Vallée Poussin’s criterion now shows the relative weak compactness of
$\\{\|\nabla Tf_{j}\|:j\geq 1\\}$. Hence, there is a subsequence of $(\nabla
Tf_{j})$, also denoted $(\nabla Tf_{j})$, and a vector field
$g\in\left(L^{1}({\mathbb{R}}^{n})+L^{\infty}({\mathbb{R}}^{n})\right)^{n}=\left({L^{\Phi_{r}}({\mathbb{R}}^{n})}\right)^{n}$,
such that $\nabla Tf_{j}$ converges weakly to $g$, i.e., for each
$h\in\left(L^{1}({\mathbb{R}}^{n})\cap
L^{\infty}({\mathbb{R}}^{n})\right)^{n}=\left(L^{{\Psi_{r}}}({\mathbb{R}}^{n})\right)^{n}$,

$\lim_{j\to\infty}\int_{{\mathbb{R}}^{n}}(\nabla
Tf_{j})(x)h(x)\,dx=\int_{{\mathbb{R}}^{n}}g(x)h(x)\,dx.$

If $h\in C^{\infty}({\mathbb{R}}^{n})$ has compact support, then $h,\partial
h/\partial x_{i}\in L^{{\Psi_{r}}}({\mathbb{R}}^{n})$ and

(6.14) $\displaystyle\int_{{\mathbb{R}}^{n}}g_{i}(x)h(x)\,dx$
$\displaystyle=\lim_{j\to\infty}\int_{{\mathbb{R}}^{n}}\frac{\partial
Tf_{j}}{\partial
x_{i}}(x)h(x)\,dx=-\lim_{j\to\infty}\int_{{\mathbb{R}}^{n}}Tf_{j}(x)\frac{\partial
h}{\partial x_{i}}(x)\,dx$
$\displaystyle=-\int_{{\mathbb{R}}^{n}}Tf(x)\frac{\partial h}{\partial
x_{i}}(x)\,dx,$

where we have used the fact that the convergence of $Tf_{j}$ to $Tf$ in
$L^{\Phi_{r}}({\mathbb{R}}^{n})$ also implies weak convergence. From (6.14) we
see that $g$ is the weak gradient of $Tf$, and the fact that $\partial
Tf/\partial x_{i}$ is locally integrable is a direct consequence of the fact
that it belongs to $L^{{\Psi_{r}}}({\mathbb{R}}^{n})$. The second claim is
proved.

Next, we claim that

(6.15) $\big{\|}\|\nabla
Tf\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}\leq\liminf_{j\to\infty}\big{\|}\|\nabla
Tf_{j}\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}.$

With this goal in mind, we first prove that if
$u\in(L^{\Phi_{r}}({\mathbb{R}}^{n}))^{n}$, then

(6.16)
$\big{\|}\|u\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}u(x)\cdot
h(x)\,dx\right|:h\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n},\big{\|}\|h\|\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}.$

A similar formula in the scalar case is proved in [30, (10), Proposition 10,
Section 3.4]; when applied with $f$ and $\Phi$ there replaced by $\|u\|$ and
$\Phi_{r}$, respectively, it becomes

(6.17)
$\big{\|}\|u\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}\|u(x)\|\,v(x)\,dx\right|:v\in{\mathcal{M}}^{{\Psi_{r}}},\|v\|^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\},$

where ${\mathcal{M}}^{{\Psi_{r}}}$ is the closure of the span of all step
functions in $L^{{\Psi_{r}}}({\mathbb{R}}^{n})$. (We warn the reader that in
[30], $\|u\|_{L^{\Phi}({\mathbb{R}}^{n})}$ and
$\|u\|^{\prime}_{{L^{\Phi}({\mathbb{R}}^{n})}}$ are denoted by $N_{\phi}(u)$
and $\|u\|_{\Phi}$, respectively.) We have already seen that
$L^{\Psi_{r}}({\mathbb{R}}^{n})=L^{1}({\mathbb{R}}^{n})\cap
L^{\infty}({\mathbb{R}}^{n})$ and that the two spaces have equivalent norms,
so ${\mathcal{M}}^{{\Psi_{r}}}=L^{{\Psi_{r}}}({\mathbb{R}}^{n})$ due to Lemma
6.1. Thus, (6.17) becomes

(6.18)
$\big{\|}\|u\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}\|u(x)\|\,v(x)\,dx\right|:v\in
L^{{\Psi_{r}}}({\mathbb{R}}^{n}),\|v\|^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}.$

Let $S_{1}$ and $S_{2}$ denote the right-hand sides of (6.16) and (6.18),
respectively. The Cauchy-Schwartz inequality yields $S_{1}\leq S_{2}$. We can
restrict the supremum in (6.18) to nonnegative $v$. If $v\in
L^{{\Psi_{r}}}({\mathbb{R}}^{n})$, $v\geq 0$,
$\|v\|^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq 1$, and

$h(x)=\begin{cases}v(x)\frac{u(x)}{\|u(x)\|},&\text{if $u(x)\neq 0$},\\\
0,&\text{if $u(x)=0$},\end{cases}$

then $h\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n}$,
$\big{\|}\|h\|\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq 1$, and

$S_{1}\geq\left|\int_{{\mathbb{R}}^{n}}u(x)\cdot
h(x)\,dx\right|=\left|\int_{{\mathbb{R}}^{n}}\|u(x)\|v(x)\,dx\right|.$

