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 be an -dimensional Riemannian manifold, . The isoperimetric property (or more commonly, isoperimetric inequality) states that there exists a constant such that for any relatively compact domain with smooth boundary, we have

The Sobolev inequality states that on a compact -dimensional Riemannian manifold (possibly with boundary), there exists a constant (different from that in (1)) such that

provided that if ; otherwise we assume either or (the completion of under the norm ).

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

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

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 , we define the function

Apply the Sobolev inequality (2) to , and take , we can get the isoperimetric property (1).

Now we prove the converse. We can without loss of generality assume . By the layercake integral formula, we have

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

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

Putting suggests that (though strictly speaking, such may not be admissible for the Sobolev inequality; nevertheless we can get around this problem by modifying to be a smooth bump function with range supported on a compact region, similar to above). Therefore, if we let , then we have to show that

We want to show that . Clearly, . By direct computation, if and only if

But this follows directly from the monotone decreasing property of (clearly is non-increasing):

Thus and so putting , we conclude that (3) is true. This completes the proof.

how can one soblev inequality to the function {f_\epsilon},

Whether is weakly differentiable is a local issue. At any point on the boundary , we can choose a local coordinates such that is mapped to 0, corresponds to and the -curves corresponds to the geodesics normal to . Under this coordinates, the corresponding function is then

This function is easily seen to be weakly differentiable, and the weak derivative is zero for and .