文中用$ M(u) $和$ N(v) $表示互余的N函数,关于N -函数的定义及其性质参见文献[1].由文献[1]知, $ M(u) $为N -函数当且仅当存在定义在$ [0, +\infty) $上的实值函数$ p(t) $,使得$ M(u) = \int_{0}^{|u|} p(t)\, {\rm d}t $,其中$ p(t) $满足下列条件

$ 1^\circ $$ p(t) $为右连续的单调递增函数;

$ 2^\circ $当$ t>0 $时, $ p(t)>0 $;

$ 3^\circ $$ p(0) = 0 $, $ p(+\infty) = +\infty $.

这时$ p(u) $为$ M(u) $的右导数.文中由N -函数$ N(u) $生成的Orlicz类$ L_{N} $是指满足

$ \rho(u, N) = \int_{0}^{2\pi} N(u(x))\, {\rm d}x<+\infty $

的可测函数的全体$ \{u(x)\} $;由N -函数$ M(u) $生成的Orlicz空间$ L^{\ast}_{M} $是指具有有限的Orlicz范数

(1.1) $\|u{{\|}_{M}}=\begin{matrix} \sup \\ \rho (v, N)\le 1 \\\end{matrix}|\int_{0}^{2\pi }{u}(x)v(x)\text{d}x|$

的可测函数的全体$ \{u(x)\} $.在$ L^{\ast}_{M} $上还可以赋予与Orlicz范数$ (1.1) $等价的Luxemburg范数

(1.2) $ \| u\|_{(M)} = {\rm inf}\bigg\{\alpha>0\bigg|\int_{0}^{2\pi} M(\frac{u(x)}{\alpha})\, {\rm d}x\leq1\bigg\}.$

以下分别用$ L^{\ast}_{M} $和$ L^{\ast}_{(M)} $表示带有Orlicz范数(1.1)和Luxemburg范数(1.2)的Orlicz空间.由文献[1]知, Orlicz范数(1.1)也可以由下式计算

(1.3) $\|u{{\|}_{M}}=\begin{matrix} \inf \\ k>0 \\\end{matrix}\ \frac{1}{k}(1+\int_{0}^{2\pi }{M}(ku(x))\text{d}x), $

且存在$ k>0 $,满足$ \int_{0}^{2\pi} N(p(k|u(x)|))\, {\rm d}x = 1 $,使得

$ {\| u\|_{M} = \frac{1}{k}\bigg(1+\int_{0}^{2\pi} M(ku(x))\, {\rm d}x\bigg)}. $

由文献[2]有如下定义.

定义1.1  设$ \Lambda\subset X $是一中心对称集, $ \Lambda $在$ X $内的n维Kolmogorov宽度定义为

(1.4) ${{d}_{n}}(\Lambda ;X)=\begin{matrix} \inf \\ {{L}_{n}} \\\end{matrix}\begin{matrix} \sup \\ x\in \Lambda \\\end{matrix}\begin{matrix} \inf \\ \varphi \in {{L}_{n}} \\\end{matrix}\|x-\varphi {{\|}_{X}}, $

其中$ \|\cdot\|_{X} $是$ X $内的范数, $ L_{n} $是$ X $中任一n维线性子空间.简称n-K宽度.能使(1.4)式内下确界实现的任一n维线性子空间$ L_{n}^{o} $称为$ \Lambda $在$ X $内的n-K宽度的极子空间.

定义1.2  设$ \Lambda\subset X $是一中心对称集, $ \Lambda $在$ X $内的n维Gelfand宽度为

(1.5) ${{d}^{n}}(\Lambda ;X)=\begin{matrix} \inf \\ {{L}^{n}} \\\end{matrix}\begin{matrix} \sup \\ f\in \Lambda \cap {{L}^{n}} \\\end{matrix}\|f{{\|}_{X}}, $

其中$ L^{n} $是$ X $中任一n余维子空间.简称n-G宽度.能使(1.5)式内下确界实现的任一n余维线性子空间都称为$ \Lambda $在$ X $内的n-G宽度的极子空间.

定义1.3  设$ \Lambda\subset X $是一中心对称集, $ \Lambda $在$ X $内的$ n $维Linear宽度为

(1.6) $ {{d}_{n}}\prime (\Lambda ;X)=\begin{matrix} \inf \\ {{L}_{n}} \\\end{matrix}\begin{matrix} \inf \\ A \\\end{matrix}\begin{matrix} \sup \\ x\in \Lambda \\\end{matrix}\|x-Ax{{\|}_{X}}, $

其中A是$ {\rm span}\{\Lambda\}\rightarrow\Lambda $的任意一个线性有界算子.简称n-L宽度.能使(1.6)式内下确界实现的任一n维线性子空间$ L_{n}^{o} $称为$ \Lambda $在$ X $内的n-L宽度的极子空间.而对一个确定的极子空间$ L_{n}^{o} $,把可以实现下确界的线性有界算子称作$ \Lambda $的n-L宽度的最佳线性算子.

