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Rudin Theorem 11.35

  1. May 26, 2008 #1
    [SOLVED] Rudin Theorem 11.35

    1. The problem statement, all variables and given/known data
    Rudin wants to show that

    [tex]\int_X |fg|d\mu \leq ||f|| ||g||[/tex]

    He says that it follows from the inequality

    [tex]0 \leq \int_X (|f| + \lambda |g|)^2 d\mu = ||f||^2 + 2 \lambda \int_X |fg| d\mu + \lambda^2 ||g||^2[/tex]

    which holds for any real lambda. Is there a special value of lambda that you are supposed to plug in here or is there a combination of values of lambda that I should use or should I take the limit as lambda goes to 0 or what...? I tried lambda = -1/2 but that didn't help at all. I don't see how you can get the norm of f and the norm of g multiplied instead of added


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 26, 2008 #2

    morphism

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    Have you tried to set lambda = something involving ||f|| and ||g||?
     
  4. May 26, 2008 #3
    we can assume that ||f|| and ||g|| are not zero since then f and g are zero except on a set of measure zero, so the integral must be zero, so we set lambda equal to -||f||/||g|| and the theorem follows, thanks.
     
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