西南大学学报 (自然科学版)  2019, Vol. 41 Issue (12): 61-68.  DOI: 10.13718/j.cnki.xdzk.2019.12.009
0
Article Options
  • PDF
  • Abstract
  • Figures
  • References
  • 扩展功能
    Email Alert
    RSS
    本文作者相关文章
    洪勇
    曾志红
    欢迎关注西南大学期刊社
     

  • 齐次核的Hilbert型级数不等式成立的充要条件及其在算子理论中的应用    [PDF全文]
    洪勇1, 曾志红2    
    1. 广东白云学院 数学教研室, 广州 510450;
    2. 广东第二师范学院 学报编辑部, 广州 510303
    摘要:参数满足什么条件时,Hilbert型级数不等式 $ \sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} K(m, n) a_{m} b_{n} \leqslant M\left(\sum\limits_{m=1}^{\infty} m^{\alpha} a_{m}^{p}\right)^{\frac{1}{p}}\left(\sum\limits_{n=1}^{\infty} n^{\beta} b_{n}^{q}\right)^{\frac{1}{q}} $ 能够成立?而当Hilbert型级数不等式成立时,其常数因子又在什么条件下是最佳的?最佳常数因子的表达式是什么?这些问题的研究无疑是具有重要意义的.利用实分析的技巧及权函数方法,对具有齐次核的Hilbert型级数不等式的形式结构及参数关系进行了分析研究,得到其成立的充分必要条件和最佳常数因子的表达式.最后讨论了其结果在算子理论中的一些应用.
    关键词Hilbert型级数不等式    齐次核    充分必要条件    最佳常数因子    有界算子    

    p>1,$\frac{1}{p}+\frac{1}{q}=1 $K(xy)≥0,αβM为常数,am≥0,bn≥0.称不等式

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le M{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} $ (1)

    为Hilbert型级数不等式.当$ K(m, n)=\frac{1}{m+n} $时,有著名的Hilbert级数不等式[1]

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{a_m}{b_n}}}{{m + n}}} } \le \frac{{\rm{ \mathsf{ π} }}}{{\sin \left( {\frac{{\rm{ \mathsf{ π} }}}{p}} \right)}}{\left( {\sum\limits_{m = 1}^\infty {a_m^p} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {b_n^q} } \right)^{\frac{1}{q}}} $

    其中的常数因子$ \frac{\pi}{\sin \left(\frac{\pi}{p}\right)} $是最佳的.

    若对∀t>0,有K(txty)=tλK(xy),则称K(xy)是λ阶的齐次函数.对于具有齐次核的Hilbert型级数或积分不等式已有充分的研究[2-14],其研究主要围绕两个方面进行:一是对一些具体形式的核,得到相应的Hilbert型不等式,并讨论最佳常数因子问题;二是对一些具有某种特点的抽象核展开研究[15-20],这类研究更为深刻且更具重要意义,有广泛的应用价值.但Hilbert型不等式成立的条件是什么?换言之,是否对任何αβ,(1)式都能够成立?若不能,则αβ应满足什么条件?这个条件是否是充分必要的?当(1)式能够成立时,其最佳常数因子是什么?对这些问题的研究无疑是更深入的探讨,对全面深刻地探究Hilbert型不等式有重要意义.

    本文对具有齐次核的Hilbert型级数不等式的结构特征进行了探讨,得到了Hilbert型级数不等式能够成立的充分必要条件.最后还讨论了其结果在算子理论中的一些应用.