定义1.4  设$ \Lambda\subset X $是一中心对称闭凸集. $ n\geq1, L_{n}\subset X $是一n维线性子空间.称

$ \beta(\Lambda;L_{n})_{X} = {\rm sup}\{\varepsilon>0|\varepsilon B_{X}\cap L_{n}\subset\Lambda\} $

是$ L_{n} $与$ \Lambda $的截集的内切球半径.当不存在正的$ \varepsilon $,使$ \varepsilon B_{X}\cap L_{n}\subset\Lambda $成立时,规定$ \beta = 0 $.变动$ L_{n} $,使$ \beta $尽量大,这就导致定义

(1.7) $ {{b}_{n}}(\Lambda ;X)=\begin{matrix} \sup \\ {{L}_{n}} \\\end{matrix}\beta {{(\Lambda ;{{L}_{n}})}_{X}} $

为$ \Lambda $在$ X $内的n维Bernstein宽度.简称n-B宽度.如果存在$ X $的一个n维线性子空间$ L_{n}^{o} $使(1.7)式中的上确界达到,就称其为$ \Lambda $在$ X $内的n-B宽度的极子空间.

给定$ r $次实系数多项式

${{P}_{r}}(\lambda )=\prod\limits_{s=1}^{k}{({{\lambda }^{2}}-2{{\alpha }_{s}}\lambda +\alpha _{s}^{2}+\beta _{s}^{2})}\prod\limits_{j=1}^{r-2k}{(\lambda -{{\lambda }_{j}});}$

$P_{r}^{*}(\lambda )=\prod\limits_{s=1}^{k}{({{\lambda }^{2}}+2{{\alpha }_{s}}\lambda +\alpha _{s}^{2}+\beta _{s}^{2})}\prod\limits_{j=1}^{r-2k}{(\lambda +{{\lambda }_{j}}), }$

其中, $ k\geq0 $, $ \lambda_{j}, \alpha_{s}, \beta_{s}\in R $, $ \beta_{s}>0 $,令$ \beta = \max_{1\leq s\leq k}\beta_{s} $.

$ \tilde{W}^r $代表满足$ f^{(r-1)} $绝对连续且以$ 2\pi $为周期的连续函数.记

$ \tilde{m}_{r, M} = \bigg\{f\in\tilde{W}^{r}|f(x) = G_{r}\ast h(x), \| h\|_{M}\leq1\bigg\}, $

这里, $ G_{r}(x) = \frac{1}{2\pi}\sum\limits^{+\infty}_{v = -\infty} \frac{{\rm e}^{{\rm i}vx}}{P_{r}({\rm i}v)}, {\rm i} = \sqrt{-1} $, $ \sum\prime $表示当$ P_{r}(0) = 0 $时,缺$ v = 0 $的项.为了计算方便,规定:除0以外的任意虚整点i$ k\ (k = \pm1, \pm2, \cdots) $都不是$ P_{r}(\lambda) = 0 $的根.则$ G_{r}(x) $为对应算子$ P_{r}(D) $的广义Bernoulli核.

房艮孙在文献[3]中给出$ m_{r, \infty} $的渐进宽度的同时,引出了$ \tilde{m}_{r, \infty} $类.随后毕宁在文献[4]中定义了$ \tilde{m}_{r, p} $类,并讨论了$ \tilde{m}_{r, p} $在$ L_{p} $空间内的Kolmogorov宽度、Gelfand宽度、Linear宽度和Bernstein宽度,得到了$ 0<\beta<\frac{1}{4} $, $ 0<p<\infty $, $ n\geq1 $时, $ \tilde{m}_{r, p} $类的2n宽度的精确估计和相应的极子空间.以上学者们都是在$ L_{p} $空间中计算函数类的宽度,本文主要在Orlicz空间中研究新定义的函数类$ \tilde{m}_{r, M} $的宽度问题,利用Bernstein宽度可以控制Kolmogorov宽度和Gelfand宽度的下方估计,以及Linear宽度可以控制Kolmogorov宽度和Gelfand宽度的上方估计的思想,通过计算Bernstein宽度的下方估计和Linear宽度的上方估计来得到该函数类在Orlicz空间内的n-K宽度, n-G宽度, n-L宽度, n-B宽度的精确值,并给出了相应的极子空间.

设$ D_{n} = \{h(x)|h(x+\frac{\pi}{n}) = -h(x), x\in \mathbb{R}\}, \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为

$ \Phi_{n, r}(x) = \int_{0}^{2\pi} G_{r}(x-t) {\rm sgn} \sin nt\, {\rm d}t $

的单个零点(参见文献[5]).

