2. Notations and preliminaries

Throughout this article, $|\cdot |$ denotes the Euclidean norm in $\mathbb{R}^n$ , while $\cdot$ is the standard Euclidean scalar product for $n\geqslant 2$ . Moreover, we denote by $|\Omega |$ the Lebesgue measure of $\Omega \subseteq \mathbb{R}^n$ , by $B_R$ the Euclidean ball centered at the origin with radius $R$ and by $\omega _n$ the measure of the unit ball.

Let $E\subseteq \mathbb{R}^n$ be a bounded, open set and $\Omega \subseteq \mathbb{R}^{n}$ be a measurable set. We recall now the definition of the perimeter of $\Omega$ in $E$ in the sense of De Giorgi, that is,

\begin{equation*} P(\Omega ;\; E)=\sup \left \{ \int _\Omega \textrm {div} \varphi \:dx \;:\;\varphi \in C^{\infty }_c(E;\;\mathbb {R}^n),\;||\varphi ||_{\infty }\leq 1 \right \}. \end{equation*}

The perimeter of $\Omega$ in $\mathbb{R}^n$ will be denoted by $P(\Omega )$ and, if $P(\Omega )\lt \infty$ , we say that $\Omega$ is a set of finite perimeter. Some results relative to the sets of finite perimeter are contained, for instance, in [Reference Maggi31]. Moreover, if $\Omega$ has Lipschitz boundary, we have that

\begin{equation*} P(\Omega )=\mathcal {H}^{n-1}(\partial \Omega ). \end{equation*}

2.1. The anisotropic norm

Let $H\;:\;\mathbb{R}^n \longrightarrow [0,+\infty [$ , $n\,{\geqslant}\, 2$ , be a $C^1(\mathbb{R}^n \setminus \{0\})$ convex function which is positively $1$ -homogeneous, that is,

(2.1) \begin{equation} H(t\xi )= |t| H(\xi ) \,\,\,\,\,\, \forall \xi \in \mathbb{R}^n \,, \,\, \forall t \in \mathbb{R}. \end{equation}

Moreover, let $0\lt \gamma \,{\leqslant}\, \delta$ be positive constants such that

(2.2) \begin{equation} \gamma |\xi | \,{\leqslant}\, H(\xi ) \,{\leqslant}\, \delta |\xi |. \end{equation}

These properties guarantee that $H$ is a norm in $\mathbb{R}^n$ . Indeed by (2.2), we have that $H(\xi )=0$ if and only if $\xi = 0$ . It is homogeneous by (2.1) and the triangular inequality is a consequence of the convexity of the function $H$ : if $\xi, \eta \in \mathbb{R}^n$ , then

\begin{equation*} \frac {H(x+y)}{2}= H\bigg (\frac {x}{2}+\frac {y}{2}\bigg )\,{\leqslant}\, \frac {H(x)}{2}+\frac {H(y)}{2}. \end{equation*}

Because of (2.1), we can assume that the set

\begin{equation*} K = \{\xi \in \mathbb {R}^n \;:\; H(\xi )\,{\leqslant}\, 1 \} \end{equation*}

is such that $|K|=\omega _n$ , where $\omega _n$ is the measure of the unit sphere in $\mathbb{R}^n$ . We can define the support function of $K$ as:

(2.3) \begin{equation} H^{\circ }(x)= \sup _{\xi \in K} \left \lt x, \xi \right \gt, \end{equation}

where $\left \lt \cdot, \cdot \right \gt$ denotes the scalar product in $\mathbb{R}^n$ . $H^{\circ }\;:\;\mathbb{R}^n \longrightarrow [0,+\infty ]$ is a convex, homogeneous function in the sense of (2.1). Moreover, $H$ and $H^{\circ }$ are polar to each other, in the sense that

\begin{equation*} H^{\circ }(x) = \sup _{\xi \neq 0 } \frac {\left \lt x, \xi \right \gt }{H(\xi )} \end{equation*}

and

\begin{equation*} H(x) = \sup _{\xi \neq 0 } \frac {\left \lt x, \xi \right \gt }{H^{\circ }(\xi )}. \end{equation*}

$H$ is the support function of the set:

\begin{equation*} K^{\circ } = \{x \in \mathbb {R}^n \;:\; H^{\circ }(x)\,{\leqslant}\, 1 \}. \end{equation*}

The set $\mathcal{W}=\{x \in \mathbb{R}^n \;:\; H^{\circ }(x)\lt 1 \}$ is the so-called Wulff shape centered at the origin. We set $k_n = |\mathcal{W}|$ . More generally, we will denote by $\mathcal{W}_R(x_0)$ the Wulff shape centered in $x_0 \in \mathbb{R}^n$ the set $R\mathcal{W}+x_0$ , and $\mathcal{W}_R(0)=\mathcal{W}_R$ .

The following properties hold for $H$ and $H^{\circ }$ :

(2.4) \begin{equation} H_{\xi }(\xi )\cdot \xi = H(\xi ), \;\;\;\; H^{\circ }_{\xi }(\xi )\cdot \xi = H^{\circ }(\xi ), \end{equation}

(2.5) \begin{equation} H(H^{\circ }_{\xi }(\xi ))= H^{\circ }(H_{\xi }(\xi ))=1 \,\,\,\,\, \forall \xi \in \mathbb{R}^n \setminus \{0\}, \end{equation}

(2.6) \begin{equation} H^{\circ }(\xi )H_{\xi }(H^{\circ }_{\xi }(\xi ))=H(\xi )H^{\circ }_{\xi }(H_{\xi }(\xi )) = \xi \,\,\, \forall \xi \in \mathbb{R}^n \setminus \{0\}. \end{equation}

