著名的Bernstein多项式定义如下^[1]: 设$ f(x)\in C[0, 1] $, 则

$ B_n(f;x) = \sum\limits^{n}_{k = 0}f(\frac{k}{n})p_{n, k}(x), $

其中$ p_{n, k}(x) = \left(^{n}_{k}\right)x^{k}(1-x)^{n-k}, k = 0, 1, 2, \cdots n. $

经典的Bernstein-Bézier多项式定义如下^[2]: 设$ \alpha >0, $$ f(x)\in C[0, 1] $, 有

(1.1) $ \begin{equation} B_{n, \alpha}(f, x) = \sum\limits^{n}_{k = 0}f(\frac{k}{n})\left[J^{\alpha}_{n, k}(x)-J^{\alpha}_{n, k+1}(x)\right] , \end{equation} $

其中$ J_{n, k}(x) = \sum\limits^{n}_{i = k}p_{n, i}(x), k = 0, 1, \cdots, n $, $ J_{n, k}(x) $是$ n $阶Bézier基函数, 且$ 1 = J_{n, 0}(x)\ge J_{n, 1}(x)\ge\cdot\cdot\cdot\ge J_{n, n}(x) = x^n $.

在计算机辅助几何设计中, Bézier型算子是非常有用的, 它们的解析性质已经被许多学者研究, 参见文献[2-9]. 2013年, Ren介绍了一类Bernstein-Bézier算子, 其定义为^[8]: 设$ f(x)\in C[0, 1], \beta\in[0, 1], $有

$ E_{n, \beta}(f;x) = f(0)p_{n, 0}(x)+\sum\limits^{n-1}_{k = 1}p_{n, k}(x)F_{n, k}^{(\beta)}(f)+f(1)p_{n, n}(x), $

其中

$ F_{n, k}^{(\beta)}(f) = \frac{1}{B(nk, n(n-k))}\int^{1}_{0}t^{nk-1}(1-t)^{n(n-k)-1}f(\beta t+(1-\beta)\frac{k}{n}){\rm d}t, \ k = 1, 2, \cdots, n-1, $

$ B(\cdot , \cdot) $为Beta函数.

2017年, Ren^[10]又引入参数$ \alpha $, 改进了上述算子为$ E_{n, \beta}^{\alpha}(f;x) $, 并研究了这些算子逼近的Jackson型定理. 为了得到该类算子逼近的Bernstein型定理, 我们对这类算子再次进行了修正, 本文所研究的新算子定义如下: 设$ \beta\in[0, 1], \alpha\geq1, n\geq 2, $有

$ L_{n, \beta}^{(\alpha)}(f;x) = f(0)Q_{n, 0}^{(\alpha)}(x)+\sum\limits^{n-1}_{k = 1}Q_{n, k}^{(\alpha)}(x)F_{n-1, k}^{(\beta)}(f)+f(1)Q_{n, n}^{(\alpha)}(x) , $

其中$ Q_{n, k}^{(\alpha)}(x) = J^{\alpha}_{n, k}(x)-J^{\alpha}_{n, k+1}(x), $$ J_{n, k}(x) $, $ F_{n-1, k}^{(\beta)}(f) $定义如上.

易知, $ L_{n, \beta}^{(\alpha)}(f;x) $是C$ [0, 1] $上的有界正算子，当$ \alpha = 1 $时, $ L_{n, \beta}^{(\alpha)}(f;x) $即为$ E_{n, \beta}(f;x) $; 当$ \beta = 0 $时, $ L_{n, \beta}^{(\alpha)}(f;x) $即为$ B_{n, \alpha}(f, x) $.

首先介绍本文所涉及的一些定义. 我们考虑$ f(x)\in C[0, 1] $: = $ \{f:f(x) $在$ [0, 1] $上连续$ \} $, $ f(x) $的范数定义为: $ \|f\| = \max\{|f(x)|: x\in [0, 1]\}. $

定义1.1   设$ f(x)\in C[0, 1] $, $ \varphi(x) = \sqrt{x(1-x)} $, $ \lambda\in[0, 1] $, 连续模定义为

$ \omega_{\varphi^{\lambda}}(f;t) = \sup\limits_{0< h\leq t} \sup\limits_{x\pm\frac{h\varphi^{\lambda}(x)}{2}\in[0, 1]}\left|f(x+\frac{h\varphi^{\lambda}(x)}{2})-f(x-\frac{h\varphi^{\lambda}(x)}{2})\right|. $

定义1.2   设$ f(x)\in C[0, 1] $, $ \varphi(x) = \sqrt{x(1-x)} $, $ \lambda\in[0, 1] $, $ K $-泛函定义为

$ K_{\varphi^{\lambda}}(f;t) = \inf\limits_{g\in W_{\lambda}} \{||f-g||+t||\varphi^{\lambda} g^{'}||\}, $

其中$ W_{\lambda} = \{g|g\in AC_{loc [0, 1]}, \|\varphi^{\lambda} g^{'}\|<\infty\}. $

注1.1   由文献[1]可知: $ K_{\varphi^{\lambda}}(f;t)\sim \omega_{\varphi^{\lambda}}(f;t). $

本文将得到所修正的广义含参量Bernstein-Bézier算子在$ C[0, 1] $上的逼近正逆定理及等价定理.

