On the ${ L}^{ p}$ norm of the torsion function

Abstract

Bounds are obtained for the $L^p$ norm of the torsion function $v_{\varOmega }$, i.e. the solution of $-\varDelta v=1,\, v\in H_0^1(\varOmega ),$ in terms of the Lebesgue measure of $\varOmega$ and the principal eigenvalue $\lambda _1(\varOmega )$ of the Dirichlet Laplacian acting in $L^2(\varOmega )$. We show that these bounds are sharp for $1\le p\le 2$.

1 Introduction

Let $\varOmega$ be a non-empty open set in Euclidean space $\mathbb {R}^m$ with boundary $\partial \varOmega$. It is well-known [2, 3] that if the bottom of the Dirichlet Laplacian defined by

$$\begin{aligned} \lambda _1(\varOmega )=\inf _{\varphi \in H_0^1(\varOmega )\setminus \{0\}}\frac{\displaystyle \int _\varOmega |D\varphi |^2}{\displaystyle \int _\varOmega \varphi ^2} \end{aligned}$$

(1)

is bounded away from 0, then

$$\begin{aligned} -\varDelta v=1,\, v\in H_0^1(\varOmega ) \end{aligned}$$

(2)

has a unique solution denoted by $v_{\varOmega }$. The function $v_{\varOmega }$ is non-negative, pointwise increasing in $\varOmega$, and satisfies,

$$\begin{aligned} \lambda _1(\varOmega )^{-1}\le \Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}\le (4+3m\log 2)\lambda _1(\varOmega )^{-1}. \end{aligned}$$

(3)

The m-dependent constant in the right-hand side of (3) has subsequently been improved [9, 16]. We denote the optimal constant in the right-hand side of (3) by

$$\begin{aligned} \mathfrak {F}_{\infty }=\sup \{\lambda _1(\varOmega ) \Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}:\varOmega \,\text { open in}\,\mathbb {R}^m,\,|\varOmega |<\infty \}, \end{aligned}$$

(4)

suppressing the m-dependence. The torsional rigidity of $\varOmega$ is defined by

$$\begin{aligned} T_1(\varOmega )=\int _{\varOmega }v_{\varOmega }. \end{aligned}$$

It plays a key role in different parts of analysis. For example the torsional rigidity of a cross section of a beam appears in the computation of the angular change when a beam of a given length and a given modulus of rigidity is exposed to a twisting moment [1, 14]. It also arises in the definition of gamma convergence [7] and in the study of minimal submanifolds [12]. Moreover, $T_1(\varOmega )/|\varOmega |$ equals $\mathbb {E}_x(\tau _{\varOmega }),$ the expected lifetime $\tau _{\varOmega }$ of Brownian motion in $\varOmega$, when averaged with respect to the uniform distribution over all starting points $x\in \varOmega$.

The torsion function has been studied extensively and numerous works have been written on this subject. We just mention the paper [6], and the references therein. There the Kohler–Jobin rearrangement technique has been applied to the p-torsional rigidity, involving the p-Laplacian, and its first Dirichlet eigenvalue.

A classical inequality, e.g. [14], asserts that the function $F_1$ defined on the open sets in $\mathbb {R}^m$ with finite Lebesgue measure

$$\begin{aligned} F_1(\varOmega )=\frac{T_1(\varOmega )\lambda _1(\varOmega )}{\vert \varOmega \vert } \end{aligned}$$

(5)

satisfies

$$\begin{aligned} F_1(\varOmega )\le 1. \end{aligned}$$

(6)

Since $\varOmega$ has finite Lebesgue measure $|\varOmega |$, (3) implies that $v\in L^p(\varOmega )$ for $1\le p \le \infty$. Moreover $\lambda _1(\varOmega )$ is in that case the principal eigenvalue of the Dirichlet Laplacian. Motivated by (5) and (6) we make the following

Definition 1

• (i) For $\varOmega$ open in $\mathbb {R}^m$ with $0<|\varOmega |<\infty$ and $1\le p<\infty$,

  $$\begin{aligned} F_p(\varOmega )=\frac{T_p(\varOmega )\lambda _1(\varOmega )}{|\varOmega |^{1/p}}, \end{aligned}$$

  (7)

  where

  $$\begin{aligned} T_p(\varOmega )=\Vert v_{\varOmega }\Vert _{L^p(\varOmega )}=\bigg (\int _{\varOmega }v_{\varOmega }^p\bigg )^{1/p}. \end{aligned}$$

  (8)

• (ii) For $\varOmega$ open in $\mathbb {R}^m$ with $\lambda _1(\varOmega )>0$,

  $$\begin{aligned} F_{\infty }(\varOmega )=\Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}\lambda _1(\varOmega ). \end{aligned}$$

It follows from the Faber–Krahn inequality that if $|\varOmega |<\infty$ then $\lambda _1(\varOmega )>0$. The converse does not hold for if $\varOmega$ is the union of infinitely many disjoint balls of radii 1 then $\lambda _1(\varOmega )>0$ but $\varOmega$ has infinite measure. Note that $2T_2^2(\varOmega )/|\varOmega |$ equals the second moment of the expected lifetime of Brownian motion in $\varOmega$, when averaged with respect to the uniform distribution over all starting points $x\in \varOmega$.

