Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was reading Rudin and I stumbled upon a proof that I do not seem to understand. It is on page 325 of Baby Rudin $3^{rd}$ edition.

In case you do not have a copy I shall write some background information:

Suppose $\mu$ is a measure on $X$, and $f$ is a complex measurable function on $X$. Then, $$\int |f|\;d\mu < +\infty$$ and we define $$\int f \;d\mu = \int u \; d\mu + i\int v \; d\mu$$.

Now onto, my question. He wants to prove the following $$ \left| \int f \; d\mu \right| \leq \int |f| \; d\mu. $$

He begins as follows:

If $f \in \mathscr{L}(\mu)$, there is a complex number $c$, $|c| =1$, such that $$ c\int f \; d\mu \geq 0 $$

Put $g = cf = u +iv$ where $u$ and $v$ are real.Then

$$ \left| \int f \; d\mu \right| = c\int f \;d\mu = \int g \;d\mu = \int u \;d\mu \leq \int |f| \; d\mu.$$

The thing that bothers me is the following equality $$\int g \;d\mu = \int u\;d\mu .$$

How are these two functions equal? If we had assumed that $g = u +iv$, so should not g be $$\int g \;d\mu = \int u \; d\mu + i\int v \; d\mu?$$.

share|improve this question
    
$\left | \int f d \mu \right |$ is real so $\int v$ has to be zero otherwise you get something complex on the RHS. –  Matt N. Apr 9 '12 at 7:23
    
What's $\mathscr L (\mu)$? –  Matt N. Apr 9 '12 at 7:24
    
@MattN. If I remember correctly from my reading of Rudin's textbook, then I believe that $\cal L(\mu)$ is the $L^1$-space of the positive measure $\mu$. I think Rudin's notation is not standard in this regard. –  Amitesh Datta Apr 9 '12 at 13:21
    
@AmiteshDatta Cool, thank you very much! –  Matt N. Apr 9 '12 at 13:30

1 Answer 1

up vote 3 down vote accepted

$$ \left | \int f d \mu \right |$$ is real so $$ \left| \int f \; d\mu \right| = \int u \; d\mu + i\int v \; d\mu$$

has to be real too, which means that $\int v d \mu $ has to vanish.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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