定理1.1 (Korovkin定理)   设$ f(x)\in C[0, 1] $, $ \alpha\geq1 $, $ \beta\in[0, 1], n\ge 2, $则$ L_{n, \beta}^{(\alpha)}(f;x) $在$ [0, 1] $上一致收敛于$ f(x) $.

定理1.2  设$ f(x)\in C[0, 1] $, $ \alpha\geq1 $, $ \varphi(x) = \sqrt{x(1-x)} $, $ \beta\in[0, 1], 0\leq\lambda\leq1, n\ge 2, $则

$ |L_{n, \beta}^{(\alpha)}(f;x)-f(x)|\leq M\omega_{\varphi^{\lambda}}(f;\frac{\varphi^{1-\lambda}(x)}{\sqrt{n}}). $

定理1.3  设$ f(x)\in C^{1}[0, 1] $: = $ \{f:f'(x) $在$ [0, 1] $上连续$ \} $, $ \alpha\geq1, \beta\in[0, 1], n\ge 2 $, 则

$ |L_{n, \beta}^{(\alpha)}(f;x)-f(x)| \leq\sqrt{\frac{(14+\beta^{2})\alpha}{4n}}\cdot\left\{\|f'\|+\omega(f';\frac{1}{\sqrt{n}})\cdot (1+\sqrt{\frac{(14+\beta^{2})\alpha}{4}})\right\}. $

定理1.4   设$ f(x)\in C[0, 1] $, $ \varphi(x) = \sqrt{x(1-x)} $, $ \beta\in[0, 1], 0\leq\gamma\leq1, 0\leq\lambda\leq1, $$ \alpha\geq1, $$ n\ge 2 $, 若

$ \|L_{n, \beta}^{(\alpha)}(f;x)-f(x)\| = O(n^{-\frac{\gamma}{2}}), $

则$ \omega_{\varphi^{\lambda}}(f;t) = O(t^{\gamma}). $

由定理1.2和定理1.4, 我们可以得到如下的等价定理.

定理1.5  设$ f(x)\in C[0, 1] $, $ \varphi(x) = \sqrt{x(1-x)} $, $ \beta\in[0, 1], 0\leq\gamma\leq1, 0\leq\lambda\leq1, $$ \alpha\geq1, $$ n\ge 2 $, 有

$ \|L_{n, \beta}^{(\alpha)}(f;x)-f(x)\| = O(n^{-\frac{\gamma}{2}}) \Leftrightarrow \omega_{\varphi^{\lambda}}(f;t) = O(t^{\gamma}). $

注1.2   本文中, $ M $是一个不依赖于$ n $和$ x $的正常数, 不同地方$ M $的取值可能不同.

注1.3^[1]   对于$ B_n(t^{j};x), j = 0, 1, 2, $有

$ \begin{eqnarray*} &(1)&B_n(1;x) = 1;\\ &(2)&B_n(t;x) = x;\\ &(3)&B_n(t^{2};x) = x^{2}+\frac{x(1-x)}{n}. \end{eqnarray*} $

利用文献[2, 7, 10] 中的方法, 可以得到本文所研究的算子矩的估计, 我们将省略具体推算过程.

引理2.1^[10]   对于$ F_{n-1, k}^{(\beta)}(t^{i}) $, $ i = 0, 1, 2, \beta\in[0, 1], n\ge 2, $有

$ \begin{eqnarray*} &(1)&F_{n-1, k}^{(\beta)}(1) = 1;\\ &(2)&F_{n-1, k}^{(\beta)}(t) = \frac{k}{n-1};\\ &(3)&F_{n-1, k}^{(\beta)}(t^{2}) = \frac{\beta^{2}}{(n-1)^{2}+1}\cdot \frac{k}{n-1}+\left(1-\frac{\beta^{2}}{(n-1)^{2}+1}\right)\cdot\frac{k^{2}}{(n-1)^{2}}. \end{eqnarray*} $

引理2.2  对于$ L_{n, \beta}(t^{i};x) $, $ i = 0, 1, 2, \beta\in[0, 1], n\ge 2, $有

$ \begin{eqnarray*} &(1)&L_{n, \beta}(1;x) = 1;\\ &(2)&L_{n, \beta}(t;x) = x+\frac{1}{n-1}(x-x^{n});\\ &(3)&L_{n, \beta}(t^{2};x) = \frac{n^{2}}{(n-1)^{2}}\cdot x^{2}+ \bigg[\frac{n}{(n-1)^{2}}+\frac{\beta^{2}}{(n-1)^{2}+1}\cdot \frac{n}{n-1}\bigg]\cdot \varphi^{2}(x)\\ &&{\quad}-\frac{\beta^{2}}{(n-1)^{2}+1}\cdot \frac{n}{(n-1)^{2}} \cdot x+ \bigg[\frac{\beta^{2}}{(n-1)^{2}+1}\cdot \frac{n}{(n-1)^{2}}-\frac{2n-1}{(n-1)^{2}}\bigg]\cdot x^{n}. \end{eqnarray*} $

引理2.3   令$ n\ge 2 $, 则

$ \begin{eqnarray*} &(1)&\sum\limits^{n-1}_{k = 1}J_{n, k}(x) = nx-x^{n};\\ &(2)&\frac{1}{n-1}\sum\limits^{n-1}_{k = 1}kJ_{n, k}(x) = \frac{n}{2}x^{2}. \end{eqnarray*} $

