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Let $f:(0,1) \to \mathbb{R}$ be continuous, and let $\delta:(0,1) \to \mathbb{R}$ be continuous and positive.

Does there always exist a polynomial $p(x)$ satisfying $|f(x)-p(x)| < \delta(x)$ for every $x \in (0,1)$?

Edit: I should have written $[0,1]$ (the closed interval) as the domain instead of $(0,1)$. (to rule out problems which come from the fact $f$ is not bounded, or uniformly continuous; if $f$ is not bounded, then it cannot be approximated by polynomials).


I guess that the answer is negative, but I don't see how to build a "sufficiently bad" $\delta$.

When $\delta$ is constant, this is just the Weierstrass approximation theorem.

Moreover, if we allow $p(x)$ to be an arbitrary smooth function, then we can always achieve a $\delta$-approximation, via a partition of unity argument.

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  • $\begingroup$ You can do Weierstrass again provided that $f$ is uniformly continuous and $\delta$ is bounded below away from zero (because in this case you're really just applying Weierstrass to the extension of $f$ to $[0,1]$ and using $\inf \delta$ as your tolerance). If $\delta$ is not bounded below away from zero (and $f$ is still uniformly continuous) then I think you can do it again, by splitting $p$ into an interpolant of $f$ at the endpoints plus a corrector, but I'd have to mess with the estimates to make sure things still work out in a neighborhood of the endpoints. $\endgroup$ – Ian Mar 28 '19 at 14:59
  • $\begingroup$ If $f$ is not uniformly continuous then things can certainly break when $f$ is not bounded, and I suspect they can break when $f$ has an oscillatory singularity at one of the endpoints as well. $\endgroup$ – Ian Mar 28 '19 at 14:59
  • $\begingroup$ It's easy to show that $\sin(1/x)$ can't be approximated within $\delta$ on $(0,1)$ by a polynomial if $\delta < 1/2$. $\endgroup$ – Robert Israel Mar 28 '19 at 15:02
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On $(0,1)$, no, even if $\delta$ is constant, because $f$ might not be bounded while polynomials are. Weierstrass requires a closed, bounded interval.

EDIT: With the interval as $[0,1]$, $\min_{x \in [0,1]} \delta(x)$ exists and is positive, so we might as well replace $\delta$ by that constant, and then we can use the Weierstrass theorem.

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  • $\begingroup$ Thanks, you are right. I really should have written the closed interval as the domain, not the open one. (The point was not to see that Weierstrass theorem fails on non-compact domains, but to investigate how power polynomials have to approximate-better than uniformly, that is). $\endgroup$ – Asaf Shachar Mar 28 '19 at 15:04
  • $\begingroup$ @AsafShachar Unless $\delta$ on $[0,1]$ has zeros, it becomes sufficient to just do Weierstrass again, once you switch over to the compact case. $\endgroup$ – Ian Mar 28 '19 at 15:06
  • $\begingroup$ But if "positive" means $> 0$, $\delta$ can't have zeros. $\endgroup$ – Robert Israel Mar 28 '19 at 15:07
  • $\begingroup$ @RobertIsrael Indeed; I'm just pointing out a possible generalization that is actually nontrivial. Still, I think given an interpolant $q$ at the zeros of $\delta$ you can then take $p=q(1+r)$ where $r$ is a Weierstrass-type approximation of $\frac{f-q}{q}$. $\endgroup$ – Ian Mar 28 '19 at 15:08
  • $\begingroup$ @Ian Thanks, you are right. I should have thought more about a careful formulation of the question. I guess that we could try to make the question less trivial in one of two ways: (1) allow $\delta$ to have finitely many zeros. (2) Keep the domain open, but assume that $f$ is bounded. $\endgroup$ – Asaf Shachar Mar 28 '19 at 15:12

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