引理2.1  对$ n\geq1 $,存在$ h_{n}\in D_{n} $, $ \| h_{n}\|_{M} = 1 $, $ \lambda_{n}>0 $,使

$ \begin{eqnarray*} \lambda_{n}& = &\| G_{r}\ast h_{n}\|_{M} = {\rm sup}\{\| G_{r}\ast h\|_{M}| \|h\|_{M}\leq1, h\in D_{n}\}\\ & = &{\rm sup}\bigg\{\frac{\| G_{r}\ast h\|_{M}}{\| h\|_{M}}\bigg| h\in D_{n}\cap L_{M, 2\pi}^{\ast}, h\neq0\bigg\}, \end{eqnarray*} $

且

(2.1) $ \int_{0}^{2\pi} p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\| G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))G_{r}(x-y)\, {\rm d}x = \lambda_{n}p\bigg(\frac{|h_{n}(y)|}{\|h_{n}\|_{(M)}}\bigg){\rm sgn}(h_{n}(y)), y\in \mathbb{R}.$

证  设$ \lambda_{n} = {\rm sup}\{\| G_{r}\ast h\|_{M}| \|h\|_{M}\leq1, h\in D_{n}\} $,则存在$ \{f_{m}\}_{m = 1}^{\infty}\subset D_{n} $,且$ \|f_{m}\|_{M}\leq1 $,满足

$\begin{matrix} \lim \\ m\to \infty \\\end{matrix}\|{{G}_{r}}*{{f}_{m}}{{\|}_{M}}={{\lambda }_{n}}.$

由文献[6]中性质2,性质4知:存在子序列$ \{f_{m_{j}}\}_{j = 1}^{\infty} $及$ g\in L_{M}^{\ast}[0, 2\pi] $, $ \|g\|_{M}\leq1 $,使得对任意$ s\in L_{(N)}^{\ast}[0, 2\pi] $,有

$ \begin{matrix} \lim \\ m\to \infty \\\end{matrix}\int_{0}^{2\pi} f_{m_{j}}(y)s(y)\, {\rm d}y = \int_{0}^{2\pi} g(y)s(y)\, {\rm d}y. $

这里,特别取$ s(y) = G_{r}(x-y) $,结合Lebesgue控制收敛原理,得

$\begin{matrix} \lim \\ m\to \infty \\\end{matrix}\| G_{r}\ast f_{m_{j}}\|_{M} = \| G_{r}\ast g\|_{M} = \lambda_{n}. $

现把$ g(x) $以$ 2\pi $为周期延拓至$ \mathbb{R} $,依然记作$ g(x) $,要证$ g\in D_{n} $.

取任意$ s\in L_{(N)}^{\ast}[-2\pi, 2\pi] $,由

$ \int_{-2\pi}^{2\pi} f_{m_{j}}(x)s(x)\, {\rm d}x = \int_{-2\pi+\frac{\pi}{n}}^{2\pi+\frac{\pi}{n}} f_{m_{j}}\bigg(x-\frac{\pi}{n}\bigg)s\bigg(x-\frac{\pi}{n}\bigg)\, {\rm d}x = -\int_{-2\pi}^{2\pi} f_{m_{j}}(x)s\bigg(x-\frac{\pi}{n}\bigg)\, {\rm d}x, $

两边同时取极限,再做变量替换,令$ x = x+\frac{\pi}{n} $,有

$ \int_{-2\pi}^{2\pi} g(x)s(x)\, {\rm d}x = \int_{-2\pi}^{2\pi} g(x)s\bigg(x-\frac{\pi}{n}\bigg)\, {\rm d}x = -\int_{-2\pi}^{2\pi} g\bigg(x+\frac{\pi}{n}\bigg)s(x)\, {\rm d}x, $

即

$ \int_{-2\pi}^{2\pi} \bigg[g(x)+g\bigg(x+\frac{\pi}{n}\bigg)\bigg]s(x)\, {\rm d}x = 0, $

特别取$ s(x) = {\rm sgn}\big(g(x)+g\big(x+\frac{\pi}{n}\big)\big) $,有

$ \int_{-2\pi}^{2\pi} \bigg|g(x)+g\bigg(x+\frac{\pi}{n}\bigg)\bigg|\, {\rm d}x = 0\Rightarrow g\bigg(x+\frac{\pi}{n}\bigg) = -g(x), {\rm a.e.}. $

因此可得$ g\in D_{n} $.取$ h_{n}(x) = g(x) $,显然有$ \|h_{n}\|_{M} = 1 $.任意$ h(x)\in D_{n} $,考虑

$ F(t) = \frac{\|G_{r}\ast (h_{n}+th)\|_{M}}{\|h_{n}+th\|_{M}}, t\in \mathbb{R}, $

