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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.

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    Is $f$ supposed to be real or complex valued?2012-03-01
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    For the forward direction : if $\int f = 0$ but $f \neq 0$, how do you believe $|f| - \beta f \ge 0$ holds almost everywhere?2012-03-01
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    @Nate : If OP uses inequalities on the values of $f$ it makes more sense to suppose that everything is real valued here.2012-03-01
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    @PatrickDaSilva: But if everything is real valued then $|f| = \beta f$ is a strange way to write that either $f \ge 0$ or $f \le 0$.2012-03-01
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    @Nate : Well then it is a strange way to do it.2012-03-01
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    related http://math.stackexchange.com/q/87636/82712012-03-01

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