Ngô Quốc Anh

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\}|.

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