Abstract: In this paper, the approximation properties of the Baskakov-Kantorovich operators in the complex space are studied according to the definition and properties of the operator in the complex space. We obtain the simultaneous approximation order for complex Baskakov-Kantorovich operators attached to entire functions or to analytic functions in compact disks.

1. 引言

${K}_{n}\left(f,z\right)=\underset{j=0}{\overset{\infty }{\sum }}{v}_{n,j}\left(z\right)\underset{0}{\overset{1}{\int }}f\left(\frac{j+t}{n+1}\right)\text{d}t,$

${v}_{n,j}\left(z\right)=\left(\begin{array}{c}n+j-1\\ j\end{array}\right){z}^{j}{\left(1+z\right)}^{-n-j}.$ [1] - [10]

$f\left(z\right)=\frac{1}{2\text{π}i}\underset{c}{\int }\frac{f\left(\xi \right)}{\xi -z}\text{d}\xi \text{}\left(z\in D\right).$

$f\left(z\right)=\underset{n=0}{\overset{\infty }{\sum }}{c}_{n}{\left(z-a\right)}^{n},$

${c}_{n}=\frac{1}{2\text{π}i}\underset{{\tau }_{\rho }}{\int }\frac{f\left(\xi \right)}{{\left(\xi -a\right)}^{n+1}}\text{d}\xi =\frac{{f}^{\left(n\right)}\left(a\right)}{n!},$

$\left({\tau }_{\rho }:|\xi -a|=\rho ,0<\rho

$|{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)|\le \frac{p!{r}_{1}{C}_{{r}_{1}}\left(f\right)}{n{\left({r}_{1}-r\right)}^{p+1}},$

${C}_{{r}_{1}}\left(f\right)=\frac{3}{2}\underset{m=1}{\overset{\infty }{\sum }}|{a}_{m}|m\left(m+1\right)\left(m+1\right)!{\left(2{r}_{1}\right)}^{m}<+\infty .$

${‖{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)‖}_{r}\ge \frac{1}{n}{B}_{{r}_{1}}\left(f\right),$

${‖{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)‖}_{r}~\frac{1}{n}{N}_{{r}_{1}}\left(f\right),$

2. 重要引理

$|{V}_{n}^{\ast }\left(f,z\right)-f\left(z\right)|\le \frac{3}{2n}\underset{m=1}{\overset{\infty }{\sum }}|{a}_{m}|m\left(m+1\right)\left(m+1\right)!{\left(2r\right)}^{m},$

${C}_{r}\left(f\right)=\frac{3}{2}\underset{m=1}{\overset{\infty }{\sum }}|{a}_{m}|m\left(m+1\right)\left(m+1\right)!{\left(2r\right)}^{m}<+\infty .$

$|{V}_{n}^{\ast }\left(f,z\right)-f\left(z\right)-\frac{1-2z}{2\left(n+1\right)}{f}^{\prime }-\frac{z\left(1+z\right)}{2\left(n+1\right)}{f}^{″}\left(z\right)|\le \frac{11}{{n}^{2}}\underset{m=2}{\overset{\infty }{\sum }}|{a}_{m}|m{\left(m-1\right)}^{2}\left(m+1\right)!{\left(2r\right)}^{m},$

${A}_{r}\left(f\right)=11\underset{m=2}{\overset{\infty }{\sum }}|{a}_{m}|m{\left(m-1\right)}^{2}\left(m+1\right)!{\left(2r\right)}^{m}<+\infty .$

$‖{V}_{n}^{\ast }\left(f\right)-f‖\ge \frac{1}{n}{B}_{r}\left(f\right),$

3. 定理的证明

$\begin{array}{c}|{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)|=\frac{p!}{\text{2π}}|\underset{\gamma }{\int }\frac{{K}_{n}\left(f,v\right)-f\left(v\right)}{{\left(v-z\right)}^{p+1}}\text{d}v|\\ \le \frac{{C}_{{r}_{1}}\left(f\right)}{n}\cdot \frac{p!}{\text{2π}}\cdot \frac{2\text{π}{r}_{1}}{{\left({r}_{1}-r\right)}^{p+1}}=\frac{p!{r}_{1}{C}_{{r}_{1}}\left(f\right)}{n{\left({r}_{1}-r\right)}^{p+1}}.\end{array}$

