A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension

Abstract

Using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel is obtained, and a best extension with two interval variables is given. The equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2000 Mathematics Subject Classification: 26D15; 47A07.

1 Introduction

Assuming that p > 1, $\frac{1}{p}+\frac{1}{q}=1$, f (≥ 0) ∈ L^p (R₊), g(≥ 0) ∈ L^q (R₊), ${∥f∥}_p={\left\{{\int }_0^{\infty }{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}>0$, || g || _q > 0, we have the following Hardy-Hilbert's integral inequality [1]:

$$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}\mathsf{\text{d}}x\mathsf{\text{d}}y<\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}\left|\right|f{\left|\right|}_p\left|\right|g{\left|\right|}_q,$$

(1)

where the constant factor $\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}$ is the best possible. If a _m , b _n ≥ 0, $a={\left\{a_m\right\}}_{m=1}^{\infty }\in l^p$, $b={\left\{b_n\right\}}_{n=1}^{\infty }\in l^q$, $\left|\right|a|{|}_p={\left\{{\sum }_{m=1}^{\infty }{a}_m^p\right\}}^{\frac{1}{p}}>0$, || b || _q > 0, then we still have the following discrete Hardy-Hilbert's inequality with the same best constant factor $\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}$:

$$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{b}_n}{m+n}<\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}\left|\right|a{\left|\right|}_p\left|\right|b{\left|\right|}_q.$$

(2)

For p = q = 2, the above two inequalities reduce to the famous Hilbert's inequalities. Inequalities (1) and (2) are important in analysis and its applications [2–4].

In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang [5] gave an extension of (1) for p = q = 2. Refinement and generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows: If λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative homogeneous function of degree - λ satisfying for any x, y, t > 0, k_λ(tx, ty) = t^-λ k_λ (x, y), $k\left({\lambda }_1\right)={\int }_0^{\infty }{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-1}\mathsf{\text{d}}t\in R_{+}$, $\varphi \left(x\right)=x^{p\left(1-{\lambda }_1\right)-1}$, $\psi \left(x\right)=x^{q\left(1-{\lambda }_2\right)-1}$, $f(\ge 0)\in L_{p,\varphi }(R_{+})=\left\{f\right|\left|\right|f{\left|\right|}_{p,\varphi }:={\{{\displaystyle {\int }_0^{\infty }}\varphi (x)|f(x{)|}^p\text{d}x\}}^{\frac{1}{p}}<\infty \}$, g(≥ 0) ∈ L _{q , ψ} , || f || _{p , ϕ} , || g || _{q , ψ} > 0, then we have

$$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y<k\left({\lambda }_1\right)\left|\right|f{\left|\right|}_{p,\varphi }\left|\right|g{\left|\right|}_{q,\psi },$$

(3)

where the constant factor k(λ₁) is the best possible. Moreover, if k_λ(x, y) is finite and $k_{\lambda }\left(x,y\right)x^{{\lambda }_1-1}\left(k_{\lambda }\left(x,y\right)y^{{\lambda }_2-1}\right)$ is decreasing with respect to x > 0(y > 0), then for a _m ,b _n ≥ 0, $a={\left\{a_m\right\}}_{m=1}^{\infty }\in l_{p,\varphi }=\left\{a\left|\right|\left|a\right|{|}_{p,\varphi }:={\left\{{\sum }_{n=1}^{\infty }\varphi \left(n\right)|a_n{|}^p\right\}}^{\frac{1}{p}}<\infty \right\}$, $b={\left\{b_n\right\}}_{n=1}^{\infty }\in l_{q,\psi }$, || a || _{p , ϕ} , || b || _{q , Ψ} > 0, we have

$$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{k}_{\lambda }\left(m,n\right)a_m{b}_n<k\left({\lambda }_1\right)\left|\right|a{\left|\right|}_{p,\varphi }\left|\right|b{\left|\right|}_{q,\psi },$$

(4)

with the best constant factor k(λ₁). Clearly, for λ = 1, $k_1\left(x,y\right)=\frac{1}{x+y}$, ${\lambda }_1=\frac{1}{q}$, ${\lambda }_2=\frac{1}{p}$ (3) reduces to (1), and (4) reduces to (2). Some other results about Hilbert-type inequalities are provided by [7–15].

On half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. And, Yang [16] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [17] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ₁, λ₂)(λ, λ₁ > 0, 0 < λ₂ ≤ 1, λ₁ + λ₂ = λ):

$$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{a_n}{{\left(x+n\right)}^{\lambda }}\mathsf{\text{d}}x<B\left({\lambda }_1,{\lambda }_2\right)\left|\right|f{\left|\right|}_{p,\phi }\left|\right|a{\left|\right|}_{q,\psi }.$$

(5)

In this article, using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel and a best constant factor is given as follows:

$$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }{k}_{\lambda }\left(x,n\right)a_n\mathsf{\text{d}}x<k\left({\lambda }_1\right)\left|\right|f{\left|\right|}_{p,\phi }\left|\right|a{\left|\right|}_{q,\psi },$$

(6)

which is a generalization of (5). A best extension of (6) with two interval variables, some equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2 Some lemmas

We set the following conditions:

Condition (i) v(y)(y ∈ [n₀ - 1, ∞)) is strictly increasing with v(n₀ - 1) ≥ 0 and for any fixed x ∈ (b, c), f(x, y) is decreasing for y ∈ (n₀ - 1, ∞) and strictly decreasing in an interval of (n₀ - 1, ∞).

Condition (ii) $v\left(y\right)\left(y\in \left[n_0-\frac{1}{2},\infty \right)\right)$ is strictly increasing with $v\left(n_0-\frac{1}{2}\right)\ge 0$ and for any fixed x ∈ (b, c), f(x, y) is decreasing and strictly convex for $y\in \left(n_0-\frac{1}{2},\infty \right)$.

Condition (iii) There exists a constant β ≥ 0, such that v(y)(y ∈ [n₀ - β, ∞)) is strictly increasing with v(n₀ - β) ≥ 0, and for any fixed x ∈ (b, c), f(x, y) is piecewise smooth satisfying

$$R\left(x\right):=\underset{n_0-\beta }{\overset{n_0}{\int }}f\left(x,y\right)\mathsf{\text{d}}y-\frac{1}{2}f\left(x,n_0\right)-\underset{n_0}{\overset{\infty }{\int }}\rho \left(y\right){f^{\prime }}_y\left(x,y\right)\mathsf{\text{d}}y>0,$$

where $\rho \left(y\right)\left(=y-\left[y\right]-\frac{1}{2}\right)$ is Bernoulli function of the first order.

