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'm studying section 3.3 of Analysis by Lieb and Loss, about symmetric-decreasing rearrangement of functions.

Let $A\subset \mathbb{R}^n$ a Borel set of finite Lebesgue measure. They define $A^*$ to be the ball centered at 0 with the same measure that $A$.

The symmetric-decreasing rearrangement of a measurable function $f:\mathbb{R}^n \to \mathbb{R}$ is then defined by

$$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt,$$

by comparison to the "layercake" representation of $f$, namely $$f(x)=\int_0^{\infty} \chi_{\{f>t\}}(x)dt.$$

They say that it is then an obvious property that

$$\{x: f^*(x)>t\}=\{x: |f(x)|>t\}^* .$$

But I can't see why/how...

share|improve this question
    
Your statement is a bit suspect: if $A^*$ has the same measure as $A$, $A^*$ most likely will not be the unit ball. It will just be a ball. –  Willie Wong Jan 16 '12 at 14:47
    
I think that some $t$'s in your definitions should be replaced by $x$'s. –  Christian Blatter Jan 16 '12 at 14:53
    
Fixed, thanks and sorry for that. –  Klaus Jan 16 '12 at 14:57
add comment

1 Answer

up vote 1 down vote accepted

Hint 1: show that if $t_1 > t_2$, then $\{ |f| > t_1\}^* \subseteq \{ |f| > t_2\}^*$.

Hint 2: use this to show that if $y\in \{ |f| > t \}^*$, then $y \in \{|f| > s\}^*$ for every $0 \leq s \leq t$. Notice that this implies that $f^*(y) \geq t$ by the definition.

Hint 3: use hint 1 again to show that if $y\not\in \{|f| > t\}^*$, $$ \sup \left\{ s \geq 0 ~~|~~ y \in \{|f| > s\}^*\right\} \leq t $$ this implies in particular $f^*(y) \leq t$ (why?).

share|improve this answer
    
hint 3 is unnecessary. You already proved that the level set of $f^{*}$ is subset of the star set. –  TKM Feb 10 at 23:48
add comment

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.