Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help understanding proof of isoperimetric inequality

  1. Jul 13, 2007 #1
    Hello All,

    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,

  2. jcsd
  3. Jul 13, 2007 #2
    I think that is because we established above that,
    [tex]A=\pi \left| \sum_{-\infty}^{\infty} n(a_n \bar b_n -\bar a_n b_n) \right| \leq \pi \sum_{-\infty}^{\infty} |n|^2(|a_n|^2+|b_n|^2)[/tex]

    Now we are considering the case when instead of [tex]\leq [/tex] we have [tex]=[/tex]. In that case the LHS is equal to preciselt to the RHS. Notice that the LHS has a factor of [tex]n[/tex] and the RHS has a factor of [tex]n^2[/tex] so they cannot be possibly equal if [tex]|n|\geq 2[/tex] because in that case [tex]|n|<|n|^2[/tex] which will violated equality.
  4. Jul 14, 2007 #3
    I think I understand their reasoning now...Thank you!

  5. Jul 14, 2007 #4
    The approach to this proof is very nice. Especially how Parseval's Identity appeared.

    However, I still like to see Lagrange's Original proof via Calculus of Variations. Do you know where I can find one?
  6. Jul 14, 2007 #5
    I haven't studied the Calculus of Variations yet, but I did have a look around the Internet. I came across a books called Introduction To The Calculus Of Variations, and found a "limited preview" of it on Google Books:

    (Sorry about the hideous link)

    A section (with proof, I think) on the isoperimetric inequality starts on page 153 (which, luckily, is in the "limited preview". However, the proof there does not appear to be that of Lagrange (it actually looks very much like the proof from Stein and Shakarchi above!), and, because of my lack of knowledge of the Calculus of Variations, this information might not be very relevant. Perhaps it will be useful though...

    You might also want to have a look at the document "Inequalities that Imply the Isoperimetric Inequality" by Andrejs Treibergs: http://www.math.utah.edu/~treiberg/isoperim/isop.pdf" [Broken]

    Some of the proofs omitted from this document are listed, and unfortunately, a "Calculus of Variations" proof is among them. However, a reference in which you can find such proofs are given, which might prove useful.

    Last edited by a moderator: May 3, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook