1
$\begingroup$

Consider a loop (let it be a continuous map $\gamma:S^1\to\mathbb C$) and a point $z=\gamma(0)$ on this loop.

Does there exist a (non-constant) polynomial $P$ so that $|P(z)|=\max_{x\in \gamma(S^1)}|P(x)|$?

Remark: This is obviously not true if one takes a continuous curve that fills some open set. So a restriction like piece-wise smooth/differentiable is in order.

  • 0
    A more interesting question is what happens for Jordan curves. In the real analytic case the claim is true. Not sure in general. – 2017-01-08
  • 0
    @Moishe What do you mean with real analytic? Do you mean analytic curves in $\Bbb R^n$ and real polynomials with $n$ indeterminants? – 2017-01-08
  • 0
    I meant real analytic $\gamma$ WIh target the complex plane, but P is still complex as in your question. – 2017-01-08

1 Answers 1

2

Generally, there doesn't exist such a polynomial. The complement of the trace of $\gamma$ has one unbounded component $U$. Let $K = \mathbb{C}\setminus U$. For $z \in \overset{\Large\circ}{K}$, and a non-constant polynomial $P$, we always have

$$\lvert P(z)\rvert < \max \Bigl\{ \lvert P(w)\rvert : w \in \partial\overset{\Large\circ}{K}\Bigr\} \leqslant \max \{ \lvert P(w)\rvert : w \in \gamma(S^1)\}.$$

If we take a smooth curve winding around $0$ twice (or more often) with one large loop, and another inside it, no point on an inner loop can maximise a non-constant polynomial on $\gamma(S^1)$.

  • 0
    A moral continuation of the conditions might be so that the $\gamma$ does not wind around any point of itself, or that $\gamma = \partial U$ where $U$ is an open, bounded simply connected set. Would one expect it to still be false in such cases? – 2017-01-08
  • 0
    Yes, but it's even more restrictive. The loop mustn't have any inward-pointing corners. At such a point, every non-constant polynomial attains values of larger absolute value at nearby points in the interior of the loop, so by the maximum modulus principle even larger values elsewhere on the loop. But I don't know a nice characterisation of points where a maximum can be attained. – 2017-01-08
  • 0
    What is an inward-pointing corner? Is it like the inner corner of a capital **L**? – 2017-01-08
  • 0
    I mean a cusp like in the [Cardioid](https://en.wikipedia.org/wiki/Cardioid). If it points outward, no problem. But inward like for the cardioid doesn't work. – 2017-01-08
  • 0
    @DanielFischer Smooth Jordan curves might be a reasonable class. – 2017-01-08
  • 0
    @MoisheCohen Yes, at least they're promising candidates (assuming they're smooth in the sense that $\gamma'(t) \neq 0$ for all $t$). But I don't know whether smoothness suffices. – 2017-01-08
  • 1
    @s.harp In my previous comment, I probably was a bit inaccurate. Cusp means that the angle of the corner is $2\pi$, but any angle $> \pi$ is sufficient. If you have a direction $e^{i\varphi}$ pointing inward with $\operatorname{Re} \bigl(e^{i\varphi}\cdot P'(z)\bigr) > 0$, you get larger values in the interior along that direction. When $P'(z)\neq 0$ and the angle is $> \pi$, that is guaranteed. If $P'(z) = 0$, we also have larger values in the interior. – 2017-01-08