Lemma 1 If λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree - λ in ${\mathsf{\text{R}}}_{+}^2,u\left(x\right)\left(x\in \left(b,c\right),-\infty \le b<c\le \infty \right)$ and v(y)(y ∈ [n₀, ∞), n₀ ∈ N) are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, setting K(x, y) = k_λ(u(x), v(y)), then we define weight functions ω(n) and ϖ(x) as follows:

$$\omega (n):={[v(n)]}^{{\lambda }_2}{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)[u(x{)]}^{{\lambda }_1-1}{u}^{\prime }(x)\text{d}x,n\ge n_0(n\in \mathbf{N}),}$$

(7)

$$\varpi (x):={[u(x)]}^{{\lambda }_1}{\displaystyle \sum _{n=n_0}^{\infty }K(x,n)[v(n{)]}^{{\lambda }_2-1}{v}^{\prime }(n),x\in (b,c)}.$$

(8)

It follows

$$\omega \left(n\right)=k\left({\lambda }_1\right):=\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-1}\mathsf{\text{d}}t.$$

(9)

Moreover, setting $f(x,y):={[u(x)]}^{{\lambda }_1}K(x,y)[v(y{)]}^{{\lambda }_2-1}{v}^{\prime }(y)$, if k(λ₁) ∈ R₊ and one of the above three conditions is fulfilled, then we still have

$$\varpi \left(x\right)<k\left({\lambda }_1\right)\left(x\in \left(b,c\right)\right).$$

(10)

Proof . Setting $t=\frac{u\left(x\right)}{v\left(n\right)}$ in (7), by calculation, we have (9).

• (i) If Condition (i) is fulfilled, then we have

  $$\begin{array}{l}\varpi (x)={\displaystyle \sum _{n=n_0}^{\infty }f(x,n)<{[u(x)]}^{{\lambda }_1}}{\displaystyle \underset{n_0-1}{\overset{\infty }{\int }}K(x,y)[v(y{)]}^{{\lambda }_2-1}{v}^{\prime }(y)\text{d}y}\\ \stackrel{t=u(x)\text{/}v(y)}{=}{\displaystyle \underset{0}{\overset{\frac{u(x)}{v(n_0-1)}}{\int }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}\text{d}t\le k({\lambda }_1).}\end{array}$$

• (ii) If Condition (ii) is fulfilled, then by Hadamard's inequality [18], we have

  $$\begin{array}{c}\varpi \left(x\right)=\sum _{n=n_0}^{\infty }f\left(x,n\right)<\underset{n_0-\frac{1}{2}}{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y\\ \stackrel{t=u\left(x\right)\mathsf{\text{/}}v\left(y\right)}{=}\underset{0}{\overset{\frac{u\left(x\right)}{v\left(n_0-\frac{1}{2}\right)}}{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-1}\mathsf{\text{d}}t\le k\left({\lambda }_1\right).\end{array}$$

• (iii) If Condition (iii) is fulfilled, then by Euler-Maclaurin summation formula [6], we have

  $$\begin{array}{ll}\hfill \varpi \left(x\right)& =\sum _{n=n_0}^{\infty }f\left(x,n\right)\\ =\underset{n_0}{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y+\frac{1}{2}f\left(x,n_0\right)+\underset{n_0}{\overset{\infty }{\int }}\rho \left(y\right){f^{\prime }}_y\left(x,y\right)\mathsf{\text{d}}y\\ =\underset{n_0-\beta }{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y-R\left(x\right)\\ =\underset{0}{\overset{\frac{u\left(x\right)}{v\left(n_0-\beta \right)}}{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-1}\mathsf{\text{d}}t-R\left(x\right)\\ \le k\left({\lambda }_1\right)-R\left(x\right)<k\left({\lambda }_1\right).\end{array}$$

The lemma is proved.    ■

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 0(p ≠ 1), $\frac{1}{p}+\frac{1}{q}=1,$ a _n ≥ 0, n ≥ n₀(n ∈ N), f (x) is a non-negative measurable function in (b, c). Then, (i) for p > 1, we have the following inequalities:

$$\begin{array}{ll}\hfill J_1:& ={\left\{\sum _{n=n_0}^{\infty }\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-p{\lambda }_2}}{\left[\underset{d}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^p\right\}}^{\frac{1}{p}}\\ \le {\left[k\left({\lambda }_1\right)\right]}^{\frac{1}{q}}{\left\{\underset{d}{\overset{c}{\int }}\varpi \left(x\right)\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_1\right)-1}}{{\left[u^{\prime }\left(x\right)\right]}^{p-1}}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}},\end{array}$$

(11)

$$\begin{array}{ll}\hfill L_1:& ={\left\{\underset{b}{\overset{c}{\int }}\frac{{\left[\varpi \left(x\right)\right]}^{1-q}{u}^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-q{\lambda }_1}}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^q\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\\ \le {\left\{k\left({\lambda }_1\right)\sum _{n=n_0}^{\infty }\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_2\right)-1}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q\right\}}^{\frac{1}{q}};\end{array}$$

(12)

(ii) for 0 < p < 1, we have the reverses of (11) and (12).

Proof(i) By Hölder's inequality with weight [18] and (9), it follows

$$\begin{array}{c}{\left[{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)f(x)\text{d}x}\right]}^p=\{{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)}\left[\frac{{[u(x)]}^{(1-{\lambda }_1)\text{/}q}}{{[v(n)]}^{(1-{\lambda }_2)\text{/}p}}\frac{{[v^{\prime }(n)]}^{1\text{/}p}}{{[u^{\prime }(x)]}^{1\text{/}q}}f(x)\right]\\ \times {\left[\frac{{[v(n)]}^{(1-{\lambda }_2)\text{/}p}}{{[u(x)]}^{(1-{\lambda }_1)\text{/}q}}\frac{{[u^{\prime }(x)]}^{1\text{/}q}}{{[v^{\prime }(n)]}^{1\text{/}p}}\right]\text{d}x\}}^p\\ \le {\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)\frac{{[u(x)]}^{(1-{\lambda }_1)(p-1)}{v}^{\prime }(n)}{{[v(n)]}^{1-{\lambda }_2}{[u^{\prime }(x)]}^{p-1}}{f}^p(x)\text{d}x}\\ \times {\left\{{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)\frac{{[v(n)]}^{(1-{\lambda }_2)(q-1)}{u}^{\prime }(x)}{{[u(x)]}^{1-{\lambda }_1}{[v^{\prime }(n)]}^{q-1}}\text{d}x}\right\}}^{p-1}\\ ={\left\{\frac{\omega (n)[v(n{)]}^{q(1-{\lambda }_2)-1}}{{[v^{\prime }(n)]}^{q-1}}\right\}}^{p-1}{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)\frac{{[u(x)]}^{(1-{\lambda }_1)(p-1)}{v}^{\prime }(n)f^p(x)}{{[v(n)]}^{1-{\lambda }_2}{[u^{\prime }(x)]}^{p-1}}\text{d}x}\\ =\frac{{[k({\lambda }_1)]}^{p-1}}{{[v(n)]}^{p{\lambda }_2-1}{v}^{\prime }(n)}{\displaystyle \underset{b}{\overset{c}{\int }}K(x,n)\frac{{[u(x)]}^{(1-{\lambda }_1)(p-1)}{v}^{\prime }(n)}{{[v(n)]}^{1-{\lambda }_2}{[u^{\prime }(x)]}^{p-1}}{f}^p(x)\text{d}x.}\end{array}$$

Then, by Lebesgue term-by-term integration theorem [19], we have

$$\begin{array}{ll}\hfill J_1& \le {\left[k\left({\lambda }_1\right)\right]}^{\frac{1}{q}}{\left\{\sum _{n=n_0}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_1\right)\left(p-1\right)}{v}^{\prime }\left(n\right)f^p\left(x\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_2}{\left[u^{\prime }\left(x\right)\right]}^{p-1}}\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}\\ ={\left[k\left({\lambda }_1\right)\right]}^{\frac{1}{q}}{\left\{\underset{b}{\overset{c}{\int }}\sum _{n=n_0}^{\infty }K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_1\right)\left(p-1\right)}{v}^{\prime }\left(n\right)f^p\left(x\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_2}{\left[u^{\prime }\left(x\right)\right]}^{p-1}}\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}\\ ={\left[k\left({\lambda }_1\right)\right]}^{\frac{1}{q}}{\left\{\underset{b}{\overset{c}{\int }}\varpi \left(x\right)\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_1\right)-1}}{{\left[u^{\prime }\left(x\right)\right]}^{p-1}}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}},\end{array}$$

and (11) follows.