This proves that $S_{1}\geq S_{2}$ and concludes the proof of (6.16). Now
(6.16) with $u$ replaced by $\nabla Tf$ and $\nabla Tf_{j}$ gives

$\displaystyle\big{\|}\|\nabla Tf\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}$
$\displaystyle=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}\nabla Tf(x)\cdot
h(x)\,dx\right|:h\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n},\big{\|}\|h\|\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}$
$\displaystyle=\sup\left\\{\lim_{j\to\infty}\left|\int_{{\mathbb{R}}^{n}}\nabla
Tf_{j}(x)\cdot
h(x)\,dx\right|:h\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n},\big{\|}\|h\|\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}$
$\displaystyle\leq\liminf_{j\to\infty}\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}\nabla
Tf_{j}(x)\cdot
h(x)\,dx\right|:h\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n},\big{\|}\|h\|\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}$ $\displaystyle=\liminf_{j\to\infty}\big{\|}\|\nabla
Tf_{j}\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}.$

This proves (6.15).

From (6.12), (6.13), and (6.15), we conclude that

(6.19) $\big{\|}\|\nabla
Tf\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}\leq\big{\|}\|\nabla
f\|\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}.$

Our fourth claim is that (6.11) holds when $\Phi$ is replaced by $\Phi_{r}$.
To see this, note that (6.19) holds if $\Phi_{r}$ is replaced by $a\,\Phi_{r}$
for any $a>0$, because all the preceding arguments are valid with this
replacement, due to $\int_{{\mathbb{R}}^{n}}a\,\Phi_{r}(\|\nabla
f(x)\|)\,dx<\infty$. If we choose $a$ so that
$\int_{{\mathbb{R}}^{n}}a\,\Phi_{r}(\|\nabla f(x)\|)\,dx=1$, then
$\big{\|}\|\nabla f\|\big{\|}_{L^{a\,\Phi_{r}}({\mathbb{R}}^{n})}=1$, by [14,
Proposition 2.1.10(4)] and the fact that $\|\nabla f\|\in
H^{a\Phi_{r}}({\mathbb{R}}^{n})$. Thus (6.19) becomes

$\big{\|}\|\nabla Tf\|\big{\|}_{L^{a\Phi_{r}}({\mathbb{R}}^{n})}\leq 1.$

By [14, Proposition 2.1.10(2)], the previous inequality holds if and only if

(6.20) $\int_{{\mathbb{R}}^{n}}a\,\Phi_{r}(\|\nabla Tf(x)\|)\,dx\leq
1=\int_{{\mathbb{R}}^{n}}a\,\Phi_{r}(\|\nabla f(x)\|)\,dx.$

This proves the fourth claim.

Since the nonnegative function $\Phi_{r}$ increases to $\Phi$ as $r\to\infty$,
one may apply the monotone convergence theorem to both sides of (6.20) and
obtain (6.11) when ${\mathcal{H}}^{n}({\mathrm{supp}}\,{f})<\infty$. To remove
the latter restriction, let $f\in
W^{1,1}_{loc}({\mathbb{R}}^{n})\cap{\mathcal{V}}({\mathbb{R}}^{n})$ and let
$f_{c}=\max\\{f-c,0\\}$ for $c>0$. Then $f_{c}$ satisfies the hypotheses of
the theorem and ${\mathcal{H}}^{n}({\mathrm{supp}}\,{f_{c}})<\infty$.
Moreover, $\nabla f_{c}=\nabla f$ on $\\{x:f(x)>c\\}$ and
$Tf_{c}=\max\\{Tf-c,0\\}$, by Proposition 3.6 with $\varphi(t)=(t-c)^{+}$, so
$\nabla Tf_{c}=\nabla Tf$ on $\\{x:Tf(x)>c\\}$. By (6.11) with $f$ replaced by
$f_{c}$, we have

$\displaystyle\int_{\\{x:Tf(x)>c\\}}\Phi(\|\nabla Tf(x)\|)\,dx$
$\displaystyle=$ $\displaystyle\int_{{\mathbb{R}}^{n}}\Phi(\|\nabla
Tf_{c}(x)\|)\,dx$ $\displaystyle\leq$
$\displaystyle\int_{{\mathbb{R}}^{n}}\Phi(\|\nabla
f_{c}(x)\|)\,dx=\int_{\\{x:f(x)>c\\}}\Phi(\|\nabla f(x)\|)\,dx.$

Letting $c\to 0$ and applying the monotone convergence theorem, we obtain
(6.11). ∎

######  Lemma 6.4.

Let $X=\mathcal{S}({\mathbb{R}}^{n})$ or ${\mathcal{V}}({\mathbb{R}}^{n})$ and
let $T:X\to X$ be a smoothing rearrangement. If $f\in
W^{1,\infty}({\mathbb{R}}^{n})\cap X$, then $Tf\in
W^{1,\infty}({\mathbb{R}}^{n})$ and

(6.21) $\operatorname*{ess\,sup}_{x\in{\mathbb{R}}^{n}}\|\nabla
Tf(x)\|\leq\operatorname*{ess\,sup}_{x\in{\mathbb{R}}^{n}}\|\nabla f(x)\|.$

