Let $u\colon\Omega\subset\mathbb{R}^N\to\mathbb{R}$ be a non negative measurable function, and $\Omega$ open and bounded. Consider $u^*$ the spherical rearrangement $$ u^*(x)=\sup\{t\geq0 : \mu\{x: u(x)\geq t \} > \omega_N |x|^N\} $$ where $\mu$ is the Lebesgue measure in $\mathbb{R}^N$, $w_N=\mu\{B(0,1)\}$ of the unitary ball in $\mathbb{R}^N$ (i.e. to the definition in http://www.math.toronto.edu/almut/rearrange.pdf). Consider a continuous function $F\colon \mathbb{R}\to \mathbb{R}$, then prove \begin{equation}\tag{1} \int_\Omega{F(u(x))dx}=\int_\Omega{F(u^*(x))dx}. \end{equation} I try to consider $$ \int_\Omega{F(u(x))dx}=\int_{-\infty}^{+\infty}{\mu\{x: F(u(x))\geq t \}dt}=\int_{-\infty}^{+\infty}{\mu\{x: F(u(x))\geq F(s) \}F'(s)ds} $$ but in this case i need a hyphotesis of derivability. Also (1) is said to be true with $\Omega=\mathbb{R}^N$, but i think that is necessary that $F\in L^1(\mathbb{R}^N)$.
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Are you sure about $u$ begin a function on $\Omega$? If $u$ has support on $\Omega$, the rearrangement $u^*$ will have support on the ball centered at the origin of volume $\mu(\Omega)$. – felipeh Jun 27 '14 at 14:34
1 Answers
I think you're on the right track, but you need the following ingredient:
For almost all $a, b\in \mathbb{R}$ with $a<b$, $$ \mu\{x; u(x)\in (a,b)\} = \mu\{x; u^*(x) \in (a,b)\}. $$
To prove this, observe that, for almost all $a, b$, $u^*(x) \in (a,b)$ is equivalent to: $$ \mu\{x; u(x)\geq b\} \leq \omega_N |x|^N \leq \mu\{x; u(x)\geq a\}. $$ This defines an annulus of measure $\mu\{x; u(x)\geq a\} - \mu\{x; u(x)\geq b\}$, which is for almost all $a,b$ equal to the left hand side of our claim. Specifically, so long as $\{x; u(x) = a\}$ and $\{x; u(x) = b\}$ are sets of measure zero, we're fine.
It's not hard to get from here to $$ \mu\{x; F(u(x)) > t\} = \mu\{x; F(u^*(x)) > t\} $$ for almost all $t$. Indeed since $F$ is continuous, $F^{-1}((t,\infty))$ is open and thus a union of open intervals. I think you can work with weaker hypotheses, but I'm not completely sure. For example, if $F$ had a discontinuity at some $a$ where $\mu\{x; u(x) = a\} > 0$, perhaps there could be a problem? My intuition says that $F$ being measurable should be fine though.
Anyway, you can complete the proof just how you started it: \begin{align*} \int_{\Omega} F(u(x))\,dx &= \int_{-\infty}^\infty \mu\{x; F(u(x)) > t\} \,dt \\ &= \int_{-\infty}^\infty \mu\{x; F(u^*(x)) > t\} \,dt \\ &= \int_{\Omega} F(u^*(x))\,dx. \end{align*} Like I mentioned in the comment, I don't think this works when $\Omega$ is not a ball centered at the origin. Also, if $\Omega = \mathbb{R}^N$, then I believe what you need is $F\circ u\in L^1$, which depends on the behavior of $F(t)$ as $t\to\infty$ and $t\to 0$, as well as the conditions on $u$.
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