若p > 1, $\frac{1}{p} + \frac{1}{q} = 1$, a_m, b_n≥0(m, n∈N={1, 2, …}), 0 < $\sum\limits_{m = 1}^\infty {{a_m}^p} $ < ∞, 0 < $\sum\limits_{n = 1}^\infty {{b_n}^p} $ < ∞, 则有如下具有最佳常数因子$\text{ }\!\!\pi\!\!\text{ }/\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{p} \right)$的Hardy-Hilbert不等式^[1]:

$ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{a_m}{b_n}}}{{m + n}}} } < \frac{\pi }{{\sin \left( {\frac{\pi }{p}} \right)}}{\left( {\sum\limits_{m = 1}^\infty {a_m^p} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {b_n^p} } \right)^{\frac{1}{q}}}. $

(1)

设{μ_m}_m^∞=1, {ν_n}_n^∞=1为正数列, U_m=$\sum\limits_{i=1}^{m}{{{\mu }_{i}}} $, V_n=$\sum\limits_{j=1}^{n}{{{\nu }_{j}}} $，则有式(1) 的推广式(在文献[2]定理321中, 置换μ_m^1/qa_m, ν_n^1/pb_n为a_m, b_n)：

$ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{a_m}{b_n}}}{{{U_m} + {V_n}}}} } < \frac{\pi }{{\sin \left( {\frac{\pi }{p}} \right)}}{\left( {\sum\limits_{m = 1}^\infty {\frac{1}{{\mu _m^{p - 1}}}a_m^p} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {\frac{1}{{v_n^{q - 1}}}b_n^p} } \right)^{\frac{1}{q}}}. $

(2)

当μ_i=ν_i=1(i∈N)时, 式(2) 变为式(1)(文献[2]并没有证明式(2) 及确定常数因子的最佳性).

文献[3]引入参数α, λ > 0, 将式(1) 推广为:若

$ \begin{array}{l} 0 < {\lambda _i} \le 1\left( {i = 1,2} \right),{\lambda _1} + {\lambda _2} = {\lambda _\alpha },{a_m},{b_n} \ge 0,\\ 0 < \sum\limits_{m = 1}^\infty {{m^{p\left( {1 - {\lambda _1}} \right) - 1}}a_m^p} < \infty ,0 < \sum\limits_{n = 1}^\infty {{n^{q\left( {1 - {\lambda _2}} \right) - 1}}b_n^q} < \infty , \end{array} $

则有以下不等式:

$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{a_m}{b_n}}}{{{{\left( {{m^\alpha } + {n^\alpha }} \right)}^\lambda }}}} } < \frac{1}{\alpha }B\left( {\frac{{{\lambda _1}}}{\alpha },\frac{{{\lambda _2}}}{\alpha }} \right) \times \\ {\left( {\sum\limits_{m = 1}^\infty {{m^{p\left( {1 - {\lambda _1}} \right) - 1}}a_m^p} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^{q\left( {1 - {\lambda _2}} \right) - 1}}b_n^q} } \right)^{\frac{1}{q}}}, \end{array} $

(3)

这里, 常数因子$\frac{1}{\alpha }{\mathit{B}}\left( {\frac{{{\lambda _1}}}{\alpha }, \frac{{{\lambda _2}}}{\alpha }} \right)$是最佳值, B(u, v)为beta函数^[4]:

$ B\left( {u,v} \right): = \int_0^\infty {\frac{1}{{{{\left( {t + 1} \right)}^{u + v}}}}{t^{u - 1}}{\rm{d}}t} \left( {u,v > 0} \right). $

(4)

当α=λ=1, λ₁=$\frac{1}{q}$, λ₂= $\frac{1}{p}$时, 式(3) 变为式(1).文献[5-6]系统论述了参量化离散的Hilbert型不等式理论.文献[7]给出了式(2) 及式(3) 的具有最佳常数因子的推广式:

$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{a_m}{b_n}}}{{{{\left( {U_m^\alpha + V_n^\alpha } \right)}^\lambda }}}} } < \frac{1}{\alpha }B\left( {\frac{{{\lambda _1}}}{\alpha },\frac{{{\lambda _2}}}{\alpha }} \right) \times \\ \;\;\;\;\;\;\;{\left[ {\sum\limits_{m = 1}^\infty {\frac{{U_m^{p\left( {1 - {\lambda _1}} \right) - 1}}}{{\mu _m^{p - 1}}}a_m^p} } \right]^{\frac{1}{p}}}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - {\lambda _2}} \right) - 1}}}{{v_n^{q - 1}}}b_n^q} } \right]^{\frac{1}{q}}}. \end{array} $

(5)

当μ_i=ν_i=1(i∈N)时, 式(5) 变为式(3);当α=λ=1, λ₁=$\frac{1}{q}$, λ₂=$\frac{1}{p}$时, 式(5) 变为式(2).

