For every $\varepsilon > 0$, is there a polynomial of $x^4$ without constant term, i.e., $p(x^4) = a_1 x^4 + a_2 x^8 + \cdots +a_n x^{4n}$, such that $$\p(x^4)x^2  x\ < \varepsilon $$ for every $x \in [0,1]$?

$\begingroup$ Can we start off by approximating $\sqrt{x}$ with a polynomial (Weierstrass), perturb the polynomial so it has zero constant term, and then do some kind of bootstrap or iteration? $\endgroup$ – Yemon Choi Jun 10 at 4:44

4$\begingroup$ Muentz‘ theorem $\endgroup$ – user131781 Jun 10 at 8:01

$\begingroup$ @user131781 Indeed, this occurred to me after my original comment, I'm getting even more forgetful I guess ... $\endgroup$ – Yemon Choi Jun 10 at 15:40
Of course there is. Let $P$ approximate on $[0, 1]$ with error no greater than $\varepsilon$ the function $$f(x) = \min\{\varepsilon^{5}, x^{5/4}\} ,$$ and define $p(x) = x P(x)$. If $x \geqslant \varepsilon^4$, then $f(x^4) = x^{5}$ and hence $$p(x^4) x^2  x = x^6 P(x^4)  f(x^4) \leqslant x^6 \varepsilon \leqslant \varepsilon .$$ On the other hand, if $x < \varepsilon^4$, we simply have $f(x^4) = \varepsilon^{5}$, and hence $$\begin{aligned}p(x^4)x^2  x & \leqslant x^6 P(x^4)  f(x^4) + x^6 f(x^4) + x \\ & \leqslant \varepsilon^{24} \varepsilon + \varepsilon^{24} \varepsilon^{5} + \varepsilon^4 \leqslant \varepsilon ,\end{aligned}$$ provided that $\varepsilon$ is small enough.

$\begingroup$ How are you justifying the first step? The function $f$ you are approximating depends on $\epsilon$. $\endgroup$ – Yuval Peres Jun 10 at 13:18

$\begingroup$ @YuvalPeres I don't see a problem (perhaps I am being slow)  one fixes epsilon, defines $f=f_\epsilon$, and then chooses some $P$ that approximates $f_\epsilon$ to with $\epsilon$. What goes wrong? $\endgroup$ – Yemon Choi Jun 10 at 19:53

$\begingroup$ @YuvalPeres: Neither do I see a problem here: $f$ and $P$ do depend on a (fixed) $\varepsilon$. (But still it is far simpler to apply the Müntz–Szász theorem, as user131781 pointed out in their comment.) $\endgroup$ – Mateusz Kwaśnicki Jun 10 at 20:59
