In the previous post we discussed that the classical FaberKrahn inequality for the first eigenvalue of the $p$Laplacian
\[\lambda_1(\Omega) \geq \lambda_1(B)\]can be refined in several ways, for instance, in the form
\[\lambda_1(\Omega) \geq \lambda_1(B)(1 + C \alpha),\]where $C>0$ is some coefficient and $\alpha$ is some deficiency factor.
However, in almost all improvements known for me, the refinement is given in a nonconstructive way. That is, at least the coefficient $C$ it not quantified. This causes some troubles if one wants to use such refinements for some particular domains $\Omega$ in order to get better lower estimates for $\lambda_1(\Omega)$. Nevertheless, in the linear case $p=2$ such quantification was done in [_{1}] (see also [_{2}] for an improvement).
In this post, I would like to transpose the arguments from [_{1}] to the nonlinear case $p \geq 2$. We will see how $C$ and $\alpha$ look like. Withal, the post could be served as a material to study some new techniques (at least for me :).
So, our main result is the following.
Theorem. Let $p \geq 2$. Let $\Omega \subset \mathbb{R}^2$ be a bounded simplyconnected domain. Let $B$ be a ball such that $\vert\Omega\vert = \vert B\vert$ and $r_0(\Omega)$ be its radius. Let $r_i(\Omega)$ be the inradius of $\Omega$. Define the interior deficiency of $\Omega$ as
\[d_i(\Omega) := 1  \frac{r_i(\Omega)}{r_0(\Omega)}.\]Then the following inequality is satisfied:
\begin{equation}\label{eq:FKimp} \lambda_1(\Omega) \geq \lambda_1(B) \left(1 + \frac{\left(\sqrt{p^2+1}p\right)^p}{(p+1)(p+2)\pi^{p/2}} \, d_i(\Omega)^{p+1}\right). \end{equation}
Although the fraction in \eqref{eq:FKimp} is quite small and decreases to zero as $p \to \infty$, the expression in the brackets can be explicitly computed (estimated) for a given domain $\Omega$ and for any $p>1$.
Proof of Theorem. Let us assume, without loss of generality, that the first eigenfunction $u$ associated with $\lambda_1(\Omega)$ is normalized such that $\int_\Omega \vert u\vert^p \, dx = 1$. That is, we have
\[\lambda_1(\Omega) = \int_\Omega \vert\nabla u\vert^p \, dx.\]Denote by $u_0$ the spherical rearrangement of $u$, i.e., $u_0: B \to \mathbb{R}$ is rotation invariant, $u_0(\vert x\vert) \geq u_0(\vert y\vert)$ for $\vert x\vert<\vert y\vert$, and $\vert \{u_0 > t\}\vert = \vert \{u > t\}\vert $ for all $t \in [0, T]$, where $T$ is a maximum of $u$. Let us also denote $\varphi(t) := \vert \{u_0 > t\}\vert = \vert \{u > t\}\vert $ for $t \in [0, T]$.
Using the coarea formula (see, e.g., Theorem 2 in Section 3.4.3 on p. 117 in [_{3}])
\[\int_\Omega g(x) \vert\nabla u\vert \, dx = \int_0^\infty \int_{\{u(x)=t\}} g(x) \, d\sigma(x) \, dt,\]we deduce that
\[\int_\Omega \vert\nabla u\vert^p \, dx = \int_\Omega \vert\nabla u\vert^{p1} \vert\nabla u\vert \, dx = \int_0^T \int_{\{u=t\}} \vert\nabla u\vert^{p1} \, d\sigma \, dt.\]Applying now the formula (31) on p. 161 of the wellknown paper of Brothers & Ziemer [_{4}] with $A(s) = s^p$ (or just using the Holder inequality), we get
\[\int_0^T \int_{\{u=t\}} \vert\nabla u\vert^{p1} \, d\sigma(x) \, dt \geq \int_0^T \frac{\left(\int_{\{u=t\}} \, d\sigma\right)^p}{(\varphi'(t))^{p1}} \, dt \equiv \int_0^T \frac{\sigma^p(\{u=t\})}{(\varphi'(t))^{p1}} \, dt.\]Here we denote by $\sigma(\{u=t\})$ the $(N1)$dimensional Haussdorf measure of the set $\{u=t\}$. We also used the fact that (see Lemma 2.3 (iii) and formula (21) in [_{4}])
\[\varphi'(t) = \int_{\{u=t\}} \frac{d\sigma}{\vert\nabla u\vert},\]since the singular set $C:=\{ x:~ \vert \nabla u \vert(x) = 0 \}$ has zero Lebesgue measure, see the article of Lou [_{5}].
