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I am trying to prove if $u\in W^{1,p}(\Omega)$ has support compactly inside $\Omega$, then $u\in W^{1,p}_0(\Omega)$, where $\Omega\subset \mathbb R^N$ is open.

Intuitively this is true. Assume $V:=\text{spt}(u)$, since $u\equiv 0$ outside $V$ and there always has some room between $\partial \Omega$ and $\partial V$, hence $u$ should have $0$ trace. But I met some difficulties to understand the proof given by H.Brezis book.

Here is the argument. we choose another set $W$ such that $V\subset\subset W\subset\subset \Omega$ and choose $\alpha\in C_c^\infty(W)$ such that $\alpha\equiv 1$ over $V$. Hence $\alpha u = u$.

Next, we could have a sequence $(u_n)\subset C_c^\infty(\mathbb R^N)$ such that $u_n\to u$ in $L^p(\Omega)$ and $\nabla u_n\to \nabla u$ over $(L^p(W))^N$. So far so good.

Then the book states that $\alpha u_n\to \alpha u$ in $W^{1,p}(\Omega)$. I feel very uncomfortable with this sentence. Of course we have $\alpha u_n\to\alpha u$ in $W^{1,p}(W)$ but why in $\Omega$. Yes, I know it should be just $0$ outside $W$ but I feel uncomfortable if I use this idea. Because if I use this idea, then I can just say outside $V$ $u$ is just $0$ and hence $0$ trace and I done. No need for this argument at all.

So, please help me to understand how sentence marked with bold is true. Thank you!

spatially
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1 Answers1

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Notice that $\text{supp}(\alpha) \subset W$, so that both $\alpha u_n$ and $\alpha u$ vanish outside $W$, and then so do their gradients. Therefore $$\| \alpha u_n -\alpha u\|_{W^{1,p}(W)}=\| \alpha u_n -\alpha u\|_{W^{1,p}(\Omega)}.$$

Be careful, trace zero only works when the domain is nice enough. In a general domain you have to show that $u$ is approximable by test functions in $\Omega$. But yes, it's a pretty 'obvious' result.

Jose27
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  • I see. Even if it is very "obvious" that $u\equiv 0$ on $\partial\Omega$ but since the boundary may not good so we can not just use Trace. However, if I assume $\Omega$ is of smooth boundary, can i just conclude that $u\in W_0^{1,p}$ without the argument I post? – spatially Jan 06 '15 at 01:43
  • Sure. Of course the 'heavy-lifting' would be done by the theorem that says $u\in W^{1,p}_0$ iff $u$ has zero trace. – Jose27 Jan 06 '15 at 03:08
  • @Jose27 I see only that $|\alpha u_n-\alpha u|_{W^{1,p}(V)}\to 0$. But in $W\setminus V$ this convergence seems not to be obvious ? – Svetoslav Jan 03 '16 at 15:03
  • @Svetoslav: You need the product rule: Clearly $\alpha u_n\to \alpha u$ in $L^p(W)$, and $\nabla (\alpha u_n)=\nabla \alpha u_n+\alpha \nabla u_n$ and $\nabla (\alpha u)=\nabla \alpha u+\alpha \nabla u$ (all these formulas holding in $\Omega$). The desired convergence follows from the convergence $u_n\to u$ in $W^{1,p}(\Omega)$. – Jose27 Jan 03 '16 at 18:40
  • @Jose27 Thanks for your comment. I thought for the product rule, but somehow missed the point. – Svetoslav Jan 05 '16 at 17:01