二元Weierstrass逼近定理及其证明

二元$Weierstrass$逼近定理：设$f(x,y)$在区域$0\le x\le 1,0\le y\le 1$上连续，称
$$B_{m,n}(f,;x,y)=\sum _{\nu =0}^m\sum _{\mu =0}^n{C}_m^{\nu }{C}_n^{\mu }{x}^{\nu }(1-x{)}^{m-\nu }{y}^{\mu }(1-y{)}^{n-\mu }f\left(\frac{\nu }{m},\frac{\mu }{n}\right)$$
为$(m,n)$阶$Bernstein$多项式，则当$m,n\to \infty$时，$B_{m,n}(f;x,y)\to f(x,y),(x,y)\in [0,1]\times [0,1]$

证明：记$b_{m,\nu ,n,\mu }(x,y)=C_m^{\nu }{C}_n^{\mu }{x}^{\nu }(1-x{)}^{m-\nu }{y}^{\mu }(1-y{)}^{n-\mu }=b_{m,\nu }(x)b_{n,\mu }(y)$

因为 $f(x,y)$是闭区间上的连续函数

所以$f(x,y)$一致连续，即$\forall \epsilon >0,\exists \delta >0$，对$\forall (x_1,y_1),(x_2,y_2)\in [0,1]\times [0,1]$，满足$|x_1-x_2|<\delta ,|y_1-y_2|<\delta$，有$|f(x_1,y_1)-f(x_2,y_2)|<\frac{\epsilon }{4}$

注意到$\sum _{\nu =0}^m\sum _{\mu =0}^n{b}_{m,\nu ,n,\mu }(x,y)=1$，对任意$(x,y)\in [0,1]\times [0,1]$，有
∣ f ( x , y ) − B m , n ( f ; x , y ) ∣ = ∣ ∑ ν = 0 m ∑ μ = 0 n f ( x , y ) b m , ν , n , μ ( x , y ) − ∑ ν = 0 m ∑ μ = 0 n f ( ν m , μ n ) b m , ν , n , μ ( x , y ) ∣ ≤ ∑ ν = 0 m ∑ μ = 0 n ∣ f ( x , y ) − f ( ν m , μ n ) ∣ b m , ν , n , μ ( x , y )

∣f(x,y)−Bm,n​(f;x,y)∣​=∣ ∣​ν=0∑m​μ=0∑n​f(x,y)bm,ν,n,μ​(x,y)−ν=0∑m​μ=0∑n​f(mν​,nμ​)bm,ν,n,μ​(x,y)∣ ∣​≤ν=0∑m​μ=0∑n​∣ ∣​f(x,y)−f(mν​,nμ​)∣ ∣​bm,ν,n,μ​(x,y)​​
将上述方程右边的求和项分为四部分
$$\sum _{\nu =0}^m\sum _{\mu =0}^n\le \sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|<\delta }+\sum _{|x-\nu /m|\ge \delta }\sum _{|y-\mu /n|<\delta }+\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|\ge \delta }+\sum _{|x-\nu /m|\ge \delta }\sum _{|y-\mu /n|\ge \delta }$$
对于第一部分，有
$$\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|<\delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)<\frac{\epsilon }{4}\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|<\delta }{b}_{m,\nu ,n,\mu }(x,y)<\frac{\epsilon }{4}$$
对于第二部分，注意到$(y-\mu /n{)}^2/{\delta }^2\ge 1,\sum _{\nu =0}^m{b}_{m,\nu }(x)=1$，于是有
$$\begin{array}{rl}\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|\ge \delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)& \le \sum _{\nu =0}^m\sum _{|y-\mu /n|\ge \delta }2M{b}_{m,\nu }(x)b_{n,\mu }(y)\\ & \le \sum _{|y-\mu /n|\ge \delta }2M\frac{(y-\mu /n{)}^2}{{\delta }^2}{b}_{n,\mu }(y)\\ & \le \frac{2M}{n^2{\delta }^2}\sum _{\mu =0}^n(\mu -ny{)}^2{b}_{n,\mu }(y)\end{array}$$
其中$M=\underset{0\le x\le 1,0\le y\le 1}{\max }|f(x,y)|$

