文献[1]中, Turky学者Gülsüm Ulusoy*和Ali Aral研究了lbragimov-Gadjiev-Durrmeyer算子在$ L_p $空间内的逼近问题，但是在Orlicz空间内至今没有发现有人研究过该算子的逼近问题, 我们在本文中尝试做这方面的工作.由于Orlicz空间是$ L_p $空间的实质性的扩充和提升, 本文的研究内容具有一定的拓展意义.

设$ \left(\varphi_n(t)\right)_{n\in N}, \left(\psi_n(t)\right)_{n\in N} $是$ R^+=[0, \infty) $上的连续函数列, 使得$ \varphi_n(0)=0 $, 对所有的$ t $有$ \psi_n(t)>0 $且$ \lim\limits_ {n\rightarrow \infty}\frac{1}{n^2\psi_n(0)}=0 $.

令$ (\alpha_n)_{n\in N} $为一列正数列且满足$ \lim\limits_{n\rightarrow \infty}\frac{\alpha_n}{n}=1. \lim\limits_{n\rightarrow \infty}\alpha_n\psi_n(0)=l_1, l_1>0. $

lbragimov-Gadjiev-Durrmeyer算子是指：

$ \begin{align*} M_n(f;x)= & (n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \\ & \times \int_{0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y, \end{align*} $

其中$ K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)=\frac{\partial^v}{\partial u^v}K_n(x, t, u)\Big|_{u=\alpha_n\psi_n(t), t=0}, $且$ \left\{K_{n}^{(v)}\left(x, 0, \alpha_n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\right\}_{v=0}^\infty $ 为lbragimov-Gadjiev-Durrmeyer基函数系.

由文献[2]直接得出, $ K_n(x, t, u) $是关于$ u $的全纯函数, 且满足以下条件:

(1) 对固定的$ x, t\in R^+, $基函数系的每个函数都是关于$ u $的全纯函数, 且满足对$ \forall x\in R^+, n\in N, $有$ K_n(x, 0, 0)=1. $

(2) 对取定的$ u=u_1, x\in R^+ $有$ \left[(-1)^{v}\frac{\partial^v}{\partial u^v}K_n(x, t, u)\Big|_{u=u_1, t=0}\right]\geq0, v=0, 1, \ldots, . $

(3) 对所有的$ x\in R^+, n\in N $

$ \frac{\partial^v}{\partial u^v}K_n(x, t, u)\Big|_{u=u_1, t=0}=-nx \left[\frac{\partial^{v-1}}{\partial u^{v-1}}K_{m+n}(x, t, u)\Big|_{u=u_1, t=0}\right], v=0, 1, \ldots, . $

其中$ m $满足$ m+n=0 $或自然数.

(4) 对$ \forall u\in R, K_n(0, 0, u)=1, $且对$ \forall p\in N, u=u_1, \lim\limits_{n\rightarrow \infty}x^pK_n^{(v)}(x, 0, u_1)=0. $

(5) 对于任何固定的$ t, u, $函数$ K_n(x, t, u) $关于变量$ x $是连续微分的, 且满足对取定的$ u=u_1 $有

$ \frac{ \mathrm{d}}{ \mathrm{d} x}K_n(x, 0, u_1)=-nu_1K_{m+n}(x, 0, u_1). $

(6) 对所有的$ x\in R^+, n\in N, $取定的$ u=u_1, \frac{n+vm}{1+u_1mx}K_n^{(v)}(x, 0, u_1)=nK_{m+n}^{(v)}(x, 0, u_1). $

(7) 由文献[1], $ K_n(x, t, u) $还满足$ (1+u_1mx)^{-r}K_n^{(v)}(x, 0, u_1)=K_{n+rm}^{(v)}(x, 0, u_1)\alpha_{n, r}, $其中$ r $为自然数, $ \alpha_{n, r} $是收敛到一个正实数的关于$ n $的数列.

