# Triangle inequality for integrals with complex valued integrand

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!

• I allowed myself to change the title of your question, so that it can be found easier using the search function. – PhoemueX Aug 4 '14 at 21:30

## 1 Answer

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.

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