关于半离散Hilbert不等式的一些最新结果, 可参阅文献[8-11].

下文引入独立参数, 应用权函数法及实分析技巧, 建立一个类似于式(5) 的具有最佳常数因子的半离散非齐次核Hilbe rt型式, 同时考虑其具有最佳常数因子的等价式.

引理1  设α, λ > 0, 0 < σ < αλ, σ≤1, k_α(σ):=$\frac{1}{\alpha }{\mathit{B}}\left( {\frac{{\alpha \lambda -\sigma }}{\alpha }, \frac{\sigma }{\alpha }} \right)$, μ(x) > 0在R₊连续, U(x):=$\int_0^x {\mu (t){\rm{d}}t} $有限(x≥0), {ν_n}_n=1^∞为正数列, V_n=$\sum\limits_{j = 1}^n {{\nu _j}} $.定义权函数：

$ {\omega _\alpha }\left( {\sigma ,x} \right): = \sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \frac{{{U^\sigma }\left( x \right)}}{{V_n^{1 - \sigma }}},x \in {{\bf{R}}_ + }, $

(6)

$ {{\tilde \omega }_\alpha }\left( {\sigma ,x} \right): = \int_0^\infty {\frac{{\mu \left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}\frac{{V_n^\sigma }}{{{U^{1 - \sigma }}\left( x \right)}}{\rm{d}}x} ,n \in {\bf{N}}, $

(7)

则有以下不等式:

$ {\omega _\alpha }\left( {\sigma ,x} \right) < {k_\alpha }\left( \sigma \right)\left( {x \in {{\bf{R}}_ + };0 < \sigma < \alpha \lambda ,\sigma \le 1} \right), $

(8)

$ {{\tilde \omega }_\alpha }\left( {\sigma ,n} \right) \le {k_\alpha }\left( \sigma \right)\left( {x \in {\bf{N}};0 < \sigma < \alpha \lambda } \right). $

(9)

证明  设v(t):=ν_n, t∈(n-1, n](n=1, 2, …), V(y):=$\int_0^y {\nu (t){\rm{d}}t} $ (y≥0)，则V(n)=V_n.当x > 0时, U′(x)=μ(x); 当y∈(n-1, n]时, V′(y)=ν(y)=ν_n.由V(y)(y > 0) 的严格递增性及1-σ≥0, 有

$ \begin{array}{l} {\omega _\alpha }\left( {\sigma ,x} \right) = \sum\limits_{n = 1}^\infty {\int_{n - 1}^n {\frac{{V'\left( t \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}\frac{{{U^\sigma }\left( x \right)}}{{V_n^{1 - \sigma }}}{\rm{d}}t} } < \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\int_{n - 1}^n {\frac{{V'\left( t \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right){V^\alpha }\left( t \right)} \right)}^\lambda }}}\frac{{{U^\sigma }\left( x \right)}}{{V_n^{1 - \sigma }}}{\rm{d}}t} } . \end{array} $

对上式做变换：u=U^α(x)V^α(t), 有

$ \begin{array}{l} {\omega _\alpha }\left( {\sigma ,x} \right) < \frac{1}{\alpha }\sum\limits_{n = 1}^\infty {\int_{{U^\alpha }\left( x \right){V^\alpha }\left( {n - 1} \right)}^{{U^\alpha }\left( x \right){V^\alpha }\left( n \right)} {\frac{{{u^{\left( {\sigma /a} \right) - 1}}}}{{{{\left( {1 + u} \right)}^\lambda }}}{\rm{d}}u} } = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{\alpha }\int_0^{{U^\alpha }\left( x \right){V^\alpha }\left( \infty \right)} {\frac{{{u^{\left( {\sigma /a} \right) - 1}}}}{{{{\left( {1 + u} \right)}^\lambda }}}{\rm{d}}u} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{\alpha }\int_0^\infty {\frac{{{u^{\left( {\sigma /a} \right) - 1}}}}{{{{\left( {1 + u} \right)}^\lambda }}}{\rm{d}}u} = {k_\alpha }\left( \sigma \right). \end{array} $

故式(8) 成立.