引理2.4^{[2, 10]}    令$ \alpha\geq1, n\ge 2 $, 则

$ \begin{eqnarray*} &(1)&\lim\limits_{n\rightarrow \infty}\frac{1}{n-1}\sum\limits^{n-1}_{k = 1}J^{\alpha}_{n, k}(x) = x \rm{于}[0, 1]\rm{上一致成立};\\ &(2)&\lim\limits_{n\rightarrow \infty}\frac{1}{(n-1)^{2}}\sum\limits^{n-1}_{k = 1}kJ^{\alpha}_{n, k}(x) = \frac{x^{2}}{2} \rm{于}[0, 1]\rm{上一致成立}. \end{eqnarray*} $

引理2.5^{[7, 10]}    令$ x\in[0, 1] $, $ \alpha\geq1 $, $ k = 0, 1, \cdots n-1, $则

$ 0\leq Q_{n, k}^{(\alpha)}(x)\leq \alpha P_{n, k}(x). $

引理2.6   令$ \alpha\geq1, \beta\in[0, 1], $则

$ \begin{eqnarray*} &(1)&L_{n, \beta}^{(\alpha)}(1;x)\equiv 1, x\in [0, 1];\\ &(2)&\lim\limits_{n\rightarrow \infty}L_{n, \beta}^{(\alpha)}(t;x) = x \rm{于}[0, 1]\rm{上一致成立};\\ &(3)&\lim\limits_{n\rightarrow \infty}L_{n, \beta}^{(\alpha)}(t^{2};x) = x^{2} \rm{于}[0, 1]\rm{上一致成立}. \end{eqnarray*} $

引理2.7   令$ \alpha\geq1 $, $ \beta\in[0, 1], \varphi(x) = \sqrt{x(1-x)}, n\ge 2 $, 则

$ \begin{eqnarray*} &(1)&L_{n, \beta}^{(\alpha)}((t-x)^{2};x)\leq\frac{\alpha}{4}(14+\beta^{2})\cdot \frac{1}{n-1}, \quad x\in[0, 1];\\ &(2)&L_{n, \beta}^{(\alpha)}((t-x)^{2};x)\leq \frac{M\varphi^{2}(x)}{n-1}, \quad x\in E_{n} = \left[\frac1n, 1-\frac1n\right]; \end{eqnarray*} $

特别地, 当$ \alpha = 1, x\in E_{n} $时, 有$ L_{n, \beta}((t-x)^{2};x)\leq \frac{3\varphi^{2}(x)}{n-1}. $

为了得到逆定理, 需要下面两个重要的引理, 这两个引理的证明较长，我们把具体的推导过程放在文章的最后一节.

引理2.8  设$ f(x)\in C[0, 1] $, $ \alpha\geq1, \varphi(x) = \sqrt{x(1-x)} $, $ \beta\in[0, 1], 0\leq\lambda\leq1, n\ge 2 $, 则

$ |\varphi^{\lambda}(x)\cdot(L_{n, \beta}^{(\alpha)}(f;x))'|\leq 15\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\| . $

引理2.9   设$ f(x)\in W_{\lambda} $, $ \varphi(x) = \sqrt{x(1-x)}, \alpha\geq 1, \beta\in[0, 1], 0\leq\lambda\leq 1, n\ge 2 $, 则

$ |\varphi^{\lambda}(x)\cdot(L_{n, \beta}^{(\alpha)}(f;x))'|\leq 104\alpha\|\varphi^{\lambda}f'\| . $

定理1.1的证明  根据Korovkin定理^[11], 结合引理2.6, 定理1.1得证.

定理1.2的证明  令$ g\in W_{\lambda} $, 则$ |L_{n, \beta}^{(\alpha)}(f;x)-f(x)|\leq 2\|f-g\|+|L_{n, \beta}^{(\alpha)}(g;x)-g(x)|. $由于$ g(t) = \int^{t}_{x}g'(u){\rm d}u+g(x), $$ L_{n, \beta}^{(\alpha)}(1;x) = 1, $我们可以得到

$ |L_{n, \beta}^{(\alpha)}(g;x)-g(x)|\leq\|\varphi^{\lambda}g'\|L_{n, \beta}^{(\alpha)}\left(\left|\int^{t}_{x} \varphi^{-\lambda}(u){\rm d}u\right|;x\right). $

结合Hölder不等式，可得

(3.1) $ \begin{equation} \left|\int^{t}_{x}\varphi^{-\lambda}(u){\rm d}u\right|\leq 4\varphi^{-\lambda}(x)\cdot|t-x|, \end{equation} $

因此

$ |L_{n, \beta}^{(\alpha)}(g;x)-g(x)|\leq 4\|\varphi^{\lambda}g'\|\cdot\varphi^{-\lambda}(x)\cdot L_{n, \beta}^{(\alpha)}(|t-x|;x). $

根据$ L_{n, \beta}^{(\alpha)}(f;x) $的定义和引理2.7(2) 可得

$ |L_{n, \beta}^{(\alpha)}(g;x)-g(x)|_{E_{n}}\leq M\|\varphi^{\lambda}g'\|\cdot\frac{\varphi^{1-\lambda}(x)}{\sqrt{n}}. $

结合文献[1]中的定理8.4.8, 有

$ |L_{n, \beta}^{(\alpha)}(f;x)-f(x)|\leq M\left(\|f-g\|+\|\varphi^{\lambda}g'\|\cdot\frac{\varphi^{1-\lambda}(x)}{\sqrt{n}}\right), $

上式右边对所有$ g\in W_{\lambda} $取下确界并注意到注1.1, 即可得到定理1.2.

