1
$\begingroup$

Let $(X,\Sigma_x,\mu)$ be a measurable space ($X$-the set, $\Sigma_x$ - is a $\sigma$ algebra over X, and $\mu$ is a positive measure).

Let $f\in L^1(\mu)$ (meaning $\int_X|f|d\mu<\infty$), $f:X\to \mathbb{C}$.

How do I prove $|\int_X{}fd\mu|\leq\int_X{}|f|d\mu$?


my thoughts:

Denote $z=\int_Xfd\mu$, let $a\in\mathbb{C}$ be such that $az=|z|$ and then:

$|\int_X{}fd\mu|=|z|=az=a\int_X{}fd\mu$

but...then what?...

any help? :)

$\endgroup$
0

2 Answers 2

2
$\begingroup$

Hope this helps:

Proposition: If $f\in L^{1}$, then $\left|\int f\right| \leq \int |f|$.

Proof - If $f$ is real then $$\left|\int f\right| = \left|\int f^{+} - \int f^{-}\right| \leq \int f^{+} + \int f^{-} = \int |f|$$ If $f$ is complex-valued and $\int f\neq 0$, then $\int f = \left|\int f\right|e^{-i\theta}$. So $$\left|\int f\right| = e^{-i\theta}\int f = \int e^{-i\theta}f$$ In particular, $\int e^{-i\theta}f$ is real, so we have \begin{align*} \left|\int f\right| = Re\int e^{-i\theta}f = \int Re(e^{-i\theta} f) &\leq \int |Re(e^{-i\theta}f)|\\ &\leq \int |e^{-i\theta} f|\\ &= \int |f| \end{align*} Some of the notation above is from Real Analysis by Folland, if you have any questions please let me know.

$\endgroup$
0
2
$\begingroup$

In spirit, this is just the triangle inequality, and that's how we'll prove it. First assume $f$ is real valued. Remember that

$\int f d\mu = \sup_g \int g d\mu$

where the supremum is taken over all simple functions $g \leq f$, i.e. all $g$ of the form

$g(x) = \sum_j c_j 1_{A_j}$

for some finitely many real numbers $c_j$ and measurable sets $A_j$. The inequality holds for any such $g$, since

$\Big| \int g d\mu \Big| = \Big| \sum_j c_j \mu(A_j) \Big| \leq \sum_j |c_j|\mu(A_j) = \int |g| d\mu,$

by the triangle inequality. Thus

$\Big| \int f d\mu \Big| \leq \sup_g \Big| \int g d\mu \Big| \leq \sup_g \int |g| d\mu = \int |f| d\mu.$

Note that the assumption $f \in L^1(\mu)$ is unnecessary, since in the case that $f$ is not integrable, the right hand side is infinite, so the inequality holds trivially.

If $f$ is complex valued, then take $\alpha \in \mathbb{C}$ so that $\alpha \int f \in \mathbb{R}$, and write

$\Big| \int f d\mu \Big| = \int Re(\alpha f) d\mu \leq \int |Re(\alpha f)| d\mu \leq \int |\alpha f| d\mu = \int |f| d\mu$.

Edit: There are a couple of not-completely-trivial facts I've used. It would be good to check these to make sure you understand what's going on!

$\endgroup$
2
  • $\begingroup$ but how can you say something like $g\leq f$ when f is a complex function? $\endgroup$
    – Daniel
    Feb 17, 2017 at 9:21
  • $\begingroup$ Good point! This works for real valued functions: now I think you can get it to work by rotating, as you suggested. I think Hermes' suggestion leads to an easier proof though... $\endgroup$
    – J Richey
    Feb 17, 2017 at 9:35

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.