Therefore, we get
\begin{equation}\label{eq:1} \lambda_1(\Omega) = \int_\Omega \vert\nabla u\vert^p \, dx \geq \int_0^T \frac{\sigma^p(\{u=t\})}{(\varphi’(t))^{p1}} \, dt. \end{equation}
On the other hand, from the remark after formula (31) in [_{4}], we know that
\[\int_0^T \int_{\{u_0=t\}} \vert\nabla u_0\vert^{p1} \, d\sigma(x) \, dt = \int_0^T \frac{\sigma^p(\{u_0=t\})}{(\varphi'(t))^{p1}} \, dt,\]which yields
\begin{equation}\label{eq:2} \int_\Omega \vert\nabla u_0\vert^p \,dx = \int_0^T \frac{\sigma^p(\{u_0=t\})}{(\varphi’(t))^{p1}} \, dt. \end{equation}
In particular, from the isoperimetric inequality $\sigma(\{u=t\}) \geq \sigma(\{u_0=t\})$ we deduce the classial FaberKrahn inequality $\lambda_1(\Omega) \geq \lambda_1(B)$:
\[\lambda_1(\Omega) = \int_\Omega \vert\nabla u\vert^p \, dx \geq \int_0^T \frac{\sigma^p(\{u=t\})}{(\varphi'(t))^{p1}} \, dt \geq \int_0^T \frac{\sigma^p(\{u_0=t\})}{(\varphi'(t))^{p1}} \, dt = \int_\Omega \vert\nabla u_0\vert^p \,dx \geq \lambda_1(B).\]Our aim now is to improve the inequality $\sigma^p(\{u=t\}) \geq \sigma^p(\{u_0=t\})$. Assume, without loss of generality due to rescaling, that $\vert \Omega\vert = 1$.
We use the Bonnesen inequality (see, e.g., [_{6}] or [_{7}]) for a closed Jordan curve, which connects the lenght $L$ of the curve, bounded area $A$, inradius $r_i$ and circumradius $R$ as
\[L^2  4\pi A \geq 4\pi (Rr_i)^2.\]Note that circumradius of $\Omega$ is always greater or equal to $r_0(\Omega)$. Therefore, we have
\[\sigma^2(\{u=t\}) \geq \sigma^2(\{u_0=t\}) + 4 \pi (r_0(\{u>t\})r_i(\{u>t\}))^2.\]Let us chose $s \in (0, T)$ such that
\[\vert \{u > s\} \vert = 1  \frac{1}{2} d_i(\Omega).\]Then, by Lemma 5.2 of [_{1}], we have for any $t \in (0,s]$ that
\[r_0(\{u>t\})r_i(\{u>t\}) \geq \frac{1}{2} (r_0(\Omega)r_i(\Omega)),\]which implies that
\[\sigma^2(\{u=t\}) \geq \sigma^2(\{u_0=t\}) + \pi (r_0(\Omega)r_i(\Omega))^2,\]or, in an equivalent way,
\[\sigma(\{u=t\}) \geq \left(\sigma^2(\{u_0=t\}) + \pi (r_0(\Omega)r_i(\Omega))^2\right)^\frac{1}{2}.\]Since $p \geq 2$, it holds $(A+B)^{p/2} \geq A^{p/2} + B^{p/2}$, and hence we deduce that
\[\sigma^p(\{u=t\}) \geq \sigma^p(\{u_0=t\}) + \pi^{p/2} \left(r_0(\Omega)r_i(\Omega)\right)^p\]for any $t \in (0,s]$.