定理1.3的证明   对于任意的$ \delta>0 $, $ t, x\in[0, 1] $, 当$ |t-x|<\delta $时, 通过泰勒展式, 有

$ \begin{eqnarray*} |f(t)-f(x)-f'(x)(t-x)|&\leq&\left|\int^{t}_{x}|f'(u)-f'(x)|{\rm d}u\right| \\ &\leq&\omega(f';\delta)(|t-x|+\delta^{-1}(t-x)^{2}), \end{eqnarray*} $

应用Cauchy-Schwarz不等式可得

$ \begin{eqnarray*} &&|L_{n, \beta}^{(\alpha)}(f(t)-f(x)-f'(x)(t-x);x)|\\ &\leq& \omega(f';\delta)(L_{n, \beta}^{(\alpha)}(|t-x|;x)+\delta^{-1}L_{n, \beta}^{(\alpha)}((t-x)^{2};x))\\ &\leq&\omega(f';\delta)\cdot\left[\sqrt{L_{n, \beta}^{(\alpha)}(1;x)}+\delta^{-1}\sqrt{L_{n, \beta}^{(\alpha)}((t-x)^{2};x)}\right]\cdot\sqrt{L_{n, \beta}^{(\alpha)}((t-x)^{2};x)}. \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&|L_{n, \beta}^{(\alpha)}(f;x)-f(x)|\\ &\leq&\|f'\|\cdot L_{n, \beta}^{(\alpha)}(|t-x|;x)+ \omega(f';\delta)\cdot\left[1+\delta^{-1}\sqrt{L_{n, \beta}^{(\alpha)}((t-x)^{2};x)} \right]\cdot\sqrt{L_{n, \beta}^{(\alpha)}((t-x)^{2};x)}. \end{eqnarray*} $

取$ \delta = \frac{1}{\sqrt{n}} $, 结合引理2.7(1) 可得定理1.3.

定理1.4的证明   由$ K $ -泛函定义及引理2.8, 引理2.9可知

$ \begin{eqnarray*} K_{\varphi^{\lambda}}(f;t)&\leq&\|f-L_{n, \beta}^{(\alpha)}(f;x)\|+t\|\varphi^{\lambda}(x) (L_{n, \beta}^{(\alpha)}(f;x))'\|\\ &\leq& Mn^{-\frac{\gamma}{2}}+t\sqrt{n}(\|f-g\|+\frac{1}{\sqrt{n}}\|\varphi^{\lambda}g'\|)\\ &\leq& M(n^{-\frac{\gamma}{2}}+t\sqrt{n}\cdot K_{\varphi^{\lambda}}(f;n^{-\frac{1}{2}})). \end{eqnarray*} $

应用Berens-Lorentz引理^[1], 并结合关系式$ \omega_{\varphi^{\lambda}}(f;t)\sim K_{\varphi^{\lambda}}(f;t), $可得到定理1.4.

引理2.8的证明   我们记

(4.1) $ \begin{equation} \left(L_{n, \beta}^{(\alpha)}(f;x)\right)' = f(0)(Q_{n, 0}^{(\alpha)}(x))'+\left(\sum\limits^{n-1}_{k = 1}Q_{n, k}^{(\alpha)}(x)F_{n-1, k}^{(\beta)}(f)\right)'+f(1)(Q_{n, n}^{(\alpha)}(x))' = I_{1}+I_{2}+I_{3}, \end{equation} $

下面分别估计$ I_{1} $, $ I_{2} $和$ I_{3} $.

首先, 注意到$ J_{n, 0}'(x) = 0, $可得

$ |I_{1}| = |\alpha f(0)J_{n, 0}^{\alpha-1}(x)\cdot J_{n, 0}'(x)-\alpha f(0)J_{n, 1}^{\alpha-1}(x)\cdot J_{n, 1}'(x)|\leq\alpha\sqrt{n}\|f\|\cdot\sqrt{n}. $

当$ x\in [0, \frac{1}{n}) $时, 有

(4.2) $ \begin{equation} \varphi^{\lambda}(x)\cdot |I_{1}|\leq \alpha\varphi^{\lambda-1}\sqrt{n}\|f\|\cdot\sqrt{n}\cdot\sqrt{x(1-x)}\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. \end{equation} $

当$ x\in (1-\frac{1}{n}, 1] $时, 有

(4.3) $ \begin{equation} \varphi^{\lambda}(x)\cdot |I_{1}|\leq \alpha\varphi^{\lambda-1}\sqrt{n}\|f\|\cdot\sqrt{n}\cdot\sqrt{x(1-x)}\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. \end{equation} $

当$ x\in E_{n} = [\frac{1}{n}, 1-\frac{1}{n}] $时, 有

(4.4) $ \begin{equation} |I_{1}| = |-\alpha f(0)J_{n, 1}^{\alpha-1}(x)J_{n, 1}'(x)|\leq|-\alpha f(0)\cdot[1-p_{n, 0}(x)]'| = |\alpha f(0)\cdot p_{n, 0}'(x)|, \end{equation} $

结合

$ p_{n, k}'(x) = \frac{n}{\varphi^{2}(x)}\left(\frac{k}{n}-x\right)\cdot p_{n, k}(x) \quad , \quad B_{n}((t-x)^{2};x) = \frac{\varphi^{2}(x)}{n}, $