###### Proof.

If $f\in W^{1,\infty}({\mathbb{R}}^{n})\cap X$, then by [3, Proposition 3.17],
$f$ coincides ${\mathcal{H}}^{n}$-almost everywhere with a Lipschitz function
in $X$. We may therefore assume that $f$ is Lipschitz, and then, as in the
proof of Theorem 5.8, also assume that $Tf$ is Lipschitz. By [3, Corollary
3.4], (6.21) is equivalent to $L_{1}\leq L_{2}$, where $L_{1}$ and $L_{2}$ are
the Lipschitz constants of $Tf$ and $f$, respectively. Since

$L_{1}=\sup_{d>0}\omega_{d}(Tf)/d\leq\sup_{d>0}\omega_{d}(f)/d=L_{2},$

the proof is complete. ∎

######  Corollary 6.5.

Let $T:{\mathcal{V}}({\mathbb{R}}^{n})\to{\mathcal{V}}({\mathbb{R}}^{n})$ be a
smoothing rearrangement and let $1\leq p\leq\infty$. If $f\in
W^{1,p}({\mathbb{R}}^{n})\cap{\mathcal{V}}({\mathbb{R}}^{n})$, then $Tf\in
W^{1,p}({\mathbb{R}}^{n})$ and

(6.22) $\big{\|}\|\nabla Tf(x)\|\big{\|}_{p}\leq\big{\|}\|\nabla
f(x)\|\big{\|}_{p},$

where $\|\cdot\|_{p}$ denotes the $L^{p}$ norm when $1\leq p<\infty$ and the
essential supremum over ${\mathbb{R}}^{n}$ when $p=\infty$.

###### Proof.

The case when $p=\infty$ corresponds to Lemma 6.4. Suppose that $1\leq
p<\infty$ and let $f\in
W^{1,p}({\mathbb{R}}^{n})\cap{\mathcal{V}}({\mathbb{R}}^{n})$. As
$W^{1,p}({\mathbb{R}}^{n})\subset W^{1,1}_{loc}({\mathbb{R}}^{n})$, Theorem
6.3 with $\Phi(t)=t^{p}$ gives (6.22). Our assumptions on $f$ show that the
right-hand side of (6.22) is finite, implying that $Tf\in
W^{1,p}({\mathbb{R}}^{n})$. ∎

The following example shows that the assumption that $T$ is smoothing in
Theorems 5.8 and 6.3 and Corollary 6.5 cannot be dropped in general. It also
shows that when $\Phi(t)=t$ in Theorem 5.8 or 6.3, or $p=1$ in Corollary 6.5,
the smoothing property is not necessary for the stated inequalities to hold.

######  Example 6.6.

Let $K\in{\mathcal{K}}^{n}_{(o)}$ and ${\mathcal{H}}^{n}(K)=\kappa_{n}$ and
let $T:{{\mathcal{V}}}({\mathbb{R}}^{n})\to{{\mathcal{V}}}({\mathbb{R}}^{n})$
be the rearrangement defined in Example 4.3(ii). We make the following two
claims.

(i) Inequalities (5.21) and (6.11) for $\Phi(t)=t$, and (6.22) for $p=1$, each
hold if and only if $K=x+B^{n}$ for some $x\in D^{n}$.

(ii) Inequalities (5.21) and (6.11) for strictly convex real-valued $\Phi$,
and (6.22) for $1<p\leq\infty$, each hold if and only if $K=B^{n}$.

Suppose these claims are true. If $K$ is not a ball, then $T$ is not smoothing
by Example 4.3(ii) and Theorems 5.8 and 6.3 (for real-valued strictly convex
$\Phi$), and Corollary 6.5 fail by (i) and (ii). If $K=x+B^{n}$ for some
$o\neq x\in D^{n}$, then $T$ is not smoothing by Example 4.3(ii) but
nevertheless Theorems 5.8 and 6.3 with $\Phi(t)=t$, and Corollary 6.5 with
$p=1$, hold by (i).

To prove the two claims, we note first that if
$f\in{{\mathcal{V}}}({\mathbb{R}}^{n})$ and
$\alpha_{f,t}={\mathcal{H}}^{n}(\\{x:f(x)\geq t\\})$ for $t\geq 0$, the
definition of $T$ yields

(6.23) $\\{x:Tf(x)\geq
t\\}=\left(\frac{\alpha_{f,t}}{\kappa_{n}}\right)^{1/n}K$

for $t>0$.

Now let $M\in{\mathcal{K}}^{n}_{(o)}$ and ${\mathcal{H}}^{n}(M)=\kappa_{n}$,
and let $f_{M}(x)=(1-h_{M^{\circ}}(x))^{+}$ for $x\in{\mathbb{R}}^{n}$, where
$M^{\circ}$ is the polar body of $M$ and $s^{+}$ is the nonnegative part of
$s\in{\mathbb{R}}$. Then $\\{x:f_{M}(x)\geq t\\}=(1-t)^{+}M$ for $t>0$. Since
$\alpha_{f_{M},t}=\left((1-t)^{+}\right)^{n}\kappa_{n}$, (6.23) with $f=f_{M}$
implies that

(6.24) $Tf_{M}=f_{K}.$

When $M=B^{n}$, we have $f_{B^{n}}(x)=(1-\|x\|)^{+}$ for
$x\in{\mathbb{R}}^{n}$ and hence

(6.25) $\|\nabla f_{B^{n}}(x)\|=\begin{cases}1,&\text{if $x\in
D^{n}\setminus\\{o\\}$,}\\\ 0,&\text{if $x\not\in B^{n}$}.\end{cases}$