Note that $\varOmega \mapsto T_p(\varOmega )$ is increasing while $\varOmega \mapsto \lambda _1(\varOmega )$ and $\varOmega \mapsto \vert \varOmega \vert ^{-1/p}$ are decreasing. It is straightforward to verify that $F_p,\,1\le p\le \infty$ is invariant under homotheties. That is, if $\alpha >0,\, \alpha \varOmega =\{x\in \mathbb {R}^m: x/\alpha \in \varOmega \}$, then $F_p(\alpha \varOmega )= F_p(\varOmega )$.

Our main results are the following.

Theorem 1

Let $\varOmega$ be an open set in $\mathbb {R}^m,\,m=1,2,3,\ldots$ with $|\varOmega |<\infty$.

• (i) If $1\le p\le q\le \infty$ then,

  $$\begin{aligned} F_p(\varOmega )\le F_q(\varOmega )\le \mathfrak {F}_{\infty }. \end{aligned}$$

  (9)

• (ii) If $1\le p\le 2$ then,

  $$\begin{aligned} F_p(\varOmega )\le F_1(\varOmega )^{1/p}\le 1. \end{aligned}$$

  (10)

Definition 2

For $1\le p\le \infty$,

• (i)
  $$\begin{aligned} \mathfrak {F}_p=\sup \{F_p(\varOmega ):\varOmega \, \text {open in}\, \mathbb {R}^m, |\varOmega |<\infty \}, \end{aligned}$$

  (11)
• (ii)
  $$\begin{aligned} \mathfrak {G}_p=\inf \{F_p(\varOmega ):\varOmega \, \text {open in}\, \mathbb {R}^m, |\varOmega |<\infty \}, \end{aligned}$$
• (iii)
  $$\begin{aligned} \mathfrak {F}_p^{\text {convex}}=\sup \{F_p(\varOmega ):\varOmega \, \text {open, convex in}\, \mathbb {R}^m, |\varOmega |<\infty \}, \end{aligned}$$
• (iv)
  $$\begin{aligned} \mathfrak {G}_p^{\text {convex}}=\inf \{F_p(\varOmega ):\varOmega \, \text {open, convex in}\, \mathbb {R}^m, |\varOmega |<\infty \}. \end{aligned}$$

It was shown in [5] that $\mathfrak {G}_{\infty }=1$.

Theorem 2

If $m=1,2,3,\ldots$, and if $1\le p < \infty ,$ then

• (i)
  $$\begin{aligned} \mathfrak {G}_p=0. \end{aligned}$$

• (ii) The mapping $p\mapsto \mathfrak {G}_p^{\text {convex}}$ is non-decreasing, and

  $$\begin{aligned} \mathfrak {G}_p^{\text {convex}}\ge 2^{-3}\pi ^2m^{-(m+2p)/p}\bigg (\frac{\varGamma (\frac{m}{2}+1)\varGamma (p+1)}{\varGamma (\frac{m}{2}+p+1)}\bigg )^{1/p}. \end{aligned}$$

  (12)

It follows from (12) that $\lim _{p\rightarrow \infty }\mathfrak {G}_p^{\text {convex}}\ge \pi ^2/8.$ This jibes with the result of [13] that

$$\begin{aligned} \mathfrak {G}_{\infty }^{\text {convex}}=\frac{\pi ^2}{8}. \end{aligned}$$

A monotone increasing sequence of cuboids which exhausts the open connected set bounded by two parallel $(m-1)$-dimensional hyperplanes is a minimising sequence for $\mathfrak {G}_{\infty }^{\text {convex}}$. See also Theorem 2 in [5].

Theorem 3

Let $m=2,3,\ldots$.

• (i) The mappings $p\mapsto \mathfrak {F}_p$, and $p\mapsto \mathfrak {F}_p^{\text {convex}}$ are non-decreasing on $[1,\infty ]$.