我们得到

(4.5) $ \begin{eqnarray} \varphi^{\lambda}(x)\cdot |I_{1}| \leq\alpha\|f\|\cdot\sum\limits^{n}_{k = 0}\left|p_{n, k}'(x)\right|\cdot\varphi^{\lambda-1}(x) \leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. \end{eqnarray} $

综合(4.2)–(4.5)式, 可得$ \varphi^{\lambda}(x)\cdot |I_{1}|\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. $

其次, 注意到$ J_{n, n+1}'(x) = 0, $则

$ \begin{eqnarray*} |I_{3}| = |\alpha f(1)J_{n, n}^{\alpha-1}(x)\cdot J_{n, n}'(x)-\alpha f(1)J_{n, n+1}^{\alpha-1}(x)\cdot J_{n, n+1}'(x)| \leq \alpha\sqrt{n}\|f\|\cdot\sqrt{n}. \end{eqnarray*} $

当$ x\in [0, \frac{1}{n})\bigcup(1-\frac{1}{n}, 1] $时, 用估计(4.2)式和(4.3)式的方法可得

(4.6) $ \begin{equation} \varphi^{\lambda}(x)\cdot |I_{3}|\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. \end{equation} $

当$ x\in E_{n} = [\frac{1}{n}, 1-\frac{1}{n}] $时, 结合$ |p_{n, n}'(x)|\leq \sum\limits^{n}_{k = 0}|p_{n, k}'(x)| $, 用估计(4.5)式的方法也可得相应于(4.6)式的结果.

最后, 考虑$ I_{2} $：由于$ |F_{n-1, k}^{(\beta)}(f)| \leq \|f\|, $故

$ \begin{eqnarray*} I_{2}& = &\alpha \sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(f)J_{n, k}^{\alpha-1}(x)\cdot J_{n, k+1}'(x)+ \alpha \sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(f)J_{n, k}^{\alpha-1}(x)\cdot p_{n, k}'(x)\\ &&-\alpha \sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(f)J_{n, k+1}^{\alpha-1}(x)\cdot J_{n, k+1}'(x)\\ & = &\alpha \sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(f)\left\{[J_{n, k}^{\alpha-1}(x)-J_{n, k+1}^{\alpha-1}(x)]J_{n, k+1}'(x)+J_{n, k}^{\alpha-1}(x)\cdot p_{n, k}'(x)\right\}, \end{eqnarray*} $

由此

$ \begin{eqnarray*} |I_{2}|&\leq&\alpha\|f\|\bigg(\sum\limits^{n-1}_{k = 1}[J_{n, k}^{\alpha-1}(x) -J_{n, k+1}^{\alpha-1}(x)]J_{n, k+1}'(x)+\sum\limits^{n-1}_{k = 1}J_{n, k}^{\alpha-1}(x)\cdot |p_{n, k}'(x)|\bigg) \\ & = &\alpha\|f\|(Q_{1}+Q_{2}). \end{eqnarray*} $

注意到$ J_{n, 1}'(x)>0, J_{n, 0}'(x) = 0, J_{n, n+1}(x) = 0 $, 可得

$ \begin{eqnarray*} Q_{1}& = &-\sum\limits^{n-1}_{k = 1}J_{n, k}^{\alpha-1}(x)\cdot p_{n, k}'(x)-J_{n, n}^{\alpha-1}(x)p_{n, n}'(x)+J_{n, 1}^{\alpha-1}(x)J_{n, 1}'(x), \\ &\leq&\sum\limits^{n-1}_{k = 1}J_{n, k}^{\alpha-1}(x)\cdot |p_{n, k}'(x)|+J_{n, n}^{\alpha-1}(x)|p_{n, n}'(x)|+J_{n, 1}^{\alpha-1}(x)J_{n, 1}'(x)\\ &\leq&Q_{2}+p_{n, n}'(x)+|p_{n, 0}'(x)|. \end{eqnarray*} $

又因为$ \alpha\|f\|\cdot p_{n, n}'(x)\leq\alpha\|f\|\sqrt{n}\cdot\sqrt{n}, $且$ \alpha\|f\|\cdot |p_{n, 0}'(x)|\leq\alpha\|f\|\sqrt{n}\cdot\sqrt{n}, $用估计$ I_{1} $相同的方法, 有

$ \varphi^{\lambda}(x)\cdot\alpha\|f\|\cdot p_{n, n}'(x)\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|, $

$ \varphi^{\lambda}(x)\cdot\alpha\|f\|\cdot |p_{n, 0}'(x)|\leq\alpha\varphi^{\lambda-1}(x)\sqrt{n}\|f\|. $

接下来我们考虑$ Q_{2}. $

当$ x = 0 $时

$ p_{n, k}'(0) = \left\{\begin{array}{ll} n, & \quad k = 1;\\ 0, & \quad 2\leq k\leq n-1; \end{array}\right. $

当$ x = 1 $时

$ p_{n, k}'(1) = \left\{\begin{array}{ll} 0, & \quad 1\leq k\leq n-2;\\ -n, & \quad k = n-1; \end{array}\right. $

当$ 0 < x < 1 $时

$ p_{n, k}'(x) = \frac{n}{\varphi^{2}(x)}\left(\frac{k}{n}-x\right)\cdot p_{n, k}(x), $

