let $A$ be the set of all polynomials that contain only terms of even degree. Then the uniform closure of $A$ is $B$. Let $b >0$ , let $B= \{ f \in  C^r([-b,b]) : f(x) = f(-x) for \ \ 0\leq x\leq b\}$, and let $A$ be the set of all polynomials that contain only terms of even degree (with domains restricted to $[-b,b]$). Then the uniform closure of $A$ is $B$.
I am not getting any clue how to solve the problem. Help Needed. 
I think we have to use Weierstrass Approximation Theorem.
 A: First you need to show that all uniform limits $f$ of sums of even-degree polynomials satisfy $f(x)=f(-x)$ and are continuous. The first is because even-degree polynomials and their sums all satisfy this relation, so any limit should too. The second part, that uniform limits of continuous functions are continuous, is standard. It uses a so-called $\frac{\epsilon}{3}$ trick: if a sequence $f_n$ converge uniformly to $f$, then find some large $N$ so that $|f_N(x)-f(x)|<\frac{\epsilon}{3}$, which we can do by uniform convergence. Then we find $\delta$ so that if $|x-y|<\delta$ then $|f_N(x)-f_N(y)|<\frac{\epsilon}{3}$. Then if $|x-y|<\delta$, we have
$|f(x)-f(y)|\leq|f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|<3\cdot\frac{\epsilon}{3}=\epsilon$.
So $f$ is (uniformly) continuous.
Now if we want to approximate $f$ by polynomials in $x^2$, cannot we try approximating $g(x)=f(\sqrt{x})$ in polynomials of $x$? $g$ is continuous. What does Weierstrass tell us if we try to approximate $g$ by polynomials on $[0, b^2]$?
A: I think the following observation would be enough to show this result

If $P$ is a polynomial,  $Q$ is the part of $P$ which contains only even degree terms and $R$ is the part of $P$ which contains only odd degree terms we can write $Q(x)=(P(x)+P(-x))/2$

Now by using Weierstrass we can find a sequence $(P_n)_{n=1}^\infty$ of polynomials such that $P_n\to f$ uniformly. Now for each $n$ and for each $t$
$$
Q_n(t)=\frac{P_n(t)+P_n(-t)}{2}\to \frac{f(t)+f(-t)}{2}=f(t)
$$
and this convergence is uniform as well.
A: We apply Weierstrass' approximation theorem to the auxiliary function
$$g(t):=f\bigl(\sqrt{t}\bigr)\qquad(0\leq t\leq 1)\ ,$$
which is continuous on $[0,1]$: For any given $\epsilon>0$ we can find a polynomial $t\mapsto p(t)$ such that
$$\bigl|g(t)-p(t)\bigr|<\epsilon\qquad(0\leq t\leq1)\ .$$
For $-1\leq x\leq1$ one has $0\leq x^2\leq 1$ and $$f(x)=f\bigl(|x|\bigr)=f\bigl(\sqrt{x^2}\bigr)=g(x^2)\ .$$ We therefore obtain
$$\bigl|f(x)-p(x^2)\bigr|<\epsilon\qquad(-1\leq x\leq1)\ .$$
