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.
Best Answer
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.