    引理1  设p>1,$ \frac{1}{p}+\frac{1}{q}=1$$ \frac{\alpha}{p}+\frac{\beta}{q}-(\lambda+1)=c $K(xy)≥0是λ阶的齐次可测函数,s+r=1(0<r<1),$K(1, t) t^{-\frac{\beta+1}{q}+c r} $$ K(t, 1) t^{-\frac{\alpha+1}{p}+c s} $都在(0,+∞)上递减,记

    $ {W_1}(\beta ,r) = \int_0^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q} + cr}}{\rm{d}}t $
    $ {W_2}(\alpha ,s) = \int_0^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t $

    W1(βr)=W2(αs),且

    $ {\omega _1}(m,\beta ,r) = \sum\limits_{n = 1}^\infty K (m,n){n^{ - \frac{{\beta + 1}}{q} + cr}} \le {m^{\lambda - \frac{{\beta + 1}}{q} + cr + 1}}{W_1}(\beta ,r) $ (2)
    $ {\omega _2}(n,\alpha ,s) = \sum\limits_{m = 1}^\infty K (m,n){m^{ - \frac{{\alpha + 1}}{p} + cs}} \le {n^{\lambda - \frac{{\alpha + 1}}{p} + cs + 1}}{W_2}(\alpha ,s) $ (3)

      由于K(xy)是λ阶的齐次函数,于是

    $ \begin{array}{l} {W_1}(\beta ,r) = \int_0^{ + \infty } {{t^{\lambda - \frac{{\beta + 1}}{q} + cr}}} K\left( {{t^{ - 1}},1} \right){\rm{d}}t = \int_0^{ + \infty } {{u^{ - \lambda + \frac{{\beta + 1}}{q} - cr - 2}}} K(u,1){\rm{d}}u = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t = {W_2}(\alpha ,s) \end{array} $

    又利用$ K(1, t) t^{-\frac{\beta+1}{q}+c r} $的递减性,有

    $ \begin{array}{l} {{\omega }_{1}}(m,\beta ,r) = {m^\lambda }\sum\limits_{n = 1}^\infty K \left( {1,\frac{n}{m}} \right){n^{ - \frac{{\beta + 1}}{q} + cr}} \le {m^{\lambda - \frac{{\beta + 1}}{q} + cr}}\int_0^{ + \infty } K \left( {1,\frac{u}{m}} \right){\left( {\frac{u}{m}} \right)^{ - \frac{{\beta + 1}}{q} + cr}}{\rm{d}}u = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{m^{\lambda - \frac{{\beta + 1}}{q} + cr + 1}}\int_0^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q} + cr}}{\rm{d}}t = {m^{\lambda - \frac{{\beta + 1}}{q} + cr + 1}}{W_1}(\beta ,r) \end{array} $

    故(2)式成立.

    同理可证(3)式也成立.

    1 主要定理及证明

    定理1  设p>1,$ \frac{1}{p}+\frac{1}{q}=1 $λαβ是常数,K(xy)>0是λ阶的齐次可测函数,$ \frac{\alpha}{p}+\frac{\beta}{q}-(\lambda+ 1)=c$s+r=1(0<r<1),$ K(1, t) t^{-\frac{\beta+1}{q}+c r} $$ K(t, 1) t^{-\frac{\alpha+1}{p}+c s} $在(0,+∞)上递减,且

    $ {W_1}(\beta ,r) = \int_0^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q} + cr}}{\rm{d}}t $

    收敛.那么:

    (ⅰ)当且仅当c≥0时,存在常数M>0,对∀am>0,∀bn≥0,有

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le M{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} $ (4)

    (ⅱ)设M0是(4)式的最佳常数因子,则$ {M_0} \le \mathop {\inf }\limits_{r \in (0, 1)} \left\{ {{W_1}(\beta , r)} \right\} $.当c=0时,有

    $ {M_0} = {W_0} = \int_0^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q}}}{\rm{d}}t $

      (ⅰ)设(4)式成立.如果$ \frac{\alpha}{p}+\frac{\beta}{q}-(\lambda+ 1)=c$<0,令

    $ {a_m} = {m^{\frac{{ - a - 1 + cps}}{p}}}\;\;\;\;m = 1,2, \cdots $
    $ {b_n} = {n^{\frac{{ - \beta - 1 + cqr}}{q}}}\;\;\;\;\;n = 1,2, \cdots $

