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I've been self-studying Fourier Analysis: An Introduction by E.M. Stein and R. Shakarchi. I'm currently stuck on trying to understand part of the proof of the isoperimetric inequality in Chapter 4 (Some Applications of Fourier Series).

For those who have the book, I'm stuck around the top of page 105, where it is proved that if equality holds in the isoperimetric inequality, then the curve under consideration must be a circle. I'm having trouble seeing how they determine the formulas for x(s) and y(s) (how they determined that the Fourier coefficients of x(s) and y(s) are zero for all [tex] |n| \geq 2[/tex]).

For those who don't have the book, here's the proof (paraphrased a bit), and where I'm stuck:

Theorem:Let [tex]\Gamma[/tex] be a simple closed curve in [tex]\mathbb{R}^2[/tex], with length [tex]\ell[/tex] and enclosed area [tex]\mathcal{A}[/tex]. Then:

[tex] \mathcal{A} \leq \frac{\ell^2}{4\pi} [/tex]

(Equality iff [tex]\Gamma[/tex] is a circle)

Assume WLOG that [tex] \ell = 2\pi [/tex], and let [tex]\gamma : [0,2\pi] \rightarrow \mathbb{R}^2,~\gamma (s) = ( x(s), y(s) )[/tex] be a parametrization by arc-length of the curve [tex] \Gamma [/tex]. ( [tex]( x'(s)^2 + y'(s)^2 = 1 [/tex] for all [tex] s \in [0,2\pi] [/tex] ). Also, assume [tex] \gamma [/tex] is at least once continuously differentiable.

This implies:

(1) [tex] \frac{1}{2\pi} \int_0^{2\pi}( x'(s)^2 + y'(s)^2 ) ds = 1 [/tex]

Because [tex] \Gamma [/tex] is simple and closed, x(s) and y(s) are [tex] 2 \pi [/tex] periodic, and so they have Fourier series representations:

[tex]x(s) = \sum_{-\infty}^{\infty} a_n e^{ins} ~ {\rm and} ~ y(s) = \sum_{-\infty}^{\infty} b_n e^{ins} [/tex]

The Fourier series of the functions' derivatives are:

[tex]x'(s) = \sum_{-\infty}^{\infty} a_n i n e^{ins} ~ {\rm and} ~ y'(s) = \sum_{-\infty}^{\infty} b_n in e^{ins} [/tex]

They then apply Parseval's identity to (1), yielding:

(2) [tex] \sum_{-\infty}^{\infty} |n|^2 ( |a_n|^2 + |b_n|^2 ) = 1 [/tex]

By Green's theorem,

[tex] \mathcal A = \frac{1}{2}\left| \int_0^{2\pi} ( x(s) y'(s) - y(s)x'(s) ) ds \right| [/tex]

They apply "the bilinear form" of Parseval's identity to the area integral:

[tex] \mathcal A = \pi \left| \sum_{-\infty}^{\infty} n( a_n \bar{b_n} - \bar{a_n} b_n ) \right| [/tex] ( [tex] \bar [/tex] denotes the complex conjugate ).

Using the facts that [tex] ~~\left| a_n \bar{b_n} - \bar{a_n} b_n \right| \leq 2|a_n| |b_n| \leq |a_n|^2 + |b_n|^2 ~~[/tex] and [tex] ~|n| \leq |n^2| [/tex], they then conclude that:

[tex] \mathcal A \leq \pi \sum_{-\infty}^{\infty} |n|^2 ( |a_n|^2 + |b_n|^2 ) [/tex]

So, using (1):

[tex] \mathcal A \leq \pi [/tex]

Now here's where I get confused. They say:

"When [tex] \mathcal A = \pi [/tex], we see from the above argument that:

[tex] x(s) = a_{-1} e^{-is}~ + ~ a_0 ~ + ~ a_1 e^{is} ~{\rm and}~ y(s) = b_{-1} e^{-is} ~ + ~ b_0 ~ + ~ b_1 e^{is} [/tex]

because [tex] |n| < |n|^2 [/tex] as soon as [tex] |n| \geq 2 [/tex]."

I can't see how they found that the Fourier coefficients of x(s) and y(s) are zero for [tex] |n| \geq 2 [/tex]. I've tried hard to figure it out, but I haven't really got very far.

If I've left any important information out above, please let me know.

Thanks a ton in advance,

-Deltinu

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# Help understanding proof of isoperimetric inequality

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