首先介绍Orlicz空间.基于 $N$函数 $M(u)$是幂函数 $|u|^{p}(p>1)$的推广, Orlicz空间是熟知的 $L_{p}(p>1)$空间的推广.

定义1.1^[1] 称定义在 $\mathbf{R}=(-\infty, +\infty)$上的实值函数 $M(u)$为 $N$函数, 假设它具有下列性质

(1) $M(u)$为偶的连续凸函数且 $M(0)=0$;

(2) 当 $u>0$时 $M(u)>0$;

(3) $\lim\limits_{u\rightarrow 0}\frac{M(u)}{u}=0, \quad \lim\limits_{u\rightarrow \infty}\frac{M(u)}{u}=\infty.$

对于给定的 $N$函数 $M(u)$, 其余 $N$函数记为 $N(v)$.

定义1.2^[1] 称 $N$函数 $M(u)$满足 $\triangle_{2}$条件(简记为 $M(u)\in\triangle_{2}$)是指存在 $k>0$, $u_{0}>0$, 使当 $u\geq u_{0}$时, 有

$ M(2u)\leq kM(u). $

由 $N$函数 $M(u)$生成的Orlicz类 $L_{M}[0, 1]$是指满足 $ \rho(u, M)=\displaystyle\int_{0}^{1}M(u(x))dx<+\infty $的可测函数 $ u(x)$的全体.

定义1.3^[1] Orlicz空间 $L_{M}^{*}[0, 1]$是指有限的Orlicz范数

$ \|u\|_{M}= \sup\limits_{\rho(v, N)\leq1}\Big|\int_{0}^{1} u(x)v(x)dx\Big| $

的可测函数 $u(x)$的全体.

Orlicz空间还赋予与Orlicz范数等价的Luxemburg范数

$ \begin{eqnarray} &&\|u\|_{(M)}=\inf\limits_{\lambda>0}\{\lambda: \int_{0}^{1}M (|\frac{u(x)}{\lambda}|)dx\leq1\}, \nonumber\\ &&\|u\|_{(M)}\leq\|u\|_{M}\leq 2 \|u\|_{(M)}. \end{eqnarray} $

(1.1)

对于 $f\in L_{M}^{*}[0, 1]$, $r$阶带权函数 $\varphi(x)=\sqrt{x(1-x)}$的Ditzian-Totik连续模 $\omega_{r, \varphi}(f, t)_{M}$, $K$ -泛函 $K_{r, \varphi}(f, t^{r})_{M}$和修正的 $K$-泛函 $\bar{K}_{r, \varphi}(f, t^{r})_{M}$的定义^[2]为

$ \begin{eqnarray*} &&\omega_{r, \varphi}(f, t)_{M}=\sup\limits_{0<h\leq t}\|\triangle_{h\varphi(x)}^{r}f(x)\|_{M}, \\ &&K_{r, \varphi}(f, t^{r})_{M}=\inf\limits_{g^{(r-1)}\in A.C._{loc}}\Big\{\|f-g\|_{M}+t^{r}\|\varphi^{r} g^{(r)}\|_{M}\Big\}, \\ &&\bar{K}_{r, \varphi}(f, t^{r})_{M}=\inf\limits_{g^{(r-1)}\in A.C._{loc}}\Big\{\|f-g\|_{M}+t^{r}\|\varphi^{r} g^{(r)}\|_{M}+t^{2r}\|g^{(r)}\|_{M}\Big\}, \end{eqnarray*} $

其中 $\triangle_{h\varphi(x)}^{r}f(x)=\sum\limits_{k=0}^{r}(-1)^{k}{r \choose k}f\Big(x+(\frac{r}{2}-k)h\varphi(x)\Big)$.

连续模 $\omega_{r, \varphi}(f, t)_{M}$与两个 $K$-泛函等价^[2]的, 即存在常数 $k_{1}, k_{2}>0$, 使得

$ \begin{eqnarray*} &&k_{1}^{-1} \omega_{r, \varphi}(f, t)_{M} \leq K_{r, \varphi}(f, t^{r})_{M}\leq k_{1} \omega_{r, \varphi}(f, t)_{M}, \\ &&k_{2}^{-1} \omega_{r, \varphi}(f, t)_{M} \leq \bar{K}_{r, \varphi}(f, t^{r})_{M}\leq k_{2} \omega_{r, \varphi}(f, t)_{M}. \end{eqnarray*} $

对于 $f\in L_{M}^{*}[0, 1]$, Bernstein-Durrmeyer算子的定义为

$ B_{n}(f, x)=(n+1)\sum\limits_{k=0}^{n}P_{n, k}(x)\int_{0}^{1}P_{n, k}(t)f(t)dt, $

其中 $P_{n, k}(x)={n \choose k}x^{k}(1-x)^{n-k}$.