    那么有

    $ \begin{array}{l} {\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} = {\left( {\sum\limits_{m = 1}^\infty {{m^{ - 1 + cps}}} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^{ - 1 + cqr}}} } \right)^{\frac{1}{q}}} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left( {1 + \int_1^\infty {{t^{ - 1 + cps}}} {\rm{d}}t} \right)^{\frac{1}{\rho }}}{\left( {1 + \int_1^\infty {{t^{ - 1 + cqr}}} {\rm{d}}t} \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left( {1 + \frac{1}{{ - cps}}} \right)^{\frac{1}{p}}}{\left( {1 + \frac{1}{{ - cqr}}} \right)^{\frac{1}{q}}} \end{array} $

    同时,由于$ K(t, 1) t^{-\frac{\alpha+1}{p}+c s} $在(0,+∞)上递减,又有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} = \sum\limits_{n = 1}^\infty {{n^{\frac{{ - \beta - 1 + cqr}}{q}}}} \left( {\sum\limits_{m = 1}^\infty K (m,n){m^{ - \frac{{\alpha + 1}}{p} + cs}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {{n^{\lambda + \frac{{ - \beta - 1 + cqr}}{q}}}} \left( {\sum\limits_{m = 1}^\infty K \left( {\frac{m}{n},1} \right){m^{ - \frac{{\alpha + 1}}{p} + cs}}} \right) \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {{n^{ - 2}}} \left( {\int_1^{ + \infty } K \left( {\frac{u}{n},1} \right){{\left( {\frac{u}{n}} \right)}^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}u} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {{n^{ - 1}}} \left( {\int_{\frac{1}{n}}^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t} \right) \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{n = 1}^\infty {{n^{ - 1}}} \left( {\int_1^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t} \right) \end{array} $

    于是得到

    $ \sum\limits_{n = 1}^\infty {{n^{ - 1}}} \int_1^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t \le M{\left( {1 + \frac{1}{{ - cps}}} \right)^{\frac{1}{p}}}{\left( {1 + \frac{1}{{ - cqr}}} \right)^{\frac{1}{q}}} $

    但因$ \sum\limits_{n = 1}^\infty {{n^{ - 1}}} $发散到+∞,得到矛盾.因而c<0不能成立,故c≥0.

    反之,设c≥0.记

    $ a = \frac{{\alpha - cps + 1}}{{pq}}\quad b = \frac{{\beta - cqr + 1}}{{pq}} $

    根据Hölder不等式及引理1,有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} = \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\left[ {\frac{{{m^a}}}{{{n^b}}}{a_m}} \right]} } \left[ {\frac{{{n^b}}}{{{m^a}}}{b_n}} \right]K(m,n) \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left( {\sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{m^{ap}}}}{{{n^{bp}}}}a_m^pK(m,n)} } } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\frac{{{n^{bq}}}}{{{m^{aq}}}}b_n^qK(m,n)} } } \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left( {\sum\limits_{m = 1}^\infty {{m^{\frac{{a - cps + 1}}{q}}}} a_m^p{\omega _1}(m,\beta ,r)} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^{\frac{{\beta - cqr + 1}}{p}}}} b_n^q{\omega _2}(n,\alpha ,s)} \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{W_1}(\beta ,r){\left( {\sum\limits_{m = 1}^\infty {{m^{\alpha - cps}}} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^{\beta - cqr}}} b_n^q} \right)^{\frac{1}{q}}} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{W_1}(\beta ,r){\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{\rho }}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} \end{array} $

    于是得到

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le \mathop {\inf }\limits_{r \in (0,1)} \left\{ {{W_1}(\beta ,r)} \right\}{\left( {\sum\limits_{m = 1}^\infty {{m^a}} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} $ (5)

    任取$M \ge \mathop {\inf }\limits_{r \in (0, 1)} \left\{ {{W_1}(\beta , r)} \right\}$,均有(4)式成立.

    在定理1的条件下,Hilbert型级数不等式(1)的基本结构特征是$\frac{\alpha }{p} + \frac{\beta }{q} - (\lambda + 1) \ge 0$,即如果$ \frac{\alpha}{p}+\frac{\beta}{q}-(\lambda+1)<0 $,则(1)式永远不可能成立.

