定义在圆盘上的复Bernstein多项式的逼近性质是由Bernstein最早开始研究的.设$ f:{\Bbb G}\rightarrow {\Bbb C} $是开集$ {\Bbb G} $上的解析函数,这里$ {\Bbb G}\subset {\Bbb C} $, $ \overline{{\Bbb D}}_1\subset {\Bbb G} $ ($ {\Bbb D}_1: = \{z\in {\Bbb C}: |z|<1\} $). Bernstein^[16]证明了经典复Bernstein多项式$ B_{n}(f, z): = \sum\limits_{k = 0}^{n}\bigg(\begin{array}{c}n\\ k\end{array}\bigg)z^{k}\left( 1-z\right) ^{n-k}f\left(\frac{k}{n}\right) $; $ z\in{\Bbb C} $,在$ \overline{{\Bbb D}}_1 $上一致收敛于$ f $.有关Bernstein多项式逼近的精确量化估计以及Voronovaskaja型可参见文献[2-3].许多学者研究了诸如Bernstein-Stancu多项式, Kantorovich型Bernstein-Stancu多项式, Durrmeyer型Bernstein-Stancu多项式等Bernstein多项式的推广形式对圆盘上解析函数的逼近性质,建立了一些重要的结论(参见文献[1-21]).

最近,蒋和虞^[17]引入了一类新的定义在移动圆盘上的复Bernstein-Stancu多项式.设$ \alpha_{1}, \alpha_{2}, \beta_1, \beta_2 $是任意给定的常数,且满足$ 0\leq\alpha_{1}\leq\beta_{1} $, $ \alpha_{2}\in {\Bbb C} $, $ |\alpha_{2}|\leq\beta_{2} $.蒋和虞^[17]引入的新的复Bernstein-Stancu多项式$ S_{n, \alpha, \beta}(f, z) $定义如下:

$ S_{n, \alpha, \beta}(f, z): = \left(\frac{n+\beta_{2}}{n}\right)^{n} \sum\limits_{k = 0}^{n}q_{n, k}(z)f\left(\frac{k+\alpha_{1}}{n+\beta_{1}}\right), $

其中

$ q_{n, k}(z) : = \bigg(\begin{array}{c}n\\ k\end{array}\bigg)\left(z-\frac{\alpha_2}{n+\beta_2}\right)^k\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^{n-k}, k = 0, 1, \cdots, n. $

当$ \alpha_{2} = \beta_{2} = 0 $时, $ S_{n, \alpha, \beta}(f, z) $即为经典复Bernstein-Stancu多项式.蒋和虞的主要结论如下:

定理1.1  设$ R $和$ r $是满足$ 1\leq r<R $的两个常数, $ {\Bbb D}_R: = \{z\in {\Bbb C};|z|<R\} $.设$ f:{\Bbb D}_R\rightarrow {\Bbb C} $是$ {\Bbb D}_R $上的解析函数,且对所有的$ z\in{\Bbb D}_R $有$ f(z) = \sum\limits_{k = 0}^{\infty}c_{k}z^{k} $.如果$ n\in {\Bbb N} $充分大使得$ r+\frac{ |\alpha_{2}|}{n+\beta_{2}} = :\overline{r}<R $及$ 1+\frac{\beta_{2}}{n}\leq 2 $,则对所有$ \big|z-\frac{\alpha_{2}}{n+\beta_{2}}\big|\leq r $,有

$ \left|S_{n, \alpha, \beta}(f, z)-f(z)\right|\leq {\mathfrak M}_{r}^{(\alpha, \beta)}(f), $

其中

$ {\mathfrak M}_{r}^{(\alpha, \beta)}(f): = \left(1+\frac{\beta_{2}}{n} \right)^{n}\sum\limits_{k = 1}^{\infty}|c_{k}|\left[\frac{\beta_{2}}{n}k\overline{r} ^{k}+\frac{2\beta_{1}k+2k(k-1)}{n+\beta_{1}}\overline{r}^{k}+\frac{ |\alpha_{2}|}{n+\beta_{2}}k\overline{r}^{k-1}\right]<\infty. $

同时,若$ 1\leq r<r_1<R, $$ n\in{\Bbb N} $充分大使得$ r_1+\frac{|\alpha_{2}|}{n+\beta_{2}} = : \overline{r_1}<R $及$ 1+\frac{\beta_{2}}{n}\leq2 $,则对所有$ \big|z-\frac{\alpha_{2}}{n+\beta_{2}}\big|\leq r_1 $, $ n, p\in{\Bbb N} $,有

(1.1) $ \begin{equation} \left|\left(S_{n, \alpha, \beta}(f, z)\right)^{(p)}-f^{(p)}(z)\right| \leq {\mathfrak M}_{r_1}^{\left(\alpha, \beta\right)}(f)\frac{p! r_{1}}{ \left(r_{1}-r\right)^{p+1}}. \end{equation} $

应该注意的是:定理1.1中研究了$ S_{n, \alpha, \beta}(f) $在圆盘$ \left\{z\in{\Bbb C} :\big|z-\frac{\alpha_{2}}{n+\beta_{2}}\big|\leq r\right\} $上的逼近性质.这个圆盘是一个移动圆盘,即随着参数$ n, \alpha_2 $和$ \beta_2 $的改变而移动.

