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Let $\Omega$ be an open, bounded domain in $\mathbb{R}^n$ (NOT necesasrily with smooth boundary).

Now, let $C_c^\infty(\Omega)$ be the space of smooth functions on $\Omega$ with compact support.

Then

How to prove that $C_c^\infty(\Omega)$ dense in $L^p(\Omega)$ for each $p \in [1,\infty)$?

This post has almost the same question and answer, but no details of proof.

Could anyone please provide some details (or at least sketch of) how to do the approximation correctly?

Keith
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  • Perhaps use convolution with a mollifier to approximate L^p by C_c^$\infty$ – fGDu94 Mar 02 '24 at 02:36
  • @fGDu94 That is the point. How can I be sure that convolution with a mollfier belongs to $C^\infty_c(\Omega)$? – Keith Mar 02 '24 at 07:33
  • Use a mollifier from $C_c^{\infty}$ – fGDu94 Mar 02 '24 at 14:32
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    Look up Folland’s analysis book. This is one if the last theorems in the section on convolutions. – peek-a-boo Mar 02 '24 at 16:16
  • Or if you know $C_c$ is dense, then you can apply Stone-Weierstrass to get a polynomial which uniformly approximates it. Then, multiply by a smooth bump function to get something with compact support. Lastly, bounded open set is not important. – peek-a-boo Mar 02 '24 at 16:18
  • @peek-a-boo The section on convolutions mostly deal with whole $\mathbb{R}^n$. If you are talking about the $C^\infty$ Urysohn lemma, I don't think that lemma is that much relevant to this question. – Keith Mar 02 '24 at 21:05
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    understand what happens in $\Bbb{R}^n$; a minor rewording gives you the result for all open $\Omega$. – peek-a-boo Mar 02 '24 at 21:20
  • https://math.stackexchange.com/questions/156444/why-c-0-infty-is-dense-in-lp This seems to contain all the details. – Keith Mar 02 '24 at 21:22

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