• (ii) If

  $$\begin{aligned} p_m=\inf \{p\ge 1: \mathfrak {F}_p>1\}, \end{aligned}$$

  and

  $$\begin{aligned} p_m^{\text {convex}}=\inf \{p\ge 1: \mathfrak {F}_p^{\text {convex}}>1\}, \end{aligned}$$

  then

  $$\begin{aligned} 2\le p_m\le p_m^{\text {convex}}\le 8m. \end{aligned}$$

  (13)

  In particular

  $$\begin{aligned} \mathfrak {F}_p=1, \,\,1\le p\le p_m. \end{aligned}$$

• (iii) Formula (11) defining $\mathfrak {F}_p$ does not have a maximiser for $1\le p\le 2.$ The maximising sequence constructed in [4] for $\mathfrak {F}_1$ is also a maximising sequence for $\mathfrak {F}_p,\,1\le p\le p_m.$ Hence inequality (10) actually reads $F_2(\varOmega ) \le F_1(\varOmega )^{1/p} < 1$, $1 \le p \le 2.$

• (iv) The mappings $p\mapsto \mathfrak {F}_p$, and $p\mapsto \mathfrak {F}_p^{\text {convex}}$ are left-continuous on $(1,\infty ]$.

• (v) If $n\in \mathbb {N},\,1\le p,$ then

  $$\begin{aligned} \mathfrak {F}_{p+n}\le \bigg (\frac{p+n}{4^np}\prod _{j=1}^n(p+j) \bigg )^{\frac{1}{p+n}}\mathfrak {F}_p^{\frac{p}{p+n}}, \end{aligned}$$

  (14)
  $$\begin{aligned} \mathfrak {F}_{p+n}^{\text {convex}}\le \bigg (\frac{p+n}{4^np} \prod _{j=1}^n(p+j)\bigg )^{\frac{1}{p+n}} \bigg (\mathfrak {F}_p^{\text {convex}}\bigg )^{\frac{p}{p+n}}. \end{aligned}$$

  (15)

  In particular if $1\le p\le 2$, then

  $$\begin{aligned} \mathfrak {F}_{p+1}\le \bigg (\frac{(p+1)^2}{4p}\bigg )^{\frac{1}{p+1}}. \end{aligned}$$

  (16)
• (vi)
  $$\begin{aligned} \mathfrak {F}_{n}\le \bigg (\frac{n.n!}{4^{n-1}}\bigg )^{\frac{1}{n}},\, n\in \mathbb {N}, \end{aligned}$$

  (17)

  and $\mathfrak {F}_3\le 3^{2/3}/2=1.04004\ldots$.

• (vii) $p\mapsto \mathfrak {F}_p$ is differentiable at $p=2$, with $\mathfrak {F}'_2=0.$

• (viii) If $1\le p\le 2$, then

  $$\begin{aligned} \mathfrak {F}_p^{\text {convex}}\le \big (\mathfrak {F}_1^{\text {convex}}\big )^{1/p}. \end{aligned}$$

  (18)

• (ix) For $m = 2,$

  $$\begin{aligned} p_2^{convex }\ge 2.0186. \end{aligned}$$

This paper is organised as follows. In Sect. 2 we prove Theorems 1 and 2. The proof of Theorem 3 will be given in Sect. 3.

We note that a general multiplicative inequality involving $T_p(\varOmega ), \lambda _1(\varOmega )$ and $|\varOmega |$ will involve three exponents. However, the requirement that it be invariant under homotheties reduces the number of exponents to two. In Sect. 4 we briefly discuss this two-parameter family of inequalities, and determine which parameter pair yields a finite supremum.

2 Proofs of Theorems 1, 2

Proof of Theorem 1

(i) To prove (9) for $1\le p\le q<\infty$ we use Hölder’s inequality to obtain that

$$\begin{aligned} \int _{\varOmega }v_{\varOmega }^p\le \bigg (\int _{\varOmega }v_{\varOmega }^q\bigg )^{p/q}|\varOmega |^{(q-p)/q}. \end{aligned}$$

So we have that

$$\begin{aligned} \Vert v_{\varOmega }\Vert _{L^p(\varOmega )}\le \Vert v_{\varOmega }\Vert _{L^q(\varOmega )}|\varOmega |^{\frac{1}{p}-\frac{1}{q}}. \end{aligned}$$

This, together with (7), implies (9). In case $q=\infty$,

$$\begin{aligned} \Vert v_{\varOmega }\Vert _{L^p(\varOmega )}\le \Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}|\varOmega |^{1/p}. \end{aligned}$$

(ii) To prove (10) we observe that since $\varOmega$ has finite Lebesgue measure the spectrum of the Dirichlet Laplacian acting in $L^2(\varOmega )$ is discrete, and consists of an increasing sequence of eigenvalues

$$\begin{aligned} \{\lambda _1(\varOmega )\le \lambda _2(\varOmega )\le \lambda _3(\varOmega )\le \ldots .\}, \end{aligned}$$

accumulating at infinity, where we have included multiplicities. We denote a corresponding orthonormal basis of eigenfunctions by $\{\varphi _{j,\varOmega },j=1,2,3,\ldots \}$. The resolvent of the Dirichlet Laplacian acting in $L^2(\varOmega )$ is compact, and its kernel $H_{\varOmega }$ has an $L^2$-eigenfunction expansion given by

$$\begin{aligned} H_{\varOmega }(x,y)=\sum _{j=1}^{\infty }\frac{1}{\lambda _j(\varOmega )} \varphi _{j,\varOmega }(x)\varphi _{j,\varOmega }(y). \end{aligned}$$