The coarea formula for Lipschitz functions $f$ on ${\mathbb{R}}^{n}$ (see [3,
Theorem 4.19], [15, Theorem 1, p. 112]) states that

(6.26) $\displaystyle\int_{{\mathbb{R}}^{n}}\|\nabla
f(x)\|\,dx=\int_{0}^{\infty}{\mathcal{H}}^{n-1}(\\{x:f(x)=t\\})\,dt.$

Suppose that $T$ satisfies (5.21) or (6.11) with $\Phi(t)=t$, or (6.22) with
$p=1$. Then, using (6.26) with $f=f_{K}$, (6.24) with $M=B^{n}$, either
(5.21), or (6.11), or (6.22) with $f=f_{B^{n}}$, and (6.25), we obtain

$\displaystyle{\mathcal{H}}^{n-1}(\partial K)$ $\displaystyle=$ $\displaystyle
n\int_{0}^{1}(1-t)^{n-1}{\mathcal{H}}^{n-1}(\partial
K)\,dt=n\int_{0}^{\infty}{\mathcal{H}}^{n-1}(\\{x:f_{K}(x)=t\\})\,dt$
$\displaystyle=$ $\displaystyle n\int_{{\mathbb{R}}^{n}}\|\nabla
f_{K}(x)\|\,dx=n\int_{{\mathbb{R}}^{n}}\|\nabla Tf_{B^{n}}(x)\|\,dx$
$\displaystyle\leq$ $\displaystyle n\int_{{\mathbb{R}}^{n}}\|\nabla
f_{B^{n}}(x)\|\,dx=n\kappa_{n}={\mathcal{H}}^{n-1}(\partial B^{n}).$

Because ${\mathcal{H}}^{n}(K)={\mathcal{H}}^{n}(B^{n})$, equality must hold in
the isoperimetric inequality and consequently $K=x+B^{n}$ for some $x\in
D^{n}$. Conversely, suppose that $K=x+B^{n}$ for some $x\in D^{n}$ and $f\in
W^{1,1}({\mathbb{R}}^{n})\cap{\mathcal{V}}({\mathbb{R}}^{n})$. From (6.23) and
the fact that the decreasing function $t\mapsto\alpha_{f,t}$ can only have
countably many discontinuities, we get

$\\{x:Tf(x)=t\\}=\\{x:Tf(x)\geq t\\}\setminus\bigcup_{s>t}\\{x:Tf(x)\geq
s\\}\subset\left(\frac{\alpha_{f,t}}{\kappa_{n}}\right)^{1/n}\partial K$

for almost all $t>0$. Since $K$ is a translate of $B^{n}$, this and (6.26)
imply that

$\displaystyle\int_{{\mathbb{R}}^{n}}\|\nabla Tf(x)\|\,dx$
$\displaystyle=\int_{0}^{\infty}{\mathcal{H}}^{n-1}(\\{x:Tf(x)=t\\})\,dt\leq\int_{0}^{\infty}{\mathcal{H}}^{n-1}\left(\left(\frac{\alpha_{f,t}}{\kappa_{n}}\right)^{1/n}\partial
K\right)\,dt$
$\displaystyle=\int_{0}^{\infty}{\mathcal{H}}^{n-1}\left(\left(\frac{\alpha_{f,t}}{\kappa_{n}}\right)^{1/n}\partial
B^{n}\right)\,dt=\int_{{\mathbb{R}}^{n}}\|\nabla f^{\\#}(x)\|dx.$

Hence, $T$ satisfies (5.21) and (6.11) with $\Phi(t)=t$, and (6.22) with
$p=1$, as the Schwarz rearrangement does so. This proves (i).

For (ii), assume first that $\Phi$ is real-valued and strictly convex, and
that (5.21) or (6.11) holds. Let $\Phi(1)=c>0$. Note that the measure on $K$
with differential $dx/\kappa_{n}$ is a probability measure, since
${\mathcal{H}}^{n}(K)=\kappa_{n}$. We use (6.25), either (5.21) or (6.11),
Jensen’s inequality (see, e.g. [30, p. 62, Proposition 5]), (6.6), and the
isoperimetric inequality, to obtain

$\displaystyle c$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}\Phi\left(\|\nabla
f_{B^{n}}(x)\|\right)\frac{dx}{\kappa_{n}}\geq\int_{{\mathbb{R}}^{n}}\Phi\left(\|\nabla
Tf_{B^{n}}(x)\|\right)\frac{dx}{\kappa_{n}}\geq\int_{K}\Phi\left(\|\nabla
Tf_{B^{n}}(x)\|\right)\frac{dx}{\kappa_{n}}$ $\displaystyle\geq$
$\displaystyle\Phi\left(\int_{K}\|\nabla
Tf_{B^{n}}(x)\|\frac{dx}{\kappa_{n}}\right)=\Phi\left(\frac{1}{n\kappa_{n}}{\mathcal{H}}^{n-1}(\partial
K)\right)\geq\Phi\left(\frac{1}{n\kappa_{n}}{\mathcal{H}}^{n-1}(\partial
B^{n})\right)=\Phi(1)=c.$

It follows that there is equality in the isoperimetric inequality, giving
$K=x+B^{n}$ for some $x\in D^{n}$, as before. But now equality also holds in
Jensen’s inequality with the strictly convex function $\Phi$, so $\|\nabla
Tf_{B^{n}}(\cdot)\|=\|\nabla f_{K}(\cdot)\|$ must be constant
${\mathcal{H}}^{n}$-almost everywhere on $K$. This is only possible when
$x=o$. If we assume instead that (6.22) holds with $1<p<\infty$, we can apply
the same argument with $\Phi(t)=t^{p}$. Conversely, when $K=B^{n}$, the
rearrangement $T$ is the Schwarz rearrangement and therefore the Pólya-Szegő
inequality holds.

