Ngô Quốc Anh

May 15, 2012

An example of compact embedding between two Banach spaces

Filed under: PDEs — Ngô Quốc Anh @ 3:09

Assume V : \mathbb R^N \to \mathbb R is continuous and satisfies the following two conditions

V(x) \geqslant V_0 >0 in \mathbb R^N for some V_0 >0

and the function

\displaystyle \frac{1}{V(x)} \in L^\frac{1}{N-1}(\mathbb R^N).

Let us define the following space

\displaystyle E=\left\{ u \in W^{1,N}(\mathbb R^N): \int_{\mathbb R^N}V(x)|u|^N dx<+\infty\right\}.

On E, we can use the following norm

\displaystyle {\left\| u \right\|_E} = {\left( {\int_{{\mathbb{R}^N}} {(|\nabla u{|^N} + V|u{|^N})dx} } \right)^{\frac{1}{N}}}.

In this note, we prove that E is compactly embedded in L^q(\mathbb R^N) for all q \geqslant 1. This is Lemma 2.4 in a paper by Y. Yang published in JFA this 2012 although the technique used is standard.

Indeed, by the first condition on V, the standard Sobolev embedding theorem implies that the following embedding

\displaystyle E \hookrightarrow W^{1,N}(\mathbb R^N)\hookrightarrow L^q(\mathbb R^N)

is obviously continuous for any q \geqslant N. It follows from the second condition on V and the Holder inequality that

\displaystyle\int_{{\mathbb{R}^N}} {|u|dx} \leqslant {\left( {\int_{{\mathbb{R}^N}} {\frac{{dx}}{{{V^{\frac{1}{{N - 1}}}}}}} } \right)^{1 - \frac{1}{N}}}{\left( {\int_{{\mathbb{R}^N}} {V|u{|^N}dx} } \right)^{\frac{1}{N}}} \leqslant {\left( {\int_{{\mathbb{R}^N}} {\frac{{dx}}{{{V^{\frac{1}{{N - 1}}}}}}} } \right)^{1 - \frac{1}{N}}}{\left\| u \right\|_E}.

For any 1<\gamma<N, there holds

\displaystyle\int_{{\mathbb{R}^N}} {|u{|^\gamma }dx} \leqslant \int_{{\mathbb{R}^N}} {(|u| + |u{|^N})dx} \leqslant {\left( {\int_{{\mathbb{R}^N}} {\frac{{dx}}{{{V^{\frac{1}{{N - 1}}}}}}} } \right)^{1 - \frac{1}{N}}}{\left\| u \right\|_E} + \frac{1}{{{V_0}}}\left\| u \right\|_E^N.

Thus, we get continuous embedding

\displaystyle E \hookrightarrow L^q(\mathbb R^N)

for all q \geqslant 1. To prove that the above embedding is also compact, we follow definition. Take a sequence of functions \{u_k\}_k \subset E such that \|u_k\|_E \leqslant C for all k, we must prove that up to a subsequence, there exists some u \in E such that u_k converges to u strongly in L^q(\mathbb R^N) for all q \geqslant 1.

Since E is reflexive, without loss of generality, we may assume that

u_k \rightharpoonup u weakly in E and strongly in L^q_{loc}(\mathbb R^N) for all q \geqslant 1.

In view of the second condition on V, for any \varepsilon>0, there exists R>0 such that

\displaystyle {\left( {\int_{|x| > R} {\frac{{dx}}{{{V^{\frac{1}{{N - 1}}}}}}} } \right)^{1 - \frac{1}{N}}} < \varepsilon.

Hence

\displaystyle\int_{|x| > R} {|{u_k} - u|dx} \leqslant {\left( {\int_{{\mathbb{R}^N}} {\frac{{dx}}{{{V^{\frac{1}{{N - 1}}}}}}} } \right)^{1 - \frac{1}{N}}}{\left\| {{u_k} - u} \right\|_E} \leqslant \varepsilon {\left\| {{u_k} - u} \right\|_E} \leqslant \varepsilon C.

Note that this C depends on the norm of u. On the other hand, since u_k \to u strongly in L^1(B_R(0)), we get that

\displaystyle\mathop {\limsup }\limits_{k \to \infty } \int_{{\mathbb{R}^N}} {|{u_k} - u|dx} \leqslant C\varepsilon.

Since \varepsilon >0 is arbitrary, we obtain

\displaystyle\mathop {\lim}\limits_{k \to \infty } \int_{{\mathbb{R}^N}} {|{u_k} - u|dx}=0.

For q>1, it follows from the continuous embedding E \hookrightarrow L^s(\mathbb R^N), s \geqslant 1, that

\displaystyle\begin{gathered} \int_{{\mathbb{R}^N}} {|{u_k} - u{|^q}dx} = \int_{{\mathbb{R}^N}} {|{u_k} - u{|^{\frac{1}{2}}}|{u_k} - u{|^{q - \frac{1}{2}}}dx} \hfill \\ \qquad\leqslant {\left( {\int_{{\mathbb{R}^N}} {|{u_k} - u|dx} } \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^N}} {|{u_k} - u{|^{2q - 1}}dx} } \right)^{\frac{1}{2}}} \hfill \\ \qquad\leqslant C{\left( {\int_{{\mathbb{R}^N}} {|{u_k} - u|dx} } \right)^{\frac{1}{2}}} \to 0 \hfill \\ \end{gathered}

as k \to \infty.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: