Maximum Modulus path

Question

Consider any entire, non constant function $f:\Bbb C\to \Bbb C$. Choose any $z\in\Bbb C$ and define $m(r)\in\overline D(z,r)$, for any $r\ge 0$, with this property: $$|f(m(r))|\ge|f(w)|\;\forall w\in \overline D(z,r)$$

I'm aware that this definition may be ambiguous, since the maximum modulus needn't be met in a single point. I'm also aware that $|m(r)-z|=r$, by the maximum modulus principle.

Questions:

• Is it always possible to choose $m(r)$ in such a way that $m$ is continuous, as a function from $[0,\infty)$ to $\Bbb C$?
• If/when it is the case, has this $m$ any known properties? Is there some theory about this?

EDIT: I suspect that the answer to the first question is yes, since the modulus of an entire function can't have any local maxima. But I haven't anything rigorous.

Answer 1

It is not always possible.

WLOG we can take $z = 0$. That leaves me free to use $z$ as the variable for complex functions.

Consider e.g.the function $f(z) = {z}^{16}+8\,{z}^{12}+432\,{z}^{8}-640\,{z}^{4}+256$. (Note: small typo in the function fixed)

The maximum modulus on the circle $|z|=r$ turns out to be on the diagonals $x = \pm y$ for $r < 1.729363340$ approximately, then abruptly switches to the $x$ and $y$ axes.

EDIT: This function is symmetric under $z \to iz$ and its absolute value is unchanged by $z \to \overline{z}$, so it suffices to consider the sector $z = r e^{i\theta}$, $0 \le \theta \le \pi/4$. A local extremum of $|f(z)|$ on the arc $|z|=r$ must satisfy $\text{Im}(z f'(z)/f(z)) = 0$. The graph looks like this:

The curvy parts actually give local minima (in fact they pass through zeros of $f$). The axes and the diagonal $x=y$ have the local maxima. If you plot $|f(r)|/|f(r e^{i\pi/4})|$ you get this:

For $r < 1.729363340$ (approximately) this ratio is less than $1$, so the maxima are on the diagonals. For $r > 1.729363340$ the ratio is greater than $1$ and the maxima are on the axes.