又注意到$\sum _{\mu =0}^n(\mu -my{)}^2{b}_{n,\mu }(y)=ny(1-y),y(1-y)\le 1/4$

于是有
$$\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|\ge \delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)\le \frac{2M}{n{\delta }^2}y(1-y)\le \frac{M}{2n{\delta }^2}$$
令$N_1=\frac{2M}{\epsilon {\delta }^2}$，则当$n>N_1$时，有
$$\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|\ge \delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)<\frac{\epsilon }{4}$$
对于第三部分同理，有
$$\sum _{|x-\nu /m|\ge \delta }\sum _{|y-\mu /n|<\delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)<\frac{\epsilon }{4}$$
对于第四部分，注意到$(x-\nu /m{)}^2/{\delta }^2\ge 1,(y-\mu /n{)}^2/{\delta }^2\ge 1$，于是有
∑ ∣ x − ν / m ∣ ≥ δ ∑ ∣ y − μ / n ∣ ≥ δ ∣ f ( x , y ) − f ( ν m , μ n ) ∣ b m , ν , n , μ ( x , y ) ≤ ∑ ∣ x − ν / m ∣ ≥ δ ∑ ∣ y − μ / n ∣ ≥ δ 2 M b m , ν ( x ) b n , μ ( y ) ≤ ∑ ∣ x − ν / m ∣ ≥ δ ∑ ∣ y − μ / n ∣ ≥ δ 2 M ( x − ν / m ) 2 δ 2 b m , ν ( x ) ( y − μ / n ) 2 δ 2 b n , μ ( y ) ≤ 2 M m 2 n 2 δ 4 ∑ ν = 0 m ∑ μ = 0 n ( ν − m x ) 2 b m , ν ( x ) ( μ − n y ) 2 b n , μ ( y )

∣x−ν/m∣≥δ∑​∣y−μ/n∣≥δ∑​∣ ∣​f(x,y)−f(mν​,nμ​)∣ ∣​bm,ν,n,μ​(x,y)​≤∣x−ν/m∣≥δ∑​∣y−μ/n∣≥δ∑​2Mbm,ν​(x)bn,μ​(y)≤∣x−ν/m∣≥δ∑​∣y−μ/n∣≥δ∑​2Mδ2(x−ν/m)2​bm,ν​(x)δ2(y−μ/n)2​bn,μ​(y)≤m2n2δ42M​ν=0∑m​μ=0∑n​(ν−mx)2bm,ν​(x)(μ−ny)2bn,μ​(y)​​
其中$M=\underset{0\le x\le 1,0\le y\le 1}{\max }|f(x,y)|$

又注意到$\sum _{\nu =0}^m(\nu -mx{)}^2{b}_{m,\nu }(x)=mx(1-x),x(1-x)\le 1/4,\sum _{\mu =0}^n(\mu -my{)}^2{b}_{n,\mu }(y)=ny(1-y),y(1-y)\le 1/4$

于是有
$$\sum _{|x-\nu /m|<\delta }\sum _{|y-\mu /n|\ge \delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)\le \frac{2M}{mn{\delta }^4}x(1-x)y(1-y)\le \frac{M}{8mn{\delta }^2}$$
则当$mn>\frac{M}{2\epsilon {\delta }^2}$时，有
$$\sum _{|x-\nu /m|\ge \delta }\sum _{|y-\mu /n|\ge \delta }\left|f(x,y)-f\left(\frac{\nu }{m},\frac{\mu }{n}\right)\right|b_{m,\nu ,n,\mu }(x,y)<\frac{\epsilon }{4}$$
综上，有$\underset{0\le x\le 1,0\le y\le 1}{\max }|f(x,y)-B_{m,n}(f;x,y)|<\epsilon$

故$B_{m,n}(f;x,y)\to f(x,y),(x,y)\in [0,1]\times [0,1]$