令$ u=\varphi_n(t), u_1=\alpha_n\psi_n(t), t=0, $根据$ \varphi_n(t)=0, K_n(x, 0, 0)=1 $可得

$ \begin{eqnarray} \sum\limits_{v=0}^{\infty}K_n^{(v)}(x, 0, u_1)\Big|_{u_1=\alpha_n\psi_n(t), t=0}\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} &=& 1. \end{eqnarray} $

(1.1)

本文将研究lbragimov-Gadjiev-Durrmeyer算子在Orlicz空间中的逼近问题, 文章里用$ M(u) $和$ N(v) $表示互余的$ N $函数, 有关于$ N $函数的定义及性质详情可见文献[3]中的论述.定义Orlicz空间中的范数:

$ \|u\|_M=\sup\limits_{\rho(v;N)\leq1}\left|\int_{0}^{\infty}u(x)v(x) \mathrm{d} x\right|, $

由具有有限的Orlicz范数的可测函数的全体$ \{u(x)\} $构成了$ N $函数$ M(u) $生成的Orlicz空间$ L_M^* [0, \infty) $, 其中$ \rho(v;N)=\int_{0}^{\infty}N\left(v(x)\right) \mathrm{d} x $表示$ v(x) $关于$ N(v) $的模.由文献[3]知, Orlicz范数还可定义为$ \|u\|_M= \inf\limits_{\beta>0}\frac{1}{\beta}\left(1+\int_{0}^{\infty}M\left(\beta u(x)\right) \mathrm{d} x\right). $本文中用$ C $表示常数, 但在不同处取值不同.

引理1 ^[2]  设$ v $为非负整数, $ x\in R^+ $, $ m, n\in N $.则有$ \int_{0}^{\infty}K_ {n}^{(v)}(y, 0, u_1) \mathrm{d} y=(-1)^v\frac{v!}{(n-m)u_1^{v+1}}. $

引理2 ^[2]  令$ v, n\in N, $对任意自然数$ r $, 有

$ \begin{align*} M_n(t^r;x)= &\frac{n^{2r}}{(n-2m)\cdots(n-rm)\left(n-(r+1)m\right)(\alpha_n)^r\left(n^2\psi_n(0)\right)^r} \\ & \times\sum\limits_{j=0}^{r}n(n+m)\cdots\left(n+(j-1)m\right)C_{j, r}\left[\alpha_n\psi_n(0)\right]^jx^j, \end{align*} $

其中$ C_{j, r}=\frac{r!}{j!}\binom{r}{j}, $

$ M_n(1;x)=1, M_n(t;x)=\frac{n^2}{(n-2m)\alpha_n}\left(\frac{\alpha_n}{n}x+\frac{1}{n^2\psi_n(0)}\right), $

$ M_n(t^2;x)=\frac{n^4}{(n-2m)(n-3m)\alpha_n^2}\left(\left(\frac{\alpha_n}{n}x\right)^2 +\frac{\alpha_n}{n}\frac{4}{n^2\psi_n(0)}x+\frac{2}{\left(n^2\psi_n(0)\right)^2}\right). $

引理3 ^[1]  令$ K_n(x, t, u) $满足条件$ (1)\rightarrow (7) $且$ f(t)=(1+u_1mt)^{-r}, t\in R^+, n, r\in N, $对$ n>mr $有$ M_n\left((1+u_1mt)^{-r};x\right)\leq C(1+u_1mx)^{-r}, x\in R^+, $其中$ C $表示与$ n $无关的常数.