明显$ \frac{{\rm d}F}{{\rm d}t}|_{t = 0} = 0 $.结合文献[7]即可算得

$ \begin{eqnarray*} &&\int_{0}^{2\pi} p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))G_{r}\ast h(x)\, {\rm d}x\\ & = &\lambda_{n} \int_{0}^{2\pi} p\bigg(\frac{|h_{n}(x)|}{\|h_{n}\|_{(M)}}\bigg){\rm sgn}(h_{n}(x))h(x)\, {\rm d}x. \end{eqnarray*} $

令

$ H_{n}(y) = \int_{0}^{2\pi} p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))G_{r}(x-y)\, {\rm d}x- \lambda_{n}p\bigg(\frac{|h_{n}(y)|}{\|h_{n}\|_{(M)}}\bigg){\rm sgn}(h_{n}(y)), $

简单计算可得

$ \int_{0}^{2\pi} H_{n}(y)h(y)\, {\rm d}y = 0, $

同证明$ g\in D_{n} $的方法一致,易证得

$ H_{n}(y)\in D_{n}. $

取$ h(y) = H_{n}(y) $,得$ H_{n}(y) = 0, $ a.e..再连续化,有$ H_{n}(y) = 0 $,且$ h(x) $连续.

引理2.2^[4]  若$ [a, b]\subset[0, \frac{\pi}{n}] $, $ 0<\beta<1 $, $ f\in C^{r-1} $, $ f(a) = 0 $, $ f^{(i)}(b) = 0, i = 0, 1, \cdots, r-1 $,则存在$ \xi\in[a, b] $,使得

$ P_{r}(D)f(\xi) = 0, $

其中, $ D = \frac{{\rm d}}{{\rm d}x} $.

设$ S_{2n, r-1} $为广义周期样条类(参见文献[8]),令

$ \hat{G}_{r}(x) = \left\{\begin{array}{ll} G_{r}(x), &P_{r}(0)\neq0, \\ G_{r}(x)-1, &P_{r}(0) = 0, \end{array}\right. $

则

$ S_{2n, r-1} = \bigg\{f(x)\in C^{r-2}|f(x) = \sum\limits_{i = 1}^{2n} C_{i}\hat{G}_{r}(x-\frac{i-1}{n}\pi), C_{i}\in R, i = 1, \cdots, 2n\bigg\}. $

引理2.3  若$ 0<\beta<\frac{1}{4} $, $ n\geq1 $,那么引理1中的$ h_{n}(x) $在$ [0, 2\pi) $内,有

1) $ h_{n}(x) $仅以$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点;

2) $ G_{r}\ast h_{n}(x) $仅以$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点.

证  由$ S_{2n, r-1} $对周期函数在$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $插值的存在唯一性,有

${{\hat{G}}_{r}}\left( \begin{matrix} \text{ }r, r+\frac{\pi }{n}, \cdots , r+\frac{2n-1}{n}\pi \\ \text{ }0, \frac{\pi }{n}, \cdots , \frac{2n-1}{n}\pi \\\end{matrix} \right)\ne 0.$

记

$\hat{K}_{r}(x, y) = \hat{G}_{r}\left(\begin{array}{c} x, r, r+\frac{\pi}{n}, \cdots, r+\frac{2n-1}{n}\pi\\ y, 0, \frac{\pi}{n}, \cdots, \frac{2n-1}{n}\pi \end{array}\right), h_{n}^{\alpha}(x) = h_{n}(x-\alpha), $

$\hat{M}_{r}(x, y) = \hat{K}_{r}(x, y)/\hat{G}_{r}\left(\begin{array}{c} r, r+\frac{\pi}{n}, \cdots, r+\frac{2n-1}{n}\pi\\ 0, \frac{\pi}{n}, \cdots, \frac{2n-1}{n}\pi \end{array}\right), \hat{L}_{r}(x, y) = |\hat{M}_{r}(x, y)|.$

由文献[4]有,对于每个固定的$ y\notin\{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $, $ \hat{K}_{r}(x, y) $作为$ x $的函数在$ [0, 2\pi) $内仅以$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点,对于每个固定的$ x\notin\{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $, $ \hat{K}_{r}(x, y) $作为$ y $的函数在$ [0, 2\pi) $内仅以$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点.

