Approximating $f(x) = x$ with a certain class of polynomials. Let
$$P = \{\sum_{j=0}^{n}a_{j}x^{2j} | n \in \mathbb{N}\cup \{0\}, a_{0},...,a_{n} \in [0,\infty)\}.$$ So that $P$ contains polynomials in $x^2$ with non-negative coefficients. For example $1+x^2+\frac{1}{4}x^4+\pi x^{8} \in P$, $x^2 \in P$, $0 \in P$.
Given an $\epsilon > 0$ is there $p_{\epsilon} \in P$ so that
$$\max_{x \in [0,1]} |p_{\epsilon}(x) - x| \leq \epsilon \text{ }?$$

This question is inspired from Let $0\leq a<b$ be real numbers. Prove that there is no continuous function $f:[a,b]\rightarrow\mathbb{R}$
If the identity function can be well approximated from functions in $P$, then I can prove the linked question.
One cannot simply apply Stone-Weierstrass Theorem for my question as $P$ contain polynomials with non-negative coefficients (more restrictive).

The answer is no; assume the opposite and let $P_n(x)=a_{0n}+a_{1n}x^2+...$ as in the OP st $|P_n(x)-x| \le 1/n, x \in [0,1]$; this means $a_{0n} \le 1/n$ so $Q_n(x)=P_n(x)-a_{0n}$ satisfies $|Q_n(x)-x| \le 2/n, x \in [0,1]$ or $|x||\sum a_{kn}x^{2k-1} -1| \le 2/n$
For $x=1$ this means $1-2/n \le \sum a_{kn} \le 1+2/n$
But then for $x=1/2$ one has $1-\sum a_{kn}(1/2)^{2k-1} \ge  1-\sum a_{kn}/2 \ge 1/2-1/n$ so $|x||\sum a_{kn}x^{2k-1} -1| \ge 1/4-1/(2n)$ contradicting the above estimate for $n \ge 20$ say
