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
up vote 1 down vote accepted

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

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.