对式(7) 做变换：t=V_n^αU^α(x), 有

$ \begin{array}{l} {{\tilde \omega }_\alpha }\left( {\sigma ,n} \right) = \frac{1}{\alpha }\int_0^{V_n^\alpha {U^\alpha }\left( \infty \right)} {\frac{{{t^{\left( {\sigma /a} \right) - 1}}}}{{{{\left( {t + 1} \right)}^\lambda }}}{\rm{d}}t} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{\alpha }\int_0^\infty {\frac{{{t^{\left( {\sigma /a} \right) - 1}}}}{{{{\left( {t + 1} \right)}^\lambda }}}{\rm{d}}t} = {k_\alpha }\left( \sigma \right). \end{array} $

故式(9) 成立.证毕.

注1  若U(∞)=∞, 则${\tilde \omega _\alpha }$ (σ, n)=k_α(σ).

引理2  若{ν_n}_n^∞=1为递减数列, 且V(∞)=∞, 则对任意ε > 0, 有

$ \sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 + \varepsilon }}}} = \frac{1}{\varepsilon }\left( {1 + o\left( 1 \right)} \right)\left( {\varepsilon \to {0^ + }} \right). $

(10)

证明  因{ν_n}_n=1^∞具有递减性, ν_n≥ν_n+1, 且V(∞)=∞, 有

$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 + \varepsilon }}}} = \frac{1}{{v_1^\varepsilon }} + \sum\limits_{n = 2}^\infty {\int_{n - 1}^n {\frac{{V'\left( t \right)}}{{{V^{1 + \varepsilon }}\left( n \right)}}{\rm{d}}t} } \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{{v_1^\varepsilon }} + \sum\limits_{n = 2}^\infty {\int_{n - 1}^n {\frac{{V'\left( t \right)}}{{{V^{1 + \varepsilon }}\left( t \right)}}{\rm{d}}t} } = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{{v_1^\varepsilon }} + \sum\limits_{n = 2}^\infty {\int_{V\left( {n - 1} \right)}^{V\left( n \right)} {\frac{{{\rm{d}}u}}{{{u^{1 + \varepsilon }}}}} } = \frac{1}{{v_1^\varepsilon }} + \int_{{v_1}}^\infty {\frac{{{\rm{d}}u}}{{{u^{1 + \varepsilon }}}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{\varepsilon }\left[ {1 + \left( {\frac{1}{{v_1^\varepsilon }} + \frac{\varepsilon }{{v_1^\varepsilon }} - 1} \right)} \right], \end{array} $

$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 + \varepsilon }}}} \ge \sum\limits_{n = 1}^\infty {\int_n^{n + 1} {\frac{{{V_{n + 1}}}}{{V_n^{1 + \varepsilon }}}{\rm{d}}t} } = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\int_n^{n + 1} {\frac{{V'\left( t \right)}}{{V_n^{1 + \varepsilon }}}{\rm{d}}t} } > \sum\limits_{n = 1}^\infty {\int_n^{n + 1} {\frac{{{\rm{d}}V\left( t \right)}}{{{V^{1 + \varepsilon }}\left( t \right)}}} } = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {\int_{V\left( n \right)}^{V\left( {n + 1} \right)} {\frac{{{\rm{d}}u}}{{{u^{1 + \varepsilon }}}}} } = \int_{{v_1}}^\infty {\frac{{{\rm{d}}u}}{{{u^{1 + \varepsilon }}}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{\varepsilon }\left[ {1 + \left( {\frac{1}{{v_1^\varepsilon }} - 1} \right)} \right], \end{array} $

故式(10) 成立.证毕.

注2  若ε=a > 0, 则式(10) 可改写为$\sum\limits_{n = 1}^\infty {\frac{{{\nu _n}}}{{{V_n}^{1 + a}}}} $=O(1).