关于Bernstein型算子的文章不少^[3-8].Bernstein-Durrmeyer算子在 $L_{p}[0, 1]$空间的逼近速度^[6]有 $\|B_{n}f-f\|_{p}=O(n^{-\alpha})\Leftrightarrow\omega_{2, \varphi}(f, t)_{p}=O(t^{2\alpha})$.为了获得更快的逼近速度, Ditian和Ivanov在文[6]中用 $B_{n}(f)$的线性组合构造了一个算子 $O_{n}(f)$, 并证明了对于 $r>\alpha$, 有

$ \|O_{n}(f)-f\|_{p}=O(n^{-\alpha})\Leftrightarrow\omega_{2r, \varphi}(f, t)_{p}=O(t^{2\alpha}). $

最近, Sablonnière在文[9]中引进了一类所谓的拟中插式算子.设 $\Pi_{n}$表示次数至多为 $n$的多项式空间, 若 $\mathbb{B}_{n}$和 $\mathbb{A}_n=\mathbb{B}_{n}^{-1}$是 $\Pi_{n}$中的线性自同构算子, 并且能够表示成带有多项式系数的微分算子形式 $\mathbb{B}_{n}=\sum\limits_{k=0}^{n}\beta_{k}^{n}D^{k}$和 $\mathbb{A}_{n}=\sum\limits_{k=0}^{n}\alpha_{k}^{n}D^{k}$, 这里 $D=\frac{d}{dx}$, $D^{0}=id$, 则一类拟中插式算子被如下定义

$ \mathbb{B}_{n}^{(r)}=\mathbb{A}_{n}^{(r)}\circ\mathbb{B}_{n}, 0\leq r\leq n, $

这里 $\mathbb{A}_{n}^{(r)}=\sum\limits_{k=0}^{n}\alpha_{k}^{n}D^{k}$.通常在 $\Pi_{n}$上 $\mathbb{B}_{n}^{(0)}=\mathbb{B}_{n}$, $\mathbb{B}_{n}^{(n)}=id$.进而还有当 $0\leq r\leq n$时, 对于所有的 $P\in\Pi_{n}$, 有 $\mathbb{B}_{n}^{(r)}P=P$.

对于 $f\in L_{M}^{*}[0, 1]$, Bernstein-Durrmeyer算子拟中插式为

$ B_{n}^{(k)}(f, x)=\mathbb{A}_{n}^{(k)}\circ B_{n}=\sum\limits_{j=0}^{k}\alpha_{j}^{n}D^{j}B_{n}(f, x):=\sum\limits_{j=0}^{k}\alpha_{j}^{n}B_{n, j}(f, x), 0\leq k\leq n, $

其中 $B_{n, j}=D^{j}B_{n}$.显然 $ B_{n}^{(0)}=B_{n}$, $B_{n}^{(n)}=id$, 且对于 $P\in\Pi_{n}$, $0\leq k\leq n$, 有 $B_{n}^{(k)}P=P$.

郭顺生等在文[7]中给出了Bernstein-Durrmeyer算子拟中插式在 $C[0, 1]$中逼近的等价定理.在文[8]中给出了Bernstein-Durrmeyer算子拟中插式在 $L_{p}[0, 1]$中逼近的等价定理.本文得到的主要结果有

定理1.1 (正定理)设 $f\in L_{M}^{*}[0, 1]$, $\varphi(x)=\sqrt{x(1-x)}$, $n\geq2r-1$, $r\in\mathbf{N}$, $N(v)\in\Delta_{2}$, 则

$ \|B_{n}^{(2r-1)}(f)-f\|_{M}\leq C \omega_{2r, \varphi}(f, \frac{1}{\sqrt{n}})_{M}. $

本文中 $C$表示与 $f$与 $n$无关的正常数, 在不同的地方, 它的值有所不同.

