# Ngô Quốc Anh

## January 7, 2015

### The failure of compact Rellich-Kondrachov embedding: Unbounded domains and critical exponents

Filed under: Uncategorized — Ngô Quốc Anh @ 19:49

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 $\Omega \subset \mathbb R^n$ be an open, bounded Lipschitz domain, and let  $1 \leqslant p \leqslant mn$. Set

$\displaystyle p^\star := \frac{np}{n - mp}.$

Then we have

$\displaystyle W^{j+m, p} (\Omega) \hookrightarrow W^{j, q} (\Omega)$ for  $1 \leqslant q \leqslant p^\star$

and

$\displaystyle W^{j+m, p} (\Omega) \hookrightarrow \hookrightarrow W^{j,q} (\Omega)$ for  $1 \leqslant q < p^\star.$

Clearly, when $q=p^\star=\frac{np}{n - mp}$, 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 $\Omega = \mathbb R^n$. Then the embedding

$\displaystyle W^{j+m, p} (\mathbb R^n) \hookrightarrow W^{j, q} (\mathbb R^n)$

is never compact for any  $1 \leqslant q$.

Be definition, to show that such an embedding is never compact, it suffices to construct a bounded sequence of functions $(\varphi_k)_k$ in $W^{j+m, p} (\mathbb R^n)$ such that there is no convergent subsequence in $W^{j, q} (\mathbb R^n)$.

To this purpose, first, we take a positive smooth function $\varphi \in C_0^\infty(\mathbb R^n)$ in such a way that $\text{supp}(\varphi) \subset B(0,1/2)$. In addition, we may select $\varphi$ such that

$\displaystyle \|\varphi\|_{L^q} = \left( \int_{\mathbb R^n} \varphi (x)^q dx \right)^\frac{1}{q}=1.$

Then for each $k \in \mathbb N$, we define

$\displaystyle \varphi_k (x) = \varphi (x-\vec k)$

where $\vec k = (k,0,\dots, 0) \in \mathbb R^n$. Obviously, $\text{supp}(\varphi_k) \subset B(\vec k,1)$ and hence $\varphi_k \in C_0^\infty (\mathbb R^n)$. Since $\varphi_k$ are obtained by translating $\varphi$, we easily verify that $(\varphi_k)_k$ is bounded in $W^{j+m, p} (\mathbb R^n)$. However, for any $k \ne m$, there holds

$\displaystyle \|\varphi_k - \varphi_m \|_{W^{j, q}} \geqslant \|\varphi_k - \varphi_m \|_{L^q} = \left( \|\varphi_k\|_{L^q}^q+ \|\varphi_m\|_{L^q}^q \right)^{1/q} = 2^{1/q}.$

This is due to the fact that $\varphi_k$ and $\varphi_m$ have non-overlapping supports. The above estimate immediately implies that $(\varphi_k)_k$ has no convergent subsequence in $W^{j, q} (\mathbb R^n)$.

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 $q$ 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 $\Omega \subsetneq \mathbb R^n$ is a bounded domain. Then the embedding

$\displaystyle W^{j+m, p} (\Omega) \hookrightarrow W^{j, p^\star} (\Omega)$

is never compact.

We still keep the function $\varphi$ as before. Let $(a_i)_i \subset \Omega$ be a sequence of points in $\Omega$. Then we select a corresponding sequence of radius $(r_i)_i$ such that $r_i \leqslant 1$. Then we scale $\varphi$ as follows:

$\displaystyle \varphi_k (x) = r_k^{j+m-\frac np} \varphi (y)$

where $x =a_k+r_k y$. Clearly, $\text{supp}(\varphi_k) \subset B(a_k, r_k)$. Hence, we can choose $r_i$ smaller, if necessary, such that all $B(a_k, r_k)$ are disjoint.

