Geometric implication of the Sobolev embedding

Question

It is stated in section 10 of this paper that the usual Sobolev embedding $$W^{1,1}(\mathbb{R}^n) \subset L^{n/(n-1)}(\mathbb{R}^n)$$ can be interpreted in geometrical terms as an isoperimetric statement. Although the authors said that this is well-known, I can not find such a statement in Evans classical book on PDEs. Can anyone elaborate the details here? That is, how can we derive an isoperimetric-type inequality from such a Sobolev embedding?

Answer 1

$\newcommand{\R}{\mathbb{R}}$ We will use the Sobolev embedding $W^{1,1}(\R^n) \subseteq L^{n/(n-1)}(\R^n)$ in the form $$ \| u \|_{\frac{n}{n-1}} \le C \| D u \|_1 \qquad \text{for } u \in W^{1,1}(\R^n). $$ This is probably what any proof gives you. If you only have $\| u \|_{\frac{n}{n-1}} \le C (\| u \|_1 + \| D u \|_1)$, you can use arbitrage to get rid of $\| u \|_1$ on the RHS${}^1$. $\newcommand{\eps}{\varepsilon}$

The next step is generalizing the inequality to $BV(\R^n)$, the space of functions of bounded variation${}^2$. This is similar to $W^{1,1}(\R^n)$, but instead of $u,Du \in L^1(\R^n)$ we require $u \in L^1(\R^n)$ and $Du \in M(\R^n)$ - distributional partial derivatives need to be representable by finite signed measures on $\R^n$. For each such $u$, one can take the approximation by convolution $u_\eps := u * \varphi_\eps$. It should be clear that $u_\eps \to u$ in $L^1(\R^n)$, and moreover $\| Du_\eps \|_1 \le \| Du \|_{M}$, where $M$ stands for the total variation norm of a (vector valued) measure. By Sobolev embedding, $$ \| u_\eps \|_{\frac{n}{n-1}} \le C \| D u_\eps \|_1 \le C \| D u \|_M $$ for each $\eps$. In $L^{\frac{n}{n-1}}(\R^N)$ one can take a weakly convergent subsequence, whose limit has to be $u$ (thank to $L^1$ convergence). In consequence, $$ \| u \|_{\frac{n}{n-1}} \le \liminf_{\eps \to 0} \| u_\eps \|_{\frac{n}{n-1}} \le C \| D u \|_M. $$

Finally, let us look at the geometric meaning of this. Consider $u$ to be a characteristic function of some set: $\chi_A$. If $A \subseteq \R^n$ is a bounded smooth set, the divergence formula $$ \int_A \operatorname{div} \varphi(x) \, dx = \int_{\partial A} \varphi(x) \cdot \vec{n}(x) \, d \mathcal{H}^{n-1}(x) $$ can be interpreted as integration by parts. In other words, it tells us that the distributional differential of $\chi_A$ is $$ D \chi_A = \vec{n} \mathcal{H}^{n-1} \llcorner \partial A, $$ the outer normal vector field on the boundary (with surface measure on the boundary). In this case, the total variation norm is $$ \| D \chi_A \|_M = \| \mathcal{H}^{n-1} \llcorner \partial A \|_M = \mathcal{H}^{n-1} (\partial A), $$ while the $L^{\frac{n}{n-1}}$ norm of $\chi_A$ is simply $(\mathcal{H}^n(A))^{\frac{n-1}{n}}$. Hence, the Sobolev embedding gives us $$ (\mathcal{H}^n(A))^{\frac{n-1}{n}} \le C \mathcal{H}^{n-1} (\partial A). $$

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${}^1$ The same trick also shows that $\frac{n}{n-1}$ is the only possible exponent on the LHS.

${}^2$ The name comes from the 1-dimensional case. It turns out $u \in BV([0,1])$ if and only if it has a representative for which the variation $$ \sup \left\{ \sum_{k=1}^n |u(t_k)-u(t_{k-1})| : 0 = t_0 < \ldots < t_n = 1 \right\} $$ is finite.

Answer 2

When I first saw this question, I thought it was asking for the opposite: derive the Sobolev embedding from the isoperimetric inequality. This is also possible and fascinating from my point of view so here's a proof.

Suppose we already know the isoperimetric inequality: \begin{equation*} \mathcal{H}^{d-1}(\partial A) \geq C \mathcal{H}^{d}(A)^{\frac{d-1}{d}}. \end{equation*} I will use this to derive the Sobolev embedding for $W^{1,1}(\mathbb{R}^{d})$. Above, we will only need to know that the isoperimetric inequality holds in the "nice" case that $A \subseteq \mathbb{R}^{d}$ is a smooth open set that is bounded.

