Let $$E=\{\sum\limits_{k=0}^{n}a_{k}\,x^{2k+1}\, | \, n\in\mathbb{N}\cup\{0\}\}$$ be the set of polynomials of odd degree in each term defined on $[1,2]$.
(a) Show that $E$ is not closed in $\mathcal{C}^{0}([1,2])$, where $\mathcal{C}^{0}([1,2])$ is the space of continuous function from $[1,2]$ to $\mathbb{R}$ with sup norm.
(b) Is $E$ dense in $\mathcal{C}^{0}([1,2])$?
Here's my proof:
(a) Consider the partial sum of the taylor series of $\sin(x)$, then every partial sum are in $E$ obviously, i.e., there is a sequence $\{f_{i}(x)\}$ of $E$, However, the limit of the sequence $f(x)=\sin(x)$ is not in $E$. This shows that $E$ is not closed.
(b) I've tried to prove it using Stone-Weierstrass Theorem, but the theorem only states the polynomial space $\mathbb{R}[x]$ dense in $\mathcal{C}^{0}([a,b])$, i.e., $$\forall \epsilon >0 \mbox{ and } \forall f(x)\in\mathcal{C}^{0}([a,b]), \exists \, p(x)\in\mathbb{R}[x] \mbox{ s.t. } \|f(x)-p(x)\|_{\infty}<\epsilon.$$
Thus I want to use triangle inequality to find a polynomial with odd degree $p'(x)$ such that $\|p(x)-p'(x)\|<\epsilon$, but it is impossible since not every polynomial function is the odd function. If you control the distance of two polynomials when $x>0$, another side must have large distance. So I'm stuck, and I guess the answer is NOT.
Can anyone give me some hints? Thanks!
The result is true for $[1,2]$.
First let $f$ be smooth. Then we can approximate $f'$ uniformly on $[1,2]$ by a polynomial with only even powers of $x$. This is an easy consequence of Stone-Wiertsrass Theorem. Now integrate to finish the proof. The general case follows by first apllying Wierstrass Theorem and then applying above argument.
EDIT: As pointed out by Severin Schraven, there will be an extra constant term in this approach. As suggested in that comment, we can approximate the continuous function $\frac {f(x)} x$ by a polynomial with only even powers of $x$ and multiply by $x$. $E$ is dense for any interval $[a,b]$ which does not contain $0$.
In the case of intervals containing $0$, only functions vanishing at $0$ belong to the closure of $E$.