# proof of Weierstrass approximation theorem in R^n

To show that the Weierstrass Approximaton Theorem holds in
${\mathbb{R}}^{n}$, we will use induction^{} on $n$.

For the sake of simplicity, consider first the case of the cubical
region $0\le {x}_{i}\le 1$, $1\le i\le n$. Suppose that $f$ is a
continuous^{}, real valued function on this region. Let $\u03f5$ be an
arbitrary positive constant.

Since a continuous functions on compact^{} regions are uniformly
continuous^{}, $f$ is uniformly continuous. Hence, there exists an
integer $N>0$ such that $$ whenever $|a-b|\le 1/N$ and both $a$ and $b$ lie in the cubical region.

Define $\varphi :\mathbb{R}\to \mathbb{R}$ as follows:

$$ |

Consider the function $\stackrel{~}{f}$ defined as follows:

$$\stackrel{~}{f}({x}_{1},\mathrm{\dots}{x}_{n})=\sum _{m=0}^{N}\varphi (N{x}_{1}+m)f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})$$ |

We shall now show that $|f({x}_{1},\mathrm{\dots},{x}_{n})-\stackrel{~}{f}({x}_{1},\mathrm{\dots},{x}_{n})|\le \u03f5/2$ whenever $({x}_{1},\mathrm{\dots},{x}_{n})$ lies in the cubical region. By way that $\varphi $ was defined, only two of the terms in the sum defining $\stackrel{~}{f}$ will differ from zero for any particular value of ${x}_{1}$, and hence

$$\stackrel{~}{f}({x}_{1},\mathrm{\dots},{x}_{n})=(Nx-\lfloor Nx\rfloor )f(\frac{\lfloor N{x}_{1}\rfloor}{N},{x}_{2},\mathrm{\dots},{x}_{n})+(\lfloor Nx\rfloor +1-x)f(\frac{\lfloor N{x}_{1}\rfloor +1}{N},{x}_{2},\mathrm{\dots},{x}_{n}),$$ |

so

$|\stackrel{~}{f}({x}_{1},\mathrm{\dots},{x}_{n})-f({x}_{1},\mathrm{\dots},{x}_{n})|$ | $=$ | $|\stackrel{~}{f}({x}_{1},\mathrm{\dots},{x}_{n})-\{(Nx-\lfloor Nx\rfloor )+(\lfloor Nx\rfloor +1-x)\}f({x}_{1},\mathrm{\dots},{x}_{n})|$ | ||

$\le $ | $(Nx-\lfloor Nx\rfloor )|f({\displaystyle \frac{\lfloor N{x}_{1}\rfloor}{N}},{x}_{2},\mathrm{\dots},{x}_{n})-f({x}_{1},{X}_{2}\mathrm{\dots},{x}_{n})|+(\lfloor Nx\rfloor +1-x)|({\displaystyle \frac{\lfloor N{x}_{1}\rfloor +1}{N}},{x}_{2},\mathrm{\dots},{x}_{n})-f({x}_{1},{x}_{2}\mathrm{\dots},{x}_{n})|$ | |||

$\le $ | $(Nx-\lfloor Nx\rfloor ){\displaystyle \frac{\u03f5}{2}}+(\lfloor Nx\rfloor +1-x){\displaystyle \frac{\u03f5}{2}}={\displaystyle \frac{\u03f5}{2}}.$ |

Next, we will use the Weierstrass approximation theorem^{} in $n-1$
dimensions^{} and in one dimesnsion to approximate $\stackrel{~}{f}$ by a
polynomial^{}. Since $f$ is continuous and the cubical region is
compact, $f$ must be bounded^{} on this region. Let $M$ be an upper
bound for the absolute value^{} of $f$ on the cubical region. Using the
Weierstrass approximation theorem in one dimension, we conclude that
there exists a polynoial $\stackrel{\u02d8}{\varphi}$ such that $$ for all $a$ in the region. Using the
Weierstrass approximation theorem in $n-1$ dimensions, we conclude
that there exist polynomials ${p}_{m}$, $0\le m\le N$ such that $|{p}_{m}({x}_{2},\mathrm{\dots},{x}_{n})-f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})|\le \frac{\u03f5}{4N}$. Then one has the following inequality^{}:

$|\stackrel{\u02d8}{\varphi}(N{x}_{1}+m){p}_{m}({x}_{2},\mathrm{\dots}{x}_{n})-\varphi (N{x}_{1}+m)f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})|$ | $=|\stackrel{\u02d8}{\varphi}(N{x}_{1}+m){p}_{m}({x}_{2},\mathrm{\dots}{x}_{n})-\stackrel{\u02d8}{\varphi}(N{x}_{1}+m)f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})$ | |||

$+$ | $\stackrel{\u02d8}{\varphi}(N{x}_{1}+m)f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})-\varphi (N{x}_{1}+m)f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})|$ | |||

$\le $ | $|\stackrel{\u02d8}{\varphi}(N{x}_{1}+m)||{p}_{m}({x}_{2},\mathrm{\dots}{x}_{n})-f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})|$ | |||

$+$ | $|f(m/N,{x}_{2},\mathrm{\dots}{x}_{n})||\stackrel{\u02d8}{\varphi}(N{x}_{1}+m)-\varphi (N{x}_{1}+m)|$ | |||

$\le $ | $\frac{\u03f5}{4N}}+M{\displaystyle \frac{\u03f5}{4MN}}={\displaystyle \frac{\u03f5}{2N}$ |

Define

$$\stackrel{\u02d8}{f}({x}_{1},\mathrm{\dots}{x}_{n})=\sum _{m=0}^{N}\stackrel{\u02d8}{\varphi}(N{x}_{1}+m){p}_{m}({x}_{2},\mathrm{\dots}{x}_{n}).$$ |

As a finite sum of products^{} of polynomials, this is a polynomial.
From the above inequality, we conclude that $|\stackrel{\u02d8}{f}(a)-\stackrel{~}{f}(a)|\le \u03f5/2$, hence $|f(a)-\stackrel{\u02d8}{f}(a)|\le \u03f5$.

It is a simple matter of rescaling variables to conclude the Weirestrass approximation theorem for arbitrary parallelopipeds. Any compact subset of ${\mathbb{R}}^{n}$ 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 ${\mathbb{R}}^{n}$.

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 |