定理1.2 (逆定理)对于 $f\in L_{M}^{*}[0, 1]$, $\varphi(x)=\sqrt{x(1-x)}$, $n\geq4r$, $r\in\mathbf{N}$, $0<\alpha<r$, 有

$ \|B_{n}^{(2r-1)}(f)-f\|_{M}=O(n^{-\alpha}) \Rightarrow \omega_{2r, \varphi}(f, t)_{M}=O(t^{2\alpha}). $

定理1.3 (等价定理)对于 $f\in L_{M}^{*}[0, 1]$, $\varphi(x)=\sqrt{x(1-x)}$, $n\geq4r$, $r\in\mathbf{N}$, $0<\alpha<r$, $N(v)\in\Delta_{2}$, 有

$ \|B_{n}^{(2r-1)}(f)-f\|_{M}=O(n^{-\alpha})\Leftrightarrow\omega_{2r, \varphi}(f, t)_{M}=O(t^{2\alpha}). $

本文中 $\delta_{n}(x)=\max\{\varphi(x), \frac{1}{\sqrt{n}}\}$, $E_{n}=[\frac{1}{n}, 1-\frac{1}{n}]$, $E_{n}^{c}=[0, \frac{1}{n})\cup(1-\frac{1}{n}, 1].$为了证明主要结果, 下面给出一些引理.

引理2.1^[10] 对于 $f\in L_{M}^{*}[0, 1]$, $\varphi(x)=\sqrt{x(1-x)}$, 有

$ \begin{eqnarray} &&B_{n, j}(f, x)=\frac{(n+1)!n!}{(n-j)!(n+j)!}\sum\limits_{k=0}^{n-j}P_{n-j, k}(x)\int_{0}^{1}P_{n+j, k+j}(t)f^{(j)}(t)dt, \end{eqnarray} $

(2.1)

$ \begin{eqnarray} &&|P_{n, k}^{(j)}(x)|\leq C\sum\limits_{i=0}^{j}(\frac{\sqrt{n}}{\varphi(x)})^{j+i}|\frac{k}{n}-x|^{i}P_{n, k}(x), x\in E_{n}, \end{eqnarray} $

(2.2)

$ \begin{eqnarray} &&\sum\limits_{k=0}^{n}|\frac{k}{n}-x|^{i}P_{n, k}(x)\leq Cn^{-\frac{i}{2}}\varphi^{i}(x), x\in E_{n}, \end{eqnarray} $

(2.3)

$ \begin{eqnarray} &&B_{n}((t-x)^{2i}, x)\leq Cn^{-i}(\varphi^{2}(x)+\frac{1}{n})^{i}, x\in[0, 1]. \end{eqnarray} $

(2.4)

引理2.2 对于 $f\in L_{M}^{*}[0, 1]$, $r, s\in\mathbf{N_{0}}=\mathbf{N}\cup\{0\}$, 有

$ \|\delta_{n}^{s}\varphi^{2r}D^{2r+s}(B_{n}(f))\|_{M}\leq Cn^{\frac{s}{2}}\|\varphi^{2r}f^{(2r)}\|_{M}, $

其中当 $r\neq0$时, $f\in W_{\varphi}^{2r}(L_{M}^{*}[0, 1])$; 当 $r=0$时, $f\in L_{M}^{*}[0, 1]$.

证 当 $x\in E_{n}^{c}$时 $\delta_{n}(x)\sim n^{-\frac{1}{2}}$.由(1.1), (2.1) 式和Jensen不等式得到

