Approximating Continuous Functions with Polynomials

Question

Given $\epsilon \gt 0$ and $f \in C^{0}[0,1]$, Weierstrass says that I can find at least one (how many? probably a lot?) polynomial $P$ which approximates f uniformly: $$\sup_{x \in [0,1]} |f(x) – P(x)| \lt \epsilon$$

This means that, under the sup norm $||.||_{\infty}$, the polynomials are dense in $C^{0}[0,1]$. So, in analogy to approximating irrationals with the rationals, I would like to know:

• What can we say about the order of $P$? Or, turning this around, given that $P$ is of order $n$, how small can $\epsilon$ get?

I'm betting this should depend in some way on the properties of $f$: the intuition is that smoother functions should be somehow "better" approximated by lower-order polynomials, and less-well-behaved functions should require higher-order polynomials. But I am not sure how to formalize this.

This is probably all well-understood, but I'm not well-read on approximation theory. Any guidance would be wonderful.

Answer 1

According to Theorem 1.2 of this paper by Sukkrasanti and Lerdkasem, we have the following result (with impressively great generality, I might add):

Let $f: [0,1]^p \rightarrow \mathbb{R}^q$ be bounded. Then there exists $C$ such that $$ \| f - B_n(f) \|_{\infty} \le C \omega(1/\sqrt{n})$$ where $\omega(\delta) := \sup_{\|t_1 - t_2\| \le \delta} \|f(t_1) - f(t_2)\|$ and $B_n(f)$ is the $n$th Bernstein polynomial of $f$.

I do not claim to have read the paper myself. Notice that if $f$ is indeed continuous, then by uniform continuity, as $n \rightarrow \infty$, $\omega(1/\sqrt{n})$ must go to zero. So the convergence rate is related to how rapidly $f$ can vary, as intuition would already suggest.