There is a classic blowup analysis proof for Aubin’s theorem, due to Uhlenbeck’s renormalization method described below to give another proof that the Yamabe problem. See the chapter5 at Schoen & Yau ‘ *Lectures on differential geometry’.*

Yamabe’s approach was to consider first the perturbed functional

where and . Set

By using a direct minimizing procedure, it can be shown that for , there exists a smooth positive function us such that its -norm is equal to one, and us satisfies the equation

The direct method does not work when because the Sobolev embedding is continuous but not compact.

However, if one can show that is uniformly bounded, i.e.there exists a positive constant such that in for , then there exists a sequence such that and converges to a smooth positive function which satisfies the Yamabe equation .

We discuss a **blow-up** argument. Suppose that no such upper bound exists. It follows that there exist sequences and such that

As is compact, we may assume that as . For a normal coordinate system centered at and with radius , let the coordinates of be .

In the local coordinates,

satisfies the equation

The** idea** here is to consider the normalized function

where . We have and as . Here is defined on a ball in of radius and as .

By the argument of diagonal subsequence and the property of normal coordinates , one observes that a subsequence of converges to a smooth positive function which is a nonnegative solution of the equation

where ,and is the standard Laplacian on .

By the strong maximum principle, . It is known that if ; and if . Let be the diameter of . By a change of variables we have

where denotes the open ball in with center at and radius equal to . we note that

From (2) the Fatou lemma and , we obtain

A similar argument implies

Let be a cutoff function satisfies in and in

Defined , then

Multiplies (1) by and integration by parts, we obtain

Taking in above equation and thanks to (4) we get

- If , then , and (2) implies , which is a contradiction with .
- If , . (2) (5) and the best Sobolev imbedding implies

Thus

We are led to the contradiction with

Therefore, is uniformly bounded.