L2上复变函数的傅里叶级数逼近

杨刚 谷懿

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L2上复变函数的傅里叶级数逼近

    作者简介: 杨 刚(1994−),男,四川人,硕士生,主要研究函数逼近论. E-mail:3020044274@qq.com;
    通讯作者: 谷懿, guyi@yun.edu.cn
  • 中图分类号: O174.41

The Fourier series approximation of complex functions in L2

    Corresponding author: GU Yi, guyi@yun.edu.cn
  • CLC number: O174.41

  • 摘要: 考虑定义在圆盘 $ {\rm{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\} $ 上解析且原点为其本性奇点的特殊复函数类的逼近. 得到最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 分别与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模及 $ {\cal{K}} $ 泛函的精确Jackson不等式. 然后,研究最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0, h} \right) $ 上的加权积分之间关系,得到相应的精确Jackson不等式. 最后,得到关于函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模,函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0, h} \right) $ 上的加权积分和 $ {\cal{K}} $ 泛函等函数类上的最佳逼近及 $ n $ 维宽度.
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    [4] Abilov V A, Abilova F V, Kerimov M K. Sharp estimates for the convergence rate of Fourier series of complex variable functions in $ {L_2}\left( {D, p\left( z \right)} \right) $ [J]. Computational Mathematics and Mathematical Physics, 2010, 50(6): 946-950. DOI:  10.1134/S0965542510060023.
    [5] Shabozov M S, Saidusaynov M S. Approximation of functions of a complex variable by Fourier sums in orthogonal systems in $ {L_2} $ [J]. Russian Mathematics, 2020, 64(6): 56-62. DOI:  10.3103/S1066369X20060080.
    [6] Vakarchuk S B. K-functionals and exact values of n-widths for several classes from $ {L_2} $ [J]. Mathematical Notes, 1999, 66(4): 404-408. DOI:  10.1007/BF02679087.
    [7] Vakarchuk S B. Mean approximation of functions on the real axis by algebraic polynomials with Chebyshev-Hermite weight and widths of function classes[J]. Mathematical Notes, 2014, 95(5-6): 599-614. DOI:  10.1134/S0001434614050046.
    [8] Vinogradov O L, Zhuk V V. The rate of decrease of constants in Jackson type inequalities in dependence of the order of the continuity modulus[J]. Journal of Mathematical ences, 2011, 178(2): 132-143. DOI:  10.1007/s10958-011-0532-2.
    [9] Babenko V F, Konareva S V. Jackson–Stechkin-Type inequalities for the approximation of elements of Hilbert spaces[J]. Ukrainian Mathematical Journal, 2019, 70(9): 1155- 1165. DOI:  10.1007/s11253-019-01573-3.
    [10] Esmaganbetov M G. Exact Jackson-Stechkin inequalities and diameters of classes of functions from $ {L_2}\left( {{{\bf{R}}^2}, {{\text{e}}^{ - {x^2} - {y^2}}}} \right) $ [J]. Russian Mathematics, 2007, 51(2): 1- 7. DOI:  10.3103/S1066369X07020016.
    [11] Trigub R M. Approximation of smooth functions and constants by polynomials with integer and natural coefficients[J]. Mathematical Notes, 2001, 70(1): 110- 122. DOI:  10.1023/A:1010282120209.
    [12] 杨智纯, 魏舟. 关于向量值函数Riemann积分的若干研究[J]. 云南大学学报: 自然科学版, 2019, 41(5): 876- 883. DOI:  10.7540/j.ynu.20180624. Yang Z C, Wei Z. On Riemann integration of vector-valued functions[J]. Journal of Yunnan University: Natural Sciences Edition, 2019, 41(5): 876- 883.
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出版历程
  • 收稿日期:  2021-01-01
  • 录用日期:  2021-07-25
  • 网络出版日期:  2021-10-14
  • 刊出日期:  2021-11-15

L2上复变函数的傅里叶级数逼近

    作者简介:杨 刚(1994−),男,四川人,硕士生,主要研究函数逼近论. E-mail:3020044274@qq.com
    通讯作者: 谷懿, guyi@yun.edu.cn
  • 云南大学 数学与统计学院,云南 昆明 650500

摘要: 考虑定义在圆盘 $ {\rm{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\} $ 上解析且原点为其本性奇点的特殊复函数类的逼近. 得到最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 分别与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模及 $ {\cal{K}} $ 泛函的精确Jackson不等式. 然后,研究最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0, h} \right) $ 上的加权积分之间关系,得到相应的精确Jackson不等式. 最后,得到关于函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模,函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0, h} \right) $ 上的加权积分和 $ {\cal{K}} $ 泛函等函数类上的最佳逼近及 $ n $ 维宽度.

English Abstract

  • 精确的Jackson不等式研究已经有50余年的历史,为了叙述已有的研究结果,先叙述一些相应的记号;$ {L_2}\left( {\text{U}} \right) $ 表示自变量在区域 $ {\text{U}} $ 内平方可积的复函数空间. 在 $ {L_2}\left( {\text{U}} \right) $ 上定义函数 $ f $ 范数为

    $ \qquad {\left\| f \right\|_2}: = {\left( {\frac{1}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {\left| {f\left( z \right)} \right|{\text{d}}\sigma }} \right)^{\frac{1}{2}}} , $

    这里 $ \left| {f\left( z \right)} \right| $ 表示函数 $ f $ 的模,$ {P_n} $ 为次数不高于 $ n $ 的代数多项式函数空间. 函数空间 $ {P_n} $ 对函数 $ f $ 的最佳逼近表达式为

    $ \qquad {E_{n - 1}}{\left( f \right)_2}: = \inf \left\{ {{{\left\| {f - g} \right\|}_2}:\forall g \in {P_{n - 1}}} \right\} . $

    在复数域上 $ {\cal{K}} $ 泛函[1]表达式为

    $ \qquad {{\cal{K}}_m}{\left( {f,{t^m}} \right)_2}: = {\cal{K}}\left( {f;{t^m};{L_2};L_2^{\left( m \right)}} \right) = \inf \left\{ {{{\left\| {f - g} \right\|}_2} + {t^m}{{\left\| {{z^m}{g^{\left( m \right)}}} \right\|}_2}:g \in L_2^{\left( m \right)}} \right\} . $

    Saidusaynov[1]和Shabozov等[2]研究 $ {L_2}\left( {\text{U}} \right) $$ {\text{U}} = \left\{ {z \in {\bf{C}}:\left| z \right| < 1} \right\} $ 上解析函数的逼近问题,得到 $ {\cal{K}} $ 泛函与最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 的精确Jackson不等式和关于 $ {\cal{K}} $ 泛函的函数类的宽度.