我们回顾由Mahmudov和Gupta给出的本征Durrmeyer-Stancu多项式(参见文献[15]):

(1.2) $ \begin{eqnarray} U_n^{(\alpha, \beta)}(f;z) &: = &(n-1)\sum\limits_{k = 1}^{n-1}p_{n, k}(z)\int_0^1p_{n-2, k-1}(t)f\left(\frac{nt+\alpha}{n+\beta}\right){\rm d}t \\ & &+f\left(\frac{\alpha}{n+\beta}\right)p_{n, 0}(z)+f\left(\frac{n+\alpha}{n+\beta}\right)p_{n, n}(z), \end{eqnarray} $

其中$ \alpha, \beta $是满足$ 0 \leq \alpha \leq \beta $的任意实数, $ z \in {\Bbb C} $, $ n = 1, 2, \cdots; $$ p_{n, k}(z) = \bigg(\begin{array}{c}n\\ k\end{array}\bigg)z^{k}(1-z)^{n-k}. $

当$ \alpha = \beta = 0 $时,该多项式即通常的本征Bernstein-Durrmeyer算子$ U_n(f;z) = U_n^{(0, 0)}(f;z) $.这种情形更早在文献[1]中已有过研究.

Mahmudov和Gupta对$ U_n^{(\alpha, \beta)}(f;z) $在圆盘上对解析函数的逼近性质做了研究(参见文献[1]).以下引述其中一个关于$ U_n^{(\alpha, \beta)}(f;z) $逼近速度的结论:

定理1.2  设$ \alpha, \beta $是满足$ 0 \leq \alpha \leq \beta $的两个任意给定的实数, $ R $和$ r $是满足$ 1 \leq r < R $的两个常数, $ {\Bbb D}_R: = \{z\in{\Bbb C} : \left|z\right|<R\} $.设$ f:{\Bbb D}_R\rightarrow {\Bbb C} $是$ {\Bbb D}_R $上的解析函数,且对所有的$ z\in{\Bbb D}_R $有$ f(z) = \sum\limits_{m = 0}^{\infty}c_{m}z^{m}, $则对所有$ \left|z\right| \le r $,有

$ \left|M_n^{(\alpha, \beta)}(f;z)-f(z)\right| \leq \frac{A_{r, \beta}(f)}{n+\beta}+\frac{\alpha B_r(f)}{n(n+\beta)}, $

其中

$ A_{r, \beta}(f) = 2(1+r)\sum\limits_{m = 1}^\infty\left|c_m\right|\left(\frac{m(m-1)}{2}+\beta m\right)r^{m-1}, $

$ B_r(f) = \alpha(1+r)\sum\limits_{m = 2}^\infty\left|c_m\right|m(m-1)r^{m-2}. $

在本文中,我们将引入一种新的Durrmeyer-Stancu型多项式以进一步推广(1.2)所定义的多项式,并使之能够用以逼近移动圆盘上的解析函数.我们还将考查这一新多项式的逼近速度.

在下文中,记$ {\Bbb D}_R $是以$ O(0, 0) $为圆心, $ R $为半径的圆,这里$ R > 1 $为常数,即$ {\Bbb D}_R: = \{z\in{\Bbb C} : \left|z\right|<R\} $.设$ \alpha_1, \alpha_2, \beta_1, \beta_2 $是任意给定的数且满足$ \alpha_2 \in {\Bbb C}, $$ 0 \leq \alpha_1 \leq \beta_1, $$ \left|\alpha_2\right| \leq \beta_2 $.

设$ f : {\Bbb D}_R\rightarrow {\Bbb C} $是$ {\Bbb D}_R $上的解析函数,且对所有的$ z\in{\Bbb D}_R $有$ f(z) = \sum\limits_{m = 0}^{\infty}c_{m}z^{m} $.定义新的Durrmeyer-Stancu如下:

$ \begin{eqnarray*} \overline{U}_{n, \alpha, \beta}(f;z) & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)f\left(\frac{nt+\alpha_1}{n+\beta_1}\right){\rm d}t\\ & &+\left(\frac{n+\beta_2}{n}\right)^n\left(z-\frac{\alpha_2}{n+\beta_2}\right)^nf\left(\frac{n^2+n\alpha_1+n\alpha_2+\alpha_1\beta_2}{(n+\beta_1)(n+\beta_2)}\right)\\ & &+\left(\frac{n+\beta_2}{n}\right)^n\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^nf\left(\frac{n\alpha_1+n\alpha_2+\alpha_1\beta_2}{(n+\beta_1)(n+\beta_2)}\right), \end{eqnarray*} $

其中$ A_n: = [\frac{\alpha_2}{n+\beta_2}, \frac{n+\alpha_2}{n+\beta_2}] $,而

$ q_{n, k}(z) = : \bigg(\begin{array}{c}n\\ k\end{array}\bigg)\left(z-\frac{\alpha_2}{n+\beta_2}\right)^k\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^{n-k}, k = 0, 1, \cdots, n, $

$ q_{n-2, k-1}^{*}(z) = : \bigg(\begin{array}{c}{n-2}\\ {k-1}\end{array}\bigg) \left(z-\frac{\alpha_2}{n+\beta_2}\right)^{k-1}\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^{n-k-1}, k = 0, 1, \cdots, n. $

当$ \alpha_2 = \beta_2 = 0 $时, $ \overline{U}_{n, \alpha, \beta}(f;z) $即为(1.2)式中定义的多项式.

