1
$\begingroup$

Show that the isoperimetric inequality is equivalent to Wirtinger's inequality, which says that if $f$ is $2\pi$-periodic, of class $C^1$, and satisfies $\int_0^{2\pi}f(t)dt=0$, then $$\int_0^{2\pi}|f(t)|^2dt\leq \int_0^{2\pi}|f'(t)|^2dt$$ with equality if and only if $f(t) = A\sin t + B \cos t$

Part of the proof:
(Isoperimetric $\implies$ Wirtinger)
If we re-parametrize $t = ks$ with $k = T/(2\pi)$, then the change of variable shows that it is sufficient to prove the claim for the period being $2\pi$.
Given f with mean 0. we found $F(t) :=\int_0^t f(s) ds$ is a $2\pi$-periodic function. So the isoperimetric and Hölder inequalities imply $$\int_0^{2\pi}f^2(t)dt=\int_0^{2\pi}f(t)F'(t)dt$$ $$ \leq \frac 1 {4\pi}\left (\int_0^{2\pi}\sqrt {(f'(t))^2 + (F'(t))^2}dt\right )^2 \tag{1}$$ $$ \leq \frac 1 2 \int_0^{2\pi}(f'(t))^2 +f(t)^2 dt \tag{2}$$ which is equivalent to the Wirtinger's inequality (How?).
(Is $\frac 1 2 \int_0^{2\pi}(f'(t))^2 +f(t)^2 dt$ equivalent to $\int_0^{2\pi}|f'(t)|^2dt$? )

Could you also show how to arrive $(1)$ and $(2)$ ?

Some defs and thms:

  • If $\Gamma$ is parametrized by $\gamma (s) = (x(s), y(s))$, then the length of the curve $\Gamma$ is defined by $l=\int_a^b |\gamma '(s)|ds=\int_a^b{x'(s)^2+ y'(s)^2}ds$

  • (the isoperimetric inequality): Suppose that $\Gamma$ is a simple closed curve in $\mathbb R^2$ of length $l$, and let $\mathcal A$ denote the area of the region enclosed by this curve. Then $$\mathcal A\leq \frac {l^2}{4\pi}$$ with equality if and only if $\Gamma$ is a circle.

  • the area $\mathcal A $ of the region enclosed by a simple closed curve $\Gamma$ is given by $\frac12 \left |\int_a^b (x(s)y'(s)-y(s)x'(s))\,ds\right |$

$\endgroup$
0
$\begingroup$

For step 1, use isoperimetric inequality: $$\mathcal A \le \frac{l^{2}}{4\pi}$$ Which can also be written as $$\int _{a}^{b}x(s)y^{'}(s)ds \le \int _{a}^{b}(x^{'}(s)^{2}+y^{'}(s)^{2})^{1/2}ds.$$ if $\Gamma$ is parametrized by $\gamma=(x(s),y(s))$ Then substitute $f(t)=x(t)$ and $F^{'}(t)=y^{'}(t)$. For step 2, use Holder inequality: $$\int_{a}^{b}| f(x)g(x) |dx \le (\int_{a}^{b} | f(x)|^{p}dx)^{1/p}(\int_{a}^{b}| g(x)| ^{q}dx)^{1/q}$$ For $(q, p) \in (1,\infty), \frac{1}{p}+\frac{1}{q}=1$

Substitute $g(t)=1$, $f(t)=\sqrt{f^{'}(t)^{2}+F^{'}(t)^{2}}dt$ and $p=q=2$ to get

$$\int_{0}^{2\pi}\sqrt{f^{'}(t)^{2}+F^{'}(t)^{2}} \le \sqrt{2\pi} \sqrt{\int_{0}^{2\pi}(f^{'}(t))^{2}+(f(t))^{2}dt.} $$

Then

$$\int_{0}^{2\pi} f^{2}(x)dt\le \frac{1}{2}\int_{0}^{2\pi} (f^{'}(t))^{2}+(f(t))^{2}dt \to \frac{1}{2}\int_{0}^{2\pi}f^{2}(t)dt\le \frac{1}{2}\int_{0}^{2\pi} (f^{'}(t))^{2}dt.$$ Which is just Wirtinger's inequality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.