## Sobolev and Isoperimetric Inequality

In this short note I will prove the equivalence of the Sobolev inequality and the isoperimetric inequality. The material here is somewhat standard, and this note is mostly for my own benefit.

Let ${(M,g)}$ be an $n$-dimensional Riemannian manifold, $n\ge 2$. The isoperimetric property (or more commonly, isoperimetric inequality) states that there exists a constant ${C}$ such that for any relatively compact domain ${\Omega}$ with smooth boundary, we have

$\displaystyle C\mathrm{Vol}(\Omega)^{\frac{n-1}{n}}\le \mathrm{Area}(\partial \Omega). \ \ \ \ \ (1)$

The Sobolev inequality states that on a compact $n$-dimensional Riemannian manifold ${(M,g)}$ (possibly with boundary), there exists a constant ${C}$ (different from that in (1)) such that

$\displaystyle C(\int_M |f|^{\frac{n}{n-1}})^{\frac{n-1}{n}}\le \int_M |\nabla f|, \ \ \ \ \ (2)$

provided that ${\int_M f =0}$ if ${\partial M=\emptyset}$; otherwise we assume either ${\int_M f =0}$ or ${f\in W^{1,2}_0}$ (the completion of ${C_c^\infty}$ under the norm ${\|f\|^2=\int_M (|f|^2 +|\nabla f|^2) }$).

It is well-known that for a non-compact Riemannian manifold, the Sobolev inequality and the isoperimetric property may fail.

For example, let ${M}$ be the surface obtained by revolving the graph of ${1/x}$ about the ${x}$-axis. Then ${M}$ has a “cusp end” which has infinite area. On the other hand, for any ${A>0}$, we can choose ${\Omega}$ whose boundary consists of two circles of arbitrary small length on the cusp, such that the area of ${\Omega}$ is ${A}$. Therefore ${M}$ does not have the isoperimetric property. By suitably choosing ${f}$ to be a “bump function” on this ${\Omega}$, we can also see that the Sobolev inequality fails to hold on ${M}$ .

In this short note, we prove the following: for a general Riemannian manifold (in particular, non-compact), the isoperimetric property is equivalent to the Sobolev inequality.
Proof: Let us assume the Sobolev inequality. For ${\varepsilon>0}$, we define the function

$\displaystyle f_\varepsilon(x)= \begin{cases} 1, &\quad x\in \Omega, d(x, \partial \Omega)\ge \varepsilon,\\ \frac{d(x, \partial \Omega)}{\varepsilon}, &\quad x\in \Omega, d(x, \partial \Omega)\le \varepsilon,\\ 0, &\quad x\not\in \Omega. \end{cases}$

Apply the Sobolev inequality (2) to ${f_\varepsilon}$, and take ${\varepsilon\rightarrow 0}$, we can get the isoperimetric property (1).

Now we prove the converse. We can without loss of generality assume ${f\ge 0}$. By the layercake integral formula, we have

$\displaystyle \begin{array}{rl} \int_M |f|^{\frac{n}{n-1}} =&\int_0^\infty \mathrm{Vol}(f^{\frac{n}{n-1}}\ge t)dt\\ =&\int_0^\infty \mathrm{Vol}(f\ge t^{\frac{n-1}{n}})dt\\ =&\frac{n}{n-1}\int_0^\infty \mathrm{Vol}(f\ge s)s^{\frac{1}{n-1}}ds. \end{array}$

On the other hand, by the coarea formula and the isoperimetric inequality (1),

$\displaystyle \begin{array}{rl} \int_M |\nabla f| =&\int_0^\infty \mathrm{Area}(f=s)ds\\ \ge&C_1\int_0^\infty\mathrm{Vol}(f\ge s)^{\frac{n-1}{n}}ds \end{array}$

So to prove (2), it suffices to show that

$\displaystyle \left(\int_0^\infty \mathrm{Vol}(f\ge s)^{\frac{n-1}{n}}\right)^{\frac{n}{n-1}}\ge C_2 \int_0^\infty \mathrm{Vol}(f\ge s)s^{\frac{1}{n-1}}ds.$

Putting ${f=1}$ suggests that ${C_2= \frac{n}{n-1}}$ (though strictly speaking, such ${f}$ may not be admissible for the Sobolev inequality; nevertheless we can get around this problem by modifying ${f}$ to be a smooth bump function with range ${[0,1]}$ supported on a compact region, similar to ${f_\varepsilon}$ above). Therefore, if we let ${V(s)= \mathrm{Vol}(f\ge s)}$, then we have to show that

$\displaystyle \left(\int_0^\infty V(s)^{\frac{n-1}{n}}ds\right)^{\frac{n}{n-1}}\ge \frac{n}{n-1}\int_0^\infty V(s)s^{\frac{1}{n-1}}. \ \ \ \ \ (3)$

Let

$\displaystyle f_1(t)= \left(\int_0^tV(s)^{\frac{n-1}{n}}ds\right)^{\frac{n}{n-1}}\quad \mathrm{and}\quad f_2(t)=\frac{n}{n-1}\int_0^t V(s)s^{\frac{1}{n-1}}.$

We want to show that ${f_1(t)\ge f_2(t)}$. Clearly, ${f_1(0)=f_2(0)=0}$. By direct computation, ${f_1'(t)\ge f_2'(t)}$ if and only if

$\displaystyle \begin{array}{rl} \int_0^t V(s)^{\frac{n-1}{n}}ds \ge V(t)^{\frac{n-1}{n}}t. \end{array}$

But this follows directly from the monotone decreasing property of ${V(s)}$ (clearly ${V}$ is non-increasing):

$\displaystyle \int_0^ t V(s)^{\frac{n-1}{n}}ds\ge \int_0^t V(t)^{\frac{n-1}{n}}ds= V(t)^{\frac{n-1}{n}}t.$

Thus ${f_1(t)\ge f_2(t)}$ and so putting ${t=\infty}$, we conclude that (3) is true. This completes the proof.

$\Box$

Whether $f_\epsilon$ is weakly differentiable is a local issue. At any point $p$ on the boundary $\partial \Omega$, we can choose a local coordinates such that $p$ is mapped to 0, $x^1>0$ corresponds to $\Omega$ and the $x^1$-curves corresponds to the geodesics normal to $\partial \Omega$. Under this coordinates, the corresponding function is then
$\displaystyle \begin{array}{rl} \displaystyle f_\varepsilon(x)= \begin{cases} 1, & \displaystyle \quad x^1\ge \varepsilon,\\ \frac{x^1}{\varepsilon}, & \displaystyle \quad 0
This function is easily seen to be weakly differentiable, and the weak derivative is zero for $x^1>\varepsilon$ and $x^1<0$.