因为$ \hat{G}_{r}\ast h_{n}^{\alpha}(x+\frac{\pi}{n}) = -\hat{G}_{r}\ast h_{n}^{\alpha}(x) $,故存在$ \alpha_{1}\in R $,使得$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为$ \hat{G}_{r}\ast h_{n}^{\alpha_{1}}(x) $的零点,并且

$ \int_{0}^{2\pi}\hat{M}_{r}(x, y)h_{n}^{\alpha_{1}}(y)\, {\rm d}y = \int_{0}^{2\pi}\hat{G}_{r}(x-y)h_{n}^{\alpha_{1}}(y)\, {\rm d}y. $

令

$ \hat{\mu}_{n} = {\rm sup}\{\|\hat{M}_{r}h\|_{M}| \|h\|_{M}\leq1, h\in D_{n}\}, $

类似引理2.1的证明,存在$ \hat{h}_{n}\in D_{n} $, $ \|\hat{h}_{n}\|_{M} = 1 $,有

$ \hat{\mu}_{n} = \|\hat{M}_{r}\hat{h}_{n}\|_{M} = {\rm sup}\{\|\hat{M}_{r}h\|_{M}| \|h\|_{M}\leq1, h\in D_{n}\}, $

而

$ {\rm sup}\{\|\hat{M}_{r}h\|_{M}| \|h\|_{M}\leq1, h\in D_{n}\} = {\rm sup}\{\|\tilde{L}_{r}\tilde{h}\|_{M}| \tilde{h}(x) = h(x){\rm sgn} {\rm sin} nx, h\in D_{n}, \|h\|_{M}\leq1\}. $

因为前面$ {\rm sup} $达到的$ \hat{h}_{n}(x) $,满足$ \hat{h}_{n}(x)\geq0 $,所以后面$ {\rm sup} $达到的$ \tilde{h}_{n}(x) $,满足$ \tilde{h}_{n}(x){\rm sgn} {\rm sin}nx\geq0 $,即在$ [0, 2\pi) $内$ \hat{h}_{n}(x) $只在$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $处变号.又因$ \hat{G}_{r}\ast \hat{h}_{n}^{\alpha}(x+\frac{\pi}{n}) = -\hat{G}_{r}\ast \hat{h}_{n}^{\alpha}(x) $,故存在$ \alpha_{2}\in \mathbb{R} $,使得

$ \hat{M}_{r}\hat{h}_{n}^{\alpha_{2}}(x) = \hat{G}_{r}\ast \hat{h}_{n}^{\alpha_{2}}(x), $

从而$ \lambda_{n} = \hat{\mu}_{n} $.即$ h_{n}(x)\in D_{n} $,满足$ h_{n}(x){\rm sgn} {\rm sin}nx\geq0 $.由(2.1)式,有: $ h_{n}(x) $无零区间,当且仅当$ G_{r}\ast h_{n}(x) $无零区间.故若$ h_{n}(x) $无零区间,但除$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $外还有其他零点.令

$ \begin{eqnarray*} F_{r}(y)& = &\int_{0}^{2\pi}p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))G_{r}(x-y)\, {\rm d}x\\ & = & \lambda_{n}p\bigg(\frac{|h_{n}(y)|}{\|h_{n}\|_{(M)}}\bigg){\rm sgn}(h_{n}(y)), \end{eqnarray*} $

则$ F_{r}(y) $至少有$ 2n+2 $个零点,由文献[9]中Rolle定理,知

$ 2n+2\leq S_{c}^{-}(P_{r}^{\ast}(D)F_{r}) = S_{c}^{-}(G_{r}\ast h_{n}), $

其中, $ S_{c}^{-} $为循环变号数.再利用Rolle定理,有

$ 2n+2\leq S_{c}^{-}(P_{r}(D)G_{r}\ast h_{n}) = S_{c}^{-}(h_{n}), $

矛盾.若$ h_{n}(x) $存在零区间,即存在$ 0\leq a<b\leq\frac{\pi}{n} $,使$ h_{n}(x) = 0 $, $ x\in[a, b] $, $ [a, b] $必须是$ [0, \frac{\pi}{n}] $的真子空间,因此$ 0<a $或$ b<\frac{\pi}{n} $.我们讨论$ 0<a $的情况, $ b<\frac{\pi}{n} $的情况类似.

因为$ h_{n}(x)\geq0 $,所以$ x\in[0, \frac{\pi}{n}] $,故存在$ 0\leq c<a $,使$ h(x)>0 $, $ x\in(c, a) $; $ h_{n}(x) = 0 $, $ x\in[a, b] $; $ h(c) = 0 $;因为$ F_{r}^{(j)}(a) = 0, j = 0, 1, \cdots, r-1 $, $ F_{r}(c) = 0 $,所以利用引理2.2可得,存在$ c<\xi<a $,使得$ P_{r}^{\ast}(D)F_{r}(\xi) = 0 = G_{r}\ast h_{n}(\xi). $又因为

$ P_{r}^{\ast}(D)F_{r}(x) = p\left(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\right){\rm sgn}(G_{r}\ast h_{n}(x)), $

所以$ G_{r}\ast h_{n}(\xi) = 0, $$ \frac{{\rm d}^{v}}{{\rm d}x^{v}}G_{r}\ast h_{n}(x)|_{x = a} = 0, v = 0, 1, \cdots, r-1. $同理,存在$ \xi<\eta<a $,使得$ P_{r}(D)G_{r}\ast h_{n}(\eta) = h_{n}(\eta) = 0 $,这与$ h_{n}(x)>0 $, $ x\in(c, a) $矛盾.因此,在$ [0, 2\pi) $内, $ h_{n}(x) $的单零点只有$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $.