本文的主要结果为:

定理1.3  设$ R $和$ r $是满足1 $ \leq $$ r $$ \leq $$ R $的两个常数.设$ f:{\Bbb D}_R\rightarrow {\Bbb C} $是$ {\Bbb D}_R $上的解析函数,且对所有的$ z\in {\Bbb D}_R $有$ f(z) = \sum\limits_{m = 0}^{\infty}c_{m}z^{m} $.如果$ n \in {\Bbb N} $充分大使得$ r\big(1+\frac{\beta_2}{n}\big) = :\overline{r}< R, $则对所有$ \big|z-\frac{\alpha_2}{n+\beta_2}\big| \leq r $,有

(1.3) $ \begin{equation} \left|\overline{U}_{n, \alpha, \beta}(f;z)-f(z)\right| \leq K_{r, n}^{(\alpha, \beta)}(f), \end{equation} $

其中

$ \begin{eqnarray*} K_{r, n}^{(\alpha, \beta)}(f) &: = &\sum\limits_{m = 2}^\infty\left|c_m\right|\left(\frac{2m\left(m-1\right)\left(2+\overline{r}\right)+2\beta_1\left(m-1\right)\left(1+\overline{r}\right)}{2\left(n+\beta_1\right)}\overline{r}+\frac{\alpha_1m\left(m-1\right)}{2n\left(n+\beta_1\right)}\overline{r}\right.\\ & &+\frac{m\left(m-1\right)\left|\alpha_2\right|}{2\left(n+\beta_1\right)\left(n+\beta_2\right)}\overline{r}+\frac{m\left(m-1\right)\left(\alpha_1+\left|\alpha_2\right|\right)}{2\left(n+\beta_1\right)^2}+\frac{m\left(m-1\right)\alpha_1^2}{2n\left(n+\beta_1\right)^2}\\ & &\left.+\frac{m\left(m-1\right)\alpha_1\left|\alpha_2\right|}{2\left(n+\beta_1\right)^2\left(n+\beta_2\right)}\right)\overline{r}^{m-2}+\left|c_1\right|\left(\frac{\beta_1}{n+\beta_1}\overline{r}+\frac{\alpha_1}{n+\beta_1}\right). \end{eqnarray*} $

同时,若$ 1\leq r<r_1<R $,那么当$ n\in{\Bbb N} $充分大使得$ r\big(1+\frac{\beta_2}{n}\big): = \overline{r}< R $时,对所有$ \big|z-\frac{\alpha_2}{n+\beta_2}\big| \leq r_1 $, $ n, p \in {\Bbb N}, $有

(1.4) $ \begin{equation} \left|\left(\overline{U}_{n, \alpha, \beta}(f;z)\right)^{(p)}-f(z)^{(p)}\right| \leq K_{{r_1}, n}^{(\alpha, \beta)}(f)\frac{p!r_1}{(r_1-r)^{p+1}}. \end{equation} $

引理2.1  设$ m, n\in{\Bbb N} \cup{0}, r\geq 1, r\big(1+\frac{\beta_2}{n}\big): = \overline{r}<R. $那么对任意$ \big|z-\frac{\alpha_2}{n+\beta_2}\big|\leq r, $有$ \left|\overline{U}_{n, \alpha, \beta}(e_m;z)\right| \leq \overline{r}^m, $其中$ e_m(z) = z^m, \; m = 0, 1, 2, \cdots. $

证  置

$ \begin{eqnarray*} U_{n, \alpha, \beta}^*(f;z) & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)f(t){\rm d}t \\ &&+\left(\frac{n+\beta_2}{n}\right)^n\left(z-\frac{\alpha_2}{n+\beta_2}\right)^nf\left(\frac{n+\alpha_2}{n+\beta_2}\right)\\ &&+\left(\frac{n+\beta_2}{n}\right)^n\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^nf\left(\frac{\alpha_2}{n+\beta_2}\right). \end{eqnarray*} $

直接计算可得

(2.1) $ \begin{equation} \overline{U}_{n, \alpha, \beta}(e_m;z) = \sum\limits_{p = 0}^m \bigg(\begin{array}{c}m\\ p\end{array}\bigg)\frac{n^p\alpha_1^{m-p}}{(n+\beta_1)^m} U_{n, \alpha, \beta}^*(e_p;z). \end{equation} $

