A quantitative form of Faber–Krahn inequality

Abstract

The Faber–Krahn inequality states that balls are the unique minimizers of the first eigenvalue of the p-Laplacian among all sets with fixed volume. In this paper we prove a sharp quantitative form of this inequality. This extends to the case $p>1$ a recent result proved by Brasco et al. (Duke Math J 164:1777–1831, 2015) for the Laplacian.

Appendix

In this section we discuss some crucial steps in the proof of Theorem 3.23. Indeed the whole proof follows closely the one of Theorem 9.1 in [12], where the authors deal with solutions of the equation

Although our Eq. (3.20) is very similar to the one above, some arguments used in [12] must be adjusted to our situation. The main changes are discussed in detail in this section. Throughout all this section we shall always assume that u is a solution in $W^{1,2}_0(B_2)$ of (3.20) with $\{u>0\}\subset B_{3/2}$.

Remark 5.1

Observe that if $x_0\in \partial \{u>0\}$ and $r>0$, by setting

$$\begin{aligned} u_{r}(x):=\frac{u(x_0+rx)}{r}, \end{aligned}$$

(5.1)

the rescaled function $u_{r}$ satisfies the equation

(5.2)

where $q_u$ is a Lipschitz function satisfying the assumptions of Theorem 3.23. Moreover, if $u\in F(\sigma _+,\sigma _-;\tau )$ in $B_r(x_0)$ with respect to some direction $\nu$, then $u_{r}\in F(\sigma _+,\sigma _-;\tau )$ in B with respect to the same direction.

The first lemma is proved with exactly the same proof of Claim 6.7 in [12].

Lemma 5.2

If $B_r(w)$ is a ball in $\{u=0\}$ touching $\partial \{u>0\}$ at z, then

$$\begin{aligned} \limsup _{x\rightarrow z,\,u(x)>0}\frac{u(x)}{\mathrm{dist}(x,B_r(w))}=q_{u}(z). \end{aligned}$$

(5.3)

Roughly speaking, the idea of the proof of Theorem 3.23 is to show that if in a small ball $B_r(x_0)$ the free boundary $\partial \{u>0\}$ is sufficiently flat, then it becomes even flatter in smaller balls. This amounts to prove a decay estimate for the quantity $F(\sigma _+,\sigma _-;\tau )$. This goal is achieved through a certain number of intermediate steps. The first one is contained in the next lemma.

Lemma 5.3

There exists a constant $\gamma _0(p,n,\min q_u)$ such that for every $\varepsilon >0$, there exists $\sigma _\varepsilon >0$ with the property that if $\,0<\sigma <\sigma _\varepsilon$, $0<r<\gamma _0\sigma$ and $u\in F(\sigma ,1;\sigma )$ in $B_r(x_0)$ with respect to $\nu \in \mathbb {S}^{n-1}$, then $u\in F(2\sigma ,\varepsilon ;\sigma )$ in $B_{\frac{r}{2}}(x_0)$ with respect to the same $\nu$.

Proof

The proof is similar to the one of Lemma 6.5 in [12]. Yet a few changes are needed.

Up to replacing u by the function $\frac{u_{r}}{q_{u}(x_0)}$, where $u_{r}$ is defined as in (5.1), by (5.2) we may assume that u satisfies in B the equation

Moreover, up to a rotation, we may assume that $\nu =e_n$. Thus we have that

$$\begin{aligned} {\left\{ \begin{array}{ll} \text {if }x\in B\text { and }x_n\ge \sigma ,\text { then }u(x)=0 &{} \\ |\nabla u(x)|\le 1+\sigma \,\,\,\text {in }B &{}\\ \end{array}\right. } \end{aligned}$$

(5.4)

As in [12] we set, for $y\in \mathbb R^{n-1}$,

$$\begin{aligned} \eta (y):=\mathrm{exp}\bigg (-\frac{9|y|^2}{1-9|y|^2}\bigg ) \end{aligned}$$

for $|y|< 1/3$ and $\eta (y)=0$ for $|y|\ge 1/3$. Then, we denote by s the largest nonnegative number such that

$$\begin{aligned} B\cap \{u>0\}\subset D:=\{x\in B: x_n<\sigma -s\eta (x^\prime )\}, \end{aligned}$$

where $x^\prime =(x_1,\dots ,x_{n-1})$. Observe that since $0\in \partial \{u>0\}$ we have $s\le \sigma$. Moreover, there exists z such that

$$\begin{aligned} z\in B_{\frac{1}{2}}\cap \partial D\cap \partial \{u>0\},\qquad z_n=\sigma -s\eta (z'). \end{aligned}$$