    (ⅱ)首先,根据(5)式可得$ {M_0} \le \mathop {\inf }\limits_{r \in (0, 1)} \left\{ {{W_1}(\beta , r)} \right\} $.

    c=0时,(5)式化为

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le {W_0}{\left( {\sum\limits_{m = 1}^\infty {{m^a}} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} $

    此时若W0不是最佳的,则M0W0,且

    $ \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le {M_0}{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} $ (6)

    对足够小的ε>0,取${a_m} = {m^{\frac{{ - a - 1 - |\lambda {|_\varepsilon }}}{p}}}(m = 1,2, \cdots )$$ b_{n}=n^{\frac{-\beta-1-|\lambda| \varepsilon}{q}}(n=1, 2, \cdots) $,有

    $ \begin{array}{l} {\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} = {\left( {\sum\limits_{m = 1}^\infty {{m^{ - 1 - |\lambda |\varepsilon }}} } \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^{ - 1 - |\lambda |\varepsilon }}} } \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1 + \sum\limits_{m = 2}^\infty {{m^{ - 1 - |\lambda |\varepsilon }}} \le 1 + \int_1^{ + \infty } {{t^{ - 1 - |\lambda |\varepsilon }}} {\rm{d}}t = 1 + \frac{1}{{|\lambda |\varepsilon }} \end{array} $ (7)

    同时还有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} = \sum\limits_{m = 1}^\infty {{m^{\frac{{ - a - 1 - |\lambda |\varepsilon }}{p}}}} \left( {\sum\limits_{n = 1}^\infty {{n^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(m,n)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{m = 1}^\infty {{m^{\lambda + \frac{{ - a - 1 - |\lambda |\varepsilon }}{p}}}} \left( {\sum\limits_{n = 1}^\infty {{n^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K\left( {1,\frac{n}{m}} \right)} \right) \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{m = 1}^\infty {{m^{ - 2 - |\lambda |\varepsilon }}} \int_1^{ + \infty } {{{\left( {\frac{u}{m}} \right)}^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K\left( {1,\frac{u}{m}} \right){\rm{d}}u = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{m = 1}^\infty {{m^{ - 1 - |\lambda |\varepsilon }}} \int_{\frac{1}{m}}^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t \end{array} $

    δ>0足够小,存在N,当mN时,$ \frac{1}{m}<\delta $.记

    $ M(N) = \sum\limits_{m = 1}^N {{m^{ - 1 - |\lambda |\varepsilon }}} \int_{\frac{1}{m}}^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t $

    那么有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \ge M(N) + \sum\limits_{m = N + 1}^\infty {{m^{ - 1 - |\lambda |\varepsilon }}} \int_{\frac{1}{m}}^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |t}}{q}}}} K(1,t){\rm{d}}t \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M(N) + \sum\limits_{m = N + 1}^\infty {{m^{ - 1 - |\lambda {|_\varepsilon }}}} \int_\delta ^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M(N) + \int_{N + 1}^{ + \infty } {{t^{ - 1 - |\lambda |\varepsilon }}} {\rm{d}}t\int_\delta ^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M(N) + \frac{1}{{|\lambda |\varepsilon }}{(N + 1)^{ - |\lambda |\varepsilon }}\int_\delta ^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t \end{array} $ (8)

    由(6),(7),(8)式,得到

    $ \varepsilon M(N) + \frac{1}{{|\lambda |}}{(N + 1)^{ - |\lambda |\varepsilon }}\int_\delta ^{ + \infty } {{t^{\frac{{ - \beta - 1 - |\lambda |\varepsilon }}{q}}}} K(1,t){\rm{d}}t \le {M_0}\left( {\varepsilon + \frac{1}{{|\lambda |}}} \right) $

    ε→0+,得到

    $ \int_\delta ^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q}}}{\rm{d}}t \le {M_0} $

    再令δ→0+,得到

    $ {W_0} = \int_0^{ + \infty } K (1,t){t^{ - \frac{{\beta + 1}}{q}}}{\rm{d}}t \le {M_0} $

    这与M0W0矛盾,故M0=W0.

