2
$\begingroup$

I am wondering if by replacing the Lebesgue measure in the definition of symmetric-decreasing function with a weighted Lebesgue measure, Riesz rearrangement inequality still holds. For clarity, I explain below.

If $w$ is a nice weight on $\mathbb{R}^n$, let $d\mu=w(x)dx$ and for any measurable set $A$ with $\mu(A)<\infty$, let $A^*$ be the open ball centred at the origin which satisfies $\mu(A)=\mu(A^*)$. Define the weighted symmetric-decreasing rearrangement of a Borel measurable function vanishing at infinity by

$$ f^*(x)=\displaystyle\int_0^\infty 1_{\left\{|f|>t\right\}^*}(x) dt,$$

where $1_A$ is just the characteristic function of $A$. As in the classical case, this is radially symmetric and non-decreasing.

Firstly, has this been studied in more detail? I came up with the definition because it seems to work in my problem, but I haven't found much about it. Secondly, as announced above, does the Riesz rearrangement inequality hold?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.