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Math Help - A complete Metric Space

  1. #1
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    A complete Metric Space

    Dear Colleagues,

    Could you please help me in solving the following problem:

    Show that the subspace Y\subseteq[a,b] consisting of all x\in C[a,b] such that x(a)=x(b) is complete.

    Best Regards.
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  2. #2
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    Quote Originally Posted by raed View Post
    Dear Colleagues,

    Could you please help me in solving the following problem:

    Show that the subspace Y\subseteq[a,b] consisting of all x\in C[a,b] such that x(a)=x(b) is complete.

    Best Regards.
    Where are you having difficulties with this? You need to take a sequence (x_n) in Y which converges in X. If x_n\to x then you have to show that x\in Y.
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    Okay I understand that but how can I prove that x(a)=x(b) ?
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  4. #4
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    Quote Originally Posted by raed View Post
    Okay I understand that but how can I prove that x(a)=x(b) ?
    Well you know that x_n\in Y, and so x_n(a) = x_n(b), for each n. Doesn't that help?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by raed View Post
    Dear Colleagues,

    Could you please help me in solving the following problem:

    Show that the subspace Y\subseteq[a,b] consisting of all x\in C[a,b] such that x(a)=x(b) is complete.

    Best Regards.
    Perhaps simpler than pure definition (though not much) is to notice that the linear functional \varphi:C[a,b]\to F:f\mapsto f(a)-f(b) is continuous and thus \ker\varphi (our set) is a closed subspace of the complete space C[a,b] and thus complete (this is easier assuming that you know (which I've stated to you before and which is a very simple fact) that the evaluation functionals \varphi_z:C[a,b]\to F:f\mapsto f(z) is continuous and \varphi=\varphi_a-\varphi_b thus continuous).
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