$ G_{r}\ast h_{n}(x) $的情况,同理可得.

由引理2.1–2.3得命题2.4.

命题2.4  $ M(u) $是给定的N函数, $ p(u) $是$ M(u) $的右导数.当$ 0<\beta<\frac{1}{4} $, $ n\geq1 $时,存在$ h_{n}(x)\in D_{n} $, $ \lambda_{n}>0 $,满足

$ \int_{0}^{2\pi} p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))G_{r}(x-y)\, {\rm d}x = \lambda_{n}p\bigg(\frac{|h_{n}(y)|}{\|h_{n}\|_{(M)}}\bigg){\rm sgn}(h_{n}(y)), y\in\mathbb{R}, $

且

1) $ h_{n}(x) $连续;

2)在$ [0, 2\pi) $内, $ h_{n}(x) $仅以$ \{\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点, $ G_{r}\ast h_{n}(x) $仅以$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为其单零点.

定义$ A_{2n} $为$ \tilde{m}_{r, M} $在$ S_{2n, r-1} $上以$ \{r+\frac{i-1}{n}\pi\}_{i = 1}^{2n} $为插值结点的插值算子.

定理3.1  当$ 0<\beta<\frac{1}{4} $, $ n\geq1 $时,有

$ d_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) = d^{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) = d_{2n}\prime(\tilde{m}_{r, M}; L_{M}^{\ast}) = b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) = \lambda_{n}, $

且

1) $ S_{2n, r-1} $为$ d_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) $的极子空间;

2) $ L_{2n}^{\ast} = \{f\in\tilde{m}_{r, M}|f(r+\frac{i-1}{n}\pi) = 0, i = 1, \cdots, 2n\} $为$ d^{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) $的极子空间;

3) $ A_{2n} $为$ d_{2n}\prime(\tilde{m}_{r, M}; L_{M}^{\ast}) $的最佳线性算子;

4) $ Y_{2n} = {\rm span}\{\int_{\frac{i-1}{n}\pi}^{\frac{i}{n}\pi} G_{r}(x-y)|h_{n}(y)|\, {\rm d}y, i = 1, \cdots, 2n\} $为$ b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast}) $的极子空间.

取任意$ 0<\sigma<\frac{\pi}{n} $,规定$ \xi_{0} = 0 $, $ \xi_{1} = \sigma $, $ \xi_{2} = \sigma+\frac{\pi}{n}, \cdots, \xi_{2n} = \sigma+\frac{2n-1}{n}\pi $, $ \xi_{2n+1} = 2\pi $, $ h_{n}^{\sigma}(x)\triangleq h_{n}(x-\sigma) $,

$ f_{i}^{\sigma}(x) = \left\{\begin{array}{ll} |h_{n}^{\sigma}(x)|, & { \xi_{i}\leq x<\xi_{i+1}$}, \\ 0 , & { x\in[0, 2\pi]\backslash [\xi_{i}, \xi_{i+1})$}, \end{array}\right. i = 0, 1, \cdots, 2n, $

$ f_{j}(x) = \left\{\begin{array}{ll} |h_{n}(x)| , & \frac{j-1}{n}\pi\leq x<\frac{j}{n}\pi, \\ 0 , & x\in[0, 2\pi]\backslash [\frac{j-1}{n}\pi, \frac{j}{n}\pi), \end{array}\right.j = 1, \cdots, 2n, $

$ g_{i}^{\sigma}(x) = \int_{0}^{2\pi}G_{r}(x-y)f_{i}^{\sigma}(y)\, {\rm d}y, i = 0, 1, \cdots, 2n, $

$ g_{i}(x) = \int_{0}^{2\pi}G_{r}(x-y)f_{i}(y)\, {\rm d}y, i = , 1, \cdots, 2n. $

引理3.2^[4]  当$ 0<\sigma<\frac{\pi}{n} $, $ 0<\beta<\frac{1}{4} $, $ n\geq1 $时, $ g_{0}^{\sigma}(x), g_{1}^{\sigma}(x), g_{2}^{\sigma}(x), \cdots, g_{2n}^{\sigma}(x) $为一组WT系.

