In a very old entry, I talked about an extension of Rellich-Kondrachov theorem for embeddings between Sobolev spaces. For the sake of convenience, here is the statement of this extension:

Theorem(Extension of Rellich-Kondrachov for bounded domains). Let be an open, bounded Lipschitz domain, and let . SetThen we have

for

and

for

Clearly, when , the above embedding is not compact, in general. In this context, we call the failure of compact Rellich-Kondrachov embedding due to critical exponents.

There is an other example of the failure of compact Rellich-Kondrachov embedding which is basically due to the unbounded domains. In this entry, we address counter-examples for these two lacks of compactness.

**1. The case of unbounded domains**.

For simplicity, let us prove the following result.

Result 1. Assume . Then the embeddingis never compact for any .

Be definition, to show that such an embedding is never compact, it suffices to construct a bounded sequence of functions in such that there is no convergent subsequence in .

To this purpose, first, we take a positive smooth function in such a way that . In addition, we may select such that

Then for each , we define

where . Obviously, and hence . Since are obtained by translating , we easily verify that is bounded in . However, for any , there holds

This is due to the fact that and have non-overlapping supports. The above estimate immediately implies that has no convergent subsequence in .

In conclusion, the failure of compact Rellich-Kondrachov embedding on unbounded domains basically due to the translation. However, for bounded domains, such translation is no longer available. To avoid this difficulty, we use dilation and this requires certain exponents called critical exponents rather than the arbitrarily of as above.

**2. The case of critical exponents**

We now turn to the loss of compactness due to critical exponents. Hence, we now discuss the following result.

Result 2. We now suppose is a bounded domain. Then the embeddingis never compact.

We still keep the function as before. Let be a sequence of points in . Then we select a corresponding sequence of radius such that . Then we scale as follows:

where . Clearly, . Hence, we can choose smaller, if necessary, such that all are disjoint.

To make sure that is bounded in , we first let . A direct calculation first shows and

Hence, we obtain

Therefore, the sequence is bounded in . In the last step, we show that this sequence of functions has no convergent subsequence in where as always .

It is worth noting that if we still estimate the norm from below using the norm , we arrive at nothing. Indeed, we first obtain

It follows that

which then gives us nothing due to the presence of the term . Hence, it is necessary to get rid of that term. To achieve that goal, instead of using the norm , we shall estimate the norm from below using

where the multi-indexes is of order , i.e. . Again, by some simple calculation, we know that

Hence, if we choose is good enough in such a way that

,

similar arguments show that the embedding is never compact, thanks to the disjointness of the supports of the functions .

Hi,

Do you know if there is a compact embedding for $\Omega=\mathbb R^n$ with a finite measure (like a Gaussian measure)?

Thanks

Comment by Marcus — October 10, 2015 @ 21:58

Hi, at the moment, I do not know.

Comment by Ngô Quốc Anh — October 10, 2015 @ 22:53