# Geometric implication of the Sobolev embedding

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?

• In short, one generalizes the Sobolev embedding to $BV \subset L^{n/(n-1)}$, where $BV$ stands for functions of bounded variations (i.e., functions for which weak derivatives are measures). Then if the set $A$ satisfies $\chi_A \in BV$, the inequality $\| \chi_A \|_{L^{n/(n-1)}} \le C \| \chi_A \|_{BV}$ has a geometric meaning. I will try to make it an answer when I have the time. Jul 6, 2021 at 22:41
• @MichałMiśkiewicz Thank you very much, I definitely want a more detailed answer. Jul 6, 2021 at 22:48
• The 2D case can be found in this survey by Osserman: projecteuclid.org/download/pdf_1/euclid.bams/1183541466 The higher dimensional case is essentially the same. Jul 7, 2021 at 4:19
• @Deane thanks, but I do not think this is a valid answer. The paper you mentioned is way too long for my question Jul 7, 2021 at 4:40
• It's a survey paper, so it covers many aspects of the isoperimetric inequality. The 2-dimensional case of your question is answered in section 3, specifically Theorem 3.1 on page 1192. Jul 7, 2021 at 13:51

$$\newcommand{\R}{\mathbb{R}}$$ 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 $$(\mathcal{H}^n(A))^{\frac{n-1}{n}} \le C \mathcal{H}^{n-1} (\partial A).$$

$${}^1$$ 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.

• +1. Very nice answer. In the footnote, the 1D case should use the essential variation, however. Jul 7, 2021 at 10:56
• You're right. I edited to mention a representative, which I hope solves the problem. Jul 7, 2021 at 12:06
• @Deane, you beat me to it. I added an answer with the derivation.
– user711689
Jul 7, 2021 at 15:33
• @Deane I see your point. I personally like BV functions very much and I thought it's a good place to introduce them. For BV functions, "geometric" and "analytic" properties have a direct interplay. Jul 9, 2021 at 7:23
• @FeiCao It's not silly, but it's really a separate question. I suppose the real question is: why the norm of $\vec{n} \mathcal{H}^{n-1} \llcorner \partial A$ equals $\mathcal{H}^{n-1}(\partial A)$? The derivation is straightforward, but it depends on your definition of the total variation norm. Jul 9, 2021 at 7:25

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 $$\mathcal{H}^{d}(A) = 0$$.

• Thank you very much! Going from the other side is also very interesting! Jul 7, 2021 at 17:06
• Thanks! Another byproduct of the equivalence is that the best constant in Sobolev embedding is exactly the isoperimetric constant, right? Jul 9, 2021 at 7:29
• @MichałMiśkiewicz, yeah, I forgot about that. It's the same constant all the way through, and you can approximate $\chi_{B(0,1)}$ by smooth bump functions so the inequality is sharp. (Equality is never attained by a $W^{1,1}(\mathbb{R}^{d})$ function, though, since the Minkowski inequality above will have to be strict when $u$ has more than one level set.)
– user711689
Jul 9, 2021 at 14:38