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I found this statement about rearrangement from analysis Lieb and Loss in chapter 3.

Suppose f, g are nonnegative functions in $L^2(\Bbb{R^n})$, then $\|f^*-g^*\|_2 \le\|f-g\|_2$

Where $f^*$ is the symmetric- decreasing rearrangement of $f$.

$$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt$$

I can see that it is true when $p=2$

but why it is true for $1 \le p \le \infty$, ie $\|f^*-g^*\|_p \le\|f-g\|_p$

Guy Fsone
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Sam
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1 Answers1

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The case $p=\infty$ is easy: let $M=\|f-g\|_\infty$. Since $f\le g+M$, it follows that $f^*\le g^*+M$, and we are done.

For $1\le p<\infty$ the function $t\mapsto |t|^p$ is convex on $\mathbb R$, which means that we can use Theorem 3.5 on the very same page of the book you are reading.

3.5 Theorem (nonexpansivity of rearrangement). Let $J:\mathbb R\to\mathbb R$ be a nonnegative convex function such that $J(0)=0$. Let $f$ and $g$ be nonnegative functions on $\mathbb R^n$, vanishing at infinity. Then $$\int_{\mathbb R^n} J(f^*-g^*)\le \int_{\mathbb R^n} J(f-g)\tag{1}$$ If we also assume that $J$ is strictly convex, that $f=f^*$, and that $f$ is strictly decreasing [in radial direction], then equality in (1) implies that $g=g^*$.

I suspect you asked this question because of the sentence "the obvious generalization is $\|f^*-g^*\|_p \le \|f-g\|_p$". I guess the authors did not mean to say "this generalization is obviously true", but rather "obviously, this is a generalization".