Show $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is not dense in $C([-1,1])$ Let $C[-1,1]$ denote the Banach space of continuous real-valued functions on $[-1,1]$ , equipped with the supremum norm. Determine whether  $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is dense in $C([-1,1])$.
First, I checked using Stone-Weierstrass theorem. Note that if we have $1, -1$, then we can't find any function that separates the the points. So I want to show  $\mathrm{span}\{1, x^2, x^4, \cdots \}$ is not dense in $C([-1,1])$.
So we want to show that there exists some $\epsilon>0$ and $f\in C[-1,1]$ such that $\sup_{x\in [-1,1]}|g(x)-f(x)|>\epsilon$ for all $g \in \mathrm{span}\{1, x^2, x^4, \cdots \}$.
I'm thinking about using some functions like $f(x)=x,x^3, x^5, \cdots$. Also, I know $g(x)= \sum_{k=0}^n a_k x^{2k}$. But I'm stuck finding the $\epsilon$.
Any help will be appreciated!
 A: Hint: Remember something in the span is a finite combination of elements in the set. So,  let $2n$ be the largest power of $x$ present in a generic element of the span,  so a generic function looks like
$$\sum_{i=0}^{n}a_ix^{2i}$$
Let's look at the distance of a function of this form to $x$
$$\sup_{x\in [-1,1]} |x-\sum_{i=0}^{n}a_ix^{2i}| $$
At $x=0$ we have the difference is $a_0$.  At $x=1$ we have
$$|1-\sum_{i=0}^{n}a_i|$$
At $x=-1$ we have
$$|-1-\sum_{i=0}^{n}a_i|$$
Pick any small $\epsilon$ you want.  Is there any way to make this sum smaller than that at both $x=1$ and $x=-1$?
A: All the functions in $\{1, x^2,x^4,\dots\}$ are even. Therefore so are all their linear combinations, i.e., all the functions in their span. Finally, any limit (with respect to the supremum norm, or any other reasonable norm) of even functions is even. So the closure of the span, unlike the whole space $C([-1,1])$, consists only of even functions. [This is essentially the same as Alan's answer, but isolating what I consider to be the essential points.]
A: Let $\varphi$ be the linear functional defined on $C[-1,1]$ by the formula $$\varphi(f)=f(1)-f(-1)$$ Clearly this functional is bounded $(\|\varphi\|=2)$ and nonzero, as $\varphi(x)=2.$ Thus $\ker \varphi$ is a closed proper subspace of $C[-1,1]$ containing the linear span of all $x^{2n}.$ Therefore the linear span is not dense.