Substituting this inequality into \eqref{eq:1}, and recalling that $\sigma^p(\{u=t\}) \geq \sigma^p(\{u_0=t\})$ for all $t \in (s,T)$ (for all $t$, in general), we obtain
\begin{equation}\label{eq:3} \lambda_1(\Omega) \geq \int_0^T \frac{\sigma^p(\{u_0=t\})}{(\varphi’(t))^{p1}} \, dt + \pi^{p/2} \left(r_0(\Omega)r_i(\Omega)\right)^p \int_0^s \frac{dt}{(\varphi’(t))^{p1}}. \end{equation}
We want to estimate the last integral. For this end, we note that, by the Holder inequality,
\[s = \int_0^s \, dt = \int_0^s \frac{1}{(\varphi'(t))^\frac{p1}{p}} (\varphi'(t))^\frac{p1}{p} \, dt \leq \left(\int_0^s \frac{dt}{(\varphi'(t))^{p1}}\right)^\frac{1}{p} \left(\int_0^s \varphi'(t) \, dt\right)^\frac{p1}{p}.\]Therefore, noting that
\[\int_0^s \varphi'(t) \, dt = \int_0^s \frac{d}{dt} \vert \{u>t\} \vert \, dt = \vert \{u>s\} \vert + \vert \{u>0\} \vert =  \left(1 \frac{1}{2} d_i(\Omega)\right) + 1 = \frac{1}{2} d_i(\Omega),\]we get
\[\int_0^s \frac{dt}{(\varphi'(t))^{p1}} \geq \frac{s^p}{\left(\frac{1}{2} d_i(\Omega)\right)^{p1}}.\]Putting this inequality into \eqref{eq:3} and recalling \eqref{eq:2}, we obtain
\[\lambda_1(\Omega) \geq \lambda_1(B) + \pi^{p/2} \left(r_0(\Omega)r_i(\Omega)\right)^p \frac{s^p}{\left(\frac{1}{2} d_i(\Omega)\right)^{p1}}.\]Moreover, since $r_0(\Omega)r_i(\Omega) = r_0(\Omega) d_i(\Omega)$, and the assumption $\vert \Omega \vert = 1$ implies $r_0(\Omega) = \pi^{1/2}$, we get
\[\lambda_1(\Omega) \geq \lambda_1(B) + 2^{p1} d_i(\Omega) s^p.\]Now we want to put this inequality into the form we need.
For this end, let us fix some $\alpha \in (0,1)$. (We will clarify it later.) Assume first that $s \geq \alpha d_i(\Omega)$. Then
\[\lambda_1(\Omega) \geq \lambda_1(B) + 2^{p1} \alpha^p d_i(\Omega)^{p+1}.\]Using an upper bound (4) for $\lambda_1(B)$ from [_{8}] and recalling that $r_0(\Omega) = \pi^{1/2}$, we get
\[\lambda_1(B) \leq \frac{(p+1)(p+2)\pi^{p/2}}{2},\]which implies that
\begin{equation}\label{eq:part1} \lambda_1(\Omega) \geq \lambda_1(B) \left(1 + \frac{2^p \alpha^p}{(p+1)(p+2)\pi^{p/2}} d_i(\Omega)^{p+1}\right). \end{equation}
Assume now that $s \leq \alpha d_i(\Omega)$. First, we get
\begin{equation}\label{eq:5} \int_{\{u>s\}} \vert us\vert^p \, dx = \int_{\{u>s\}} \left\vert (us)^2\right\vert^\frac{p}{2} \, dx = \int_{\{u>s\}} \left\vert u^2  2 s u + s^2\right\vert^\frac{p}{2} \, dx = \int_{\{u>s\}} u^p \left\vert 1  \frac{2 s}{u} + \frac{s^2}{u^2}\right\vert^\frac{p}{2} \, dx. \end{equation}
Recalling that $p \geq 2$ and using the Bernoullitype inequality $\vert 1+x \vert^{p/2} \geq 1+\frac{p}{2}x$, $x \in \mathbb{R}$, we obtain
\begin{equation}\label{eq:6} \eqref{eq:5} \geq \int_{\{u>s\}} u^p \left(1  \frac{s p}{u} + \frac{s^2 p}{2u^2}\right) \, dx = \int_{\{u>s\}} u^p \, dx  s p \int_{\{u>s\}} u^{p1} \, dx + \frac{s^2 p}{2} \int_{\{u>s\}} u^{p2} \, dx. \end{equation}
We estimate from below the last integral by zero. Moreover, since $\vert \Omega\vert =1$ and $\int_\Omega u^p \, dx = 1$, we obtain $\int_\Omega u^{p1} \leq 1$. Therefore, we can write