在下文中, 考虑到光滑模和逼近速度之间的联系, 通过文献[4][5]中给出的光滑模, 有

$ \omega_\varphi^2(f, t)_M=\sup\limits_{0<H\leq t}\|\triangle_H^r\varphi f\|_M, t\in R^+, \varphi(x)=\sqrt{x\left(1+xm\alpha_n\psi_n(0)\right)}, $

其中

$ \triangle_H^r f(x)= \begin{cases} \sum\limits_{k=0}^{\infty}\binom{r}{k}(-1)^kf\left(x+\left(\frac{r}{2}-k\right)H\right), \quad \ \ & \left[x-\frac{r}{2}H, x+\frac{r}{2}H\right]\in R^+, \\ \quad\quad\quad\quad\quad0\quad\quad\quad\quad\quad\quad\quad\quad, \quad \ \ & 其他 . \end{cases} $

有与该光滑模等价的$ K $泛函$ \overline{K}_\varphi^2(f, x)=\inf\limits_{g\in\overline{\Omega}_M^r\left(\varphi, [0, \infty)\right)} \left\{\|f-g\|_M+t^r\|\varphi^rg^{(r)}\|_M+t^{2r}\|g^{(r)}\|_M\right\}, $其中$ \overline{\Omega}_M^r\left(\varphi, [0, \infty)\right) =\left\{g\in L_M^*[0, \infty), g^{(r-1)}\in AC_{loc}(0, \infty);\varphi^rg^{(r)}\in L_M^*[0, \infty)\right\} $表示对应的加权Sobolev空间.

在本文中, 光滑模的定义为: 对给定的整数$ k\geq1, h\neq 0, $设$ x, x+kh\in [a, b].\omega_k(f, t)_M=\sup\limits_{|h|\leq t}\|\vartriangle_h^k(f;\cdot)\|_M $称为$ f $的$ k $阶光滑模.

引理4  令$ n\in N, n>m, f\in L_M^*[0, \infty), $不等式$ \|M_nf\|_M\leq\|f\|_M $成立.

证明  由引理1和(1.1)式, 利用Jensen不等式

$ \begin{align*} & \|M_nf\|_M \\ =& \inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\int_{0}^{\infty}M\bigg(\beta(n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \\ & \times \int_{0}^{\infty}f(y)K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y\bigg) \mathrm{d} x\bigg) \\ \leq &\inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\int_{0}^{\infty}\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \\ & \times M\bigg(\beta(n-m)\alpha_n\psi_n(0)\int_{0}^{\infty}f(y)K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y\bigg) \mathrm{d} x\bigg) \\ = &\inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\int_{0}^{\infty}\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} x \\ & \times M\bigg(\beta(n-m)\alpha_n\psi_n(0)\int_{0}^{\infty}f(y)K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y\bigg)\bigg) \\ \end{align*} $

$ \begin{align*} =&\inf\limits_{\beta>0}\frac{1}{\beta}\Bigg(1+\int_{0}^{\infty}\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} x \\ & \times M\Bigg(\frac{\int_{0}^{\infty}\beta f(y)K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y}{\int_{0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y}\Bigg)\Bigg)\\ \leq & \inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\int_{0}^{\infty}\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} x\\ & \times (n-m)\alpha_n\psi_n(0)\int_{0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}M\left(\beta f(y)\right) \mathrm{d} y\bigg) \\ \leq & \inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\sum\limits_{v=0}^{\infty}\int_{0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} x \\ & \times (n-m)\alpha_n\psi_n(0)\int_{0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}M\left(\beta f(y)\right) \mathrm{d} y\bigg)\\ = & \inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\sum\limits_{v=0}^{\infty}\int_{0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}M\left(\beta f(y)\right) \mathrm{d} y\bigg) \\ = & \inf\limits_{\beta>0}\frac{1}{\beta}\bigg(1+\int_{0}^{\infty}\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}M\left(\beta f(y)\right) \mathrm{d} y\bigg) \\ =&\inf\limits_{\beta>0}\frac{1}{\beta}\left(1+\int_{0}^{\infty}M\left(\beta f(y)\right) \mathrm{d} y\right)=\|f\|_M. \end{align*} $

证毕.

