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<t\}, \{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\}|.

(more…)

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