Still by Hölder's inequality, we have

$$\begin{array}{ll}\hfill {\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^q& =\left\{\sum _{n=n_0}^{\infty }K\left(x,n\right)\left[\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_1\right)∕q}}{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_2\right)∕p}}\frac{{\left[v^{\prime }\left(n\right)\right]}^{1∕p}}{{\left[u^{\prime }\left(x\right)\right]}^{1∕q}}\right]\right.\\ \times {\left.\left[\frac{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_2\right)∕p}}{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_1\right)∕q}}\frac{{\left[u^{\prime }\left(x\right)\right]}^{1∕q}}{{\left[v^{\prime }\left(n\right)\right]}^{1∕p}}{a}_n\right]\right\}}^q\\ \le {\left\{\sum _{n=n_0}^{\infty }K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_1\right)\left(p-1\right)}}{{\left[v\left(n\right)\right]}^{1-{\lambda }_2}}\frac{v^{\prime }\left(n\right)}{{\left[u^{\prime }\left(x\right)\right]}^{p-1}}\right\}}^{q-1}\\ \times \sum _{n=n_0}^{\infty }K\left(x,n\right)\frac{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_2\right)\left(q-1\right)}}{{\left[u\left(x\right)\right]}^{1-{\lambda }_1}}\frac{u^{\prime }\left(x\right)}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q\\ =\frac{{\left[u\left(x\right)\right]}^{1-q{\lambda }_1}}{{\left[\varpi \left(x\right)\right]}^{1-q}{u}^{\prime }\left(x\right)}\sum _{n=n_0}^{\infty }K\left(x,n\right)\frac{u^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-{\lambda }_1}}\frac{{\left[v\left(n\right)\right]}^{\left(q-1\right)\left(1-{\lambda }_2\right)}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q.\end{array}$$

Then, by Lebesgue term-by-term integration theorem, we have

$$\begin{array}{ll}\hfill L_1& \le {\left\{\underset{b}{\overset{c}{\int }}\sum _{n=n_0}^{\infty }K\left(x,n\right)\frac{u^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-{\lambda }_1}}\frac{{\left[v\left(n\right)\right]}^{\left(q-1\right)\left(1-{\lambda }_2\right)}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\\ ={\left\{\sum _{n=n_0}^{\infty }\left[{\left[v\left(n\right)\right]}^{{\lambda }_2}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{u^{\prime }\left(x\right)\mathsf{\text{d}}x}{{\left[u\left(x\right)\right]}^{1-{\lambda }_1}}\right]\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_2\right)-1}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q\right\}}^{\frac{1}{q}}\\ ={\left\{\sum _{n=n_0}^{\infty }\omega \left(n\right)\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_2\right)-1}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}{a}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

and then in view of (9), inequality (12) follows.

• (ii) By the reverse Hölder's inequality [18] and in the same way, for q < 0, we have the reverses of (11) and (12). ■

3 Main results

We set $\Phi \left(x\right):=\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_1\right)-1}}{{\left[u^{\prime }\left(x\right)\right]}^{p-1}}\left(x\in \left(b,c\right)\right)$, $\Psi \left(n\right):=\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_2\right)-1}}{{\left[v^{\prime }\left(n\right)\right]}^{q-1}}\left(n\ge n_0,n\in \mathbf{N}\right)$, wherefrom ${\left[\Phi \left(x\right)\right]}^{1-q}=\frac{u^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-q{\lambda }_1}}$, ${\left[\Psi \left(n\right)\right]}^{1-p}=\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-p{\lambda }_2}}$.

Theorem 1 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree - λ¸ in ${\mathsf{\text{R}}}_{+}^2,u\left(x\right)\left(x\in \left(b,c\right),-\infty \le b<c\le \infty \right)$ and v(y)(y ∈ [n₀, ∞), n₀ ∈ N are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, ϖ(x) < k (λ₁) ∈ R₊(x ∈ (b, c)). If $p>1,\frac{1}{p}+\frac{1}{q}=1$ , f ( x ), a _n ≥ 0, f ∈ L _{p Φ} ( b , c ), $a={\left\{a_n\right\}}_{n=n_0}^{\infty }\in l_{q,\Psi },$ || f || _p ,_Φ > 0 and || a || _q ,_Ψ > 0, then we have the following equivalent inequalities:

$$\begin{array}{ll}\hfill I:& =\sum _{n=n_0}^{\infty }{a}_n\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}f\left(x\right)\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\mathsf{\text{d}}x\\ <k\left({\lambda }_1\right)\left|\right|f\left|{|}_{p,\Phi }\right|\left|a\right|{|}_{q,\Psi },\end{array}$$

(13)

$$J:={\left\{\sum _{n=n_0}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^p\right\}}^{\frac{1}{p}}<k\left({\lambda }_1\right)\left|\right|f|{|}_{p,\Phi },$$

(14)

$$L:={\left\{\underset{b}{\overset{c}{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^q\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}<k\left({\lambda }_1\right)\left|\right|a|{|}_{q,\Psi }.$$

(15)

Moreover, if $\frac{v^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge n_0\right)$ is decreasing and there exist constants δ < λ₁ and M > 0, such that$k_{\lambda }\left(t,1\right)\le \frac{M}{t^{\delta }}\left(t\in \left.\left(0,\frac{1}{v\left(n_0\right)}\right.\right]\right)$, then the constant factor k(λ₁) in the above inequalities is the best possible.

Proof By Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x) < k(λ₁) ∈ R₊, we have (14). By Hölder's inequality, we have

$$I=\sum _{n=n_0}^{\infty }\left[{\Psi }^{{\frac{-1}{q}}^{}}\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]\left[{\Psi }^{\frac{1}{q}}\left(n\right)a_n\right]\le J\left|\right|a{\left|\right|}_{q,\Psi }.$$

(16)

Then, by (14), we have (13). On the other hand, assuming that (13) is valid, setting

$$a_n:={\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p-1},n\ge n_0,$$

then J^p-1 = || a || _{q ,Ψ} . By (11), we find J < ∞. If J = 0, then (14) is naturally valid; if J > 0, then by (13), we have

$$\left|\right|a|{|}_{q,\Psi }^q=J^p=I<k\left({\lambda }_1\right)|\left|f\right|{|}_{p,\Phi }\left|\right|a|{|}_{q,\Psi },|\left|a\right|{|}_{q,\Psi }^{q-1}=J<k\left({\lambda }_1\right)\left|\right|f|{|}_{p,\Phi },$$

and we have (14), which is equivalent to (13).

