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Suppose $\Omega\subset\mathbb{R^2}$ is open and bounded, and let $f:\Omega\rightarrow [0,\infty)$ be measurable. Moreover, let $\Omega^{\ast}$ denote the closed disk with midpoint $0\in\mathbb{R}^2$ and the same area as $\Omega$ (i.e. $\vert\Omega\vert=\vert \Omega^{\ast} \vert$).

The Schwarz symmetrization is defined as the function $$f^{\ast}:\Omega^{\ast}\rightarrow[0,\infty),\quad f(x):=\sup\lbrace{ c\geq 0:x\in \Omega_c^{\ast} \rbrace},$$

where $\Omega_c:=f^{-1}([c,\infty))\subset\Omega$ and $\Omega_c^{\ast} $ denotes the closed disk with midpoint $0\in\mathbb{R}^2$ and $\vert\Omega_c^{\ast} \vert= \vert\Omega_c \vert$.

I would like to understand the following two problems:

1) Why are $f$ and $f^{\ast}$ equimeasurable, i.e. $\vert \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace \vert=\vert \Omega_{c} \vert$ for all $c\geq 0$ ?

2) Sometimes, the function $f^{\ast}$ is defined alternatively as

$$f^{\ast}(x):=\int_{0}^{\infty} \chi_{ {\lbrace f \geq t \rbrace}^{\ast} }(x) dt.$$

I am wondering if that definition is equivalent to the first one?

Unfortunately, I do not know how to answer 1) or 2). I think that if we use $\vert\Omega_c^{\ast} \vert= \vert\Omega_c \vert$, it might be helpful. Then it remains to show: $$\vert \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace \vert = \vert \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast} \vert$$

Are both sets within $\vert\cdots \vert$ in the last equation the same? That is, do we have: $$\lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace = \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast}$$

If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.

Oliver Watt
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1 Answers1

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Yes, the equality $$ \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace = \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast} $$ is true, which implies affirmative answers to your other questions. Note that the set on the right is $\Omega_c^*$. So the claim is that $$ f^*(x)\ge c \iff x\in \Omega_c^* \tag1 $$ The implication $\impliedby$ is immediate because $f^*(x):=\sup\lbrace{ c\geq 0:x\in \Omega_c^{\ast} \rbrace}$.

Conversely, if $f^*(x)\ge c$ then for every $n\in\mathbb{N}$ it holds that $x\in \Omega_{c-1/n}^{\ast}$. Note that $$ \Omega_c = \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n} $$ which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness): $$ \left|\Omega_c\right| = \inf_{n\in\mathbb{N}}|\Omega_{c-1/n}| $$ Since the measure determines the Schwarz symmetrization of these sets, we have $$ \Omega_c^* = \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n}^* $$ (the intersection of closed balls of radii $r_n\searrow r$ is the closed ball of radius $r$).

Thus $x\in \Omega_x^*$, proving (1).

  • Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ \Omega_c^* = \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n}^* $". Is the following explanation correct?: $$ \vert \Omega_c^* \vert = \vert \Omega_c \vert = \inf_{n\in\mathbb{N}} \vert \Omega_{c-1/n}\vert = \inf_{n\in\mathbb{N}} \vert \Omega_{c-1/n}^* \vert = \vert \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n}^* \vert$$. Thus, both symmetrizations must be the same. – Denilson Orr Aug 02 '18 at 18:55
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    I meant that the intersection of closed balls of radii $r_n\searrow r$ is the closed ball of radius $r$ –  Aug 02 '18 at 19:03