By the coarea formula, if $u \in C^{\infty}_{c}(\mathbb{R}^{d})$, then \begin{equation*} \int_{\mathbb{R}^{d}} \|Du(x)\| \, dx = \int_{-\infty}^{\infty} \mathcal{H}^{d-1}(\{u = t\}) \, dt. \end{equation*} Sard's Theorem tells us that $\{u > t\}$ is a bounded smooth open set for almost every $t \in u(\mathbb{R}^{d}) = [\min \, u, \max \, u]$, and it's boundary is $\{u = t\}$. Therefore, for all $u \in C^{\infty}_{c}(\mathbb{R}^{d})$ with $u \geq 0$, the isoperimetric inequality gives \begin{equation*} \int_{\mathbb{R}^{d}} \|Du(x)\| \, dx \geq C \int_{0}^{\infty} \mathcal{H}^{d}(\{u > t\})^{\frac{d-1}{d}}\,dt \end{equation*} Finally, we use Minkowski's integral inequality: \begin{align*} \int_{0}^{\infty} \mathcal{H}^{d}(\{u > t\})^{\frac{d-1}{d}} \, dt &= \int_{0}^{\infty} \left[ \int_{\mathbb{R}^{d}} \chi_{\{u > t\}}(x)^{\frac{d}{d-1}} \, dx \right]^{\frac{d-1}{d}} \,dt \\ &\geq \left(\int_{\mathbb{R}^{d}} \left[ \int_{0}^{\infty} \chi_{\{u > t\}}(x) \,dt \right]^{\frac{d}{d-1}} \, dx \right)^{\frac{d-1}{d}} \\ &= \left(\int_{\mathbb{R}^{d}} u(x)^{\frac{d}{d-1}} \, dx \right)^{\frac{d-1}{d}}. \end{align*} Thus, \begin{equation*} \|Du\|_{L^{1}(\mathbb{R}^{d})} \geq C \|u\|_{L^{\frac{d}{d-1}}(\mathbb{R}^{d})} \quad \text{if} \, \, u \in C^{\infty}_{c}(\mathbb{R}^{d}), \, \, u \geq 0. \end{equation*} At the same time, if $u \in W^{1,1}(\mathbb{R}^{d})$, then $|u| \in W^{1,1}(\mathbb{R}^{d})$ and we can approximate it by a sequence of non-negative functions $(u_{n})_{n \in \mathbb{N}} \subseteq C^{\infty}_{c}(\mathbb{R}^{d})$ (e.g. by cut-off and mollification). This implies the Sobolev inequality in general since $\|D|u|\|_{L^{1}(\mathbb{R}^{d})} = \|Du\|_{L^{1}(\mathbb{R}^{d})}$.

Finally, notice that this shows that the best constant in the $W^{1,1}(\mathbb{R}^{d})$ Sobolev inequality is precisely the isoperimetric constant $C_{0}$, given by \begin{equation*} C_{0} = \sup \left\{ \frac{\mathcal{H}^{d-1}(A)}{\mathcal{H}^{d}(A)^{\frac{d-1}{d}}} \, \mid \, A \, \, \text{smooth and bounded} \right\} \end{equation*} which, incidentally, is attained whenever $A$ is a Euclidean ball. To see this, note that we could have used $C = C_{0}$ above so $C_{0}$ serves in the Sobolev inequality. On the other hand, if we set $u = \chi_{B(0,1)}$, then it is not hard to build smooth bump functions $(u_{n})_{n \in \mathbb{N}} \subseteq C^{\infty}_{c}(\mathbb{R}^{d})$ such that \begin{gather*} \|Du_{n}\|_{L^{1}(\mathbb{R}^{d})} \to \|Du\|_{TV(\mathbb{R}^{d})} = \mathcal{H}^{d-1}(B(0,1)), \\ \|u_{n}\|_{L^{\frac{d}{d-1}}(\mathbb{R}^{d})} \to \|u\|_{L^{\frac{d}{d-1}}(\mathbb{R}^{d})} = \mathcal{H}^{d}(B(0,1))^{\frac{d-1}{d}}. \end{gather*}
and, thus, \begin{equation*} C_{0} = \sup \left\{ \frac{\|Du\|_{L^{1}(\mathbb{R}^{d})}}{\|u\|_{L^{\frac{d}{d-1}}(\mathbb{R}^{d})}} \, \mid \, u \in W^{1,1}(\mathbb{R}^{d}) \right\}. \end{equation*}

I guess it is not possible to do this for $W^{1,p}(\mathbb{R}^{d})$ with $p > 1$ since, in that case, a characteristic function $\chi_{A}$ cannot have finite $W^{1,p}$ norm unless $\mathcal{H}^{d}(A) = 0$.