$ \begin{eqnarray*} &&\|\delta_{n}^{s}(x)\varphi^{2r}(x)D^{2r+s}(B_{n}(f, x))\|_{M, E_{n}^{c}} \\ &=&\|\delta_{n}^{s}(x)\varphi^{2r}(x)\frac{(n+1)!n!}{(n+2r)!(n-2r)!}\sum\limits_{k=0}^{n-2r}P_{n-2r, k}^{(s)}(x) \int_{0}^{1}P_{n+2r, k+2r}(t)f^{(2r)}(t)dt \|_{M, E_{n}^{c}}\\ &\leq&2\inf_{\lambda>0} \Big\{\lambda:\int_{E_{n}^{c}}M\Big(\frac{1}{\lambda}\delta_{n}^{s}(x)\varphi^{2r}(x) \sum\limits_{k=0}^{n-2r}\frac{(n-2r)!}{(n-2r-s)!}P_{n-2r-s, k}(x)\sum\limits_{i=0}^{s}(n+1)\\ &&\int_{0}^{1} P_{n+2r, k+2r+i}(t)\varphi^{-2r}(t)|\varphi^{2r}(t)f^{(2r)}(t)|dt\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0} \Big\{\lambda:\int_{E_{n}^{c}}M\Big(\frac{Cn^{\frac{s}{2}}}{\lambda} \sum\limits_{k=0}^{n-2r-s}P_{n-s, k+r}(x)(n+1)\\ &&\int_{0}^{1} P_{n, k+r+i}(t)|\varphi^{2r}(t)f^{(2r)}(t)|dt\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0} \Big\{\lambda:\int_{E_{n}^{c}}\sum\limits_{k=0}^{n-s}P_{n-s, k}(x) (n+1)\int_{0}^{1}P_{n, k+i}(t)M\Big(\frac{Cn^{\frac{s}{2}}}{\lambda}|\varphi^{2r}(t)f^{(2r)}(t)|\Big) dtdx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0} \Big\{\lambda:\int_{0}^{1}M\Big(\frac{Cn^{\frac{s}{2}}}{\lambda}|\varphi^{2r}(t)f^{(2r)}(t)|\Big) dt\leq1\Big\}\\ &\leq& Cn^{\frac{s}{2}}\|\varphi^{2r}f^{(2r)}\|_{(M)}\leq Cn^{\frac{s}{2}}\|\varphi^{2r}f^{(2r)}\|_{M}. \end{eqnarray*} $

当 $x\in E_{n}$时, $\delta_{n}(x)\sim \varphi(x)$.由引理2.1得

$ \begin{eqnarray*} &&|\delta_{n}^{s}(x)\varphi^{2r}(x)D^{2r+s}(B_{n}(f, x))|\\ &\leq& C\varphi^{s}(x)\sum\limits_{k=0}^{n-2r}\sum\limits_{i=0}^{s}\Big(\frac{\sqrt{n-2r}}{\varphi(x)}\Big)^{s+i} \Big|\frac{k}{n-2r}-x\Big|^{i}P_{n, k+r}(x)(n+1)\int_{0}^{1}P_{n, k+r}(t)\varphi^{2r}(t)\\ &&|f^{(2r)}(t)|dt\\ &\leq& C\sum\limits_{i=0}^{s}n^{\frac{s+i}{2}}\varphi^{-i}(x)\Big(\sum\limits_{k=0}^{n-2r}\Big|\frac{k+r}{n}-x\Big|^{i}P_{n, k+r}(x) +\sum\limits_{k=0}^{n-2r}\Big|\frac{k}{n-2r}-\frac{k+r}{n}\Big|^{i}P_{n, k+r}(x)\Big)\\ &&(n+1)\int_{0}^{1}P_{n, k+r}(t)\varphi^{2r}(t)|f^{(2r)}(t)|dt\\ &\leq& C\sum\limits_{i=0}^{s}n^{\frac{s+i}{2}}\varphi^{-i}(x) \sum\limits_{k=0}^{n-2r}\Big|\frac{k+r}{n}-x\Big|^{i}P_{n, k+r}(x)(n+1)\int_{0}^{1}P_{n, k+r}(t)\varphi^{2r}(t)|f^{(2r)}(t)|dt\\ &&+ C\sum\limits_{i=0}^{s}n^{\frac{s+i}{2}}\varphi^{-i}(x)\sum\limits_{k=0}^{n-2r}\Big|\frac{k}{n-2r}-\frac{k+r}{n}\Big|^{i}P_{n, k+r}(x)\\ &&(n+1)\int_{0}^{1}P_{n, k+r}(t)\varphi^{2r}(t)|f^{(2r)}(t)|dt\\ &:=&I_{1}+I_{2}. \end{eqnarray*} $

从而 $\|\delta_{n}^{s}\varphi^{2r}D^{2r+s}(B_{n}(f))\|_{M, E_{n}}\leq\|I_{1}\|_{M, E_{n}}+\|I_{2}\|_{M, E_{n}}$. $\|I_{1}\|_{M, E_{n}}$和 $\|I_{2}\|_{M, E_{n}}$估计类似, 所以只证 $\|I_{1}\|_{M, E_{n}}$.