    Shabozov等[2]定义 $ {L_2}\left( {\text{U}} \right) $$ {\text{U}} = \left\{ {z \in {\bf{C}}:\left| z \right| < 1} \right\} $ 上解析函数的平移算子 $ {F_h} $

    $ \qquad {F_h}\left( f \right) = \frac{1}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {f\left( \xi \right)} T\left( {z,\xi ,h} \right){\text{d}}\sigma = \sum\limits_{k = 0}^\infty {{c_k}} \left( f \right){z^k}{\left( {1 - h} \right)^k} , $

    其中函数 $ T\left( {\xi ,\eta ,h} \right) = \displaystyle\sum\limits_{k = 0}^\infty {{\varphi _k}} \left( \xi \right)\overline {{\varphi _k}} \left( \eta \right){\left( {1 - h} \right)^k} $$ \left\{ {{\varphi _k}\left( z \right) = {z^k}} \right\}_{k = 0}^\infty $ 为正交系统. 得到函数 $ f $ 的一阶差分为

    $ \qquad \Delta _h^1f\left( z \right) = {F_h}\left( {f\left( z \right)} \right) - f\left( z \right) = \sum\limits_{k = 0}^\infty {{c_k}} \left( f \right){z^k}\left( {{{\left( {1 - h} \right)}^k} - 1} \right) . $

    计算可得函数 $ f $$ m $ 阶差分为

    $ \qquad \Delta _h^mf\left( z \right) = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{c_k}} \left( f \right){z^k}{\left( {1 - {{\left( {1 - h} \right)}^k}} \right)^m} . $

    根据平移算子定义了 $ m $ 阶连续模

    $ \qquad {\Omega }_{m}{\left(f,t\right)}_{2}:=\sup\left\{{\Vert {\Delta }_{h}^{m}f(\cdot )\Vert }_{2}:0 < h\leqslant t\right\} . $

    给出最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模 $ {\Omega _m}{\left( {{z^r}{f^{\left( r \right)}},t} \right)_2} $ 的精确Jackson不等式和关于 $ {\Omega _m}{\left( {{z^r}{f^{\left( r \right)}},t} \right)_2} $ 的函数类的宽度(也可参考文献[3]~[4]). Shabozov等[5]得到分别关于 $ m $ 阶连续模和 $ {\cal{K}} $ 泛函这2个函数类的最佳逼近. Abilov[4]等得到关于最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $$ m $ 阶连续模的Jackson不等式的逆定理.

    在实数域 $ {\bf{R}} $ 上,Vakarchuk[6]给出在 $ {L_2}\left( {\bf{R}} \right) $ 上的关于 $ {\cal{K}} $ 泛函的函数类宽度. Vakarchuk[7]和Vinogeadov等[8]$ {L_2}\left( {\bf{R}} \right) $ 上得到最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 和连续模的精确Jackson不等式(也可参考文献[9]);Esmaganbetov[10]$ {L_2}\left( {{{\bf{R}}^2},{{\text{e}}^{ - {x^2} - {y^2}}}} \right) $ 上得到最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 与连续模的精确Jackson不等式以及函数类的宽度. Trigub[11]$ {L_2}[0, 1] $ 上用多项式函数逼近光滑函数得到最佳逼近的估计. 参考文献[12]~[13]讨论了函数以及算子的一些相关性质. 在文献[14]中,孙永生介绍了经典的精确Jackson不等式.

    • 基于以上研究成果,本文考虑 $ {L_2}\left( {\text{U}} \right) $${\text{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\}$ 上的函数空间,在 ${\text{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\}$ 上函数 $ f $ 解析且有 $f = \displaystyle\sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}}$ 的洛朗展式. 我们给出了平移算子,构造了关于函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模. 得到关于最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 分别与函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模和 $ {\cal{K}} $ 泛函的精确Jackson不等式;引入权函数 $ q\left( t \right) $,得到函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0,h} \right) $ 上的加权积分与最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 的精确Jackson不等式. 再分别定义了关于函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模、$ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0,h} \right) $ 上加权积分和 $ {\cal{K}} $ 泛函的函数类,最后得到了3种函数类的最佳逼近及宽度.

      我们取 ${\text{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\}$,序列函数 $ {\varphi _k} = {z^k},k \in {\bf{Z}} $,则

      $ \qquad \frac{1}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {\varphi {}_k}\left( z \right)\overline {{\varphi _l}\left( z \right)} {\text{d}}\sigma = \frac{1}{{\text{π}}}\int_{\frac{1}{2}}^1 {\int_0^{2\pi } {{r^{k + l + 1}}} } {{\text{e}}^{{\text{i}}\left( {k - l} \right)t}}{\text{d}}t{\text{d}}r = 0,k \ne l , $

      $ \qquad \frac{1}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {{{\left| {\varphi {}_k\left( z \right)} \right|}^2}}{\text{d}}\sigma = \frac{1}{{\text{π}}}\int_{\frac{1}{2}}^1 {\int_0^{2\pi } {{r^{2k + 1}}} } {\text{d}}t{\text{d}}r = \frac{1}{{k + 1}}\left( {1 - {{\left( {\frac{1}{2}} \right)}^{2k + 2}}} \right),k \ne - 1 , $

      $ \qquad \frac{{\text{1}}}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {{{\left| {\varphi {}_k\left( z \right)} \right|}^2}}{\text{d}}\sigma = \frac{1}{{\text{π}}}\int_{\frac{1}{2}}^1 {\int_0^{2\pi } {{r^{2k + 1}}} } {\text{d}}t{\text{d}}r = 2\ln 2,k = - 1 . $

      所以系统 $\varphi _k^ * \left( z \right) = {\left( {\dfrac{1}{{k + 1}}\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{2k + 2}}} \right)} \right)^{ - \frac{1}{2}}}{z^k}\left( {k \ne - 1} \right)\bigcup {\varphi _{ - 1}^ * \left( z \right)} = {\left( {2\ln 2} \right)^{ - \frac{1}{2}}}{z^{ - 1}}$ 是正交的.

      如果 $ f \in {L_2}\left( {\text{U}} \right) $,则我们取 $ f $ 在正交系统 $ \left\{ {\varphi _{ - k}^ * \left( z \right)} \right\}_{k = 0}^\infty $ 下的傅里叶级数几乎处处有

      $ \qquad f = \sum\limits_{k = 0}^\infty {{a_{ - k}}} \left( f \right)\varphi _{ - k}^ * \left( z \right) . $

      ${a_{ - k}}\left( f \right) = \dfrac{{\text{1}}}{{\text{π}}}\displaystyle\iint_{\left( {\text{U}} \right)} {f\left( z \right)}\overline {\varphi _{ - k}^ * \left( z \right)} {\text{d}}\sigma$ 为傅里叶系数. 令 $ A\left( {\text{U}} \right) $ 是在区域 $ {\text{U}} $ 的解析函数的集合,则本文考虑特殊复函数 $ f \in A\left( {\rm{U}} \right) $ 且洛朗展式为 $f = \displaystyle\sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}}$. 这里的 $ {c_{ - k}}\left( f \right) $$ f $ 的洛朗展式系数. 从而得出 $ {a_{ - k}}\left( f \right) $$ {c_{ - k}}\left( f \right) $ 的关系 ${a_{ - k}}\left( f \right) = {\left( {\dfrac{1}{{k + 1}}\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{2k + 2}}} \right)} \right)^{\frac{1}{2}}}{c_{ - k}}\left( f \right)\left( {k \ne - 1} \right)$${a_{ - 1}}\left( f \right) = {\left( {2\ln 2} \right)^{\frac{1}{2}}}{c_{ - 1}}\left( f \right)$. 根据2者系数的关系可以得到

      $ \qquad f = \sum\limits_{k = 0}^\infty {{a_{ - k}}} \left( f \right)\varphi _{ - k}^ * \left( z \right) = \sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}}, $

      $ \qquad \left\| f \right\|_2^2 = \sum\limits_{k = 2}^\infty {\frac{{{{\left| {c{}_{ - k}\left( f \right)} \right|}^2}\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{ - 2k + 2}}} \right)}}{{ - k + 1}}} + 2\ln 2{\left| {{c_{ - 1}}\left( f \right)} \right|^2} + {\left| {{c_0}\left( f \right)} \right|^2}. $

