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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...

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    Fixed, thanks and sorry for that.2012-01-16

4 Answers 4

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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?).

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    hint 3 is unnecessary. You already proved that the level set of $f^{*}$ is subset of the star set.2014-02-10
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Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. One can check that for every, $0 one has $\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}$ this entails that, \begin{equation}\label{eq-inclu t-s}\tag{I} \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}~~~\textrm{for all $s\in ]0,t[$}. \end{equation} this implies that,$ \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) =1 ~~~s\in (0,t)$

Therefore, from definition of $f^{*}$, if $y\in \{|f|>t\}^*$ then we have $\begin{align*} f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ &= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ & = \int_{0}^{t} ds+\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds \\ &>t. \end{align*}$

Whence, $\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{ x \in \mathbb{R}^n:f^{*}(x)> t \right\}.$ On the other hand, if we suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$ then for all $s>0$ such that $ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}$ one has $0.

Indeed, $t>s $ then from \eqref{eq-inclu t-s} $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$ which is contradiction since we assumed that the converse is true. this means that,

$\sup\left\{s>0 : y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}\right\}\leq t. $ We then deduce that, $\begin{align*} f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ &= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \underbrace{\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds}_{=0}\leq=t \end{align*}$ that is $f^*(y)\leq t$ or that $y\notin \left\{x \in \mathbb{R}^n: f^*(x) > t \right\}$. We've just prove that,

\begin{equation}\label{eq}\tag{II} \Bbb R^n\setminus \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \Bbb R^n\setminus\left\{x \in \mathbb{R}^n: f^*(x) > s \right\}~~~\textrm{for all $s\in ]0,t[$}. \end{equation}

Which end the prove by taking the complementary.

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I have been working on this subject for some months now and your question was one of the main obstacles I encountered. I think what makes this particular problem hard to attack is that it seems so obvious and geometrically correct that most would prefer not to bother with a formal proof. Anyway, what worked for me was proving first that $ r(t) = C_n\sqrt[n]{\mathcal{L}^n\left(\{ |f| > t \}\right)} $ where $C_n = \left( \frac{n\Gamma(n/2)}{2\pi^{n/2}} \right)^{\frac{1}{n}}$, is a lower semicontinuous function. Then you can prove that the parameter $t$ in the integral which defines $f^*$ will define a set that contains $x$ just in an OPEN interval. Showing that it is an interval is pretty easy, but the open part is where you need the lower semicontinuity of the function $r$.

Best regards!