由(1.1) 式及Jensen不等式得到

$ \begin{eqnarray*} &&\|I_{1}\|_{M, E_{n}}\\ &\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}}M\Big(\frac{C}{\lambda}\sum\limits_{i=0}^{s}n^{\frac{s+i}{2}}\varphi^{-i}(x)\sum\limits_{k=0}^{n-2r} \Big|\frac{k+r}{n}-x\Big|^{i}P_{n, k+r}(x)(n+1)\\ &&\int_{0}^{1}P_{n, k+r}(t)\varphi^{2r}(t)|f^{(2r)}(t)|dt\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}}M\Big(\frac{C}{\lambda}\sum\limits_{k=0}^{n}\Big|\frac{k}{n}-x\Big|^{i}P_{n, k}(x) n^{\frac{s+i}{2}}\varphi^{-i}(x)(n+1)\\ &&\int_{0}^{1}P_{n, k}(t)\varphi^{2r}(t)|f^{(2r)}(t)|dt\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}}\sum\limits_{k=0}^{n}P_{n, k}(x)M\Big(\sum\limits_{i=0}^{s}\Big|\frac{k}{n}-x\Big|^{i} n^{\frac{i}{2}}\varphi^{-i}(x)(n+1)\\ &&\int_{0}^{1}P_{n, k}(t)n^{\frac{s}{2}}\varphi^{2r}(t)\frac{C}{\lambda}|f^{(2r)}(t)|dt\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{0}^{1}\sum\limits_{k=0}^{n}P_{n, k}(x)(n+1)\int_{0}^{1}P_{n, k}(t)M\Big(\frac{C}{\lambda}n^{\frac{s}{2}}\varphi^{2r}(t) |f^{(2r)}(t)|\Big)dtdx\leq1\Big\}\\ &\leq& Cn^{\frac{s}{2}}\|\varphi^{2r}f^{(2r)}\|_{(M)}\leq Cn^{\frac{s}{2}}\|\varphi^{2r}f^{(2r)}\|_{M}. \end{eqnarray*} $

引理2.3^[8] 对于 $j\geq0$, $x\in[0, 1]$, 有 $ |\alpha_{j}^{n}(x)|\leq Cn^{-\frac{j}{2}}\delta_{n}^{j}(x). $

引理2.4 对于 $f\in L_{M}^{*}[0, 1]$, $r\in\mathbf{N_{0}}$, 有 $ \|B_{n}^{(r)}(f)\|_{M}\leq C\|f\|_{M}.$

证 由引理2.2, 引理2.3得到

$ \begin{eqnarray*} \|B_{n}^{(r)}(f)\|_{M}&=&\|\sum\limits_{j=0}^{r}\alpha_{j}^{n}(x)D^{j}B_{n}(f)\|_{M} \leq\sum\limits_{j=0}^{r}\|\alpha_{j}^{n}(x)D^{j}B_{n}(f)\|_{M}\\ &\leq& Cn^{-\frac{j}{2}}\sum\limits_{j=0}^{r}\|\delta_{n}^{j}(x)D^{j}B_{n}(f)\|_{M}\leq C\|f\|_{M}. \end{eqnarray*} $

引理2.5^[11] 若 $N$函数 $N(v)$满足 $\Delta_{2}$条件, 则函数 $f\in L_{M}^{*}[0, 1]$的Hardy-Littlewood极大函数

$ G(x)=\sup\limits_{0\leq t\leq1, t\neq x}\frac{1}{t-x}\int_{x}^{t}|f(u)|du, $

$G(x)\in L_{M}^{*}[0, 1]$且 $ \|G\|_{M}\leq C\|f\|_{M}. $

定理1.1的证明 由 $\bar{K}_{2r, \varphi}(f, t^{2r})_{M}$的定义, 可以选取 $g(t)=g_{n}(t)$, 使得

$ \begin{eqnarray} \|f-g\|_{M}+n^{-r}\|\varphi^{2r} g^{(2r)}\|_{M}+n^{-2r}\| g^{(2r)}\|_{M}\leq2\bar{K}_{2r, \varphi}(f, n^{-r})_{M}.\end{eqnarray} $

