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Math Help - Show that ||x|| is not a norm on Lp[0,1]

  1. #1
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    Show that ||x|| is not a norm on Lp[0,1]

    I could not prove this.
    Attached Thumbnails Attached Thumbnails Show that ||x|| is not a norm on Lp[0,1]-2.3.37.jpg  
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  2. #2
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    Let f be the function that is equal to 1 on the interval [0,1/2] and 0 on the interval [1/2,1], and let g=1-f.

    Then \|f\|_p = \|g\|_p = (1/2)^{1/p}<1/2, and \|f+g\|_p = 1. So \|f+g\|_p>\|f\|_p+\|g\|_p.
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  3. #3
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    need to show violation of sclar multiplication

    Opalq, thank you for your response. That is a very good proof. But I need to use the hint given in the question. That is I need to show the violation of

    \|cx\|_p=|c| \|x\|_p where c is a scalar, 0 < c < 1
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  4. #4
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    Quote Originally Posted by oztuserd View Post
    Opalq, thank you for your response. That is a very good proof. But I need to use the hint given in the question. That is I need to show the violation of

    \|cx\|_p=|c| \|x\|_p where c is a scalar, 0 < c < 1
    I think you are misunderstanding the hint. You will never find a function x such that \|cx\|_p \ne |c| \|x\|_p, because \|cx\|_p = \left[\int_0^1|cx(t)|^pdt\right]^{1/p} = \left[|c|^p\int_0^1|x(t)|^pdt\right]^{1/p} = |c|\|x\|_p.

    In fact, I used the hint in my example, in the inequality (1/2)^{1/p}<1/2, which is equivalent to 1/2<(1/2)^p.
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