注意到

$ \begin{eqnarray*} &&\int_{A_n}q_{n-2, k-1}^{*}(t)t^p{\rm d}t\\ & = &\int_{A_n}\bigg(\begin{array}{c}n-2\\ k-1\end{array}\bigg)\left(t-\frac{\alpha_2}{n+\beta_2}\right)^{k-1}\left(\frac{n+\alpha_2}{n+\beta_2}-t\right)^{n-1-k} \left(t-\frac{\alpha_2}{n+\beta_2}+\frac{\alpha_2}{n+\beta_2}\right)^p{\rm d}t\\ & = &\int_{A_n} \bigg(\begin{array}{c} {n-2}\\ {k-1}\end{array}\bigg)\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(t-\frac{\alpha_2}{n+\beta_2}\right)^{k-1+j} \left(\frac{n+\alpha_2}{n+\beta_2}-t\right)^{n-1-k}{\rm d}t, \end{eqnarray*} $

令$ u = \frac{n+\beta_2}{n}t-\frac{\alpha_2}{n}, $则有

$ \begin{eqnarray*} &&\int_{A_n}q_{n-2, k-1}^{*}(t)t^p{\rm d}t\\ & = &\bigg(\begin{array}{c}n-2\\ k-1\end{array}\bigg)\sum\limits_{j = 0}^p \bigg(\begin{array}{c}p\\ j\end{array}\bigg) \bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^{n+j-1}\int_0^1u^{k-1+j}(1-u)^{n-1-k}{\rm d}u \\ & = &\bigg(\begin{array}{c}n-2\\ k-1\end{array}\bigg)\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^{n+j-1}\frac{\left(k-1+j\right)!\left(n-k-1\right)!}{\left(n+j-1\right)!}\\ & = &\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^{n+j-1}\frac{\left(n-2\right)!}{\left(n+j-1\right)!}F_j(k), \end{eqnarray*} $

其中$ F_j(k) = \prod\limits_{s = 0}^{j-1}(k+s), k\geq 0, j \geq 1. $因此

$ \begin{eqnarray*} &&U_{n, \alpha, \beta}^*(e_p;z)\\ & = &\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^j\frac{(n-1)!}{(n+j-1)!} \left(\frac{n+\beta_2}{n}\right)^n\sum\limits_{k = 1}^{n-1}q_{n, k}(z)F_j(k) \\ &&+\left(\frac{n+\beta_2}{n}\right)^n\left(z-\frac{\alpha_2}{n+\beta_2}\right)^n\left(\frac{n+\alpha_2}{n+\beta_2}\right)^p +\left(\frac{n+\beta_2}{n}\right)^n\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^n\left(\frac{\alpha_2}{n+\beta_2}\right)^p. \end{eqnarray*} $

令$ \widetilde{z} = \frac{n+\beta_2}{n}z-\frac{\alpha_2}{n} $.则$ \left(\frac{n+\beta_2}{n}\right)^nq_{n, k}(z) = p_{n, k}(\widetilde{z}), k = 0, 1, 2, \cdots, $从而

(2.2) $ \begin{eqnarray} U_{n, \alpha, \beta}^*(e_p;z) & = &\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^j \\ & &\times\left(\frac{(n-1)!}{(n+j-1)!}\sum\limits_{k = 1}^{n-1}p_{n, k}(\widetilde{z})F_j(k)+p_{n, n}(\widetilde{z})e_j(1)+p_{n, 0}(\widetilde{z})e_j(0)\right) \\ & = &\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\alpha_2}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^jU_n^{(0, 0)}(e_j;\widetilde{z}). \end{eqnarray} $

若$ 1\le r<R, $则对所有$ \left|z\right|\le r, $有^[1]

$ \left|U_n^{(0, 0)}(e_j;z)\right| \le r^j, j = 0, 1, 2, \cdots , $

结合下式

$ \left|\widetilde{z}\right| = \left|\frac{n+\beta_2}{n}z-\frac{\alpha_2}{n}\right| = \left|\frac{n+\beta_2}{n}\left(z-\frac{\alpha_2}{n+\beta_2}\right)\right| \leq \overline{r}, $

得$ \left|U_n^{(0, 0)}(e_j;\widetilde{z})\right| \leq \overline{r}^j, j = 0, 1, 2, \cdots . $因此

$ \begin{eqnarray*} \left|U_{n, \alpha, \beta}^*(e_p;z)\right| &\leq&\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\left|\alpha_2\right|}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^j\overline{r}^j\\ &\leq&\overline{r}^p\sum\limits_{j = 0}^p\bigg(\begin{array}{c}p\\ j\end{array}\bigg)\left(\frac{\left|\alpha_2\right|}{n+\beta_2}\right)^{p-j}\left(\frac{n}{n+\beta_2}\right)^j\\ & = &\overline{r}^p\left(\frac{n+\left|\alpha_2\right|}{n+\beta_2}\right)^p\leq \overline{r}^p. \end{eqnarray*} $

由(2.1)式,有

$ \begin{eqnarray*} \left|\overline{U}_{n, \alpha, \beta}(e_m;z)\right| &\leq&\sum\limits_{p = 0}^m\bigg(\begin{array}{c}m\\ p\end{array}\bigg)\frac{n^p\alpha_1^{m-p}}{(n+\beta_1)^m}\overline{r}^p\leq\overline{r}^m\sum\limits_{p = 0}^m\bigg(\begin{array}{c}m\\ p\end{array}\bigg)\frac{n^p\alpha_1^{m-p}}{(n+\beta_1)^m}\\ & = &\overline{r}^m\left(\frac{n+\alpha_1}{n+\beta_1}\right)^m\leq\overline{r}^m. \end{eqnarray*} $

引理2.1证毕.

