Sharp and quantitative estimates for the p-Torsion of convex sets

Abstract

Let $\Omega \subset \mathbb {R}^n$, $n\ge 2$, be a non-empty, bounded, open and convex set and let f be a positive and non-increasing function depending only on the distance from the boundary of $\Omega$. We consider the p-torsional rigidity associated to $\Omega$ for the Poisson problem with Dirichlet boundary conditions, denoted by $T_{f,p}(\Omega )$. Firstly, we prove a Pólya type lower bound for $T_{f,p}(\Omega )$ in any dimension; then, we consider the planar case and we provide two quantitative estimates in the case $f\equiv 1$.

1 Introduction

Let $\Omega \subset \mathbb {R}^n$, $n\ge 2$, be a non-empty, bounded, open and convex set and let $p\in (1,+\infty )$. We consider the Poisson equation for the $p-$Laplace operator, defined as

$$\begin{aligned} -\Delta _p u:=-\textrm{div}\left( |\nabla u|^{p-2}\nabla u\right) , \end{aligned}$$

with Dirichlet boundary condition:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _p u(x) = f(d(x,\partial \Omega )) &{} \text { in } \Omega \\ u=0 &{} \text { on } \partial \Omega , \end{array}\right. } \end{aligned}$$

(1)

where $f: [0 , R_\Omega ] \rightarrow [0, +\infty [$ is a continuous, non-increasing and not identically zero function, $d(\cdot , \partial \Omega ):\Omega \rightarrow [0,+\infty [$ is the distance function from the boundary defined as

$$\begin{aligned} d(x,\partial \Omega ):=\inf _{y\in \partial \Omega }{\left| x-y\right| } \end{aligned}$$

(2)

and $R_\Omega$ is the inradius of $\Omega$, i.e.

$$\begin{aligned} R_\Omega =\sup _{x\in \Omega } d(x,\partial \Omega ). \end{aligned}$$

(3)

This class of functions, depending only on the distance, are the so called web functions, see as a reference [7]. A function $u\in W^{1,p}_0(\Omega )$ is a weak solution to (1) if and only if

$$\begin{aligned} \int _\Omega |\nabla u(x)|^{p-2} \nabla u(x) \nabla \varphi (x) \, dx=\int _\Omega f(d(x,\partial \Omega ))\varphi (x ) \, dx \quad \forall \varphi \in W^{1,p}_0(\Omega ). \end{aligned}$$

The (f, p)-torsional rigidity of $\Omega$, that we denote by $T_{f,p}(\Omega )$, is defined as

$$\begin{aligned} T_{f,p}(\Omega ) = \max _{\begin{array}{c} \varphi \in W^{1,p}_0(\Omega )\\ \varphi \not \equiv 0 \end{array} } \frac{\displaystyle {\left( \int _\Omega f(d(x,\partial \Omega )){\left| \varphi (x)\right| } \, dx\right) ^{\frac{p}{p-1}}}}{\left( \displaystyle {\int _\Omega {\left| \nabla \varphi (x)\right| }^p \, dx}\right) ^{\frac{1}{p-1}}} \end{aligned}$$

(4)

and, if $u_p\in W^{1,p}_0(\Omega )$ is the unique solution to (1), we have

$$\begin{aligned} T_{f,p}(\Omega )= \int _\Omega f u_p \, dx. \end{aligned}$$

For the sake of simplicity, when $f\equiv 1$ in $\Omega$, we set $T_p(\Omega ):=T_{1,p}(\Omega )$ and, if we are also in the case $p=2$, we set $T(\Omega ):=T_{1,2}(\Omega )$. We recall that the quantities $T(\Omega )$ and $T_p(\Omega )$ are usually called, respectively, torsional rigidity and $p-$torsional rigidity and so, by analogy, we have chosen the above terminology for $T_{f,p}(\Omega )$.

In what follows, we denote by $|\Omega |$ and $P(\Omega )$ respectively the Lebesgue measure and the perimeter of $\Omega$ in the sense of De Giorgi. In [17] the author gives some estimates on the torsional rigidity $T(\Omega )$. In particular, he proves that, among all bounded, open and convex planar sets, the following inequality holds

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{{\left| \Omega \right| }^3}\ge \frac{1}{3}, \end{aligned}$$

(5)

and equality is asymptotically achieved by a sequence of thinning rectangles. Moreover, Makai in [15] proves that among all bounded, open and convex planar sets, the following upper bound holds

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{{\left| \Omega \right| }^3}\le \frac{2}{3}, \end{aligned}$$

(6)

which is sharp on a sequence of thinning triangles (for the exact definition of thinning domains see Definition 11). Estimates (5) and (6) are generalized to the $p-$Laplacian in [9]. More precisely, the authors prove that, in the class of bounded, open and convex planar sets,

$$\begin{aligned} \frac{1}{q+1}< \frac{T_p(\Omega ) P^q(\Omega )}{{\left| \Omega \right| }^{q+1}}< \frac{2^{q+1}}{(q+2)(q+1)}, \qquad q=\frac{p}{p-1}, \end{aligned}$$

(7)

where the lower and the upper bounds hold asymptotically on a sequence of thinning rectangles and on a sequence of thinning isosceles triangles, respectively. In [8] the authors generalize the lower bound (7) in every dimensions, proving that for open, bounded and convex sets $\Omega \subseteq \mathbb {R}^n$, it holds

$$\begin{aligned} \frac{T_p(\Omega ) P^q(\Omega )}{{\left| \Omega \right| }^{q+1}}> \frac{1}{q+1}, \end{aligned}$$

(8)

and they extend this result also to the anisotropic case.

We also recall that in [6] the authors consider the functional

$$\begin{aligned} H_k(\Omega )=\dfrac{P(\Omega ) T^{k}(\Omega )}{|\Omega |^{\alpha _k}}, \qquad \alpha _k=1+k+\frac{2k-1}{n}, \end{aligned}$$

and prove that, among bounded, open and convex sets in $\mathbb {R}^n$, this functional is bounded if and only if $k=1/2$. More precisely, they prove the following:

$$\begin{aligned} \frac{1}{\sqrt{3}}\le H_{\frac{1}{2}}(\Omega ) \le \frac{2^n n^{3n/2}}{\omega _n }\left( \frac{n}{n+2}\right) ^{\frac{1}{2}}, \end{aligned}$$

(9)

where $\omega _n$ is the Lebesgue measure of the unit ball. We note that, in the planar case, the lower bound in (9) coincides with the one given in (5), while the upper bound is strictly larger than the one given in (6). It is conjectured that, in the higher dimensional case, the upper bound is

$$\begin{aligned} H_{\frac{1}{2}}(\Omega )\le n \left( \frac{2}{(n+1)(n+2)}\right) ^{\frac{1}{2}}. \end{aligned}$$

Moreover, we observe that the lower bound in (9) is asymptotically achieved by a sequence of thinning cylinders. More precisely, denoting by $w_\Omega$ the minimal width and by ${{\,\mathrm{{diam}}\,}}(\Omega )$ the diameter of the set (see Section 2 for the exact definitions and Fig. 1), we give the following

Definition 1.1

Let $\Omega _l$ be a sequence of non-empty, bounded, open and convex sets of $\mathbb {R}^n$. We say that $\Omega _l$ is a sequence of thinning domains if

$$\begin{aligned} \dfrac{w_{\Omega _l}}{{{\,\mathrm{{diam}}\,}}(\Omega _l)}\xrightarrow {l \rightarrow 0}0, \end{aligned}$$

(10)

Fig. 1

Minimal width and diameter of a convex set.