    定理2  设p>1,$ \frac{1}{p}+\frac{1}{q}=1 $λαγ是常数,K(xy)>0是λ阶的齐次可测函数,$ \frac{\alpha-\gamma}{p}-(\lambda+1)=c $r+s=1(0<s<1),$ K(1, t) t^{\frac{\gamma+1}{p}-1+c r} $$ K(t, 1) t^{-\frac{\alpha+1}{p}+c s} $在(0,+∞)上递减,且

    $ {W_2}(\alpha ,s) = \int_0^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p} + cs}}{\rm{d}}t $

    收敛.那么:

    (ⅰ)当且仅当c≥0时,存在常数$ \widetilde{M}>0 $,对∀am≥0,有

    $ \sum\limits_{n = 1}^\infty {{n^\gamma }} {\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)^p} \le \tilde M\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p $ (9)

    (ⅱ)若$ \widetilde{M}_{0} $是(9)式的最佳常数因子,则$ {\widetilde M_0} \le \mathop {\inf }\nolimits_{s \in (0, 1)} \left\{ {W_2^{\frac{1}{\rho }}(\alpha , s)} \right\} $.当c=0时,

    $ {{\tilde M}_0} = {\left( {\int_0^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{p}}}{\rm{d}}t} \right)^p} $

      记$\beta=\frac{\gamma}{1-p} $,则由$ \frac{\alpha-\gamma}{p}-(\lambda+1)=c $,可得

    $ \frac{\alpha }{p} + \frac{\beta }{q} - (\lambda + 1) = c\;\;\;\;\;\frac{{\gamma + 1}}{p} - 1 + cr = - \frac{{\beta + 1}}{q} + cr $

    于是只需证明(9)式与(4)式等价即可.

    若存在常数$\widetilde M$使得(9)式成立,则由Hölder不等式,对∀am≥0,∀bn≥0,有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} = \sum\limits_{n = 1}^\infty {\left( {{n^{ - \frac{\beta }{q}}}\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)} \left( {{n^{\frac{\beta }{q}}}{b_n}} \right) \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left[ {\sum\limits_{n = 1}^\infty {{n^{(1 - p)\beta }}} {{\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)}^p}} \right]^{\frac{1 }{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left[ {\sum\limits_{n = 1}^\infty {{n^\gamma }} {{\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)}^p}} \right]^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\tilde M}^{\frac{1}{\rho }}}{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} \end{array} $

    $\widetilde{M}^{\frac{1}{p}}=M $,则得到(4)式.

    反之,若存在常数M使得(4)式成立,令

    $ {b_n} = {n^\gamma }{\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)^{p - 1}}\;\;\;\;n = 1,2, \cdots $

    则有

    $ \begin{array}{l} \sum\limits_{n = 1}^\infty {{n^\gamma }} {\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)^p} = \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty K } (m,n){a_m}{b_n} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left( {\sum\limits_{n = 1}^\infty {{n^\beta }} b_n^q} \right)^{\frac{1}{q}}} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)^{\frac{1}{p}}}{\left[ {\sum\limits_{n = 1}^\infty {{n^\gamma }} {{\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)}^p}} \right]^{\frac{1}{q}}} \end{array} $

    由此得到

    $ \sum\limits_{n = 1}^\infty {{n^\gamma }} {\left( {\sum\limits_{m = 1}^\infty K (m,n){a_m}} \right)^p} \le {M^p}\sum\limits_{m = 1}^\infty {{m^a}} a_m^p $

    $M^{p}=\widetilde{M} $,得到(9)式.

