# Define $A$ to be the family of polynomials containing only even powers of $x$. Is $A$ dense in $C[0,1]$?

Define $$A$$ to be the family of polynomials containing only even powers of $$x$$. Is $$A$$ dense in $$C[0,1]$$?

How to approach these kind of problems? Would the same technique apply to $$B$$, the family of polynomials containing only odd powers of $$x$$ in $$C[-1,1]$$?

• For which norm you want the density ? Commented Apr 12, 2018 at 4:21
• the infinite norm Commented Apr 12, 2018 at 4:24
• For the second statement it is false because for $P \in B$, $P(0) = 0$ and if we had density, $f(0) = 0$ for every $f\in C([-1,1])$ which is not true. Commented Apr 12, 2018 at 4:28
• Are the constants allowed in the second part? Commented Apr 12, 2018 at 4:41

Here is a simple argument which does not require Stone-Weierstrass Theorem: if $f$ is continuous so is $f(\sqrt x)$. If $|f(\sqrt x)-p(x)|<\epsilon$ then $|f(x)-p(x^{2})|<\epsilon$ and $p(x^{2})$ has only even powers of $x$. For the second part note that if a sequence of polynomials with only odd powers of $x$ converges pointwise on $[-1,1]$ then the limit is an odd function. So functions (like the constant function 1) which are not odd cannot be approximated by polynomials with only odd powers.
Stone-Weierstrass Theorem says that if $A$ is an algebra of $C(X)$ that separates points in $X$ and contains the constant functions, then $A$ is dense in $C(X)$, here $X$ is compact.
Realizing to the set $A$ and $X=[0,1]$ in question, even powers of polynomials contain those of constants, and clearly it is an algebra which separates points, consider for example, $x^{2}$, this is an one-to-one function.
• Algebra: $f,g\in A$, $f+g, fg\in A$, $cf\in A$, $c\in{\bf{R}}$. Separate points: Given $x,y\in X$, $x\ne y$, there exists some $f\in A$ such that $f(x)\ne f(y)$. Commented Apr 12, 2018 at 5:16