(see Fig. 2). In particular, if $l>0$ and C is a bounded, open and convex set of ${{\,\mathrm{\mathbb {R}}\,}}^{n-1}$ with unitary $(n-1)$-dimensional measure, then, if $l \rightarrow 0$, the sequence

$$\begin{aligned} \Omega _l = l^{-\frac{1}{n-1}}C \times \left[ -\frac{l}{2}, \frac{l}{2}\right] \end{aligned}$$

(11)

is called a sequence of thinning cylinders. Moreover, in the case $n=2$, the sequence (11) is called sequence of thinning rectangles.

Fig. 2

Thinning cylinders.

The first result that we prove is a lower bound for the (f, p)-torsional rigidity, which generalizes the lower bound in (7). The hypothesis that f is a web function allows us to use the method of proof contained in [17].

Theorem 1.1

Let $\Omega$ be a non-empty, bounded, open and convex set of ${{\,\mathrm{\mathbb {R}}\,}}^n$, $n\ge 2$, and let $f: [0 , R_\Omega ] \rightarrow [0, +\infty [$ be a continuous and non-increasing function such that $f\not \equiv 0$. Then,

$$\begin{aligned} T_{f,p}(\Omega )\ge c_p\frac{\mu ^{q+1}_f(\Omega )}{f(0)P^{q}(\Omega )}, \end{aligned}$$

(12)

where

$$\begin{aligned} c_p =\frac{p-1}{2p-1}, \qquad q=\frac{p}{p-1}, \end{aligned}$$

and

$$\begin{aligned} \mu _f(\Omega )= \int _\Omega f(d(x,\partial \Omega )) \, dx. \end{aligned}$$

Moreover, the equality sign is asymptotically achieved by a sequence of thinning cylinders.

We stress that both the estimate and the constant in Theorem 1.1 are independent of n.

In the second part of the present paper, we focus our study on the case $f\equiv 1$ and we obtain some quantitative estimates. We define the following functional

$$\begin{aligned} \mathcal {F}_p(\Omega )=\frac{T_p(\Omega )P^{q}(\Omega )}{|\Omega |^{q+1}}, \qquad q=\frac{p}{p-1}, \end{aligned}$$

(13)

which is scaling invariant, since for every $t>0$

$$\begin{aligned} {\left| t\Omega \right| }= t^n {\left| \Omega \right| }, \qquad P(t\Omega )= t^{n-1}P(\Omega ) \end{aligned}$$

and

$$\begin{aligned} T_{p}(t\Omega )= t^{n+q}T_{p}(\Omega ). \end{aligned}$$

We can rewrite inequality (12), in the case $f\equiv 1$, as follows

$$\begin{aligned} \mathcal {F}_p(\Omega ) \ge c_p. \end{aligned}$$

From Theorem 1.1, it follows that along a sequence of thinning cylinders $\{\Omega _l\}_{l\in \mathbb {N}}$ defined in (11), we have

$$\begin{aligned} \mathcal {F}_p(\Omega _l) \xrightarrow {l \rightarrow 0} c_p . \end{aligned}$$

This leads to the following stability issue: if $\mathcal {F}_p(\Omega )$ is close to $c_p$, can we say that $\Omega$ is close in some sense to a cylinder? The following result gives us information on the nature of the geometry of $\Omega$: when $\mathcal {F}_p(\Omega )-c_p$ is sufficiently small, the set $\Omega$ is a thin domain, in the sense that the ratio $w_\Omega /\textrm{diam}(\Omega )$ is small.

The main novelty of the present paper consists indeed in the following quantitative results of the Pólya estimates (5) and the Pólya type lower bound in (7) by means of suitable deficits. For completeness, we recall some standard references about isoperimetric quantitative results, see for example [4, 5, 10,11,12, 16]. The main difference between these results and ours is that the equality in Pólya’s estimates is achieved asymptotically for a sequence of thinning cylinders. Hence, the proof of quantitative result must take into account that we do not have a minimum, as in the classical isoperimetric stability results.

Theorem 1.2

Let $\Omega$ be a non-empty, bounded, open and convex set of ${{\,\mathrm{\mathbb {R}}\,}}^n$ and let $f\equiv 1$. Then,

$$\begin{aligned} \mathcal {F}_p(\Omega ) -c_p \ge K(n,p) \bigg (\frac{w_{\Omega }}{{{\,\mathrm{{diam}}\,}}(\Omega )}\bigg )^{n-1}, \end{aligned}$$

(14)

where K(n, p) is a positive constant depending only on p and the dimension of the space n. In particular, in the case $n=2$, the exponent of the quantity $\displaystyle \frac{w_{\Omega }}{{{\,\mathrm{{diam}}\,}}(\Omega )}$ is sharp.

We prove a second quantitative result in the case $p=2$ and $n=2$.

Theorem 1.3

Let $\Omega$ be a non-empty, bounded, open and convex set in ${{\,\mathrm{\mathbb {R}}\,}}^2$, let $f\equiv 1$ and let $p=2$. Then, there exists a positive constant $\tilde{K}$ such that

$$\begin{aligned} \mathcal {F}_2(\Omega ) -c_2= \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \tilde{K} \left( \frac{|\Omega \bigtriangleup Q|}{|\Omega |}\right) ^3, \end{aligned}$$

(15)

where $\Omega \bigtriangleup Q$ denotes the symmetric difference between $\Omega$ and a rectangle Q with sides $P(\Omega )/ 2$ and $w_{\Omega }$ containing $\Omega$.

We prove the first quantitative result (14) starting from the proof of Theorem 1.1, estimating the remainder term with geometric quantities. Finally, in order to prove Theorem 1.3, we use Steiner formulas (see Section 2.2 and the references therein).

The paper is organized as follows. In Section 2 we recall some basic notions, definitions and we recall some classical results, focusing in particular on the class of convex sets. In Section 3 we prove Theorem 1.1 and, finally, Section 4 is dedicated to the proof of the quantitative results when $f\equiv 1$.

2 Notations and Preliminaries

2.1 Notations and basic facts

Throughout this article, $|\cdot |$ will denote the Euclidean norm in $\mathbb {R}^n$, while $\cdot$ is the standard Euclidean scalar product for $n\ge 2$. By $\mathcal {H}^k(\cdot )$, for $k\in [0,n)$, we denote the $k-$dimensional Hausdorff measure in $\mathbb {R}^n$.

The perimeter of $\Omega$ in $\mathbb {R}^n$ will be denoted by $P(\Omega )$ and, if $P(\Omega )<\infty$, we say that $\Omega$ is a set of finite perimeter. In our case, $\Omega$ is a bounded, open and convex set; this ensures us that $\Omega$ is a set of finite perimeter and that $P(\Omega )=\mathcal {H}^{n-1}(\partial \Omega )$. Moreover, if $\Omega$ is an open set with Lipschitz boundary, it holds

Theorem 2.1

(Coarea formula) Let $\Omega \subset \mathbb {R}^n$ be an open set. Let $f\in W^{1,1}_{\text {loc}}(\Omega )$ and let $u:\Omega \rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ be a measurable function. Then,

$$\begin{aligned} {\displaystyle \int _{\Omega }u(x)|\nabla f(x)|dx=\int _{\mathbb {R} }dt\int _{\Omega \cap f^{-1}(t)}u(y)\, d\mathcal {H}^{n-1}(y)}. \end{aligned}$$

(16)

Some references for results relative to the sets of finite perimeter and for the coarea formula are, for instance, [1, 14].

We give now the definition of the support function of a convex set and minimal width (or thickness) of a convex set.

