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Thread: Non-continuous operator

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

    Hi,

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

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

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

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

    For a similar but alternative point of view, note that $\displaystyle A^{-1}$ continuous implies $\displaystyle A^{-1}$ bounded. Then $\displaystyle A^{-1}A$ is compact, which is possible only when $\displaystyle E$ 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 $\displaystyle E$ is infinite dimensional. The only thing I know is that if the Hilbert space $\displaystyle E$ is infinite dimensional and separable then we can find a Hilbertbasis.
    Since $\displaystyle A$ is bijective I think that $\displaystyle A^{-1}A(e_n) = e_n$.

    Suppose that $\displaystyle A^{-1}$ is continuous. Since $\displaystyle A$ is compact and $\displaystyle e_n \to 0$ (weak) it follows that $\displaystyle Ae_n \to 0$, hence $\displaystyle A^{-1}A(e_n) = e_n \to A^{-1}(0) = 0$ (the only element in the kernel is 0 as f is injective). Now, I have proved that weak convergence implies convergence and thus $\displaystyle E$ has to be finite dimensional. Which is a contradiction is if we have given that $\displaystyle E$ is infinite dimensional.
    Last edited by Siron; Jun 3rd 2013 at 12:57 AM.
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