定理1   令$ n\in N, n>3m, \forall f\in L_M^*[0, \infty), $有

$ \|M_nf-f\|_M\leq C\left\{\omega_\varphi^2\left(f, \left((n-3m)u_1\right)^{-\frac{1}{2}}\right)_M+ \left((n-3m)u_1\right)^{-1}\|f\|_M\right\}. $

证   由$ \omega_\varphi^2\left(f, \left((n-3m)u_1\right)^{-\frac{1}{2}}\right)_M $和$ \overline{K}_\varphi^2\left(f, \left((n-3m)u_1\right)^{-1}\right)_M $的等价性, 只需证明以下不等式

$ \|M_nf-f\|_M\leq C\left\{\overline{K}_\varphi^2\left(f, \left((n-3m)u_1\right)^{-1}\right)_M +\left((n-3m)u_1\right)^{-1}\|f\|_M\right\}. $

对所有的$ g\in \overline{\Omega}_M^2(\varphi;R^+), $由引理4, 有

$ \begin{align} \begin{split} \|M_nf-f\|_M& =\|M_n(f-g+g)-(f-g+g)\|_M \\ & \leq\|M_n(f-g)-(f-g)+M_ng-g\|_M \\ & \leq\|M_n(f-g)\|_M+\|f-g\|_M+\|M_ng-g\|_M \\ & \leq\|f-g\|_M+\|f-g\|_M+\|M_ng-g\|_M\\ & \leq2\|f-g\|_M+\|M_ng-g\|_M. \end{split} \end{align} $

(3.1)

下面估计上式中的$ \|M_ng-g\|_M, $利用$ g $的泰勒展式

$ g(t)=g(x)+g'(x)(t-x)+R_2(g, t, x), $

其中$ R_2(g, t, x)= \int_{x}^{t}(t-u)g''(u) \mathrm{d} u, $则$ M_n(g;x)-g(x)=M_n\left[(t-x)g'(x)\right]+M_n\left(R_2(g, t, x);x\right), $下面证明以下不等式的有效性

$ \left\|M_n\left(R_2(g, t, \cdot);\cdot\right)\right\|_M\leq C\left(n-3m)u_1\right)^{-1}\left\|\left(\varphi^2+\left(n-3m)u_1\right)^{-1}\right)g''\right\|_M. $

由Orlicz空间中的$ H\ddot{o}lder $不等式

$ \begin{align*} & |(R_2(g, t, x)|\\= &|\int_{x}^{t}(t-u)g''(u) \mathrm{d} u| \\ =&\left|\int_{x}^{t}(t-u)\left[\varphi^2(u)+ \left((n-3m)u_1\right)^{-1}\right]g''(u)\left[\varphi^2(u)+ \left((n-3m)u_1\right)^{-1}\right]^{-1} \mathrm{d} u\right| \\ \leq &\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M\left\|(t-u)\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]^{-1}\right\|_N \\ = & \left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M\sup\limits_{\rho(\gamma;N)\leq1}\left|\int_{x}^{t}(t-u)\left[\varphi^2(u)+ \left((n-3m)u_1\right)^{-1}\right]^{-1}\gamma(u) \mathrm{d} u\right| \\ \leq &C \left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M\int_{x}^{t}(t-u)\left[\varphi^2(u)+ \left((n-3m)u_1\right)^{-1}\right]^{-1} \mathrm{d} u \\ \leq & \left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M\int_{x}^{t}|t-u|\left[\varphi^2(u)+ \left((n-3m)u_1\right)^{-1}\right]^{-1} \mathrm{d} u. \end{align*} $

由文献[1]可得当$ x<t $时,

$ |R_2(g, t, x)|\leq C\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M(t-x)^2\left((n-3m)u_1\right), $

当$ x>t $时,

$ \begin{align*} |R_2(g, t, x)|\leq &C\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g"\right\|_M \\ & \times \frac{(t-x)^2}{x}\left\{\frac{x}{\varphi^2(x)+ \left((n-3m)u_1\right)^{-1}}+\frac{t}{\varphi^2(t)+ \left((n-3m)u_1\right)^{-1}}\right\}. \end{align*} $

