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 $|r(t)|\le1$.

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.


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.