Given a continuous map $f:S^1\to \mathbb{C}$ from the unit circle to the complex numbers, one can form its Fourier series $\sum_{n=-\infty}^\infty a_n\exp(in\theta)$. I want to stick with those $f$ that give simple closed curves, bounding a closed topological disk, going round the disk in a counter-clockwise direction, and parametrized proportional to arclength. I am happy to add the hypothesis that $f'(t)$ is a continuous function of $t$ and that, for $t\in S^1$, $|f'(t)| = 1$.

Is it then true that $a_1\neq 0$?

If this is true, is $|a_1|$ bounded away from zero as $f$ varies? It may be that some other normalization might make the second question more tractable: for example, instead of normalizing the length to be $2\pi$ by a change of scale, as I have done above, one could require that a disk of unit radius be contained in the disk bounded by $f$. Any such normalization of $f$ would be highly acceptable.

I'm motivated by trying to describe the ``space of shapes'' in the plane, by using Fourier descriptors, a topic of interest both in machine vision and in microscopy in biology.

  • $\begingroup$ Is $a_1 \neq 0$ the final goal or are you interested in necessary (and possible sufficient) conditions on the sequence $\{a_n\}$ for $$ f(\theta) := \sum a_n e^{in\theta} $$ to be a Jordan curve? $\endgroup$ – alvarezpaiva Mar 19 '12 at 10:28

OK, let's modify Sean's construction to remove any doubts (it won't look the same, but it is based on the same idea). We will consider the curves symmetric with respect to the real axis and parametrized so that $f(-\theta)=\bar f(\theta)$, so we are sure that all Fourier coefficients are real. Now take $a\in\mathbb R$ and draw any continuous family of nice symmetric counterclockwise shapes $\Gamma_a$ that visit the points $1$, $a+i$, $a-i$ in this order. Note that the shapes will be necessarily non-convex for $a\ge 1$. Take small neighborhoods of these three points and replace the quick almost straight passages that are there by some "drunken walks" without self-intersections that have huge lengths but move essentially nowhere so that the whole length of the curve becomes essentially concentrated at those 3 points and the corresponding "wasted time" intervals are close to $(-\pi/2,\pi/2)$, $(\pi/2,\pi)$, $(-\pi,-\pi/2)$. Now, $2\pi a_1$ for the corresponding function is essentially $\int_{-\pi/2}^{\pi/2}\cos\theta\\, d\theta+2\Re\left[(a+i)\int_{\pi/2}^{\pi}e^{-i\theta}\\,d\theta\right]$, which is positive for large negative $a$ and negative for large positive $a$. However, the family of curves we created is continuous and so is the family of their parametrizations, so the intermediate value theorem finishes the story.

As usual, the existence of a counterexample most likely merely means that what you asked for is not what you need. So, what's the actual goal?


Dear David,

This is just a reflection on your question:

Since you assume that the curve is parametrized by arc-length, applying Plancharel's formula to $f'$ yields $$ \sum n^2|a_n|^2 = 1. $$ Moreover, you also assume that the map $f' : S^1 \rightarrow S^1$ has degree 1 and Brezis's formula for the degree of a $C^1$ map from the circle to the circle (Google Kahane's paper Winding number and Fourier series for the formula and the amusing story behind it) yields $$ \sum n^3|a_n|^2 = 1. $$

Averaging these two equations and assuming $a_1 = 0$ one gets that $$ \sum_{|n|> 1} {n^2(1 + n) \over 2}|a_n|^2 = 1 $$ but I don't see right now if this and could lead to a contradiction with $\sum n^2|a_n|^2 = 1$.

I don't know if this helps with your precise question, but I think Brezis's formula for the degree in terms of Fourier coefficients could come in useful.

  • $\begingroup$ Actually Brezis formula holds for maps in the Sobolev space $H^{1/2}(S^1,S^1)$ although I assumed that $f'$ is $C^1$. $\endgroup$ – alvarezpaiva Mar 17 '12 at 12:50

The answer to your first question is 'No'.

Let $g$ be the function $S^1\to S^1$ which starts at $1$, moves anticlockwise to $-1$, then moves clockwise $1 + \sqrt{2}$ times as fast once round the circle back to $-1$, and then moves anticlockwise back to $1$ again. This function $g$ has degree $0$ and $\hat{g}(0)=0$. The function $f(\theta) = g(\theta) e^{i\theta}$, which moves at a constant speed, therefore has degree $1$ and $\hat{f}(1)=0$.

You may reasonably complain at this point that $f$ is not differentiable and certainly not simple, but $f$ can be deformed very slightly so that it bounds a topological disc and makes smooth turns.

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    $\begingroup$ I am missing why $g(\theta) e^{i \theta}$ has constant speed. The function $g$ makes one total loop clockwise and one total loop anti-clockwise. It makes the first loop 3 times as fast as the second, so it must spend $1/4$ of its time traveling clockwise. So $g$ is of the form $e^{4 i \theta}$ on the clockwise portions, and $e^{-(4/3) i \theta}$ on the anti-clockwise portions. So $g$ sometimes has speed $4-1=3$ and sometime has speed $1-(-4/3) = 7/3$. $\endgroup$ – David E Speyer Mar 17 '12 at 17:05
  • $\begingroup$ You're absolutely right. I guess what I want is the following: suppose that $g$ moves clockwise $x$ times as fast as it moves anticlockwise. Then it spends $1/(x+1)$ of its time clockwise and $x/(x+1)$ of its time anticlockwise. Its speed clockwise is therefore $1+x$, anticlockwise $1+1/x$. I want then that $1+x−1=1+1/x+1$, i.e., $x^2−2x−1=0$. $\endgroup$ – Sean Eberhard Mar 17 '12 at 17:41
  • $\begingroup$ Editted solution to reflect this. $\endgroup$ – Sean Eberhard Mar 17 '12 at 17:43
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    $\begingroup$ Sean's example is beautifully simple. It definitely shows that $|a_1|$ cannot be bounded away from zero. However, deforming $f$ so that it becomes a simple closed curve, with $a_1=0$ exactly, seems delicate to me. Perhaps I'm not looking at things the right way. Because I'm still hoping for a definitive answer that satisfies all of my conditions, I'm not yet ready to mark Sean's as RIGHT, even though I'm full of admiration for its simplicity and brevity. $\endgroup$ – David Epstein Mar 18 '12 at 21:54
  • $\begingroup$ I agree with you that this is a little delicate. I suppose it is more of a belief that $f$ can be deformed appropriately while maintaining $a_1=0$. If I think of a simple argument I will relate it. $\endgroup$ – Sean Eberhard Mar 18 '12 at 22:16

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