(5.5)

Then, given $\xi \in B_{\frac{3}{4}}$, with $\xi _n\le -1/2$, let us denote by v the unique solution of the following problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -{\text {div}}(|\nabla v|^{p-2}\nabla v)=\displaystyle \frac{r}{q_{u}(x_0)^{p-1}} &{}\quad \text {in } \,D{\setminus }{\overline{B}}_\varrho (\xi ), \\ v(x)=0 &{}\quad \text {if }\,x\in \partial D\cap B, \\ v(x)=(1+\sigma )(\sigma -x_n) &{}\quad \text {if }\,x\in \partial D{\setminus } B, \\ v(x)=-(1-\kappa \sigma )x_n &{}\quad \text {if }\,x\in \partial B_{\varrho }(\xi ), \\ \end{array}\right. } \end{aligned}$$

(5.6)

where $\varrho <1/4$ and $\kappa$ are two positive constants, to be chosen later independently of $\sigma$ and r. In Lemma 5.4 below we shall prove that there exist two positive constants $C(\varrho )$ and $c(\varrho )$, depending only on $\varrho$, such that if $\sigma <\sigma (\kappa ,\varrho )$, then

$$\begin{aligned} |\nabla v(z)|\le 1+C(\varrho )\sigma -c(\varrho )\kappa \sigma . \end{aligned}$$

Observe that if $u\le v$ on $\partial B_\varrho (\xi )$, then by construction and by the second condition in (5.4) we would have that $u\le v$ on $\partial (D{\setminus } B_\varrho (\xi ))$, thus concluding, by the comparison principle, that $u\le v$ in $D{\setminus } B_\varrho (\xi )$. Therefore, choosing $\kappa (\varrho )$ sufficiently large and $0<\sigma <\sigma (\varrho ,\kappa )$ we would have

$$\begin{aligned} |\nabla v(z)|<1-\sigma , \end{aligned}$$

where z is the point defined in (5.5). But this is impossible since, from (5.3) and the assumption $u\in F(\sigma ,1;\sigma )$, we have

$$\begin{aligned} |\nabla v(z)|\ge \frac{q_{u}(x_0+rz)}{q_{u}(x_0)}\ge 1-\sigma . \end{aligned}$$

Therefore there exists a point $x_\xi \in \partial B_\varrho (\xi )$ such that

$$\begin{aligned} u(x_\xi )\ge v(x_\xi ). \end{aligned}$$

(5.7)

Hence for $\sigma$ sufficiently small, recalling the second condition in (5.4), we have

$$\begin{aligned} u(\xi )&\ge u(x_\xi )-\varrho (1+\sigma )\ge v(x_\xi )-\varrho (1+\sigma ) =-(1-\kappa \sigma )(x_\xi )_n-\varrho (1+\sigma )\\&\ge -(x_\xi )_n-\kappa \sigma -2\varrho \ge -\xi _n-4\varrho . \end{aligned}$$

Integrating this inequality along vertical lines and using again the inequality $|\nabla u|\le 1+\sigma$ we have, still for $\sigma <\sigma (\varrho )$

$$\begin{aligned} u(\xi +te_n)\ge u(\xi )-t(1+\sigma )>-\xi _n-4\varrho -t(1+\sigma )\ge -(\xi _n+t)-5\varrho . \end{aligned}$$

Choosing $\varrho =\min \{\varepsilon /5,1/5\}$, multiplying both sides of the previous inequality by $rq_{u}(x_0)$ and rescaling back to $B_r(x_0)$, we obtain immediately the assertion. $\square$

Lemma 5.4

Let v be the function defined in (5.6). There exists a positive constant $\gamma _0(p,n, \min q_u)$ such that for every $\varrho <1/4$ and every $\kappa >0$ there exist two positive constants $C(\varrho )$ and $c(\varrho )$ and a positive constant $\sigma (\kappa ,\varrho )$ such that if $0<r<\gamma _0\sigma$ and $0<\sigma <\sigma (\kappa ,\varrho )$, then

$$\begin{aligned} |\nabla v(z)|\le 1+C(\varrho )\sigma -\kappa c(\varrho )\sigma , \end{aligned}$$

where z is as in (5.5).