引理3.3

$ \lim\limits_{\sigma\rightarrow0} \inf\limits_{C_{i}\neq0\atop i = 0, \cdots, 2n} \frac{\bigg\|\sum\limits_{i = 0}^{2n}C_{i}g_{i}^{\sigma}\bigg\|_{M}} {\bigg\|\sum\limits_{i = 0}^{2n}C_{i}f_{i}^{\sigma}\bigg\|_{M}} = \inf\limits_{C_{i}\neq0\atop i = 1, \cdots, 2n} \frac{\bigg\|\sum\limits_{i = 1}^{2n}C_{i}g_{i}\bigg\|_{M}} {\bigg\|\sum\limits_{i = 1}^{2n}C_{i}f_{i}\bigg\|_{M}}. $

证  因为$ f_{i}^{\sigma}(x)\Rightarrow f_{i}(x), i = 1, \cdots, 2n $, $ f_{0}^{\sigma}(x)\Rightarrow0 (\sigma\rightarrow0) $.所以$ g_{i}^{\sigma}(x) $一致收敛到$ g_{i}(x), i = 1, \cdots, 2n $, $ g_{0}^{\sigma}(x) $一致收敛到$ 0(\sigma\rightarrow0) $.从而,得

$ \inf\limits_{C_{i}\neq0\atop i = 0, \cdots, 2n} \frac{\bigg\|\sum\limits_{i = 0}^{2n}C_{i}g_{i}^{\sigma}\bigg\|_{M}}{\bigg\|\sum\limits_{i = 0}^{2n}C_{i}f_{i}^{\sigma}\bigg\|_{M}} = \inf\limits_{C_{i}\neq0\atop i = 0, \cdots, 2n} \frac{\bigg\|\sum\limits_{i = 1}^{2n}C_{i}g_{i}\bigg\|_{M}}{\bigg\|\sum\limits_{i = 1}^{2n}C_{i}f_{i}\bigg\|_{M}}, \sigma\rightarrow0. $

证毕.

引理3.4  当$ 0<\beta<\frac{1}{4} $, $ n\geq1 $时,有

$ \inf\limits_{C_{i}\neq0\atop i = 0, \cdots, 2n} \frac{\bigg\|\sum\limits_{i = 1}^{2n}C_{i}g_{i}\bigg\|_{M}}{\bigg\|\sum\limits_{i = 1}^{2n}C_{i}f_{i}\bigg\|_{M}}\geq\lambda_{n}. $

证  设$ \inf\limits_{C_{i}\neq0\atop i = 0, \cdots, 2n} \frac{\big\|\sum\limits_{i = 1}^{2n}C_{i}g_{i}\big\|_{M}}{\big\|\sum\limits_{i = 1}^{2n}C_{i}f_{i}\big\|_{M}} = \frac{\big\|\sum\limits_{i = 1}^{2n}\bar{C}_{i}g_{i}\big\|_{M}}{\big\|\sum\limits_{i = 1}^{2n}\bar{C}_{i}f_{i}\big\|_{M}} = \bar{\mu}_{n} $, $ \bar{g}(x) = \sum\limits_{i = 1}^{2n}\bar{C}_{i}g_{i}(x) $, $ \bar{f}(x) = \sum\limits_{i = 1}^{2n}\bar{C}_{i}f_{i}(x) $, $ |\bar{C}_{i}|\leq1, i = 1, \cdots, 2n $.且存在$ 1\leq i_{0}\leq2n $,使$ \bar{C}_{i_{0}} = (-1)^{i_{0}-1} $.利用反证法,假设$ \bar{\mu}_{n}<\lambda_{n} $,因为

$ G_{r}\ast h_{n}(x)-\bar{g}(x) = \sum\limits_{i = 1}^{2n}[(-1)^{i-1}-\bar{C}_{i}]g_{i}(x), $

对$ \bar{C}_{i_{0}} $作微小$ \varepsilon $ -摄动,得$ \bar{g}^{\varepsilon}(x) $,利用Rolle定理,有

(3.1) $ S_{c}^{-}(G_{r}\ast h_{n}-\bar{g}^{\varepsilon})\leq2n-2, $

由于

$ \begin{eqnarray*} &&\int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}(x)|}{\|\bar{g}\|_{(M)}}\bigg){\rm sgn}(\bar{g}(x))\bigg]g_{k}(x)\, {\rm d}x\\ & = &\bigg[\lambda_{n}(-1)^{k-1}- \bar{\mu}_{n}p\bigg(\frac{|\bar{C}_{k}|}{\|\bar{C}_{k}\|_{(M)}}\bigg){\rm sgn}(\bar{C}_{k})\bigg]\beta_{k} \end{eqnarray*} $

严格交错变号,这里$ k = 1, \cdots, 2n $, $ 0<\beta_{k} = \int_{\frac{k-1}{n}\pi}^{\frac{k}{n}\pi} M(|h_{n}(x))|\, {\rm d}x $.而

