Dynamics of the repetitions of $f(z) = z^{2} +\frac{1}{4}$

Question

This is probably a classic and most likely an easy question, but I was not able to find an answer. If we have the iteration $z_{n+1}=g(z_n)$, where $$g(z) = z + a z^{2},$$ with $a\ne 0$, then using a suitable linear change of coordinates it can be brought to the form $$f(z) = z^{2} + \frac{1}{4}.$$ Is there a neighbourhood of the unique fixed point $z=\frac{1}{2}$ of $f$, in which the sequence of iterates remains bounded?

Answer 1

The answer in NO.

We shall show that: there is no open neighbourhood of $z=\frac{1}{2}$, where the iterates of $f(z)=z^2+\frac{1}{4}$ remain bounded.

Take $z_0=\frac{1}{2}+\varepsilon$, where $\varepsilon>0$. Then the sequence $$z_{n+1}=f(z_n), \,\,\,n\in\mathbb N,$$ is strictly increasing and thus convergent in $(0,\infty]$. But if it had a finite limit, that would be a fixed point of $f$, which is impossible, as the only fixed point of $f$ is $\frac{1}{2}$ and $\lim f(z_n)>\frac{1}{2}$.

Note. We have instability only in the positive direction (i.e., $\vartheta=0$). In fact, for every $\vartheta\in(0,2\pi)$, there exists an $\varepsilon_\vartheta$, such that, if $r\in(0,\varepsilon_\vartheta)$, then the sequence $z_0=\frac{1}{2}+r\mathrm{e}^{i\vartheta}$, $z_{n+1}=f(z_n)$, remains bounded. (See here.)

Answer 2

No, there isn't. The point $z=1/2$ is on the boundary of the Julia set.