In view of (12), for [ϖ(x) ]^1-q> [k(λ₁)]^1-q, we have (15). By Hölder's in equality, we find

$$I=\underset{b}{\overset{c}{\int }}\left[{\Phi }^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\Phi }^{\frac{-1}{p}}\left(x\right)\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]\mathsf{\text{d}}x\le \left|\right|f{\left|\right|}_{p,\Phi }L.$$

(17)

Then, by (15), we have (13). On the other hand, assuming that (13) is valid, setting

$$f\left(x\right):={\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^{q-1},x\in \left(b,c\right),$$

then L^q_-1 = || f || _{p Φ} . By (12), we find L < ∞. If L = 0, then (15) is naturally valid; if L > 0, then by (13), we have

$$\left|\right|f|{|}_{p,\Phi }^p=L^q=I<k\left({\lambda }_1\right)|\left|f\right|{|}_{p,\Phi }\left|\right|a|{|}_{q,\Psi },|\left|f\right|{|}_{p,\Phi }^{p-1}=L<k\left({\lambda }_1\right)\left|\right|a|{|}_{q,\Psi },$$

and we have (15) which is equivalent to (13).

Hence, inequalities (13), (14) and (15) are equivalent.

There exists an unified constant d ∈ (b, c), satisfying u(d) = 1. For 0 < ε < p(λ₁ - δ), setting $\tilde{f}\left(x\right)=0$, x ∈ (b, d); $\tilde{f}\left(x\right)={\left[u\left(x\right)\right]}^{{\lambda }_1-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)$, x ∈ (d, c), and ${ã}_n={\left[v\left(n\right)\right]}^{{\lambda }_2-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)$, n ≥ n₀, if there exists a positive number k(≤ k(λ₁)), such that (13) is valid as we replace k(λ₁) by k, then in particular, we find

$$\begin{array}{ll}\hfill Ĩ:& =\sum _{n=n_0}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right){ã}_n\tilde{f}\left(x\right)\mathsf{\text{d}}x<k\left|\right|\tilde{f}{\left|\right|}_{p,\Phi }\left|\right|ã{\left|\right|}_{q,\Psi }\\ =k{\left\{\underset{d}{\overset{c}{\int }}\frac{u^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{\epsilon +1}}\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+\sum _{n=n_0+1}^{\infty }\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\\ <k{\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}{\left\{\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+\underset{n_0}{\overset{\infty }{\int }}{\left[v\left(y\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(y\right)\mathsf{\text{d}}y\right\}}^{\frac{1}{q}}\\ =\frac{k}{\epsilon }{\left\{\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right\}}^{\frac{1}{q}}\end{array}$$

(18)

$$\begin{array}{l}\tilde{I}={\displaystyle \sum _{n=n_0}^{\infty }{[v(n)]}^{{\lambda }_2-\frac{\epsilon }{q}-1}{v}^{\prime }(n){\displaystyle \underset{d}{\overset{c}{\int }}K(x,n)[u(x{)]}^{{\lambda }_1-\frac{\epsilon }{p}-1}{u}^{\prime }(x)\text{d}x}}\\ \stackrel{t=u(x)\text{/}v(n)}{=}{\displaystyle \sum _{n=n_0}^{\infty }}{[v(n)]}^{-\epsilon -1}{v}^{\prime }(n){\displaystyle \underset{1/v(n)}{\overset{\infty }{\int }}{k}_{\lambda }(t,1)t^{{\lambda }_1-\frac{\epsilon }{p}-1}\text{d}t}\\ =k\left({\lambda }_1-\frac{\epsilon }{p}\right){\displaystyle \sum _{n=n_0}^{\infty }}{[v(n)]}^{-\epsilon -1}{v}^{\prime }(n)-A(\epsilon )\\ >k\left({\lambda }_1-\frac{\epsilon }{p}\right){\displaystyle \underset{n_0}{\overset{\infty }{\int }}{[v(y)]}^{-\epsilon -1}{v}^{\prime }(y)\text{d}y-A(\epsilon )}\\ =\frac{1}{\epsilon }k\left({\lambda }_1-\frac{\epsilon }{p}\right){[v(n_0)]}^{-\epsilon }-A(\epsilon ),\\ A(\epsilon ):={\displaystyle \sum _{n=n_0}^{\infty }{[v(n)]}^{-\epsilon -1}{v}^{\prime }(n)}{\displaystyle \underset{0}{\overset{1/v(n)}{\int }}{k}_{\lambda }(\mathrm{t,}1)t^{{\lambda }_1-\frac{\epsilon }{p}-1}\text{d}t}.\end{array}$$

(19)

For $k_{\lambda }\left(t,1\right)\le M\left(\frac{1}{t^{\delta }}\right)\left(\delta <{\lambda }_1;t\in \left(0,1∕v\left(n_0\right)\right]\right)$, we find

$$\begin{array}{ll}\hfill 0<A\left(\epsilon \right)& \le M\sum _{n=n_0}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)\underset{0}{\overset{1∕v\left(n\right)}{\int }}{t}^{{\lambda }_1-\delta -\frac{\epsilon }{p}-1}\mathsf{\text{d}}t\\ =\frac{M}{{\lambda }_1-\delta -\frac{\epsilon }{p}}\sum _{n=n_0}^{\infty }{\left[v\left(n\right)\right]}^{-{\lambda }_1+\delta -\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\\ =\frac{M}{{\lambda }_1-\delta -\frac{\epsilon }{p}}\left[\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{{\lambda }_1-\delta +\frac{\epsilon }{q}+1}}+\sum _{n=n_0+1}^{\infty }\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{{\lambda }_1-\delta +\frac{\epsilon }{q}+1}}\right]\\ \le \frac{M}{{\lambda }_1-\delta -\frac{\epsilon }{p}}\left[\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{{\lambda }_1-\delta +\frac{\epsilon }{q}+1}}+\underset{n_0}{\overset{\infty }{\int }}\frac{v^{\prime }\left(y\right)}{{\left[v\left(y\right)\right]}^{{\lambda }_1-\delta +\frac{\epsilon }{q}+1}}\mathsf{\text{d}}y\right]\\ =\frac{M}{{\lambda }_1-\delta -\frac{\epsilon }{p}}\left[\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{{\lambda }_1-\delta +\frac{\epsilon }{q}+1}}+\frac{{\left[v\left(n_0\right)\right]}^{-{\lambda }_1+\delta -\frac{\epsilon }{q}}}{{\lambda }_1-\delta +\frac{\epsilon }{q}}\right]<\infty ,\end{array}$$

namely A(ε) = O(1)(ε → 0⁺). Hence, by (18) and (19), it follows

$$k\left({\lambda }_1-\frac{\epsilon }{p}\right){\left[v\left(n_0\right)\right]}^{-\epsilon }-\epsilon O\left(\mathsf{\text{1}}\right)<k{\left\{\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right\}}^{\frac{1}{q}}.$$

(20)

By Fatou Lemma [19], we have $k\left({\lambda }_1\right)\le {\underset{}{\text{lim}}}_{\epsilon \to 0^{+}}k\left({\lambda }_1-\frac{\epsilon }{p}\right)$, then by (20), it follows k(λ₁) ≤ k(ε → 0⁺). Hence, k = k(λ₁) is the best value of (12).

