Decreasing rearrangement

Remarks

Example: let $X = [0, 1]$ with Lebesgue measure and $f(x) = x$. Then $d_f(s) = 1 - s$ for $0 \le s \le 1$ and $d_f(s) = 0$ for $s \ge 1$. For $0 \le t < 1$ the condition $d_f(s) \le t$ means $s \ge 1 - t$, so $f^*(t) = 1 - t$. For $t \ge 1$ every $s \ge 0$ works, so $f^*(t) = 0$.

Non-example: the function $f(x) = x$ on $[0, 1]$ is not its own decreasing rearrangement, since it is increasing. A function $g \ge 0$ on $[0, \infty)$ is its own rearrangement exactly when it is non-increasing and right-continuous.

The function $f^*$ is non-increasing and right-continuous. It has the same distribution function as $\abs{f}$ for Lebesgue measure on $[0, \infty)$. Hence $\int_0^\infty (f^*)^p \, dt = \int_X \abs{f}^p \, d\mu$ for $0 < p < \infty$. The symmetric decreasing rearrangement of a function on $\R^n$ is a related notion: it is radial, decreasing in $\abs{x}$, and equimeasurable with $\abs{f}$.