proof of Weierstrass approximation theorem in R^n


To show that the Weierstrass Approximaton Theorem holds in n, we will use inductionMathworldPlanetmath on n.

For the sake of simplicity, consider first the case of the cubical region 0xi1, 1in. Suppose that f is a continuousMathworldPlanetmathPlanetmath, real valued function on this region. Let ϵ be an arbitrary positive constant.

Since a continuous functions on compactPlanetmathPlanetmath regions are uniformly continuousPlanetmathPlanetmath, f is uniformly continuous. Hence, there exists an integer N>0 such that |f(a)-f(b)|<ϵ/2 whenever |a-b|1/N and both a and b lie in the cubical region.

Define ϕ: as follows:

ϕ(x)={0x<-11+x-1x01-x0x10x>1

Consider the function f~ defined as follows:

f~(x1,xn)=m=0Nϕ(Nx1+m)f(m/N,x2,xn)

We shall now show that |f(x1,,xn)-f~(x1,,xn)|ϵ/2 whenever (x1,,xn) lies in the cubical region. By way that ϕ was defined, only two of the terms in the sum defining f~ will differ from zero for any particular value of x1, and hence

f~(x1,,xn)=(Nx-Nx)f(Nx1N,x2,,xn)+(Nx+1-x)f(Nx1+1N,x2,,xn),

so

|f~(x1,,xn)-f(x1,,xn)| = |f~(x1,,xn)-{(Nx-Nx)+(Nx+1-x)}f(x1,,xn)|
(Nx-Nx)|f(Nx1N,x2,,xn)-f(x1,X2,xn)|+(Nx+1-x)|(Nx1+1N,x2,,xn)-f(x1,x2,xn)|
(Nx-Nx)ϵ2+(Nx+1-x)ϵ2=ϵ2.

Next, we will use the Weierstrass approximation theoremMathworldPlanetmath in n-1 dimensionsMathworldPlanetmath and in one dimesnsion to approximate f~ by a polynomialPlanetmathPlanetmath. Since f is continuous and the cubical region is compact, f must be boundedPlanetmathPlanetmathPlanetmathPlanetmath on this region. Let M be an upper bound for the absolute valueMathworldPlanetmathPlanetmathPlanetmath of f on the cubical region. Using the Weierstrass approximation theorem in one dimension, we conclude that there exists a polynoial ϕ˘ such that |ϕ˘(a)-ϕ(a)|<ϵ/(4MN) for all a in the region. Using the Weierstrass approximation theorem in n-1 dimensions, we conclude that there exist polynomials pm, 0mN such that |pm(x2,,xn)-f(m/N,x2,xn)|ϵ4N. Then one has the following inequalityMathworldPlanetmath:

|ϕ˘(Nx1+m)pm(x2,xn)-ϕ(Nx1+m)f(m/N,x2,xn)| =|ϕ˘(Nx1+m)pm(x2,xn)-ϕ˘(Nx1+m)f(m/N,x2,xn)
+ ϕ˘(Nx1+m)f(m/N,x2,xn)-ϕ(Nx1+m)f(m/N,x2,xn)|
|ϕ˘(Nx1+m)||pm(x2,xn)-f(m/N,x2,xn)|
+ |f(m/N,x2,xn)||ϕ˘(Nx1+m)-ϕ(Nx1+m)|
ϵ4N+Mϵ4MN=ϵ2N

Define

f˘(x1,xn)=m=0Nϕ˘(Nx1+m)pm(x2,xn).

As a finite sum of productsPlanetmathPlanetmathPlanetmath of polynomials, this is a polynomial. From the above inequality, we conclude that |f˘(a)-f~(a)|ϵ/2, hence |f(a)-f˘(a)|ϵ.

It is a simple matter of rescaling variables to conclude the Weirestrass approximation theorem for arbitrary parallelopipeds. Any compact subset of n can be embedded in some paralleloped and any continuous function on the compact subset can be extended to a continuous function on the parallelopiped. By approximating this extended function, we conclude the Weierstrass approximation theorem for arbitrary compact subsets of n.

Title proof of Weierstrass approximation theorem in R^n
Canonical name ProofOfWeierstrassApproximationTheoremInRn
Date of creation 2013-03-22 15:40:03
Last modified on 2013-03-22 15:40:03
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 8
Author rspuzio (6075)
Entry type Proof
Classification msc 41A10