所以

$ \begin{eqnarray*} \varphi^{\lambda}(x)Q_{2}&\leq&\varphi^{\lambda}(x)|p_{n, 1}'(x)|+\varphi^{\lambda}(x)\sum\limits^{n-2}_{k = 2}J_{n, k}^{\alpha-1}(x)\cdot |p_{n, k}'(x)|+\varphi^{\lambda}(x)|p_{n, n-1}'(x)|\\ &\leq& 4\varphi^{\lambda-1}(x)\sqrt{n}+\frac{n}{\varphi(x)}\sum\limits^{n-2}_{k = 2}\left|\frac{k}{n}-x\right|\cdot p_{n, k}(x)\cdot\varphi^{\lambda}(x) \\ &\leq& 5\varphi^{\lambda-1}(x)\sqrt{n}. \end{eqnarray*} $

综上知

(4.7) $ \begin{equation} \varphi^{\lambda}(x)Q_{2}\leq 5\varphi^{\lambda-1}(x)\sqrt{n} \end{equation} $

和

$ \varphi^{\lambda}(x)Q_{1}\leq 5\varphi^{\lambda-1}(x)\sqrt{n}+2\sqrt{n}\cdot \varphi^{\lambda-1}(x)\cdot\frac{\sqrt{nx}}{e^{nx}}\leq 7\sqrt{n}\varphi^{\lambda-1}(x). $

所以

(4.8) $ \begin{equation} \varphi^{\lambda}(x)|I_{2}|\leq \alpha\|f\|\cdot \varphi^{\lambda}(x)\left(Q_{1}+Q_{2}\right)\leq 12\alpha\sqrt{n}\|f\|\cdot \varphi^{\lambda-1}(x). \end{equation} $

由(4.2)–(4.8)式可得所求.

引理2.9的证明  由于$ f(x)\left(L_{n, \beta}^{(\alpha)}(1;x)\right)' = 0, $有

$ \begin{eqnarray*} \left(L_{n, \beta}^{(\alpha)}(f;x)\right)' & = &f(0)\left(Q_{n, 0}^{(\alpha)}(x)\right)'+\sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(f)\left[J_{n, k}^{\alpha}(x)-J_{n, k+1}^{\alpha}(x)\right]'+f(1)\left(Q_{n, n}^{(\alpha)}(x)\right)'\nonumber\\ &&-f(x)\left\{\left(Q_{n, 0}^{(\alpha)}(x)\right)'+\sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}(1)\left[J_{n, k}^{\alpha}(x)-J_{n, k+1}^{\alpha}(x)\right]'+\left(Q_{n, n}^{(\alpha)}(x)\right)'\right\}{\nonumber}\\ & = &[f(0)-f(x)]\left(Q_{n, 0}^{(\alpha)}(x)\right)'+\sum\limits^{n-1}_{k = 1}\left[F_{n-1, k}^{(\beta)}(f)-f(x)\right] \left[J_{n, k}^{\alpha}(x)-J_{n, k+1}^{\alpha}(x)\right]'{\nonumber}\\ &&+[f(1)-f(x)]\left(Q_{n, n}^{(\alpha)}(x)\right)', \end{eqnarray*} $

我们记

(4.9) $ \begin{equation} \left(L_{n, \beta}^{(\alpha)}(f;x)\right)' = H_{1}+H_{2}+H_{3}, \end{equation} $

接下来分别估计$ H_{1} $, $ H_{2} $和$ H_{3} $. 首先由(3.1)式, 知

$ \begin{eqnarray*} \varphi^{\lambda}(x)|H_{1}|& = &\varphi^{\lambda}(x)\cdot\left|\int^{0}_{x}\frac{\varphi^{\lambda}(u)f'(u)}{\varphi^{\lambda}(u)}{\rm d}u\right|\cdot \left|\left(J_{n, 0}^{\alpha}(x)\right)'-\left(J_{n, 1}^{\alpha}(x)\right)'\right|\\ &\leq& 4\alpha\|\varphi^{\lambda} f'\|\cdot nx(1-x)^{n-1}. \end{eqnarray*} $

通过求$ nx(1-x)^{n-1} $在$ [0, 1] $上的最大值, 可知

$ nx(1-x)^{n-1}\leq (1-\frac1n)^{n-1}< 1. $

因此

(4.10) $ \begin{equation} \varphi^{\lambda}(x)|H_{1}|\leq4\alpha\|\varphi^{\lambda} f'\|. \end{equation} $

同理, 有

(4.11) $ \begin{equation} \varphi^{\lambda}(x)|H_{3}|\leq4\alpha\|\varphi^{\lambda} f'\|. \end{equation} $

接下来估计$ H_2: $

$ H_{2} = \sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))} \bigg[\int^{\beta t+(1-\beta)\frac{k}{n-1}}_{x}f'(u){\rm d}u\bigg]{\rm d}t[J_{n, k}^{\alpha}(x)-J_{n, k+1}^{\alpha}(x)]', $

由(3.1)式知

$ \varphi^{\lambda}(x)\cdot\left|\int^{\beta t+(1-\beta)\frac{k}{n-1}}_{x}f'(u){\rm d}u\right| \leq 4\|\varphi^{\lambda} f'\|\cdot \left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|, $