      对函数 $ f $$ r $ 阶导数,则有 ${f^{\left( r \right)}} = \displaystyle\sum\limits_{k = 1}^\infty {{c_{ - k}}} \left( f \right)\left( { - k} \right)\left( { - k - 1} \right) \cdots \left( { - k - r + 1} \right){z^{ - k - r}}$,令 ${b_{k,r}} =\left( { - k} \right)\left( { - k - 1} \right) \cdots \left( { - k - r + 1} \right)$,即 ${f^{\left( r \right)}} = \displaystyle\sum\limits_{k = 1}^\infty {{b_{k,r}}{c_{ - k}}} \left( f \right){z^{ - k - r}},{z^r}{f^{\left( r \right)}} = \displaystyle\sum\limits_{k = 1}^\infty {{b_{k,r}}{c_{ - k}}} \left( f \right){z^{ - k}}$. 令 ${S_{n - 1}} = \displaystyle\sum\limits_{k = 0}^{n - 1} {{c_{ - k}}} \left( f \right){z^{ - k}},{c_{ - k}}\left( f \right) \in {\bf{C}}$${P_{n - 1}} = \displaystyle\sum\limits_{k = 0}^{n - 1} {{d_{ - k}}} {z^{ - k}},{d_{ - k}} \in $$ {\bf{C}}$,则

      $ \qquad {E_{n - 1}}{\left( f \right)_2} = \inf \left\{ {{{\left\| {f - {p_{n - 1}}} \right\|}_2}:{p_{n - 1}} \in {P_{n - 1}}} \right\} = \inf \left\{ {{{\left\| {f - {p_{n - 1}}} \right\|}_2}} \right\} = {\left\| {f - {S_{n - 1}}} \right\|_2} , $

      这里 $ {E_{n - 1}}{\left( f \right)_2} $ 表示用 $ {P_{n - 1}} $ 函数来逼近 $ f $ 得到的最佳逼近. $ {P_{n - 1}} $ 表示 $ {z^{ - 1}} $ 次数小于等于 $ n - 1 $ 的函数全体.

      定义 1 函数

      $ \qquad T\left( {\xi ,\eta ,h} \right) = \sum\limits_{k = 0}^\infty {\varphi _{ - k}^ * \left( \xi \right)} \overline {\varphi _{ - k}^ * \left( \eta \right)} {\left( {1 + h} \right)^{ - k}} . $

      定义 2 平移算子 $ {F_h} $ 的表达式为

      $ \qquad {F_h}\left( {f\left( z \right)} \right) = \frac{1}{{\text{π}}}\iint_{\left( {\text{U}} \right)} {f\left( \xi \right)} T\left( {z,\xi ,h} \right){\text{d}}\sigma = \sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}}{\left( {1 + h} \right)^{ - k}} , $

      $ {F_h}\left( {f\left( z \right)} \right) = \dfrac{1}{{\text{π}}}\displaystyle\iint_{\left( {\text{U}} \right)} {f\left( \xi \right)}T\left( {z,\xi ,h} \right){\text{d}}\sigma $ 具有下列性质:

      (1)${F_h}\left( {{f_1} + {f_2}} \right) = {F_h}\left( {{f_1}} \right) + {F_h}\left( {{f_2}} \right) ; $

      (2)${F_h}\left( {\alpha f} \right) = \alpha {F_h}\left( f \right) ;$

      (3)${\left\| {{F_h}\left( f \right)} \right\|_2} \leqslant \left\| f \right\| ; $

      (4)${\left\| {{F_h}\left( f \right) - f} \right\|_2} \to 0 (h \to {0^ + } ).$

      定义 3 函数 $ f $$ m $ 阶连续模为

      $ \qquad {\Omega }_{m}{\left(f,t\right)}_{2}:=\sup\left\{{\Vert {\Delta }_{h}^{m}f(\cdot )\Vert }_{2}:0 < h\leqslant t\right\} . $

      根据(1)和(4)式可得到函数 $ f $ 的一阶差分为 $\Delta _h^1f\left( z \right) = {F_h}\left( {f\left( z \right)} \right) - f\left( z \right) = \displaystyle\sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}}\left( {{{\left( {1 + h} \right)}^{ - k}} - 1} \right)$,计算可得函数 $ f $$ m $ 阶差分为

      $ \qquad \Delta _h^mf\left( z \right) = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{c_{ - k}}} \left( f \right){z^{ - k}}{\left( {1 - {{\left( {1 + h} \right)}^{ - k}}} \right)^m} . $

      $ {L_2} $ 空间的正交系统下得出函数 $ f $$ m $ 阶差分的范数为

      $ \qquad {\Vert {\Delta }_{h}^{m}f(\cdot )\Vert }_{2}^{2}={\sum\limits_{k=2}^{\infty }{\left[1-{\left(1+h\right)}^{-k}\right]}^{2m}{\left|{c}_{-1}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}+2\mathrm{ln}2{\left[1-{\left(1+h\right)}^{-1}\right]}^{2m}{\left|{c}_{-1}\left(f\right)\right|}^{2} . $

      再根据(2)和(5)式得到函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模为

      $ \qquad {\Omega _m}{\left( {{z^r}{f^{\left( r \right)}},t} \right)_2} = {\left( {\sum\limits_{k = 2}^\infty {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - k}}} \right]}^{2m}}b_{k,r}^2{{\left| {{c_{ - k}}\left( f \right)} \right|}^2}\frac{{{4^{k - 1}} - 1}}{{k - 1}}} + {{\left[ {1 - {{\left( {1 + t} \right)}^{ - 1}}} \right]}^{2m}}b_{1,r}^22\ln 2{{\left| {{c_{ - 1}}\left( f \right)} \right|}^2}} \right)^{\frac{1}{2}}} . $

      本文中 $ L_2^{\left( r \right)}: = L_2^{\left( r \right)}\left( {\text{U}} \right) \left( {{L_2}: = {L_2}\left( {\text{U}} \right)} \right) $,函数 $ f \in L_2^{\left( r \right)} $ 表示函数 $ f \in {L_2} $ 且函数 $ {z^r}{f^{\left( r \right)}} \in {L_2} $.

    • 定理 1 对于 $n\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1$,有

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{E_{n - 1}}{{\left( {{z^r}{f^{\left( r \right)}}} \right)}_2}}} = \frac{1}{{\left| {{b_{n,r}}} \right|}}.$

      证明 根据(3)式对最佳逼近的定义知 $ E_{n - 1}^2{\left( f \right)_2} = \displaystyle\sum\limits_{k = n}^\infty {\frac{{{4^{k - 1}} - 1}}{{k - 1}}} {\left| {{c_{ - k}}\left( f \right)} \right|^2} $,化简得到

      $ \qquad {E}_{n-1}^{2}{\left(f\right)}_{2}\leqslant {\sum\limits_{k=n}^{\infty }\frac{{b}_{k,r}^{2}}{{b}_{n,r}^{2}}\frac{{4}^{k-1}-1}{k-1}}{\left|{c}_{-k}\left(f\right)\right|}^{2}=\frac{1}{{b}_{n,r}^{2}}{\sum\limits_{k=n}^{\infty }{b}_{k,r}^{2}\frac{{4}^{k-1}-1}{k-1}}{\left|{c}_{-k}\left(f\right)\right|}^{2}=\frac{1}{{b}_{n,r}^{2}}{E}_{n-1}^{2}{\left({z}^{r}{f}^{\left(r\right)}\right)}_{2} , $

      因此,对 $ f \in {L_2} $ 得到(7)式右边的上界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{E_{n - 1}}{{\left( {{z^r}{f^{\left( r \right)}}} \right)}_2}}} \leqslant \frac{1}{{\left| {{b_{n,r}}} \right|}}. $

      又因为 $ {f_0} = {z^{ - n}} $,则有 $ {z^r}f_{_0}^{\left( r \right)} = {b_{n,r}}{z^{ - n}} $,根据(3)式对最佳逼近的定义知

      $ \qquad {E_{n - 1}}{\left( {{f_0}} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}} \right)^{\frac{1}{2}}} , {E_{n - 1}}{\left( {{z^r}f_0^{\left( r \right)}} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}b_{n,r}^2} \right)^{\frac{1}{2}}} . $

      从而得到(7)式左边的下界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{E_{n - 1}}{{\left( {{z^r}{f^{\left( r \right)}}} \right)}_2}}} \geqslant \frac{{{E_{n - 1}}{{\left( {{f_0}} \right)}_2}}}{{{E_{n - 1}}{{\left( {{z^r}f_0^{\left( r \right)}} \right)}_2}}} = \frac{1}{{\left| {{b_{n,r}}} \right|}}. $

      证毕.

