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Math Help - How to prove that LP(R) does not form a Hilbert space for p different from 2.

  1. #1
    Ase
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    How to prove that LP(R) does not form a Hilbert space for p different from 2.

    Hi

    I am trying to prove that LP(R) does not form a Hilbert space for p different from 2.

    My thoughts are to try and disprove either Cauchy-Schwarz' inequality, the parallelogram law or similar. I am however not quite sure how to do this.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Ase View Post
    Hi

    I am trying to prove that LP(R) does not form a Hilbert space for p different from 2.

    My thoughts are to try and disprove either Cauchy-Schwarz' inequality, the parallelogram law or similar. I am however not quite sure how to do this.
    Cauchy–Schwarz won't work, because it involves the inner product, and if the space is not a Hilbert space you don't have an inner product to work with. But the parallelogram law approach is a good idea. You want to know whether or not \|f+g\|_p^2 + \|f-g\|_p^2 = 2(\|f\|_p^2 + \|g\|_p^2). Try taking f to be the function that is equal to 1 on the interval [0,1] and zero elsewhere, and g to be the function that is equal to 1 on the interval [1,2] and zero elsewhere. The parallelogram identity applied to these two functions reduces to the equation 2(2)^{2/p} = 4(1)^{2/p}, which holds only if p=2.
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  3. #3
    Ase
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    Hi Opalg

    Thanks a lot for the brilliant answer. Just what I needed.
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