So $v_{\varOmega }$, defined by (2), is given by

$$\begin{aligned} v_{\varOmega }(x)=\sum _{j=1}^{\infty }\frac{1}{\lambda _j(\varOmega )} \bigg (\int _{\varOmega }\varphi _{j,\varOmega }\bigg )\varphi _{j,\varOmega }(x). \end{aligned}$$

Since $v_{\varOmega }\in L^2(\varOmega )$ we have by orthonormality that

$$\begin{aligned} \int _{\varOmega }v_{\varOmega }^2&=\int _{\varOmega }dx\,\sum _{j=1}^{\infty } \sum _{k=1}^{\infty }\frac{1}{\lambda _j(\varOmega )} \bigg (\int _{\varOmega }\varphi _{j,\varOmega }\bigg ) \varphi _{j,\varOmega }(x)\frac{1}{\lambda _k(\varOmega )} \bigg (\int _{\varOmega }\varphi _{k,\varOmega }\bigg )\varphi _{k,\varOmega }(x)\nonumber \\&=\sum _{j=1}^{\infty }\frac{1}{\lambda _j^2(\varOmega )}\bigg (\int _{\varOmega } \varphi _{j,\varOmega }\bigg )^2\nonumber \\&\le \frac{1}{\lambda _1(\varOmega )}\sum _{j=1}^{\infty }\frac{1}{\lambda _j(\varOmega )}\bigg (\int _{\varOmega }\varphi _{j,\varOmega }\bigg )^2\nonumber \\&=\frac{T_1(\varOmega )}{\lambda _1(\varOmega )}. \end{aligned}$$

(19)

We conclude that

$$\begin{aligned} T_2(\varOmega )\le \bigg (\frac{T_1(\varOmega )}{\lambda _1(\varOmega )}\bigg )^{1/2}. \end{aligned}$$

(20)

Multiplying both sides of the inequality above with $\lambda _1(\varOmega )/|\varOmega |^{1/2}$ we obtain that $F_2(\varOmega )\le \big (F_1(\varOmega )\big )^{1/2}$. By (i) $\big (F_1(\varOmega )\big )^{1/2}\le \big (F_2(\varOmega )\big )^{1/2}.$ This, together with the previous inequality, implies that $F_2(\varOmega )\le 1.$ We now use Hölder’s inequality, and interpolate with $0<\alpha <1, \rho >1$ as follows.

$$\begin{aligned} \int _{\varOmega }v_{\varOmega }^p=\bigg (\int _{\varOmega }v_{\varOmega }^{\alpha p\rho }\bigg )^{1/\rho }\bigg (\int _{\varOmega }v_{\varOmega }^{(1-\alpha ) p\rho /(\rho -1)}\bigg )^{(\rho -1)/\rho }. \end{aligned}$$

Choosing $\alpha p\rho =2,(1-\alpha )p\rho /(\rho -1)=1$ gives that $\rho =(p-1)^{-1}$. Hence by (20),

$$\begin{aligned} T_p^p(\varOmega )=\int _{\varOmega }v_{\varOmega }^p\le \bigg (T_2^2(\varOmega )\bigg )^{p-1}\bigg (T_1(\varOmega )\bigg )^{2-p}\le \frac{T_1(\varOmega )}{\lambda _1(\varOmega )^{p-1}}. \end{aligned}$$

Multiplying both sides of the inequality above with $\lambda _1(\varOmega )^p/|\varOmega |$ gives that

$$\begin{aligned} F_p^p(\varOmega )\le F_1(\varOmega ). \end{aligned}$$

$\square$

Proof of Theorem 2

(i) We let $\varOmega _n$ be the disjoint union of one ball of radius 1 and n balls with radii $r_n$, with $r_n<1$. Then

$$\begin{aligned} |\varOmega _n|=\big (nr_n^m+1\big )|B_1|, \end{aligned}$$

where $B_1=\{x\in \mathbb {R}^m:|x|<1\}$. Since $r_n<1$ we have that

$$\begin{aligned} \lambda _1(\varOmega _n)=\lambda _1(B_{1}). \end{aligned}$$

Since $T^p_p$ is additive on disjoint open sets we have by scaling that

$$\begin{aligned} T^p_p(\varOmega _n)=\big (nr_n^{2p+m}+1\big )T^p_p(B_1). \end{aligned}$$