(2.5)

对于多项式 $P\in\Pi_{k}$, $B_{n}^{(k)}P=P$, 及引理2.4, 有

$ \begin{eqnarray}&&\|B_{n}^{(2r-1)}(f)-f\|_{M}\leq C\|f-g\|_{M}+\|B_{n}^{(2r-1)}(g)-g\|_{M}\nonumber\\ &=&C\|f-g\|_{M}+\|B_{n}^{(2r-1)}(R_{2r}(g, \cdot, x), x)\|_{M}:=C\|f-g\|_{M}+J, \end{eqnarray} $

(2.6)

其中 $R_{2r}(g, \cdot, x)=\frac{1}{(2r-1)!}\displaystyle\int_{x}^{t}(t-u)^{2r-1}g^{(2r)}(u)du$为 $g$的泰勒展开式的积分型余项.

下面只需估计 $J$.

$ \begin{eqnarray} J=\|\sum\limits_{j=0}^{2r-1}\alpha_{j}^{n}(x)D^{j}B_{n}(R_{2r}(g, \cdot, x), x)\|_{M}\leq\sum\limits_{j=0}^{2r-1}J_{j}, \end{eqnarray} $

(2.7)

其中 $J_{j}=\|\alpha_{j}^{n}(x)D^{j}B_{n}(R_{2r}(g, \cdot, x), x)\|_{M}$,

$ \begin{eqnarray} J_{j}&\leq&\|\alpha_{j}^{n}(x)D^{j}B_{n}(R_{2r}(g, \cdot, x), x)\|_{M, E_{n}}+ \|\alpha_{j}^{n}(x)D^{j}B_{n}(R_{2r}(g, \cdot, x), x)\|_{M, E_{n}^{c}}\nonumber\\ &:=&J_{j_{1}}+J_{j_{2}}.\end{eqnarray} $

(2.8)

先估计 $J_{j_{1}}$.由引理2.3, (1.1), (2.2) 式及不等式(当 $u$在 $t$和 $x$之间时) $ \frac{|t-u|}{\delta_{n}^{2}(u)}\leq\frac{|t-x|}{\delta_{n}^{2}(x)}, $得到

$ \begin{eqnarray*} J_{j_{1}}&\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}}M\Big(\frac{1}{\lambda} \Big|\alpha_{j}^{n}(x)\sum\limits_{k=0}^{n}P_{n, k}^{(j)}(x)(n+1)\int_{0}^{1}P_{n, k}(t) \frac{1}{(2r-1)!}\\ && \int_{x}^{t}(t-u)^{2r-1}g^{(2r)}(u)dudt\Big|\Big)dx\leq1\Big\}\\ &\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}}M\Big(\frac{C}{\lambda}\Big|n^{-\frac{j}{2}}\varphi^{j}(x) \sum\limits_{k=0}^{n}\sum\limits_{i=0}^{j}\Big(\frac{\sqrt{n}}{\varphi(x)}\Big)^{j+i}\Big|\frac{k}{n}-x\Big|^{i} P_{n, k}(x)(n+1)\\ &&\int_{0}^{1}P_{n, k}(t)\frac{(t-x)^{2r}}{\delta_{n}^{2r}(x)}dt G(x)\Big|\Big)dx\leq1\Big\}. \end{eqnarray*} $

由Hölder不等式, (2.3), (2.4) 式, 得

$ \Big|\sum\limits_{i=0}^{j}n^{\frac{j}{2}}\varphi^{-i}(x)\delta_{n}^{-2r}(x)\sum\limits_{k=0}^{n}P_{n, k}(x) \Big|\frac{k}{n}-x\Big|^{i}(n+1)\int_{0}^{1}P_{n, k}(t)(t-x)^{2r}dt\Big|\leq Cn^{-r}. $

从而由(1.1) 式及引理2.5, 得到

$ \begin{eqnarray} J_{j_{1}}&\leq& Cn^{-r}\|G\|_{(M)}\leq Cn^{-r}\|G\|_{M}\leq Cn^{-r}\|\delta_{n}^{2r}g^{(2r)}\|_{M}\nonumber\\ &\leq& C(n^{-r}\|\varphi^{2r}g^{(2r)}\|_{M}+n^{-2r}\|g^{(2r)}\|_{M}).\end{eqnarray} $

