# Approximation in Sobolev Spaces

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• October 8th 2013, 01:11 PM
Johnyboy
Approximation in Sobolev Spaces
I just want to know how it follows that $v^{\epsilon} \in C^{\infty}(\bar{V})$? I could see how $v^{\epsilon} \in C^{\infty}(V)$ by using the translations, but I'm having difficulty seeing how it extends to $\bar{V}$, since it says that $u_{\epsilon}(x) := u(x^{\epsilon}) \text{ for } x \text{ in } V$, we are mollifying on $V$. Do you have any idea of how this follows? Thanks for the assistance.
• December 2nd 2013, 11:56 AM
Rebesques
Re: Approximation in Sobolev Spaces
What the symbolism denotes is that the limit $\lim_{x\rightarrow y \in \overline{V}}D^{\alpha}v^{\epsilon}$ necessarily exists for all $\alpha$ (as $D^{\alpha}u$ is defined on $\overline{V}$).