    2 在算子理论中的应用

    p>1,α是常数,am≥0(m=1,2,…),并记

    $ l_\alpha ^p = \left\{ {a = \left\{ {{a_m}} \right\}:{a_m} \ge 0,{{\left\| a \right\|}_{p,\alpha }} = {{\left( {\sum\limits_{m = 1}^\infty {{m^\alpha }} a_m^p} \right)}^{\frac{1}{p}}} < + \infty } \right\} $

    对∀a={am}(am≥0),定义级数算子T

    $ T{(a)_n} = \sum\limits_{m = 1}^\infty K (m,n){a_m}\;\;\;\;n = 1,2, \cdots $ (10)

    其中核K(mn)≥0.若存在常数M,对∀a={am}∈lαp,有

    $ {\left\| {T(a)} \right\|_{p,\gamma }} \le M{\left\| a \right\|_{p,\alpha }} $

    则称Tlαplγp的有界算子,此时T的(pp)-型范数定义为

    $ {\left\| T \right\|_{(p,p)}} = \mathop {\sup }\limits_{a \in l_\alpha^p} \frac{{{{\left\| {T(a)} \right\|}_{p,\gamma }}}}{{{{\left\| a \right\|}_{p,a}}}} $

    特别地,记‖ap,0=‖apl0p=lp.若Tlαp到自身的有界算子,则称Tlαp中的有界算子.

    命题1  设p>1,$ \frac{1}{p}+\frac{1}{q}=1 $λαγ是常数,K(xy)>0是λ阶的齐次可测函数,

    $ \frac{\alpha }{p} - \frac{\gamma }{p} - (\lambda + 1) = c\;\;\;\;r + s = 1\;\;\;\;0 < s < 1 $

    $ K(1, t) t^{\frac{\gamma+1}{p}-1+c r} $$ K(t, 1) t^{-\frac{\alpha+1}{p}+c s} $在(0,+∞)上递减,$ W_{2}(\alpha, s)=\int_{0}^{+\infty} K(t, 1) t^{-\frac{a+1}{p}+c s} \text{d}t$收敛,算子T由(10)式定义.那么

    (ⅰ) Tlαplγp的有界算子的充分必要条件是c≥0;

    (ⅱ) c≥0时,T的算子范数$||T||{_{(p,p)}}\mathop {\inf }\limits_{s \in (0,1)} \left\{ {{W_2}(\alpha ,s)} \right\} $.当c=0时,

    $ {\left\| T \right\|_{(p,p)}} = \int_0^{ + \infty } K (t,1){t^{ - \frac{{\alpha + 1}}{\rho }}}{\rm{d}}t $

      由定理2可证.

    注1  在命题1的条件下,当且仅当$\frac{\alpha}{p}-\frac{\gamma}{p}-(\lambda+1) \geqslant 0$,时T才是lαplγp的有界算子.特别地,Tlp中的有界算子,其基本特征是λ≤-1.

    命题2  设p>1,$ \frac{1}{p}+\frac{1}{q}=1 $λαγ是常数,满足

    $ \max \left\{ {2 + \frac{\gamma }{p} + \frac{\alpha }{q} - p,\frac{1}{{1 - p}}\left( {2 + \frac{\gamma }{p} + \frac{\alpha }{q} - p} \right)} \right\} < \lambda \le \min \left\{ {2 + \frac{\alpha }{q} + \frac{\gamma }{p},2 - \frac{q}{p}\left( {\frac{\alpha }{q} + \frac{\gamma }{p}} \right)} \right\} $ (11)

    定义算子T

    $ T{(a)_n} = \sum\limits_{m = 1}^\infty {\frac{1}{{{m^\lambda } + {n^\lambda }}}a_m} \;\;\;\;a = \left\{ {{a_m}} \right\},{a_m} \ge 0 $

    则当且仅当$ \lambda \geqslant 1+\frac{\gamma-\alpha}{p}$时,Tlαplγp的有界算子.