Definition 2.1

Let $\Omega$ be a bounded, open and convex set of $\mathbb {R}^n$. The support function of $\Omega$ is defined as

$$\begin{aligned} h_\Omega (y)=\sup _{x\in \Omega }\left( x\cdot y\right) , \qquad y\in \mathbb {R}^n . \end{aligned}$$

Definition 2.2

Let $\Omega$ a bounded, open and convex set of $\mathbb {R}^n$, the width of $\Omega$ in the direction $y \in \mathbb {R}$ is defined as

$$\begin{aligned} \omega _{\Omega }(y)=h_{\Omega }(y)+h_{\Omega }(-y) \end{aligned}$$

and the minimal width of $\Omega$ as

$$\begin{aligned} w_\Omega =\min \{ \omega _{\Omega }(y)\,|\; y\in \mathbb {S}^{n-1}\}. \end{aligned}$$

We recall the following estimate, which is proved in [2] in the planar case and is generalized in [3] to all dimensions.

Proposition 2.2

Let $\Omega$ be a non-empty bounded, open and convex set of $\mathbb {R}^n$. Then,

$$\begin{aligned} \dfrac{1}{n} \le \dfrac{|\Omega |}{P(\Omega ) R_\Omega } < 1. \end{aligned}$$

(17)

The upper bound is sharp on a sequence of thinning cylinders, while the lower bound is sharp, for example, on balls. Moreover, for $n=2$, any circumscribed polygon, that is a polygon whose incircle touches all the sides, verifies the lower bound with the equality sign.

In the planar case the following inequalities hold true (see as a reference [18, 20, 21]).

Proposition 2.3

Let $\Omega$ be a bounded, open and convex set of $\mathbb {R}^2$. Then,

$$\begin{aligned} 2\le \frac{w_\Omega }{R_{\Omega }} \le 3. \end{aligned}$$

(18)

The upper bound is achieved by equilateral triangles and the lower bound is achieved by disks.

Moreover,

$$\begin{aligned} \left( w_{\Omega }-2R_{\Omega } \right) P(\Omega )\le \dfrac{2}{\sqrt{3}}w^2_{\Omega }, \end{aligned}$$

(19)

with equality holding for equilateral triangles, and

$$\begin{aligned} |\Omega |\le R_{\Omega }\left( P(\Omega )-\pi R_{\Omega } \right) , \end{aligned}$$

(20)

with equality holding for the stadii (convex hull of two identical disjoint balls).

Eventually,

$$\begin{aligned} 2{{\,\mathrm{{diam}}\,}}(\Omega )<P(\Omega )\le \pi {{\,\mathrm{{diam}}\,}}(\Omega ), \end{aligned}$$

(21)

where the lower bound is asymptotically achieved by a sequence of thinning rectangles and the upper bound by sets of constant width.

2.2 Inner parallel sets

Let $\Omega$ be a non-empty, bounded, open and convex set of ${{\,\mathrm{\mathbb {R}}\,}}^n$. We defined the distance function from the boundary in (2) and we will denote it by $d(\cdot )$. We remark that the distance function is concave, as a consequence of the convexity of $\Omega$.

The superlevel sets of the distance function

$$\begin{aligned} \Omega _t=\{x\in \Omega \, : \, d(x)>t \}, \qquad t\in [0, R_\Omega ] \end{aligned}$$

(22)

are called inner parallel sets, where $R_\Omega$ is the inradius of $\Omega$, and we use the following notations:

$$\begin{aligned} \mu (t)= {\left| \Omega _t\right| }, \qquad P(t)=P(\Omega _t)\qquad t\in [0, R_\Omega ]. \end{aligned}$$

(23)

By coarea formula (16), recalling that ${\left| \nabla d\right| }=1$ almost everywhere, we have

$$\begin{aligned} \mu (t)= & {} \int _{\{d>t\}} \, dx = \int _{\{d>t\}} \frac{{\left| \nabla d\right| }}{{\left| \nabla d\right| }} \, dx= \int _{t}^{R_\Omega } \frac{1}{{\left| \nabla d\right| }} \int _{\{d=s\}} d\mathcal {H}^{n-1} \, ds \\= & {} \int _{t}^{R_\Omega } P(s)\; ds; \end{aligned}$$

hence, the function $\mu (t)$ is absolutely continuous, decreasing and its derivative is $\mu '(t)=-P(t)$ almost everywhere. Moreover, it is possible to prove that the perimeter P(t) is non increasing and absolutely continuous, as a consequence of the concavity of the distance function and the Brunn-Minkowski inequality for the perimeter (see [19] as a reference).

Finally, let us consider the case $n=2$. For $\Omega$ non-empty bounded, open and convex set of $\mathbb {R}^2$, the Steiner formulas for the inner parallel sets hold (see [22]):

$$\begin{aligned} P(t)\le P(\Omega )-2\pi t \qquad \forall t \in [0,R_\Omega ], \end{aligned}$$

(24)

$$\begin{aligned} \mu (t)\ge |\Omega |-P(\Omega ) t+\pi t^2 \qquad \forall t \in [0,R_\Omega ], \end{aligned}$$

(25)

equality holding in both (24) and (25) for the stadii (see [9]).

As a consequence of the Alexandrov-Fenchel inequality and the isoperimetric inequality for the quermassintegrals (see [19]), we have

$$\begin{aligned} -P'(t) \ge n(n-1)\omega _n^{\frac{1}{n-1}}\bigg (\frac{P(t)}{n}\bigg )^{\frac{n-2}{n-1}}, \end{aligned}$$

(26)

that, for $n=2$, reads

$$\begin{aligned} -P'(t) \ge 2\pi , \end{aligned}$$

(27)

with equality if $\Omega$ is a ball or a stadium.

3 Proof of Theorem 1.1

In this Section we prove Theorem 1.1. Since the proof is quite long, we split it in two parts: firstly we prove inequality (12) and, then, we prove its sharpness.

3.1 Step 1: proof of inequality (12) in Theorem 1.1

Proof

Let us choose in the variational characterization (4) $\varphi (x)= g(d(x))$ as a test function, where g is a positive and non-decreasing function in $W^{1,p}([0,R_\Omega ])$ such that $g(0)=0$. Then, by coarea formula (16),

$$\begin{aligned} \int _\Omega f(d(x,\partial \Omega ))\varphi (x) \, dx = \int _0^{R_\Omega }f(t)g(t) P(t) \, dt \end{aligned}$$

(28)

and

$$\begin{aligned} \int _\Omega {\left| \nabla \varphi (x)\right| }^p \, dx = \int _0^{R_\Omega }g'^p(t) P(t) \, dt. \end{aligned}$$

(29)

By (4), (28) and (29) we have

$$\begin{aligned} T_{f,p}(\Omega ) \ge \frac{\displaystyle {\left( \int _0^{R_\Omega }f(t)g(t) P(t) \, dt\right) ^\frac{p}{p-1}}}{\left( \displaystyle {\int _0^{R_\Omega }g'^p(t) P(t) \, dt}\right) ^{\frac{1}{p-1}}}. \end{aligned}$$

(30)

Now, if we define the following measure

$$\begin{aligned} \mu _f(E) = \int _E f(d(x)) \, dx, \end{aligned}$$

we have

$$\begin{aligned} \mu _f(t):=\mu _f(\Omega _t) = \int _{\Omega _t} f(d(x)) \, dx= \int _t^{R_\Omega } f(s) P(s)\, ds. \end{aligned}$$

(31)

Since f(s)P(s) is a decreasing function, we get

$$\begin{aligned} \mu _f(t) \le (R_\Omega - t) f(t) P(t). \end{aligned}$$

(32)

From (31), we have

$$\begin{aligned} -\mu '_f(t)= f(t) P(t) \qquad \text { a.e. } t \in [0, R_\Omega ]. \end{aligned}$$

(33)

Using (28), (33) and integrating by parts, we obtain

$$\begin{aligned} \int _0^{R_\Omega }f(t)g(t) P(t) \, dt=-\int _0^{R_\Omega }g(t)\mu '_f(t) \, dt=\int _0^{R_\Omega }g'(t)\mu _f(t)\, dt. \end{aligned}$$

