A proof of Pólya-Szegő inequality

Question

Denote by $|A|$ the $N-$dimensional Lebesgue measure of a Borel set $A \subset \mathbb{R}^N$ and define $$ A^\ast := B_{R}(0), \quad R = \left(\frac{N}{\omega_N}|A| \right)^{\frac{1}{N}}, $$ where $\omega_N$ is the surface measure of $S^{N-1}$. It is known that $|A| = |A^\ast|$. The symmetric-decreasing rearrangement of the characteristic function $\chi_A$ is defined by $$ (\chi_A)^\ast = \chi_{A^\ast}. $$ Now, for $f \in C^\infty_0(\mathbb{R}^N)$ we define, for $t > 0$ the set $A_t = \{x \in \mathbb{R}^N : |f(x)| > t\}$ and the function $$ f^\ast(x) = \int_0^\infty (\chi_{A_t})^\ast (x) dt. $$ In other words, $$ f^\ast(x) = \int_0^\infty \chi_{A_t^\ast} (x) dt. $$ It is also also known that $f^\ast$ is nonnegative, radially symmetric and nonincreasing. As a reference I am using the book Analysis from Elliott H. Lieb and Michael Loss, page 81. I am looking for a proof of the inequality $$ \|\nabla f ^\ast\|_{L^2(\mathbb{R}^N)} \leq \|\nabla f\|_{L^2(\mathbb{R}^N)} $$ whenever $f \in H^1(\mathbb{R}^N)$. Or more general, $$ \int_{\mathbb{R}^N} |\nabla f^\ast|^p \leq \int_{\mathbb{R}^N} |\nabla f|^p, $$ for any $p > 1$. Any help is welcome.

Answer 1

That is known as the Polya-Szego inequality whose proof uses the co-area formula see for instance: Wikipedia.

See also the monographs: Symmetric Decreasing Rearrangement Is Sometimes Continuous Frederick J. Almgren, Jr., Elliott H. Lieb

For more details see the book: Baernstein II, Albert: Symmetrization in analysis, 2019.

Answer 2

This proof is due to a brilliant friend of mine, Joel, but of course that any mistakes are entirely my own.

Let us start by proving that if $u\geq 0$ is such that $u\in W^{1,\infty}(\mathbb{R}^n)$, then $u^*\in W^{1,\infty}(\mathbb{R}^n)$ with: $$\|\nabla u\|_{L^\infty(\mathbb{R}^n)}\leq \|\nabla u\|_{L^\infty(\mathbb{R}^n)}$$

Because $|\{u>t\}|=|\{u^*>t\}|$ (this follows from this other post) for any $t>0$, it follows by choosing $t=\|u\|_{L^\infty(\mathbb{R}^n)}$ that $\|u^*\|_{L^\infty(\mathbb{R}^n)}\leq \|u\|_{L^\infty(\mathbb{R}^n)}$. Similarly, by choosing $t=\|u\|_{L^\infty(\mathbb{R}^n)}$ we have the other inequality and it follows that $\|u\|_{L^\infty(\mathbb{R}^n)}=\|u^*\|_{L^\infty(\mathbb{R}^n)}$.

Now we will try to prove that $u^*$ is Lipschitz with constant $L=\|\nabla u\|_{L^{\infty}}$. This will be enough because then $(u^*(x+te_i )-u^*(x))/t$ will be bounded in $L^2(B_N)$ for any compact $N\in \mathbb{N}$. Using that $L^2(B_N)$ is reflexive, there will exists a sequence such that the quotients converge weakly. We call the weak limit $\partial_i (u^*)^N$. It is clear that for different $N$ our functions coincide in the smaller domain. All in all, one is able to define a $\partial_i u^*$ which will be a weak derivative for $u^*$. Suppose $\|\nabla u^*\|_{L^\infty(\mathbb{R}^n)}>L $. By construction there is weak convergence of some subsequence of $(u^*(x+te_i )-u^*(x))/t$ in $L^2(K)$ for a fixed compact set $K\subseteq \{u>L\}$ of positive measure (this is possible by inner regularity of Lebesgue's measure). But then, by weak convergence $L<\|\nabla u^*\|_{L^2(K)}\leq\liminf_k\|[|(u^*(x+t_ke_i )-u^*(x))/t_k)_{i=1}^n|\|_{L^2(K)}\leq L$, which is a contradiction. Therefore, $\|\nabla u^*\|_{L^\infty(\mathbb{R}^n)}\leq L$.

