Lipschitz property of the symmetric rearrangement I'm currently reading Talenti's paper "Best constant in Sobolev inequality" and am rather stuck on an argument on pg 363 (or pg 11 if you're reading the pdf). In this section of the paper, Talenti is proving the Polya-Szego inequality. More specifically, I don't see how (23a) implies the Lipschitzness of $u^*$. For some background, recall that for a non-negative function $u \in C^{\infty}_{c}(\mathbb{R}^n)$, the symmetric decreasing rearrangement of $u$, denoted by $u^*$ is defined as follows $$ u^*(x) := \sup\{t\geq0 : \mu(t) >  C_{n}\ |x|^n \} ,$$ where here $C_{n}$ is the volume of the unit ball in $\mathbb{R}^n$ and $\mu(t) = |\{x: u(x)>t\}|$ is the Lebesgue measure of the set where $u$ exceeds t. Then, evidently, $u^*$ is radially symmetric, is a decreasing function of the radius, and satisfies $$ |\{x: u(x)>t\}| =|\{x: u^*(x)>t\}| .$$ Using the coarea formula, together with the isoperimetric inequality, one can show the inequality:
$$Ch\mu(t)^{\frac{n-1}{n}} \leq \mu(t-h)-\mu(t),$$ where $C$ is a dimensional constant and $h>0.$ I'd like to show that this implies that $u^*$ is a Lipschitz function provided $u \in C^{\infty}_{c}(\mathbb{R}^n).$ So far I've tried plugging in $u^*(x)$ and $u^*(y)$ into the above inequality taking the roles of $t$ and $t-h$ to try and bound their difference but one gets a problematic factor of $|x|^{n-1}$ on the left hand side. Also note that if one can prove the Polya-Szego inequality for the $p = \infty$ case then one can bypass Talenti's argument entirely. Appreciate if anyone could take a look at this.
 A: $\newcommand{\R}{\mathbb R}$By the continuity of measure, the nonincreasing function $\mu$ is right-continuous. The function $u^*$ is radial, that is,
\begin{equation*}
    u^*(x)=U(|x|)
\end{equation*}
for some nonincreasing function $U\colon[0,\infty)\to[0,\infty)$ and all $x\in\R^n$. So,
\begin{equation*}
    U(a)=\sup\{t\ge0\colon\mu(t)>C_n a\}=\sup E_a \tag{1}\label{1}
\end{equation*}
for all real $a\ge0$, where
\begin{equation*}
    E_a:=\{t\ge0\colon\nu(t)>B_n a\}, \quad \nu:=\mu^{1/n},\quad B_n:=C_n^{1/n}. \tag{2}\label{2}
\end{equation*}
Note that the function $\nu$ is nondecreasing and right-continuous.
We want to show that the function $U$ is Lipschitz.
Take any real $s$ and $t$ such that $0\le s<t$. Then
\begin{equation*}
\begin{aligned}
    \nu(s)-\nu(t)&=\mu(s)^{1/n}-\mu(t)^{1/n} \\ 
    &\ge(\mu(s)-\mu(t))\frac1n\,\mu(s)^{1/n-1} \\ 
    &\ge C(t-s)\mu(t)^{1-1/n}\frac1n\,\mu(s)^{1/n-1}.  
\end{aligned}
\end{equation*}
The first inequality in the above display holds because $0\le s<t$ and $\mu$ is nonincreasing, and the second inequality in the above display holds by the last displayed inequality in the OP.
So,
\begin{equation*}
    \frac{\nu(t)-\nu(s)}{t-s}\le-\frac Cn\mu(t)^{1-1/n}\,\mu(s)^{1/n-1}.
\end{equation*}
Letting now $t\downarrow s$ and recalling that $\mu$ is right-continuous, we get
\begin{equation*}
    D^+\nu(s)\le-\frac Cn
\end{equation*}
for all real $s\ge0$, where $D^+\nu(s):=\limsup_{t\downarrow s}\frac{\nu(t)-\nu(s)}{t-s}$, the upper right derivative of $\nu$ at $s$.
Since $\nu$ is nonincreasing, it follows that
\begin{equation*}
    \nu(t)-\nu(s)\le-\frac Cn\,(t-s) \tag{3}\label{3}
\end{equation*}
for any real $s$ and $t$ such that $0\le s\le t$; cf., for instance, Example (v), Section 11.3.
Let now
\begin{equation*}
    L:=nB_n/C. \tag{4}\label{4}
\end{equation*}
To obtain a contradiction, suppose that
\begin{equation*}
    U(a)-U(b)>L(b-a)
\end{equation*}
for some real $a$ and $b$ such that $0\le a<b$. Then for some $t\in[0,U(a))$ and $s=U(b)$ we have
$t-s>L(b-a)$ and hence, by \eqref{3},
\begin{equation}
\nu(t)-\nu(s)<-\frac Cn\,(t-s)<-\frac Cn\,L(b-a)=\frac Cn\,L(a-b).  \tag{5}\label{5}
\end{equation}
On the other hand, by \eqref{1}--\eqref{2}, the conditions $t\in[0,U(a))$ and $s=U(b)$ imply that $t\in E_a$ and $s'\notin E_b$ for any real $s'>s$, so that $\nu(t)>B_na$ and $\nu(s)=\nu(s+)\le B_n b$. So,
\begin{equation*}
B_n(a-b)=B_na-B_nb<\nu(t)-\nu(s)<\frac Cn\,L(a-b) 
\end{equation*}
by \eqref{5},
which indeed contradicts \eqref{4}.
Thus,
\begin{equation*}
    U(a)-U(b)\le L(b-a)
\end{equation*}
for all real $a$ and $b$ such that $0\le a<b$. Since the function $U$ is nonincreasing, this shows that $U$ is Lipschitz. $\quad\Box$