$ \begin{eqnarray*} && \int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{k}^{\sigma}(x)\, {\rm d}x\\ & = &\int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}(x)|}{\|\bar{g}\|_{(M)}}\bigg){\rm sgn}(\bar{g}(x))\bigg]g_{k}^{\sigma}(x)\, {\rm d}x\\ &&+\int_{0}^{2\pi} \bigg[p\bigg(\frac{|\bar{g}(x)|}{\|\bar{g}\|_{(M)}}\bigg){\rm sgn}(\bar{g}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{k}^{\sigma}(x)\, {\rm d}x, \end{eqnarray*} $

$ k = 1, \cdots, 2n $.因为$ g_{k}^{\sigma}(x) $一致收敛到$ g_{k}(x) $, $ \sigma\rightarrow0 $, $ k = 1, \cdots, 2n $, $ \bar{g}^{\varepsilon}(x) $一致收敛到$ \bar{g}(x) $, $ \varepsilon\rightarrow0 $,所以由$ \sigma $、$ \varepsilon $的任意性,令它们充分小,有

$ \int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{k}^{\sigma}(x)\, {\rm d}x, k = 1, \cdots, 2n $

严格交错变号.如果

$ \int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{0}^{\sigma}(x)\, {\rm d}x = 0, $

则取$ g_{0}^{\ast}(x) = g_{0}^{\sigma}(x)+g_{k}^{\ast}(x) $, $ g_{k}^{\ast}(x) = g_{k}^{\sigma}(x) $, $ k = 1, \cdots, 2n $.如果

$ \int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{0}^{\sigma}(x)\, {\rm d}x\neq0, $

则取$ g_{0}^{\ast}(x) = g_{0}^{\sigma}(x) $或$ -g_{0}^{\sigma}(x) $, $ g_{k}^{\ast}(x) = g_{k}^{\sigma}(x) $, $ k = 1, \cdots, 2n $.使得

$ \int_{0}^{2\pi} \bigg[p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}}\bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg]g_{k}^{\ast}(x)\, {\rm d}x, k = 0, 1, \cdots, 2n $

严格交错变号,其中$ \{g_{0}^{\ast}(x), \cdots, g_{2n}^{\ast}(x)\} $为WT系.根据引理3.2和文献[9]中的性质3.6(2),有

$ \begin{eqnarray*} 2n&\leq&\bar{S}_{c}\bigg(p\bigg(\frac{|G_{r}\ast h_{n}(x)|}{\|G_{r}\ast h_{n}\|_{(M)}}\bigg){\rm sgn}(G_{r}\ast h_{n}(x))- p\bigg(\frac{|\bar{g}^{\varepsilon}(x)|}{\|\bar{g}^{\varepsilon}\|_{(M)}} \bigg){\rm sgn}(\bar{g}^{\varepsilon}(x))\bigg)\\ & = & \bar{S}_{c}(G_{r}\ast h_{n}(x)-\bar{g}^{\varepsilon}(x)), \end{eqnarray*} $

与(3.1)式矛盾.即证得$ \bar{\mu}_{n}\geq\lambda_{n} $.

引理3.5  如果满足定理条件,则

$ b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast})\geq\lambda_{n}. $

证  反证法,假设$ b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast})<\lambda_{n} $,存在$ \tau>0 $,使得$ b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast})<\lambda_{n}-\tau $,根据引理3.3,引理3.4,存在$ 0<\sigma<\frac{\pi}{n} $,使得

$\begin{matrix} \inf \\ \begin{matrix} {{C}_{i}}\ne 0 \\ i=0, \cdots , 2n \\\end{matrix} \\\end{matrix}\frac{\|\sum\limits_{i=0}^{2n}{{{C}_{i}}}g_{i}^{\sigma }{{\|}_{M}}}{\|\sum\limits_{i=0}^{2n}{{{C}_{i}}}f_{i}^{\sigma }{{\|}_{M}}}\ge {{\lambda }_{n}}-\tau .$

从而$ b_{2n}(\tilde{m}_{r, M}; L_{M}^{\ast})\geq\lambda_{n}-\tau $,出现矛盾.即证.

引理3.6^[9]  设$ \Lambda $是线性赋范空间$ X $上的一个中心对称的闭凸子集,则

$ b_{n}(\Lambda; X)\leq d^{n}(\Lambda; X). $

引理3.7^[2]  对任意$ \Lambda\subset X $,有

$ d_{n}(\Lambda; X)\leq d_{n}\prime(\Lambda; X); $

$ d^{n}(\Lambda; X)\leq d_{n}\prime(\Lambda; X). $

引理3.8  如果满足定理条件,则

$ {\rm sup}\{\|f-A_{2n}(f)\|_{M}|f\in\tilde{m}_{r, M}\}\leq\lambda_{n}. $

证  仿照文献[9],再结合文献[10]即可证.

综合引理2.1–2.3及引理3.2–3.8,即可完成定理的证明.关于极子空间的论断只需一些简单的计算便可得.