      在定理1成立的基础之上,下面给出最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $$ {\Omega _m}{\left( {{z^r}{f^{\left( r \right)}},t} \right)_2} $ 的精确Jackson不等式.

      定理 2 对于 $ n,m\in {\rm{N}},r\in {{\rm{Z}}}_{+},n > r\geqslant 1,t\in \left(0,1\right) $,则有

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{\Omega _m}{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}}} = \frac{1}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}\left| {{b_{n,r}}} \right|}}. $

      证明 根据函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模、(3)和(7)式得

      $ \qquad \begin{split} {\Omega }_{m}^{2}{\left({z}^{r}{f}^{\left(r\right)},t\right)}_{2}\geqslant & {\sum\limits_{k=n}^{\infty }{\left[1-{\left(1+t\right)}^{-k}\right]}^{2m}{b}_{k,r}^{2}{\left|{c}_{-k}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}\geqslant {\left[1-{\left(1+t\right)}^{-n}\right]}^{2m}{\sum\limits_{k=n}^{\infty }{b}_{k,r}^{2}{\left|{c}_{-k}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}=\\ &{\left[1-{\left(1+t\right)}^{-n}\right]}^{2m}{E}_{n-1}^{2}{\left({z}^{r}{f}^{\left(r\right)}\right)}_{2}\geqslant {\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]^{2m}}b_{n,r}^2E_{n - 1}^2{\left( f \right)_2} . \end{split} $

      从而得(8)式右边的上界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{\Omega _m}{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}}} \leqslant \frac{1}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}\left| {{b_{n,r}}} \right|}}. $

      因为 $ {f_0} = {z^{ - n}} \in {L_2} $,则有 $ {z^r}f_0^{\left( r \right)} = {b_{n,r}}{z^{ - n}} $,根据(3)和(6)式知

      $ \qquad {E_{n - 1}}{\left( {{f_0}} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}} \right)^{\frac{1}{2}}} , {\Omega _m}{\left( {{z^r}f_0^{\left( r \right)},t} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}b_{n,r}^2} \right)^{\frac{1}{2}}}{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]^m} , $

      从而得(8)式左边的下界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{\Omega _m}{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}}} \geqslant \frac{{{E_{n - 1}}{{\left( {{f_0}} \right)}_2}}}{{{\Omega _m}{{\left( {{z^r}f_0^{\left( r \right)},t} \right)}_2}}} = \frac{1}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}\left| {{b_{n,r}}} \right|}}.$

      证毕.

      下面得到函数 $ {z^r}{f^{\left( r \right)}} $$ m $ 阶连续模在区间 $ \left( {0,h} \right) $ 上加权积分与最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $ 的精确Jackson不等式.

      定义 4[2] 加权函数 $ q\left( t \right) $ 是区间 $ \left( {0,h} \right) $ 上实的非负可测函数,且不恒等于0.

      定理 3 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,h\in \left(0,1\right),0 < p\leqslant 2$$ q\left( t \right) $ 是区间 $ \left( {0, h} \right) $ 上的加权函数(见定义4),则

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\left( {\displaystyle\int_0^h {\Omega _m^p{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}q\left( t \right){\rm{d}}t} } \right)}^{\frac{1}{p}}}}} = \frac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\rm{d}}t} } \right)}^{\frac{1}{p}}}\left| {{b_{n,r}}} \right|}}.$

      证明 闵可夫斯基不等式为

      $ \qquad {\left( {\int _0^h{{\left( {\sum\limits_{k = n}^\infty {{{\left| {{g_k}\left( t \right)} \right|}^2}} } \right)}^{\frac{p}{2}}}{\rm{d}}t} \right)^{\frac{1}{p}}} \geqslant {\left( {\sum\limits_{k = n}^\infty {{{\left( {\int _0^h{{\left| {{g_k}\left( t \right)} \right|}^p}{\rm{d}}t} \right)}^{\frac{2}{p}}}} } \right)^{\frac{1}{2}}}, $

      $0 < p\leqslant 2,h\in {{\bf{R}}}^{+}$ 上恒成立. 令 $ {g_k} = {f_k}{q^{\frac{1}{p}}} $,在 $ 0 < p\leqslant 2,h\in {{\bf{R}}}^{+} $ 上,则有

      $ \qquad \begin{split} &{\left( {\int _0^h{{\left( {\sum\limits_{k = n}^\infty {{{\left| {{f_k}\left( t \right)} \right|}^2}} } \right)}^{\frac{p}{2}}}q\left( t \right){\rm{d}}t} \right)^{\frac{1}{p}}} \geqslant {\left( {\sum\limits_{k = n}^\infty {{{\left( {\int _0^h{{\left| {{f_k}\left( t \right)} \right|}^p}q\left( t \right){\rm{d}}t} \right)}^{\frac{2}{p}}}} } \right)^{\frac{1}{2}}},\\ &{\left( {\int_0^h {\Omega _m^p{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}q\left( t \right){\rm{d}}t} } \right)^{\frac{1}{p}}} = {\left( {\int_0^h {{{\left( {\Omega _m^2{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}} \right)}^{\frac{p}{2}}}q\left( t \right){\rm{d}}t} } \right)^{\frac{1}{p}}}. \end{split} $

      再根据连续模的定义知

      $ \qquad {\left( {\int _0^h{{\left( {\Omega _m^2{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}} \right)}^{\frac{p}{2}}}q\left( t \right){\rm{d}}t} \right)^{\frac{1}{p}}} \geqslant {\left( {\int _0^h{{\left( {\sum\limits_{k = n}^\infty {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - k}}} \right]}^{2m}}} b_{k,r}^2{{\left| {{c_{ - k}}\left( f \right)} \right|}^2}\frac{{{4^{k - 1}} - 1}}{{k - 1}}} \right)}^{\frac{p}{2}}}q\left( t \right){\rm{d}}t} \right)^{\frac{1}{p}}}. $

      由(3)和(6)式得

      $ \qquad {\left( {\int _0^h{{\left( {\Omega _m^2{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}} \right)}^{\frac{p}{2}}}q\left( t \right){\rm{d}}t} \right)^{\frac{1}{p}}} \geqslant {E_{n - 1}}{\left( {{z^r}{f^{\left( r \right)}}} \right)_2}{\left( {\int _0^h{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\rm{d}}t} \right)^{\frac{1}{p}}},$

      再由(7)式,得到(9)式右边的下界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\left( {\displaystyle\int _0^h\Omega _m^p{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}q\left( t \right){\rm{d}}t} \right)}^{\frac{1}{p}}}}} \leqslant \frac{1}{{{{\left( {\displaystyle\int _0^h{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\rm{d}}t} \right)}^{\frac{1}{p}}}\left| {{b_{n,r}}} \right|}}. $

      因为 $ {f_0} = {z^{ - n}} \in {L_2} $,则有 $ {z^r}f_0^{\left( r \right)} = {b_{n,r}}{z^{ - n}} $,根据(3)和(6)式得

      $ \qquad \begin{split} & {E_{n - 1}}{\left( {{f_0}} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}} \right)^{\frac{1}{2}}} , \\ &{\left( {\int_0^h {\Omega _m^p{{\left( {{z^r}f_0^{\left( r \right)},t} \right)}_2}q\left( t \right){\text{d}}t} } \right)^{\frac{1}{p}}} = {\left( {\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)^{\frac{1}{p}}}{\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}b_{n,r}^2} \right)^{\frac{1}{2}}} , \end{split} $

      从而得到(9)式左边的下界

      $ \qquad \begin{split} \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\left( {\displaystyle\int _0^h\Omega _m^p{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}q\left( t \right){\rm{d}}t} \right)}^{\frac{1}{p}}}}} \geqslant & \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( {{f_0}} \right)}_2}}}{{{{\left( {\displaystyle\int _0^h\Omega _m^p{{\left( {{z^r}f_0^{\left( r \right)},t} \right)}_2}q\left( t \right){\rm{d}}t} \right)}^{\frac{1}{p}}}}} =\\ &\frac{1}{{{{\left( {\displaystyle\int _0^h{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\rm{d}}t} \right)}^{\frac{1}{p}}}\left| {{b_{n,r}}} \right|}}.\end{split} $

      证毕.

