I found two interesting formulas related to co-area formula while reading some tricks done by Talenti regarding to the best constant of the Sobolev inequality. The first result is to derive a representation of

and the second result is to deal with differentiation of level sets. Having all these stuffs, I will derive a very short and beautiful proof concerning the lower bound of where , a positive solution to the following PDE

This proof I firstly learned from a paper published in *Duke Math. J.* in 1991 by W. Cheng and C. Li [here].

**Co-area formula**. Suppose that is an open set in , and is a real-valued Lipschitz function on . Then, for an integrable function

where is the -dimensional Hausdorff measure.

**The Sard theorem**. Let be , times continuously differentiable, where . Let be the critical set of , the set of points in at which the Jacobian matrix of has . Then has Lebesgue measure in .

The Sard theorem has some useful applications. For example, if the space of test functions where , then for almost every in the range of , we have that on the level set . Thus that level set will be an -dimensional surface. Furthermore

and

.

Theorem. Let be an open set and . If then for any , we havewhere over .

*Proof*. Two cases are possible

**If **. Define

then for all we have

.

Here is the trick. Let . Set

.

Using this test function and observing that is supported on the set and that on this set , we get

.

Thus, differentiating with respect to , we get

.

Integrating this relation over the range of , i.e. on we get

.

For such that on the level set , the Green theorem implies

since on the level set the tangential derivatives of vanish and since inside this surface, we have

.

**If **. Due to the presence of singularity, we need to use an approximation technique. Let . Define

then for all we have

.

Choose the same test function as in the previous case we arrive at

.

We now pass the limit as on both sides. On the LHS, the integrand coverges pointwise to ; further, since , we have

which is integrable. Thus, by the Dominated Convergence Theorem, we get

.

On the RHS, one again observes that

which is integrable on the set . So again by the Dominated Convergence Theorem,

as .

Further,

and

since . Thus another application of the Dominated Convergence Theorem yields

.

The proof follows.

**Remark**. Concerning the Schwarz symmetrization, the following

holds.

See also: *Symmetrization And Applications* (Series in Analysis) by S. Kesavan.

In case 1, the second from the last equation, the right most integral should be , right?

Comment by Chao — August 15, 2010 @ 19:13

You are right, Chao. The correct one reads as follows

.

Thanks for pointing out this misprint.

Comment by Ngô Quốc Anh — August 15, 2010 @ 19:21