Let's say we have an integrable function $f:\Omega \rightarrow \mathbb{C}$ on some measure space $(\Omega, \Sigma, \mu )$ for which the triangle inequality for integrals holds, i.e.: $$ \lvert \int_\Omega f d\mu \rvert = \int_\Omega \lvert f \rvert d\mu $$

If $f$ was real valued we could just write:

$$ \lvert \int_{\{ f > 0 \}} f d\mu + \int_{\{ f < 0 \}} f d\mu \rvert = \lvert \int_\Omega f d\mu \rvert = \int_\Omega \lvert f \rvert d\mu = \int_\Omega sgn(f) f d\mu = \int_{\{ f > 0 \}} f d\mu - \int_{\{ f < 0 \}} f d\mu $$

and look at the cases $ \int_{\{ f > 0 \}} f d\mu > -\int_{\{ f < 0 \}} f d\mu $ and $ \int_{\{ f > 0 \}} f d\mu < -\int_{\{ f < 0 \}} f d\mu $ to see that $f$ has to have the same sign almost everywhere.

I can't really see a way to extend the proof to the complex case. I read that equality in the first equation implies $f=e^{i \theta} \lvert f \rvert$ (almost everywhere) for some real constant $\theta$ - which sounds very natural. But how do I prove this?

  • $\begingroup$ Choose a $\theta$ such that $$e^{-i\theta} \int_{\Omega} f\,d\mu \geqslant 0.$$ $\endgroup$ – Daniel Fischer Oct 21 '15 at 19:58
  • $\begingroup$ But how do I proceed from there? If we knew that $f=e^{i \theta} \lvert f \rvert$ such a $\theta$ would also have this property. $\endgroup$ – NiU Oct 22 '15 at 17:38

We reduce the complex case to the real case by a rotation. We choose a $\theta \in \mathbb{R}$ with

$$e^{-i\theta} \int_{\Omega} f\,d\mu \geqslant 0.$$

If the integral vanishes, $\theta$ is completely arbitrary, otherwise it is determined modulo $2\pi$. In any case, we have

\begin{align} 0 &= \biggl\lvert \int_{\Omega} f\,d\mu\biggr\rvert - \int_{\Omega} \lvert f\rvert\,d\mu\\ &= e^{-i\theta}\int_{\Omega} f\,d\mu - \int_{\Omega} \lvert f\rvert\,d\mu\\ &= \int_{\Omega} e^{-i\theta} f - \lvert f\rvert\,d\mu\\ &= \int_{\Omega} \operatorname{Re}\, \bigl(e^{-i\theta} f\bigr) - \lvert f\rvert\,d\mu\\ &= \int_{\Omega} \operatorname{Re}\, \bigl(e^{-i\theta} f\bigr) - \lvert e^{-i\theta} f\rvert\,d\mu. \end{align}

For every $z \in \mathbb{C}$ we have $\operatorname{Re} z \leqslant \lvert z\rvert$, so the integrand of the last integral is a non-positive function. Since the integral is $0$, it follows that

$$E = \bigl\{ \omega \in \Omega :\operatorname{Re}\,\bigl(e^{-i\theta} f(\omega)\bigr) \neq \lvert e^{-i\theta} f(\omega)\rvert \bigr\}$$

is a $\mu$ null set. Since $\operatorname{Re} z = \lvert z\rvert$ implies $\operatorname{Im} z = 0$, we have

$$e^{-i\theta} f(\omega) = \operatorname{Re} \,\bigl(e^{-i\theta} f(\omega)\bigr) = \lvert e^{-i\theta}f(\omega)\rvert = \lvert f(\omega)\rvert,$$


$$f(\omega) = e^{i\theta} \lvert f(\omega)\rvert$$

on $\Omega \setminus E$, that is, $\mu$ almost everywhere.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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