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Let $\Omega \subset \mathbb{R}^n$. The space $W_0^{k,p}(\Omega)$ is defined as the closure of $C_0^{\infty}(\Omega)$ in $W^{k,p}(\Omega)$.

I can't say I fully understand the rationale of this definition. Would it be wrong to just say that $W_0^{k,p}(\Omega)$ is the set of compactly supported $W^{k,p}(\Omega)$ functions?

un umile appassionato
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That's not true. For example, when $n=1$, $\sin x$ is an element in $W^{1,p}_0(0,\pi)$ since it is a $W^{1,p}$-limit of elements in $C^\infty_c((0,\pi))$. However, $\sin x$ does not have compact support $(0,\pi)$. In general, if $\Omega$ has a nice boundary (for example if it's $C^1$), $W^{k,p}_0(\Omega)$ are those in $W^{k,p}(\Omega)$ which vanishes at the boundary. To be precise, there is a bounded trace operator

$$T : W^{k,p}(\Omega) \to L^p(\partial \Omega)$$

so that $Tu = u|_{\partial \Omega)}$ if $u\in W^{k,p}(\Omega) \cap C(\overline\Omega)$. And we have

$$ W^{k,p}_0(\Omega) = \{ f\in W^{k,p}(\Omega): Tu = 0\}.$$

The proof can be found in Evans' PDE for example.

  • Just to be sure: any open interval in $\mathbb{R}$ is $C^1$? – un umile appassionato Jul 17 '16 at 11:37
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    @unumileappassionato Yes ${}{}{}{}$ –  Jul 17 '16 at 12:46
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    A nice answer. Just a small addition. If $u \in W_0^{k,p}(\Omega)$ then $\partial_i u \in W_0^{k-1,p}(\Omega)$. Thus all partial derivatives of order less than or equal to $k-1$ are in $W_0^{1,p}(\Omega)$. So if your boundary is $C^1$ the trace operator of all such partial derivatives are 0.

    So if you have a problem, where you need two weak derivatives and the boundary condition $u_{\partial \Omega}=0$ the natural space is $W_0^{1,p} (\Omega)\cap W^{2,p}(\Omega)$. If the first derivatives also need to be zero then the natural space is $W_0^{2,p} (\Omega)$.

     – Martin Jul 17 '16 at 14:57