Therefore

$$\begin{aligned} F^p_p(\varOmega _n)&=\frac{\big (nr_n^{2p+m}+1\big )T^p_p(B_1) \lambda ^p_1(B_1)}{\big (nr_n^m+1\big )|B_1|}\nonumber \\&=\frac{nr_n^{2p+m}+1}{nr_n^m+1}F^p_p(B_1)\nonumber \\&\le \big (r_n^{2p}+n^{-1}r_n^{-m}\big )F^p_p(B_1). \end{aligned}$$

(21)

We now choose $r_n$ as to minimise the right-hand side of (21),

$$\begin{aligned} r_n=\bigg (\frac{m}{2pn}\bigg )^{1/(2p+m)}. \end{aligned}$$

This gives that

$$\begin{aligned} F^p_p(\varOmega _n)\le \bigg (1+\frac{2p}{m}\bigg )\bigg (\frac{m}{2p}\bigg )^{2p/(2p+m)} n^{-2p/(2p+m))}F^p_p(B_1), \end{aligned}$$

which implies the assertion.

(ii) The first part of the assertion follows directly by (9). To prove the second part we recall John’s ellipsoid theorem [10, 11] which asserts the existence of an ellipsoid $\varUpsilon$ with centre c such that $\varUpsilon \subset \varOmega \subset c + m(\varUpsilon - c).$ Here $c + m(\varUpsilon - c)=\{c + m(x - c) : x \in \varUpsilon \}.$ This is the dilation of $\varUpsilon$ by the factor m. $\varUpsilon$ is the ellipsoid of maximal volume in $\varOmega$. By translating both $\varOmega$ and $\varUpsilon$ we may assume that

$$\begin{aligned} \varUpsilon =\left\{ x\in \mathbb {R}^m: \sum _{i=1}^m \frac{x_i^2}{a_i^2}<1\right\} , \qquad a_i>0,\quad i=1,\dots ,m. \end{aligned}$$

It is easily verified that the unique solution of (2) for $\varUpsilon$ is given by

$$\begin{aligned} v_{\varUpsilon }(x)=2^{-1}\left( \sum _{i=1}^m\frac{1}{a_i^2}\right) ^{-1}\left( 1-\sum _{i=1}^m\frac{x_i^2}{a_i^2}\right) . \end{aligned}$$

By changing to spherical coordinates, we find that

$$\begin{aligned} \int _{\varUpsilon } v_{\varUpsilon }^p=2^{-p}\omega _m\frac{\varGamma (\frac{m}{2}+1)\varGamma (p+1)}{\varGamma (\frac{m}{2}+p+1)}\left( \sum _{i=1}^m\frac{1}{a_i^2}\right) ^{-p}\prod _{i=1}^ma_i, \end{aligned}$$

where $\omega _m=|B_1|$. Since $\varOmega \mapsto v_{\varOmega }$ is increasing we have by (8) that $\varOmega \mapsto T_p(\varOmega )$ is increasing, and

$$\begin{aligned} T_p(\varOmega )&\ge T_p(\varUpsilon )\nonumber \\&=2^{-1}\omega _m^{1/p}\bigg (\frac{\varGamma (\frac{m}{2}+1)\varGamma (p+1)}{\varGamma (\frac{m}{2}+p+1)}\bigg )^{1/p}\left( \sum _{i=1}^m\frac{1}{a_i^2}\right) ^{-1}\bigg (\prod _{i=1}^ma_i\bigg )^{1/p}. \end{aligned}$$

(22)

Since $\varOmega \subset m\varUpsilon$,

$$\begin{aligned} |\varOmega |\le \int _{m\varUpsilon }dx=\omega _mm^m\prod _{i=1}^ma_i. \end{aligned}$$

(23)

By the monotonicity of Dirichlet eigenvalues, we have that $\lambda _1(\varOmega )\ge \lambda _1(m\varUpsilon )$. The ellipsoid $m\varUpsilon$ is contained in a cuboid with lengths $2ma_1,\ldots ,2ma_m.$ So we have that

$$\begin{aligned} \lambda _1(\varOmega )\ge \frac{\pi ^2}{4m^2}\sum _{i=1}^m\frac{1}{a_i^2}. \end{aligned}$$

(24)

Combining (22), (23), (24), and (8) gives (12). $\square$

3 Proof of Theorem 3

(i) It follows from the second inequality in (9) that $\mathfrak {F}_q\le \mathfrak {F}_{\infty }$. Hence $F_p(\varOmega )\le \mathfrak {F}_q\le \mathfrak {F}_{\infty }$. Taking subsequently the supremum over all $\varOmega$ with finite measure we obtain the first assertion under (i). As (9) holds for all open sets with finite measure, it also holds for all bounded convex sets. Then, the preceding argument gives the second assertion under (i).