故当$ x\in \left[0, \frac{1}{u_1(n-3m)}\right] $时, 由文献[1]有

$ \begin{align*} |M_n\left(R_2(g, t, x);x\right)|\leq & C\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_MM_n\left((t-x)^2\left((n-3m)u_1\right)\right) \\ \leq & C\left((n-3m)u_1\right)^{-1}\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M, \end{align*} $

当$ x\in\Big[\frac{1}{u_1(n-3m)}, \infty\Big) $时, 由文献[1]有

$ \begin{align*} |M_n\left(R_2(g, t, x);x\right)|\leq& C\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M\Big\{C\left((n-3m)u_1\right)^{-1} \\ & + \frac{1}{x}\left[M_n\left((t-x)^4;x\right)\right]^\frac{1}{2}\left[M_n\left((1+u_1mt)^{-2};x\right)\right]^\frac{1}{2}\Big\} \\ \leq &C\left((n-3m)u_1\right)^{-1}\left\|\left[\varphi^2+ \left((n-3m)u_1\right)^{-1}\right]g''\right\|_M, \end{align*} $

从而

$ \|M_n\left(R_2(g, t, \cdot);\cdot\right)\|_M\leq C\left(n-3m)u_1\right)^{-1}\left\|\left(\varphi^2+\left(n-3m)u_1\right)^{-1}\right)g''\right\|_M. $

另一方面, 对所有的$ g\in \overline{\Omega}_M^2(\varphi;R^+), $有

$ \|M_ng-g\|_M\leq C\left((n-3m)u_1\right)^{-1}\left\{\|g\|_M+\|\varphi^2g^{''}\|_M+ \left\|\left(\varphi^2+\left((n-3m)u_1\right)^{-1}\right)g^{''}\right\|_M\right\}, $

上式可由文献[4]的定理9.5.3(a)(c)式的证明得. 结合(3.1)式可得

$ \begin{align*} \|M_nf-f\|_M\leq& 2\|f-g\|_M+C\left((n-3m)u_1\right)^{-1} \\ &\times \left\{\|f-g\|_M+\|f\|_M+\|\varphi^2g^{''}\|_M+\left\|\left(\varphi^2 +\left((n-3m)u_1\right)^{-1}\right)g^{''}\right\|_M\right\}\\ \leq& C\Big\{\|f-g\|_M+\left((n-3m)u_1\right)^{-1}\|f\|_M\\ &+ \left((n-3m)u_1\right)\|\varphi^2g^{''}\|_M+\left((n-3m)u_1\right)\|g^{''}\|_M\Big\}. \end{align*} $

对所有的$ g\in \overline{\Omega}_M^2(\varphi;R^+), $取下确界

$ \|M_nf-f\|_M\leq C\left\{\overline{K}_\varphi^2\left(f, \left((n-3m)u_1\right)^{-1}\right)_M +\left((n-3m)u_1\right)^{-1}\|f\|_M\right\}. $

证毕.

定理2  令$ n\in N, n>3m, $则$ \lim\limits_{n\rightarrow \infty}\|M_nf-f\|_M=0. $

证  由$ \omega_\varphi^2\left(f, \left((n-3m)u_1\right)^ {-\frac{1}{2}}\right)_M $和$ \overline{K}_\varphi^2\left(f, \left((n-3m)u_1\right)^{-1}\right)_M $的等价性可证.

定义

$ L_{M, 2r}^*[0, \infty)=\left\{f:R^+\rightarrow R;\|f\|_{M, 2r} =\sup\limits_{\rho(\gamma;N)\leq1}\left|\int_{0}^{\infty}\frac{|f(t)|}{1+t^{2r}} \gamma(t) \mathrm{d} t\right|\right\}, $

即可得$ \|f\|_{M, 2r}=\left\|\frac{f}{1+\cdot^{2r}}\right\|_M. $

引理5  令$ f\in L_{M, 2r}^*[0, \infty), r\in N. $有$ \|M_nf-f\|_{M, 2r}\leq C\|f\|_{M, 2r}. $