Proof

The proof goes as the one of Claim 6.8 in [12] with some extra work needed to estimate a few quantities in a more precise way.

As in [12], for every $\xi \in \mathbb R^n$ we set

$$\begin{aligned} a_{ij}(\xi ):=\delta _{ij}+(p-2)\frac{\xi _i\xi _j}{|\xi |^2}\qquad \text { for all }i,j=1,\dots ,n\text { and }\xi \in \mathbb R^n{\setminus }\{0\}. \end{aligned}$$

(5.8)

Observe that

$$\begin{aligned} \frac{1}{\lambda }|\zeta |^2\le a_{ij}(\xi )\zeta _i\zeta _j\le \lambda |\zeta |^2\qquad \text {for every } \,\xi ,\zeta \in \mathbb R^n, \end{aligned}$$

with $\lambda (p)>0$. We are going to construct a comparison function w of the form $w=v_1-\kappa \sigma v_2$. Let $\eta$ and D be as in the proof of Lemma  5.3 and define for $x\in D$

$$\begin{aligned} v_1(x):=\frac{\gamma _1}{\mu _1}\big (1-\mathrm{exp}(-\mu _1d_1(x))\big ), \end{aligned}$$

where

$$\begin{aligned} d_1(x):=-x_n+\sigma -s\eta (x^\prime ) \end{aligned}$$

and $\mu _1<1$ and $\gamma _1$ are positive constants depending on $\sigma$ to be chosen later. Observe that for $\sigma$ small

$$\begin{aligned} v_1(x)\ge \gamma _1d_1(x)(1-2\mu _1)\qquad \text {for } x\in \partial (D{\setminus } B_\varrho (\xi )). \end{aligned}$$

Therefore, if we impose that

$$\begin{aligned} \gamma _1(1-2\mu _1)\ge 1+2\sigma , \end{aligned}$$

(5.9)

we have

$$\begin{aligned} v_1(x)\ge (1+2\sigma )d_1(x)\ge v(x)\qquad \text {for } x\in \partial (D{\setminus } B_\varrho (\xi )). \end{aligned}$$

(5.10)

Observe that there exists an absolute constant $C_1>0$ such that

$$\begin{aligned} 1\le |\nabla d_1|\le 1+C_1\sigma ,\qquad |D^2d_1|\le C_1\sigma . \end{aligned}$$

(5.11)

Therefore, recalling (5.9), we have

$$\begin{aligned} 1\le \gamma _1(1-2\mu _1)\le \gamma _1\mathrm{exp}(-\mu _1d_1)\le |\nabla v_1|\le \gamma _1(1+C_1\sigma ). \end{aligned}$$

(5.12)

Next, we estimate ${\text {div}}(|\nabla v_1|^{p-2}\nabla v_1)$. We have

$$\begin{aligned} {\text {div}}(|\nabla v_1|^{p-2}\nabla v_1)&=|\nabla v_1|^{p-2}a_{ij}(\nabla v_1)D_{ij}v_1\\&=\gamma _1|\nabla v_1|^{p-2}\mathrm{exp}(-\mu _1d_1)a_{ij}(\nabla v_1)(D_{ij}d_1-\mu _1D_id_1D_jd_1). \end{aligned}$$

From (5.11) it follows that there exist two positive constants $C_2(p), c_2(p)$ such that

$$\begin{aligned} a_{ij}(\nabla v_1)(D_{ij}d_1-\mu _1D_id_1D_jd_1)\le C_2\sigma -c_2\mu _1. \end{aligned}$$

Therefore, if we choose

$$\begin{aligned} \mu _1=2\frac{C_2}{c_2}\sigma , \end{aligned}$$

from (5.12) we get that for $\sigma$ sufficiently small

$$\begin{aligned} {\text {div}}(|\nabla v_1|^{p-2}\nabla v_1)\le -\frac{1}{2}\gamma _1\min \big \{1,\gamma _1^{p-2}(1+C_1\sigma )^{p-2}\big \}C_2\sigma . \end{aligned}$$

