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Math Help - Hilbert spaces

  1. #1
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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
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  2. #2
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    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)
    Last edited by Jose27; October 24th 2009 at 10:20 PM.
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    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) ?
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  4. #4
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    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.
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