# Isoperimetric inequality implies Wirtinger's inequality

Let $C: x=x(t), y=y(t), a\le t\le b$ be a $C^1$ closed curve (not necessarily simple).The isoperimetric inequality says that $$A\le \frac{\ell^2}{4\pi},$$ where $$A=\left|\int_C y(t)x'(t) dt\right|$$ is the area enclosed by $C$, and $\ell=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2} dt$ is the arc length of $C$. My question is how to use this theorem to prove Wirtinger theorem: If $f(t)$ is a $T$-periodic $C^1$ real-valued function such that $\int_0^T f(t) dt=0,$ then $$\int_0^T |f(t)|^2 dt\le \frac{T^2}{4\pi^2}\int_0^T |f'(t)|^2 dt.$$

-

First thing to note that if we re-parametrize $t = ks$, we have, by the chain rule, $\frac{d}{ds}f = k (\frac{d}{dt}f)\circ t$. Choose $k = T/(2\pi)$, then the change of variable shows that it is sufficient to prove the claim for the period being $2\pi$. That is, if we can show for any $2\pi$ periodic function with mean 0

$$\int_0^{2\pi} f^2 ds \leq \int_0^{2\pi} |f'|^2 ds$$

we'll be done by a re-scaling argument.

Since $f$ has mean 0, you can write $f = F'$ for $F$ another $2\pi$ periodic function. So the isoperimetric inequality implies

$$\left|\int_0^{2\pi} f F' ds \right| \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{ (F')^2 + (f')^2 } ds \right)^2$$

or

$$\int_0^{2\pi} f^2 ds \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{f^2 + (f')^2} ds \right)^2$$

Now use Holder's inequality on the finite interval $[0,2\pi]$, we get

$$\int_0^{2\pi} \sqrt{f^2 + (f')^2} ds \leq \sqrt{2\pi} \left( \int f^2 + (f')^2 ds \right)$$

So we get

$$\int_0^{2\pi} f^2 ds \leq \frac{1}{2}\int_0^{2\pi} f^2 + (f')^2 ds$$

which, subtracting $\frac12 \int f^2$ from both sides yields the desired inequality.

Now, a short remark on why it is necessary to first use the scaling argument. The principle is the following: Wirtinger's inequality, as discussed in the first paragraph above, is scale invariant: changing the scale of parameter $t$ changes the terms $f^2$ and $T^2 (f')^2$ equally.

The isoperimetric inequality, however, is not scale invariant in the same way: using the ansatz where $x$ corresponds to $F$ and $y$ corresponds to $F' = f$, you see that a change of parametrization $t \to ks$ will leave $x$ the same while changing $y$ by a multiple factor.

In particular, this means that the depending on scale, the inequality may not be sharp. In other words, the changing of scale corresponds, morally speaking, to changing the $x$ and $y$ directions in different proportions. So if we start out with a circle, which is a maximizer of the isoperimetric inequality, after this funny reparametrisation which scales $x$ and $y$ differently, we end up with an ellipse, which no longer is a maximizer.

The change of scale in the first paragraph allows us to "use a version of the isoperimetric inequality as close to the circle version as possible". In other words, by isolating the scale you can most efficiently use the isoperimetric inequality, which then allows for a simple proof of Wirtinger's inequality.

-
To clarify a bit on the paragraphs after the rule: you can still prove, by brute-force, Wirtinger from Isoperimetric without the normalisation of $T = 2\pi$. But in the step where I applied Holder inequality (or Cauchy Schwarz if you prefer) to $\sqrt{f^2 + (f')^2} = \sqrt{f^2 + (f')^2}\cdot 1$, it will be necessary to use a different, likely much more complicated splitting to properly optimize the final result. – Willie Wong Dec 21 '10 at 3:32
can you please elaborate how did you got the second inequality? The trick is to define $x(t)$ and $y(t)$, but obvious choice $x(t)=t$, $y(t)=f(t)$ clearly does not work. – mpiktas Dec 21 '10 at 8:07
@mpiktas. Define $x(t)=\int_0^t f(s) ds, y(t)=f(t), 0\le t\le 2\pi$. The condition $\int_0^{2\pi}f(t) dt=0$ implies that the curve is closed. – TCL Dec 21 '10 at 14:28
@mpiktas: right, @TCL already explained it. I thought it was obvious that the introduction of the $F$ means I am setting $x = F$. I suppose it is not. Let me add a line to clarify. Also, note that setting $x(t) = t$ contradicts the assumption that you are describing a closed curve, since $x(2\pi) = 2\pi \neq x(0) = 0$, and so in fact is not an obvious choice. – Willie Wong Dec 21 '10 at 15:31
@TCL, @Willie Wong, thanks for explanation. – mpiktas Dec 21 '10 at 21:17