(ii) It follows from (10) that $\mathfrak {F}_p\le 1, \, 1\le p\le 2$. In Theorem 1.2 of [4] it was shown that the bound $F_1(\varOmega )\le 1$ is sharp. That is $\mathfrak {F}_1=1$. This, together with (i), then implies that $\mathfrak {F}_p=1$ for $1\le p\le 2$. Hence $p_m\ge 2$. Since $\mathfrak {F}_p^{\text {convex}}\le \mathfrak {F}_p$ we conclude the second inequality in (13). To prove the upper bound on $p_m^{convex }$ we recall that

$$\begin{aligned} v_{B_1}(x)=\frac{1-|x|^2}{2m}. \end{aligned}$$

Hence, denoting by $(f)_{+}$ the positive part of a real-valued function f, we have that

$$\begin{aligned} T_p(B_1)&=\bigg (\int _{[0,1]}dr\,m\omega _m\bigg (\frac{1-r^2}{2m}\bigg )^pr^{m-1}\bigg )^{1/p}\nonumber \\&=\frac{(m\omega _m)^{1/p}}{2^{(p+1)/p}m}\bigg (\int _{[0,1]}d\theta (1-\theta )^p\theta ^{(m-2)/2}\bigg )^{1/p}\nonumber \\&\ge \frac{(m\omega _m)^{1/p}}{2^{(p+1)/p}m}\bigg (\int _{[0,1]}d\theta (1-p\theta )_+\theta ^{(m-2)/2}\bigg )^{1/p}\nonumber \\&\ge \frac{2^{1/p}\omega _m^{1/p}}{2m(m+2)^{1/p}p^{m/(2p)}}\nonumber \\&\ge \frac{\omega _m^{1/p}}{2m^{(p+1)/p}p^{m/(2p)}}. \end{aligned}$$

(25)

It follows that

$$\begin{aligned} \mathfrak {F}_p\ge F_p(B_1)\ge \frac{j^2_{(m-2)/2}}{2m^{(p+1)/p}p^{m/(2p)}}, \end{aligned}$$

where $\lambda _1(B_1)=j^2_{(m-2)/2}$, and $j_{(m-2)/2}$ is the first positive zero of the Bessel function $J_{(m-2)/2}$. Hence

$$\begin{aligned} \mathfrak {F}_{8m}\ge \frac{j^2_{(m-2)/2}}{2m^{1+\frac{1}{8m}}(8m)^{\frac{1}{16}}}\ge \frac{j^2_{(m-2)/2}}{2^{\frac{19}{16}}m^{\frac{9}{8}}}. \end{aligned}$$

(26)

One verifies numerically that for $m=2,\ldots ,19,$ the right-hand side of (26) is strictly greater than 1. Since $j^2_{(m-2)/2}\ge ((m-2)/2)^2$ (see inequality (1.6) in [8]) we have for $m\ge 20$ that $j^2_{(m-2)/2}> m^2/5$. But $m^{\frac{7}{8}}\ge 5\cdot 2^{19/16}, m\ge 20$.

(iii) It was shown in [4] that the formula defining $\mathfrak {F}_1$ in (11) does not have a maximiser. Since by (10), $F_p(\varOmega )\le F_1(\varOmega )^{1/p}\le 1$ for any $1\le p \le 2$ and any open subset $\varOmega \subset \mathbb {R}^m$ with $| \varOmega | < \infty$, none of the formulae defining $\mathfrak {F}_p,\, 1\le p\le 2$, have maximisers. Clearly, the maximising sequence constructed in [4] for $\mathfrak {F}_1$ is a maximising sequence for $\mathfrak {F}_p,\, 1\le p\le p_m$.

(iv) To prove left-continuity we first fix $1<q<\infty$, and let $\epsilon >0$ be arbitrary. There exists an open set $\varOmega _{q,\epsilon }\subset \mathbb {R}^m$ such that

$$\begin{aligned} \mathfrak {F}_q\ge F_q(\varOmega _{q,\epsilon })\ge \mathfrak {F}_q-\frac{\epsilon }{2}. \end{aligned}$$

(27)

