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, 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 $k \to \infty$.