An example of an anisotropic norm that satisfies the above-mentioned properties is the following. Let $r\in (1,+\infty )$ and let us consider

\begin{equation*} H(\xi ) = \bigg (\sum _{i=1}^n \left \vert \xi _i\right \vert ^r\bigg )^{\frac {1}{r}}, \end{equation*}

known in literature also as $r$ -norm. With this choice, the highly nonlinear operator $\mathcal{L}_p(u)$ , defined in (1.3), becomes

\begin{equation*} \mathcal {L}_p(u)= \sum _{k=1}^n\frac {\partial }{\partial x_k} \bigg (\bigg ( \sum _{i=1}^n \left \vert \frac {\partial u}{\partial x_i}\right \vert ^r\bigg )^{\frac {p-r}{r}}\left \vert \frac {\partial u}{\partial x_k}\right \vert ^{r-2}\frac {\partial u}{\partial x_k}\bigg ). \end{equation*}

We stress that for $r=p$ , $\mathcal{L}_p(u)$ is the so called pseudo- $p$ -Laplace operator. Examples of non-smooth anisotropic norm can be found in [Reference Bellettini, Caselles, Chambolle and Novaga8] and references therein, where the authors consider a crystalline anisotropy and the associated Wulff shape is a polyhedron.

If $E \subset \mathbb{R}^n$ is an open, bounded set with Lipschitz boundary and $\Omega$ is an open subset of $\mathbb{R}^n$ , we can give a generalized definition of perimeter of $\Omega$ with respect to the anisotropic norm as follows (see for instance [Reference Amar and Bellettini6]):

\begin{equation*} P_H(\Omega, E)= \int _{\partial ^* \Omega \cap E} H(\nu )\,d\mathcal {H}^{n-1}, \end{equation*}

where $\partial ^*\Omega$ is the reduced boundary of $\Omega$ (for the definition see [Reference Evans and Gariepy21]), $\nu$ is its Euclidean outer normal, and $\mathcal{H}^{n-1}$ is the $(n-1)-$ dimensional Hausdorff measure in $\mathbb{R}^n$ . Clearly, if $E$ is open, bounded and Lipschitz, then the outer unit normal exists almost everywhere and

(2.7) \begin{equation} P_H(E,\mathbb{R}^n)\;:\!=\;P_H (E) = \int _{\partial E} H(\nu )\,d\mathcal{H}^{n-1}. \end{equation}

By (2.2), we have that

\begin{equation*} \gamma P(E) \,{\leqslant}\, P_H(E)\,{\leqslant}\, \delta P(E). \end{equation*}

In [Reference Alvino, Trombetti and Lions5], it is shown that if $u\in W^{1,1}(\Omega )$ , then for, that is, $t\gt 0$

(2.8) \begin{equation} -\frac{d}{dt}\int _{\{u\gt t\}} H(\nabla u)\,dx = P_H(\{u\gt t\},\Omega ) = \int _{\partial ^*\{u\gt t\}\cap \Omega } \frac{H(\nabla u)}{|\nabla u|}\,d\mathcal{H}^{n-1}. \end{equation}

Moreover, an isoperimetric inequality for the anisotropic perimeter holds (for instance see [Reference Alvino, Ferone, Lions and Trombetti2, Reference Busemann12, Reference Dacorogna and Pfister18, Reference Figalli, Maggi and Pratelli23, Reference Fonseca and Müller24])

(2.9) \begin{equation} P_H( E ) \,{\geqslant}\, n k^{\frac{1}{n}}_n |E|^{1-\frac{1}{n}}. \end{equation}

2.2. Convex symmetrization

Let $\Omega \subset \mathbb{R}^n$ be an open, bounded, and connected set. Let $f: \Omega \longrightarrow [0,+\infty ]$ be a measurable function. The decreasing rearrangement $f^*$ of $f$ is defined as follows:

\begin{equation*} f^*(s)= \inf \{t\,{\geqslant}\, 0 \;:\; \mu (t)\lt s\} \,\,\,\,\,\, s\in [0,|\Omega |], \end{equation*}

where

\begin{equation*} \mu (t) = |\{x\in \Omega \;:\; |f(x)|\gt t\}|, \end{equation*}

is the distribution function of $f$ . We recall that the Schwarz symmetrand of $f$ is a radially spherically function defined as follows:

\begin{equation*} f^{\sharp }(x) = f^*(\omega _n |x|^n) \,\,\,\,\,\,\,\, x\in \Omega ^{\sharp }. \end{equation*}

where $\Omega ^{\sharp }$ is the ball centered at the origin such that $|\Omega ^{\sharp }|=|\Omega |$ . The convex symmetrization $f^{\star }$ of $f$ , instead, is a function symmetric with respect to $H^{\circ }$ defined as follows:

\begin{equation*} f^{\star }(x) = f^*(k_n (H^{\circ }(x))^n) \,\,\,\,\,\,\,\, x\in \Omega ^{\star }, \end{equation*}

where $\Omega ^\star$ is a Wulff shape centered at the origin and such that $|\Omega ^{\star }|=|\Omega |$ (see [Reference Alvino, Ferone, Lions and Trombetti2]). We stress that both $f^{\star }$ and $f^{\sharp }$ are defined by means the decreasing rearrangement $f^*$ , but they have different symmetry. In particular, it is well known that the functions $f$ , $f^*$ , $f^{\sharp }$ , and $f^{\star }$ are equimeasurable, that is,

\begin{equation*} |\{f\gt t\}| =|\{f^{\sharp } \gt t\}| = |\{f^* \gt t\}| = |\{f^{\star }\gt t\}| \,\,\,\,\, t\,{\geqslant}\, 0. \end{equation*}

As a consequence, if $f\in L^p(\Omega )$ , $p\,{\geqslant}\, 1$ , then

(2.10) \begin{equation} \| f\|_{L^p(\Omega )}\| =\| f^{\sharp } \|_{L^p(\Omega ^{\sharp })} = \| f^* \|_{L^p([0,|\Omega |])} = \|f^{\star }\|_{L^p(\Omega ^{\star })}. \end{equation}

Regarding the norm of the gradient, a generalized version of the well-known Pólya–Szegö inequality holds and it states (see for instance [Reference Alvino, Ferone, Lions and Trombetti2])

Theorem 2.1. (Pólya–Szegö principle). If $w \in W^{1,p}_0(\Omega )$ for $p\,{\geqslant}\, 1$ , then we have that

\begin{equation*} \int _{\Omega }H(\nabla u)^p\,dx \,{\geqslant}\, \int _{\Omega ^{\star }} H(\nabla u^{\star })^p\,dx. \end{equation*}

where $\Omega ^\star$ is the Wulff Shape such that $|\Omega ^\star |=|\Omega |$ .

