Hilbert spaces

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October 24th 2009, 08:38 PM
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Hilbert spaces

Is it possible that a fonction u(x) element of C[0,1], u(0)=u(1)=0 is not element of H°1 (0,1) , the closure of the H1 Hilbert space ?

I believe it is not possible, but I cannot manage to justify my answer.

Thank you
October 24th 2009, 08:45 PM
Jose27

What is H°1 (0,1) is it $H_0 ^1 (0,1):= \overline {C_0 ^{\infty} (0,1)} \subseteq H^1(0,1)$ ? Are you talking about Sobolev spaces?

Edit: Assuming this is what you meant, the strongest I could find is that if $u \in H^1(0,1) \cap C[0,1]$ and $u(0)=u(1)=0$ then $u\in H_0 ^1 (0,1)$
October 25th 2009, 07:18 AM
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Quote:

Originally Posted by Jose27

What is H°1 (0,1) is it $H_0 ^1 (0,1):= \overline {C_0 ^{\infty} (0,1)} \subseteq H^1(0,1)$ ? Are you talking about Sobolev spaces?

Edit: Assuming this is what you meant, the strongest I could find is that if $u \in H^1(0,1) \cap C[0,1]$ and $u(0)=u(1)=0$ then $u\in H_0 ^1 (0,1)$

Yes, indeed, that was I meant. That's exactly my reasoning, but I was not sure of one conclusion :

Even if $u \in C[0,1]$ and not $u \in C^{\infty}[0,1]$ , we can have the same conclusions ?

The exact question I have is that if $u \in C[0,1]$ and $u(0)=u(1)=0$ , then is it possible that $u \notin H_0 ^1 (0,1)$ ?

We never have the hypothesis that $u \in H^1(0,1) \cap C[0,1]$ and $u(0)=u(1)=0$ . So, must I conclude that without that stronger hypothesis, we can have a function that is $C_0 [0,1]$ and not $u \in H^1(0,1)$ ?
October 25th 2009, 09:31 AM
Jose27

Tough one... I don't really know since working with functions in Sobolev spaces is messy as it is, but maybe trying to characterize these functions in easier terms is the best approach. For example: Is a function that is nowhere differentiable weakly differentiable? If the answer is no, then you have the desired function.