# 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, $\overline u_j^\varepsilon \in C^\infty(\overline \Omega)$.

## February 24, 2012

### The pullback metric under scaling

Filed under: Uncategorized — Ngô Quốc Anh @ 20:11

Today, let us discuss a very simple question. Assume that $(M,g)$ is a Riemannian manifold and $\lambda>0$ is constant. We study the following so-called pullback metric $\lambda^\star g$ of $g$ under the scaling $x \mapsto \lambda x$. Precisely, we aim to compare $\lambda^\star g$ and $g$.

For simplicity, we follow definition of the pullback metric. Besides, by $\lambda$ we mean the map $x \mapsto \lambda x$.

The pushforward $\lambda_\star$ by $\lambda$. This is the first step. Suppose $X \in T_xM$ for some $x \in M$. By definition, we have $\lambda_\star X \in T_{\lambda (x)}M$ which is just

$\displaystyle (\lambda_\star X) f = X ( f \circ \lambda), \quad \forall f \in C^\infty(M,\mathbb R).$

By the definition of derivation, in local coordinates, we clearly have

$\displaystyle X(f \circ \lambda ) = X(f \circ (\lambda {\rm id})) = \lambda X(f \circ {\rm id}) = \lambda X(f).$

The pullback metric $\lambda^\star g$. By definition, we have

$\displaystyle \lambda^\star g (X, Y) = g(\lambda_\star X, \lambda_\star Y), \quad \forall X, Y \in T_xM.$

Obviously,

$\displaystyle g({\lambda _ \star }X,{\lambda _ \star }Y) = {\lambda ^2}g({\rm id}_*X,{\rm id}_ \star Y)= {\lambda ^2}g(X,Y).$

Thus, we can conclude that

$\displaystyle \lambda^\star g=\lambda^2 g.$

## February 6, 2012

### A note on the maximum principle

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

Today, let us discuss a very interesting stuff. Let say $M$ is a compact manifold without boundary of dimention $n \geqslant 3$.

On $M$, we consider the following simple PDE

$\displaystyle -\Delta u + hu = a$

where $h$ and $a$ are smooth functions with $h>0$ and $a \geqslant 0$. Since $h>0$, it is well-known that the operator $-\Delta +h$ is coercive, see here. A standard variation method tells us that there exists a weak solution $u$ to the above PDE. By regularity theorem, $u$ is at least a $C^2$ function, thus, a strong solution (in the classical sense).

Next, we claim that $u \geqslant 0$. To this purpose, assume that the solution $u$ achieves its minimum at some point $x_0 \in M$. In particular, there holds

$-\Delta u (x_0) \leqslant 0$.

This, together with the fact that $h>0$ and $a \geqslant 0$, implies that $u(x_0) \geqslant 0$. Thus, we have shown that $u \geqslant 0$ in $M$.

Once we have the non-negativity of $u$, in view of the strong maximum principle, either $u \equiv 0$ in $M$ or $u>0$. In other words, the solution $u$ cannot achieves its minimum inside the manifold. Since the manifold has no boundary, it is natural to think that the solution $u$ cannot achieve its minimum although $u>0$. This is clear a contradiction to the fact that the manifold is compact and $u$ is of class $C^2(M)$.

So something went wrong but what and why?

In fact, we have made a small mistake. In view of the strong maximum principle, we can only claim that the solution can only achieve its non-positive minimum value on the boundary. Therefore, there are cases so that $u$ may achieve its positive minimum inside $M$. Thus, there is no contradiction here.

In order to see this, let us go back to a proof of the strong maximum principle. Roughly speaking, it starts with the following simple one.

Lemma 1. If $-\Delta u +hu >0$ at any point in $M$, then $u$ cannot have non-positive minimum value in $M$.

Proof. The proof is standard. Assume that at $x_0 \in M$, the function $u$ realizes its minimum, besides, $u(x_0) \leqslant 0$. In particular, $-\Delta u(x_0) \leqslant 0$ and $h(x_0)u(x_0) \leqslant 0$. These force $-\Delta u + hu \leqslant 0$ at $x_0$. A contradiction.

Form the proof above, if $u(x_0)>0$, we cannot get any contradiction. This is why we can claim either $u \equiv 0$ or $u>0$ since $u$ can achieve its positive minimum in $M$.

Notice that, if we don’t have any $h$ (in the operator), the non-positivity can be dropped.