λ B max m λ , n λ .

3.26
By 3.12 and using the same method of Theorem 2.5, we obtain 3.24 .We may show that the constant factor in 3.24 is the best possible and inequality 3.12 is equivalent to 3.24 .

Theorem 3 . 5 . 1 ,
Under the assumption ofTheorem 3.3,    where the constant factor C λ A, B p is the best possible.Inequalities 3.12 and 3.24 are equivalent.Proof.Setting b n n λp/2−1 ∞ m 1 a m A min m λ , n λ B max m λ , n λ p−

Theorem 3 . 6 .
Under the assumption ofTheorem 3.4,    where the constant factor C λ A, B p is the best possible.Inequalities 3.18 and 3.27 are equivalent.

Lemma 2.2. For p > 1 or 0
For ε → 0 , it follows that C λ A, B ≥ H which contradicts the fact that H > C λ A, B .Hence, the constant factor C λ A, B in 2.22 is the best possible.The theorem is proved.
∞ 0 33Therefore, by 2.29 we have 2.11 , and inequalities 2.29 and 2.11 are equivalent.If the constant factor in 2.29 is not the best possible, then by 2.33 we can get a contradiction that the constant factor in 2.11 is not the best possible.The theorem is proved.