所以

$ \begin{eqnarray*} \varphi^{\lambda}(x)|H_{2}| &\leq&4\|\varphi^{\lambda} f'\|\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}} {B((n-1)k, (n-1)(n-1-k))}\nonumber\\ &&\times\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\left|\left[J_{n, k}^{\alpha}(x)-J_{n, k+1}^{\alpha}(x)\right]'\right|\\ & = &4\alpha\|\varphi^{\lambda} f'\|\left\{\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}} {B((n-1)k, (n-1)(n-1-k))}\right.\\ &&\times\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\left|J_{n, k}^{\alpha-1}(x)-J_{n, k+1}^{\alpha-1}(x)\right|J_{n, k+1}'(x)\\ &&+\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))} \\ &&\times \left.\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}tJ_{n, k}^{\alpha-1}(x)|p_{n, k}'(x)|\right\}, \end{eqnarray*} $

记

(4.12) $ \begin{equation} \varphi^{\lambda}(x)|H_{2}|\leq4\alpha\|\varphi^{\lambda} f'\|\cdot(A+B), \end{equation} $

我们将在$ E_n^C $和$ E_n $上分别估计$ A $和$ B. $

(I) 当$ x\in E_{n}^{c} = [0, \frac{1}{n})\bigcup(1-\frac{1}{n}, 1] $时, 有

$ \begin{eqnarray*} B&\leq&\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\cdot n \cdot p_{n-1, k-1}(x)\\ &&+\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\cdot n\cdot p_{n-1, k}(x), \end{eqnarray*} $

记

(4.13) $ \begin{equation} B\leq L_{1}+L_{2}. \end{equation} $

由于$ E_{n-1, \beta}((t-x)^{2};x)\leq \frac{2\varphi^{2}(x)}{n-1}, $应用Cauchy-Schwarz不等式可得

$ \begin{eqnarray*} L_{2}& = &n\bigg(\sum\limits^{n-1}_{k = 1}F_{n-1, k}^{(\beta)}((t-x)^{2};x)\cdot p_{n-1, k}(x)\bigg)^{\frac{1}{2}}\cdot \bigg(\sum\limits^{n-1}_{k = 1}p_{n-1, k}(x)\bigg)^{\frac{1}{2}}\\ &\leq& 2n\cdot \frac{\sqrt{x(1-x)}}{\sqrt{n}}. \end{eqnarray*} $

当$ x\in [0, \frac{1}{n}) $时, 有

(4.14) $ \begin{equation} L_{2}\leq 2n\cdot \frac{\sqrt{\frac{1}{n}}\cdot\sqrt{1-x}}{\sqrt{n}}\leq 2; \end{equation} $

当$ x\in (1-\frac{1}{n}, 1] $时, 有

(4.15) $ \begin{equation} L_{2}\leq 2n\cdot \frac{\sqrt{x}\cdot\sqrt{\frac{1}{n}}}{\sqrt{n}}\leq 2. \end{equation} $

接下来我们估计$ L_{1} $.

$ \begin{eqnarray*} L_{1} &\leq& \sum\limits^{n-2}_{j = 0}\frac{\int^{1} _{0}t^{(n-1)(j+1)-1}(1-t)^{(n-1)[n-1-(j+1)]-1}}{B((n-1)(j+1), (n-1)[n-1-(j+1)])}\left|\beta t+(1-\beta)\frac{j+1}{n-1}-x\right|{\rm d}t\cdot np_{n-1, j}(x)\nonumber\\ &\leq& \frac{1}{n-1}\cdot n\cdot (1-x)^{n-1}+x\cdot n\cdot (1-x)^{n-1}+L_{3}+\frac{n}{n-1}\cdot (1-\beta)\sum\limits^{n-2}_{j = 1}p_{n-1, j}(x)\\ &\leq& 4+n\cdot x(1-x)^{n-1}+L_{3}, \end{eqnarray*} $

通过求$ nx(1-x)^{n-1} $在$ [0, 1] $上的最大值, 可知$ nx(1-x)^{n-1}\leq (1-\frac1n)^{n-1}< 1 $, 应用递推公式$ B(p, q) = \frac{p-1}{q}B(p-1, q+1), \ (p, q>0) $, 当$ j\geq1 $时, 有

$ \begin{eqnarray*} &&\int^{1} _{0}t^{(n-1)j-1+(n-1)}(1-t)^{(n-1)(n-1-j)-1-(n-1)}\cdot|\beta t+(1-\beta)\frac{j}{n-1}-x|{\rm d}t\nonumber\\ &\leq&\frac{(n-1)j+(n-1)}{(n-1)(n-1-j)-(n-1)}\cdot \frac{(n-1)j+(n-2)}{(n-1)(n-1-j)-(n-2)}\cdot\cdot\cdot\frac{(n-1)j+1}{(n-1)(n-1-j)-1}\\ &&\times\left[\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}\cdot|\beta t+(1-\beta)\frac{j}{n-1}-x|{\rm d}t\right.\\ &&\left.+\frac{n}{(n-1)^{2}}\cdot \beta\cdot\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}{\rm d}t\right], \end{eqnarray*} $

进而

$ \begin{eqnarray*} L_{3} &\leq& \sum\limits^{n-2}_{j = 1}\frac{\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}}{B((n-1)j, (n-1)(n-1-j))}\left|\beta t+(1-\beta)\frac{j}{n-1}-x\right|{\rm d}t\cdot n\cdot p_{n-1, j}(x)\\ &&+\sum\limits^{n-2}_{j = 1}\frac{\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}}{B((n-1)j, (n-1)(n-1-j))}{\rm d}t\cdot\frac{n}{(n-1)^{2}}\cdot \beta\cdot p_{n-1, j}(x)\\ &\leq& L_{2}+\sum\limits^{n-2}_{j = 1}\frac{\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}}{B((n-1)j, (n-1)(n-1-j))}{\rm d}t\cdot \beta\cdot p_{n-1, j}(x)\leq 3, \end{eqnarray*} $

