Ngô Quốc Anh

May 14, 2010

Symmetrization: Schwarz symmetrization

Filed under: Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:34

Given a measurable subset $E \subset \mathbb R^N$, we denote its $N$-dimensional Lebesgue measure by $|E|$. We will denote by $E^\star$ the open ball centered at the origin and having the same measure as $E$, i.e. $|E^\star|=|E|$. The norm of vector $x \in \mathbb R^n$ will be denoted by $|x|$. Finally, we will denote by $\omega_N$ the volume of the unit ball in $\mathbb R^N$. It is worth recalling that

$\displaystyle \omega_N=\frac{\pi^\frac{N}{2}}{\Gamma \left(\frac{N}{2}+1\right)}$

where $\Gamma$ us the usual gamma function.

Definition (Schwarz symmetrization). Let $\Omega \subset \mathbb R^N$ be a bounded domain. Let $u : \Omega \to \mathbb R$ be a measurable function. Then, its Schwarz symmetrization (or the spherically symmetric and decreasing rearrangement) is the function $u^\star : \Omega^\star \to \mathbb R$ defined by

$u^\star(x)=u^\sharp (\omega_N|x|^N), \quad x \in \Omega^\star$.

Observe that if $R$ is the radius of $\Omega^\star$, then

$\displaystyle\begin{gathered} \int_{{\Omega ^ \star }} {{u^ \star }(x)dx} = \int_{{\Omega ^ \star }} {{u^\sharp }({\omega _N}{{\left| x \right|}^N})dx} \hfill \\ \qquad= \int_0^R {{u^\sharp }({\omega _N}{{\left| x \right|}^N})N{\omega _N}{\tau ^{N - 1}}d\tau } \hfill \\ \qquad= \int_0^{|{\Omega ^ \star }|} {{u^\sharp }(s)ds} \hfill \\ \qquad= \int_0^{|\Omega |} {{u^\sharp }(s)ds} . \hfill \\ \end{gathered}$

We obviously have the following properties of Schwarz symmetrization (more…)

May 2, 2010

Symmetrization: The Decreasing Rearrangement

Given a measurable subset $E \subset \mathbb R^N$, we denote its $N$-dimensional Lebesgue measure by $|E|$.

Let $\Omega$ be a bounded measurable set. Let $u :\Omega \to \mathbb R$ be a measurable function. For $t \in \mathbb R$, the level set $\{u>t\}$ is defined as

$\displaystyle \{u>t\}=\{x\in \Omega: u(x)>t\}$.

The sets $\{u, $\{u \geqslant t\}$, $\{u=t\}$ and so on are defined by analogy. Then the distribution function of $u$ is given by

$\displaystyle \mu_u(t)=|\{u>t\}|$.

This function is a monotonically decreasing function of $t$ and for $t \geq {\rm esssup}(u)$ we have $\mu_u(t)=0$ while for $t\leqslant {\rm essinf}(u)$, we have $\mu_u(t)=|\Omega|$. Thus the range of $\mu_u$ is the interval $[0, |\Omega|]$.

Definition (Decreasing rearrangement). Let $\Omega \subset \mathbb R^N$ be bounded and let $u :\Omega \to \mathbb R$ be a measurable function. Then the (unidimensional) decreasing rearrangement of $u$, denoted by $u^\sharp$, is defined on $[0, |\Omega|]$ by

$\displaystyle {u^\sharp }(s) = \begin{cases} {\rm esssup} (u),& s = 0, \hfill \\ \mathop {\inf }\limits_t \left\{ {t:{\mu _u}(t) < s} \right\}, & s > 0. \hfill \\ \end{cases}$

Essentially, $u^\sharp$ is just the inverse function of the distribution function $\mu_u$ of $u$. The following properties of the decreasing rearrangement are immediate from its definition.

Proposition 1. Let $u : \Omega \to \mathbb R^N$ where $\Omega \subset \mathbb R^N$ is bounded. Then $u^\sharp$ is a nonincreasing and left-continuous function.

Proposition 2. The mapping $u \mapsto u^\sharp$ is non-decreasing, i.e. if $u\leqslant v$ in the sense that $u(x) \leqslant v(x)$ for all $x$, where $u$ and $v$ are real-valued functions on $\Omega$ then $u^\sharp \leqslant v^\sharp$.

We now see that $u^\sharp$ is indeed a rearrangement of $u$.

Proposition 3. The function $u : \Omega \to \mathbb R$ and $u^\sharp : [0,|\Omega|] \to \mathbb R$ are equimeasurable (i.e. they have the same distribution function), i.e. for all $t$

$\displaystyle |\{u >t\}|=|\{u^\sharp >t\}|$.