For the sake of completeness, we will state the result concerning the equality case of the Pólya–Szegö inequality, whose proof is contained in [Reference Ferone and Volpicelli22] for the generic anisotropic case and in [Reference Ziemer and Brothers38] for the Euclidean case.

Theorem 2.2. Let $u$ be a non-negative function in $W^{1,p}(\mathbb{R}^n)$ , for $1\lt p\lt +\infty$ , such that

\begin{equation*} |\{|\nabla u^\star |=0\}\cap \{0\lt u^\star \lt \operatorname {ess\;sup} u\}|=0. \end{equation*}

Then

\begin{equation*} \int _{\mathbb {R}^n}H(\nabla u)^p\,dx = \int _{\mathbb {R}^n} H(\nabla u^{\star })^p\,dx \end{equation*}

if and only if $u=u^\star$ a.e. in $\mathbb{R}^n$ , up to translations.

Obviously, Theorem 2.2 can holds true in the case of a $W^{1,p}_0(\Omega )$ function.

We conclude this section by recalling some known properties about rearrangements that we will use in the proof of the main theorem. The following result is the well-known Hardy–Littlewood inequality (see [Reference Kesavan30]):

(2.11) \begin{equation} \int _{\Omega } |f(x)g(x)| \,dx \,{\leqslant}\, \int _0^{|\Omega |} f^*(s)g^*(s)\,ds. \end{equation}

So, if we consider $g$ as the characteristic function of the set $\{x\in \Omega \;:\; u(x)\gt t\}$ , for some measurable function $u\;:\;\Omega \rightarrow \mathbb{R}$ and $t\,{\geqslant}\, 0$ , then we get

(2.12) \begin{equation} \int _{\{u\gt t\}} f(x)\,dx \,{\leqslant}\, \int _0^{\mu (t)} f^*(s)\,ds, \end{equation}

where, again, $\mu (t)$ is the distribution function of $u$ . Finally, we recall the definition of dominated rearrangements (see for instance [Reference Alvino, Lions and Trombetti4] and [Reference Chong and Rice17]).

Definition 2.3. Let $f,g \in L^1(\Omega )$ be nonnegative functions. We say that $g$ is dominated by $f$ and write $g\prec f$ if the following two statements hold

1. (i) $\displaystyle \int _{0}^s g^* (t)\,dt \,{\leqslant}\, \displaystyle \int _{0}^s f^* (t)\,dt$ ;

2. (ii) $\displaystyle \int _{0}^{|\Omega |} g^* (t)\,dt =\displaystyle \int _{0}^{|\Omega |} f^* (t)\,dt$ .

In [Reference Alvino, Lions and Trombetti4], the authors prove the following result:

Proposition 2.4. Let $f,g,h$ be positive and such that $hf, hg\in L^1(\Omega )$ . Let $F$ be a convex, nonnegative function such that $F(0)=0$ . If $hg \prec hf$ Then

\begin{equation*} \int _0^{|\Omega |}h^* F(g^*) \, dt \,{\leqslant}\, \int _0^{|\Omega |}h^* F(f^*) \, dt. \end{equation*}

Moreover, if $F$ is strictly convex, the equality holds if and only if $f^*\equiv g^*$ , that is, in $[0,|\Omega |]$ .

3. The (p,q)-anisotropic Laplacian

In this section, we study the main properties of (1.1) and the corresponding minimizers. Let $\Omega \subset \mathbb{R}^n$ , $n\,{\geqslant}\, 2$ be an open, bounded, and connected set. Let $m\in L^{\infty }(\Omega )$ be a positive function and $p, q$ be such that $1\lt p\lt \infty$ and $1\lt q\lt p^*$ , where $p^*=np/(n-p)$ , if $p\lt n$ , and $p^*=\infty$ , if $p\,{\geqslant}\, n$ . A function $v\in W_0^{1,p}(\Omega )$ is a weak solution to the problem (1.2) corresponding to $\lambda$ if

(3.1) \begin{equation} \int _{\Omega }(H(\nabla v))^{p-1}H_{\xi }(\nabla v) \cdot \nabla \varphi \, dx=\lambda \|v\|_{q,m}^{p-q}\int _{\Omega }m(x)\,|v|^{q-2}\,v \,\varphi \,dx, \end{equation}

for every $\varphi \in W_0^{1,p}(\Omega )$ . By standard argument of calculus of variations, it is not difficult to prove the following result:

As regard the simplicity, we have

The proof of the previous result is contained in [Reference Kawohl, Lucia and Prashanth29], where the authors consider a more general class of quasilinear operators. We stress that this result was already proved when $H=\mathcal{E}$ and $m\equiv 1$ in the paper [Reference Kawohl28] and in the case of a positive and essentially bounded weight in [Reference Nazarov33]. Finally, we have the following:

Proof. The proof is similar to the one contained in [Reference Brasco and Franzina10, Theorem 5.1], and it follows standard arguments and a general Picone inequality. For the reader convenience and sake of completeness, we write the main steps. Let $v$ be a non-negative weak solution to the problem (1.2) corresponding to $\lambda$ . By the strong maximum principle, we have that $v\gt 0$ in $\Omega .$ Let $u$ be the first positive eigenfunction corresponding to $\lambda ^H_{p,q}(\Omega )$ such that

