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Show subspace of H^1[0,1] is closed

  1. Apr 5, 2008 #1


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    [SOLVED] Show subspace of H^1[0,1] is closed

    1. The problem statement, all variables and given/known data
    I have an assignment that deals with some Sobolev spaces but I have never worked with them before. Only the definitions are given.

    Consider the Sobolev space

    [tex]W^{1,2}([0,1])=H^1([0,1])=\{u\in C([0,1]): \mbox{ there exists } u'\in L^2[0,1] \mbox{ such that } u(t)-u(0)=\int_0^tu'(s)ds \ \ \forall t \in [0,1]\}[/tex]

    It is a Hilbert space with the inner product


    I am trying to show that the following subspace of H^1 is closed:

    [tex]E=\{u\in H^1([0,1]):u(0)=u(1)\}[/tex]

    3. The attempt at a solution

    So I say, let u_n-->u (in H^1); we must show that u(0)=u(1).

    u_n-->u means that ||u_n - u||-->0 in the norm induced by the above inner product. So it means

    [tex]\int_0^1[(u_n-u)'^2+(u_n-u)^2] \rightarrow 0[/tex]

    or, using the fact that [tex]||u_n-u||^2=(u_n-u,u_n-u)=(u_n,u_n)-2(u_n,u)+(u,u)[/tex],

    [tex]\int_0^1[(u_n')^2+u_n^2] - 2\int_0^1[u_n'u'+u_nu]+\int_0^1[(u')^2+u^2]\rightarrow 0[/tex]

    I don't see how to retrieve u(0)=u(1).
    Last edited: Apr 5, 2008
  2. jcsd
  3. Apr 6, 2010 #2

    how to construct an example of a non-separable Hilbert Space?

    can anyone suggest me something please?
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