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