Consequently, (30) becomes

$$\begin{aligned} T_{f,p}(\Omega ) \ge \frac{\displaystyle {\left( \int _0^{R_\Omega }g'(t)\mu _f(t)\, dt\right) ^{\frac{p}{p-1}}}}{\left( \displaystyle {\int _0^{R_\Omega }g'^p(t) P(t) \, dt}\right) ^{\frac{1}{p-1}}}. \end{aligned}$$

We can choose

$$\begin{aligned} g(t)=\int _0^t\left( \dfrac{\mu _f(s)}{P(s)}\right) ^{1/(p-1)}ds \end{aligned}$$

and we observe that $g\in W^{1,p}([0, R_{\Omega }])$, since, using (32), we have

$$\begin{aligned} g(t) \le \int _0^{R_\Omega } (R_\Omega -s)^{\frac{1}{p-1}} f(s)^{\frac{1}{p-1}}\,ds \le {\left\| f\right\| }_{L^\infty }^{\frac{1}{p-1}} R_\Omega ^{\frac{p}{p-1}}\in L^p([0, R_{\Omega }]), \\ g'(t) \le {\left\| f\right\| }_{L^\infty }^{\frac{1}{p-1}} R_\Omega ^{\frac{1}{p-1}} \in L^p([0, R_{\Omega }]). \end{aligned}$$

So, we have

$$\begin{aligned} T_{f,p}(\Omega ) \ge \int _0^{R_\Omega }\frac{\mu _f^{\frac{p}{p-1}}(t)}{P^{\frac{1}{p-1}}(t)}\, dt =-\frac{p-1}{2p-1}\int _0^{R_\Omega }\frac{(\mu ^{\frac{2p-1}{p-1}}_f(t))'}{f(t)P^{\frac{p}{p-1}}(t)}\, dt. \end{aligned}$$

(34)

Let us set $c_p =(p-1)/(2p-1)$. Since f(s) is a non-negative and non-increasing function, integrating by parts in (34), we get

$$\begin{aligned} \begin{aligned} T_{f,p}(\Omega )&\ge -c_p\int _0^{R_\Omega }\frac{(\mu ^{\frac{2p-1}{p-1}}_f(t))'}{f(t)P^{\frac{p}{p-1}}(t)}\, dt =-c_p\frac{\mu ^{\frac{2p-1}{p-1}}_f(t)}{f(t)P^{\frac{p}{p-1}}(t)} \Bigg |_0^{R_\Omega } + \\&\quad -c_p\int _0^{R_\Omega }\frac{\mu ^{\frac{2p-1}{p-1}}_f(t)}{f^2(t)P^{\frac{2p}{p-1}}(t)} \left( f'(t)P^{\frac{p}{p-1}}(t)+ \frac{p}{p-1} f(t)P^{\frac{1}{p-1}}(t)P'(t)\right) \, dt\\&\ge c_p\frac{\mu ^{\frac{2p-1}{p-1}}_f(\Omega )}{f(0)P^{\frac{p}{p-1}}(\Omega )} + \frac{c_p}{ P^{\frac{p}{p-1}}(\Omega )} \int _0^{R_\Omega }\frac{\mu ^{\frac{2p-1}{p-1}}_f(t)}{f^2(t)} (-f'(t))\, dt, \end{aligned} \end{aligned}$$

(35)

where in the last inequality we use (32) and the fact that $P'(t)\le 0$. Now, since f(s) is non-increasing, we obtain the desired estimate

$$\begin{aligned} T_{f,p}(\Omega )\ge c_p\frac{\mu ^{\frac{2p-1}{p-1}}_f(\Omega )}{f(0)P^{\frac{p}{p-1}}(\Omega )}. \end{aligned}$$

(36)

$\square$

3.2 Step 2: proof of the sharpness of (12)

Proof

We prove that inequality (12) is sharp and that the optimum is asymptotically achieved by the sequence of thinning cylinders $\Omega _l$ with unitary measure, as defined in (11), that is

$$\begin{aligned} \Omega _l = l^{-\frac{1}{n-1}}C \times \left( -\frac{l}{2}, \frac{l}{2}\right) \end{aligned}$$

where $C\subseteq {{\,\mathrm{\mathbb {R}}\,}}^{n-1}$ is a bounded, open and convex set with unitary $(n-1)-$measure. It is easy to verify that, for $n\ge 3$,

$$\begin{aligned} \begin{aligned} P(\Omega _l)&= 2 \mathcal {H}^{n-1} (l^{-\frac{1}{n-1}} C) + l \mathcal {H}^{n-2} (\partial ( l^{-\frac{1}{n-1}}C) )\\&=2l^{-1}+ l^{\frac{1}{n-1}} \mathcal {H}^{n-2}(\partial C), \end{aligned} \end{aligned}$$

(37)

and we observe that, in the case $n=2$, we have that $\mathcal {H}^{n-2}(\partial C)=2$.

Let u be the solution to the following p-torsion problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _p u = 1 &{}\text {in } \Omega _l\\ u=0 &{}\text {on } \partial \Omega _l, \end{array}\right. } \end{aligned}$$

such that

$$\begin{aligned} \int _{\Omega _l} u\;dx=T_p(\Omega _l), \end{aligned}$$

and let us consider the following function, depending only on the last component $x_n$ of $x\in {{\,\mathrm{\mathbb {R}}\,}}^n$,

$$\begin{aligned} v(x) = \frac{p-1}{p} \left[ \left( \frac{l}{2}\right) ^{\frac{p}{p-1}} - {\left| x_n\right| }^{\frac{p}{p-1}}\right] , \end{aligned}$$

satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _p v = 1 &{}\text {in } \Omega _l\\ v\ge 0 &{}\text {on } \partial \Omega _l. \end{array}\right. } \end{aligned}$$

The comparison principle, see [13], ensures that $u\le v$ in $\Omega _l$ and, as a consequence,

$$\begin{aligned} \begin{aligned} T_{p}(\Omega _l)&= \int _{\Omega _l} u\,dx \le \int _{\Omega _l}v\,dx \\ {}&=\frac{p-1}{p} \int _{l^{-\frac{1}{n-1} }C} \int _{-\frac{l}{2}}^{\frac{l}{2}}\left[ \left( \frac{l}{2}\right) ^{\frac{p}{p-1}} - {\left| x_n\right| }^{\frac{p}{p-1}}\right] \,dx_n\,d\mathcal {H}^{n-1} \\&=2\; \frac{p-1}{p} l^{-1} \int _0^{\frac{l}{2}} \left[ \left( \frac{l}{2}\right) ^{\frac{p}{p-1}} - x_n^{\frac{p}{p-1}}\right] \,dx_n\\&=2\; \frac{p-1}{p} \left[ 1 - \frac{p-1}{2p-1}\right] l^{-1} \left( \frac{l}{2}\right) ^{\frac{2p-1}{p-1}} = 2c_p l^{-1} \left( \frac{l}{2}\right) ^{\frac{2p-1}{p-1}}. \end{aligned} \end{aligned}$$

(38)

By (38) and (37), we have

$$\begin{aligned}{} & {} T_p(\Omega _l) P^{\frac{p}{p-1}}(\Omega _l) \le 2c_p l^{-1} \left( \frac{l}{2}\right) ^{\frac{2p-1}{p-1}} \left( 2l^{-1}+ l^{\frac{1}{n-1}} \mathcal {H}^{n-2}(\partial C)\right) ^{\frac{p}{p-1}} \\{} & {} \quad = c_p \left( 1+ \frac{l^{\frac{n}{n-1}}}{2}\mathcal {H}^{n-2}(\partial C)\right) ^{\frac{p}{p-1}}. \end{aligned}$$