定理1  在引理1的条件下, 若p > 1, $\frac{1}{p} + \frac{1}{q} = 1$, f(x), a_n≥0, 0 < $\int_{0}^{\infty }{\frac{{{(U(x))}^{p(1-\sigma )-1}}}{{{\mu }^{p-1}}(x)}}{{f}^{p}}(x)\text{d}x$ < ∞, 0 < $\sum\limits_{n = 1}^\infty {\frac{{{V_n}^{q(1 - \sigma ) - 1}}}{{{\nu _n}^{q - 1}}}} {a_n}^q$ < ∞, 则有以下等价不等式:

$ \begin{array}{l} I: = \sum\limits_{n - 1}^\infty {\int_0^\infty {\frac{{{a_n}f\left( x \right)}}{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}}{\rm{d}}x} } < \frac{1}{\alpha }B\left( {\frac{{\alpha \lambda - \sigma }}{\alpha },\frac{\sigma }{\alpha }} \right) \times \\ \;\;\;{\left[ {\int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}a_n^q} } \right]^{\frac{1}{q}}}, \end{array} $

(11)

$ \begin{array}{l} {J_1}: = {\left\{ {\sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 - p\sigma }}}{{\left[ {\int_0^\infty {\frac{{f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } \right]}^p}} } \right\}^{\frac{1}{p}}} < \\ \;\;\;\;\;\frac{1}{\alpha }B\left( {\frac{{\alpha \lambda - \sigma }}{\alpha },\frac{\sigma }{\alpha }} \right){\left[ {\int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}, \end{array} $

(12)

$ \begin{array}{*{20}{c}} {{J_2}: = {{\left\{ {\int_0^\infty {\frac{{\mu \left( x \right)}}{{{{\left( {U\left( x \right)} \right)}^{1 - q\sigma }}}}\sum\limits_{n - 1}^\infty {{{\left[ {\frac{{{a_n}}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \right]}^q}} {\rm{d}}x} } \right\}}^{\frac{1}{q}}} < }\\ {\frac{1}{\alpha }B\left( {\frac{{\alpha \lambda - \sigma }}{\alpha },\frac{\sigma }{\alpha }} \right){{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}a_n^q} } \right]}^{\frac{1}{q}}}.} \end{array} $

(13)

证明  配方, 并由带权的Hölder不等式^[12], 有

$ \begin{array}{l} {\left[ {\int_0^\infty {\frac{{f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } \right]^p} = \\ \;\;\;\;\;\;\left[ {\int_0^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \left( {\frac{{V_n^{\left( {1 - \sigma } \right)/p}{\mu ^{1/q}}\left( x \right)}}{{{U^{\left( {1 - \sigma } \right)/q}}\left( x \right)v_n^{1/p}}}} \right) \times } \right.\\ \;\;\;\;\;\;{\left. {\left( {\frac{{{U^{\left( {1 - \sigma } \right)/q}}\left( x \right)v_n^{1/p}}}{{V_n^{\left( {1 - \sigma } \right)/p}{\mu ^{1/q}}\left( x \right)}}f\left( x \right)} \right){\rm{d}}x} \right]^p} \le \\ \;\;\;\;\;\;\left[ {\int_0^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \frac{{{U^{\left( {1 - \sigma } \right)p/q}}\left( x \right){v_n}}}{{V_n^{\left( {1 - \sigma } \right)}{\mu ^{p/q}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} \right] \times \\ \;\;\;\;\;\;{\left[ {\int_0^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \frac{{V_n^{\left( {1 - \sigma } \right)q/p}\mu \left( x \right)}}{{{U^{1 - \sigma }}\left( x \right)v_n^{q/p}}}{\rm{d}}x} \right]^{p/q}} = \\ \;\;\;\;\;\;{\left( {{{\tilde \omega }_\alpha }\left( {\sigma ,n} \right)} \right)^{p - 1}}\frac{{V_n^{1 - p\sigma }}}{{{v_n}}}\int_0^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} \times \\ \;\;\;\;\;\;\frac{{{{\left( {U\left( x \right)} \right)}^{\left( {1 - \sigma } \right)p/q}}{v_n}}}{{V_n^{1 - \sigma }{\mu ^{p/q}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x. \end{array} $

(14)