      首先由(11)式左边可知λ>0.设$K(x, y)=\frac{1}{x^{\lambda}+y^{\lambda}}$,则K(xy)是λ阶的齐次非负函数.又令

    $ f(t) = K(t,1){t^\sigma } = K(1,t){t^\sigma } = \frac{{{t^\sigma }}}{{{t^\lambda } + 1}}\;\;\;\;t \in (0, + \infty ) $

    求导数可得

    $ f'(t) = \frac{{{t^{\sigma - 1}}}}{{{{\left( {{t^\lambda } + 1} \right)}^2}}}\left[ {(\sigma - \lambda ){t^\lambda } + \sigma } \right] $

    易知σ<0时,f′(t)<0,从而f(t)在(0,+∞)上递减.根据(11)式右边,经计算可知

    $ - \frac{{\alpha + 1}}{p} + \frac{c}{p} \le 0\;\;\;\;\frac{{\gamma + 1}}{p} - 1 + \frac{c}{q} \le 0 $

    其中

    $ c = \frac{\alpha }{p} - \frac{\gamma }{p} - ( - \lambda + 1) $

    于是$ K(t, 1) t^{-\frac{\alpha+1}{p}+\frac{c}{p}} $$ K(1, t) t^{\frac{\gamma+1}{p}-1+\frac{c}{q}} $都在(0,+∞)上递减.

    再由(11)式左边,经计算可知

    $ \frac{1}{\lambda }\left( { - \frac{{\alpha + 1}}{p} + \frac{c}{p} + 1} \right) > 0\;\;\;\;1 - \frac{1}{\lambda }\left( { - \frac{{\alpha + 1}}{p} + \frac{c}{p} + 1} \right) > 0 $

    从而

    $ \int_0^{ + \infty } {K(t,1){t^{ - \frac{{\alpha + 1}}{p} + \frac{c}{p}}} = \frac{1}{\lambda }B\left( {\frac{1}{\lambda }\left( { - \frac{{\alpha + 1}}{p} + \frac{c}{p} + 1} \right),1 - \frac{1}{\lambda }\left( { - \frac{{\alpha + 1}}{p} + \frac{c}{p} + 1} \right)} \right)} $

    收敛.其中B(·,·)是Beta函数.

    在命题1中取$ r=\frac{1}{q}, s=\frac{1}{p} $,可知命题2成立.

    注2  在命题2中,取α=γ=0,则(11)式化为

    $ \max \left\{ {2 - p,\frac{{p - 2}}{{p - 1}}} \right\} < \lambda \le 2 $

    $ \lambda \geqslant 1+\frac{\gamma-\alpha}{p}$化为λ≥1.再注意到$ \max \left\{2-p, \frac{p-2}{p-1}\right\}<1 $,于是可知,当且仅当1≤λ≤2时,Tlp中的有界算子.