Now, since $f(x)\le f(0)$, we have that, for every bounded, open and convex set $\Omega$,

$$\begin{aligned} T_{f,p}(\Omega )\le f^{\frac{p}{p-1}}(0) T_p(\Omega ). \end{aligned}$$

(39)

It follows that

$$\begin{aligned} \begin{aligned} T_{f,p}(\Omega _l)P^{\frac{p}{p-1}}(\Omega _l)&\le f^{\frac{p}{p-1}}(0) T_p(\Omega _l) P^{\frac{p}{p-1}}(\Omega _l)\\&\le c_p f^{\frac{p}{p-1}}(0) \left( 1+ \frac{l^{\frac{n}{n-1}}}{2}\mathcal {H}^{n-2}(\partial C)\right) ^{\frac{p}{p-1}}. \end{aligned} \end{aligned}$$

(40)

Moreover we observe that, if f never vanishes, we can use its monotonicity property to bound $\mu _f$ from below in the following way:

$$\begin{aligned} \mu _f(\Omega ) = \int _{\Omega }f(d(x))\,dx \ge f(R_{\Omega })|\Omega |, \end{aligned}$$

obtaining

$$\begin{aligned} T_{f,p}(\Omega )\ge c_p\frac{f^{\frac{2p-1}{p-1}}(R_{\Omega })|\Omega |^{\frac{2p-1}{p-1}}}{f(0)P^{\frac{p}{p-1}}(\Omega )}. \end{aligned}$$

(41)

Joining (41) and (40), we obtain

$$\begin{aligned} c_p\frac{f^{\frac{2p-1}{p-1}}(R_{\Omega _l})}{f(0)}\le T_{f,p}(\Omega _l)P^{\frac{p}{p-1}}(\Omega _l)\le c_p f^{\frac{p}{p-1}}(0) \left( 1+ \frac{l^{\frac{n}{n-1}}}{2}\mathcal {H}^{n-2}(\partial C)\right) ^{\frac{p}{p-1}}. \end{aligned}$$

Eventually, passing to the limit when $l\rightarrow 0$, observing that $\displaystyle \lim _{l\rightarrow 0} R_{\Omega _l}=0$ and that f is continuous, we have

$$\begin{aligned} T_{f,p}(\Omega _l)P^{\frac{p}{p-1}}(\Omega _l) \longrightarrow c_p f^{\frac{p}{p-1}}(0). \end{aligned}$$

$\square$

Remark 3.1

If we assume that $f:[0,R_\Omega ]\rightarrow [0,+\infty [$ is a function in $L^{\infty }([0,R_\Omega ])$, then, using the variational characterization (4) and the result (8) proved in [8], we have

$$\begin{aligned} T_{f,p}(\Omega )\ge \left( \inf _{t \in [0, R_\Omega ]} f(t)\right) ^{\frac{p}{p-1}} T_p(\Omega )\ge \left( \inf _{t \in [0, R_\Omega ]} f(t)\right) ^{\frac{p}{p-1}} c_p \dfrac{|\Omega |^{\frac{2p-1}{p-1}}}{P(\Omega )^{\frac{p}{p-1}}} \end{aligned}$$

(42)

and the sharpness of (42) can be proved in an analogous way as in (12).

4 The quantitative results

4.1 Proof of Theorem 1.2

Proof

Let us start by proving (14) in the case $n=2$. If $f\equiv 1$, (35) becomes

$$\begin{aligned} T_p(\Omega ) \ge c_p \frac{|\Omega |^{\frac{2p-1}{p-1}}}{P^{\frac{p}{p-1}}(\Omega )}+ c_p \frac{p}{p-1}\int _0^{R_{\Omega }} \left( \frac{\mu (t)}{P(t)}\right) ^{\frac{2p-1}{p-1}}(-P'(t))\,dt. \end{aligned}$$

(43)

Joining (17), (18), (27) and (43), we have that

$$\begin{aligned} \begin{aligned} \frac{T_p(\Omega )P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}} -c_p&> c_p \frac{p}{p-1}\frac{P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}}\int _0^{R_{\Omega }} \left( \frac{\mu (t)}{P(t)}\right) ^{\frac{2p-1}{p-1}}(-P'(t))\,dt \\&\ge \frac{\pi }{2^{\frac{p}{p-1}}} \frac{p}{2p-1} \frac{P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}} \int _0^{R_{\Omega }} \left( R_\Omega -t\right) ^{\frac{2p-1}{p-1}}\,dt \\&\ge \frac{\pi }{2^{\frac{p}{p-1}}} \frac{(p-1)p}{(3p-2)(2p-1)} \frac{R_\Omega }{P(\Omega )}\left( \frac{R_\Omega P(\Omega )}{|\Omega |} \right) ^{{\frac{2p-1}{p-1}}} \\&\ge \frac{\pi }{2^{\frac{p}{p-1}}} \frac{(p-1)p}{(3p-2)(2p-1)} \frac{R_\Omega }{P(\Omega )}. \end{aligned} \end{aligned}$$

Hence, by applying (21) and (18) we get

$$\begin{aligned} \mathcal {F}_p(\Omega ) -c_p \ge K(2,p) \frac{w_\Omega }{{{\,\mathrm{{diam}}\,}}(\Omega )}, \end{aligned}$$

(44)

where

$$\begin{aligned} K(2,p)=\frac{(p-1)p}{ 2^{\frac{p}{p-1}}3(3p-2)(2p-1)}. \end{aligned}$$

(45)

We now prove that the exponent of $\displaystyle \frac{w_\Omega }{{{\,\mathrm{{diam}}\,}}(\Omega )}$ in (44) is sharp. In order to do that, we only need to find a sequence $\{\Omega _l\}_{l\in \mathbb {N}}$ of convex sets with fixed measure such that

$$\begin{aligned} M \frac{w_{\Omega _l}}{{{\,\mathrm{{diam}}\,}}(\Omega _l)} \ge \mathcal {F}_p(\Omega _l) -c_p , \end{aligned}$$

for some positive constant M. Let $0<l<1$, we consider the following rectangle

$$\begin{aligned} \Omega _l = \left( -\frac{1}{2l},\frac{1}{2l}\right) \times \left( -\frac{l}{2}, \frac{l}{2}\right) \end{aligned}$$

and we notice that its inradius and area are $R_{\Omega _l}= \frac{l}{2}$ and ${\left| \Omega _l\right| }=1$. Let u be the unique solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _p u = 1 \quad \text {in } \Omega _l \\ u=0 \qquad \text {on } \partial \Omega _l \end{array}\right. } \end{aligned}$$

and let us consider the following function

$$\begin{aligned} v(y) = \frac{p-1}{p}\left[ \left( \frac{l}{2}\right) ^{\frac{p}{p-1}} -{\left| y\right| }^{\frac{p}{p-1}}\right] , \end{aligned}$$

which solves

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta v = 1 &{}\text {in } \Omega _l\\ v\ge 0 &{}\text {on } \partial \Omega _l. \end{array}\right. } \end{aligned}$$

The comparison principle gives $u\le v$ in $\Omega _l$ and

$$\begin{aligned} T_p(\Omega _l) = \int _{\Omega _l}u_p\,dx \le \int _{\Omega _l}v\,dx. \end{aligned}$$

Arguing as in (38), we have

$$\begin{aligned} \int _{\Omega _l}v\,dx = c_p\left( \frac{l}{2}\right) ^{\frac{p}{p-1}}. \end{aligned}$$

On the other hand, the perimeter of the rectangle is given by

$$\begin{aligned} P(\Omega _l) = \frac{2}{l}\left( 1+l^2\right) \end{aligned}$$

and its Taylor expansion with respect to $l>0$ is

$$\begin{aligned} P^{{\frac{p}{p-1}}}(\Omega _l) = \left( \frac{2}{l}\right) ^{{\frac{p}{p-1}}}\left( 1+l^2\right) ^{{\frac{p}{p-1}}} = \left( \frac{2}{l}\right) ^{{\frac{p}{p-1}}}\left( 1+\frac{p}{p-1}l^2 + o(l^2)\right) . \end{aligned}$$