Finally, for (ii) when $p=\infty$, take $f=f_{B^{n}}$ in (6.22). We have
$Tf_{B^{n}}=f_{K}$ by (6.24), and it is clear that
$\operatorname*{ess\,sup}_{x\in{\mathbb{R}}^{n}}\|\nabla f_{B^{n}}(x)\|=1$,
while $\operatorname*{ess\,sup}_{x\in{\mathbb{R}}^{n}}\|\nabla f_{K}(x)\|>1$
if and only if $K\neq B^{n}$. This completes the proof of (ii).

##  7\. The anisotropic case

The following result generalizes Theorem 5.8, which corresponds to the case
when $K=B^{n}$.

######  Theorem 7.1.

Let $X={\mathcal{S}}({\mathbb{R}}^{n})$ or ${\mathcal{V}}({\mathbb{R}}^{n})$,
let $K\in{\mathcal{K}}^{n}_{(o)}$, let $T:X\to X$ be a rearrangement, and let
$\Phi$ be a Young function. If $T$ is $K$-smoothing and $f\in X$ is Lipschitz,
then $Tf$ coincides with a Lipschitz function ${\mathcal{H}}^{n}$-almost
everywhere on ${\mathbb{R}}^{n}$, and

(7.1) $\int_{{\mathbb{R}}^{n}}\Phi\left(h_{-K}(\nabla
Tf(x))\right)\,dx\leq\int_{{\mathbb{R}}^{n}}\Phi\left(h_{-K}(\nabla
f(x))\right)\,dx,$

where the integrals may be infinite.

###### Proof.

We can argue as in the proof of Theorem 5.8, with very few changes. With
arguments as at the beginning of that proof, we may assume without loss of
generality that $\Phi$ is a nontrivial real-valued Young function. Let $f\in
X$ be Lipschitz with Lipschitz constant $L$. Choosing $0<r\leq R$ such that
$rB^{n}\subset K\subset RB^{n}$, we may use Lemma 5.7 to conclude that $Tf$
coincides ${\mathcal{H}}^{n}$-almost everywhere with a Lipschitz function with
Lipschitz constant at most $LR/r$. Inequality (5.22) follows as before.
Instead of (5.23) we have

$\max\left\\{\|\nabla f(x)\|,\|\nabla Tf(x)\|\right\\}\leq LR/r$

for ${\mathcal{H}}^{n}$-almost all $x\in{\mathbb{R}}^{n}$. Then, if $C$ is the
convex body from Lemma 5.6 corresponding to $M=LR/r$ and $\Phi$, we have

$h_{C}(y,1)-1=b\,\Phi(h_{K}(y))=b\,\Phi(h_{-K}(-y))$

for $y\in{\mathbb{R}}^{n}$ with $h_{K}(y)\leq M$ and some $b>0$. As before,
this leads to (7.1). ∎

Finally, we present in Theorem 7.3 an anisotropic version of Theorem 6.3,
which again corresponds to the case when $K=B^{n}$. We shall need the
following lemma. Recall that a convex body is smooth if all its boundary
points are regular and strictly convex if it does not contain a line segment
in its boundary; see [33, pp. 83, 87].

######  Lemma 7.2.

Let $L\in{\mathcal{K}}^{n}_{(o)}$ be a smooth and strictly convex body, and
let $u:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}$ and
$v:{\mathbb{R}}^{n}\to{\mathbb{R}}$ be measurable. Then there is a measurable
$w:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}$ such that

(i) $h_{L^{\circ}}(w(x))=v(x)$ for all $x$ such that $u(x)\neq o$, and

(ii) $u(x)\cdot w(x)=h_{L}(u(x))\,h_{L^{\circ}}(w(x))$ for all
$x\in{\mathbb{R}}^{n}$.

###### Proof.

If $A=\\{x:u(x)\neq 0\\}$, then $A$ is ${\mathcal{H}}^{n}$-measurable. As is
observed in [33, Remark 1.7.14], it follows easily from (2.3) that since $L$
is smooth and strictly convex, the same is true of $L^{\circ}$. Let
$n(L^{\circ},y)$ denote the unit outer normal to $L^{\circ}$ at $y\in\partial
L^{\circ}$. Note that $u/h_{L}(u)=\rho_{L^{\circ}}(u)u\in\partial L^{\circ}$
by (2.3), and define $\eta:S^{n-1}\to S^{n-1}$ by

$\eta(u)=n\left(L^{\circ},u/h_{L}(u)\right).$

The map $\eta$ is continuous, since $L^{\circ}$ is a convex body of class
$C^{1}$ by [33, Theorem 2.2.4]. The map $f:{\mathbb{R}}^{n}\setminus\\{o\\}\to
S^{n-1}$ defined by $f(z)=z/\|z\|$ is also continuous. The composition
$\eta\circ f\circ u:A\to S^{n-1}$ of measurable functions is therefore also
measurable, and so is its composition with the continuous support function
$h_{L^{\circ}}:{\mathbb{R}}^{n}\to{\mathbb{R}}$ (see [33, p. 115]). Define

