Folland 2.22 absolute value of Lebesgue integral less than integral of absolute value

complex-analysislebesgue-integralmeasure-theoryreal-analysis

I have been learning real analysis but I am a bit shaky with complex numbers. In the following proposition, Folland writes $|\int f|=\overline{\operatorname{sgn}(\int f)}\int f$? Why is this true and what is the motivation for doing this? I understand that in the real numbers, $x=\operatorname{sgn}(x)|x|$? I could understand then getting $|x|=\frac{1}{\operatorname{sgn}(x)}x$. However why does the above hold in the complex numbers using modulus instead of absolute value?

EDIT: One more question, why is it that $\int \alpha f$ is real?

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Best Answer

By definition $\operatorname{sgn}(z) = z/|z|$ is a unit vector. As such, $|z| = z/\operatorname{sgn}(z) = \overline{\operatorname{sgn}(z)} * z$.

This is because for any unit vector $u \in \mathbb{C}$, we have $1/u = \overline{u}$.

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