By the equivalence, the constant factor k(λ₁) in (14) and (15) is the best possible, otherwise we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. ■

Remark 1 (i) Define a half-discrete Hilbert's operator T : $L_{p,\Phi }\left(b,c\right)\to l_{p,{\Psi }^{1-p}}$ as: for f ∈ L _{p Φ} (b, c), we define $Tf\in l_{p,{\Psi }^{1-p}}$, satisfying

$$Tf\left(n\right)=\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)dx,n\ge n_0.$$

Then, by (14), it follows $\left|\right|Tf|{|}_{p.{\Psi }^{1-p}}\le k\left({\lambda }_1\right)|\left|f\right|{|}_{p,\Phi }$ and then T is a bounded operator with || T || ≤ k(λ₁). Since, by Theorem 1, the constant factor in (14) is the best possible, we have || T || = k(λ₁).

• (ii) Define a half-discrete Hilbert's operator $\tilde{T}:l_{q,\Psi }\to L_{q,{\Phi }^{1-q}}\left(b,c\right)$ as: for a ∈ l _{q Ψ} , we define $\tilde{T}a\in L_{q,{\Phi }^{1-q}}\left(b,c\right)$, satisfying

  $$\tilde{T}a\left(x\right)=\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n,x\in \left(b,c\right).$$

Then, by (15), it follows $\left|\right|\tilde{T}a|{|}_{q,{\Phi }^{1-q}}\le k\left({\lambda }_1\right)|\left|a\right|{|}_{q,\Psi }$ and then $\tilde{T}$ is a bounded operator with $\left|\right|\tilde{T}\left|\right|\le k\left({\lambda }_1\right)$. Since, by Theorem 1, the constant factor in (15) is the best possible, we have $\left|\right|\tilde{T}\left|\right|=k\left({\lambda }_1\right)$.

In the following theorem, for 0 < p < 1, or p < 0, we still use the formal symbols of $\left|\right|f|{|}_{p,\tilde{\Phi }}$ and ||a|| _q ,_Ψ and so on. ■

Theorem 2 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree -λ in $R_{+}^2$, u(x)(x ∈ (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y ∈ [n₀, ∞), n₀ ∈ N) are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, k(λ₁) ∈ R₊, θ_λ(x) ∈ (0, 1), k(λ₁)(1 - θ_λ(x)) < ϖ(x) < k(λ₁)(x ∈ (b, c)). If 0 < p < 1, $\frac{1}{p}+\frac{1}{q}=1$, f(x), a _n ≥ 0, $\tilde{\Phi }\left(x\right):=\left(\mathsf{\text{1}}-{\theta }_{\lambda }\left(x\right)\right)\Phi \left(x\right)\left(x\in \left(b,c\right)\right)$, $0<\left|\right|f|{|}_{q,\tilde{\Phi }}<\infty$ and 0 < ||a|| _q ,_Ψ < ∞. Then, we have the following equivalent inequalities:

$$\begin{array}{ll}\hfill I:& =\sum _{n=n_0}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right)a_{n}f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}\sum _{n=n_0}^{\infty }K\left(x,n\right)a_{n}f\left(x\right)\mathsf{\text{d}}x\\ >k\left({\lambda }_1\right)\left|\right|f\left|{|}_{p,\tilde{\Phi }}\right|\left|a\right|{|}_{q,\Psi },\end{array}$$

(21)

$$J:={\left\{\sum _{n=n_0}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^p\right\}}^{\frac{1}{p}}>k\left({\lambda }_1\right)\left|\right|f|{|}_{p,\tilde{\Phi }},$$

(22)

$$\tilde{L}:={\left\{\underset{b}{\overset{c}{\int }}{\left[\tilde{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^q\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}>k\left({\lambda }_1\right)\left|\right|a|{|}_{q,\Psi }.$$

(23)

Moreover, if $\frac{v^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge n_0\right)$ is decreasing and there exist constants δ, δ₀ > 0, such that ${\theta }_{\lambda }\left(x\right)=O\left(\frac{1}{{\left[u\left(x\right)\right]}^{\delta }}\right)\left(x\in \left(d,c\right)\right)$ and k(λ₁ - δ₀) ∈ R₊, then the constant factor k(λ₁) in the above inequalities is the best possible.

Proof. In view of (9) and the reverse of (11), for ϖ(x) > k(λ₁)(1 - θ_λ(x)), we have (22). By the reverse Hölder's inequality, we have

$$I=\sum _{n=n_0}^{\infty }\left[{\Psi }^{\frac{-1}{q}}\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]\left[{\Psi }^{\frac{1}{q}}\left(n\right)a_n\right]\ge J\left|\right|a|{|}_{q,\Psi }.$$

(24)

Then, by (22), we have (21). On the other hand, assuming that (21) is valid, setting a _n as Theorem 1, then J^p-1 = ||a|| _{q Ψ}. By the reverse of (11), we find J > 0. If J = ∞, then (24) is naturally valid; if J < ∞, then by (21), we have

$$\left|\right|a|{|}_{q,\Psi }^q=J^p=I>k\left({\lambda }_1\right)|\left|f\right|{|}_{p,\tilde{\Phi }}\left|\right|a|{|}_{q,\Psi },|\left|a\right|{|}_{q,\Psi }^{q-1}=J>k\left({\lambda }_1\right)\left|\right|f|{|}_{p,\tilde{\Phi }},$$

and we have (22) which is equivalent to (21).

In view of (9) and the reverse of (12), for [ϖ(x)]^1-q> [k(λ₁)(1 - θ _λ (x))]^1-q(q < 0), we have (23). By the reverse Hölder's inequality, we have

$$I=\underset{b}{\overset{c}{\int }}\left[{\tilde{\Phi }}^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\tilde{\Phi }}^{\frac{-1}{p}}\left(x\right)\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]\mathsf{\text{d}}x\ge \left|\right|f|{|}_{p,\tilde{\Phi }}\tilde{L}.$$

(25)

Then, by (23), we have (21). On the other hand, assuming that (21) is valid, setting

$$f\left(x\right):={\left[\tilde{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^{q-1},x\in \left(b,c\right),$$

then ${\tilde{L}}^{q-1}=\left|\right|f|{|}_{p,\tilde{\Phi }}$. By the reverse of (12), we find $\tilde{L}>0$. If $\tilde{L}=\infty$, then (23) is naturally valid; if $\tilde{L}<\infty$, then by (21), we have

$$\left|\right|f|{|}_{p,\tilde{\Phi }}^p={\tilde{L}}^q=I>k\left({\lambda }_1\right)|\left|f\right|{|}_{p,\tilde{\Phi }}\left|\right|a|{|}_{q,\Psi },|\left|f\right|{|}_{p,\tilde{\Phi }}^{p-1}=\tilde{L}>k\left({\lambda }_1\right)\left|\right|a|{|}_{q,\Psi },$$

and we have (23) which is equivalent to (21).    ■

Hence, inequalities (21), (22) and (23) are equivalent.