Using (18) and (21), we get

$$\begin{aligned} \begin{aligned} T_p(\Omega _l)P^{\frac{p}{p-1}}(\Omega _l) - c_p&\le c_p \left( \frac{l}{2}\right) ^{\frac{p}{p-1}} \left( \frac{2}{l}\right) ^{{\frac{p}{p-1}}}\left( 1+\frac{p}{p-1}l^2+ o(l^2)\right) - c_p\\&\le 2c_p \frac{p}{p-1} l^2 \le 16 c_p \frac{ p}{p-1} \frac{R_{\Omega _l}}{P(\Omega _l)}\\&\le 4 c_p \frac{ p}{p-1} \frac{w_{\Omega _l}}{{{\,\mathrm{{diam}}\,}}(\Omega _l)} \end{aligned} \end{aligned}$$

and this concludes the proof in dimension $n=2$.

Let us now prove (14) in the case $n>2$. If we choose $f\equiv 1$, (35) becomes

$$\begin{aligned} T_p(\Omega ) \ge c_p \frac{|\Omega |^{\frac{2p-1}{p-1}}}{P^{\frac{p}{p-1}}(\Omega )}+ c_p \frac{p}{p-1}\int _0^{R_{\Omega }} \left( \frac{\mu (t)}{P(t)}\right) ^{\frac{2p-1}{p-1}}(-P'(t))\,dt. \end{aligned}$$

(46)

Hence, combining (26) and (46), we have

$$\begin{aligned} \frac{T_p(\Omega )P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}} -c_p \ge k(n,p) \frac{P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}}\int _0^{R_{\Omega }} \left( \frac{\mu (t)}{P(t)}\right) ^{\frac{2p-1}{p-1}}P(t)^{\frac{n-2}{n-1}}\,dt. \end{aligned}$$

(47)

Moreover, from (17), we obtain that

$$\begin{aligned} P(t)\ge n\omega _n (R_{\Omega }-t)^{n-1}, \end{aligned}$$

(48)

and so, using (48) in (47), we get

$$\begin{aligned} \begin{aligned} \frac{T_p(\Omega )P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}} -c_p&\ge k(n,p) \frac{P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}}\int _0^{R_{\Omega }} \left( R_{\Omega }-t\right) ^{\frac{2p-1}{p-1}+n-2}\,dt \\&=k(n,p) \bigg (\frac{R_{\Omega }P(\Omega )}{|\Omega |}\bigg )^{\frac{2p-1}{p-1}}\frac{R_{\Omega }^{n-1}}{P(\Omega )}. \end{aligned} \end{aligned}$$

(49)

If we combine (49) with (17), with the following estimate (that can be found in [2]):

$$\begin{aligned} \displaystyle R_{\Omega }\ge {\left\{ \begin{array}{ll} w_{\Omega } \displaystyle {\frac{\sqrt{n+2}}{2n+2}} &{} n \,\, \text {even}\\ \\ w_{\Omega } \displaystyle {\frac{1}{2\sqrt{n}}} &{} n \,\, \text {odd}, \end{array}\right. } \end{aligned}$$

and with

$$\begin{aligned} \displaystyle {P(\Omega )\le n\omega _n \left( \frac{n}{2n+2}\right) ^{\frac{n-1}{2}} {{\,\mathrm{{diam}}\,}}(\Omega )^{n-1}}, \end{aligned}$$

we finally get

$$\begin{aligned} \frac{T_p(\Omega )P^{\frac{p}{p-1}}(\Omega )}{|\Omega |^{\frac{2p-1}{p-1}}} -c_p \ge K(n,p) \bigg (\frac{w_{\Omega }}{{{\,\mathrm{{diam}}\,}}(\Omega )}\bigg )^{n-1}. \end{aligned}$$

$\square$

Remark 4.1

As far as the sharpness of (14) in the case $n>2$, we conjecture that the sharp exponent is 1 as in the planar case. Indeed, the minimizing sequence $\{\Omega _l\}$ satisfies

$$\begin{aligned} T_p(\Omega _l)P^{\frac{p}{p-1}}(\Omega _l)-c_p \approx C \frac{w_{\Omega _l}}{{{\,\mathrm{{diam}}\,}}(\Omega _l)}. \end{aligned}$$

Remark 4.2

As already remarked in the Introduction, inequality (14) gives information on the set $\Omega$. Indeed, if

$$\begin{aligned} \mathcal {F}_p(\Omega ) -c_p \end{aligned}$$

is small, then the ratio between $w_\Omega$ and ${{\,\mathrm{{diam}}\,}}(\Omega )$ has to be necessarily small, i.e. $\Omega$ must be a thin domain. Moreover, inequality (14) tells us also that the infimum of $\mathcal {F}_p(\Omega )$ is not achieved among bounded, open and convex sets. Assuming by contradiction that there exists a bounded, open and convex set $\tilde{\Omega }$ such that

$$\begin{aligned} \mathcal {F}_p(\tilde{\Omega }) = c_p, \end{aligned}$$

we have that

$$\begin{aligned} \frac{w_{\tilde{\Omega }}}{{{\,\mathrm{{diam}}\,}}({\tilde{\Omega }})} < \varepsilon \,\,\,\,\,\,\,\,\,\,\, \forall \varepsilon >0, \end{aligned}$$

which is impossible.

Theorem 1.2 only tells us that any minimizing sequence of $\mathcal {F}_p(\cdot )$ is a sequence of thinning domains. On the other hand, Theorem 1.3 gives us more precise information on the geometry of such minimizing sequence in the planar case.

4.2 Proof of Theorem 1.3

Proof

Let $\Omega$ be a non-empty, bounded, open and convex set in $\mathbb {R}^2$ and let us consider a rectangle Q with sides $P(\Omega )/2$ and $w_{\Omega }$ containing $\Omega$ (see Fig. 3). Such a rectangle exists, since it is enough to choose the shorter side of Q parallel to the direction of $w_{\Omega }$ and to recall the lower bound in (21).

Fig. 3

Rectangle with sides $P(\Omega )/2$ and $w_{\Omega }$ containing $\Omega$.

Now, let $\sigma >0$ be such that

$$\begin{aligned}{} & {} \frac{1}{ 4^3\cdot 6} - \frac{\pi ^2}{ 2^3\cdot 3^3}\frac{\sigma ^2}{K^2(2)}\ge 0; \end{aligned}$$

(50)

$$\begin{aligned}{} & {} \frac{1}{ 3^3\cdot 6} -\frac{\pi }{48}\frac{\sigma }{K(2))} -\frac{\pi ^2}{ 2^5\cdot 3}\frac{\sigma ^2}{K^2(2)}\ge 0; \end{aligned}$$

(51)

$$\begin{aligned}{} & {} \frac{\pi }{4}-\frac{\pi }{2\sqrt{3}}\frac{\sigma }{K(2)}\ge \frac{4}{3 \sqrt{3}}, \end{aligned}$$

(52)

where $K(2):=K(2,2)$ is the constant defined in (45). If

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \sigma , \end{aligned}$$

then, by (18) and (17), we have

$$\begin{aligned} \frac{{\left| Q\bigtriangleup \Omega \right| }}{{\left| \Omega \right| }}=\left( \frac{P(\Omega ) w_\Omega }{2{\left| \Omega \right| }}-1\right) \le \left( \frac{3}{2}\frac{P(\Omega ) R_\Omega }{{\left| \Omega \right| }}-1\right) \le 2. \end{aligned}$$

So, it follows that

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \frac{\sigma }{2^3}2^3\ge \frac{\sigma }{2^3}\left( \frac{{\left| Q\bigtriangleup \Omega \right| }}{|\Omega |} \right) ^3. \end{aligned}$$

On the other hand, let us assume that

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} <\sigma . \end{aligned}$$

(53)

By Theorem 1.2, we have that

$$\begin{aligned} \frac{w_\Omega }{{{\,\mathrm{{diam}}\,}}(\Omega )} \le \frac{1}{K(2)}\left[ \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \right] < \frac{\sigma }{K(2)}, \end{aligned}$$

(54)

and we observe that, by the choice of $\sigma$ made in (50)-(52), a ball cannot satisfy (53).

