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I have a problem I need help in solving.

Suppose that $f\in L^1(\mu)$. I would like to show that $\left|\int_X f~d\mu\right| = \int_X |f|~d\mu$ if and only if $\exists$ a constant $\beta$ such that $|f|=\beta f$ a.e. on $X$.

My Attempt(for the forward direction)

$\left|\int_X f~d\mu\right| =\beta\int_X f~d\mu$ for some constant $\beta$. So $$\int_X |f|~d\mu = \beta \int_X f~d\mu \implies \int_X \left(|f|-\beta f\right)~d\mu =0$$ But since $|f|-\beta f \geq 0$, $|f| =\beta f$ a.e.

For the backward direction, this is what I have so far:
Suppose $|f|=\beta f$ a.e. I know that $\left|\int_X f~d\mu\right| \leq \int_X |f|~d\mu$. So, I must show the reverse inequality and I need some help with it.

Also is what I did for the forward direction ok?
Thanks very much.

Cindy
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1 Answers1

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The forward part is on the right track. You need a couple more steps though in the complex case. Taking the real part of $\int_X |f| - \beta f = 0$ gives $\int_X (|f| - Re(\beta f)) = 0$. Since $|\beta| = 1$ here, for a given $x$, $Re(\beta f(x)) \leq |f(x)|$ with equality if and only if $\beta f(x) = |f(x)|$ (Prove this). Then $\int_X (|f| - Re(\beta f)) = 0$ implies that $\beta f(x) = |f(x)|$ for almost every $x$.

The other direction is easier: use that $\int_X \beta f(x) = \beta \int_X f(x)$. So taking absolute values of both sides....

Zarrax
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