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I'm studying complex dynamics and I am struggling to verify the following facts.

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a complex-valued polynomial of degree $d\geq 2$.

That is, define $f(z)=a_dz^d+...+a_0$. Take $R=\sup(1,\frac{1+|a_{d-1}+...+|a_0|}{|a_d|})$.

Show that $|f(z)|\geq |z|^d/R$ whenever $|z|>R$ ($\star$).

[I guess I need the reverse triangle inequality but I'm still lost]

Now, we define the filled Julia set of $f$ to be the set $K_f$ such that under the iterations of $f^n(z)$ do not tend to $\infty$ as $n$ tends to $\infty$.

Thus the statement $(\star)$ shows that $K_f$ is compact.

Now, we want to prove that $K_f$ is in fact $\bigcap_{n\in\mathbb{N}} f^{-n}(\overline{D}_R)$ (where $R$ is defined as above).

Nothingone
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1 Answers1

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I hope the second statement is trivial to see, if $f^n(z)\to \infty$ then $|f^n(z)|>R$ for some $n$, so that this $z$ is not contained in $f^{-n}(\bar D_R)$?

As for the root radius, the important thing to remember is that $R\ge 1$. Then if $|z|>R\ge 1$ one gets, indeed by the triangle inequality $|a-b|\le|a|+|b|\implies |a|\ge|a-b|-|b|$, \begin{align} |f(z)|&\ge|a_dz^d|-(|a_{d-1}z^{d-1}|+...+|a_1z|+|a_0|)\\ &\ge|a_d||z|^d-|z|^{d-1}(|a_{d-1}|+...+|a_1|+|a_0|)\\ &\ge|a_d||z|^d-\frac{|z|^{d}}R(|a_{d-1}|+...+|a_1|+|a_0|)\\ &=\frac{|z|^{d}}R(|a_d|R-(|a_{d-1}|+...+|a_1|+|a_0|))\\ \end{align} Now look at the construction of $R$ to find that $$ |a_d|R\ge 1+|a_{d-1}|+...+|a_1|+|a_0|, $$ so that indeed the last factor is larger than $1$. Compare this to the usual root radius estimates $$ R=\max\left(1,\frac{|a_{d-1}|+...+|a_1|+|a_0|}{|a_d|}\right) $$ and $$ R=1+\max_{k=0,..,d-1}\frac{|a_k|}{|a_d|}. $$ You get the root radius used if you replace $|a_{d-1}|$ with $(1+|a_{d-1}|)$ in both formulas. See "every polynomial of odd degree has one real root" for a different way to enforce a non-zero bound. However, this is not as tight as the bound here.

Between the first and second step you could put, using $\frac{|z|}R>1$, the step $$ |f(z)|\ge... \ge |z|^d(|a_d|-|a_{d-1}|R^{-1}-|a_{d-2}|R^{-2}-...-|a_1|R^{1-d}-|a_0|R^{-d}). $$ What you need for the claim is some number $u=\frac1R>0$ with $$ |a_d|-|a_{d-1}|u-|a_{d-2}|u^2-...-|a_1|u^{d-1}-|a_0|u^d\ge u \\\iff |a_d|-(1+|a_{d-1}|)u-|a_{d-2}|u^2-...-|a_1|u^{d-1}-|a_0|u^d\ge 0. $$ The left side in the last inequality is a falling concave polynomial, so Newton's method converges without constraints on the positive half axis to the unique positive root, giving rapidly a much better value for $R$. Note that the tangent roots of the Newton steps lie on the "wrong" side of the root, so that one needs to shift the last approximation in the direction of the Newton step to get the right inequality.

Lutz Lehmann
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