The purpose of this note is to talk about the following so-called Sobolev trace inequality

where is a smooth -dimensional, compact, Riemannian manifold with a smooth boundary with and . The constant appearing from the above inequality is called the best constant.

In fact, this is just a weak type of the true Sobolev trace inequality, which can be stated as follows

where and are positive constant. It is know that in order to prove the above Sobolev inequality, a weaker version is needed.

We now talk about its motivation. First, we start with the standard Sobolev inequality appearing when we talk about the the following embedding

More precise, the following

holds. The idea of the Sobolev inequality is the following: if , by definition, it is clear to see that . One now wishes to find the biggest constant such that . The Sobolev embedding (or the Sobolev inequality) can answer this question. Indeed, for , the biggest is nothing but known as the critical exponent in the Sobolev embedding. It is important to note that in the case , the critical exponent is no longer available, in this case, we have the so-called Trudinger inequality.

Since the embedding is continuous, by definition, there exists a constant such that

By squaring both sides and using the standard norms, we get the above form for the Sobolev inequality.

In the theory of the Sobolev spaces, the trace of any is defined to be where is the unique linear bounded operator independent of . Again, we wish to find the biggest number such that . Once this procedure is done, by the continuous embedding, the corresponding inequality is called the Sobolev trace inequality. In other words, we wish to find the following

Clearly, in order to find the above Sobolev trace inequality, the following weak version is needed

This is equivalent to showing that the integral on the left hand side of the Sobolev inequality can be reduced to the integral on the boundary . Indeed, it was shown by P. L. Lions that

is achieved. But this is clear by partition of unity. Moreover, by the work of Lions, we find that .

In the last part of this note, we make use of this weak version to show that one can change the term

to the term

in order to get the following weak version

Indeed, by an argument by contradiction, there exists some such that for all , there holds

Next, we can show that is achieved by some non-negative function , and by normalizing, there holds

For a proof, we refer the reader to the reference below. Once we have this, we can see that the sequence is bounded by some constant independent of . Therefore, we have some weak limit . This leads to

.

Using the constraint for and the strong convergence in , that is,

there holds on by sending . In particular,

On the other hand, by the Brezis-Lieb lemma, there holds

which implies that, for each and large,

where we have used the weak form of the Sobolev inequality since, due to the strong convergence in ,

Sending to we get

Sending to we get a contradiction.

Regarding to the Sobolev trace inequality for the embedding , with , the exponent will be changed to , that is,

It is worth recalling that the critical exponent for the embedding with is . The only difference is that the dimensional constant appearing in the denominator, , represents the preimage space , while the dimensional constant appearing in the numerator, either or , represents the image space, either or . In this manner, we should remember these inequalities in the following way

If you want a good reference, the following is a good one: *Sharp Sobolev trace inequalities on Riemannian manifolds with boundaries*.

Nice post. Keep going Quốc Anh. How is your thesis going?

Comment by doanchi — May 11, 2012 @ 9:36

Thanks doanchi for your interest in my post. The first draft of my thesis had been done, thanks God, I still have almost three months to go before the submission.

Comment by Ngô Quốc Anh — May 11, 2012 @ 12:16