By scaling we may assume that $|\varOmega _{q,\epsilon }|=1$. Let $p\in [1,q)$. Then

$$\begin{aligned} \mathfrak {F}_q\ge \mathfrak {F}_p\ge F_p(\varOmega _{q,\epsilon }), \end{aligned}$$

and

$$\begin{aligned} \int _{\varOmega _{q,\epsilon }}v^q_{\varOmega _{q,\epsilon }}&\le \Vert v_{\varOmega _{q,\epsilon }}\Vert ^{q-p}_{L^{\infty }(\varOmega _{q,\epsilon })} \int _{\varOmega _{q,\epsilon }}v^p_{\varOmega _{q,\epsilon }}\nonumber \\&\le \mathfrak {F}_{\infty }^{q-p}\lambda _1^{p-q}(\varOmega _{q,\epsilon }) \int _{\varOmega _{q,\epsilon }}v^p_{\varOmega _{q,\epsilon }}, \end{aligned}$$

implying that

$$\begin{aligned} F_p(\varOmega _{q,\epsilon })\ge \mathfrak {F}_{\infty }^{(p-q)/p}F_q(\varOmega _{q,\epsilon })^{q/p}. \end{aligned}$$

Since $p\mapsto F_p$ is increasing we have that

$$\begin{aligned} \lim _{p\uparrow q}F_p(\varOmega _{q,\epsilon })\ge \mathfrak {F}_{\infty }^{(p-q)/p}F_q(\varOmega _{q,\epsilon })^{q/p}. \end{aligned}$$

(28)

Since $p<q$, we have by the continuity of the right-hand side of (28) in p, (26) and (27) that

$$\begin{aligned} \mathfrak {F}_q\ge \lim _{p\uparrow q}\mathfrak {F}_p\ge F_q(\varOmega _{q,\epsilon })\ge \mathfrak {F}_q-\frac{\epsilon }{2}. \end{aligned}$$

Letting $\epsilon \downarrow 0$ concludes the proof for $1<q<\infty$.

To prove left-continuity at $q=\infty$ we let $\epsilon >0$ be arbitrary. By (11) there exists an open set $\varOmega _{\infty ,\epsilon }$ such that

$$\begin{aligned} \mathfrak {F}_{\infty }\ge F_{\infty }(\varOmega _{\infty ,\epsilon })\ge \mathfrak {F}_{\infty }-\frac{\epsilon }{2}. \end{aligned}$$

(29)

Without loss of generality we may assume by scaling that $|\varOmega _{\infty ,\epsilon }|=1$. Then $v_{\varOmega _{\infty ,\epsilon }}\in L^p(\varOmega _{\infty ,\epsilon }),\, 1\le p\le \infty$, and

$$\begin{aligned} F_{\infty }(\varOmega _{\infty ,\epsilon })=\lim _{p\rightarrow \infty } F_{p}(\varOmega _{\infty ,\epsilon }). \end{aligned}$$

Hence there exists $p(\epsilon )<\infty$ such that

$$\begin{aligned} |F_{\infty }(\varOmega _{\infty ,\epsilon })- F_{p}(\varOmega _{\infty ,\epsilon })|\le \frac{\epsilon }{2},\quad p\ge p(\epsilon ). \end{aligned}$$

(30)

This implies, by (29) and (30), that

$$\begin{aligned} \mathfrak {F}_p&\ge F_p(\varOmega _{\infty ,\epsilon }) \nonumber \\ {}&\ge F_{\infty }(\varOmega _{\infty ,\epsilon })-\frac{\epsilon }{2}\nonumber \\&\ge \mathfrak {F}_{\infty }-\epsilon ,\quad p\ge p(\epsilon ). \end{aligned}$$

(31)

Hence by (i) and (31),

$$\begin{aligned} \mathfrak {F}_{\infty }\ge \lim _{p\uparrow \infty }\mathfrak {F}_p\ge \mathfrak {F}_{\infty }-\epsilon . \end{aligned}$$

The left-continuity at $\infty$ now follows since $\epsilon >0$ was arbitrary.

(v) Let $n\in \mathbb {N},\, p\ge 1$. Without loss of generality we may assume that $|\varOmega |=1$. An integration by parts shows that

$$\begin{aligned} \int _{\varOmega }v_{\varOmega }^p=-\int _{\varOmega }v_{\varOmega }^p\varDelta v_{\varOmega }=p\int _{\varOmega }v_{\varOmega }^{p-1} |Dv_{\varOmega }|^2=\frac{4p}{(p+1)^2}\int _{\varOmega }|Dv_{\varOmega }^{(p+1)/2}|^2. \end{aligned}$$

(32)

By (1)

$$\begin{aligned} \lambda _1(\varOmega )\le \frac{\displaystyle \int _\varOmega | Dv_{\varOmega }^{(p+1)/2}|^2}{\displaystyle \int _\varOmega v_{\varOmega }^{p+1}}. \end{aligned}$$

(33)

By (32) and (33) we have that

$$\begin{aligned} \int _{\varOmega }v_{\varOmega }^p\ge \frac{4p}{(p+1)^2}\lambda _1(\varOmega )\int _{\varOmega }v_{\varOmega }^{p+1}. \end{aligned}$$

(34)