引理2.2  对所有$ m, n\in{\Bbb N} $, $ z\in {\Bbb C} $,成立

(2.3) $ \begin{eqnarray} \overline{U}_{n, \alpha, \beta}(e_{m+1};z) & = &\frac{n\left(z-\frac{\alpha_2}{n+\beta_2}\right)\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)}{\left(n+\beta_1\right)\left(n+m\right)}\left(\overline{U}_{n, \alpha, \beta}(e_{m};z)\right)^\prime \\ & &+\left(\frac{\alpha_1\left(n+2m\right)+n^2z}{\left(n+\beta_1\right)\left(n+m\right)}+\frac{mn\left(n+2\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)\left(n+m\right)}\right)\overline{U}_{n, \alpha, \beta}(e_{m};z) \\ & &-\left(\frac{mn\alpha_1\left(n+2\alpha_2\right)}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)\left(n+m\right)}+\frac{m\alpha_1^2}{\left(n+\beta_1\right)^2\left(n+m\right)}\right. \\ & &+\left.\frac{mn^2\alpha_2\left(n+\alpha_2\right)}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)^2\left(n+m\right)}\right)\overline{U}_{n, \alpha, \beta}(e_{m-1};z). \end{eqnarray} $

证  令

$ W_{n, \alpha, \beta}(e_m, z) = \left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)e_{m}\left(\frac{nt+\alpha_1}{n+\beta_1}\right), $

$ \begin{eqnarray*} V_{n, \alpha, \beta}(e_m, z) & = &\left(\frac{n+\beta_2}{n}\right)^n\left(z-\frac{\alpha_2}{n+\beta_2}\right)^ne_{m}\left(\frac{n^2+n\alpha_1+n\alpha_2+\alpha_1\beta_2}{(n+\beta_1)(n+\beta_2)}\right)\\ & &+\left(\frac{n+\beta_2}{n}\right)^n\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)^ne_{m}\left(\frac{n\alpha_1+n\alpha_2+\alpha_1\beta_2}{(n+\beta_1)(n+\beta_2)}\right). \end{eqnarray*} $

由计算可得

$ \begin{eqnarray*} & &\left(z-\frac{\alpha_2}{n+\beta_2}\right)\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)\left(W_{n, \alpha, \beta}(e_m, z)\right)^\prime\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m\left(\frac{kn+n\alpha_2}{n+\beta_2}-nz\right){\rm d}t\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m\\ &&\times\left(\frac{kn+n\alpha_2}{n+\beta_2}-nt+2t-\frac{n+2\alpha_2}{n+\beta_2}\right){\rm d}t\\ & &+\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m\left(nt-2t-nz\right){\rm d}t\\ & &+\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m\left(\frac{n+2\alpha_2}{n+\beta_2}\right){\rm d}t\\ & = :&J_1+J_2+J_3. \end{eqnarray*} $

对于$ J_1 $,有

$ J_1 = \left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}\left(q_{n-2, k-1}^{*}(t)\right)^\prime\left(t-\frac{\alpha_2}{n+\beta_2}\right)\left(\frac{n+\alpha_2}{n+\beta_2}-t\right) $

$ \begin{eqnarray*} & &\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\left(2t-\frac{n+2\alpha_2}{n+\beta_2}\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m\right.\\ & &\left.-\frac{mn}{n+\beta_1}\left(\frac{n+2\alpha_2}{n+\beta_2}t-t^2-\frac{\left(n+\alpha_2\right)\alpha_2}{\left(n+\beta_2\right)^2}\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m-1}\right){\rm d}t\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(2t-\frac{n+2\alpha_2}{n+\beta_2}\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t \\ & &-\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t )\frac{mn}{n+\beta_1}\\ && \times\left(\frac{n+2\alpha_2}{n+\beta_2}t-t^2-\frac{\left(n+\alpha_2\right)\alpha_2}{\left(n+\beta_2\right)^2}\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m-1}{\rm d}t\\ & = :&K_1-K_2. \end{eqnarray*} $

对于$ K_1 $,有

$ \begin{eqnarray*} K_1 & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)2t\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t\\ & &-\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{n+2\alpha_2}{n+\beta_2}\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)2\left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n}-\frac{\alpha_1}{n}\right)\\ & &\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t-\frac{n+2\alpha_2}{n+\beta_2}W_{n, \alpha, \beta}(e_m, z)\\ & = &\frac{2\left(n+\beta_1\right)}{n}\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m+1}{\rm d}t\\ & &-\frac{2\alpha_1}{n}\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^m{\rm d}t\\ & &-\frac{n+2\alpha_2}{n+\beta_2}W_{n, \alpha, \beta}(e_m, z)\\ & = &\frac{2\left(n+\beta_1\right)}{n}W_{n, \alpha, \beta}(e_{m+1}, z)-\frac{2\alpha_1}{n}W_{n, \alpha, \beta}(e_m, z)-\frac{n+2\alpha_2}{n+\beta_2}W_{n, \alpha, \beta}(e_m, z). \end{eqnarray*} $