From the above choice of $\mu _1$ it is clear that there exists a constant $C_3$, depending only on p such that, if we set

$$\begin{aligned} \gamma _1:=1+C_3\sigma , \end{aligned}$$

then $\gamma _1$ satisfies the constraint (5.9) and we have

$$\begin{aligned} {\text {div}}(|\nabla v_1|^{p-2}\nabla v_1)\le -C_4\sigma , \end{aligned}$$

for a positive constant $C_4$ depending only on p. Therefore, we may conclude that if $0<r<C_4\sigma \min q_u^{p-1}$, then

$$\begin{aligned} {\text {div}}(|\nabla v_1|^{p-2}\nabla v_1)\le -\frac{r}{q_{u}(x_0)^{p-1}}\qquad \text {in }D{\setminus } B_{\varrho }(\xi ). \end{aligned}$$

(5.13)

From this estimate, using the comparison principle and recalling (5.10) we have that

$$\begin{aligned} v_1(x)\ge v(x)\qquad \text {for all }x\in D{\setminus } B_{\varrho }(\xi ). \end{aligned}$$

Moreover, from the above choice of $\gamma _1$ and (5.12) we have that there exists a positive constant $C_5(p)$ depending only on p such that

$$\begin{aligned} 1\le |\nabla v_1|\le 1+C_5\sigma . \end{aligned}$$

Taking $\sigma <1/10$ we now define a function $v_2$, by setting

$$\begin{aligned} v_2(x):=\frac{\gamma _2}{\mu _2}(\mathrm{exp }(\mu _2d_2(x))-1)\qquad \text {for }x\in \widetilde{D}{\setminus } B_{\varrho }(\xi ), \end{aligned}$$

where $\mu _2$ a positive constant to be chosen later,

$$\begin{aligned} \gamma _2:=\frac{1}{4}\frac{\mu _2}{\mathrm{e}^{\mu _2}-1} \end{aligned}$$

$\widetilde{D}\subset D$ is a domain containing $D{\setminus }\mathcal N_{\frac{1}{10}}(\partial B\cap \{x_n=0\})$ and $d_2$ is a function in $C^2(D{\setminus } B_{\varrho }(\xi ))$ with values in (0, 1), satisfying the following conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} d_2=0 &{}\quad \text {on }\partial \widetilde{D} \\ d_2=1 &{}\quad \text {on }\partial B_\varrho (\xi ) \\ \frac{1}{\widetilde{C}}\le |\nabla d_2|\le \widetilde{C} &{}\quad \text {in }\widetilde{D}{\setminus } B_\varrho (\xi ), \end{array}\right. } \end{aligned}$$

for a positive constant $\widetilde{C}(n,\varrho )$. Arguing as above, it is clear that we may choose $\mu _2$ large enough, depending only on p, n and $\varrho$, so that for all $\xi \in \mathbb R^n$

$$\begin{aligned} a_{ij}(\xi )D_{ij}v_2=\gamma _2\mathrm{exp}(\mu _2d_2)a_{ij}(\xi )(D_{ij}d_2+\mu _2D_id_2D_jd_2)\ge \overline{\mu }>0, \end{aligned}$$

(5.14)

for some positive constant $\overline{\mu }$. Moreover, since $\nabla v_2=\gamma _2\mathrm{exp}(\mu _2d_2)\nabla d_2$, we have

$$\begin{aligned} \frac{\gamma _2}{\widetilde{C}}\le |\nabla v_2|\le \gamma _2\mathrm{e}^{\mu _2}\widetilde{C}. \end{aligned}$$