(3.2) \begin{equation} \|u\|_{L^q(\Omega, m)}=\|v\|_{L^q(\Omega, m)}. \end{equation}

Then,

(3.3) \begin{equation} \int _{\Omega }(H(\nabla u))^p\,dx=\lambda ^H_{p,q}(\Omega )\left (\int _{\Omega }m(x) \,u^q \,dx \right )^{\frac pq}. \end{equation}

Being $v$ a weak positive solution to (1.2) corresponding to $\lambda$ , we can chose $\varphi =\displaystyle \frac{u^q}{v^{q-1}}$ as test function in (3.1) obtaining

(3.4) \begin{align} \int _{\Omega }\left ((H(\nabla v)\right )^{p-1}H_{\xi }(\nabla v ) \cdot \nabla \left ( \frac{u^q}{v^{q-1}}\right )\,dx &=\lambda \|m^{\frac{1}{q}} v\|_q^{p-q}\int _{\Omega }m(x)\,u^q\,dx \nonumber \\[5pt] &= \lambda \left (\int _{\Omega }m(x) \,u^q \,dx \right )^{\frac pq}, \end{align}

where last equality follows by (3.2). In the left-hand side, we can apply the general Picone inequality (see Proposition 2.9 in [Reference Brasco and Franzina10]) and we have

\begin{equation*} \int _{\Omega }\left (H(\nabla v)\right )^q \left (H(\nabla u)\right )^{p-q} \,dx \,{\geqslant}\, \lambda \left (\int _{\Omega }m(x) \,u^q \,dx \right )^{\frac pq}. \end{equation*}

By the Hölder inequality, the normalization (3.2) and (3.3) we get that $\lambda ^{H}_{p,q}(\Omega )\,{\geqslant}\, \lambda$ , that implies $u=v$ .

3.1. The case $\Omega = \mathcal{W}_R$

In this subsection, we study the problem (1.2) when $\Omega$ is a Wulff shape. In this case, the eigenfunctions inherit some symmetry properties. Let be $\Omega = \mathcal{W}_R$ and let $m\in L^{\infty }(\mathcal{W}_R)$ be a positive function such that $m(x)=m^{\star }(x)$ . Then problem (1.2) becomes

(3.5) \begin{equation} \begin{cases} -\mathcal{L}_{p}(v)=\lambda m^{\star }(x) \| v\|_{q, m^\star }^{p-q}|v|^{q-2}v & \mbox{in}\ \mathcal{W}_R\\[5pt] v=0&\mbox{on}\ \partial \mathcal{W}_R. \end{cases} \end{equation}

The following result holds

Proposition 3.4. Let $1\lt p\lt \infty$ and $1\lt q\,{\leqslant}\, p$ . Let $v \in C^1(\overline{\Omega })\cap C^{1,\alpha }(\Omega )$ be a first positive eigenfunction to the problem (3.5). Then there exists a decreasing function $\rho (r)$ , $r \in [0,R]$ , such that $\rho \in C^{\infty }((0,R))\cap C^1([0,R])$ , $\rho ^{\prime }(0)=0$ , and $v(x)=\rho (H^o(x))$ .

Proof. By the simplicity, we can assume that $\|v\|_{L^q(\mathcal W_R,m^{\star })}=1$ . Let $B_R$ be the ball centered at the origin with radius $R\gt 0$ , and let us consider the weighted p-Laplace eigenvalue problem in $B_R$ :

(3.6) \begin{equation} \begin{cases} -\Delta _p z=\lambda \tilde m(|x|)\|z\|_{q,m^{\star }}^{p-q}|z|^{q-2}z & \mbox{in}\ B_R\\[5pt] z=0&\mbox{on}\ \partial B_R, \end{cases} \end{equation}

where $\tilde m(r)=m^*(k_n r^n)$ , $0 \,{\leqslant}\, r\,{\leqslant}\, R$ . Let $z$ be the positive eigenfunction corresponding to the first eigenvalue $\lambda ^{\mathcal E}_{p,q}(B_R)$ to the problem (3.6), such that $\|z\|_{L^q(B_R,\tilde{m})}=\|v\|_{L^q(\mathcal W_R,m^{\star })}=1$ . Then uniqueness guarantees that $z$ is radially symmetric, which means that there exists a positive one-dimensional function $\rho _p \;:\; r\in [0,R]\to \mathbb{R}^+$ such that $z(x)=\rho _p(|x|)$ , and $\rho _p$ solves the following problem:

(3.7) \begin{equation} \begin{cases} -(p-1)|\rho ^{\prime }_p|^{p-2}\rho ^{\prime \prime }_p+\frac{n-1}{r}|\rho ^{\prime }_p|^{p-1}= \lambda ^{\mathcal E}_{p,q}(B_R)\tilde m|\rho _p|^{q-2}\rho _p,& r \in (0,R) \\[5pt] \rho ^{\prime }_p(0)=\rho _p(R)=0. \end{cases} \end{equation}

In particular, integrating equation (3.7), it is possible to see that $\rho ^{\prime }_p$ is zero only when $r=0$ and consequently that $\rho _p$ is strictly decreasing in $[0,R]$ . Now we can come back to the anisotropy. Indeed if we consider $w=\rho _p(H^\circ (x))$ , then using properties (2.4)-(2.6) and the regularity of $H$ , by construction, we obtain that $w(x)$ is a solution to problem (3.5), which is positive and radial with respect to the anisotropic norm. The simplicity and Theorem 3.3 imply that $v=w$ , and this concludes the proof.

Remark 3.5. We stress that the proof of the previous result shows that the first eigenvalue $\lambda ^H_{p,q}(\mathcal W_R)$ coincides with the first eigenvalue of problem (3.6).