For 0 < ε < pδ₀, setting $\tilde{f}\left(x\right)$ and ${ã}_n$ as Theorem 1, if there exists a positive number k(≥ k(λ₁)), such that (21) is still valid as we replace k(λ₁) by k, then in particular, for q < 0, in view of (9) and the conditions, we have

$$\begin{array}{ll}\hfill Ĩ:& =\underset{b}{\overset{c}{\int }}\sum _{n=n_0}^{\infty }K\left(x,n\right){ã}_n\tilde{f}\left(x\right)\mathsf{\text{d}}x>k\left|\right|\tilde{f}\left|{|}_{p,\tilde{\Phi }}\right|\left|ã\right|{|}_{q,\Psi }\\ =k{\left\{\underset{d}{\overset{c}{\int }}\left(1-O\left(\frac{1}{{\left[u\left(x\right)\right]}^{\delta }}\right)\right)\frac{u^{\prime }\left(x\right)\mathsf{\text{d}}x}{{\left[u\left(x\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=n_0}^{\infty }\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\\ =k{\left\{\frac{1}{\epsilon }-O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+\sum _{n=n_0+1}^{\infty }\frac{v^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\\ >k{\left\{\frac{1}{\epsilon }-O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+\underset{n_0}{\overset{\infty }{\int }}\frac{v^{\prime }\left(y\right)}{{\left[v\left(y\right)\right]}^{+1}}\mathsf{\text{d}}y\right\}}^{\frac{1}{q}}\\ =\frac{k}{\epsilon }{\left\{1-\epsilon O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right\}}^{\frac{1}{q}},\end{array}$$

(26)

$$\begin{array}{ll}\hfill Ĩ& =\sum _{n=n_0}^{\infty }{\left[v\left(n\right)\right]}^{{\lambda }_2-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\underset{d}{\overset{c}{\int }}K\left(x,n\right){\left[u\left(x\right)\right]}^{{\lambda }_1-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)\mathsf{\text{d}}x\\ \le \sum _{n=n_0}^{\infty }{\left[v\left(n\right)\right]}^{{\lambda }_2-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,n\right){\left[u\left(x\right)\right]}^{{\lambda }_1-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)\mathsf{\text{d}}x\\ \stackrel{t=u\left(x\right)/v\left(n\right)}{=}\sum _{n=n_0}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right)t^{\left({\lambda }_1-\frac{\epsilon }{p}\right)-1}\mathsf{\text{d}}t\\ \le k\left({\lambda }_1-\frac{\epsilon }{p}\right)\left[\frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+\underset{n_0}{\overset{\infty }{\int }}{\left[v\left(y\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(y\right)\mathsf{\text{d}}y\right]\\ =\frac{1}{\epsilon }k\left({\lambda }_1-\frac{\epsilon }{p}\right)\left[\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right].\end{array}$$

(27)

Since we have $k_{\lambda }\left(t,1\right)t^{{\lambda }_1-\frac{\epsilon }{p}-1}\le k_{\lambda }\left(t,1\right)t^{{\lambda }_1-{\delta }_0-1}$, t ∈ (0, 1] and

$$\underset{0}{\overset{1}{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-{\delta }_0-1}\mathsf{\text{d}}t\le k\left({\lambda }_1-{\delta }_0\right)<\infty ,$$

then by Lebesgue control convergence theorem [19], it follows

$$\begin{array}{ll}\hfill k\left({\lambda }_1-\frac{\epsilon }{p}\right)& \le \underset{1}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-1}\mathsf{\text{d}}t+\underset{0}{\overset{1}{\int }}{k}_{\lambda }\left(t,1\right)t^{{\lambda }_1-\frac{\epsilon }{p}-1}\mathsf{\text{d}}t\\ =k\left({\lambda }_1\right)+o\left(1\right)\left(\epsilon \to 0^{+}\right).\end{array}$$

By (26) and (27), we have

$$\left(k\left({\lambda }_1\right)+o\left(1\right)\right)\left[\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right]>k{\left\{1-\epsilon O\left(1\right)\right\}}^{\frac{1}{p}}{\left[\epsilon \frac{v^{\prime }\left(n_0\right)}{{\left[v\left(n_0\right)\right]}^{\epsilon +1}}+{\left[v\left(n_0\right)\right]}^{-\epsilon }\right]}^{\frac{1}{q}},$$

and then k(λ₁) ≥ k(ε → 0⁺). Hence, k = k(λ₁) is the best value of (21).

By the equivalence, the constant factor k(λ₁) in (22) and (23) is the best possible, otherwise we can imply a contradiction by (24) and (25) that the constant factor in (21) is not the best possible.    ■

In the same way, for p < 0, we also have the following theorem.

Theorem 3 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree -λ¸ in $R_{+}^2$, u(x)(x ∈ (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y ∈ [n₀, ∞), n₀ ∈ N) are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, ϖ(x) < k(λ₁) ∈ R ₊ (x ∈ (b, c)). If p < 0, $\frac{1}{p}+\frac{1}{q}=1$, f(x), a _n ≥ 0, 0 < || f || _{p Φ} < ∞ and 0 < ||a|| _q ,_Ψ < ∞. Then, we have the following equivalent inequalities:

$$\begin{array}{c}I:=\sum _{n=n_0}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right)a_{n}f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}\sum _{n=n_0}^{\infty }K\left(x,n\right)a_{n}f\left(x\right)\mathsf{\text{d}}x\\ >k\left({\lambda }_1\right)\left|\right|f\left|{|}_{p,\Phi }\right|\left|a\right|{|}_{q,\Psi },\end{array}$$

(28)

$$J:={\left\{\sum _{n=n_0}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^p\right\}}^{\frac{1}{p}}>k\left({\lambda }_1\right)\left|\right|f|{|}_{p,\Phi },$$

(29)

$$\tilde{L}:={\left\{\underset{b}{\overset{c}{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=n_0}^{\infty }K\left(x,n\right)a_n\right]}^q\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}>k\left({\lambda }_1\right)\left|\right|a|{|}_{q,\Psi }.$$

(30)

Moreover, if $\frac{v^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge n_0\right)$ is decreasing and there exists constant δ₀ > 0, such that k(λ₁ + δ₀) ∈ R₊, then the constant factor k(λ₁) in the above inequalities is the best possible.

Remark 2 (i) For n₀ = 1, b = 0, c = ∞, u(x) = v(x) = x, if

$$\varpi \left(x\right)=x^{{\lambda }_1}\sum _{n=1}^{\infty }{k}_{\lambda }\left(x,n\right)n^{{\lambda }_2-1}<k\left({\lambda }_1\right)\in R_{+}\left(x\in \left(0,\infty \right)\right),$$

then (13) reduces to (6). In particular, for $k_{\lambda }\left(x,n\right)=\frac{1}{{\left(x+n\right)}^{\lambda }}\left(\lambda ={\lambda }_1+{\lambda }_2,{\lambda }_1>0,0<{\lambda }_2\le 1\right)$,(6) reduces to (5).

