# Stone Weierstrass Theorem (Example)

/This week we continue our discussion on the Stone Weierstrass Theorem with an example. This exercise is taken from Rudin's *Principles of Mathematical Analysis* (affectionately known as "Baby Rudin").

## Rudin, *PMA*, Chapter 7 #20

## Solution

We'll prove by contradiction. Suppose $f$ is not identically 0 on $[0,1]$. Then there is some $x_0\in[0,1]$ for which, without loss of generality, $f(x_0)$ is positive. Since $f$ is continuous, we can find an open interval $I\subset[0,1]$ containing $x_0$ such that $f$ is positive on all of $I$.

This implies that $$\int_0^1f(x)^2\;dx>0$$ (since $f(x)^2\geq0$ for all $x\in[0,1]$). Further, by the Extreme Value Theorem, we can choose a constant $M$ so that $\|f\|=\sup_{x\in[0,1]}\{|f(x)|\}\leq M$.

Now let $\epsilon>0$. By the Stone Weierstrass Theorem we know that the polynomials on $[0,1]$ are dense in $\mathscr{C}([0,1],\mathbb{R})$. Thus there exists a polynomial $p$ such that $$\|f-p\| < \epsilon/M.$$ We also know that $$\int_0^1p(x)f(x)\;dx=0.$$
This follows since we have assumed $\int_0^1x^nf(x)\;dx=0$ for all $n=0,1,2,\ldots$. By linearity of the integral, this implies $\int_0^1q(x)f(x)\;dx=0$ for *any* polynomial $q$, including the polynomial $p$ which uniformly approximates $f$.

Observe next that \begin{align*} \int_0^1f(x)^2\;dx&=\int_0^1f(x)(f(x)-p(x))\;dx\\ &\leq \int_0^1\|f\|\|f-p\|\;dx\\ &=\|f\|\|f-p\|\\ &< M\cdot\epsilon/M\\ &=\epsilon. \end{align*} Since $\epsilon>0$ was arbitrary, we conclude $\int_0^1f(x)^2\;dx=0$. But this contradicts the fact that $f$ is positive on the interval $I\subset[0,1]$! Hence we must have that $f(x)=0$ for all $x\in[0,1]$ as desired.