证  由Orlicz空间中的Hölder不等式

$ \begin{align*} |M_n(f;x)|= & (n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \\ & \times \int_{0}^{\infty}\frac{|f(y)|}{1+y^{2r}}(1+y^{2r}) K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y\\ = &\|f\|_{M, 2r} \times \Bigg\|(n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\\ & \times (1+y^{2r}) K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\Bigg\|_N. \end{align*} $

因为

$ \begin{align*} \Bigg\|(n-m)\alpha_n\psi_n(0)&\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} (1+y^{2r}) K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\Bigg\|_N \\ =&\sup\limits_{\rho(\gamma;M)\leq1}\Bigg|(n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\\ & \times\int_{0}^{\infty}(1+y^{2r}) K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\gamma(y) \mathrm{d} y\Bigg|\\ \leq& C(n-m)\alpha_n\psi_n(0)\sum\limits_{v=0}^{\infty}K_{n}^{(v)}\left(x, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!}\\ & \times\int_{0}^{\infty}(1+y^{2r}) K_{n}^{(v)}\left(y, 0, \alpha _n\psi_n(0)\right)\frac{\left[\alpha_n\psi_n(0)\right]^v}{v!} \mathrm{d} y. \end{align*} $

由文献[1]得

$ |M_n(f;x)|\leq C\|f\|_{M, 2r}\left\{M_n(1;x)+M_n(t^{2r};x)\right\}. $

则

$ \begin{align*} \|M_n(f;x)\|_M\leq & C\|f\|_{M, 2r}\sup\limits_{\rho(\gamma;N)\leq1}\left|\int_{0}^{\infty} \frac{\left\{M_n(1;x)+M_n(t^{2r};x)\right\}}{1+x^{2r}}\gamma(x) \mathrm{d} x\right| \\ \leq& C\|f\|_{M, 2r}\int_{0}^{\infty}\frac{\left\{M_n(1;x)+M_n(t^{2r};x)\right\}}{1+x^{2r}} \mathrm{d} x\\ \leq & C\|f\|_{M, 2r}, \end{align*} $

其中$ C $是独立于$ n $的常数, 且在不同位置取值不同. 证毕.

定理3   令$ w $是整个实轴上的正连续函数, 且满足条件$ \int\limits_{[0, \infty)}t^cw(t) \mathrm{d} t<\infty, $其中$ c $为某一常数. 令$ (L_n)_{n\in N} $是$ L_{M, w}^*[0, \infty)\rightarrow L_{M, w}^*[0, \infty) $的非一致有界的正线性算子, 满足条件

$ \lim\limits_{n\rightarrow \infty}\left\|L_n(t^i;x)-x^i\right\|_{M, w}=0, i=0, 1, 2. $

则对$ \forall f\in L_{M, w}^*[0, \infty) $, 有$ \lim\limits_{n\rightarrow \infty}\left\|L_nf-f\right\|_{M, w}=0. $

当$ w(x)=(1+x^{2r})^{-1} $时, 得到下述定理.

定理4  令$ f\in L_{M, w}^*[0, \infty), r\in N. $则$ \lim\limits_{n\rightarrow \infty}\left\|M_nf-f\right\|_{M, w}=0. $

证  由定理3, 只须证明以下三个条件$ \lim\limits_{n\rightarrow \infty}\left\|M_n(t^v;x)-x^v\right\|_{M, w}=0, v=0, 1, 2. $ $ v=0 $时, 因为$ M_n(1;x)=1 $, 上式显然成立. $ v=1 $时, 由引理2, 对$ n>2m $, 有