• (ii) For n₀ = 1, b = 0, c = ∞, u(x) = v(x) = x^α(α > 0), $k_{\lambda }\left(x,y\right)=\frac{1}{{\left(max\left\{x,y\right\}\right)}^{\lambda }}\left(\lambda ,{\lambda }_1>0,0<\alpha {\lambda }_2\le 1\right)$, since

  $$f\left(x,y\right)=\frac{\alpha x^{\alpha {\lambda }_1}{y}^{\alpha {\lambda }_2-1}}{{\left(max\left\{x^{\alpha },y^{\alpha }\right\}\right)}^{\lambda }}=\left\{\begin{array}{c}\hfill \alpha x^{-\alpha {\lambda }_2}{y}^{\alpha {\lambda }_2-1},y\le x\hfill \\ \hfill \alpha x^{\alpha {\lambda }_1}{y}^{-\alpha {\lambda }_1-1},y>x\hfill \end{array}\right.$$

is decreasing for y ∈ (0, ∞) and strictly decreasing in an interval of (0, ∞), then by Condition (i), it follows

$$\begin{array}{l}\varpi (x)<\alpha x^{\alpha {\lambda }_1}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{{(max\{x^{\alpha },y^{\alpha }\})}^{\lambda }}{y}^{\alpha ({\lambda }_2-1)}{y}^{\alpha -1}\text{d}y}\\ \stackrel{t=(y/x{)}^{\alpha }}{=}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{t^{{\lambda }_2-1}}{{(max\{1,t\})}^{\lambda }}\text{d}t}=\frac{\lambda }{{\lambda }_1{\lambda }_2}=k({\lambda }_1).\end{array}$$

Since for $\delta =\frac{{\lambda }_1}{2}<{\lambda }_1$, $k_{\lambda }\left(t,1\right)=1\le \frac{1}{t^{\delta }}\left(t\in \left(0,1\right]\right)$, then by (13), we have the following inequality with the best constant factor $\frac{\lambda }{\alpha {\lambda }_1{\lambda }_2}$:

$$\begin{array}{c}\sum _{n=1}^{\infty }{a}_n\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(max\left\{x^{\alpha },n^{\alpha }\right\}\right)}^{\lambda }}f\left(x\right)\mathsf{\text{d}}x\\ <\frac{\lambda }{\alpha {\lambda }_1{\lambda }_2}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-\alpha {\lambda }_1\right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-\alpha {\lambda }_2\right)-1}{a}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

(31)

• (iii) For n₀ = 1, b = β, c = ∞, $u\left(x\right)=v\left(x\right)=x-\beta \left(0\le \beta \le \frac{1}{2}\right)$, $k_{\lambda }\left(x,y\right)=\frac{ln\left(x∕y\right)}{x^{\lambda }-y^{\lambda }}\left(\lambda ,{\lambda }_1>0,0<{\lambda }_2\le 1\right)$, since for any fixed x ∈ (β, ∞),

  $$f(x,y)=(x-\beta {)}^{{\lambda }_1}\frac{ln[(x-\beta )/(y-\beta )]}{{(x-\beta )}^{\lambda }-{(y-\beta )}^{\lambda }}{(y-\beta )}^{{\lambda }_2-1}$$

is decreasing and strictly convex for $y\in \left(\frac{1}{2},\infty \right)$, then by Condition (ii), it follows

$$\begin{array}{l}\varpi (x)<{(x-\beta )}^{{\lambda }_1}{\displaystyle \underset{\frac{1}{2}}{\overset{\infty }{\int }}\frac{ln[(x-\beta )/(y-\beta )]}{{(x-\beta )}^{\lambda }-{(y-\beta )}^{\lambda }}{(y-\beta )}^{{\lambda }_2-1}\text{d}y}\\ \stackrel{t=[(y-\beta )/(x-\beta {)]}^{\lambda }}{=}\frac{1}{{\lambda }^2}{\displaystyle \underset{{[\frac{\frac{1}{2}-\beta }{x-\beta }]}^{\lambda }}{\overset{\infty }{\int }}\frac{lnt}{t-1}{t}^{({\lambda }_2/\lambda )-1}\text{d}t}\\ \le \frac{1}{{\lambda }^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{lnt}{t-1}{t}^{({\lambda }_2/\lambda )-1}\text{d}t}={\left[\frac{\pi }{\lambda sin(\frac{\pi {\lambda }_1}{\lambda })}\right]}^2.\end{array}$$

Since for $\delta =\frac{{\lambda }_1}{2}<{\lambda }_1$, $k_{\lambda }\left(t,1\right)=\frac{lnt}{t^{\lambda }-1}\le \frac{M}{t^{\delta }}\left(t\in \left(0,1\right]\right)$, then by (13), we have the following inequality with the best constant factor ${\left[\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_1}{\lambda }\right)}\right]}^2$:

$$\begin{array}{c}\sum _{n=1}^{\infty }{a}_n\underset{\beta }{\overset{\infty }{\int }}\frac{ln\left[\left(x-\beta \right)∕\left(n-\beta \right)\right]}{{\left(x-\beta \right)}^{\lambda }-{\left(n-\beta \right)}^{\lambda }}f\left(x\right)\mathsf{\text{d}}x<{\left[\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_1}{\lambda }\right)}\right]}^2\\ \times {\left\{\underset{\beta }{\overset{\infty }{\int }}{\left(x-\beta \right)}^{p\left(1-{\lambda }_1\right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\right)-1}{a}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

(32)

• (iv) For n₀ = 1, b = 1 - β = γ, c = ∞, u(x) = v(x) = (x - γ), $\gamma \le 1-\frac{1}{8}\left[\lambda +\sqrt{\lambda \left(3\lambda +4\right)}\right]$, $k_{\lambda }\left(x,y\right)=\frac{1}{x^{\lambda }+y^{\lambda }}\left(0<\lambda \le 4\right)$, ${\lambda }_1={\lambda }_2=\frac{\lambda }{2}$, we have

  $$\begin{array}{ll}\hfill k\left(\frac{\lambda }{2}\right)& =\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right)t^{\frac{\lambda }{2}-1}\mathsf{\text{d}}t=\frac{2}{\lambda }\underset{0}{\overset{\infty }{\int }}\frac{1}{t^{\lambda }+1}\mathsf{\text{d}}{t}^{\frac{\lambda }{2}}\\ =\frac{2}{\lambda }arctan{t}^{\frac{\lambda }{2}}{|}_0^{\infty }=\frac{\pi }{\lambda }\in R_{+}\end{array}$$

and

$$f\left(x,y\right)=\frac{{\left(x-\gamma \right)}^{\frac{\lambda }{2}}{\left(y-\gamma \right)}^{\frac{\lambda }{2}-1}}{{\left(x-\gamma \right)}^{\lambda }+{\left(y-\gamma \right)}^{\lambda }}\left(x,y\in \left(\gamma ,\infty \right)\right).$$

Hence, v(y)(y ∈ [γ, ∞)) is strictly increasing with v(1 - β) = v(γ) = 0, and for any fixed x ∈ (γ, ∞), f(x, y) is smooth with

$${f^{\prime }}_y(x,y)=-\left(1+\frac{\lambda }{2}\right)\frac{{(x-\gamma )}^{\frac{\lambda }{2}}{(y-\gamma )}^{\frac{\lambda }{2}-2}}{{(x-\gamma )}^{\lambda }+{(y-\gamma )}^{\lambda }}+\frac{\lambda {(x-\gamma )}^{\frac{3\lambda }{2}}{(y-\gamma )}^{\frac{\lambda }{2}-2}}{[(x-\gamma {)}^{\lambda }+{(y-{\gamma }^{\lambda }]}^2}.$$