对于$ K_2 $,有

$ \begin{eqnarray*} K_2 & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z) \int_{A_n}q_{n-2, k-1}^{*}(t)\frac{mn}{n+\beta_1}\\ &&\times\left(\frac{n+2\alpha_2}{n+\beta_2} \left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n} -\frac{\alpha_1}{n}\right) -\left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n}-\frac{\alpha_1}{n}\right)^2-\frac{\alpha_2\left(n+\alpha_2\right)}{\left(n+\beta_2\right)^2}\right) \\ &&\times\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m-1}{\rm d}t\\ & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t) \frac{mn}{n+\beta_1}\\ &&\times\left(\frac{n+2\alpha_2}{n+\beta_2}\left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n} -\frac{\alpha_1}{n}\right)\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m-1}{\rm d}t\\ && -\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t)\frac{mn}{n+\beta_1}\\ & &\left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n}-\frac{\alpha_1}{n}\right)^2\left(\frac{nt+\alpha_1} {n+\beta_1}\right)^{m-1}{\rm d}t -\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\\ &&\times \int_{A_n}q_{n-2, k-1}^{*}(t)\frac{mn}{n+\beta_1}\frac{\alpha_2\left(n+\alpha_2\right)} {\left(n+\beta_2\right)^2}\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m-1}{\rm d}t\\ & = :&L_1-L_2-L_3. \end{eqnarray*} $

容易观察到

$ \begin{eqnarray*} L_1& = &\frac{m\left(n+2\alpha_2\right)}{n+\beta_2}W_{n, \alpha, \beta}(e_m, z)-\frac{m\alpha_1\left(n+2\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)}W_{n, \alpha, \beta}(e_{m-1}, z), \\ L_2& = &\frac{m\left(n+\beta_1\right)}{n}W_{n, \alpha, \beta}(e_{m+1}, z)-\frac{2m\alpha_1}{n}W_{n, \alpha, \beta}(e_{m}, z)+\frac{m\alpha_1^2}{n\left(n+\beta_1\right)}W_{n, \alpha, \beta}(e_{m-1}, z), \\ L_3& = &\frac{mn\alpha_2\left(n+\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)^2}W_{n, \alpha, \beta}(e_{m-1}, z). \end{eqnarray*} $

对于$ J_2 $和$ J_3 $,有

$ \begin{eqnarray*} J_2 & = &\left(\frac{n+\beta_2}{n}\right)^{2n-1}(n-1)\sum\limits_{k = 1}^{n-1}q_{n, k}(z)\int_{A_n}q_{n-2, k-1}^{*}(t) \\ &&\times\left(\left(n-2\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\frac{n+\beta_1}{n}-\frac{\alpha_1}{n}\right)-nz\right)\left(\frac{nt+\alpha_1}{n+\beta_1}\right)^{m}{\rm d}t\\ & = &\frac{\left(n-2\right)\left(n+\beta_1\right)}{n}W_{n, \alpha, \beta}(e_{m+1}, z)-\left(\frac{\left(n-2\right)\alpha_1}{n}+nz\right)W_{n, \alpha, \beta}(e_{m}, z), \end{eqnarray*} $

而

$ J_3 = \frac{n+2\alpha_2}{n+\beta_2}W_{n, \alpha, \beta}(e_{m}, z). $

因此

$ \begin{eqnarray*} &&\left(z-\frac{\alpha_2}{n+\beta_2}\right)\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)\left(W_{n, \alpha, \beta}(e_m, z)\right)^\prime\\ & = &\frac{\left(n+\beta_1\right)\left(m+n\right)}{n}W_{n, \alpha, \beta}(e_{m+1}, z) -\left(\frac{2m\alpha_1+n\alpha_1}{n}+nz+\frac{m\left(n+2\alpha_2\right)}{n+\beta_2}\right)W_{n, \alpha, \beta}(e_{m}, z)\nonumber \\ & &+\left(\frac{m\alpha_1\left(n+2\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)}+\frac{m\alpha_1^2}{n\left(n+\beta_1\right)}+\frac{mn\alpha_2\left(n+\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)^2}\right)W_{n, \alpha, \beta}(e_{m-1}, z).\nonumber \end{eqnarray*} $

即$ W_{n, \alpha, \beta}(e_m, z) $满足等式(2.3).

由直接计算可知$ V_{n, \alpha, \beta}(e_m, z) $也满足等式(2.3).引理2.2证毕.