Now, arguing as in (34) with $f\equiv 1$ and $p=2$, we know that

$$\begin{aligned} T(\Omega ) \ge \int _0^{R_\Omega }\frac{\mu ^2(t)}{P(t)}\,dt. \end{aligned}$$

(55)

We set $\displaystyle {\rho =\dfrac{P^2(\Omega )}{4 \pi } - {\left| \Omega \right| }}$ and $p_R=P(\Omega )- 2 \pi R_\Omega$ and we observe that they are both strictly positive by the isoperimetric inequality and the monotonicity of the perimeter, respectively. Using inequalities (24) and (25) in (55), we have that

$$\begin{aligned} \begin{aligned}&T(\Omega )P^2(\Omega ) \ge P^2(\Omega ) \int _0^{R_\Omega } \frac{(|\Omega |-P(\Omega )t+\pi t^2)^2}{P(\Omega )-2\pi t}\,dt\\&\quad = P^2(\Omega ) \int _0^{R_\Omega } \frac{1}{P(\Omega ) -2 \pi t}\left( \frac{\left( P(\Omega ) -2 \pi t\right) ^2}{4\pi } -\left( \frac{P^2(\Omega )}{4 \pi } - {\left| \Omega \right| }\right) \right) ^2 \,dt\\&\quad = P^2(\Omega ) \int _0^{R_\Omega } \left( \frac{\left( P(\Omega ) -2 \pi t\right) ^3}{(4\pi )^2} - \frac{\rho }{2 \pi } \left( P(\Omega ) -2 \pi t\right) +\frac{\rho ^2}{P(\Omega ) -2 \pi t}\right) \, dt\\&\quad = \frac{P^2(\Omega ) }{2\pi } \left( \frac{P^4(\Omega )-p_R^4}{4(4\pi )^2} - \frac{\rho }{4 \pi } \left( P^2(\Omega ) -p_R^2\right) -\rho ^2 \log \left( 1-\frac{2\pi R_\Omega }{P(\Omega )}\right) \right) , \end{aligned} \end{aligned}$$

(56)

and, using Newton’s formula and the Taylor series for the logarithm, we get

$$\begin{aligned} \begin{aligned} P^2(\Omega )-p^2_R&= 4\pi R_\Omega P(\Omega )-4\pi ^2 R^2_\Omega ;\\ P^4(\Omega )-p^4_R&= 8\pi R_\Omega P^3(\Omega ) -24 \pi ^2 R_\Omega ^2 P^2(\Omega )+32 \pi ^3 R^3_\Omega P(\Omega )- 16 \pi ^4 R^4_\Omega ;\\&\quad -\log \left( 1-\frac{2\pi R_\Omega }{P(\Omega )}\right) =\sum _{i=1}^{\infty }\frac{1}{i} \left( \frac{2\pi R_\Omega }{P(\Omega )}\right) ^i \\&\ge \frac{2\pi R_\Omega }{P(\Omega )}+ \frac{2\pi ^2 R^2_\Omega }{P^2(\Omega )} + \frac{8}{3}\frac{\pi ^3 R^3_\Omega }{P^3(\Omega )} + \frac{4\pi ^4 R^4_\Omega }{P^4(\Omega )}. \end{aligned} \end{aligned}$$

(57)

By (57) and (56), dividing by ${\left| \Omega \right| }^3$ and subtracting 1/3, we have

$$\begin{aligned} \begin{aligned} \frac{T(\Omega )P^2(\Omega )}{{\left| \Omega \right| }^3} - \frac{1}{3} \ge&\frac{1}{3}\bigg (\frac{P(\Omega )R_\Omega }{{\left| \Omega \right| }}-1\bigg )^3 +\pi \frac{R_\Omega ^2}{{\left| \Omega \right| }^2} \left( {\left| \Omega \right| }-\frac{2}{3} P(\Omega )R_\Omega \right) \\&+\frac{4}{3}\pi ^2 \frac{R_\Omega ^3}{P(\Omega ){\left| \Omega \right| }^2} \left( {\left| \Omega \right| }-\frac{3}{4} P(\Omega )R_\Omega \right) . \end{aligned} \end{aligned}$$

(58)

As an intermediate step we want to prove the following inequality:

$$\begin{aligned}{} & {} \frac{1}{3}\bigg (\frac{P(\Omega )R_\Omega }{{\left| \Omega \right| }}-1\bigg )^3 +\pi \frac{R_\Omega ^2}{{\left| \Omega \right| }^2} \left( {\left| \Omega \right| }-\frac{2}{3} P(\Omega )R_\Omega \right) \nonumber \\{} & {} \quad +\frac{4}{3}\pi ^2 \frac{R_\Omega ^3}{P(\Omega ){\left| \Omega \right| }^2} \left( {\left| \Omega \right| }-\frac{3}{4} P(\Omega )R_\Omega \right) \ge \frac{1}{6}\bigg (\frac{P(\Omega )R_\Omega }{{\left| \Omega \right| }}-1\bigg )^3, \end{aligned}$$

(59)

that, combined with (58), implies

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \frac{1}{6} \left( \frac{P(\Omega )R_\Omega }{|\Omega |}-1 \right) ^3, \end{aligned}$$

(60)

where we choose the constant 1/6 as an arbitrary constant less then 1/3. In particular, (59) is equivalent to

$$\begin{aligned}{} & {} \frac{1}{6}\big (P(\Omega )R_\Omega -|\Omega |\big )^3 +\pi R_\Omega ^2 {\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{2}{3} P(\Omega )R_\Omega \right) \nonumber \\{} & {} \quad +\frac{4}{3}\pi ^2 \frac{R_\Omega ^3}{P(\Omega )}{\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{3}{4} P(\Omega )R_\Omega \right) \ge 0. \end{aligned}$$

(61)

In order to prove (61), we distinguish three cases:

• 1) if $\displaystyle {{\left| \Omega \right| }\ge \frac{3}{4} P(\Omega )R_\Omega }$, then (61) is trivial, since the left hand side is the sum of positive quantities;

• 2) if $\displaystyle {\frac{2}{3}P(\Omega )R_\Omega \le {\left| \Omega \right| }< \frac{3}{4} P(\Omega )R_\Omega }$, using (18), (21), (50) and (54), we have

  $$\begin{aligned} \begin{aligned} \frac{1}{6}\big (P(\Omega )R_\Omega -|\Omega |\big )^3&+\pi R_\Omega ^2 {\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{2}{3} P(\Omega )R_\Omega \right) \\&\quad +\frac{4}{3}\pi ^2 \frac{R_\Omega ^3}{P(\Omega )}{\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{3}{4} P(\Omega )R_\Omega \right) \\ {}&\ge P^3(\Omega )R^3_\Omega \left( \frac{1}{ 4^3\cdot 6} -\frac{2\pi ^2}{3^3}\frac{R^2_\Omega }{P^2(\Omega )}\right) \\ {}&\ge P^3(\Omega )R^3_\Omega \left( \frac{1}{ 4^3\cdot 6} - \frac{\pi ^2}{ 2^3\cdot 3^3}\frac{w^2_\Omega }{{{\,\mathrm{{diam}}\,}}^2(\Omega )}\right) \\&\ge P^3(\Omega )R^3_\Omega \left( \frac{1}{ 4^3\cdot 6} - \frac{\pi ^2}{ 2^3\cdot 3^3}\frac{\sigma ^2}{K^2(2)}\right) \ge 0. \end{aligned} \end{aligned}$$

