# Is the space $C(0,1)$ dense in Sobolev space $H^2(0,1)$? And $C^\infty_c(\Omega) = C^\infty_0(\Omega)$?

I have two questions:

1. Is $C(0,1)$ dense in $H^2(0,1)$? Yes or no? Could you give me a proof or an explanation?

2. $C^\infty_c(\Omega) = C^\infty_0(\Omega)$? from the different textbooks, I see that the spaces $C^\infty_0(\Omega)$ consists of all those functions in $C^\infty(\Omega)$ that have compact support in $\Omega$. But it seems that $C^\infty_c(\Omega)$ has the same meaning. the little $c$ means compact support? or it depends on the set $\Omega$?

Please help me. Thanks.

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• $C(0,1)$ isn't even contained in $H^2(0,1)$... – Nate Eldredge Apr 12 '17 at 4:11
• thanks. can you give some details? in addition, what about the second one? – I. Melissa Apr 12 '17 at 6:12
• The second one depends on your definitions of $C_c^\infty(\Omega)$ and $C_0^\infty(\Omega)$. – gerw Apr 12 '17 at 6:54
• What do you mean by $H^2$? Hardy space or Sobolev space? – haemi Apr 12 '17 at 7:10

## 1 Answer

The meaning of subscript $0$ in $C_0^\infty$ depends on where you see this notation. Some authors use it to mean "compact support". Others use it to mean "vanishing on the boundary", and if they need notation for smooth functions with compact support, it's $C_c^\infty$.

The space of continuous functions, such as $C(0,1)$, is not contained in $H^2$, so one can't really talk about it being dense there. But it's true that its intersection with $H^2$ is dense in $H^2$, since it contains all smooth functions with square-integrable second derivatives, and those are dense in $H^2$.