    注3  在命题2中,取α=(p-1)(1-λ),γ=λ-1,则可得到:当0<λ≤max{pq}时,有

    $ \sum\limits_{n = 1}^\infty {{n^{\lambda - 1}}} {\left( {\sum\limits_{m = 1}^\infty {\frac{{{a_m}}}{{{m^\lambda } + {n^\lambda }}}} } \right)^p} \le {\left( {\frac{{\rm{ \mathsf{ π} }}}{{\lambda \sin \frac{{\rm{ \mathsf{ π} }}}{p}}}} \right)^p}\sum\limits_{m = 1}^p {{m^{(p - 1)(1 - \lambda )}}} a_m^p $
    参考文献
    [1]
    HARDY G H, LITTLEWOOD J E, POLYA G. Inequalities[M]. Cambridge: Cambridge University Press, 1952.
    [2]
    洪勇. Hardy-Hilbert积分不等式的全方位推广[J]. 数学学报, 2001, 44(4): 619-626. DOI:10.3321/j.issn:0583-1431.2001.04.007
    [3]
    YANG B C. On a Hilbert-Hardy-Type Integral Operator and Application[J]. J Ineq Appl, 2010, 2010: 1-10.
    [4]
    ZHAO C J, DEBNATH L. Some New Inverse Type Hilbert Integral Inequalities[J]. J Math Anal Appl, 2001, 262(1): 411-418.
    [5]
    和炳, 杨必成. 一个核带超几何函数的0次齐次的Hilbert型积分不等式[J]. 数学的实践与认识, 2010, 40(18): 203-211.
    [6]
    洪勇. 涉及多个函数的Hardy型积分不等式[J]. 数学学报, 2006, 49(1): 39-44.
    [7]
    匡继昌. 常用不等式[M]. 济南: 山东科学技术出版社, 2004.
    [8]
    YANG B C. On the Norm of a Hilbert's Type Linear Operator and Applications[J]. J Math Anal Appl, 2007, 325(1): 529-541.
    [9]
    胡克. 解析不等式的若干问题[M]. 武汉: 武汉大学出版社, 2007.
    [10]
    ČIŽMESIJA A, KRNIČ M, PEČARIC J. General Hilbert-Type Inequalities with Non-Conjugate Exponents[J]. Math Inequal Appl, 2008(2): 237-269.
    [11]
    BMETĆ I, PEČARIĆ J. Generalization of Hilbert's Integral Inequality[J]. Mathematical Inequalities and Applications, 2004(2): 199-205. DOI:10.7153/mia-07-22
    [12]
    吕中学, 谢鸿政. 关于Hilbert积分不等式新的改进和推广[J]. 数学进展, 2003, 32(4): 419-424.
    [13]
    杨必成, 陈强. 一个较为精确的半离散非齐次核的Hilbert不等式[J]. 西南师范大学学报(自然科学版), 2013, 38(8): 29-34. DOI:10.3969/j.issn.1000-5471.2013.08.008
    [14]
    杨必成, 陈强. 一个核为双曲正割函数的半离散Hilbert型不等式[J]. 西南师范大学学报(自然科学版), 2015, 40(2): 26-32.
    [15]
    洪勇. 一类带齐次核的奇异重积分算子的范数及其应用[J]. 数学年刊(A辑), 2011, 32(5): 599-606.
    [16]
    洪勇. 带对称齐次核的级数算子的范数刻画及其应用[J]. 数学学报, 2008, 51(2): 365-370.
    [17]
    杨必成. 参量化的Hilbert型不等式研究综述[J]. 数学进展, 2009, 38(3): 257-268. DOI:10.11845/sxjz.2009.38.03.0257
    [18]
    KTNIĆ M, GAO M Z, PECĂRIČ J, et al. On the Best Constant in Hilbert's Inequality[J]. Math Inequal Appl, 2005(2): 317-329.
    [19]
    YANG B C, KRNIĆ M. Hilbert-Type Inequalities and Related Operators with Homogeneous Kernel of Degree 0[J]. Mathematical Inequalities and Applications, 2010(4): 817-839. DOI:10.7153/mia-13-59
    [20]
    杨必成. 算子范数与Hilbert型不等式[M]. 北京: 科学出版社, 2009.
    The Necessary and Sufficient Condition for Hilbert-Type Series Inequality with a Homogeneous Kernel to Hold and Its Applications in the Operator Theory
    HONG Yong1, ZENG Zhi-hong2    
    1. Department of Mathematics, Guangdong Baiyun University, Guangzhou 510450, China;
    2. Editorial Department of Journal, Guangdong Second Normal University, Guangzhou 510303, China
    Abstract: What conditions should the parameters satisfy for Hilbert's type series inequality $ \sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} K(m, n) a_{m} b_{n} \leqslant M\left(\sum\limits_{m=1}^{\infty} m^{\alpha} a_{m}^{p}\right)^{\frac{1}{p}}\left(\sum\limits_{n=1}^{\infty} n^{\beta} b_{n}^{q}\right)^{\frac{1}{q}} $ to hold? Under what conditions is the constant factor best when Hilbert's type series inequality is established? What is the expression of the best constant factor? The study of these problems is undoubtedly of great significance. In this paper, using the techniques of real analysis and the method of weight function, the authors study and analyze the formal structure and parameter relation of Hilbert's type series inequality with a homogeneous kernel, and obtain the necessary and sufficient condition for it to hold and the expression of its best constant factor. Finally, a discussion is presented of its applications in the operator theory.
    Key words: Hilbert's type series inequality    homogeneous kernel    necessary and sufficient condition    the best constant factor    bounded operator    
    X