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${\bf Global\ Approximation\ Theorem}$(251 page inEvans's PDE book) : If $U$ is ${\bf bounded}$ in ${\bf R}^n$, then for $u\in W^{k,p}(U)$, there exists $u_n \in C^\infty (U)\cap W^{k,p}(U)$ such that $$u_n \rightarrow u\ in\ W^{k,p}(U)$$

For the proof, we used partition of unity argument. But I cannot understand why we need the assumption of boundedness.

Proof : First recall that fact,

$$ u^\epsilon \rightarrow u \ in \ W^{k,p}(V)$$ where $V$ has compact closure and $u^\epsilon =\eta_\epsilon\ast u$.

For the convenience we let $U={\bf R}^n$ and $V_i$ to be an open ball $B(i,0)$.

On $ U_i=V_{i+3} - V_{i+1}$ we can have subordinate partition of unity :

$$ f_i \in C_c^\infty(U_i) $$

Hence $$ f_i u\in W^{k,p}(U) $$ If $u^i = (f_iu)^{\epsilon_i}$ then $$ |u^i - f_iu|_{W^{k,p}(U) }< \frac{\delta }{2^{i+1}} $$ for small $\epsilon_i$.

Hence for any compact set $V$ in $U$ since $u=\sum f_iu$, $$ |\sum_{i=1}^\infty u^i - u|_{W^{k,p}(V)} \leq \delta$$

HK Lee
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    With the boundedness assumtpion, you get global convergence while without it, the only thing you can get is local convergence, i.e. for each compact set contained in $U$ you have convergence. It is true that there exist functions for which the convergence is global, but this is not always the case. Also note that your proof does not implies global convergence. – Tomás Nov 02 '13 at 11:24
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    Hi Tomas could you expand on your answer of why you get global convergence if U is bounded. You are saying that $u_{m} \rightarrow u \text{ in } W^{k,p}(U)$ if $U$ is bounded. But if $U$ is not assumed bounded then we can only get $u_{m} \rightarrow u \text{ in } W_{0}^{k,p}(U)$? – Alex Nov 08 '13 at 13:07
  • @HeeKwon Send me an email if you want to discuss Evans PDE book, I am currently studying it as well. – Alex Nov 08 '13 at 14:42
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    @Alex, I am saying that if $U$ is not bounded then, $u_m\to u$ in $W_{loc}^{k,p}(U)$ – Tomás Nov 08 '13 at 14:59
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    Yes I made a mistake, I meant $W_{\text{loc}}^{k,p}$. What is the motivation for this difference in convergence depending on whether $U$ is bounded? – Alex Nov 08 '13 at 15:56
  • @Alex, nice meet to you. But I can not find an e-mail on your profile site. – HK Lee Nov 09 '13 at 02:21
  • It's [email protected] – Alex Nov 09 '13 at 09:09

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Edit: One problem with Evans's proof is that he defines $U_i$ as $$U_i=\{x\in U\big|\mathrm{dist}(x,\partial U)<1/i\},$$ and $V_i=U_{i+3}\setminus\overline{U}_i$. But if, for example, $U=\mathbb R^n$, what are those $V$'s? So, Evans's proof will not work here.

In your proof, you use the geometry of $\mathbb R^n$, to say that it is covered by your $U_i$'s. This proves that the theorem holds for $\mathbb R^n$, but might not apply in other cases.

detnvvp
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