3.2. A Faber–Krahn type inequality

Theorem 3.6. Let $\Omega \in \mathbb{R}^n$ , $n\,{\geqslant}\, 2$ , be an open, bounded, and connected set and let $1\lt q\,{\leqslant}\, p$ . Then

(3.8) \begin{equation} \lambda ^H_{p,q}(\Omega )\,{\geqslant}\, \lambda ^H_{p,q}(\Omega ^{\star }), \end{equation}

where $\Omega ^\star$ is the Wulff shape such that $|\Omega ^\star |=|\Omega |$ . The equality case holds if and only if $\Omega =\Omega ^{\star }$ and $m=m^{\star }$ , that is, in $\Omega$ , up to translations, where $m^{\star }$ is the convex symmetrization of $m$ .

Proof. We argue as in [Reference Belloni, Ferone and Kawohl9]. We observe that $\lambda ^H_{p,q}(\Omega ^{\star })$ has the following variational characterization:

(3.9) \begin{equation} \lambda ^H_{p,q}(\Omega ^{\star }) = \inf _{\substack{w\in W^{1,p}_0(\Omega ^{\star }), \\ w \neq 0}}\frac{\displaystyle \int _{\Omega ^{\star }} H(\nabla w)^p\,dx}{\bigg (\displaystyle \int _{\Omega ^{\star }} m^{\star }|w|^q\, dx\bigg )^{\frac{p}{q}}}. \end{equation}

The Faber–Krahn inequality is a straightforward application of the Pólya–Szegö principle and the Hardy–Littlewood inequality. Indeed if $u$ is a positive eigenfunction corresponding to $\lambda ^H_{p,q}(\Omega )$ , then

(3.10) \begin{equation} \lambda ^H_{p,q}(\Omega ) = \frac{\displaystyle \int _\Omega H(\nabla u)^p\,dx}{\bigg (\displaystyle \int _\Omega m{u}^q\,dx\bigg )^{\frac{p}{q}}} \,{\geqslant}\, \frac{\displaystyle \int _{\Omega ^{\star }} H(\nabla u^{\star })^p\,dx}{\bigg (\displaystyle \int _{\Omega ^{\star }}m^{\star }(u^{\star })^q\,dx\bigg )^{\frac{p}{q}}}\,{\geqslant}\, \lambda ^H_{p,q}(\Omega ^{\star }). \end{equation}

Let us now consider the equality case. From (3.10), Pólya–Szegö principle and Hardy–Littlewood inequality, we get

\begin{equation*} 1\,{\leqslant}\, \frac {\displaystyle \int _\Omega H(\nabla u)^p\,dx}{\displaystyle \int _{\Omega ^{\star }} H(\nabla u^{\star })^p\,dx}= \frac {\bigg (\displaystyle \int _{\Omega ^{\star }} m^{\star }(u^{\star })^q\,dx\bigg )^{\frac {p}{q}}}{\bigg (\displaystyle \int _\Omega m u^q\,dx\bigg )^{\frac {p}{q}}}\,{\leqslant}\, 1. \end{equation*}

It follows that

(3.11) \begin{equation} \int _\Omega H(\nabla u)^p\,dx\,dx=\int _{\Omega ^{\star }} H(\nabla u^{\star })^p\,dx, \end{equation}

and

(3.12) \begin{equation} \bigg (\displaystyle \int _\Omega m u^q\,dx\bigg )^{\frac{p}{q}}=\bigg (\displaystyle \int _{\Omega ^{\star }} m^{\star }(u^{\star })^q\,dx\bigg )^{\frac{p}{q}}. \end{equation}

The thesis follows from (3.11), Theorem 2.2 and (3.12).

4. A Chiti type inequality

In this section, we prove a reverse Hölder inequality for the eigenfunctions corresponding to $\lambda ^H_{p,q}(\Omega )$ . We first prove the following proposition as in the spirit of the Talenti result contained in [Reference Talenti37] (see also [Reference Alberico, Ferone and Volpicelli1–Reference Alvino, Ferone and Trombetti3, Reference Amato, Barbato, Masiello and Paoli7, Reference Sannipoli36]).

Proposition 4.1. Let $\Omega \subset \mathbb{R}^n$ , $n\,{\geqslant}\, 2$ , be an open, bounded, and connected set, $1\lt q\,{\leqslant}\, p$ , and let $m\in L^{\infty }(\Omega )$ be a positive function. Let $u$ be a positive eigenfunction corresponding to $\lambda ^H_{p,q}(\Omega )$ . Then we have that

(4.1) \begin{equation} (-{u^*}^{\prime }(s))^{p-1} \,{\leqslant}\, n^{-p} k_n^{-\frac{p}{n}}\lambda ^H_{p,q}(\Omega ) \| u\|_{q,m}^{p-q}\int _0^{s} m^*u^*(r)^{q-1}\,dr, \qquad s\in [0,|\Omega |]. \end{equation}

In particular, the equality case holds if and only if $\Omega = \Omega ^{\star }$ and $m=m^{\star }$ , that is, in $\Omega$ , up to translations, where $m^{\star }$ is the convex symmetrization of $m$ .

Proof. We argue exactly as in the proof of [Reference Belloni, Ferone and Kawohl9, Lemma 3.6]. Let $u$ be a weak solution to the problem (1.2) corresponding to the first eigenvalue $\lambda ^H_{p,q}(\Omega )$ , that is,

\begin{equation*} \begin {cases} -\mathcal {L}_{p}(u)=\lambda ^H_{p,q}(\Omega ) m\| u\|_{q,m}^{p-q}u^{q-1} & \mbox {in}\ \Omega \\[5pt] u=0&\mbox {on}\ \partial \Omega . \end {cases} \end{equation*}

Let $t,h\gt 0$ and let us choose as a test function in (3.1) the following function in $W^{1,p}_0(\Omega )$ :

\begin{equation*} \varphi _h = \begin {cases} 0 & u\,{\leqslant}\, t\\[5pt] u-t & t\lt u\,{\leqslant}\, t+h\\[5pt] h & u\gt t+h. \end {cases} \end{equation*}