\[\eqref{eq:6} \geq 1  \int_{\{u \leq s\}} u^p \, dx  sp \geq 1  s^p  sp.\]Noting that $s^p < s^2$ for $s \in (0,1)$, we see that the positive root of $1  s^p  sp=0$ can be estimated from below by $s_0 = \frac{\sqrt{p^2+4}p}{2}$. For further simplicity of calculations, it would be convenient to introduce $s_1 = \frac{\sqrt{p^2+1}p}{2}$, and hence $s_1 < s_0$. Since $1  s^p  sp$ is strictly decreasing on $(0, s_0)$, and $d_i(\Omega) < 1$, we get
\[1  s^p  sp \geq 1  s_1^p d_i(\Omega)^p  s_1 d_i(\Omega) p \geq 1  \left(\left(\frac{\sqrt{p^2+1}p}{2}\right)^p + p \frac{\sqrt{p^2+1}p}{2}\right) d_i(\Omega)\]for all $s \in (0,s_1 d_i(\Omega))$. Moreover, it is possible to show (in fact, I didn’t prove it rigorously, but the plot suggests so) that for all $p \geq 2$ it holds
\[\left(\frac{\sqrt{p^2+1}p}{2}\right)^p + p \frac{\sqrt{p^2+1}p}{2} \leq \frac{1}{4}.\]Thus, we finally obtain nice estimate
\[\int_{\{u>s\}} \vert us\vert^p \, dx \geq 1  \frac{1}{4} d_i(\Omega).\]Now, we have
\[\frac{\int_{\{u>s\}} \vert \nabla u\vert^p \, dx}{\int_{\{u>s\}} \vert us\vert^p \, dx} \geq \lambda_1(\{u > s\}) \geq \lambda_1(\{u_0 > s\}) = \frac{\lambda_1(B)}{\vert \{u > s\}\vert^\frac{p}{2}} = \frac{\lambda_1(B)}{\left(1\frac{1}{2}d_i(\Omega)\right)^\frac{p}{2}},\]and hence we conclude that
\[\lambda_1(\Omega) = \int_\Omega \vert \nabla u \vert^p \, dx \geq \int_{\{u>s\}} \vert \nabla u\vert^p \, dx \geq \lambda_1(B) \frac{1\frac{1}{4}d_i(\Omega)}{\left(1\frac{1}{2}d_i(\Omega)\right)^\frac{p}{2}}.\]It can be shown that
\[\frac{1\frac{1}{4}d_i(\Omega)}{\left(1\frac{1}{2}d_i(\Omega)\right)^\frac{p}{2}} \geq 1 + \left(2^\frac{p4}{2}3  1\right) d_i(\Omega)^{p+1}.\]Indeed, if we can try to find a constant $A>0$ such that
\[\frac{1\frac{1}{4}d_i(\Omega)}{\left(1\frac{1}{2}d_i(\Omega)\right)^\frac{p}{2}} \geq 1 + A d_i(\Omega)^{p+1},\]then it is sufficient for $A$ to satisfy
\[A \leq \frac{1\frac{1}{4}d_i(\Omega)}{d_i(\Omega)^{p+1} \left(1\frac{1}{2}d_i(\Omega)\right)^\frac{p}{2}}  \frac{1}{d_i(\Omega)^{p+1}}.\]It should be possible to show that the righthand side here is strictly decreasing with respect to $d_i(\Omega) \in (0,1)$. Hence, calculating the righthand side at $d_i(\Omega)=1$, we get $A \leq 2^\frac{p4}{2} 31$.
Thus, we counclude that
\begin{equation}\label{eq:part2} \lambda_1(\Omega) \geq \lambda_1(B) \left(1 + \left(2^\frac{p4}{2} 3  1\right) d_i(\Omega)^{p+1} \right) \end{equation}
for all $s \leq s_1 d_i(\Omega)$. Hance, we put $\alpha = s_1 = \frac{\sqrt{p^2+1}p}{2}$. Substituting this $\alpha$ into \eqref{eq:part1}, we get
\begin{equation}\label{eq:part12} \lambda_1(\Omega) \geq \lambda_1(B) \left(1 + \frac{\left(\sqrt{p^2+1}p\right)^p}{(p+1)(p+2)\pi^{p/2}} \, d_i(\Omega)^{p+1}\right). \end{equation}
Finally, one can convince himself that the expression in the brackets in \eqref{eq:part2} is strictly smaller that the corresponding one in \eqref{eq:part12}. Therefore, we conclude that \eqref{eq:part12} is valid for all $s \in (0,T)$. The proof is complete.
Remark. It is clear that almost all inequalities used in the proof are not optimal. Therefore, the fraction in \eqref{eq:part12} could be improved (possibly substantially). In particular, in [_{1}] for the case $p=2$ the coefficient for $d_i(\Omega)^3$ was estimated as $\frac{1}{250}$. If we substitute $p=2$ into \eqref{eq:part12}, then our coefficient will be worse.
Last modified: 29Jan18
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