proof of Weierstrass approximation theorem in R^n

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

For the sake of simplicity, consider first the case of the cubicalregion $0\le x_i\le 1$, $1\le i\le n$. Suppose that $f$ is acontinuous, real valued function on this region. Let $ϵ$ be anarbitrary positive constant.

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

Define $\varphi :ℝ\to ℝ$ as follows:

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

$$\tilde{f}(x_1,\mathrm{\dots }{x}_n)=\sum _{m=0}^N\varphi (N{x}_1+m)f(m/N,x_2,\mathrm{\dots }{x}_n)$$

We shall now show that $|f(x_1,\mathrm{\dots },x_n)-\tilde{f}(x_1,\mathrm{\dots },x_n)|\le ϵ/2$ whenever $(x_1,\mathrm{\dots },x_n)$ lies in thecubical region. By way that $\varphi$ was defined, only two of the termsin the sum defining $\tilde{f}$ will differ from zero for anyparticular value of $x_1$, and hence

$$\tilde{f}(x_1,\mathrm{\dots },x_n)=(Nx-\lfloor Nx\rfloor )f(\frac{\lfloor N{x}_1\rfloor }{N},x_2,\mathrm{\dots },x_n)+(\lfloor Nx\rfloor +1-x)f(\frac{\lfloor N{x}_1\rfloor +1}{N},x_2,\mathrm{\dots },x_n),$$

so

$|\tilde{f}(x_1,\mathrm{\dots },x_n)-f(x_1,\mathrm{\dots },x_n)|$$=$$|\tilde{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{ϵ}{2}}+(\lfloor Nx\rfloor +1-x){\displaystyle \frac{ϵ}{2}}={\displaystyle \frac{ϵ}{2}}.$

Next, we will use the Weierstrass approximation theorem in $n-1$dimensions and in one dimesnsion to approximate $\tilde{f}$ by apolynomial. Since $f$ is continuous and the cubical region iscompact, $f$ must be bounded on this region. Let $M$ be an upperbound for the absolute value of $f$ on the cubical region. Using theWeierstrass approximation theorem in one dimension, we conclude thatthere exists a polynoial $\breve{\varphi }$ such that for all $a$ in the region. Using theWeierstrass approximation theorem in $n-1$ dimensions, we concludethat there exist polynomials $p_m$, $0\le m\le N$ such that $|p_m(x_2,\mathrm{\dots },x_n)-f(m/N,x_2,\mathrm{\dots }{x}_n)|\le \frac{ϵ}{4N}$. Then one has the following inequality:

$|\breve{\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)|$$=|\breve{\varphi }(N{x}_1+m)p_m(x_2,\mathrm{\dots }{x}_n)-\breve{\varphi }(N{x}_1+m)f(m/N,x_2,\mathrm{\dots }{x}_n)$

$+$$\breve{\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$$|\breve{\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)||\breve{\varphi }(N{x}_1+m)-\varphi (N{x}_1+m)|$

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

Define

$$\breve{f}(x_1,\mathrm{\dots }{x}_n)=\sum _{m=0}^N\breve{\varphi }(N{x}_1+m)p_m(x_2,\mathrm{\dots }{x}_n).$$

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

It is a simple matter of rescaling variables to conclude theWeirestrass approximation theorem for arbitrary parallelopipeds. Anycompact subset of ${ℝ}^n$ can be embedded in some parallelopedand any continuous function on the compact subset can be extended to acontinuous function on the parallelopiped. By approximating thisextended function, we conclude the Weierstrass approximation theoremfor arbitrary compact subsets of ${ℝ}^n$.