Sharp planar isoperimetric inequality for finite-perimeter sets

First suppose $G\subset {\mathbb{R}}^2$ is a bounded open set with compact smooth boundary $\mathrm{\Gamma }$. The boundary is a compact smooth one-dimensional manifold without boundary. Connected coordinate arcs cover $\mathrm{\Gamma }$; compactness gives a finite subcover, so $\mathrm{\Gamma }$ has finitely many connected components. By [F5] each component is a smooth embedded circle. Parametrize component $i$ by an orientation-compatible unit-speed smooth periodic curve ${\gamma }_i:[0,L_i]\to {\mathbb{R}}^2$, where $L_i$ is its length. The divergence formula in [F3] gives $D{\chi }_G=-\nu {\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\Gamma }}$ with $\nu$ the outward normal, and [F4] therefore gives $P(G;{\mathbb{R}}^2)={\mathcal{H}}^1(\mathrm{\Gamma })={\sum }_i{L}_i$. Applying Stokes in [F3] to $\omega =(xdy-ydx)\mathrm{/}2$, whose exterior derivative is $dx\wedge dy$, gives $\mathrm{\mid }G\mathrm{\mid }=\frac{1}{2}{\sum }_i{\int }_0^{L_i}({\gamma }_{i,1}{\gamma }_{i,2}^{\mathrm{\prime }}-{\gamma }_{i,2}{\gamma }_{i,1}^{\mathrm{\prime }})ds\mathrm{.}$ For each $i$, translating ${\gamma }_i$ by its arclength mean ${\bar{\gamma }}_i=L_i^{-1}{\int }_0^{L_i}{\gamma }_i(s)ds$ does not change this integral, since each coordinate derivative integrates to zero over a period. Write $z_i={\gamma }_i-{\bar{\gamma }}_i$.

Now let $E$ be as in the Statement and put $u={\chi }_E$. By [F1], choose $u_j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ with $e_j:=\mathrm{\parallel }{u}_j-{\chi }_E{\mathrm{\parallel }}_1\to 0$ and $a_j:=\int \mathrm{\mid }\mathrm{\nabla }{u}_j\mathrm{\mid }\to P(E;{\mathbb{R}}^2)$. If $\mathrm{\mid }E\mathrm{\mid }=0$, the result is immediate. If some $e_j=0$, continuity forces $u_j$ to take only the values $0$ and $1$ everywhere; connectedness of ${\mathbb{R}}^2$ and compact support then force $u_j=0$, contradicting $\mathrm{\mid }E\mathrm{\mid }>0$. Thus $e_j>0$. Let ${\tau }_j=\sqrt{e_j}$ and discard finitely many terms so $0<{\tau }_j<\min \{1\mathrm{/}2,\mathrm{\mid }E\mathrm{\mid }\mathrm{/}2\}$. For every $t\in I_j:=[{\tau }_j,1-{\tau }_j]$, a point where ${\chi }_E$ and ${\chi }_{\{u_j>t\}}$ differ contributes at least ${\tau }_j$ to $\mathrm{\mid }{u}_j-{\chi }_E\mathrm{\mid }$; hence $\mathrm{\mid }\{u_j>t\}\mathrm{△}E\mathrm{\mid }\le e_j\mathrm{/}{\tau }_j={\tau }_j$.

For a smooth $L$-periodic scalar function $z$ of mean zero, set $v(\theta )=z(L\theta \mathrm{/}(2\pi ))$. Its zeroth Fourier coefficient vanishes. Integration by parts gives $\widehat{v^{\mathrm{\prime }}}(k)=ik\widehat{v}(k)$ for every integer $k$. Parseval [F6] and $k^2\ge 1$ for $k\mathrm{\ne }0$ imply ${\int }_0^{2\pi }\mathrm{\mid }v{\mathrm{\mid }}^{2}d\theta \le {\int }_0^{2\pi }\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}d\theta$. Changing variables yields ${\int }_0^L\mathrm{\mid }z(s){\mathrm{\mid }}^{2}ds\le {\left(\frac{L}{2\pi }\right)}^2{\int }_0^L\mathrm{\mid }{z}^{\mathrm{\prime }}(s){\mathrm{\mid }}^{2}ds\mathrm{.}$ Apply this to both coordinates of $z_i$ and add. Since $\mathrm{\mid }{z}_i^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{\gamma }_i^{\mathrm{\prime }}\mathrm{\mid }=1$, the Cauchy–Schwarz inequality gives $\mid \frac{1}{2}{\int }_0^{L_i}(z_{i,1}{z}_{i,2}^{\mathrm{\prime }}-z_{i,2}{z}_{i,1}^{\mathrm{\prime }})ds\mid \le \frac{1}{2}\mathrm{\parallel }{z}_i{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{z}_i^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\le \frac{L_i^2}{4\pi }\mathrm{.}$ Summing over the finitely many boundary components and using ${\sum }_i{L}_i^2\le ({\sum }_i{L}_i{)}^2$ proves $4\pi \mathrm{\mid }G\mathrm{\mid }\le P(G;{\mathbb{R}}^2{)}^2$.

By [F2], the measurable level-perimeter function has integral at most $a_j$ on $I_j$, whose length is $1-2{\tau }_j$. For any $\epsilon >0$, the set of $t\in I_j$ with $P(\{u_j>t\};{\mathbb{R}}^2)\le a_j\mathrm{/}(1-2{\tau }_j)+\epsilon$ has positive measure; otherwise its integral over $I_j$ would exceed $a_j$. The critical values of $u_j$ form a null set by [F2], so take $\epsilon =1\mathrm{/}(j+1)$ and choose such a regular value $t_j$. Since $t_j>0$ and $u_j$ is compactly supported, $G_j:=\{u_j>t_j\}$ is bounded and its closure is a compact smooth domain with boundary $\{u_j=t_j\}$. Thus step 2.1 applies to $G_j$, and $\mathrm{\mid }{G}_j\mathrm{△}E\mathrm{\mid }\le {\tau }_j,P(G_j;{\mathbb{R}}^2)\le \frac{a_j}{1-2{\tau }_j}+\frac{1}{j+1}\mathrm{.}$ The right side has limit $P(E;{\mathbb{R}}^2)$, while $\mathrm{\mid }{G}_j\mathrm{\mid }\to \mathrm{\mid }E\mathrm{\mid }$.

Step 2.1 gives $4\pi \mathrm{\mid }{G}_j\mathrm{\mid }\le P(G_j;{\mathbb{R}}^2{)}^2$ for every selected $j$. Taking upper limits, the convergence $\mathrm{\mid }{G}_j\mathrm{\mid }\to \mathrm{\mid }E\mathrm{\mid }$ and the perimeter upper bound from step 3.1 give $4\pi \mathrm{\mid }E\mathrm{\mid }\le P(E;{\mathbb{R}}^2{)}^2$. This proves the claimed inequality for every finite-measure finite-perimeter set.