# Ngô Quốc Anh

## February 28, 2012

### The Rellich embedding theorem on a bounded domain

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:18

Today, let us summarize steps during the standard proof of the Rellich embedding theorem on a bounded domain $\Omega \subset \mathbb R^n$. This theorem says that

Theorem (Rellich). Suppose that $\Omega \subset \mathbb R^n$ is an open, bounded domain with $C^1$ boundary,  and that $1. Then $W^{1,p}(\Omega)$ is compactly embedded in $L^q(\Omega)$ for all $1\leqslant q <\frac{np}{n-p}$.

In particular, for any sequence $\{u_j\}_j \subset W^{1,p}(\Omega)$ , there exists a subsequence $\{u_{j_k}\}_k \subset \{u_j\}_j$ such that $u_{j_k} \to u$ strongly in $L^q(\Omega)$ for some $u \in L^q(\Omega)$.

In order to prove the Rellich theorem, we need the so-called Arzela-Ascoli theorem.

Lemma (The Arzela-Ascoli theorem). Suppose that $u_j \in C^0(\overline\Omega)$, $\|u_j\|_{C^0(\overline\Omega)} \leqslant M < \infty$, and $\{u_j\}_j$ is equicontinuous. Then there exists a subsequence $u_{j_k} \to u$ uniformly on $\Omega$.

The Arzela-Ascoli theorem is well-known. To prove the Rellich theorem, we shall use the standard mollifiers. To do that, we have to extend $\Omega$ a little bit.

Step 1. Assume that $\overline\Omega \subset \mathbb R^n$ is also an open, bounded domain with $C^1$ boundary with $\Omega \subset\subset \overline \Omega$. By the Sobolev extension theorem, we can define $Eu_j$ by $\overline u_j$ with

$\sup_j \|\overline u_j\|_{W^{1,p}(\mathbb R^n)} < C<\infty.$

By the Gagliardo-Nirenberg inequality,

$\sup_j \|\overline u_j\|_{L^q(\Omega} < C<\infty.$

For $\varepsilon>0$, let $\eta_\varepsilon$ denote the standard mollifiers and set $\overline u_j^\varepsilon = \eta_\varepsilon * \overline u_j$. By choosing $\varepsilon$ sufficiently small, $\overline u_j^\varepsilon \in C^\infty(\overline \Omega)$.

Step 2. Since

$\displaystyle\overline u _j^\varepsilon = \int_{B(0,\varepsilon )} {\frac{1}{{{\varepsilon ^n}}}\eta \left( {\frac{y}{\varepsilon }} \right){{\overline u }_j}(x - y)dy} = \int_{B(0,1)} {\eta (z){{\overline u }_j}(x - \varepsilon z)dz}$

we can get that

$\displaystyle |\overline u _j^\varepsilon (x) - {\overline u _j}(x)| = \varepsilon \int_{B(0,1)} {\eta (z)\left( {\int_0^1 {|D{{\overline u }_j}(x - \varepsilon tz)|dt} } \right)dz} .$

Therefore,

$\displaystyle\int_{\overline \Omega } {|\overline u _j^\varepsilon (x) - {{\overline u }_j}(x)|dx} \leqslant \varepsilon {\left\| {D{{\overline u }_j}} \right\|_{{L^1}(\overline \Omega )}} \leqslant \varepsilon {\left\| {D{{\overline u }_j}} \right\|_{{L^p}(\overline \Omega )}} < C.$

This and the $L^p$-interpolation inequality imply that

$\displaystyle {\left\| {\overline u _j^\varepsilon - {{\overline u }_j}} \right\|_{{L^q}(\overline \Omega )}} < C\varepsilon .$

Step 3. Our goal is to employ the Arzela-Ascoli theorem. It is not hard to see the following

$\displaystyle\mathop {\sup }\limits_j {\left\| {\overline u _j^\varepsilon } \right\|_{{C^0}(\overline \Omega )}} \leqslant {\left\| {{\eta _\varepsilon }} \right\|_{{L^\infty }({\mathbb{R}^n})}}\mathop {\sup }\limits_j {\left\| {{{\overline u }_j}} \right\|_{{L^1}(\overline \Omega )}} \leqslant \frac{C}{{{\varepsilon ^n}}} < \infty$

and

$\displaystyle\mathop {\sup }\limits_j {\left\| {D\overline u _j^\varepsilon } \right\|_{{C^0}(\overline \Omega )}} \leqslant {\left\| {D{\eta _\varepsilon }} \right\|_{{L^\infty }({\mathbb{R}^n})}}\mathop {\sup }\limits_j {\left\| {{{\overline u }_j}} \right\|_{{L^1}(\overline \Omega )}} \leqslant \frac{C}{{{\varepsilon ^{n + 1}}}} < \infty.$

