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The following problem is from Stein's Fourier analysis.

This problem explores another relationship between the geometry of a curve and Fourier series. The diameter of a closed curve $\Gamma$ parametrized by $\gamma(t)=(x(t),y(t))$ on $[-\pi,\pi]$ is defined by $$d=\sup_{P,Q\in\Gamma}|P-Q|=\sup_{t_1,t_2\in[-\pi,\pi]}|\gamma(t_1)-\gamma(t_2)|.$$ If $a_n$ is the $n$th Fourier coefficient of $\gamma(t)=x(t)+iy(t)$ and $l$ denotes the length of $\Gamma$, then
(a) $2|a_n|\le d$ for all $n\ne0$.
(b) $l\le\pi d$.
Property (a) follows from the fact that $2a_n=\frac{1}{2\pi}\int_{-\pi}^\pi{[\gamma(t)-\gamma(t+\pi/n)]e^{-int}dt}$.
The equality $l=\pi d$ is satisfied when $\Gamma$ is a circle, but surprisingly, this is not the only case. In fact, one finds that $l=\pi d$ is equivalent to $2|a_1|=d$. With the normalization $a_1=1$, we the have $d=2$ if and only if the derivative of $\gamma$ takes the form $$\gamma'(t)=ie^{it}(1+r(t)),$$ where $r$ is a real-valued function which satisfies $r(t)+r(t+\pi)=0$, and $|r(t)|\le1$.
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In fact, there may be some mistakes.

First, it's easy to check that property (b) is not satisfied by all closed curves. For convex ones, it is proved here.

Second, I can deduce $l=\pi d$ from $2|a_1|=d$, but I doubt the converse. Choose a circle, for instance, the origin $(0,0)$ in its interior but not at the center. Using polar coordinates, let the angle $\theta$ be our parameter $t$. From $2a_1=\frac{1}{2\pi}\int_{-\pi}^\pi{[\gamma(t)-\gamma(t+\pi)]e^{-it}dt}$, we know that the integrand is the length of the chord through $(0,0)$ with an angle $t$ from $x$-axis. Hence, it's less than $d$ except two $t$'s. And then $2|a_1|<d$.

For the last assertion, I have only proved the "if" part.

My questions are (a) Is my counterexample for $2|a_1|<d$ correct? (b) If it is correct, what is an equivalent condition given by the Fourier coefficients? (c) How to proceed with the last assertion?

Thanks for your patience.

Note: Let's assume $\Gamma$ is regular, that is, $\gamma'$ never vanishes.

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Edited, based on comments

(1) In my copy of the book, property (b) includes the constraint that $\Gamma$ is convex

(2) Your counterexample does not work because you are not using the correct parametrization. In your parametrization $\gamma(t), \gamma(t+\pi)$ are not on opposite sides of the circle. For a circle of radius $R$ shifted by $\epsilon$, a correct parametrization would be $\gamma(t)=\epsilon+R \cos(t/R)+ i R \sin(t/R)$.

Your version of the problem is missing the following sentence: "We re-parametrize $\gamma$ so that for each $t\in [-\pi,\pi]$ the tangent to the curve makes an angle $t$ with the $y$-axis."

(3) There is one error in the statement of the problem. It should say "the tangent to the curve makes an angle $t+\pi/2$ with the $y$-axis", not "an angle $t$". Actually it is stated correctly since it is the angle with the $y$-axis, not the $x$-axis

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  • $\begingroup$ I have searched on the Internet the latest edition but it is exactly what I have quoted. Could you please post the question in your book? (Or give a link.) And I never find what "the correct parametrization" is. It is never mentioned in my book. Part (3) in your answer is never mentioned, either. $\endgroup$ – Eclipse Sun Apr 7 '15 at 15:12
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    $\begingroup$ I think the only missing information in your version regarding parametrization is the sentence "We re-parametrize $\gamma$ so that for each $t\in[-\pi,\pi]$ the tangent to the curve makes an angle $t$ with the $y$-axis." $\endgroup$ – Dunham Apr 7 '15 at 15:46
  • $\begingroup$ For a circle of radius $R$ shifted by $\epsilon$, I think a correct parametrization would be $\gamma(t) = \epsilon + R\cos(t/R) + i R\sin(t/R)$ $\endgroup$ – Dunham Apr 7 '15 at 15:52

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