By standard arguments, we have

(4.2) \begin{equation} -\frac{d}{dt}\int _{\{u\gt t\}} H(\nabla u)^p\,d\mathcal{H}^{n-1}=\lambda _{p,q}^H(\Omega ) \|u\|_{q,m}^{p-q}\int _{\{u\gt t\}}mu^{q-1}\,dx. \end{equation}

Recalling that the anisotropic perimeter can be written as follows:

\begin{equation*} -\frac {d}{dt}\int _{\{u\gt t\}} H(\nabla u)\,d\mathcal {H}^{n-1} = P_H(\{u\gt t\}), \end{equation*}

by Hölder inequality, we get

\begin{equation*} P_H(\{u\gt t\})\,{\leqslant}\, \bigg (-\frac {d}{dt}\int _{\{u\gt t\}} H(\nabla u)^p\,d\mathcal {H}^{n-1}\bigg )^{\frac {1}{p}}(-\mu ^{\prime }(t))^{1-\frac {1}{p}}. \end{equation*}

Therefore, the isoperimetric inequality (2.9) gives

\begin{equation*} (-\mu ^{\prime }(t))^{1-p}\bigg (-\frac {d}{dt}\int _{\{u\gt t\}} H(\nabla u)^p\,d\mathcal {H}^{n-1}\bigg ) \,{\geqslant}\, n^pk_n^{\frac {p}{n}}\mu (t)^{p-\frac {p}{n}} \end{equation*}

Since $\mu ^{\prime }(t)=\frac{1}{{u^*}^{\prime }(\mu (t))}$ and (4.2) holds true, we have

\begin{equation*} (-{u^*}^{\prime }(\mu (t)))^{p-1} \,{\leqslant}\, n^{-p}k_n^{-\frac {p}{n}} \lambda ^H_{p,q}(\Omega )\| u\|_{q,m}^{p-q}\mu (t)^{\frac {p}{n}-p}\int _{\{u\gt t\}} m(x)u^{q-1} \,d\mathcal {H}^{n-1}. \end{equation*}

Using (2.12) and calling $s=\mu (t)$ , we have

\begin{equation*} (-{u^*}^{\prime }(s))^{p-1} \,{\leqslant}\, n^{-p}k_n^{-\frac {p}{n}}\lambda ^H_{p,q}(\Omega )\| u\|_{q,m}^{p-q}\,s^{\frac {p}{n}-p}\int _0^{s} m^* (u^*)^{q-1} \,dr. \end{equation*}

An application of the Hardy–Littlewood inequality gives the desired result.

The main tool we use in order to prove Theorem 1.1 is a suitable comparison result between $u$ and an eigenfunction $z$ of a suitable eigenvalue problem. More precisely, let $\Omega ^{\star }_\lambda$ be the Wulff shape centered at the origin such that $\lambda ^H_{p,q}(\Omega )$ is the first eigenvalue to the following symmetric problem:

(4.3) \begin{equation} \begin{cases} -\mathcal{L}_{p}(z)=\mu m^\star \| z\|_{q,m^\star }^{p-q}z^{q-1} & \mbox{in}\ \Omega ^{\star }_\lambda \\[5pt] z=0&\mbox{on}\ \partial \Omega ^{\star }_\lambda, \end{cases} \end{equation}

We stress that the Faber–Krahn inequality (3.8) implies that

(4.4) \begin{equation} |\Omega |\,{\geqslant}\, |\Omega ^{\star }_\lambda |, \end{equation}

and hence $m^\star$ is well defined in $\Omega ^\star _\lambda$ .

Let $z$ be a positive eigenfunction for the problem (4.3) corresponding to the first eigenvalue $\lambda ^H_{p,q}(\Omega )$ , and we observe that repeating the same argument as before, by Proposition 3.4, for any $1\lt q\,{\leqslant}\, p$ we have

(4.5) \begin{equation} (-{z^*}^{\prime }(s))^{p-1} = n^{-p} k_n^{-\frac{p}{n}}\lambda ^H_{p,q}(\Omega ) \| z^*\|_{q,m^*}^{p-q}\int _0^{s} m^*z^*(r)^{q-1}\,dr. \end{equation}

The following proposition gives a comparison result between the eigenfunctions $u$ and $z$ when they are normalized with respect to the $L^{\infty }$ norm.

Proposition 4.2. Let $\Omega \subset \mathbb{R}^n$ , $n\,{\geqslant}\, 2$ , be an open, bounded, and connected set, $1\lt q\,{\leqslant}\, p$ and let $m\in L^{\infty }(\Omega )$ be a positive function. Let $u$ be a positive solution to the problem (1.2) corresponding to $\lambda ^H_{p,q}(\Omega )$ and let $z$ be a positive eigenfunction to the problem (4.3) corresponding to $\lambda ^H_{p,q}(\Omega )$ such that

\begin{equation*}\|u\|_{L^{\infty }(\Omega )}=\|z\|_{L^{\infty }(\Omega ^\star _\lambda )}\end{equation*}

Then

\begin{equation*} u^*(s)\,{\geqslant}\, z^*(s), \qquad \forall s \in [0,|\Omega ^\star _\lambda |], \end{equation*}

where $u^*$ and $z^*$ are, respectively, the decreasing rearrangements of $u$ and $z$ . The equality case holds if and only if $\Omega =\Omega ^{\star }_\lambda$ and $m=m^{\star }$ , that is, in $\Omega$ , up to translations, where $m^\star$ is the convex symmetrization of $m$ .

Proof. First of all we stress that, if $|\Omega |=|\Omega ^\star _\lambda |$ , then there is nothing to prove, since Faber–Krahn inequality implies that $u^*(s)=z^*(s)$ .