      推论 1 对于 $ n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,t\in \left(0,1\right),p=\dfrac{1}{m} $$ q\left( t \right) $ 是区间 $ \left( {0,h} \right) $ 上的权函数,则

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\left( {\displaystyle\int_0^h {\Omega _m^{\frac{1}{m}}{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}q\left( t \right){\rm{d}}t} } \right)}^{\frac{1}{p}}}}} = \frac{1}{{{{\left( {\displaystyle\int_0^h {\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]q\left( t \right){\rm{d}}t} } \right)}^{\frac{1}{p}}}\left| {{b_{n,r}}} \right|}}. $

      上述得到最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $$ {\Omega _m}{\left( {{z^r}{f^{\left( r \right)}},t} \right)_2} $ 的精确Jackson不等式. 在复数域上 $ {\cal{K}} $ 泛函[1]表达式基础上,本文定义了 $ {\cal{K}} $ 泛函.

      定义 5 $ {\cal{K}} $ 泛函的表达式为

      $ \qquad {{\cal{K}}_m}{\left( {f,{t^m}} \right)_2}: = {\cal{K}}\left( {f;{t^m};{L_2};L_2^{\left( m \right)}} \right) = \inf \left\{ {{{\left\| {f - g} \right\|}_2} + {t^m}{{\left\| {{z^m}{g^{\left( m \right)}}} \right\|}_2}:g \in L_2^{\left( m \right)}} \right\} . $

      这里 $ g $ 在圆盘 ${\text{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\}$ 上且洛朗展式为 $ g\left( z \right) = \displaystyle\sum\limits_{k = 0}^\infty {{d_{ - k}}} {z^{ - k}}; {d_{ - k}} \in {\bf{C}} $. 下面得到了最佳逼近 $ {E_{n - 1}}{\left( f \right)_2} $$ {\cal{K}} $ 泛函的精确Jackson不等式.

      定理 4 对于 $ n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r+m\geqslant 1 $,则

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\cal{K}}_m}{{\left( {{z^r}{f^{\left( r \right)}},\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right)}_2}}} = \frac{1}{{\left| {{b_{n,r}}} \right|}}. $

      证明 对任意的 $ g \in L_2^{\left( m \right)} $ 都有 $ {\left\| {g - {S_{n - 1}}\left( g \right)} \right\|_2} = {E_{n - 1}}{\left( g \right)_2} $,根据(7)式得

      $ \qquad \begin{split} &{\Vert g-{S}_{n-1}\left(g\right)\Vert }_{2}\leqslant \frac{1}{\left|{b}_{n,m}\right|}{E}_{n-1}{\left({z}^{m}{g}^{\left(m\right)}\right)}_{2}\leqslant \frac{1}{\left|{b}_{n,m}\right|}{\Vert {z}^{m}{g}^{\left(m\right)}\Vert }_{2} ,\\ &{E}_{n-1}{\left(f\right)}_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\left\{{\Vert {z}^{r}{f}^{\left(r\right)}-g\Vert }_{2}+{\Vert g-{S}_{n-1}\left(g\right)\Vert }_{2}\right\} , \end{split}$

      进一步有 $ {E}_{n-1}{\left(f\right)}_{2}\leqslant \dfrac{1}{\left|{b}_{n,r}\right|}\left\{{\Vert {z}^{r}{f}^{\left(r\right)}-g\Vert }_{2}+\frac{1}{\left|{b}_{n,m}\right|}{\Vert {z}^{m}{g}^{\left(m\right)}\Vert }_{2}\right\} $. 根据 $ {\cal{K}} $ 泛函的定义知 $ {E}_{n-1}{\left(f\right)}_{2}\leqslant \dfrac{1}{\left|{b}_{n,r}\right|}{{\cal{K}}}_{m} $$ {\left({z}^{r}{f}^{\left(r\right)},\frac{1}{\left|{b}_{n,m}\right|}\right)}_{2} $,从而得(12)式右边的上界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\cal{K}}_m}{{\left( {{z^r}{f^{\left( r \right)}},\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right)}_2}}} \leqslant \dfrac{1}{{\left| {{b_{n,r}}} \right|}}. $

      $ g = 0 $,根据 $ {\cal{K}} $ 泛函的定义得 $ {{\cal{K}}}_{m}{\left(f,{t}^{m}\right)}_{2}\leqslant {\Vert f\Vert }_{2} $;因为 $ {f_0} = {z^{ - n}} \in {L_2} $,则有 $ {z^r}f_0^{\left( r \right)} = {b_{n,r}}{z^{ - n}} $. 根据(3)和(11)式得

      $ \qquad \begin{split}&{E_{n - 1}}{\left( {{f_0}} \right)_2} = {\left( {\frac{{{4^{n - 1}} - 1}}{{n - 1}}} \right)^{\frac{1}{2}}} ,\\ &{{\cal{K}}}_{m}{\left({z}^{r}{f}_{0}^{\left(r\right)},\frac{1}{\left|{b}_{n,m}\right|}\right)}_{2}\leqslant {\Vert {z}^{r}{f}_{0}^{\left(r\right)}\Vert }_{2}=\left|{b}_{n,r}\right|{\left(\frac{{4}^{n-1}-1}{n-1}\right)}^{\frac{1}{2}} ,\end{split} $

      从而得(12)式左边的下界

      $ \qquad \mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( f \right)}_2}}}{{{{\cal{K}}_m}{{\left( {{z^r}{f^{\left( r \right)}},\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right)}_2}}} \geqslant\mathop {\sup }\limits_{\scriptstyle f \in L_2^{\left( r \right)}\atop \scriptstyle f \notin {P_{n - 1}}} \frac{{{E_{n - 1}}{{\left( {{f_0}} \right)}_2}}}{{{{\cal{K}}_m}{{\left( {{z^r}f_0^{\left( r \right)},\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right)}_2}}} = \frac{1}{{\left| {{b_{n,r}}} \right|}}. $

      证毕.

      定理 5(逆定理) 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,t\in \left(0,1\right)$,有

      $ \qquad {\Omega }_{m}^{2}{\left({z}^{r}{f}^{\left(r\right)},t\right)}_{2}\leqslant 2m{\left(\frac{2t}{1+t}\right)}^{2m}{\sum\limits_{k=1}^{\left[\frac{1+t}{t}\right]}{k}^{2m-1}}{E}_{k-1}^{2}{\left({z}^{r}{f}^{\left(r\right)}\right)}_{2} . $

      证明 令 $ N = \left[ {\dfrac{{1 + t}}{t}} \right] $,则有 $N\leqslant \dfrac{1+t}{t} < 2N \Rightarrow \dfrac{t}{t+1}\leqslant \dfrac{1}{N}< \dfrac{2t}{t+1}$. 由连续模的定义知