$\begin{array}{c}{K}_{n}\left(f,v\right)-f\left(v\right)=\frac{1}{n}\left\{\frac{1-2v}{2\left(n+1\right)}{f}^{\prime }\left(v\right)+\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\\ +\frac{1}{n}\left[{n}^{2}\left({K}_{n}\left(f,v\right)-f\left(v\right)-\frac{1-2v}{2\left(n+1\right)}{f}^{\prime }\left(v\right)-\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\right)\right]\right\}\end{array}$

$\begin{array}{l}{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)\\ =\frac{1}{n}\left\{\frac{p!}{2\text{π}i}\underset{\gamma }{\int }\frac{\left(1-2v\right){f}^{\prime }\left(v\right)}{2{\left(v-z\right)}^{p+1}}\text{d}v+\underset{\gamma }{\int }\frac{v\left(1+v\right){f}^{″}\left(v\right)}{2{\left(v-z\right)}^{p+1}}\text{d}v\begin{array}{c}\underset{}{\overset{}{}}\\ \\ \end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{n}\frac{p!}{2\text{π}i}\underset{\gamma }{\int }\frac{{n}^{2}\left[\left({K}_{n}\left(f,v\right)-f\left(v\right)-\frac{1-2v}{2\left(n+1\right)}{f}^{\prime }\left(v\right)-\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\right)\right]}{{\left(v-z\right)}^{p+1}}\text{d}v\right\}\\ =\frac{1}{n}\left\{{\left[\frac{\left(1-2z\right)}{2}{f}^{\prime }\left(z\right)\right]}^{\left(p\right)}+{\left[\frac{z\left(1+z\right)}{2}{f}^{″}\left(z\right)\right]}^{\left(p\right)}\begin{array}{c}\underset{}{\overset{}{}}\\ \\ \end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{n}\frac{p!}{2\text{π}i}\underset{\gamma }{\int }\frac{{n}^{2}\left[\left({K}_{n}\left(f,v\right)-f\left(v\right)-\frac{1-2v}{2\left(n+1\right)}{f}^{\prime }\left(v\right)-\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\right)\right]}{{\left(v-z\right)}^{p+1}}\text{d}v\right\}\end{array}$

$\begin{array}{l}|{K}_{n}^{\left(p\right)}\left(f,z\right)-{f}^{\left(p\right)}\left(z\right)|\\ \ge \frac{1}{n}\left\{|{\left[\frac{\left(1-2z\right)}{2}{f}^{\prime }\left(z\right)\right]}^{\left(p\right)}+{\left[\frac{z\left(1+z\right)}{2}{f}^{″}\left(z\right)\right]}^{\left(p\right)}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{1}{n}|\frac{p!}{2\text{π}i}\underset{\gamma }{\int }\frac{{n}^{2}\left[\left({K}_{n}\left(f,v\right)-f\left(v\right)-\frac{\left(1-2v\right)}{2\left(n+1\right)}{f}^{\prime }\left(v\right)-\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\right)\right]}{{\left(v-z\right)}^{p+1}}\text{d}v|\right\},\end{array}$

$\begin{array}{l}|\frac{p!}{2\text{π}i}\underset{\gamma }{\int }\frac{{n}^{2}\left[\left({K}_{n}\left(f,v\right)-f\left(v\right)-\frac{1-2v}{2\left(n+1\right)}{f}^{\prime }\left(v\right)-\frac{v\left(1+v\right)}{2\left(n+1\right)}{f}^{″}\left(v\right)\right)\right]}{{\left(v-z\right)}^{p+1}}\text{d}v|\\ \le \frac{p!}{2\text{π}i}\cdot \frac{2\text{π}{r}_{1}{n}^{2}}{{\left({r}_{1}-r\right)}^{p+1}}\cdot \frac{11}{{n}^{2}}\underset{m=2}{\overset{\infty }{\sum }}|{a}_{m}|m{\left(m-1\right)}^{2}\left(m+1\right)!{\left(2r\right)}^{m}\\ \le \frac{11p!{r}_{1}}{{\left({r}_{1}-r\right)}^{p+1}}\underset{m=2}{\overset{\infty }{\sum }}|{a}_{m}|m{\left(m-1\right)}^{2}\left(m+1\right)!{\left(2r\right)}^{m}.\end{array}$

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