$ \begin{align*} \|M_n(t;\cdot)-\cdot\|_{M, w}= & \sup\limits_{\rho(\gamma;N)\leq1}\left|\int_{0}^{\infty}\frac{\left|M_n(t;x)-x\right|}{1+x^{2r}}\gamma(x) \mathrm{d} x\right| \\ \leq & C\int_{0}^{\infty}\frac{\left|M_n(t;x)-x\right|}{1+x^{2r}} \mathrm{d} x \\ =& C\int_{0}^{\infty}\frac{1}{1+x^{2r}}\left\{\frac{n^2}{(n-2m)\alpha_n} \left(\frac{\alpha_n}{n}x+\frac{1}{n^2\psi_n(0)}\right)-x\right\} \mathrm{d} x, \end{align*} $

由文献[1]得

$ \|M_n(t;\cdot)-\cdot\|_{M, w}\leq C\left\{\left(\frac{n}{n-2m}-1\right)\int_{0}^{\infty}\frac{x}{1+x^{2r}} \mathrm{d} x+\frac{1}{(n-2m)\alpha_n\psi_n(0)}\int_{0}^{\infty}\frac{1}{1+x^{2r}} \mathrm{d} x\right\} $

当$ n\rightarrow \infty $时成立. $ v=2 $时, 由引理2, 对$ n>3m $, 由文献[1], 有

$ \begin{align*} \|M_n(t^2;\cdot)-\cdot^2\|_{M, w}=& \sup \limits_{\rho(\gamma;N)\leq 1} \left|\int_{0}^{\infty}\frac{\left|M_n(t^2;x)-x^2\right|}{1+x^{2r}}\gamma(x) \mathrm{d} x\right| \\ \leq& C\int_{0}^{\infty}\frac{\left|M_n(t^2;x)-x^2\right|}{1+x^{2r}} \mathrm{d} x \\ \leq& C\left(\frac{n(m+n)}{(n-2m)(n-3m)-1}\right)\int_{0}^{\infty }\frac{x^2}{1+x^{2r}} \mathrm{d} x \\ &+C\left(\frac{4n}{(n-2m)(n-3m)\alpha_n\psi_n(0)}\right)\int_{0}^{\infty} \frac{x}{1+x^{2r}} \mathrm{d} x \\ &+C\left(\frac{2}{(n-2m)(n-3m)\left(\alpha_n\psi_n(0)\right)^2}\right)\int_{0}^{\infty}\frac{1}{1+x^{2r}} \mathrm{d} x. \end{align*} $

当$ n\rightarrow \infty $时成立. 证毕.

算子$ M_n(f) $在以下情况下可以转化为我们所熟知的算子

(1) 若令$ K_n(x, t, u)=[1+t+ux]^{-1}, \alpha_n=n, \psi_n=\frac{1}{n}, m=1 $, 算子降低为Baskakov-Durrmeyer算子$ B_n(f;x)=(n-1)\sum_{k=0}^{\infty}v_{n, k}(x)\int_{0}^{\infty}v_{n, k}(x)f(t) \mathrm{d} t, $其中$ v_{n, k}(x)=\binom{n+k-1}{k}\frac{x^k}{(1+x)^{n+k}}. $

(2) 若令$ K_n(x, t, u)= e^{-n(t+ux)}, \alpha_n=n, \psi_n(0)=\frac{1}{n}, m=0 $, 算子降低为Szasz-Durrmeyer算子$ S_n(f;x)=n\sum_{k=0}^{\infty}p_{n, k}(x)\int_{0}^{\infty}p_{n, k}(t)f(t) \mathrm{d} t, $其中$ p_{n, k}(x)=e^{-nx\frac{(nx)^k}{k!}}. $

定理5  若$ f\in L_{M, w}^*[0, \infty), $则$ \|S_nf-f\|_M\leq C\left\{\omega_\varphi^2\left(f, n^{-\frac{1}{2}}\right)_M+n^{-1}\|f\|_M\right\}. $

若$ f\in L_{M, w}^*[0, \infty), n>3, $则$ \|B_nf-f\|_M\leq C\left\{\omega_\varphi^2\left(f, (n-3)^{-\frac{1}{2}}\right)_M+(n-3)^{-1}\|f\|_M\right\}. $