      $ \qquad \begin{split}{\Omega }_{m}^{2}{\left({z}^{r}{f}^{\left(r\right)},t\right)}_{2}=&{\sum\limits_{k=2}^{\infty }{\left[1-{\left(1+t\right)}^{-k}\right]}^{2m}{b}_{k,r}^{2}{\left|{c}_{-k}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}+{\left[1-{\left(1+t\right)}^{-1}\right]}^{2m}{b}_{1,r}^{2}2\mathrm{ln}2{\left|{c}_{-1}\left(f\right)\right|}^{2}\leqslant \\ &\frac{1}{{N}^{2m}}\left({\sum\limits_{k=2}^{N}{k}^{2m}{b}_{k,r}^{2}{\left|{c}_{-k}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}+{b}_{1,r}^{2}2\mathrm{ln}2{\left|{c}_{-1}\left(f\right)\right|}^{2}\right)+{\sum\limits_{k=N+1}^{\infty }{b}_{k,r}^{2}{\left|{c}_{-k}\left(f\right)\right|}^{2}\frac{{4}^{k-1}-1}{k-1}}=\\ &\frac{1}{{N}^{2m}}\left({\sum\limits_{k=2}^{N}\left({k}^{2m}-{\left(k-1\right)}^{2m}\right){\sum\limits_{l=k}^{\infty }{b}_{l,r}^{2}{\left|{c}_{-l}\left(f\right)\right|}^{2}}\frac{{4}^{l-1}-1}{l-1}}+{b}_{1,r}^{2}2\mathrm{ln}2{\left|{c}_{-1}\left(f\right)\right|}^{2}\right)\leqslant \\ &\frac{1}{{N}^{2m}}\left(2m{\sum\limits_{k=2}^{N}{k}^{2m}{\sum\limits_{l=k}^{\infty }{b}_{l,r}^{2}{\left|{c}_{-l}\left(f\right)\right|}^{2}}\frac{{4}^{l-1}-1}{l-1}}+{b}_{1,r}^{2}2\mathrm{ln}2{\left|{c}_{-1}\left(f\right)\right|}^{2}\right)\leqslant \\ &\frac{1}{{N}^{2m}}\left(2m{\sum\limits_{k=2}^{N}{k}^{2m}{\sum\limits_{l=k}^{\infty }{b}_{l,r}^{2}{\left|{c}_{-l}\left(f\right)\right|}^{2}}\frac{{4}^{l-1}-1}{l-1}}+{E}_{0}^{2}{\left({z}^{r}{f}^{\left(r\right)}\right)}_{2}\right)\leqslant \\ &2m{\left( {\frac{{2t}}{{1 + t}}} \right)^{2m}}\sum\limits_{k = 1}^{\left[ {\frac{{1 + t}}{t}} \right]} {{k^{2m - 1}}} E_{k - 1}^2{\left( {{z^r}{f^{\left( r \right)}}} \right)_2} .\end{split} $

      证毕.

    • 定义 6 函数 $ \Phi $ 定义在 $ {{\bf{R}}_ + } $ 上单调递增的实函数,且满足 $ \Phi (0) = 0 $$\Phi (t) \to 0 ( t \to 0)$.

      定义 7 3种函数类:

      $ \qquad {W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right):=\left\{f\in {L}_{2}^{\left(r\right)};{\Omega }_{m}{\left({z}^{r}{f}^{\left(r\right)},t\right)}_{2}\leqslant \Phi \left(t\right),t\in \left(0,1\right),r\in {{\bf{Z}}}_+,m\in {\bf{N}}\right\} ; $

      $ \qquad {W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right):=\left\{f\in {L}_{2}^{\left(r\right)};{{\int }_{0}^{h}{\Omega }_{m}^{p}{\left({z}^{r}{f}^{\left(r\right)},t\right)}_{2}q\left(t\right)\text{d}t}\leqslant 1,t\in \left(0,h\right),h\in \left(0,1\right),r\in {{\bf{Z}}}_+,m\in {\bf{N}};0 < p\leqslant 2\right\} ; $

      $ \qquad {W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right):=\left\{f\in {L}_{2}^{\left(r\right)};{{\cal{K}}}_{m}{\left({z}^{r}{f}^{\left(r\right)},{t}^{m}\right)}_{2}\leqslant \Phi \left({t}^{m}\right),t\in \left(0,1\right),r\in {{\bf{Z}}}_+,m\in {\bf{N}}\right\} ; $

      $ \Phi $ 被称为函数类 $ W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $$ W_2^{\left( r \right)}\left( {E,\Phi } \right) $$ W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $ 上的限制函数.

      定义 8[5] $ {M^{\left( r \right)}} $ 表示 $ L_2^{\left( r \right)} $ 的子类,函数类 $ {M^{\left( r \right)}} $ 的最佳逼近表达式为

      $ \qquad {E_{n - 1}}{\left( {{M^{\left( r \right)}}} \right)_2}: = \sup \left\{ {{E_{n - 1}}{{\left( f \right)}_2}:f \in {M^{\left( r \right)}}} \right\} . $

      定理 6 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,t\in \left(0,1\right)$$ \Phi $ 是函数类 $ W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $ 上的限制函数(见定义6,7),则

      $ \qquad {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} . $

      证明 由(8)式知,对任意的 $ f \in W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $,有

      $ \qquad {E_{n - 1}}{\left( f \right)_2} \leqslant \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{{{\Omega _m}{{\left( {{z^r}{f^{\left( r \right)}},t} \right)}_2}}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} , $

      再由(14)式,得到(15)式右边的上界

      $ \qquad {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right)} \right)_2} \leqslant \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} . $

      $ {f_0}\left( z \right) = {\left( {\dfrac{{n - 1}}{{{4^{n - 1}} - 1}}\dfrac{1}{{b_{n,{\rm{r}}}^2}}} \right)^{\frac{1}{2}}}\dfrac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}}{z^{ - n}} \in L_2^{\left( r \right)} $,根据(3)和(6)式得

      $ \qquad {E_{n - 1}}{\left( {{f_0}} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} , {\Omega }_{m}{\left({z}^{r}{f}_{0}^{\left(r\right)},t\right)}_{2}\leqslant \Phi \left(t\right), $

      从而有 $ {f_0}\left( z \right) \in W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $,因此得到(15)式左边的下界

      $ \qquad {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right)\right)}_{2}\geqslant {E}_{n-1}{\left({f}_{0}\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\frac{\Phi \left(t\right)}{{\left[1-{\left(1+t\right)}^{-n}\right]}^{m}} . $

      证毕.

      定理 7 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,0 < p\leqslant 2,h\in \left(0,1\right)$$ q\left( t \right) $ 是定义在区间 $ \left( {0,h} \right) $ 上的加权函数(见定义4),有

      $ \qquad {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)}^{\frac{1}{p}}}}} . $

      证明 由(9)式知,对任意的 $ f \in W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right) $,有

      $ \qquad {E}_{n-1}{\left(f\right)}_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}} , $

      再由(14)式,得到(16)式右边的上界

      $ \qquad {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right)\right)}_{2}\leqslant \dfrac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}} . $

      ${f_0}\left( z \right) = {\left( {\dfrac{{n - 1}}{{{4^{n - 1}} - 1}}\frac{1}{{b_{n,r}^2}}} \right)^{\frac{1}{2}}}\dfrac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)}^{\frac{1}{p}}}}}{z^{ - n}} \in L_2^{\left( r \right)}$,根据(3)和(6)式得

      $ \qquad \begin{split} &{E_{n - 1}}{\left( {{f_0}} \right)_2} = \dfrac{1}{{\left| {{b_{n,r}}} \right|}}\dfrac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)}^{\frac{1}{p}}}}} , \\ &{\left( {\displaystyle\int_0^h {\Omega _m^p{{\left( {{z^r}f_0^{\left( r \right)},t} \right)}_2}q\left( t \right){\text{d}}t} } \right)^{\frac{1}{p}}} = 1 , \end{split}$

      从而有 $ {f_0}\left( z \right) \in W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right) $,得(16)式左边的下界

      $ \qquad {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right)\right)}_{2}\geqslant {E}_{n-1}{\left({f}_{0}\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}} . $

      证毕.

