# An inequality inspired by the isoperimetric inequality

Let us consider the simplest isoperimetric inequality. Consider a smooth simple closed curve given by $r=\rho(\theta)$ in polar coordinates, where $\rho(\theta)>0$ can be regarded as a smooth periodic function with period $2\pi$.

As $dA=rdrd\theta$, the area of the region $\Omega$ enclosed by the closed curve is \begin{align*} A=\iint_{\Omega}rdrd\theta =\frac{1}{2}\int_0^{2\pi}\rho(\theta)^2 d\theta. \end{align*} On the other hand, it is an easy exercise in calculus that the length of the curve is given by \begin{align*} L=&\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta. \end{align*}

By the isoperimetric inequality, we have $L^2\ge 4\pi A$, so we must have $$2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le \left(\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta\right)^2.$$

This looks similar to the Wirtinger's inequality, which states that $$\begin{split} \int_{0}^{2\pi}\rho'(\theta)^2d\theta \ge \int_{0}^{2\pi} \left(\rho(\theta)-\overline \rho\right)^2 d\theta = \int_{0}^{2\pi} \rho(\theta)^2 d\theta-\frac{1}{2\pi} \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2. \end{split}$$

So \begin{align*} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le& \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2+2\pi\int_{0}^{2\pi}\rho'(\theta)^2d\theta, \end{align*} which isn't quite what I want.

It doesn't seem right to me that we can apply Wirtinger's inequality directly to prove it because the equality case in Wirtinger's inequality holds when $\rho(\theta)=C+a\cos \theta+ b\sin \theta$, but the equality in our inequality can only hold when $\rho$ is constant the equality in our inequality doesn't hold in this case. (However, by geometric consideration, the equality does hold for, say, $\rho(\theta)=\sin \theta$, but only if we restrict $\theta$ to $[0, \pi]$.)

This is where I am stuck. So my question is: Can we show this inequality without using the isoperimetric inequality (say by Fourier analysis or using Wirtinger's inequality more carefully)? Can it be used to prove (at least a simple case of) the isoperimetric inequality? If not, why? (If nothing else, at least I obtain an inequality on circle :-) )

To elaborate further, it is well-known that we can apply Wirtinger's inequality (or Fourier type argument) to prove the isoperimetric inequality on the plane. Indeed, the Wirtinger's inequality and the isoperimetric inequality are equivalent (e.g. Osserman's paper on isoperimetric inequality). Usually, these kinds of proofs involve shifting the center of mass to $0$, applying the Green's theorem and the Wirtinger's (or Cauchy-Schwarz) inequality on some combination of two functions (say $x(s), y(s)$). So as a subquestion, why is there no such argument involving only a single function (say $\rho(\theta)$)?

• but the equality in our inequality can only hold when $\rho$ is constant Not really: it holds for circles containing the origin but not all such circles are centered at the origin. You have actually noticed it yourself in the next sentence. – fedja Jul 22 '17 at 3:17
• @user50396 One dimensional inequality: you can always try the standard Bellman function approach like I discussed here before mathoverflow.net/questions/199418/… using Hamilton--Jacobi--Bellman PDE – Paata Ivanishvili Jul 22 '17 at 22:44
• If you are interested I can elaborate little bit more: it reduces the problem to finding a certain Bellman function of 4 variables with a very simple obstacle condition. – Paata Ivanishvili Jul 22 '17 at 22:46
• I think your problem is that an arbitrary (even convex) simple, closed curve doesn't have a polar representation $r(\theta) e^{i\theta}$ the points on the curve are one to one with $\theta \in [0, 2\pi)$. For that you need to do something like translate the curve so it encloses the origin in such a way that the resulting curve becomes star shaped around the origin (so convex if fine). Then I suspect you are just working backwards from Wirtinger $\Rightarrow$ Isoperimetric inequality modulo translation. – Paul Bryan Jul 23 '17 at 1:17
• @PaataIvanisvili I am not familiar with this approach. Can you elaborate more? – user50396 Jul 24 '17 at 12:57