Moreover, we have $u^*(|\Omega ^\star _\lambda |)\gt z^*(|\Omega ^\star _\lambda |)=0$ . Then, the following definition is well posed:

\begin{equation*} s_0=\inf \{s\in [0,|\Omega ^\star _\lambda |]\,:\; u^*(t)\,{\geqslant}\, z^*(t), \, \forall t \in [s,|\Omega ^\star _\lambda |]\}. \end{equation*}

By definition, $u^*(s_0)=z^*(s_0)$ , and we want to prove that $s_0=0$ . We proceed by contradiction supposing that $s_0\gt 0$ . Then under this assumption, $u^*$ and $z^*$ coincide in $0$ and $s_0$ and we have

(4.6) \begin{equation} \begin{cases} u^*(s)\lt z^*(s) & s\in (0,s_0)\\[5pt] u^*(s)\,{\geqslant}\, z^*(s) & s\in (s_0,|\Omega ^\star _\lambda |). \end{cases} \end{equation}

By (4.1), (4.5), and (4.6), we have that

\begin{equation*} -{u^*}^{\prime }(t)\,{\leqslant}\, -{z^*}^{\prime }(t), \quad \text { for every } t\in (0,s_0). \end{equation*}

Integrating between $(0,s)$ , with $s\in (0,s_0)$ , being $u^*(0)=z^*(0)$ , we get

\begin{equation*} u^*(s)\,{\geqslant}\, z^*(s), \qquad \forall s \in (0,s_0), \end{equation*}

which is in contradiction with the definition of $s_0$ . Hence, $s_0=0$ , and the proof is completed.

As an immediately consequence of the previous result, we get the following scale-invariant inequality for any $r\gt 0$ :

(4.7) \begin{equation} \frac{\|u\|_{L^r(\Omega, m)}}{\|u\|_{L^\infty (\Omega )}}\,{\geqslant}\, \frac{\|z\|_{L^r(\Omega ^{\star }_{\lambda },m^\star )}}{\|z\|_{L^\infty (\Omega _{\lambda }^{\star })}}. \end{equation}

When the functions $u$ and $z$ are normalized with respect to the weighted $L^q$ -norm, we get the following comparison result.

Theorem 4.3. Let $\Omega \subset \mathbb{R}^n$ be an open, bounded, and connected set, $1\lt q\,{\leqslant}\, p$ and let $m\in L^{\infty }(\Omega )$ be a positive function. Let $u$ be a positive solution to the problem (1.2) corresponding to $\lambda ^H_{p,q}(\Omega )$ and let $z$ be a positive eigenfunction to the problem (4.3) corresponding to $\lambda ^H_{p,q}(\Omega )$ such that

(4.8) \begin{equation} \int _{\Omega } m\,u^{q}\,dx = \int _{\Omega ^{\star }_\lambda } m^\star z^{q}\,dx. \end{equation}

Then we have

(4.9) \begin{equation} \int _0^{s} m^* \,(u^*)^{r}\,dt\,{\leqslant}\, \int _0^{s} m^*\,(z^*)^{r}\,dt, \quad s\in [0,|\Omega ^\star _\lambda |], \quad q\,{\leqslant}\, r \end{equation}

where $u^*$ , $m^*$ , and $z^*$ are, respectively, the decreasing rearrangements of $u$ , $m$ , and $z$ , and $m^\star$ is the convex symmetrization of $m$ . The equality case holds if and only if $\Omega =\Omega ^{\star }$ , $z=u=u^*$ , and $m=m^{\star }$ , that is, $\Omega$ , up to translations.

Proof. If $|\Omega |=|\Omega ^{\star }_\lambda |$ , the conclusion is trivial. Let be $|\Omega |\gt |\Omega ^{\star }_\lambda |$ , since $u$ and $z$ verify (4.8), by (4.7) it holds that

\begin{equation*} u^*(0)=\|u\|_{L^{\infty }(\Omega )} \,{\leqslant}\, \|z\|_{L^{\infty }(\Omega ^{\star }_{\lambda })}= z^*(0), \end{equation*}

If $u^*(0)=z^*(0)$ , then Proposition 4.2 and the normalization (4.8) imply that $u^*(s)=z^*(s)$ for every $s\in [0,|\Omega ^{\star }_\lambda |]$ and than the claim follows trivially.

Let $u^*(0)\lt z^*(0)$ . Since $u^*(|\Omega ^{\star }_\lambda |)\gt z^*(|\Omega ^{\star }_\lambda |)$ , we can consider

\begin{equation*} s_0 = \sup \{s\in (0,|\Omega ^{\star }_\lambda |) \;:\; u^*(t)\,{\leqslant}\, z^*(t) \text { for } t \in [0,s]\}. \end{equation*}

Obviously, $0\lt s_0\lt |\Omega ^{\star }_\lambda |$ , $u^*(s_0)=z^*(s_0)$ and $u^*\,{\leqslant}\, z^*$ in $[0,s_0]$ . We want to show that $u^*\gt z^*$ in $[s_0,|\Omega ^{\star }_\lambda |]$ . Indeed, if we suppose by contradiction that there exists $s_1\gt s_0$ such that $u^*(s_1)=z^*(s_1)$ and $u^*(s)\gt z^*(s)$ for $s\in (s_0,s_1)$ , we can construct the following function:

\begin{equation*} w^*(s)= \begin {cases} z^*(s) & s \in [0,s_0]\cup [s_1,|\Omega ^{\star }_\lambda |]\\[5pt] u^*(s) & s \in [s_0,s_1]. \end {cases} \end{equation*}

It is straightforward to check that

\begin{equation*} \int _{\Omega }H(\nabla w)^p\,dx = n^pk^p_n\int _{\Omega } (-{w^*}^{\prime }(k_n H^{\circ }(x)^n))^pH^{\circ }(x)^{p(n-1)}\,dx. \end{equation*}

Applying Coarea Formula and considering the change of variables $s=k_n t^n$ , we get

\begin{equation*} \int _{\Omega } H(\nabla w)^p\,dx = n^p k_n^{\frac {p}{n}}\int _0^{|\Omega ^{\star }_\lambda |} s^{p-\frac {p}{n}}(-{w^*}^{\prime }(t))^p\,dt. \end{equation*}