$w(x)=\begin{cases}v(x)\frac{\eta(f(u(x)))}{h_{L^{\circ}}(\eta(f(u(x))))},&{\text{if
$x\in A$,}}\\\ 0,&{\text{if $x\not\in A$}}.\end{cases}$

Then $w:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}$ is measurable and clearly
satisfies (i). By its definition, $w(x)$ is an outer normal to $L^{\circ}$ at
$f(u(x))/h_{L}(f(u(x)))=u(x)/h_{L}(u(x))$, so (ii) holds due to the equality
condition for (2.7) stated immediately after it. ∎

######  Theorem 7.3.

Let $K\in{\mathcal{K}}^{n}_{(o)}$, let
$T:{\mathcal{V}}({\mathbb{R}}^{n})\to{\mathcal{V}}({\mathbb{R}}^{n})$ be a
$K$-smoothing rearrangement, and let $\Phi$ be a Young function. If $f\in
W^{1,1}_{loc}({\mathbb{R}}^{n})\cap{\mathcal{V}}({\mathbb{R}}^{n})$ and
$\int_{{\mathbb{R}}^{n}}\Phi\left(h_{-K}(\nabla f(x))\right)\,dx<\infty$, then
$Tf\in W^{1,1}_{loc}({\mathbb{R}}^{n})$ and

(7.2) $\int_{{\mathbb{R}}^{n}}\Phi\left(h_{-K}(\nabla
Tf(x))\right)\,dx\leq\int_{{\mathbb{R}}^{n}}\Phi\left(h_{-K}(\nabla
f(x))\right)\,dx.$

###### Proof.

The proof follows that of Theorem 6.3, up to the end of the proof of the
second claim, verbatim except that the role of Theorem 5.8 is now played by
Theorem 7.1 and that $\|\nabla f\|$ and $\|\nabla Tf\|$ are replaced by
$h_{-K}(\nabla f)$ and $h_{-K}(\nabla Tf)$, respectively. Instead of the third
claim, that (6.15) holds, we claim that

(7.3) $\big{\|}h_{-K}(\nabla
Tf)\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}\leq\liminf_{j\to\infty}\big{\|}h_{-K}(\nabla
Tf_{j})\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}.$

The argument is similar to the one given for (6.15). We first prove that if
$u\in(L^{\Phi_{r}}({\mathbb{R}}^{n}))^{n}$, then

(7.4)
$\big{\|}h_{-K}(u)\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}u(x)\cdot
w(x)\,dx\right|:w\in(L^{{\Psi_{r}}}({\mathbb{R}}^{n}))^{n},\big{\|}h_{-K^{\circ}}(w)\big{\|}^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\},$

where $-K^{\circ}=(-K)^{\circ}$ is the polar body of $-K$; see (2.2). For
this, we apply [30, (10), Proposition 10, Section 3.4] to $h_{-K}(u)$, which
yields

(7.5)
$\big{\|}h_{-K}(u)\big{\|}_{L^{\Phi_{r}}({\mathbb{R}}^{n})}=\sup\left\\{\left|\int_{{\mathbb{R}}^{n}}h_{-K}(u(x))\,v(x)\,dx\right|:v\in
L^{{\Psi_{r}}}({\mathbb{R}}^{n}),\|v\|^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq
1\right\\}.$

(Here we have used the fact that
${\mathcal{M}}^{{\Psi_{r}}}=L^{{\Psi_{r}}}({\mathbb{R}}^{n})$, where
${\mathcal{M}}^{{\Psi_{r}}}$ is the closure of the span of all linear step
functions in $L^{{\Psi_{r}}}({\mathbb{R}}^{n})$; this was explained after
(6.17), along with a warning about the different notation employed in [30].)
Let $S_{1}$ and $S_{2}$ denote the right-hand sides of (7.4) and (7.5),
respectively. From (2.7) with $K$, $x$, and $y$ replaced by $-K$, $u$, and
$w$, respectively, we obtain $S_{1}\leq S_{2}$.

For the converse, we can restrict the supremum in (7.5) to nonnegative $v$.
Let $u\in(L^{\Phi_{r}}({\mathbb{R}}^{n}))^{n}$ and let $v\in
L^{{\Psi_{r}}}({\mathbb{R}}^{n})$, $v\geq 0$, and
$\|v\|^{\prime}_{L^{{\Psi_{r}}}({\mathbb{R}}^{n})}\leq 1$. By [33, Theorem
2.7.1], the set of smooth and strictly convex bodies is dense in
$\mathcal{K}^{n}$ with the Hausdorff metric, while the compact convex sets
strictly contained in $K$ and strictly containing $aK$ for a fixed $0<a<1$
form an open set in $\mathcal{K}^{n}$. We may therefore choose a sequence
$(K_{m})$ of smooth and strictly convex bodies converging to $K$ as
$m\to\infty$ in the Hausdorff metric and such that

(7.6) $\frac{1}{2}K\subset K_{m}\subset K$

for $m\in{\mathbb{N}}$. Let $w_{m}:{\mathbb{R}}^{n}\to S^{n-1}$ be the
measurable vector field supplied by Lemma 7.2 with $L=-K_{m}$. Since $v\in
L^{\Psi_{r}}({\mathbb{R}}^{n})$, we have $h_{-K_{m}^{\circ}}(w_{m})\in
L^{\Psi_{r}}({\mathbb{R}}^{n})$; this is equivalent to $\|w_{m}\|\in
L^{\Psi_{r}}({\mathbb{R}}^{n})$ and hence
$w_{m}\in(L^{\Psi_{r}}({\mathbb{R}}^{n}))^{n}$. By (7.6), $-K^{\circ}\subset-
K_{m}^{\circ}$, so from Lemma 7.2(i), we get

