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 $\frac12 \left |\int_a^b (x(s)y'(s)-y(s)x'(s))\,ds\right |$