From this estimate, setting

$$\begin{aligned} w:=v_1-\kappa \sigma v_2 \end{aligned}$$

and recalling (5.14) we may easily conclude, arguing as in the proof of (5.13), that for $\sigma <\sigma (\kappa ,\varrho )$, with $\sigma (\kappa ,\varrho )$ sufficiently small,

$$\begin{aligned} {\text {div}}(|\nabla w|^{p-2}\nabla w)\le -C_6\sigma , \end{aligned}$$

for some positive constant $C_6$ depending only $p,n,\varrho$. Hence, if $0<r<\gamma _0\sigma$, for some constant $\gamma _0$ depending only on $p,n,\varrho$ and $\min q_u^{p-1}$, we may conclude that

$$\begin{aligned} {\text {div}}(|\nabla w|^{p-2}\nabla w)\le -\frac{r}{q_{u}(x_0)^{p-1}}. \end{aligned}$$

(5.15)

Finally observe that from the definition of $d_2$ and (5.10) we have that

$$\begin{aligned} w= v_1\ge v\qquad \text {on }\,\partial \widetilde{D}, \end{aligned}$$

while, observing that $v_2\equiv 1/4$ on $\partial B_\varrho (\xi )$ and using (5.10) again, we have that for all $x\in \partial B_\varrho (\xi )$

$$\begin{aligned} w(x)\ge d_1(x)-\frac{\kappa \sigma }{4}\ge -(1-\kappa \sigma )x_n=v(x), \end{aligned}$$

since $\xi _n\le -1/2$ and $\varrho <1/4$. Thus, recalling (5.15), we may conclude that $w\ge v$ in $\widetilde{D}{\setminus } B_\varrho (\xi )$. Therefore, if z is as in (5.5) and $\sigma <\sigma (\kappa ,\varrho )$,

$$\begin{aligned} |\nabla v(z)|\le |\nabla w(z)|= |\nabla v_1(z)|-\kappa \sigma |\nabla v_2(z)|\le 1+C_5\sigma -\kappa \sigma \frac{\gamma _2}{\widetilde{C}}, \end{aligned}$$

where the equality in the above formula follows by observing that $\nabla d_1(z)$ and $\nabla d_2(z)$ have the same direction. This estimate concludes the proof of the lemma. $\square$

Next lemma provides an estimate from below on $\nabla u_j$ near the free boundary.

Lemma 5.5

For every $\varepsilon ,\delta >0$ there exists $\sigma _{\varepsilon ,\delta }$ such that if $\,0<\sigma <\sigma _{\varepsilon ,\delta }$ and $0<r<\gamma _0\sigma$, with $\gamma _0$ as in Lemma 5.3, and $u\in F(\sigma ,1;\sigma )$ in $B_r(x_0)$ with respect to some direction $\nu$, then $|\nabla u|\ge q_{u}(x_0)(1-\delta )$ in $B_{\frac{19r}{20}}(x_0)\cap \{(x-x_0)\cdot \nu \le -\varepsilon r\}$.

Proof

We argue as in the proof of Lemma 6.6 in [12] with the some technical modifications.

First, up to a translation and a rotation we may assume that $x_0=0$ and that $\nu =e_n$. Then, we argue by contradiction, assuming that there exist a sequence of Lipschitz functions $u_{k}$, satisfying the assumptions of Theorem 3.23 with constants uniformly bounded above and away from zero, and a sequence of radii $0<r_k<\frac{\gamma _0}{k}$ such that $u_{k}\in F(\frac{1}{k},1;\frac{1}{k})$ with respect to $e_n$ in $B_{r_k}$ and

$$\begin{aligned} |\nabla u_{k}(\widetilde{x}_k)|< q_{u_{k}}(0)(1-\delta )\qquad \text {for some }\,\widetilde{x}_k\in B_{\frac{19r_k}{20}}\cap \{x_n\le -\varepsilon r_k\}. \end{aligned}$$

By rescaling $u_{k}$ as in the proof of Lemma 5.3, we set for all $x\in B$

$$\begin{aligned} v_k(x)=\frac{u_{k}(r_kx)}{r_kq_{u_{k}}(0)} \end{aligned}$$

and we have that

and that

$$\begin{aligned} |\nabla v_{k}( x_k)|\le 1-\delta \qquad \text {for some }\,x_k\in B_{\frac{19}{20}}\cap \{x_n\le -\varepsilon \}. \end{aligned}$$

