Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:46:48Z http://mathoverflow.net/feeds/question/91454 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91454/simple-closed-curves-and-the-coefficent-of-expi-theta-in-the-associated-four Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series David Epstein 2012-03-17T10:06:33Z 2012-03-21T02:33:11Z <p>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 $t\in S^1$, $|f'(t)| = 1$.</p> <blockquote> <p>Is it then true that $a_1\neq 0$?</p> </blockquote> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/91454/simple-closed-curves-and-the-coefficent-of-expi-theta-in-the-associated-four/91461#91461 Answer by alvarezpaiva for Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series alvarezpaiva 2012-03-17T12:19:12Z 2012-03-17T13:44:26Z <p>Dear David,</p> <p>This is just a reflection on your question:</p> <p>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 <em>Winding number and Fourier series</em> for the formula and the amusing story behind it) yields $$\sum n^3|a_n|^2 = 1.$$</p> <p>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 $\sum n^2|a_n|^2 = 1$.</p> <p>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.</p> http://mathoverflow.net/questions/91454/simple-closed-curves-and-the-coefficent-of-expi-theta-in-the-associated-four/91465#91465 Answer by Sean Eberhard for Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series Sean Eberhard 2012-03-17T13:55:44Z 2012-03-17T17:43:28Z <p>The answer to your first question is 'No'.</p> <p>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 $\hat{f}(1)=0$.</p> <p>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.</p> http://mathoverflow.net/questions/91454/simple-closed-curves-and-the-coefficent-of-expi-theta-in-the-associated-four/91784#91784 Answer by fedja for Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series fedja 2012-03-21T02:33:11Z 2012-03-21T02:33:11Z <p>OK, let's modify Sean's construction to remove any doubts (it won't <em>look</em> the same, but it <em>is based</em> 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.</p> <p>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? </p>