From:
Sobolev inequality: For all $u\in C_c^{\infty}(\mathbb{R}^n)$ $$\|u\|_{L^{\frac{n}{n-1}}(\mathbb{R}^n)}\leq C \|\nabla u\|_{L^1(\mathbb{R}^n)}.$$
I want to prove:
Isoperimetric inequality: For every bounded open set $E\subseteq\mathbb{R}^n$ with boundary $\partial E$ of class $C^1$, $$|E|^{\frac{n-1}{n}}\leq C|\partial E|.$$
The idea is to put $u=\chi_E$ in the Sobolev inequality, because in such a case $\|u\|_{L^{\frac{n}{n-1}}(\mathbb{R}^n)}=|E|^{\frac{n-1}{n}}$ is the left hand-side of the isoperimetric inequality. The problem is that $\chi_E$ is not in the Sobolev space $W^{1,1}(\mathbb{R}^n)$.
Thus, we regularize via convolutions: define $u_\epsilon=\chi_E\ast \rho_\epsilon$, where $\rho_\epsilon(x)=(1/\epsilon^n)\rho(x/\epsilon)$, $\rho\in C_c^\infty(\mathbb{R}^n)$, $\text{support}(\rho)\subseteq B(0,1)$, $\rho\geq0$ and $\int \rho=1$. We have: $$C\|\nabla u_\epsilon\|_{L^1(\mathbb{R}^n)}\geq \|u_\epsilon\|_{L^{\frac{n}{n-1}}(\mathbb{R}^n)}\stackrel{\epsilon\rightarrow0^+}{\longrightarrow} \|\chi_E\|_{L^{\frac{n}{n-1}}(\mathbb{R}^n)}=|E|^{\frac{n-1}{n}}.$$ We want to prove that $\|\nabla u_\epsilon\|_{L^1(\mathbb{R}^n)}\leq |\partial E|$ for all $\epsilon>0$. We use the following result:
For all $v\in C_c^{\infty}(\mathbb{R}^n)$ $$\int_{\mathbb{R}^n}|\nabla v|\,dx=\sup\left\{-\int_{\mathbb{R}^n}\nabla v\cdot X\,dx:\,X\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\,|X(x)|=\left(\sum_{i=1}^n (X^i(x))^2\right)^{\frac12}\leq 1\right\}.$$
It suffices then $$-\int_{\mathbb{R}^n}\nabla u_\epsilon\cdot X\,dx\leq |\partial E|$$ for all $X\in C_c^1(\mathbb{R}^n,\mathbb{R}^n)$ and $|X|=(\sum_{i=1}^n (X^i)^2)^{\frac12}\leq 1$. We have, using integration by parts, $$-\int_{\mathbb{R}^n}\nabla u_\epsilon\cdot X\,dx=\int_{\mathbb{R}^n}u_\epsilon\,\text{Div}X\,dx=\int_E \text{Div}X_\epsilon\,dx=\int_{\partial E}X_\epsilon\cdot\nu\,d\sigma,$$ where $(X_\epsilon)^i=\rho_\epsilon\ast X^i$. It suffices to see that $|X_\epsilon(x)|\leq 1$ using $|X(x)|\leq 1$. I was not able to prove this. Any idea?
Hölder's inequality/the trivial bound $ \lvert \int fg \rvert \leq \lVert f \rVert_{\infty} \lVert g \rVert_{1} $ gives \begin{align} \lVert X_{\epsilon} \rVert_{\infty} &= \sup_x \left\lvert \int \rho_{\epsilon}(x-y)X(y) \, dy \right\rvert \\ &\leq \sup_x \left( \sup_y \lvert X(y) \rvert \right)\left\lvert \int \rho_{\epsilon}(x-y) \, dy \right\rvert \\ &= \sup_x \left( \sup_y \lvert X(y) \rvert \right)\left\lvert \int \rho_{\epsilon}(y) \, dy \right\rvert \\ &\leq 1 \end{align} since both terms are now independent of $x$ and $\int \rho_{\epsilon} = \int \rho = 1$.