(2.9)

现在估计 $J_{j_{2}}$.由引理2.3, (1.1), (2.1) 式及 $\delta_{n}(x)\sim n^{-\frac{1}{2}}, x\in E_{n}^{c}$, 得

$ \begin{eqnarray*} J_{j_{2}}&\leq&2\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}^{c}}M\Big(\frac{1}{\lambda} \alpha_{j}^{n}(x)\frac{n!}{(n-j)!}\sum\limits_{k=0}^{n-j}P_{n-j, k}(x)(n+1)\int_{0}^{1}\sum\limits_{i=0}^{j} (-1)^{j-i}{j \choose i}P_{n, k+i}(t)\\ &&\frac{1}{(2r-1)!}\int_{x}^{t}(t-u)^{2r-1}g^{(2r)}(u)dudt\Big|\Big)dx\leq1\Big\}\\ &&\le 2\mathop {{\rm{inf}}}\limits_{\mathit{\lambda } > 0} \{ \mathit{\lambda }:\int_{\mathit{E}_\mathit{n}^\mathit{c}} \mathit{M} (\frac{\mathit{C}}{\mathit{\lambda }}\sum\limits_{k = 0}^{n - j} {{P_{n - j,k}}} (x)(n + 1)\int_0^1 {\sum\limits_{i = 0}^j {\left( \begin{array}{c} j\\ i \end{array} \right)} } {P_{n,k + i}}(t){(t - x)^{2r}}\\ &&dtG(x)\delta _n^{ - 2r}(x))dx\leq1\Big\}\\ &=&\inf\limits_{\lambda>0}\Big\{\lambda:\int_{E_{n}^{c}}M\Big(\frac{C}{\lambda}\sum\limits_{k=0}^{n-j}P_{n-j, k}(x) (n+1)\int_{0}^{1}\frac{{n \choose k+i}}{{n-j \choose k}}P_{n-j, k}(t)t^{i}(1-t)^{j-i}(t-x)^{2r}dt G(x)\\ &&\delta_{n}^{-2r}(x)\Big)dx\leq1\Big\}. \end{eqnarray*} $

令 $K=\frac{{n \choose k+i}}{{n-j \choose k}}$, 利用Hölder不等式两次和Beta函数, 有

$ \begin{eqnarray*} &&\sum\limits_{k=0}^{n-j}P_{n-j, k}(x) (n+1)\int_{0}^{1}\frac{{n \choose k+i}}{{n-j \choose k}}P_{n-j, k}(t)t^{i}(1-t)^{j-i}(t-x)^{2r}dt\delta_{n}^{-2r}(x)\\ &\leq&\Big(\sum\limits_{k=0}^{n-j}P_{n-j, k}(x) (n+1)\int_{0}^{1}K^{2}t^{2i}(1-t)^{2(j-i)}P_{n-j, k}(t)dt\Big)^{2}\cdot\Big(\sum\limits_{k=0}^{n-j}P_{n-j, k}(x) (n+1)\\ &&\int_{0}^{1}P_{n-j, k}(t)(t-x)^{4r}dt)^{\frac{1}{2}}\delta_{n}^{-2r}(x)\Big)\\ &\leq& Cn^{-r}(\varphi^{2}(x)+\frac{1}{n})^{r}\delta_{n}^{-2r}(x)\leq Cn^{-r}. \end{eqnarray*} $

从而

$ \begin{eqnarray} J_{j_{2}}&\leq& Cn^{-r}\|\delta_{n}^{2r}g^{(2r)}\|_{(M)}\leq Cn^{-r}\|\delta_{n}^{2r}g^{(2r)}\|_{M}\nonumber\\&\leq& C(n^{-r}\|\varphi^{2r}g^{(2r)}\|_{M}+n^{-2r}\|g^{(2r)}\|_{M}). \end{eqnarray} $

(2.10)

由(2.8), (2.9), (2.10) 式知

$ J_{j}\leq C(n^{-r}\|\varphi^{2r}g^{(2r)}\|_{M}+n^{-2r}\|g^{(2r)}\|_{M}). $

(2.11)

再由(2.5)-(2.7), (2.11) 式及修正的 $K$ -泛函与连续模的等价性得

$ \begin{eqnarray*} && \|B_{n}^{(2r-1)}(f)-f\|_{M}\leq C(\|f-g\|_{M}+n^{-r}\|\varphi^{2r}g^{(2r)}\|_{M}+n^{-2r}\|g^{(2r)}\|_{M})\\ &\leq& C\omega_{2r, \varphi}(f, \frac{1}{\sqrt{n}})_{M}. \end{eqnarray*} $

从而正定理得证.

