Regarding the absolute value inequality for functions $f: I \subseteq \mathbb{R} \to \mathbb{C}$

Question

I'm studying the following simple proof of the inequality: $$ \left|\int_I f\,\right| \leq \int_I |f| $$

We consider $\alpha\in[0,2\pi]$ such that $e^{i\alpha}\int f$ is real. I understand how this is possible. Now, it follows that

$$ \left|\int f\,\right| = \left|e^{i\alpha}\int f\,\right| = \left|\operatorname{Re}\biggl(e^{i\alpha}\int f\biggr)\right| = \left|\int \operatorname{Re}(e^{i\alpha}f)\right| \leq \int \lvert\operatorname{Re}(e^{i\alpha}f)\rvert \leq \int |f| $$ which to me seems pretty transparent except for the third equality, $$ \left|\operatorname{Re}\biggl(e^{i\alpha}\int f\biggr)\right| = \left|\int \operatorname{Re}(e^{i\alpha}f)\right| $$

I've fiddled with this expression for some time with no luck. Any ideas? I'm convinced I am just missing some obvious property of taking real parts, but so far I haven't been able to see it. Thanks in advance.

Answer 1

$|Re(e^{i\alpha}\int f)| = |Re(\int e^{i\alpha}f)|$ due to $e^{i\alpha}$ being a constant.

$|Re(\int e^{i\alpha}f)| = |\int Re(e^{i\alpha}f)|$, because for $g(t) = u(t) + iv(t)$ and $I$ a real subset we have the definition $$\int_I g(t) \, dt = \int_I \big( u(t) + i v(t) \big) \, dt = \int_I u(t) \, dt + i \int_I v(t) \, dt,$$ so $Re(\int_I g) = \int_I u= \int_I Re(g)$.

Answer 2

You can remove several absolute values: given a complex number $z$ one can choose $\alpha$ such that $|z|=e^{i\alpha}z$. Thus, with this choice for $z=\int f$, $$ \left|\int f\,\right| = e^{i\alpha}\int f = \operatorname{Re}\biggl(e^{i\alpha}\int f\biggr) = \operatorname{Re}\biggl(\int e^{i\alpha}f\biggr) \color{red}{=} \int \operatorname{Re}(e^{i\alpha}f) \leq \int \lvert\operatorname{Re}(e^{i\alpha}f)\rvert \leq \int |f| $$ The red equality is the definition of integration of function with complex values (of a real variable).