综上知

(4.16) $ \begin{equation} L_{1}\leq L_{3}+5\leq 8, \end{equation} $

由(4.13)–(4.16)式, 知$ B\leq 10. $注意到$ J_{n, 0}'(x) = 0 $且$ x\in E_{n}^{C} $, 有

$ \begin{eqnarray*} A& = &\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\cdot J_{n, k}^{\alpha-1}(x)J_{n, k}'(x)\nonumber\\ &&-\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\cdot J_{n, k}^{\alpha-1}(x)p_{n, k}'(x)\\ &&-\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\cdot J_{n, k+1}^{\alpha-1}(x)J_{n, k+1}'(x)\\ &\leq&L_{4}+B-\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))}\left|\beta t+(1-\beta)\frac{k}{n-1}-x\right|{\rm d}t\\ &&\times J_{n, k+1}^{\alpha-1}(x)J_{n, k+1}'(x), \end{eqnarray*} $

令$ k = j+1 $, 则

$ \begin{eqnarray*} L_{4}&\leq& \frac{n}{n-1}\cdot (1-x)^{n-1}+nx\cdot (1-x)^{n-1}+\frac{n-2}{n-1}\nonumber\\ &&+\sum\limits^{n-2}_{j = 1}\frac{\int^{1} _{0}t^{(n-1)(j+1)-1}(1-t)^{(n-1)[n-1-(j+1)]-1}}{B((n-1)(j+1), (n-1)[n-1-(j+1)])}\left|\beta t+(1-\beta)\frac{j}{n-1}-x\right|{\rm d}t\\ &&\times J_{n, j+1}^{\alpha-1}(x)J_{n, j+1}'(x)\\ &\leq& 4+ \sum\limits^{n-2}_{j = 1}\frac{\int^{1} _{0}t^{(n-1)j-1}(1-t)^{(n-1)(n-1-j)-1}}{B((n-1)j, (n-1)(n-1-j))}\left|\beta t+(1-\beta)\frac{j}{n-1}-x\right|{\rm d}tJ_{n, j+1}^{\alpha-1}(x)J_{n, j+1}'(x), \end{eqnarray*} $

因此$ A\leq 14. $结合$ A $和$ B $可知

(4.17) $ \begin{equation} |\varphi^{\lambda}(x)H_{2}|\leq 96\alpha\|\varphi^{\lambda} f'\|. \end{equation} $

由(4.10)式, (4.11)式, (4.17)式, 对于$ x\in E_{n}^{C} $, 有

$ |\varphi^{\lambda}(x)\cdot(L_{n, \beta}^{(\alpha)}(f;x))'|\leq 104\alpha\|\varphi^{\lambda}f'\| . $

(Ⅱ) 当$ x\in E_{n} = [\frac{1}{n}, 1-\frac{1}{n}] $时, 注意到$ p_{n, k}'(x) = \frac{n}{\varphi^{2}(x)}(\frac{k}{n}-x)\cdot p_{n, k}(x) $, 应用Cauchy-Schwarz, 不等式可得

$ \begin{eqnarray*} B&\leq& \bigg(\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}}{B((n-1)k, (n-1)(n-1-k))} \Big(\beta t+(1-\beta)\frac{k}{n-1}-x\Big)^{2}{\rm d}t\cdot p_{n, k}(x)\bigg)^{\frac{1}{2}}\\ &&\times\bigg(\sum\limits^{n-1}_{k = 1}\frac{\int^{1} _{0}t^{(n-1)k-1}(1-t)^{(n-1)(n-1-k)-1}{\rm d}t} {B((n-1)k, (n-1)(n-1-k))}\cdot p_{n, k}(x)\cdot\Big(\frac{k}{n}-x\Big)^{2}\bigg)^{\frac{1}{2}}\cdot \frac{n}{\varphi^{2}(x)}\\ &\leq&\bigg(\sum\limits^{n-1}_{k = 1}p_{n, k}(x)F_{n-1, k}^{(\beta)}((t-x)^{2};x)\bigg)^{\frac{1}{2}} \cdot\left(B_{n}((t-x)^{2};x)\right)^{\frac{1}{2}}\cdot\frac{n}{\varphi^{2}(x)}<5, \ \ (n\geq 2 ), \end{eqnarray*} $

最后的一个不等关系中用到了引理2.7(2).

当$ x\in E_{n}^{C} $时, 用相同的方法可得$ A\leq 4+B\leq 9, $那么

$ |\varphi^{\lambda}(x)H_{2}|\leq 56\alpha\|\varphi ^{\lambda}f'\|. $

因此, 当$ x\in E_{n}, n\ge 2 $时, 有

$ |\varphi^{\lambda}(x)\cdot(L_{n, \beta}^{(\alpha)}(f;x))'|\leq|\varphi^{\lambda}(x)H_{1}|+|\varphi^{\lambda}(x)H_{2}|+|\varphi^{\lambda}(x)H_{3}| \leq 64\alpha\|\varphi ^{\lambda}f'\| . $

证毕.