I will illustrate the Bellman function approach to prove Wirtinger's inequality which, of course, is simpler than the original problem. The advantage of the approach is that it does not use any Fourier analysis (which apparently is the best thing to do for this particular problem since $f$ is periodic), Cauchy--Schwarz, or variational calculus such as Euler--Lagrange equation. If somebody finds the approach interesting you can try to use it to prove the original problem (or maybe I will try to do it myself but later), and as Paul Bryan noticed, unfortunately it will only give you isoperimetric inequality for the sets whose boundary has a nice 1-1 parametrization in polar coordinate systems (for example, star shaped sets).

Consider the function of 4 variables

$$M(t,x,y,z):=\frac{2tx^{2}(\cos(t)-1)-y^{2}\sin(t) +z^{2}(t\cos(t)-\sin(t))+2(1-\cos(t))(2xy+xzt-yz)}{2-2\cos(t)-t\sin(t)}$$

defined in the domain $(0,2\pi)\times \mathbb{R}^{3}$. In what follows $M_{t}, M_{x}, M_{y}, M_{z}$ denote partial derivatives of $M$.

Claim 1: For any $v \in \mathbb{R}$ and all $(t,x,y,z)\in (0,2\pi)\times \mathbb{R}^{3}$ we have $$x^{2}-v^{2}\leq M_{t}+v M_{x}+xM_{y}+vM_{z}. \qquad (*).$$

Proof: Optimize the inequality over all $v$. The optimal value is attained when $v=-\frac{M_{x}+M_{z}}{2}$. Therefore it is enough to have $$x^{2}\leq -\left(\frac{M_{x}+M_{z}}{2}\right)^{2}+M_{t}+xM_{y}. \qquad (**)$$ After straightforward calculations one notices that the inequality $(**)$ is in fact equality! The details are left to the reader.

Claim 2: For any $f\in C^{1}([0,2\pi])$ we have $$\int_{0}^{2\pi} f^{2}-(f')^{2}\leq \limsup_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right) - \liminf_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right).$$

Proof: Notice that $$f^{2}(t)-(f'(t))^{2}\leq \frac{d}{dt} M\left(t, f(t), \int_{0}^{t}f, \int_{0}^{t}f' \right). \qquad (***)$$ Indeed, after taking the derivative the inequality $(***)$ simplifies to $$f^{2}(t)-(f'(t))^{2}\leq M_{t}+M_{x}\, f'(t)+M_{y}\,f(t)+M_{z}\,f'(t).$$ The latter follows from $(*)$ where we take $v=f'(t), \;x=f(t), \;y=\int_{0}^{t} f$ and $z=\int_{0}^{t}f'$. Finally we just integrate $(***)$ in $t$ from $t_{1}$ to $t_{2}$ where $t_{1}, t_{2}\in (0,2\pi)$, and take the lower and upper limits $t_{1}\to 0$ and $t_{2} \to 2\pi$.

Claim 3: Let $f\in C^{1}([0,2\pi])$ be such that $f(0)=f(2\pi)$ and $\int_{0}^{2\pi}f=0$. Then $$\int_{0}^{2\pi} f^{2}-(f')^{2}\leq 0$$ Proof: Indeed, using Claim 2 it will be enough to show that \begin{align*} &\limsup_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)= \lim_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right) =0\\ &\liminf_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)=\lim_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)=0. \end{align*} These inequalities roughly speaking follow from the following observations \begin{align*} &\lim_{\delta \to 0}\; M(2\pi-\delta, f(0)-\delta f'(0),-\delta f(0), -\delta f'(0))=0\\ &\lim_{\varepsilon \to 0}\; M(\varepsilon, f(0)+\varepsilon f'(0), \varepsilon f(0), \varepsilon f'(0))=0 \end{align*} for any $f(0)$ and $f'(0)$. The end.