Hence there exists a subsequence $u_{j_k}$ which converges uniformly on $\overline \Omega$ so that

$\displaystyle\mathop {\lim \sup }\limits_{k,l \to \infty } {\left\| {\overline u _{{j_k}}^\varepsilon - {{\overline u }_{{j_l}}}} \right\|_{{L^q}(\overline \Omega )}} = 0.$

By the triangle inequality, we easily get that

$\displaystyle\mathop {\lim \sup }\limits_{k,l \to \infty } {\left\| {\overline u _{{j_k}} - {{\overline u }_{{j_l}}}} \right\|_{{L^q}(\overline \Omega )}} < C \varepsilon .$

Step 4. By choosing $\varepsilon=\frac{1}{2}, \frac{1}{4},...$ and using the diagonal argument to extract further subsequences, we can arrange to find a subsequence $u_{j_{k_l}} \subset u_{j_k}$ such that

$\displaystyle\mathop {\lim \sup }\limits_{l,m \to \infty } {\left\| {\overline u _{j_{k_l}} - {{\overline u }_{j_{k_m}}}} \right\|_{{L^q}(\overline \Omega )}} < C\varepsilon.$

Thus,

$\displaystyle\mathop {\lim \sup }\limits_{l,m \to \infty } {\left\| u _{j_{k_l} - u _{j_{k_m}}} \right\|_{{L^q}(\Omega )}} =0.$

This concludes the proof.

We know discuss a simple application of the idea of the above proof. Assume that the sequence of functions $v_j \in [0,1]$ solves the following

$-\Delta v_j = h_jv_j + f_jv_j^{2^\star} \quad \text{ in } \mathbb R^n,$

where $f_j$ and $h_j$ are smooth functions and uniformly bounded in $\mathbb R^n$. We claim that, up to subsequences, $v_j \to v$ strongly in $C^2_{\rm loc}(\mathbb R^n)$. In other words, for fix $R>0$, the sequence $\{v_j\}_j$ is always compact in $C^2(B_R(0))$. To see this, we use the standard elliptic regularity theory.

Step 1. Indeed, since $v_j$ are uniformly bounded in $\mathbb R^n$, we can see that $v_j$ are uniformly bounded in $L_{\rm loc}^p(\mathbb R^n)$ for any $p \geqslant 1$. Since all coefficients of the PDE are uniformly bounded (and even smooth) in $\mathbb R^n$, we get via the standard $L^p$-estimates that $v_j \in W_{\rm loc}^{2,q}(\mathbb R^n)$ for any $q \geqslant 1$. Using the Sobolev embedding theorem, the following $W_{\rm loc}^{2,q}(\mathbb R^n) \hookrightarrow C_{\rm loc}^{0,\alpha}(\mathbb R^n)$ is compact for some $\alpha \in (0,1)$. By the Schauder estimates, we can conclude that $v_j \in C_{\rm loc}^{2,\alpha}(\mathbb R^n)$.

Step 2. Since all coefficients of the PDE are bounded, we can conclude that the sequence $\{v_j\}$ is bounded in $C_{\rm loc}^{2,\alpha}(\mathbb R^n)$.

Step 3. To complete the proof, let say we have $K$ a compact subset in $\mathbb R^n$. It is obvious to see that the embedding $C^{0,\alpha}(K) \hookrightarrow C^0(K)$ is compact. Indeed, the uniformly boundedness and the equicontinuous come from the following estimates

$\displaystyle |{D^2}{v_j}(x) - {D^2}{v_j}(y)| = \frac{{|{D^2}{v_j}(x) - {D^2}{v_j}(y)|}}{{|x - y{|^\alpha }}}|x - y{|^\alpha } \leqslant 2 {\rm diam}{(K)^\alpha }{\left\| {{D^2}{v_j}} \right\|_{{C^{0,\alpha }}(K)}}$

and

$\displaystyle |{D^2}{v_j}(x) - {D^2}{v_j}(y)| \leqslant 2 {\rm diam}{(K)^{\alpha-1} }{\left\| {{D^2}{v_j}} \right\|_{{C^{0,\alpha }}(K)}}|x-y|.$

By the Arzela-Ascoli theorem, there is a subsequence $D^2v_{j_k}$ converges uniformly in $K$. In order to use this compact embedding, we need some boundedness of $v_j$ in $C_{\rm loc}^{2,\alpha}(\mathbb R^n)$. Thanks to Step 2, we can conclude that $v_j \to v$ strongly in $C^2_{\rm loc}(\mathbb R^n)$ up to subsequences.