We set

$$R\left(x\right):=\underset{\gamma }{\overset{1}{\int }}f\left(x,y\right)\mathsf{\text{d}}y-\frac{1}{2}f\left(x,1\right)-\underset{1}{\overset{\infty }{\int }}\rho \left(y\right){f^{\prime }}_y\left(x,y\right)\mathsf{\text{d}}y.$$

(33)

For x ∈ (γ, ∞), 0 < λ ≤ 4, by (33) and the following improved Euler-Maclaurin summation formula [6]:

$$\frac{-1}{8}g(1)<{\displaystyle \underset{1}{\overset{\infty }{\int }}\rho (y)g(y)\text{d}y}<0((-{1)}^i{g}^{(i)}(y)>0,g^{(i)}(\infty )=0,i=0,1),$$

we have

$$\begin{array}{l}R(x)={\displaystyle \underset{\gamma }{\overset{1}{\int }}\frac{{(x-\gamma )}^{\frac{\lambda }{2}}{(y-\gamma )}^{\frac{\lambda }{2}-1}}{{(x-\gamma )}^{\lambda }+{(y-\gamma )}^{\lambda }}\text{d}y-\frac{1}{2}\frac{{(x-\gamma )}^{\frac{\lambda }{2}}{(1-\gamma )}^{\frac{\lambda }{2}-1}}{{(x-\gamma )}^{\lambda }+{(1-\gamma )}^{\lambda }}}\\ +\left(1+\frac{\lambda }{2}\right){\displaystyle \underset{1}{\overset{\infty }{\int }}\rho (y)\frac{{(x-\gamma )}^{\frac{\lambda }{2}}{(y-\gamma )}^{\frac{\lambda }{2}-2}}{{(x-\gamma )}^{\lambda }+{(y-\gamma )}^{\lambda }}\text{d}y}\\ -{\displaystyle \underset{1}{\overset{\infty }{\int }}\rho (y)\frac{\lambda {(x-\gamma )}^{\frac{3\lambda }{2}}{(y-\gamma )}^{\frac{\lambda }{2}-2}}{[(x-\gamma {)}^{\lambda }+{(y-{\gamma }^{\lambda }]}^2}\text{d}y}\\ >\frac{2}{\lambda }\arctan {\left(\frac{1-\gamma }{x-\gamma }\right)}^{\frac{\lambda }{2}}-\frac{{(1-\gamma )}^{\frac{\lambda }{2}-1}{(x-\gamma )}^{\frac{\lambda }{2}}}{2[(1-\gamma {)}^{\lambda }+(x-{\gamma }^{\lambda }]}\\ -\frac{1}{8}\left(1+\frac{\lambda }{2}\right)\frac{{(1-\gamma )}^{\frac{\lambda }{2}-2}{(x-\gamma )}^{\frac{\lambda }{2}}}{{(1-\gamma )}^{\lambda }+{(x-\gamma )}^{\lambda }}+0\\ =h(x):=\frac{2}{\lambda }\arctan {\left(\frac{1-\gamma }{x-\gamma }\right)}^{\frac{\lambda }{2}}\\ -\left[\frac{1-\gamma }{2}+\frac{1}{8}(1+\frac{\lambda }{2})\right]\frac{{(1-\gamma )}^{\frac{\lambda }{2}-2}{(x-\gamma )}^{\frac{\lambda }{2}}}{{(1-\gamma )}^{\lambda }+{(x-\gamma )}^{\lambda }}.\end{array}$$

Since for $\gamma \le 1-\frac{1}{8}\left[\lambda +\sqrt{\lambda \left(3\lambda +4\right)}\right]$, i.e. $1-\gamma \ge \frac{1}{8}\left[\lambda +\sqrt{\lambda \left(3\lambda +4\right)}\right]\left(0<\lambda \le 4\right)$,

$$\begin{array}{l}{h}^{\prime }(x)=\frac{{(1-\gamma )}^{\frac{\lambda }{2}}{(x-\gamma )}^{-\frac{\lambda }{2}-1}}{{(1-\gamma )}^{\lambda }+{(x-\gamma )}^{\lambda }}-\left[\frac{1-\gamma }{2}+\frac{1}{8}(1+\frac{\lambda }{2})\right]{(1-\gamma )}^{\frac{\lambda }{2}-2}\\ \times \left\{\frac{\lambda {(x-\gamma )}^{\frac{\lambda }{2}-1}}{2[(1-\gamma {)}^{\lambda }+{(x-\gamma )}^{\lambda }]}-\frac{\lambda {(x-\gamma )}^{\frac{3\lambda }{2}-1}}{{[(1-\gamma {)}^{\lambda }+{(x-\gamma )}^{\lambda }]}^2}\right\}\\ =-\left[{(1-\gamma )}^2-\frac{\lambda }{4}(1-\gamma )-\frac{\lambda }{16}(1+\frac{\lambda }{2})\right]\frac{{(1-\gamma )}^{\frac{\lambda }{2}-2}{(x-\gamma )}^{\frac{\lambda }{2}-1}}{{(1-\gamma )}^{\lambda }+{(x-\gamma )}^{\lambda }}\\ -\left[\frac{1-\gamma }{2}+\frac{1}{8}(1+\frac{\lambda }{2})\right]\frac{\lambda {(1-\gamma )}^{\frac{3\lambda }{2}-2}{(x-\gamma )}^{\frac{\lambda }{2}-1}}{{[(1-\gamma {)}^{\lambda }+{(x-\gamma )}^{\lambda }]}^2}<0,\end{array}$$

then h(x) is strictly decreasing and R(x) > h(x) > h(∞) = 0.

Then, by Condition (iii), it follows $\varpi \left(x\right)<k\left(\frac{\lambda }{2}\right)=\frac{\pi }{\lambda }\left(x\in \left(\gamma ,\infty \right)\right)$. For $\delta =0<\frac{\lambda }{2}$, it follows

$$k_{\lambda }\left(t,1\right)=\frac{1}{t^{\lambda }+1}\le 1=\frac{1}{t^{\delta }},t\in \left(0,\frac{1}{1-\gamma }\right),$$

and by (13), we have the following inequality with the best constant factor $\frac{\pi }{\lambda }$:

$$\begin{array}{l}\sum _{n=1}^{\infty }{a}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{1}{{\left(x-\gamma \right)}^{\lambda }+{\left(n-\gamma \right)}^{\lambda }}f\left(x\right)\mathsf{\text{d}}x\\ <\frac{\pi }{\lambda }{\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{p\left(1-\frac{\lambda }{2}\right)-1}{f}^p\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\infty }{\left(n-\gamma \right)}^{q\left(1-\frac{\lambda }{2}\right)-1}{a}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(34)

where $\gamma \le 1-\frac{1}{8}\left[\lambda +\sqrt{\lambda \left(3\lambda +4\right)}\right]\left(0<\lambda \le 4\right)$.