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This is a step in a lecture note I'm reading. It should be simple because the author considers it obvious but I can't see it. What am I missing?

Suppose $U$ and $V$ are integrable over measure space $(\Omega,\mathcal{F},\mu)$. Claim: $$ \int_\Omega\sqrt{U^2+V^2}d\mu\geq\sqrt{\left(\int_\Omega Ud\mu\right)^2+\left(\int_\Omega Vd\mu\right)^2}. $$ Alternative phrasing: with $Z=U+iV$, claim that $$ \int_\Omega|Z|d\mu\geq\left|\int_\Omega Zd\mu\right|. $$ Thank you for your time!

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  • $\begingroup$ I allowed myself to change the title of your question, so that it can be found easier using the search function. $\endgroup$ – PhoemueX Aug 4 '14 at 21:30
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The second can be shown as follows (I like the "trick" used here):

Choose $\alpha \in \Bbb{C}$, $|\alpha| = 1$ such that (why is there such an $\alpha$?)

$$ \alpha \cdot \int_\Omega Z\,d\mu = \left|\int_\Omega Z\,d\mu\right|. $$

Then (because the left hand side is a real number)

$$ \left| \int_\Omega Z\,d\mu\right| = \rm{Re}\left(\int_\Omega \alpha Z \,d\mu\right) = \int_\Omega \rm{Re}(\alpha Z)\,d\mu \leq \int_\Omega |\alpha Z|\,d\mu = \int_\Omega |Z|\,d\mu. $$

The first form follows.

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  • $\begingroup$ Neat trick. I will remember it. Thanks. :) $\endgroup$ – Kim Jong Un Jun 21 '14 at 16:07

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