Letting $k\rightarrow \infty$, from Lemma 5.3 we obtain that

$$\begin{aligned} v_k\rightarrow v_0\qquad \text {uniformly in }\,\overline{B}, \end{aligned}$$

where $v_0$ is the p-harmonic function such that $v_0(x)=0$ if $x_n\ge 0$, $v_0(x)=-x_n$ if $x_n<0$. Note that by elliptic regularity $v_k\rightarrow v_0$ locally in $C^{1,\alpha }$ in $B\cap \{x_n<0\}$. Therefore, if $\overline{x}$ is a limit point of the sequence $x_k$ we have that $|\nabla v_0(\overline{x})|\le 1-\delta$ thus getting a contradiction since $|\nabla v_0(\overline{x})|=1$. $\square$

Next step in the proof of the regularity of the free boundary is to improve the decay estimates contained in Lemmas 5.3 and 5.5. This is the content of the following two results.

Proposition 5.6

There exist $\sigma _0,C_0>0$, depending only on p, n and $\min q_u$ such that if $u\in F(\sigma ,1;\sigma )$ in $B_r$, then $u\in F(2\sigma ,C_0\sigma ;\sigma )$ in $B_{\frac{r}{2}}$, provided that $0<\sigma <\sigma _0$ and $0<r<\gamma _0\sigma$, where $\gamma _0$ is as in Lemma 5.3.

Proof

Following the proof of Theorem 6.3 in [12], we start with the same rescaling as in the proof of Lemma 5.3 and with the same definition of the function v as in (5.6). Then, we choose $\varrho =1/10$, and $\kappa$ accordingly, so that (5.7) holds. Then, we set for $x\in D$

$$\begin{aligned} w(x):=(1+\sigma )(\sigma -x_n)-u(x). \end{aligned}$$

Then, the second condition in (5.4) implies that $w(x)\ge 0$ in $B_{2\varrho }(\xi )$, while from (5.7) and the definition of v we get

$$\begin{aligned} w(x_\xi )\le (1+\sigma )(\sigma -(x_\xi )_n)-v(x_\xi )\le C_1(\kappa )\sigma , \end{aligned}$$

for some positive constant $C_1$ depending only on the choice of $\kappa$. If $\sigma$ is sufficiently small we have from Lemma 5.5 that $|\nabla u|>\frac{1}{2}$ in $B_{2\varrho }(\xi )$, hence u satisfies the linear equation in nondivergence form

$$\begin{aligned} a_{ij}(\nabla u(x))D_{ij}u(x)=-\frac{r}{q_{u}(x_0)^{p-1}|\nabla u(x)|^{p-2}}, \end{aligned}$$

where the coefficients $a_{ij}$ are defined as in (5.8). From the Harnack inequality, see Corollary 9.21 and Theorem 9.22 of [18], and the bounds on $|\nabla u|$ we may conclude that

$$\begin{aligned} w(\xi )\le C(w(x_\xi )+r), \end{aligned}$$

where C depends on p, n and $\min q_{u}$. Therefore, for a possibly different constant C, we have

$$\begin{aligned} u(\xi )\ge -\xi _n-C\sigma \quad \text {on }\,\{\xi \in \partial B_{\frac{3}{4}}:\,\xi _n\le -1/2\}. \end{aligned}$$

The assertion then follows integrating on vertical lines as at the end of the proof of Lemma 5.3. $\square$

Next result improves the statement of Lemma 5.5. Its proof follows from the above Proposition 5.6 with exactly the same proof of Theorem 6.4 in [12], by using the same rescaling used in the proof of Lemma 5.5.

Proposition 5.7

For every $\delta \in (0,1)$ there exist $\sigma _{\delta }, C_\delta >0$ such that if $\,0<\sigma <\sigma _{\delta }$ and $0<r<\gamma _0\sigma$, with $\gamma _0$ as in Lemma 5.3, and $u\in F(\sigma ,1;\sigma )$ in $B_r(x_0)$ with respect to some direction $\nu$, then $|\nabla u|\ge q_{u}(x_0)(1-\delta )$ in $B_{\frac{r}{2}}(x_0)\cap \{(x-x_0)\cdot \nu \le -C_\delta r\}$.

From this proposition on the proof of Theorem 3.23 goes as in [12] with the same kind of technical changes that we have performed in the proofs given in this section. Further details on this remaining part of the proof can be found in [17].