引理3.1^[8]  $j\geq0$, $r\leq j$, 有 $ |D^{r}\alpha_{j}^{n}(x)|\leq Cn^{-\frac{j-r}{2}}\delta_{n}^{j-r}(x), x\in[0, 1]. $

引理3.2 对于 $n\geq4r$, $r\in\mathbf{N}$且 $r\geq2$, 有

$ \begin{eqnarray} &&\|\varphi^{2r}D^{2r}B_{n}^{(2r-1)}(f)\|_{M}\leq Cn^{r}\|f\|_{M}, f\in L_{M}^{*}[0, 1], \end{eqnarray} $

(3.1)

$ \begin{eqnarray} &&\|\varphi^{2r}D^{2r}B_{n}^{(2r-1)}(f)\|_{M}\leq C\|\varphi^{2r}f^{(2r)}\|_{M}, f\in W_{\varphi}^{2r}(L_{M}^{*}[0, 1]). \end{eqnarray} $

(3.2)

证 先证(3.1) 式.由引理2.2 $(r=0)$和引理3.1得

$ \begin{eqnarray*} &&\|\varphi^{2r}D^{2r}B_{n}^{(2r-1)}(f)\|_{M}=\|\varphi^{2r}\sum\limits_{j=0}^{2r-1}\sum\limits_{i=0}^{j} {2r \choose i} D^{i}\alpha_{j}^{n}(x)D^{j+2r-i}B_{n}(f)\|_{M}\\ &\leq& C \sum\limits_{j=0}^{2r-1}\sum\limits_{i=0}^{j}\|\delta_{n}^{2r}(x)n^{-\frac{j-i}{2}}\delta_{n}^{j-i}(x)D^{j+2r-i}B_{n}(f)\|_{M}\leq Cn^{r}\|f\|_{M}. \end{eqnarray*} $

再证(3.2) 式.由引理2.2, 引理3.1得

$ \begin{eqnarray*} &&\|\varphi^{2r}D^{2r}B_{n}^{(2r-1)}(f)\|_{M}=\|\varphi^{2r}\sum\limits_{j=0}^{2r-1}\sum\limits_{i=0}^{j} {2r \choose i} D^{i}\alpha_{j}^{n}(x)D^{j+2r-i}B_{n}(f)\|_{M}\\ &\leq& C \sum\limits_{j=0}^{2r-1}\sum\limits_{i=0}^{j}n^{-\frac{j-i}{2}}\|\varphi^{2r}(x)\delta_{n}^{j-i}(x)D^{j+2r-i}B_{n}(f)\|_{M}\leq C\|\varphi^{2r}f^{(2r)}\|_{M}. \end{eqnarray*} $

定理1.2的证明 由 $K$ -泛函的定义和引理3.2得

$ \begin{eqnarray*} K_{2r, \varphi}(f, t^{2r})_{M}&\leq&\|f-B_{n}^{(2r-1)}(f)\|_{M}+t^{2r}\|\varphi^{2r}D^{2r}B_{n}^{(2r-1)}(f)\|_{M}\\ &\leq& C[n^{-\alpha}+t^{2r}n^{r}K_{2r, \varphi}(f, n^{-r})_{M}]\\ &=&C \Big[(n^{-\frac{1}{2}})^{2\alpha}+\Big(\frac{t}{n^{-\frac{1}{2}}}\Big)^{2r}K_{2r, \varphi}(f, (n^{-\frac{1}{2}})^{2r})_{M}\Big]. \end{eqnarray*} $

由Berens-Lorentz引理^[10]知对 $0<\alpha<r$, 有 $ K_{2r, \varphi}(f, t^{2r})_{M}=O(t^{2\alpha}). $从而由 $K$ -泛函与连续模的等价性知 $ \omega_{2r, \varphi}(f, t)_{M}=O(t^{2\alpha}). $

由定理1.1和定理1.2得到定理1.3.