Solve Rudin Theorem 11.35: Norm Inequality

[ehrenfest]

[SOLVED] Rudin Theorem 11.35

Homework Statement

Rudin wants to show that

\int_X |fg|d\mu \leq ||f|| ||g||

He says that it follows from the inequality

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

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

 

[morphism]

Have you tried to set lambda = something involving ||f|| and ||g||?

 

[ehrenfest]

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.