As remarked, it suffices to prove $u^*$ is Lipschitz almost everywhere. Take the full measure event which yields the $u^*$ and the $u$ supremumns to coincide with the $L^\infty$ upper bounds. Because $u^*$ is radial and decreasing we need only consider what happens to $0<u^*(y_1)-u^*(y_2)$ when $\|y_1\|<\|y_2\|$ and $y_1$ and $y_2$ are in the full measure event we are considering above. Because our functions share the same $L^\infty$ norm, then there are points $x_2$ with $u^*(y_2)\leq u(x_2)$ (as $u^*(y_2)<u^*(y_1)\leq \|u\|_{L^\infty(\mathbb{R}^n)}$). Now we have that if $y\in x_2+B(0,u^*(y_2)-u^*(y_1)/L)$, then:

$$ u^*(y_2) \leq u(x_2)< u(y)+L\frac{u^*(y_2)-u^*(y_1)}{L} $$ or equivalently, $u(y)> u^*(y_1)$ and we conclude that: $$ \{x: u(x)\geq u^*(y_2)\}+B(0,u^*(y_2)-u^*(y_1)/L)\subseteq\{x: u(x)>u^*(y_1)\} $$ By Brunn-Minkowski (applied with compact sets and then with a final approximation argument), yields that: $$|u\geq u^* (y_2)|^{1/n}+ w_n^{1/n}\frac{u^*(y_2)-u^*(y_1)}{L}\leq |u>u^*(y_1)|^{1/n} \tag{1}$$

Now, notice that: $$u^*(y)=\int_0^\infty \chi_{\{f>t\}^*}(y) dt=\sup\left\{0<t: \|y\|<\left(\frac{|\{u>t\}|}{w_n}\right)^{1/n}\right\}$$ $$=\inf\left\{0<t: \|y\|\geq\left(\frac{|\{u>t\}|}{w_n}\right)^{1/n}\right\}$$ If we take sequences revelaing the infimum and the supremum and using continuity of the measures we obtain that: $$\|y\|\leq \left(\frac{|\{u\geq u^*(y)\}|}{w_n}\right)^{1/n}\quad \text{and}\quad \|y\|\geq \left(\frac{|\{u>u^*(y)\}|}{w_n}\right)^{1/n} $$

Using this in equation (1) yields that: $$w_n^{1/n}\frac{u^*(y_2)-u^*(y_1)}{L}\leq w_n^{1/n}\|y_1\|-w_n^{1/n}\|y_2\|$$ $$0<u^*(y_2)-u^*(y_1)\leq L|\|y_1\|-\|y_2\||\leq L\|y_1-y_2\|$$

This proves $u^*\in W^{1,\infty}(\mathbb{R}^n)$.

If $p>1$ and $u\in W^{1,p}(\mathbb{R}^n)$ for $u\geq 0$, then, if $u\in W^{1,p}(\mathbb{R}^n)$ we obtain that $u^*\in W^{1,p}(\mathbb{R}^n)$. Furthermore, >$$\|\nabla u^*\|_{L^p(\mathbb{R}^n)}\leq \|\nabla u\|_{L^p(\mathbb{R}^n)}$$

Take a sequence of $u_n\in C_0^\infty(\mathbb{R}^n)$ such that $u_n\rightarrow u$ in $W^{1,p}(\mathbb{R}^n)$. Because $u_n\in W^{1,\infty}(\mathbb{R}^n)$, then $u^*_n\in W^{1,\infty}(\mathbb{R}^n)\subseteq W^{1,p}(B_N)$ for any $N\in \mathbb{N}$. Because $ W^{1,p}(B_1)$ is reflexive we guarantee that there exists a subsequence such that $u_{n,1}^*\rightarrow u^{\#, 1}$ in $W^{1,p}(B_1)$. Taking a sub-subsequence we may consider $u_{n,2}^*\rightarrow u^{\#, 2}$ in $W^{1,p}(B_2)$. Proceeding with a diagonal argument, $u_{N,N}^*\rightarrow u^{\#, N}$ in $W^{1,p}(B_N)$ for any $N\in \mathbb{N}$.

Integrating agains a compactly supported smooth function is the smaller unit ball is it not difficult to see that the $u^{\#, N}$ are extensions of each other. Therefore, one may define $u^\#$ in $\mathbb{R}^n$. Now we notice that because $u_n$ converges to $u_n$ in $L^p(\mathbb{R}^n)$, then because the rearrangement decreases $L^p$ norms ( this is not exactly trivial and is an exercise with hints here), we have that $u^{\#}=u^*$. All in all, $u$ being in $W^{1,p}(\mathbb{R}^n)$ implies $u^*\in W^{1,p}(\mathbb{R}^n)$. Because there is weak convergence in $B_N$ we obtain that: $$\|\nabla u^*\|_{L^p(B_N)}\leq \liminf \|\nabla u_n^*\|_{L^p(B_N)}\leq \liminf \|\nabla u_n\|_{L^p(B_N)}=\|\nabla u\|_{L^p(B_N)}$$ where because the $u_n$ are smooth of compact support $\|\nabla u_n^*\|_{L^p(B_N)}\leq \|\nabla u_n\|_{L^p(B_N)}$ follows from the Wikipedia proof using the coarea formula.