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Rudin Theorem 11.35

  • Thread starter ehrenfest
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  • #1
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[SOLVED] Rudin Theorem 11.35

Homework Statement


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


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
morphism
Science Advisor
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Have you tried to set lambda = something involving ||f|| and ||g||?
 
  • #3
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Have you tried to set lambda = something involving ||f|| and ||g||?
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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