What is the most elementary but still correct according to the most rigorous standard proof of the isoperimetric inequality?

Question

Can you write the most elementary proof of the isoperimetric inequality (but still correct according to the most rigorous standard )?

$$l^2> 4πA$$

Answer 1

The most elementary proof can be found in the book by Burago and Zalgaller:

Burago, Yu. D.; Zalgaller, V. A. Geometric inequalities. Translated from the Russian by A. B. Sosinskiĭ. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 285. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1988.

This proof is completely combinatorial and therefore "elementary". In particular, it does not use Fourier series. The proof replaces the Jordan curve by an approximating polygon and then applies straightforward techniques on area growth to get the result.

Answer 2

My favorite one uses only a bit of Fourier series; I consider it more elementary than variational approach or Steiner symmetrization method. Think of the plane as complex plane. Parametrize the boundary curve $\Gamma$ counterclockwise with constant speed: that is, take a map $f:[0,1]\to \mathbb C$ that traces the curve once and satisfies $|f'|=L$ almost everywhere. Expand it into Fourier series: $$f(t) = \sum_{n\in\mathbb Z} c_n e^{2\pi int}\tag{1}$$ Then $$f'(t) = \sum_{n\in\mathbb Z} c_n (2\pi i n) e^{2\pi int}\tag{2}$$ (I don't claim pointwise convergence, but since $f'\in L^2$, we have convergence in $L^2$.) Since the functions $e^{2\pi int}$ form an orthonormal basis for $L^2[0,1]$, it follows that $$ L^2 = \int_0^1 |f'(t)|^2 \,dt = 4\pi^2 \sum_{n\in\mathbb Z} n^2 |c_n|^2 \tag{3}$$ On the other hand, the area bounded by $\Gamma$ can be computed with Green's formula, $A=\frac12 \int (y\,dx-x\,dy)$. In our notation, $$y\,dx-x\,dy =\operatorname{Im}(x+iy)(dx-idy)) = \operatorname{Im}(f(t)\overline{f'(t)}) \,dt$$ Using (1) and (2) again, we obtain $$ A = \frac12 \int_0^1 \operatorname{Im}(f(t)\overline{f'(t)}) \,dt = \pi \sum_{n\in \mathbb Z} n |c_n|^2 \tag{4} $$ Since $n\le n^2$ for all $n\in\mathbb Z$, comparison of (3) and (4) shows that $$ \frac{A}{\pi}\le \frac{L^2}{4\pi^2} $$