  (62)

• 3) if $\displaystyle {\frac{1}{2}P(\Omega )R_\Omega \le {\left| \Omega \right| }< \frac{2}{3} P(\Omega )R_\Omega }$, arguing as before, we have

  $$\begin{aligned} \begin{aligned} \frac{1}{6}\big (P(\Omega )R_\Omega -|\Omega |\big )^3&+\pi R_\Omega ^2 {\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{2}{3} P(\Omega )R_\Omega \right) \\&\quad +\frac{4}{3}\pi ^2 \frac{R_\Omega ^3}{P(\Omega )}{\left| \Omega \right| } \left( {\left| \Omega \right| }-\frac{3}{4} P(\Omega )R_\Omega \right) \\ {}&\ge P^3(\Omega )R^3_\Omega \left( \frac{1}{ 3^3\cdot 6} -\frac{\pi }{48}\frac{w_\Omega }{{{\,\mathrm{{diam}}\,}}(\Omega )} -\frac{\pi ^2}{ 2^5\cdot 3}\frac{w^2_\Omega }{{{\,\mathrm{{diam}}\,}}^2(\Omega )}\right) \\&\ge P^3(\Omega )R^3_\Omega \left( \frac{1}{ 3^3\cdot 6} -\frac{\pi }{48}\frac{\sigma }{K(2)} -\frac{\pi ^2}{ 2^5\cdot 3}\frac{\sigma ^2}{K^2(2)}\right) \ge 0. \end{aligned} \end{aligned}$$

  (63)

So, we have proved the intermediate step (60). Now, by combining (60) and (19), we deduce

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \frac{1}{6} \left[ \frac{P(\Omega )R_\Omega }{|\Omega |}-1 \right] ^3\ge \frac{1}{6} \left[ \frac{P(\Omega ) w_\Omega }{2|\Omega |}-1 -\frac{1}{\sqrt{3}}\frac{w_\Omega ^2}{{\left| \Omega \right| }} \right] ^3. \end{aligned}$$

(64)

Using (20), (19), (54) and (52), we have

$$\begin{aligned} \begin{aligned} \frac{P(\Omega ) w_\Omega }{2{\left| \Omega \right| }}-1&\ge \frac{P(\Omega ) R_\Omega }{{\left| \Omega \right| }}-1\ge \pi \frac{R_\Omega ^2}{{\left| \Omega \right| }}\ge \frac{\pi }{{\left| \Omega \right| }}\left( \frac{w_\Omega }{2}- \frac{w_\Omega ^2}{\sqrt{3}P(\Omega )}\right) ^2 \\&=\frac{w_\Omega ^2}{{\left| \Omega \right| }}\left( \frac{\pi }{4}-\frac{\pi }{\sqrt{3}}\frac{w_\Omega }{P(\Omega )}+ \frac{\pi }{3}\frac{w_\Omega ^2 }{P(\Omega )^2}\right) \\&\ge \frac{w_\Omega ^2}{{\left| \Omega \right| }}\left( \frac{\pi }{4}-\frac{\pi }{2\sqrt{3}}\frac{w_\Omega }{{{\,\mathrm{{diam}}\,}}(\Omega )}\right) \\&\ge \frac{w_\Omega ^2}{{\left| \Omega \right| }}\left( \frac{\pi }{4}-\frac{\pi }{2\sqrt{3}}\frac{\sigma }{K(2)}\right) \\&\ge \frac{4}{3\sqrt{3}} \frac{w_\Omega ^2}{{\left| \Omega \right| }}. \end{aligned} \end{aligned}$$

(65)

Finally, by combining (64) and (65), we get the conclusion

$$\begin{aligned} \frac{T(\Omega )P^2(\Omega )}{|\Omega |^3}-\frac{1}{3} \ge \frac{1}{6}\left[ \frac{P(\Omega )R_\Omega }{{\left| \Omega \right| }} -1\right] ^3\ge \tilde{K}\left[ \frac{{\left| Q\bigtriangleup \Omega \right| }}{{\left| \Omega \right| }}\right] ^3. \end{aligned}$$

(66)

$\square$

The next remark shows that a sequence of thinning triangles is not sharp for (14) in the case $n=2$ and this is the reason for which we need Theorem 1.3 to obtain more precise information.

Remark 4.3

Let us consider a sequence of isosceles triangles $\mathcal {T}_l$ of base L and height l such that ${\left| \mathcal {T}_l\right| }=1$ (see Fig. 4).

Fig. 4

Isosceles triangle $\mathcal {T}_l$ of base L and height l.

If we compute (60) on the sequence $\mathcal {T}_l$ and we use (17), we get, for every l,

$$\begin{aligned} \dfrac{T(\mathcal {T}_l) P^2(\mathcal {T}_l)}{|\mathcal {T}_l|^3}-\frac{1}{3}\ge \frac{1}{6}\left( \frac{P(\mathcal {T}_l)R_{\mathcal {T}_l}}{{\left| \mathcal {T}_l\right| }}-1\right) ^3=\frac{1}{6} \end{aligned}$$

(67)

and, so, the quantity on the left-hand side of (67) is bounded away from zero.

Remark 4.4

We point out that

$$\begin{aligned} \frac{P(\Omega )R_\Omega }{{\left| \Omega \right| }} -1\ge K \frac{{\left| Q\bigtriangleup \Omega \right| }}{{\left| \Omega \right| }}, \end{aligned}$$

in (66) is a quantitative version of the inequality in the right hand side of (17).

Open problem 4.5

We conclude by listing the following open problems:

• We believe that the exponent 3 in the inequality (15) is not sharp: we expect it to be 1. We clarify that in Example (4.6).

• We conjecture that the sharp exponent in (14) in the case $n>2$ is 1 (see Remark 4.1).

• The results contained in Theorem 1.3 could be studied in higher dimension and extended to the f, p-torsional rigidity. Our proof cannot be adapted to higher dimension because in dimension $n>2$ we do not have any more Steiner formulas (24) and (25) for inner parallel sets.

Example 4.6

Let $\displaystyle {\Omega _l= \left( -\frac{1}{2l},\frac{1}{2l} \right) \times \left( -\frac{l}{2},\frac{l}{2}\right) }$ be a sequence of rectangles of measure 1. It is possible to give an explicit upper bound to the functional $\mathcal {F}_2(\Omega _l)$. Hence, following the computations in (38), we have

$$\begin{aligned} \mathcal {F}_2 (\Omega _l)- c_2 \le 4 l^2. \end{aligned}$$

Considering the rectangle Q with sides $P(\Omega _l)/ 2$ and $w_{\Omega }$ containing $\Omega _l$, that is

$$\begin{aligned} \displaystyle {Q= \left( -\frac{1+l^2}{2l},\frac{1+l^2}{2l} \right) \times \left( -\frac{l}{2},\frac{l}{2}\right) }, \end{aligned}$$

it is straightfoward to compute

$$\begin{aligned} {\left| \Omega _l \Delta Q\right| }= 2l^2. \end{aligned}$$

Hence, we have

$$\begin{aligned} 2 \ge \frac{\mathcal {F}_2(\Omega _l) - c_2}{{\left| \Omega _l \Delta Q\right| }} \ge \tilde{K}\,{\left| \Omega _l \Delta Q\right| }^2 =\tilde{K}\, l^4. \end{aligned}$$