- #1
ehrenfest
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[SOLVED] Rudin Theorem 11.35
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 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