A Hilbert-type integral inequality in the whole plane related to the kernel of exponent function

By using real analysis and weight functions, we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and the operator expressions.

In this paper, by using real analysis and weight functions we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and operator expressions.

Lemma 2
We have the following inequalities:

Lemma 3 If there exists a constant M such that, for any nonnegative measurable functions
f (x) and g(y) in R, then we have σ 1 = σ .

Lemma 4
If there exists a constant M such that, for any nonnegative measurable functions f (x) and g(y) in R, then we have K Proof For σ 1 = σ , by (8) we have In view of the presented results, we find Since e -ρu γ u 2σ is continuous in (0, ∞), and e -ρu γ u 2σ → 0 (u → ∞), there exists a positive constant M 1 such that e -ρu γ u 2σ ≤ M 1 (u ∈ [m α,β , ∞)). By (4) it follows that By (15) it follows that In the same way, we have By (14) (for f = f n , g = g n ), we have For n → ∞, by Fatou lemma (see [27]), (16), and (17) we find The lemma is proved.

Lemma 5
We define the following weight functions: Then we have Proof For fixed y ∈ R\{0}, setting u = x δ α y β , we find for fixed x ∈ R\{0}, setting u = x δ α y β , it follows that Hence we have (20). The lemma is proved.

Theorem 2
If M is a constant, then the following statements (i), (ii), and (iii) are equivalent: (25) and (26) is the best possible.
Proof For σ 1 = σ , under and the assumption of statement (i), if (24) takes the form of equality for y ∈ R\{0}, then there exist constants A and B such that they are not both zero and (see [28]) We suppose that A = 0 (otherwise, B = A = 0). Then it follows that Hence (24) takes the form of strict inequality, and so does (21). Hence (25) and (26) are valid.
In view of Theorem 1, we still can conclude that statements (i), (ii), and (iii) in Theorem 2 are equivalent.
The theorem is proved.
In view of (31), it follows that Tf p,ψ 1-p = h 1 p,ψ 1-p ≤ M f p,ϕ , and then the operator T is bounded and satisfies Tf p,ψ 1-p f p,ϕ ≤ M.