On principal frequencies, volume and inradius in convex sets

Abstract

We provide a sharp double-sided estimate for Poincaré–Sobolev constants on a convex set, in terms of its inradius and $N-$dimensional measure. Our results extend and unify previous works by Hersch and Protter (for the first eigenvalue) and of Makai, Pólya and Szegő (for the torsional rigidity), by means of a single proof.

Appendix A. Some technical results

The following simple one-dimensional result was an essential ingredient for the proof of the lower bound (1.6).

Lemma A.1

Let $a>0$ and let $\xi :[0,a]\rightarrow {\mathbb {R}}$ be an absolutely continuous function such that

$$\begin{aligned} \xi (0)=0\quad \text{ and } \quad t\mapsto \frac{\xi (t)}{t} \text{ is } \text{ non-decreasing }. \end{aligned}$$

Let $\psi :[0,a]\rightarrow [0,+\infty )$ be a non-increasing function. Then we have

$$\begin{aligned} \int \nolimits _0^a \xi '(t)\,\psi (t)\,dt\le \frac{\xi (a)}{a}\,\int \nolimits _0^a \psi (t)\,dt. \end{aligned}$$

Proof

Without loss of generality we can suppose that $\psi$ is smooth. By integrating by parts and observing that $\xi (0)=0$, we have

$$\begin{aligned} \begin{aligned} \int \nolimits _0^a \xi '(t)\,\psi (t)\,dt&=\xi (a)\,\psi (a)+\int \nolimits _0^{a} \xi (t)\,(-\psi '(t))\,dt\\&=\xi (a)\,\psi (a)+\int \nolimits _0^a \frac{\xi (t)}{t}\,t\,(-\psi '(t))\,dt\\&\le \xi (a)\,\psi (a)+\int \nolimits _0^a \frac{\xi (a)}{a}\,t\,(-\psi '(t))\,dt, \end{aligned} \end{aligned}$$

where we also used the monotonicity of both $\xi (t)/t$ and $\psi (t)$. We now further integrate by parts the last integral, so to get

$$\begin{aligned} \int \nolimits _0^a \xi '(t)\,\psi (t)\,dt\le \xi (a)\,\psi (a)-\frac{\xi (a)}{a}\,a\,\psi (a)+\frac{\xi (a)}{a}\,\int \nolimits _0^a \psi (t)\,dt. \end{aligned}$$

This concludes the proof. $\square$

Lemma A.2

Let $N\ge 2$ and let $1\le q<2^*$, for every $L>0$ we set

$$\begin{aligned} \Omega _L=\left( -\frac{L}{2},\frac{L}{2}\right) ^{N-1}\times (0,1). \end{aligned}$$

Then we have:

• 1. for $1\le q\le 2$

  $$\begin{aligned} \lim _{L\rightarrow +\infty } L^{(N-1)\,\frac{2-q}{q}}\,\lambda _{2,q}(\Omega _L)=\Big (\pi _{2,q}\Big )^2,\quad \text{ as } L\rightarrow +\infty . \end{aligned}$$

• 2. for $2<q<2^*$

  $$\begin{aligned} \lim _{L\rightarrow +\infty } \lambda _{2,q}(\Omega _L)=\lambda _{2,q}({\mathbb {R}}^{N-1}\times (0,1))>0. \end{aligned}$$

Proof

We distinguish again the two cases.

Case$1\le q\le 2$. For $q=2$ this is contained for example in [5, Lemma A.2], we thus focus on the case $q<2$. By [5, equation (3.6)], we have

$$\begin{aligned} \lim _{L\rightarrow +\infty }\lambda _{2,q}(\Omega _L)\,\left( \frac{|\Omega _L|^{\frac{1}{2}+\frac{1}{q}}}{P(\Omega _L)}\right) ^2=\left( \frac{\pi _{2,q}}{2}\right) ^2, \end{aligned}$$

where $P(\Omega _L)$ stands for the perimeter of $\Omega _L$. If we now use that

$$\begin{aligned} |\Omega _L|= L^{N-1}\quad \text{ and } \quad P(\Omega _L)\sim 2\,L^{N-1},\quad \text{ as } L\rightarrow +\infty , \end{aligned}$$

we get the desired result.

Case$2<q<2^*$. By monotonicity of $\lambda _{2,q}$ with respect to set inclusion, we have that

$$\begin{aligned} \lambda _{2,q}(\Omega _L)\ge \lambda _{2,q}({\mathbb {R}}^{N-1}\times (0,1))\quad \text{ and }\quad L\mapsto \lambda _{2,q}(\Omega _L) \text{ is } \text{ monotone } \text{ decreasing }. \end{aligned}$$

Thus we get that

$$\begin{aligned} \lim _{L\rightarrow +\infty }\lambda _{2,q}(\Omega _L)\ge \lambda _{2,q}({\mathbb {R}}^{N-1}\times (0,1)). \end{aligned}$$

On the other hand, for every $\varepsilon >0$ we can take $\varphi _\varepsilon \in C^\infty _0({\mathbb {R}}^{N-1}\times (0,1))$ such that

$$\begin{aligned} \lambda _{2,q}({\mathbb {R}}^{N-1}\times (0,1))+\varepsilon \ge \frac{\displaystyle \int \nolimits _{{\mathbb {R}}^{N-1}\times (0,1)} |\nabla \varphi _\varepsilon |^2\,dx}{\displaystyle \left( \int \nolimits _{{\mathbb {R}}^{N-1}\times (0,1)} |\varphi _\varepsilon |^q\,dx\right) ^\frac{2}{q}}. \end{aligned}$$

Since $\varphi _\varepsilon$ has compact support, for L large enough we get that $\varphi _\varepsilon \in C^\infty _0(\Omega _L)$, as well. This shows that for every $\varepsilon >0$

$$\begin{aligned} \lambda _{2,q}({\mathbb {R}}^{N-1}\times (0,1))+\varepsilon \ge \lim _{L\rightarrow +\infty }\lambda _{2,q}(\Omega _L), \end{aligned}$$

and thus the claimed convergence of $\lambda _{2,q}(\Omega _L)$ follows.

We are only left with showing that ${\mathbb {R}}^{N-1}\times (0,1)$ has a non-trivial Poincaré–Sobolev constant $\lambda _{2,q}$. By recalling that for a open convex set $\Omega \subset {\mathbb {R}}^N$ we have (see [6, Proposition 6.3])

$$\begin{aligned} \lambda _{2,q}(\Omega )\ge \frac{C_{N,q}}{R_\Omega ^{2+\frac{2-q}{q}\,N}}, \end{aligned}$$

we get the desired assertion. $\square$