Is there a function $f:\mathbb N\to \mathbb R$ such that $\lim_{n\to\infty} f(n) = 0$ and for every $c\in\mathbb C$:
If $z_0=0$, $z_{n+1}=z_n^2+c$ and $|z_k|<2$, then there exists a point $c'$ in the Mandelbrot set satisfying $|c-c'|<f(k)$?
Such fact would be very useful for rendering the Mandelbrot set, as it would give the exact count of iterations needed for given resolution.
In principle, yes. The set of points whose orbits stay within $|z|< 4$ for the first $n$ iterations is a bounded open subset $M_n\subset\mathbb C$. We have $M_{n+1}\Subset M_n$ and $M=\bigcap _{n\in\mathbb N}M_n$. Let $f(n)=\max\{d(z,M)\mid z\in \overline{M_n}\}$. Then $z\in M_n$ implies there is a point $c\in M$ with $d(c,z)\le f(n)$. The sequence $(f(n))$ is nonincreasing, hence $a:=\lim_{n\to\infty} f(n)$ exists. Then for all $n$, we get a point $z_n\in M_n$ with $d(z_n,M)=a$. A subsequence of $(z_n)$ converges to some $\zeta$ with $d(\zeta,M)=a$, but we also have $\zeta\in\bigcap \overline{M_n}=M$. However, this mere existence of $f$ won't help you in practice.