      定理 8 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r+m\geqslant 1,t\in \left({\left|{b}_{n.m}\right|}^{-\frac{1}{m}},1\right)$$ \Phi $ 是函数类 $ W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $ 上的限制函数(见定义6,7),有

      $ \qquad {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\Phi \left( {\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right) . $

      证明 由(12)式知,对任意的 $ f \in W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $,有

      $ \qquad {E}_{n-1}{\left(f\right)}_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) , $

      再由(14)式,得(17)式右边的上界

      $ \qquad {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right)\right)}_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) . $

      $ {f_0}\left( z \right) = {\left( {\dfrac{{n - 1}}{{{4^{n - 1}} - 1}}\dfrac{1}{{b_{n,{\rm{r}}}^2}}} \right)^{\frac{1}{2}}}\Phi \left( {\dfrac{1}{{\left| {{b_{n,m}}} \right|}}} \right){z^{ - n}} \in L_2^{\left( r \right)} $,根据(3)式得 $ {E_{n - 1}}{\left( {{f_0}} \right)_2} = \dfrac{1}{{\left| {{b_{n,r}}} \right|}}\Phi \left( {\dfrac{1}{{\left| {{b_{n,m}}} \right|}}} \right) $;当 $ g = 0 $$ {\cal{K}}{\left( {f,{t^m}} \right)_2} = $$ {\left\| f \right\|_2} $. 把函数 $ {z^r}f_0^{\left( r \right)} $ 带入 $ {\cal{K}} $ 泛函里得

      $ \qquad {{\cal{K}}}_{m}{\left({z}^{r}{f}_{0}^{\left(r\right)},{t}^{m}\right)}_{2}\leqslant {\Vert {z}^{r}{f}_{0}^{\left(r\right)}\Vert }_{2}=\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) , $

      因为 $ t \in \left( {{{\left| {{b_{n.m}}} \right|}^{ - \frac{1}{m}}},1} \right) $,所以

      $ \qquad {{\cal{K}}}_{m}{\left({z}^{r}{f}_{0}^{\left(r\right)},{t}^{m}\right)}_{2}\leqslant {\Vert {z}^{r}{f}_{0}^{\left(r\right)}\Vert }_{2}=\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right)\leqslant \Phi \left({t}^{m}\right) , $

      从而有 $ {f_0}\left( z \right) \in W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $. 得(17)式左边的下界

      $ \qquad {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right)\right)}_{2}\geqslant {E}_{n-1}{\left({f}_{0}\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) . $

      证毕.

    • 定义 9[3] $ B $ 是在 $ {L_2} $ 空间下的单位球,假设 $ \Lambda {}_n \subset {L_2} $$ n $ 维子空间;$ {\Lambda ^n} \subset {L_2} $$ n $ 维余子空间;$ \sigma :{L_2} \to {\Lambda _n} $ 是线性连续算子,$ {\sigma ^ \bot }:{\Lambda _n} \to {L_2} $ 是线性连续算子,$ M $$ {L_2} $ 下的凸对称子集,定义:

      $ \qquad {b_n}\left( {M,{L_2}} \right) = \sup \left\{ {\sup \left( {\varepsilon ;\varepsilon B \cap {\Lambda _{n + 1}} \subset M} \right):{\Lambda _{n + 1}} \subset {L_2}} \right\} ; $

      $ \qquad {d_n}\left( {M,{L_2}} \right) = \inf \left\{ {\sup \left\{ {\inf \left\{ {{{\left\| {f - g} \right\|}_2}:g \in {\Lambda _n}} \right\}:f \in M} \right\}:{\Lambda _n} \subset {L_2}} \right\} ; $

      $ \qquad {\delta _n}\left( {M,{L_2}} \right) = \inf \left\{ {\inf \left\{ {\sup \left\{ {{{\left\| {f - \sigma f} \right\|}_2}:f \in M} \right\}:\sigma {L_2} \in {\Lambda _2}} \right\}:{\Lambda _n} \subset {L_2}} \right\} ; $

      $ \qquad {d^n}\left( {M,{L_2}} \right) = \inf \left\{ {\sup \left\{ {{{\left\| f \right\|}_2}:f \in M \cap {\Lambda ^n}} \right\}:{\Lambda ^n} \subset {L_2}} \right\} ; $

      $ \qquad {\Pi _n}\left( {M,{L_2}} \right) = \inf \left\{ {\inf \left\{ {\sup \left\{ {{{\left\| {f - {\sigma ^ \bot }f} \right\|}_2}:f \in M} \right\}:{\sigma ^ \bot }{L_2} \subset {\Lambda _n}} \right\}:{\Lambda _n} \subset {L_2}} \right\} . $

      在希尔伯特空间中有

      $ \qquad {b}_{n}\left(M,{L}_{2}\right)\leqslant {d}^{n}\left(M,{L}_{2}\right)\leqslant {d}_{n}\left(M,{L}_{2}\right)={\delta }_{n}\left(M,{L}_{2}\right)={\Pi }_{n}\left(M,{L}_{2}\right) . $

      定理 9 对于 $n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,t\in \left(0,1\right)$$ \Phi $ 是函数类 $ W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $ 上的限制函数(见定义6,7),有

      $ \qquad {\lambda _n}\left( {W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right)} \right) = {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} , $

      这里的 $ {\lambda }_{n}(\cdot ) $ 表示 $ {b_n}\left( {M,{L_2}} \right),{d^n}\left( {M,{L_2}} \right),{d_n}\left( {M,{L_2}} \right),{\delta _n}\left( {M,{L_2}} \right),{\Pi _n}\left( {M,{L_2}} \right) $ 中任意一种宽度.

      证明 由(15)式知,$ {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right)} \right)_2} = \dfrac{1}{{\left| {{b_{n,r}}} \right|}}\dfrac{{\Phi \left( t \right)}}{{{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^m}}} $. 再根据 $ n $ 维宽度的定义和(18)式,得(19)式右边的上界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right),{L}_{2}\right)\leqslant {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right)\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\frac{\Phi \left(t\right)}{{\left[1-{\left(1+t\right)}^{-n}\right]}^{m}} . $

      $ n + 1 $ 维函数为 $ {Q_{n + 1}} = \left\{ {{p_n} \in {P_n}:{p_n} = \displaystyle\sum\limits_{k = 0}^n {{a_k}{z^{ - k}};{a_k} \in {\bf{C}}} } \right\} $,这里 $ n + 1 $ 维球体是

      $ \qquad {B}_{n+1}=\left\{{p}_{n}\in {Q}_{n+1}:{\Vert {p}_{n}\Vert }_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\frac{\Phi \left(t\right)}{{\left[1-{\left(1+t\right)}^{-n}\right]}^{m}}\right\} , $

      $ \qquad \Omega _m^2{\left( {{z^r}p_n^{\left( r \right)},t} \right)_2} = \sum\limits_{k = 2}^\infty {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - k}}} \right]}^{2m}}b_{k,r}^2{{\left| {{c_{ - k}}\left( f \right)} \right|}^2}\frac{{{4^{k - 1}} - 1}}{{k - 1}}} {\left| {{a_k}} \right|^2} + {\left[ {1 - {{\left( {1 + t} \right)}^{ - 1}}} \right]^{2m}}b_{1,r}^22\ln 2{\left| {{c_{ - 1}}\left( f \right)} \right|^2}{\left| {{a_1}} \right|^2} . $

      化简得 $ {\Omega }_{m}^{2}{\left({z}^{r}{p}_{n}^{\left(r\right)},t\right)}_{2}\leqslant {\left[1-{\left(1+t\right)}^{-n}\right]}^{2m}{b}_{{\rm{n}},r}^{2}{\Vert {p}_{n}\Vert }_{2}^{2}\leqslant {\Phi }^{2}\left(t\right) $,从而有 $ {B_{n + 1}} \subset W_2^{\left( r \right)}\left( {{\Omega _m},\Phi } \right) $. 再根据(18)式,得(19)式左边的下界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right),{L}_{2}\right)\geqslant {b}_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},\Phi \right),{L}_{2}\right)\geqslant {b}_{n}\left({B}_{n+1},{L}_{2}\right)\geqslant \frac{1}{\left|{b}_{n,r}\right|}\frac{\Phi \left(t\right)}{{\left[1-{\left(1+t\right)}^{-n}\right]}^{m}} . $

      证毕.

