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Math Help - Non-continuous operator

  1. #1
    MHF Contributor Siron's Avatar
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    Non-continuous operator

    Hi,

    Let E be a Hilbert space and let A: E \to F be a linear, bijective and compact operator. Prove that B: F \to E, B = A^{-1} is not continuous.
    Hint: (e_n)_n is an orthonormal and total sequence in E which convergens weak to 0 then Ae_n \to 0 as A is compact.

    I don't know how I can use the hint?

    Anyone?
    Thanks in advance.
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  2. #2
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    Re: Non-continuous operator

    Quote Originally Posted by Siron View Post
    Hi,

    Let E be a Hilbert space and let A: E \to F be a linear, bijective and compact operator. Prove that B: F \to E, B = A^{-1} is not continuous.
    Hint: (e_n)_n is an orthonormal and total sequence in E which convergens weak to 0 then Ae_n \to 0 as A is compact.

    I don't know how I can use the hint?

    Anyone?
    Thanks in advance.
    I believe you need E to be an infinite dimensional Hilbert space?

    If Ae_n \to 0, what is A^{-1}Ae_n?

    For a similar but alternative point of view, note that A^{-1} continuous implies A^{-1} bounded. Then A^{-1}A is compact, which is possible only when E is finite dimensional.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Non-continuous operator

    It's not explicitly given that E is infinite dimensional. The only thing I know is that if the Hilbert space E is infinite dimensional and separable then we can find a Hilbertbasis.
    Since A is bijective I think that A^{-1}A(e_n) = e_n.

    Suppose that A^{-1} is continuous. Since A is compact and e_n \to 0 (weak) it follows that Ae_n \to 0, hence A^{-1}A(e_n) = e_n \to A^{-1}(0) = 0 (the only element in the kernel is 0 as f is injective). Now, I have proved that weak convergence implies convergence and thus E has to be finite dimensional. Which is a contradiction is if we have given that E is infinite dimensional.
    Last edited by Siron; June 3rd 2013 at 01:57 AM.
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