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Math Help - Not compact embedding

  1. #1
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    Not compact embedding

    Hi,

    how to prove, that for n=1 and \Omega:=(0,1) the embedding C^0(\overline{\Omega}) \to L^2(\Omega) is not compact?

    I've got a hint, to use the sequence  f_k(x)=\sin(\pi k x) with k \in \mathbb N. So, I should prove, that the space is not sequentially compact.

    Please can you help me?

    Bye,
    Alexander
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  2. #2
    Super Member girdav's Avatar
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    Re: Not compact embedding

    The sequence \{f_k\} is weakly convergent in C^0(\overline{\Omega}) but it's not strongly convergent in L^2(\Omega).
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  3. #3
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    Re: Not compact embedding

    An argument without weak topologies would be: \| f_k\| _{C^0} = 1, but \| f_k\|_{L^2} = 2\pi (I think, haven't checked this is the correct value, the fact is that this norm is constant) and f_k is orthogonal in L^2.
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  4. #4
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    Re: Not compact embedding

    Hi,
    thank you for your hints. By definition, a sequence \{x_k\} in a banach space X is weakly convergent to an element x\in X , if  f(x_k) \to f(x) for every  f \in X^* , where X^* denotes the dual space of X.

    The dual space of C^0([0,1]) is the space of all linear functions C^0([0,1]) \to \mathbb R. So I have to prove that there exists a f, with  g(f_k)= g(\sin (\pi k))\rightarrow g(f) for all g:C^0([0,1]) \to \mathbb R ?
    ( f_k(x)=\sin(\pi k x))
    How to find this f?

    Now to the other statement:
    Do I have to prove that there is no \widetilde{f} with \left \| \sin(\pi k) - \widetilde{f}\right \|_2 \to 0 ?

    Bye,
    Alexander
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  5. #5
    Super Member girdav's Avatar
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    Re: Not compact embedding

    You can show that \{f_k\} converges weakly to 0. Then use the characterization of the dual space of C^0(\overline{\Omega}).
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