      定理 10 对于 $ n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r\geqslant 1,0 < p\leqslant 2,h\in \left(0,1\right) $$ q\left( t \right) $ 是区间 $ \left( {0,h} \right) $ 上的加权函数,有

      $ \qquad {\lambda _n}\left( {W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right),{L_2}} \right) = {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\frac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)}^{\frac{1}{p}}}}} . $

      这里的 $ {\lambda }_{n}(\cdot ) $ 表示 $ {b_n}\left( {M,{L_2}} \right),{d^n}\left( {M,{L_2}} \right),{d_n}\left( {M,{L_2}} \right),{\delta _n}\left( {M,{L_2}} \right),{\Pi _n}\left( {M,{L_2}} \right) $ 中任意一种宽度.

      证明 由(16)式知

      $ \qquad {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right)} \right)_2} = \dfrac{1}{{\left| {{b_{n,r}}} \right|}}\dfrac{1}{{{{\left( {\displaystyle\int_0^h {{{\left[ {1 - {{\left( {1 + t} \right)}^{ - n}}} \right]}^{pm}}q\left( t \right){\text{d}}t} } \right)}^{\frac{1}{p}}}}};$

      再根据 $ n $ 维宽度的定义和(18)式,得(20)式右边的上界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right),{L}_{2}\right)\leqslant {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right)\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}} . $

      $ n + 1 $ 维球体为

      $ \qquad {B}_{n+1}=\left\{{p}_{n}\in {Q}_{n+1}:{\Vert {p}_{n}\Vert }_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}}\right\} , $

      这里 $ {Q_{n + 1}} $ 为定理9证明过程中的 $ {Q_{n + 1}} $. 由连续模的定义知 $ {{\displaystyle\int }_{0}^{h}{\Omega }_{m}^{p}{\left({z}^{r}{p}_{n}^{\left(r\right)},t\right)}_{2}q\left(t\right)\text{d}t}\leqslant 1 $,则 $ {B_{n + 1}} \subset W_2^{\left( r \right)}\left( {{\Omega _m},h,q} \right) $,再根据(18)式,得(20)式左边的下界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right),{L}_{2}\right)\geqslant {b}_{n}\left({W}_{2}^{\left(r\right)}\left({\Omega }_{m},h,q\right),{L}_{2}\right)\geqslant {b}_{n}\left({B}_{n+1},{L}_{2}\right)\geqslant \frac{1}{\left|{b}_{n,r}\right|}\frac{1}{{\left({{\displaystyle\int }_{0}^{h}{\left[1-{\left(1+t\right)}^{-n}\right]}^{pm}q\left(t\right)\text{d}t}\right)}^{\frac{1}{p}}} . $

      证毕.

      定理 11 对于 $ n,m\in {\bf{N}},r\in {{\bf{Z}}}_{+},n > r+m\geqslant 1,t\in \left({\left|{b}_{n.m}\right|}^{-\frac{1}{m}},1\right) $$ \Phi $ 是函数类 $ W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $ 上的限制函数(见定义6,7),有

      $ \qquad {\lambda _n}\left( {W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right),{L_2}} \right) = {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right)} \right)_2} = \frac{1}{{\left| {{b_{n,r}}} \right|}}\Phi \left( {\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right) . $

      这里的 $ {\lambda }_{n}(\cdot ) $ 表示 $ {b_n}\left( {M,{L_2}} \right),{d^n}\left( {M,{L_2}} \right),{d_n}\left( {M,{L_2}} \right),{\delta _n}\left( {M,{L_2}} \right),{\Pi _n}\left( {M,{L_2}} \right) $ 中任意一种宽度.

      证明 根据(17)式知 $ {E_{n - 1}}{\left( {W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right)} \right)_2} = \dfrac{1}{{\left| {{b_{n,r}}} \right|}}\Phi \left( {\dfrac{1}{{\left| {{b_{n,m}}} \right|}}} \right) $. 再根据 $ n $ 维宽度的定义和(18)式,得(21)式右边的上界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right),{L}_{2}\right)\leqslant {E}_{n-1}{\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right)\right)}_{2}=\frac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) . $

      $ n + 1 $ 维球体为

      $ \qquad {B}_{n+1}=\left\{{p}_{n}\in {Q}_{n+1}:{\Vert {p}_{n}\Vert }_{2}\leqslant \frac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right)\right\} , $

      这里 $ {Q_{n + 1}} $ 为定理9证明过程中的 $ {Q_{n + 1}} $. 根据 $ {\cal{K}} $ 泛函的定义知,当 $ g = 0 $ 时,${\cal{K}}{\left( {f,{t^m}} \right)_2} \leqslant {\left\| f \right\|_2}$. 把函数 $ {z^r}p_n^{\left( r \right)} $ 带入 $ {\cal{K}} $ 泛函里得

      $ \qquad {{\cal{K}}_m}{\left( {{z^r}p_n^{\left( r \right)},{t^m}} \right)_2} \leqslant {\left\| {{z^r}p_n^{\left( r \right)}} \right\|_2} = \Phi \left( {\frac{1}{{\left| {{b_{n,m}}} \right|}}} \right) , $

      又因为 $ t \in \left( {{{\left| {{b_{n.m}}} \right|}^{ - \frac{1}{m}}},1} \right) $,所以 $ {{\cal{K}}}_{m}{\left({z}^{r}{p}_{n}^{\left(r\right)},{t}^{m}\right)}_{2}\leqslant {\Vert {z}^{r}{p}_{n}^{\left(r\right)}\Vert }_{2}=\Phi \left(\dfrac{1}{\left|{b}_{n,m}\right|}\right)\leqslant \Phi \left({t}^{m}\right) $,从而有 $ {B_{n + 1}} \in W_2^{\left( r \right)}\left( {{{\cal{K}}_m},\Phi } \right) $,得(21)式左边的下界

      $ \qquad {\lambda }_{n}\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right),{L}_{2}\right)\geqslant {b}_{n}\left({W}_{2}^{\left(r\right)}\left({{\cal{K}}}_{m},\Phi \right),{L}_{2}\right)\geqslant {b}_{n}\left({B}_{n+1},{L}_{2}\right)\geqslant \dfrac{1}{\left|{b}_{n,r}\right|}\Phi \left(\frac{1}{\left|{b}_{n,m}\right|}\right) . $

      证毕.

    • 本文在复数域上研究了 $ {L_2}\left( {\text{U}} \right) $$ {\text{U}} = \left\{ {z \in {\bf{C}}:\dfrac{1}{2} < \left| z \right| < 1} \right\} $ 上的一类特殊函数空间,要求函数 $ f $ 解析,原点为其本性奇点,且有 $ f = \displaystyle\sum\limits_{k = 0}^\infty {{c_{ - k}}} \left( f \right){z^{ - k}} $ 的洛朗展式. 我们给出此空间上精确的逼近正定理、逆定理及一些宽度的估计. 在此基础上,我们下一步将研究形如 $ f = \displaystyle\sum\limits_{k = - \infty }^\infty {{c_k}} \left( f \right){z^k} $ 洛朗展式的函数空间相应的逼近问题.

      致谢:感谢审稿人给予的宝贵意见. 通信作者感谢国家留学基金委员会给予的资助.

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