Complex Lebesgue integral, property

Question

Lets say that you for real functions have proved that:

$|\int_{\Omega}fd\mu|\le \int_{\Omega}|f|d\mu$.

How do I then prove that it also holds for complex-valued functions?

I guess this amounts to showing that if $f=u+iv$, then:

$\sqrt{(\int_{\Omega}ud\mu)^2+(\int_{\Omega}vd\mu)^2}\le\int_{\Omega}\sqrt{u^2+v^2}d\mu$.

I guess we see that the property then also holds if the function f is purely imaginary, but what about when it is complex?

Answer 1

Choose $\varphi \in \mathbb{R}$ such that

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

Then

$$\begin{aligned} \left\lvert \int_\Omega f\,d\mu\right\rvert &= e^{i\varphi}\int_\Omega f\,d\mu\\ &= \int_\Omega e^{i\varphi}f\,d\mu\\ &= \int_\Omega \operatorname{Re} (e^{i\varphi}f)\,d\mu\\ &\leqslant \int_\Omega \lvert \operatorname{Re}(e^{i\varphi}f)\rvert\,d\mu\\ &\leqslant \int_\Omega \lvert e^{i\varphi}f\rvert\,d\mu\\ &= \int_\Omega \lvert f\rvert\,d\mu. \end{aligned}$$

Answer 2

HINT: Use Jensen's inequality.

Works for (semi)norms and vector valued functions.

Indeed: take a function $f$ with values in a (finite dimensional) normed vector space $(V, ||\cdot||)$. The function $||\cdot||$ is convex so we have the Jensen inequality:

$$ || \frac{1}{\mu(\Omega)} \cdot \int_{\Omega} f\, d\mu || \le \frac{1}{\mu(\Omega)} \cdot \int_{\Omega} ||f||\, d\mu $$ and since $||\cdot ||$ is positively homogenous of degree $1$ we get $$ || \int_{\Omega} f\, d\mu || \le \int_{\Omega} ||f||\, d\mu $$

To handle the case of infinite measure $\mu(\Omega) = \infty$ approximate the integrals with integrals on domains of finite measure.

Answer 3

We have:

$$\int_{\Omega}fd\mu=\left|\int_{\Omega}fd\mu\right|(i\sin \theta + \cos \theta)=\left|\int_{\Omega}fd\mu\right|\lambda$$

for $\theta \in (0,2\pi]$ and $\lambda \in \mathbb{C}$, $|\lambda|=1$.Then:

$$\left|\int_{\Omega}fd\mu\right|=\overline{\lambda}\int_{\Omega}fd\mu$$

Next:

$$\Re \left( \left|\int_{\Omega}fd\mu\right|\right)=\Re \left(\overline{\lambda}\int_{\Omega}fd\mu\right)=\int_{\Omega}\Re \overline{\lambda}fd\mu \leq \int_{\Omega}|\Re \overline{\lambda}f|d\mu \leq \int_{\Omega}| \overline{\lambda}f|d\mu=\int_{\Omega}|f|d\mu$$

Because $|\overline{\lambda}|=1$.