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Math Help - counterexample to parallelogram law

  1. #1
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    counterexample to parallelogram law

    This is a counterexample to the parallelogram law to show that the L1 norm is not induced by an inner product in [0,1]:

    f(x) = 1
    g(x) = 1

    |f|=|g|=1 and |f+g|=|f-g|=2

    Why is |f-g|=2??
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  2. #2
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    edit: stupid mistake

    Those two do not contradict the parallelogram law. Try the indicators on [0,1/2] and [1/2,1].
    Last edited by maddas; April 25th 2010 at 10:23 AM.
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  3. #3
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    f(x) = 1 on [0,1/2] and f(x) = 0 on [1/2,1]
    g(x) = 1 on [1/2,1] and f(x) = 0 on [0,1/2]

    |f|=|g|=1/2 and |f+g|=|f-g|=1

    But again then why is |f-g|=1??

    I am confused Hope you can help M. Thanks
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  4. #4
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    Um, I don't really understand... ;[

    The parallelogram law states that 2||f||^2 + 2||g||^2 - ||f+g||^2 = ||f-g||^2. Now ||f||=||g|| = 1/2 and ||f+g||=1. Therefore the LHS is 0, so the RHS is too. But ||f-g|| = 1, a contradiction. Therefore the parallelogram law does not hold in L1([0,1]).
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  5. #5
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    Everything makes sense there except why is
    ?

    Surely that would be 0?
    Thanks ever so much for your help.
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  6. #6
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    Let f:=1_{[0,1/2)},\;g:=1_{[1/2,1]}.

    |f-g| = |1_{[0,1/2)} - 1_{[1/2,1]}|. Since the indicators are on disjoint sets, the absolute value of their sum is the sum of their absolute values vis. |1_{[0,1/2)} - 1_{[1/2,1]}| = |1_{[0,1/2)}| + |-1_{[1/2,1]}| = |1_{[0,1/2)}| + |1_{[1/2,1]}| = 1. Integrating this gives 1.
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  7. #7
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    Maddas - you are absolutely awesome! This makes perfect sense. I have an exam after 2 days and has lowered my stress a bit! Thank you!!
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