To make sure that $(\varphi_k)_k$ is bounded in $W^{j+m, p} (\Omega)$, we first let $|\alpha| \leqslant j+m$. A direct calculation first shows $dx=r_k^ndy$ and

$\displaystyle ({D^\alpha }{\varphi _k})({a_k} + {r_k}y) = r_k^{j + m - \frac{n}{p}}({D^\alpha }\varphi )({a_k} + {r_k}y) = r_k^{j + m - \frac{n}{p} - |\alpha |}({D^\alpha }\varphi )(y).$

Hence, we obtain

$\displaystyle \int_\Omega |D^\alpha \varphi_k |^p (x) dx= r_k^{(j+m-|\alpha|)p} \int_\Omega |D^\alpha \varphi |^p (y) dy \leqslant \int_\Omega |D^\alpha \varphi |^p (y) dy.$

Therefore, the sequence $(\varphi_k)_k$ is bounded in $W^{j+m, p} (\Omega)$. In the last step, we show that this sequence of functions has no convergent subsequence in $W^{j, p^\star} (\Omega)$ where as always $p^\star := np/(n - mp)$.

It is worth noting that if we still estimate the norm $\|\cdot\|_{W^{j, p^\star}}$ from below using the norm $\|\cdot\|_{L^{p^\star}}$, we arrive at nothing. Indeed, we first obtain

$\begin{array}{lcl} \displaystyle\int_\Omega {|{\varphi _k}{|^{\frac{{np}}{{n - mp}}}}dx} &=& \displaystyle\int_\Omega {|{\varphi _k}{|^{\frac{{np}}{{n - mp}}}}({a_k} + {r_k}y)d({a_k} + {r_k}y)} \hfill \\ &=& \displaystyle\int_\Omega {r_k^{(j + m - \frac{n}{p})\frac{{np}}{{n - mp}}}|\varphi {|^{\frac{{np}}{{n - mp}}}}(y)r_k^ndy} \hfill \\ &=& \displaystyle\int_\Omega {r_k^{(j + m - \frac{n}{p})\frac{{np}}{{n - mp}}}|\varphi {|^{\frac{{np}}{{n - mp}}}}(y)r_k^ndy} \hfill \\ &=& \displaystyle r_k^{j\frac{{np}}{{n - mp}}}\int_\Omega {|\varphi {|^{\frac{{np}}{{n - mp}}}}(y)dy} . \end{array}$

It follows that

$\displaystyle \|\varphi_k\|_{L^{p^\star}} = r_k^j \|\varphi_k\|_{L^{p^\star}},$

which then gives us nothing due to the presence of the term $r_k^j$. Hence, it is necessary to get rid of that term. To achieve that goal, instead of using the norm $\|\cdot\|_{L^{p^\star}}$, we shall estimate the norm $\|\cdot\|_{W^{j, p^\star}}$ from below using

$\displaystyle \|f \|_{{W^{j,m}}} \geqslant \left( \int_\Omega {|{D^\alpha }f{|^{{p^ \star }}}(y)dy} \right)^\frac{1}{p^\star},$

where the multi-indexes $\alpha$ is of order $j$, i.e. $|\alpha|=j$. Again, by some simple calculation, we know that

$\begin{array}{lcl} \displaystyle\int_\Omega {|{D^\alpha }{\varphi _k}{|^{\frac{{np}}{{n - mp}}}}dx} &=& \displaystyle\int_\Omega {|{D^\alpha }{\varphi _k}{|^{\frac{{np}}{{n - mp}}}}({a_k} + {r_k}y)d({a_k} + {r_k}y)} \hfill \\ &=& \displaystyle\int_\Omega {r_k^{(j + m - \frac{n}{p} - |\alpha |)\frac{{np}}{{n - mp}}}|{D^\alpha }\varphi {|^{\frac{{np}}{{n - mp}}}}(y)r_k^ndy} \hfill \\ &=& \displaystyle\int_\Omega {r_k^{(m - \frac{n}{p})\frac{{np}}{{n - mp}}}|{D^\alpha }\varphi {|^{\frac{{np}}{{n - mp}}}}(y)r_k^ndy} \hfill \\ &=& \displaystyle\int_\Omega {|{D^\alpha }\varphi {|^{\frac{{np}}{{n - mp}}}}(y)dy} . \end{array}$

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

$\displaystyle\int_\Omega {|{D^\alpha } \varphi |^{{p^ \star }} (y)dy} > 0$,

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

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