Proposition
Let $\mathcal{P_e}$ be the set of functions $p_e(x) = a_o + a_2x^2 +
\cdots + a_{2n}x^{2n}$, $p_e : \mathbb{R} \to \mathbb{R}$Show that all $f:[0,1]\to\mathbb{R}$ can be uniformly approximated by
elements in $\mathcal{P_e}$
Attempt:
Since we are talking about polynomials let's try play with Weierstrass Theorem
By definition $\mathcal{P_e}$ is dense in $C^0([0,1], \mathbb{R})$ , if for every funciton $f \in C^0([0,1], \mathbb{R})$, $\exists p_e \in \mathcal{P_e}$ such that $\forall \epsilon > 0, \forall x \in [0,1], ||p_e-f||< \epsilon$
So we want to show $\|p_e – f\| < \epsilon$
Try something like…since every continuous function is approximated by polynomials i.e. $\forall \epsilon > 0,\ \|f – p\| < \epsilon$, therefore let $p$ be a polynomial, $p = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$
$$\|p_e – f\| \leq \|f – p\| + \|p – p_e\|$$
$$\Rightarrow \|p_e – f\| < \epsilon + \|p_o\|,$$ where $p_o$ is an odd polynomial
$$\Rightarrow \|p_e – f\| < \epsilon + \sup_x|a_1+a_3+\cdots+a_{2_n+1}|$$
Stuck.
In any case, $p$ and $p_o$ is poorly defined. What would be the standard approach to prove the proposition?
Best Answer
Two standard approaches would be either to use the Stone-Weierstraß theorem, and note that the algebra of even polynomials satisfies the premises of that theorem, or to look at the isometry $S\colon C^0([0,1],\mathbb{R}) \to C^0([0,1],\mathbb{R})$ given by
$$S(f) \colon t \mapsto f(\sqrt{t}).$$
Approximate $S(f)$ with a polynomial $q$, and set $p = S^{-1}(q)$ to obtain an aproximation of $f$ by an even polynomial.