由式(9) 及Lebesgue逐项积分定理^[13], 有

$ \begin{array}{l} {J_1} \le {\left( {{k_\alpha }\left( \sigma \right)} \right)^{\frac{1}{q}}}\left[ {\sum\limits_{n = 1}^\infty {\int_0^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} } \times } \right.\\ \;\;\;\;\;\;\;\;{\left. {\frac{{{{\left( {U\left( x \right)} \right)}^{\left( {1 - \sigma } \right)\left( {p - 1} \right)}}{v_n}}}{{V_n^{1 - \sigma }{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} \right]^{\frac{1}{p}}} = \\ \;\;\;\;\;\;\;\;{\left( {{k_\alpha }\left( \sigma \right)} \right)^{\frac{1}{q}}}\left[ {\int_0^\infty {\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}} } \times } \right.\\ \;\;\;\;\;\;\;\;{\left. {\frac{{{{\left( {U\left( x \right)} \right)}^{\left( {1 - \sigma } \right)\left( {p - 1} \right)}}{v_n}}}{{V_n^{1 - \sigma }{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} \right]^{\frac{1}{p}}} = \\ \;\;\;\;\;\;\;\;{\left( {{k_\alpha }\left( \sigma \right)} \right)^{\frac{1}{q}}}{\left[ {\int_0^\infty {{\omega _\alpha }\left( {\sigma ,x} \right)\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}. \end{array} $

(15)

再由式(8)，有式(12).配方并由Hölder不等式^[12], 有

$ \begin{array}{l} I = \sum\limits_{n = 1}^\infty {\left[ {\frac{{v_n^{1/p}}}{{V_n^{\frac{1}{p} - \sigma }}}\int_0^\infty {\frac{{f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } \right]\left[ {\frac{{V_n^{\frac{1}{p} - \sigma }}}{{v_n^{1/p}}}{a_n}} \right]} \le \\ \;\;\;\;\;\;{J_1}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}a_n^q} } \right]^{\frac{1}{q}}}. \end{array} $

(16)

由式(12), 有式(11).反之, 设式(11) 成立.置

$ {a_n} = \frac{{{v_n}}}{{V_n^{1 - p\sigma }}}{\left[ {\int_0^\infty {\frac{{f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } \right]^{p - 1}},n \in {\bf{N}}, $

则J₁=${[\sum\limits_{n = 1}^\infty {\frac{{{V_n}^{q(1-\sigma )-1}}}{{{\nu _n}^{q-1}}}} {a_n}^q]^{\frac{1}{p}}}$.若J₁=0, 则式(12) 自然成立; 若J=∞, 因ω_α(σ, x) < k_α(σ), 则式(15) 不成立.可设0 < J₁ < ∞.由式(11), 有

$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}a_n^q} = J_1^p = I < \\ {k_\alpha }\left( \sigma \right){\left[ {\int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}b_n^q} } \right]^{\frac{1}{q}}},\\ {J_1}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}a_n^q} } \right]^{\frac{1}{q}}} < \\ \;\;\;\;{k_\alpha }\left( \sigma \right){\left[ {\int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{f^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}. \end{array} $

故式(12) 成立, 且与式(11) 等价.

同理可证式(13) 成立, 且其与式(11) 等价.因而式(11)、(12) 与式(13) 齐等价.证毕.

定理2  在定理1的条件下, 若{ν_n}_n=1^∞为递减数列, V(∞)=∞及U(∞)=∞, 则式(11)~(13) 的常数因子k_α(σ)都为最佳值.

证明  对任意0 < ε < p(αλ-σ), 设$\tilde{\sigma }=\sigma +\frac{\varepsilon }{p}(<\alpha \lambda )$, $\tilde f(x) = {U^{\tilde \sigma - 1}}(x)\mu (x)$, x∈(0, U^-1(1)); $\tilde f(x) = 0$, x∈[U^-1(1), ∞), ${\tilde a_n} = {V_n}^{\tilde \sigma- \varepsilon- 1}{\nu _n}$, 则由式(10) 及引理1、引理2的注, 有

