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I'm trying to prove an isoperimetric inequality, but I have absolutely no idea how to go about it.

let $\Gamma$ be a closed plane curve parametrized by $\gamma(t) = (x(t), y(t))$ on $[-\pi, \pi]$. Let $l$ denote the length of the curve and $d = \sup_{t_1, t_2 \in [-\pi, \pi]} |\gamma(t_1) - \gamma(t_2)|$.

If $\Gamma$ is convex then $l\leq \pi d$.

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Generally speaking, isoperimetric refers to an inequality between area and perimeter, or between their analogues in higher dimensions. There is also isodiametric inequality between area and diameter. Both of those hold for quite general sets, as long as one can make sense of area and perimeter.

The inequality you stated, $l\le \pi d$, is special to convex curves. Calling it an isoperimetric inequality is somewhat misleading (it misled two commentators so far). I don't know if there is a standard name for perimeter-diameter inequalities. Any descriptive name such as length-diameter inequality for convex curves would do.

Proof. Let $z(t)=x(y)+iy(t)$, using complex notation. For every $\theta\in [0,\pi]$ the function $f_\theta(t)=\operatorname{Re} (e^{i\theta} z(t))$ takes values within some interval of length at most $d$. By convexity, this function has two intervals of monotonicity on the circle (I think of $[-\pi,\pi]$ as the circle, gluing the endpoints together). Therefore, its total variation is at most $2d$. The total variation is the integral of absolute value of derivative, provided that we chose a sensible (absolutely continuous) parametrization $\gamma$: the arclength parametrization would do, for example. Hence, $$\int_{-\pi}^{\pi} |\operatorname{Re} (e^{i\theta} z'(t))|\,dt \le 2d \tag1$$ Integrate (1) with respect to $\theta \in [0,\pi]$ and exchange the order of integration: $$\int_{-\pi}^{\pi} \int_0^\pi |\operatorname{Re} (e^{i\theta} z'(t))|\,d\theta\,dt \le 2\pi d \tag2$$ A little calculation shows that for any complex number $\zeta$ we have $$ \int_0^\pi |\operatorname{Re} (e^{i\theta} \zeta)|\,d\theta = 2|\zeta| \tag3$$ (This boils down to $\int_0^\pi \sin \theta\,d\theta=2$.)

From (2) and (3) it follows that $$\int_{-\pi}^{\pi} |z'(t)|\,dt \le \pi d $$ which was to be proved. $\quad \Box$


Remarks

  1. The above proof does not use Fourier analysis. I don't know of any way to prove $l\le \pi d$ with Fourier analysis.
  2. It's not necessary to use complex notation. It just makes for a convenient formula for orthogonal projection onto a line: $z\mapsto \operatorname{Re} (e^{i\theta} z)$
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Here's my attempt at a proof using Fourier Analysis. This follows the problem in Stein and Shakarchi's Fourier Analysis text, chapter 4 problem 1.

First show that $2|a_n|\le d$ for all $n\ne 0$, where $a_n$ is the $n$th Fourier coefficient. To do this, we note that $$\begin{array}{ll} a_n &= \frac 1 {2\pi} \int_{-\pi}^{\pi} \! \gamma(t) e^{-int}\, dt\\ &= \frac {-1} {2\pi} \int_{-\pi}^{\pi}\gamma(t+\tfrac{\pi}{n}) e^{-int}\,dt \end{array}$$ so we have $$2a_n=\frac{1}{2\pi} \int_{-\pi}^{\pi} \! [\gamma(t)-\gamma(t+\tfrac{\pi}{n}]e^{-int}\, dt$$ and therefore $$2|a_n|\le \frac{1}{2\pi} \int_{-\pi}^{\pi} \! |\gamma(t)-\gamma(t+\tfrac{\pi}{n})|\, dt\le d.$$

Now suppose $\Gamma$ is convex. Then (this is the part that I need some additional justification on) we can parametrize $\Gamma$ by a curve $\gamma$ such that $$\gamma'(t)=ie^{it}z(t)\quad\text{where }z(t):[-\pi,\pi]\to\mathbb{R}_{\ge0}.$$ Assuming we have this, we use that $\widehat{f'}(n)=in\hat f(n)$ to conclude

$$\frac{1}{2\pi}\int_{-\pi}^\pi ie^{it}z(t)e^{-it}\,dt = i\frac{1}{2\pi}\int_{-\pi}^\pi z(t) \, dt = ia_1.$$

Now $$\ell=\int_{-\pi}^\pi\!|\gamma'(t)|\,dt=\int_{-\pi}^\pi z(t)\,dt=2\pi a_1 \le \pi d$$ by the prior work.

If someone could verify the validity of this approach, or point out any errors, I would appreciate it.

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