$\|h_{-K^{\circ}}(w_{m})\|^{\prime}_{L^{\Psi_{r}}({\mathbb{R}}^{n})}\leq\|h_{-K_{m}^{\circ}}(w_{m})\|^{\prime}_{L^{\Psi_{r}}({\mathbb{R}}^{n})}=\|v\|^{\prime}_{L^{\Psi_{r}}({\mathbb{R}}^{n})}\leq
1.$

Using (i) and (ii) of Lemma 7.2 with $L=-K_{m}$, we obtain

(7.7) $S_{1}\geq\left|\int_{{\mathbb{R}}^{n}}u(x)\cdot
w_{m}(x)\,dx\right|=\left|\int_{{\mathbb{R}}^{n}}h_{-K_{m}}(u(x))\,h_{-K_{m}^{\circ}}(w_{m}(x))\,dx\right|=\left|\int_{{\mathbb{R}}^{n}}h_{-K_{m}}(u(x))\,v(x)\,dx\right|.$

By (7.6),

$0\leq h_{-K_{m}}(u(x))\,v(x)\leq h_{-K}(u(x))\,v(x)$

and the function on the right-hand side is integrable. Taking the limit as
$m\to\infty$ in (7.7), the dominated convergence theorem yields

$S_{1}\geq\lim_{m\to\infty}\int_{{\mathbb{R}}^{n}}h_{-K_{m}}(u(x))\,v(x)\,dx=\int_{{\mathbb{R}}^{n}}h_{-K}(u(x))\,v(x)\,dx.$

This proves that $S_{1}\geq S_{2}$ and concludes the proof of (7.4). Now (7.3)
follows from (7.4) by the same argument that showed that (6.15) follows from
(6.16).

The remainder of the proof of the theorem is a repetition of the last part of
the proof of Theorem 6.3, from the point where (6.15) has been established
onwards. ∎

Suppose that $K\subset{\mathbb{R}}^{n}$ is an $o$-symmetric convex body. Then

$h_{K}(x)=h_{-K}(x)=\|x\|_{K^{\circ}}$

for $x\in{\mathbb{R}}^{n}$, where $\|\cdot\|_{K^{\circ}}$ is the norm for
which the unit ball is $K^{\circ}$, the polar body of $K$, defined by (2.2).
In this case, (7.1) and (7.2) may be rewritten in the form

$\int_{{\mathbb{R}}^{n}}\Phi\left(\|\nabla
Tf(x)\|_{K^{\circ}}\right)\,dx\leq\int_{{\mathbb{R}}^{n}}\Phi\left(\|\nabla
f(x)\|_{K^{\circ}}\right)\,dx.$

##  8\. Appendix

The purpose of this appendix is to provide a proof of Proposition 3.7. This
follows easily from Lemma 8.1 below, first proved by Crowe, Zweibel, and
Rosenbloom [11, Theorem 3] for Schwarz rearrangement without the assumption
that $F(s,0)$ and $F(0,t)$ decrease with $s\geq 0$ and $t\geq 0$,
respectively. Variants of Proposition 3.7 are stated for general
rearrangements by Brock and Solynin [7, Theorem 3.1] and by Van Schaftingen
and Willem [43, Corollary 1], whose approaches to rearrangements differ from
ours; see [6, Appendix] for a comparison. Brock and Solynin refer to [11] for
a proof, but do not explain why it should apply to general rearrangements,
while [43, Corollary 1] is stated with the extra assumption that the function
$j$ is even. The proof of [43, Corollary 1] is based on that of [39,
Proposition 3.3.9], which does not assume that $j$ is even, or that it is
nonnegative, but which requires a considerable amount of preliminary
observations and terminology. For this reason, we prefer to follow the
argument in [11].

######  Lemma 8.1.

Let $F:{\mathbb{R}}^{2}\to{\mathbb{R}}$ be continuous with $F(0,0)=0$ and such
that $F(s,0)$ and $F(0,t)$ decrease with $s\geq 0$ and $t\geq 0$,
respectively. Suppose that for all coordinate rectangles $R=[a,b]\times[c,d]$,
where $a\leq b$, $c\leq d$,

(8.1) $G(R)=F(b,d)+F(a,c)-F(b,c)-F(a,d)\geq 0.$

Let $T:{\mathcal{V}}({\mathbb{R}}^{n})\to{\mathcal{V}}({\mathbb{R}}^{n})$ be a
rearrangement. If $f,g\in{\mathcal{V}}({\mathbb{R}}^{n})$, then

(8.2)
$\int_{{\mathbb{R}}^{n}}F(f(x),g(x))\,dx\leq\int_{{\mathbb{R}}^{n}}F(Tf(x),Tg(x))\,dx.$

###### Proof.

The function $G$ is additive on coordinate rectangles, that is, if $R$, $S$,
and $R\cup S$ are non-overlapping coordinate rectangles, then $G(R\cup
S)=G(R)+G(S)$. This allows $G$ to be extended to a measure $\nu$ on
${\mathbb{R}}^{2}$ such that each coordinate rectangle $R$ is $\nu$-measurable
and $\nu(R)=G(R)$; see [31, pp. 64–68] (where the union of sets is denoted by
$+$).

