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

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- Oct 24th 2009, 08:38 PMTellHilbert 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 - Oct 24th 2009, 08:45 PMJose27
What is H°1 (0,1) is it $\displaystyle H_0 ^1 (0,1):= \overline {C_0 ^{\infty} (0,1)} \subseteq H^1(0,1)$? Are you talking about Sobolev spaces?

Edit: Assuming this is what you meant, the strongest I could find is that if $\displaystyle u \in H^1(0,1) \cap C[0,1]$ and $\displaystyle u(0)=u(1)=0$ then $\displaystyle u\in H_0 ^1 (0,1)$ - Oct 25th 2009, 07:18 AMTell
Yes, indeed, that was I meant. That's exactly my reasoning, but I was not sure of one conclusion :

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

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

We never have the hypothesis that $\displaystyle u \in H^1(0,1) \cap C[0,1]$ and $\displaystyle u(0)=u(1)=0$. So, must I conclude that without that stronger hypothesis, we can have a function that is $\displaystyle C_0 [0,1] $and not $\displaystyle u \in H^1(0,1) $ ? - Oct 25th 2009, 09:31 AMJose27
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.