Inequalities Similar to Hilbert′s Inequality

Theorem A. If p₁, p₂ > 1 such that 1/p₁   + 1/p₂   ≥ 1 and 0 < λ = 2 − 1/p₁   − 1/p₂   = 1/q₁   + 1/q₂   ≤ 1, where, as usual, q₁ and q₂ are the conjugate exponents of p₁ and p₂, respectively,  then

where K = K(p₁, p₂) depends on p₁ and p₂ only.

Theorem 1. Let p > 1 be constant and 1/p + 1/q = 1. If a(s) and b(t) are real-valued functions defined for {0,1, …, m} and {0,1, …, n}, respectively, and a(0) = b(0) = 0. Moreover, define the operators ∇ by ∇u(t) = u(t) − u(t − 1). Then,

Theorem 2. Let p > 1 be constants, and 1/p + 1/q = 1. For i = 1,2, let a_i(s_i, t_i) be real-valued functions defined for (s_i, t_i), where s_i = 1,2, …, m_i;  t_i = 1,2, …, n_i, and let m_i, n_i be natural numbers. Let a_i(0, t_i) = a_i(s_i, 0) = 0, and define the operators ∇₁, ∇₂ by

Then,

where

Remark 3. Inequality (4) is just a similar version of the following inequality established by Pachpatte [11]:

Theorem B. Let p₁, p₂, q₁, q₂, and λ be as in Theorem A. If f ∈ L^p(0, ∞) and g ∈ L^q(0, ∞), then

where K = K(p, q) depends on p and q only.

Theorem 4. Let p > 1 be constants, and 1/p + 1/q = 1. If f(s) and g(t) are real-valued continuous functions defined on [0, x) and [0, y), respectively, and let f(0) = g(0) = 0. Then,

Theorem 5. Let p > 1, and 1/p + 1/q = 1. For i = 1,2, let h_i ≥ 1,  f_i(s_i, t_i) be real-valued differentiable functions defined on [0, x_i)×[0, y_i), where x_i ∈ (0, ∞),   y_i ∈ (0, ∞), and f_i(0, t_i) = f_i(s_i, 0) = 0. As usual, partial derivatives of f_i are denoted by D₁f_i, D₂f_i, D₁₂f_i = D₂₁f_i, and so forth. Let

Then, where and S(h) is as in (6).

Remark 6. Inequality (13) is just a similar version of the following inequality established by Pachpatte [11]:

Proof of Theorem 2. From the hypotheses of Theorem 2, we have

By using Hölder’s inequality and noticing the reverse Young’s inequality [19], for positive real numbers s₁, s₂ and 1/α + 1/β = 1, α > 1, where S(h) is as in (6). Hence, Dividing both sides of (20) by taking the sum of both sides of (20) over t_i and s_i from 1 to m_i and n_i  (i = 1,2), respectively, and making use of Hölder’s inequality, we have

This completes the proof.

Proof of Theorem 5. From the hypotheses of  Theorem 5, we obtain for i = 1,2:

From (23), Hölder’s integral inequality and in view of the reverse Young’s inequality (19), we have Integrating both sides of (24) over s_i and t_i from 1 to x_i and y_i (i = 1,2), respectively, and by using Hölder’s integral inequality, we arrive at

This completes the proof.