由引理2.2可得

$ \begin{eqnarray*} && \overline{U}_{n, \alpha, \beta}(e_{m};z)-e_m(z)\\ & = &\frac{n\left(z-\frac{\alpha_2}{n+\beta_2}\right)\left(\frac{n+\alpha_2}{n+\beta_2}-z\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\left(\overline{U}_{n, \alpha, \beta}(e_{m-1};z)\right)^\prime\\ & &+\frac{n\left(m-1\right)+n^2z+\alpha_1\left(n+m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\left(\overline{U}_{n, \alpha, \beta}(e_{m-1};z)-e_{m-1}\right)\\ & &+\frac{\alpha_1\left(m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-1};z) +\frac{\left(m-1\right)n\left(n+\alpha_2\right)}{\left(n+\beta_1\right)\left(n+\beta_2\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-1};z) \\ & &+\frac{\left(m-1\right)n\alpha_2}{\left(n+\beta_1\right)\left(n+\beta_2\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-1};z) -\frac{n\left(m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-1};z)\\ & &-\frac{\left(m-1\right)n\alpha_1\left(n+\alpha_2\right)}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-2};z)\\ & &-\frac{\left(m-1\right)n\alpha_1\alpha_2}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-2};z) -\frac{\left(m-1\right)\alpha_1^2}{\left(n+\beta_1\right)^2\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-2};z)\\ & &-\frac{\left(m-1\right)n^2\alpha_2\left(n+\alpha_2\right)}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)^2\left(n+m-1\right)}\overline{U}_{n, \alpha, \beta}(e_{m-2};z)\\ & &+\frac{\left(m-1\right)n\left(1-z\right)+\left(\alpha_1-z\beta_1\right)\left(n+m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}e_{m-1}(z). \end{eqnarray*} $

注意到

$ |z|\leq\big|z-\frac{\alpha_{2}}{n+\beta_{2}}\big| +\left|\frac{\alpha_{2}}{n+\beta_{2}}\right|\leq r+\frac{|\alpha_{2}|}{n+\beta_{2}}\leq \bar{r}, $

有

$ \begin{eqnarray*} \left|\frac{n\left(m-1\right)+n^2z+\alpha_1\left(n+m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\right| &\leq&\frac{n}{n+\beta_1}\frac{m-1+n|z|}{n+m-1}+\frac{\alpha_1}{n+\beta_1} \leq\frac{n\overline{r}+\alpha_1}{n+\beta_1}\leq\overline{r}. \end{eqnarray*} $

同理可得

$ \begin{eqnarray*} && \left|\frac{\left(m-1\right)n\left(1-z\right)+\left(\alpha_1-z\beta_1\right)\left(n+m-1\right)}{\left(n+\beta_1\right)\left(n+m-1\right)}\right| \\ &\leq&\frac{(m-1)(1+\overline{r})}{n+\beta_1}+\left|\frac{\alpha_1-z\beta_1}{n+\beta_1}\right| \leq\frac{(m-1)(1+\overline{r})}{n+\beta_1}+\frac{\beta_1(1+\overline{r})}{n+\beta_1} = \frac{(m-1+\beta_1)(1+\overline{r})}{n+\beta_1}. \end{eqnarray*} $

因此,对于$ m\ge $ 2,有

$ \begin{eqnarray*} && \left|\overline{U}_{n, \alpha, \beta}(e_{m};z)-e_m(z)\right|\\ &\leq&\overline{r}\left|\overline{U}_{n, \alpha, \beta}(e_{m-1};z)-e_{m-1}(z)\right| +\frac{\left(m-1\right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-1} +\frac{\alpha_1\left(m-1\right)}{n\left(n+\beta_1\right)}\overline{r}^{m-1} +\frac{m-1}{n+\beta_1}\overline{r}^{m-1}\\ &&+\frac{\left(m-1\right)\left|\alpha_2\right|} {\left(n+\beta_1\right)\left(n+\beta_2\right)}\overline{r}^{m-1} +\frac{m-1}{n+\beta_1}\overline{r}^{m-1} +\frac{\left(m-1\right)\alpha_1}{\left(n+\beta_1\right)^2}\overline{r}^{m-2}+\frac{\left(m-1\right)\alpha_1\left|\alpha_2\right|}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)}\overline{r}^{m-2} \\ &&+\frac{\left(m-1\right)\alpha_1^2}{n\left(n+\beta_1\right)^2}\overline{r}^{m-2} +\frac{\left(m-1\right) \left|\alpha_2\right|}{\left(n+\beta_1\right)^2}\overline{r}^{m-2} +\frac{\left(m-1+\beta_1\right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-1}. \end{eqnarray*} $