$ \begin{array}{l} \int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}} {{\tilde f}^p}\left( x \right){\rm{d}}x = \\ \;\;\;\;\;\;\;\;\;\;\;\;\int_0^{{U^{ - 1}}\left( 1 \right)} {{{\left( {U\left( x \right)} \right)}^{ - 1 + \varepsilon }}{\rm{d}}U\left( x \right)} = \frac{1}{\varepsilon },\\ \;\;\;\;\;\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}\tilde a_n^q} = \sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 + \varepsilon }}}} = \frac{1}{\varepsilon }\left( {1 + o\left( 1 \right)} \right),\\ \tilde I: = \sum\limits_{n = 1}^\infty {\int_0^\infty {\frac{{{{\tilde a}_n}f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } = \\ \sum\limits_{n = 1}^\infty {\left[ {\int_0^{{U^{ - 1}}\left( 1 \right)} {\frac{{\mu \left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}\frac{{V_n^{\tilde \sigma }}}{{{{\left( {U\left( x \right)} \right)}^{1 - \tilde \sigma }}}}{\rm{d}}x} } \right]\frac{{{v_n}}}{{V_n^{\varepsilon + 1}}}} = \\ \sum\limits_{n = 1}^\infty {\left[ {{{\tilde \omega }_\alpha }\left( {\tilde \sigma ,n} \right) - \int_{{U^{ - 1}}\left( 1 \right)}^\infty {\frac{{\mu \left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}\frac{{V_n^{\tilde \sigma }}}{{{{\left( {U\left( x \right)} \right)}^{1 - \tilde \sigma }}}}{\rm{d}}x} } \right]\frac{{{v_n}}}{{V_n^{\varepsilon + 1}}}} \ge \\ \sum\limits_{n = 1}^\infty {\left[ {{k_\alpha }\left( {\tilde \sigma } \right) - \int_{{U^{ - 1}}\left( 1 \right)}^\infty {\frac{1}{{{U^{\alpha \lambda }}\left( x \right)V_n^{\alpha \lambda }}}\frac{{V_n^{\tilde \sigma }}}{{{{\left( {U\left( x \right)} \right)}^{1 - \tilde \sigma }}}}{\rm{d}}U\left( x \right)} } \right]} \frac{{{v_n}}}{{V_n^{\varepsilon + 1}}} = \\ {k_\alpha }\left( {\tilde \sigma } \right)\left[ {\sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{1 + \varepsilon }}}} - \frac{1}{{\left( {\alpha \lambda - \tilde \sigma } \right){k_\alpha }\left( {\tilde \sigma } \right)}}\sum\limits_{n = 1}^\infty {\frac{{{v_n}}}{{V_n^{\left( {\varepsilon /q} \right) + \alpha \lambda - \sigma + 1}}}} } \right] = \\ \frac{1}{\varepsilon } \cdot \frac{1}{\alpha }B\left( {\frac{{\alpha \lambda - \tilde \sigma }}{\alpha },\frac{{\tilde \sigma }}{\alpha }} \right)\left( {1 + o\left( 1 \right) - \varepsilon O\left( 1 \right)} \right). \end{array} $

(17)

若用正常数K(≤k_α(σ))取代式(11) 的常数因子k_α(σ)后，式(11) 仍成立, 则有

$ \begin{array}{l} \varepsilon \sum\limits_{n = 1}^\infty {\int_0^\infty {\frac{{{{\tilde a}_n}\tilde f\left( x \right)}}{{{{\left( {1 + {U^\alpha }\left( x \right)V_n^\alpha } \right)}^\lambda }}}{\rm{d}}x} } < \\ \varepsilon K{\left[ {\int_0^\infty {\frac{{{{\left( {U\left( x \right)} \right)}^{p\left( {1 - \sigma } \right) - 1}}}}{{{\mu ^{p - 1}}\left( x \right)}}{{\tilde f}^p}\left( x \right){\rm{d}}x} } \right]^{\frac{1}{p}}}{\left[ {\sum\limits_{n = 1}^\infty {\frac{{V_n^{q\left( {1 - \sigma } \right) - 1}}}{{v_n^{q - 1}}}\tilde a_n^q} } \right]^{\frac{1}{q}}}. \end{array} $

代入式(17), 有

$ \begin{array}{l} \frac{1}{\alpha }B\left( {\frac{{\alpha \lambda - \tilde \sigma }}{\alpha },\frac{{\tilde \sigma }}{\alpha }} \right)\left( {1 + o\left( 1 \right) - \varepsilon O\left( 1 \right)} \right) < K{\left( {1 + o\left( 1 \right)} \right)^{\frac{1}{q}}},\\ 即有\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{k_\alpha }\left( \sigma \right) \le K\left( {\varepsilon \to {0^ + }} \right). \end{array} $

故K=k_α(σ)为式(11) 的最佳值.

式(12) 的常数因子必是最佳值.不然, 由式(16), 必导出式(11) 的常数因子亦非最佳值的矛盾.同理, 由等价性, 可证式(13) 的常数因子为最佳值.证毕.