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Let $J_c$ be the Julia set for the quadratic polynomial $f_c(z) = z^2 + c$, and the Mandelbrot set is $M = \{ c \in \mathbb{C} : J_c \text{ is connected} \}$. Call the closed disc of radius $2$ centered at the origin $D = \{ c \in \mathbb{C} : |c| \le 2 \}$, now $M \subseteq D$.

Question: does $c \in M$ imply $J_c \subseteq D$ ?

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I think I can get this using the triangle inequality and the fact that all possible $c$ in the mandelbrot set are bounded in the disk of radius $2$.

Let $f_c(z)=z^2+c$ and pick a some $r=|z|>2$.

$|z^2|=|f_c(z)-c| \leq |f_c(z)|+|c|$

which means basically that $|f_c(z)| \geq |z^2|-|c| \geq r|z|-|z|>(r-1)|z|$.

Since $(r-1)>1$, iterates are greater than $r^n|z| \to \infty$ so $z \notin J_c$.

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  • $\begingroup$ It was not obvious to me at first glance where you used $2 \ge |c|$, but it's in the step from $\ldots - |c| \ge \ldots - |z|$ using $|z| \gt 2 \ge |c|$. Thanks! $\endgroup$
    – Claude
    Aug 16, 2018 at 3:52
  • $\begingroup$ yes! Sorry about that. I usually omit that kind of thing because I assume that analysis-types live and breathe that kind of stuff. I too forget it often, and I had to write the inequality a few times to make sure I was doing the right thing myself (I knew that I needed a damn 2....) Anyway, no problem :) $\endgroup$
    – Andres Mejia
    Aug 16, 2018 at 3:53

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