# Applying the Stone-Weierstrass Theorem to approximate even functions

Let $f:[-1,1] \rightarrow \mathbb{R}$ be any even continuous function on $[-1,1]$ (i.e. $f(-x)=f(x)$ $\forall x \in [-1,1]$). Let $\epsilon>0$. Prove that there exists an even polynomial $p$ such that $$|f(x)-p(x)|< \epsilon$$ $$\forall x \in [-1,1]$$

Here, "even polynomial" means that $p(-x)=p(x)$, not simply that it has even degree.

I think I should use the Stone-Weierstrass theorem to show that the subalgebra of even polynomials, call it $\mathcal{A}$, over this interval is dense, from which the result follows immediately.

For this to work I require that $\mathcal{A}$ contains the constants (obviously true) and separates points...which is not true, unfortunately. Anyone have any hints? I would prefer hints only, rather than solutions.

Oh yes, and I should mention that the version of the Stone-Weierstrass theorem that I can use says that if a subalgebra of $C(\mathbb{R})$ contains the constants and separates points, then it is dense in $C(\mathbb{R})$.

• By S-W there is certainly a polynomial $p$ that does what you want, so it would seem you then use the "evenness" of $f$ to show that $p$ must also be even (i.e. no odd polynomial could be within epsilon of $f$).
– user452
Commented Jan 28, 2015 at 1:35
• Right, this seems like the most straightforward option, though it still eludes me Commented Jan 28, 2015 at 2:14
• @trb456: It isn't true that $p$ must be even, but its even part will still approximate $f$. Commented Jan 28, 2015 at 2:43
• @trb456: Starting with a polynomial function $p$ on $[-1,1]$, one can define another polynomial $q$ on $[-1,1]$ by the formula $q(x) = \frac12(p(x)+p(-x))$. This $q$ is called the even part of $p$, and you can see that what it does is remove all the odd degree terms from $p$. It is easy to check that $q$ is even, whether or not $p$ is even. Typically, $q$ and $p$ are different polynomials. In case $f$ is even, it turns out that if $p$ approximates $f$ on $[-1,1]$, then so does $q$ (in a sense made precise in orangekid's answer). But this does not mean that $p$ is even. Commented Jan 28, 2015 at 2:59
• @JonasMeyer: Yes, duh! The answer shows that if there is some polynomial approximation, then there is some even approximation because $f$ is even, but it does not have to be the same approximation. Thanks!
– user452
Commented Jan 28, 2015 at 3:05

Hint: If $\sup_{x \in [-1,1]} |\,p(x) - f(x)|<\epsilon$ then $\sup_{x \in [-1,1]} |\,\frac{1}{2}(\,p(x)+ p(-x)\,)- f(x)\,|<\epsilon$

$\bf{Added:}$ Let $p(x)$ a polynomial so that for every $x \in [-1,1]$ we have $|\,p(x) - f(x)|<\epsilon$. Note that if $x \in [-1,1]$ then also $-x \in [-1,1]$. So we have $$|\,p(x) - f(x)|<\epsilon\\ |\,p(-x) - f(-x)|<\epsilon$$

Add up the inequality and divide by $2$:

$$\frac{1}{2} ( |\,p(x) - f(x)| + |\,p(-x) - f(-x)|) < \epsilon$$

Note that we have $|a+b| \le |a|+|b|$ for all numbers. Therefore we get

$$\frac{1}{2} \cdot |(p(x) + p(-x) ) - (f(x) + f(-x) ) | < \epsilon$$ or $$|\frac{1}{2}\cdot (p(x) + p(-x) ) - \frac{1}{2}\cdot(f(x) + f(-x) ) | < \epsilon$$ $\tiny{\text{ the averages also satisfy the inequality }}$.

Now since $f$ is even we have $\frac{1}{2}\cdot(f(x) + f(-x) ) = f(x)$. Moreover, $\frac{1}{2}\cdot (p(x) + p(-x) )$ is an even polynomial already. We are done.

• I can see that this is true if $p$ is even, but I'm not seeing how this helps the proof (excuse my ignorance). Commented Jan 28, 2015 at 2:15

Just for reference, here is another trick: Since $x\mapsto f(\sqrt{x})$ is continuous on $[0, 1]$, for $\epsilon > 0$ there exists a polynomial $q(x)$ such that

$$\sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| < \epsilon. \tag{*}$$

Now let $p(x) = q(x^2)$ and notice that $p$ is even and

$$\sup_{x\in[-1,1]}|f(x) - p(x)| = \sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| \stackrel{\text{by (*)}}{<} \epsilon$$

A little late, but I'd like to add a short proof for possible future readers. The subspace $E$ of even polynomials on $[0,1]$ is dense in $C([0,1])$ by Stone-Weierstrass. Approximate $f|_{[0,1]}$ by $p\in E$ to within $\epsilon$; then $x\mapsto p(|x|)$ on $[-1,1]$ is an even polynomial, and $|f(x)-p(|x|)|<\epsilon$ on $[-1,1]$.

• Well, $p(|x|)$ even needs not be a polynomial. Commented Jun 14, 2017 at 1:59
• $E$ was defined (possibly not clearly) so that if $p\in E$, then $p(x)=\sum_0^n a_k x^{2k}$. Commented Jun 14, 2017 at 2:02
• Ah, you are right. I hesitatingly jumped into the last sentence. Commented Jun 14, 2017 at 2:04