Thanks to the normalization (4.8) and the definition of $w$ , we have that

\begin{equation*} \| u\|_{L^q(\Omega, m)} = \| z\|_{L^q(\Omega ^\star _\lambda, m^\star )} \,{\leqslant}\, \| w\|_{L^q(\Omega ^\star _\lambda, m^\star )}, \end{equation*}

then by (4.1) and (4.5), we have that

(4.10) \begin{equation} (-{w^*}^{\prime }(s))^{p-1} \,{\leqslant}\, n^{-p} k_n^{-\frac{p}{n}}\lambda ^H_{p,q}(\Omega ) \| w^*\|_{q,m^*}^{p-q}s^{\frac{p}{n}-p}\int _0^{s} m^*(r)({w^*})^{q-1}(r)\,dr. \end{equation}

Multiplying (4.10) by $-w^{\prime }$ , rearranging the terms and integrating between $0$ and $|\Omega ^{\star }_\lambda |$ , we get

\begin{equation*} \begin {aligned} n^pk_n^{\frac {p}{n}}&\int _0^{|\Omega ^{\star }_\lambda |} s^{p-\frac {p}{n}}(-{w^*}^{\prime }(s))^p\,ds\,{\leqslant}\, \\[5pt] &\,{\leqslant}\, \lambda ^H_{p,q}(\Omega ) \| w^*\|_{q,m^*}^{p-q} \int _0^{|\Omega ^{\star }_\lambda |}(-{w^*}^{\prime }(s))\int _0^sm^*(r)({w^*})^{q-1}(r)\,dr\,ds. \end {aligned} \end{equation*}

An integration by parts allows us to conclude that

\begin{equation*} \frac {\displaystyle \int _{\Omega ^\star _\lambda } H(\nabla w)^p\,dx}{\bigg (\displaystyle \int _{\Omega ^\star _\lambda }m^\star w^q\,dx\bigg )^{\frac {p}{q}}}=\frac {\displaystyle n^pk_n^{\frac {p}{n}}\int _0^{|\Omega ^{\star }_\lambda |} s^{p-\frac {p}{n}}(-w^{\prime }(s))^p\,ds}{\bigg (\displaystyle \int _0^{|\Omega ^{\star }_\lambda |}m^*(s)({w^*})^q(s)\,ds\bigg )^{\frac {p}{q}}}\,{\leqslant}\, \lambda ^H_{p,q}(\Omega )=\lambda ^H_{p,q}(\Omega ^{\star }_\lambda ). \end{equation*}

By the minimality and the simplicity of $\lambda ^H_{p,q}$ , and the definition of $w^*$ , it must be $w^*(s)=z^*(s)$ for every $s\in [0,|\Omega ^{\star }_\lambda |]$ , but this is a contradiction since in $(s_0,s_1)$ we have that $u^*(s)\gt z^*(s)$ .

In this way, we have proved that there exists a unique point $s_0$ where $u^*$ and $z^*$ can cross each other, and such that

(4.11) \begin{equation} \begin{cases} u^*(s)\,{\leqslant}\, z^*(s) & s\in [0,s_0]\\[5pt] u^*(s)\,{\geqslant}\, z^*(s) & s\in [s_0,|\Omega ^{\star }_\lambda |]. \end{cases} \end{equation}

If we extend $z^*$ to be zero in $[|\Omega ^{\star }_\lambda |, |\Omega |]$ , by (4.8) and (4.11) then we have that for every $s\in [0,|\Omega |]$

(4.12) \begin{equation} \int _0^{s} m^*(t)({u^*(t)})^{q}\,dt\,{\leqslant}\, \int _0^{s} m^*(t)({z^*(t)})^{q}\,dt. \end{equation}

Indeed (4.11) implies that the function:

\begin{equation*}G(s)=\int _0^{s} {m^*}(t)((z^*))^{q}-(u^*)^q\,dt, \qquad s \in [0,|\Omega |] \end{equation*}

has a maximum in $s_0$ and cannot be negative in any point. This proves (4.12). Finally, inequality (4.9) follows easily by (4.12) by using Proposition 2.4 being $m^{\star }(u^{\star })^q \prec m^{\star }z^q$ .

Proof of Theorem 1.1. The proof of statement i) follows directly from (4.8) and (4.9), indeed we have

\begin{equation*} \bigg (\int _{\Omega } m u^{r}\,dx\bigg )^{\frac {1}{r}}\,{\leqslant}\, \bigg (\int _{\Omega ^{\star }_\lambda } m^\star z^{r}\,dx\bigg )^{\frac {1}{r}} = \frac {\displaystyle \bigg (\int _{\Omega ^{\star }_\lambda } m^\star z^{r}\,dx\bigg )^{\frac {1}{r}} }{\displaystyle \bigg (\int _{\Omega ^{\star }_\lambda } m^\star z^{q}\,dx\bigg )^{\frac {1}{q}} } \bigg (\int _{\Omega } m u^{q}\,dx\bigg )^{\frac {1}{q}}. \end{equation*}

Therefore, we have

\begin{equation*} \|u \|_{L^r(\Omega, m)} \,{\leqslant}\, C\,\,\| u\|_{L^q(\Omega, m)}, \quad q \,{\leqslant}\, r\lt +\infty ; \end{equation*}

with

\begin{equation*} C = \displaystyle \frac {\|z \|_{L^r(\Omega ^{\star }_{\lambda },m^{\star })}}{\|z \|_{L^q(\Omega ^{\star }_{\lambda },m^{\star })}}. \end{equation*}

The proof of statement ii) follows immediately by Proposition 4.2 with $C= \frac{\|z\|_{\infty }}{\|z\|_{L^r(\Omega ^\star, m^\star )}}$ and the theorem is completely proved.