因此,对于$ m = 2, 3, \cdots $,可推得

(3.1) $ \begin{eqnarray} &&\left|\overline{U}_{n, \alpha, \beta}(e_{m};z)-e_m(z)\right|\\ &\leq&\overline{r}\left(\overline{r}\left|\overline{U}_{n, \alpha, \beta}(e_{m-2};z)-e_{m-2}(z)\right|+\frac{\left(m-2\right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-2}\right. +\frac{\alpha_1\left(m-2\right)}{n\left(n+\beta_1\right)}\overline{r}^{m-2} \\ &&+\frac{m-2}{n+\beta_1}\overline{r}^{m-2}+ \frac{\left(m-2\right)\left|\alpha_2\right|}{\left(n+\beta_1\right)\left(n+\beta_2\right)}\overline{r}^{m-2}+\frac{m-2}{n+\beta_1}\overline{r}^{m-2}+\frac{\left(m-2\right)\alpha_1}{\left(n+\beta_1\right)^2}\overline{r}^{m-3} \\ &&+ \frac{\left(m-2\right)\alpha_1\left|\alpha_2\right|}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)}\overline{r}^{m-3} +\frac{\left(m-2\right)\alpha_1^2}{n\left(n+\beta_1\right)^2}\overline{r}^{m-3}+\frac{\left(m-2\right)\left|\alpha_2 \right|}{\left(n+\beta_1\right)^2}\overline{r}^{m-3} \\ &&+\left.\frac{\left(m-2+\beta_1\right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-2}\right)+\frac{\left(m-1 \right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-1} +\frac{\alpha_1\left(m-1\right)}{n\left(n+\beta_1\right)}\overline{r}^{m-1} \\ &&+\frac{m-1}{n+\beta_1}\overline{r}^{m-1}+ \frac{\left(m-1\right)\left|\alpha_2\right|}{\left(n+\beta_1\right)\left(n+\beta_2\right)}\overline{r}^{m-1} +\frac{m-1}{n+\beta_1}\overline{r}^{m-1}+\frac{\left(m-1\right)\alpha_1}{\left(n+\beta_1\right)^2}\overline{r}^{m-2} \\ && +\frac{\left(m-1\right)\alpha_1\left|\alpha_2\right|}{\left(n+\beta_1\right)^2\left(n+\beta_2\right)}\overline{r}^{m-2} +\frac{\left(m-1\right)\alpha_1^2}{n\left(n+\beta_1\right)^2}\overline{r}^{m-2}+\frac{\left(m-1\right)\left|\alpha_2\right|}{\left(n+\beta_1\right)^2}\overline{r}^{m-2} \\ &&+\frac{\left(m-1+\beta_1\right)\left(1+\overline{r}\right)}{n+\beta_1}\overline{r}^{m-1} \\ &\leq&\cdots\leq\frac{m\left(m-1\right)\left(1+\overline{r}\right)}{2\left(n+\beta_1\right)}\overline{r}^{m-1}+\frac{m\left(m-1\right)}{2\left(n+\beta_1\right)}\overline{r}^{m-1} +\frac{\alpha_1m\left(m-1\right)}{2n\left(n+\beta_1\right)}\overline{r}^{m-1} \\ &&+\frac{m\left(m-1\right)\left|\alpha_2\right|}{2\left(n+\beta_1\right)\left(n+\beta_2\right)}\overline{r}^{m-1} +\frac{m\left(m-1\right)}{2\left(n+\beta_1\right)}\overline{r}^{m-1} +\frac{m\left(m-1\right)\alpha_1}{2\left(n+\beta_1\right)^2}\overline{r}^{m-2} \\ && +\frac{m\left(m-1\right)\alpha_1\left|\alpha_2\right|}{2\left(n+\beta_1\right)^2\left(n+\beta_2\right)}\overline{r}^{m-2} +\frac{m\left(m-1\right)\alpha_1^2}{2n\left(n+\beta_1\right)^2}\overline{r}^{m-2}+\frac{m\left(m-1\right)\left|\alpha_2\right|}{2\left(n+\beta_1\right)^2}\overline{r}^{m-2} \\ &&+\frac{\left(m\left(m-1\right)+2\beta_1\left(m-1\right)\right)\left(1+\overline{r}\right)}{2\left(n+\beta_1\right)}\overline{r}^{m-1}. \end{eqnarray} $

由(3.1)式及$ \overline{U}_{n, \alpha, \beta}(e_0;z) = 1; $$ \overline{U}_{n, \alpha, \beta}(e_1;z) = \frac{n}{n+\beta_1}z+\frac{\alpha_1}{n+\beta_1} $,得

$ \left|\overline{U}_{n, \alpha, \beta}(f;z)-f(z)\right|\le\sum\limits_{m = 0}^\infty\left|{c_m}\right| \left|\overline{U}_{n, \alpha, \beta}(e_{m};z)-e_m(z)\right| = K_{r, n}^{(\alpha, \beta)}(f), $

(1.3)式得证.

记$ \Gamma $是以$ z_0 : = \frac{\alpha_2}{n+\beta_2} $为圆心, $ r_1 $为半径的圆周,且$ r_1 > r. $因为对于任意的$ \big|z-\frac{\alpha_2}{n+\beta_2}\big|\leq r, $$ v \in \Gamma $,有$ \left|v-z\right|\geq r_1-r. $由柯西公式,对于任意$ \big|z-\frac{\alpha_2}{n+\beta_2}\big| \leq r, n, p \in {\Bbb N}, $有

$ \begin{eqnarray*} \left|\left(\overline{U}_{n, \alpha, \beta}(f;z)\right)^{(p)}-f^{(p)}(z)\right| & = &\frac{p!}{2\pi}\left|\int_{\Gamma}\frac{\overline{U}_{n, \alpha, \beta}(f;z)-f(z)}{(v-z)^{p+1}}{\rm d}v\right|\\ &\leq&K_{r_1, n}^{(\alpha, \beta)}(f)\frac{p!}{2\pi}\frac{2\pi r_1}{(r_1-1)^{p+1}}\\ & = &K_{r_1, n}^{(\alpha, \beta)}(f)\frac{p!r_1}{(r_1-r)^{p+1}}. \end{eqnarray*} $

这就证得了(1.4)式.