Let $H$ denote the Heaviside function, i.e., $H(x)=1$ if $x\geq 0$ and
$H(x)=0$ if $x<0$. Then for $b,d,s,t\geq 0$,

$1_{[0,b]\times[0,d]}(s,t)=H(b-s)H(d-t).$

It follows that

$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}H(b-s)H(d-t)\,d\nu(s,t)$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}1_{[0,b]\times[0,d]}(s,t)\,d\nu(s,t)$
$\displaystyle=$ $\displaystyle\nu([0,b]\times[0,d])=G([0,b]\times[0,d])$
$\displaystyle=$ $\displaystyle F(b,d)-F(b,0)-F(0,d).$

From this we obtain

$F(b,d)=F(b,0)+F(0,d)+\int_{0}^{\infty}\int_{0}^{\infty}H(b-s)H(d-t)\,d\nu(s,t).$

On setting $b=f(x)$ and $d=g(x)$ and integrating, this gives

(8.3)
$\int_{{\mathbb{R}}^{n}}F(f(x),g(x))\,dx=\int_{{\mathbb{R}}^{n}}F(f(x),0)\,dx+\int_{{\mathbb{R}}^{n}}F(0,g(x))\,dx+I(f,g),$

where by Fubini’s theorem,

(8.4) $\displaystyle I(f,g)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}\int_{0}^{\infty}\int_{0}^{\infty}H(f(x)-s)H(g(x)-t)\,d\nu(s,t)\,dx$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\int_{{\mathbb{R}}^{n}}H(f(x)-s)H(g(x)-t)\,dx\,d\nu(s,t).$

Our assumptions on $F$ imply that $-F(r,0)\geq 0$ increases with $r\geq 0$. We
can therefore apply (3.11), with $\varphi(r)=-F(r,0)$ for $r\geq 0$, to obtain
$-F(Tf(x),0)=T(-F(f(x),0))$, for each $x\in{\mathbb{R}}^{n}$. With this, the
layer-cake representation formula, and the equimeasurability of $T$, we
conclude that

(8.5) $\displaystyle\int_{{\mathbb{R}}^{n}}F(Tf(x),0)\,dx$ $\displaystyle=$
$\displaystyle-\int_{{\mathbb{R}}^{n}}T(-F(f(x),0))\,dx$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}{{\mathcal{H}}}^{n}(\\{x:T(-F(f(x),0))>t\\})\,dt$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}{{\mathcal{H}}}^{n}(\\{x:-F(f(x),0)>t\\})\,dt=\int_{{\mathbb{R}}^{n}}F(f(x),0)\,dx.$

Similarly,

(8.6)
$\int_{{\mathbb{R}}^{n}}F(0,Tg(x))\,dx=\int_{{\mathbb{R}}^{n}}F(0,g(x))\,dx.$

Since $\diamondsuit_{T}$ is monotonic, we have

(8.7) ${{\mathcal{H}}}^{n}(\diamondsuit_{T}(A\cap
B))\leq{{\mathcal{H}}}^{n}((\diamondsuit_{T}A)\cap\diamondsuit_{T}B)$

whenever $A,B\in{\mathcal{L}}^{n}$. Consequently, using the measure-preserving
property of $\diamondsuit_{T}$, (8.7), and (3.9), we obtain

$\displaystyle\int_{{\mathbb{R}}^{n}}H(f(x)-s)H(g(x)-t)\,dx$ $\displaystyle=$
$\displaystyle{{\mathcal{H}}}^{n}(\\{x:f(x)\geq s\\}\cap\\{x:g(x)\geq t\\})$
$\displaystyle=$
$\displaystyle{{\mathcal{H}}}^{n}(\diamondsuit_{T}(\\{x:f(x)\geq
s\\}\cap\\{x:g(x)\geq t\\}))$ $\displaystyle\leq$
$\displaystyle{{\mathcal{H}}}^{n}((\diamondsuit_{T}\\{x:f(x)\geq
s\\})\cap\diamondsuit_{T}\\{x:g(x)\geq t\\})$ $\displaystyle=$
$\displaystyle{{\mathcal{H}}}^{n}(\\{x:Tf(x)\geq s\\}\cap\\{x:Tg(x)\geq t\\})$
$\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}}H(Tf(x)-s)H(Tg(x)-t)\,dx.$

By (8.4), this yields $I(f,g)\leq I(Tf,Tg)$. Then (8.2) follows from (8.3),
(8.3) with $f$ and $g$ replaced by $Tf$ and $Tg$, respectively, (8.5), and
(8.6). ∎

Proof of Proposition 3.7. Let $F(s,t)=-j(s-t)$ for $s,t\geq 0$. Since $j$ is
convex, for $r\in{\mathbb{R}}$ and $s,t\geq 0$, we have

$j(r)-j(r-s)\leq j(r+t)-j(r-s+t).$

If $r=b-d$, $s=b-a$, and $t=d-c$, this gives

$j(b-d)-j(a-d)\leq j(b-c)-j(a-c),$

yielding (8.1). Moreover, $F(r,0)=-j(r)$ and $F(0,r)=-j(-r)$ both decrease
with $r\geq 0$ since $j\geq 0$ and $j(0)=0$. Applying Lemma 8.1 with this
choice of $F$, we obtain (3.12).

The $L^{p}$-contracting property results from taking $j(r)=|r|^{p}$, $p\geq
1$. ∎

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