Multiplying both sides of (34) by $\lambda ^p_1(\varOmega )$ gives that

$$\begin{aligned} \frac{4p}{(p+1)^2}F_{p+1}^{p+1}(\varOmega )\le F_p^p(\varOmega ). \end{aligned}$$

(35)

Taking the supremum over all open $\varOmega \subset \mathbb {R}^m$ with measure 1, in the right-hand side of (35), and subsequently in the left-hand side of (35) gives that

$$\begin{aligned} \mathfrak {F}_{p+1}^{p+1}\le \frac{(p+1)^2}{4p}\mathfrak {F}_p^p. \end{aligned}$$

(36)

Iterating (36) $n-1$ times we find (14). The same calculation carries over when $\varOmega$ is an open, bounded convex set. This proves (15). By part (ii) we have that for $1\le p\le p_m,\, \mathfrak {F}_p=1$. This, together with (14), gives (16).

(vi) Since $\mathfrak {F}_1=1$, we consider the case $n\in N,\,n\ge 2$. Put $p=1$ in (14), and replace n by $n-1$. This gives (17).

(vii) Substituting $p=1+\delta ,\,0<\delta \le 1$ in (16) gives that

$$\begin{aligned} \delta ^{-1}\big (\mathfrak {F}_{2+\delta }-\mathfrak {F}_2\big )\le \delta ^{-1}\bigg (\bigg (1+\frac{\delta ^2}{4}\bigg )^{\frac{1}{2+\delta }} -1\bigg ),\, \end{aligned}$$

and the assertion follows by L’ Hôpital’s rule.

(viii) Taking suprema in (10) over all bounded convex open sets $\varOmega$ yields (18).

(ix) Let $m = 2$. By (18), and the numerical estimate (1.10) in [4] we have for $p=1+\delta ,0<\delta \le 1$, that

$$\begin{aligned} \mathfrak {F}_{1+\delta }^{\text {convex}}\le \bigg (1-\frac{1}{11560}\bigg )^{\frac{1}{1+\delta }}. \end{aligned}$$

(37)

By (15) for $n=1$ we have that

$$\begin{aligned} \mathfrak {F}_{2+\delta }^{\text {convex}}&\le \bigg (\frac{(2+\delta )^2}{4+4\delta }\bigg )^{\frac{1}{2+\delta }} \bigg (\mathfrak {F}_{1+\delta }^{\text {convex}}\bigg )^{\frac{1+\delta }{2+\delta }}\nonumber \\&\le \bigg (1+\frac{\delta ^2}{4}\bigg )^{\frac{1}{2+\delta }} \bigg (\mathfrak {F}_{1+\delta }^{\text {convex}}\bigg )^{\frac{1+\delta }{2+\delta }}\nonumber \\&\le \bigg (1+\frac{\delta ^2}{4}\bigg )^{\frac{1}{2+\delta }} \bigg (1-\frac{1}{11560}\bigg )^{\frac{1}{2+\delta }}, \end{aligned}$$

(38)

where we have used (37) in the last inequality. Since the right-hand side of (38) is equal to 1 for $\delta ^*=\frac{2}{(11559)^{1/2}}$, we conclude that

$$\begin{aligned} p_2^{\text {convex}}\ge 2+\delta ^*, \end{aligned}$$

which proves the assertion in (ix). $\square$

4 A two-parameter family of inequalities

As mentioned at the end of the Introduction one can define a two-parameter family of products involving $T_p(\varOmega ),$$\lambda _1(\varOmega ),$ and $|\varOmega |$, which is invariant under homotheties.

Definition 3

For an open set $\varOmega \subset \mathbb {R}^m$ with finite Lebesgue measure, $p\ge 1,\, q\in \mathbb {R}$,

• (i)
  $$\begin{aligned} F_{p,q}(\varOmega )=\frac{T_p(\varOmega )\lambda ^q_1(\varOmega )}{|\varOmega |^{\frac{1}{p}+\frac{2}{m}(1-q)}}, \end{aligned}$$

  (39)
• (ii)
  $$\begin{aligned} F_{\infty ,q}(\varOmega )=\frac{\Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}\lambda ^q_1(\varOmega )}{|\varOmega |^{\frac{2}{m}(1-q)}}, \end{aligned}$$

  (40)
• (iii)
  $$\begin{aligned} \mathfrak {F}_{p,q}=\sup \{F_{p,q}(\varOmega ):\varOmega \,\, \text {open in}\, \mathbb {R}^m, \, |\varOmega |<\infty \}, \end{aligned}$$

  (41)
• (iv)
  $$\begin{aligned} \mathfrak {F}_{\infty ,q}=\sup \{F_{\infty ,q}(\varOmega ):\varOmega \,\, \text {open in}\, \mathbb {R}^m, \, |\varOmega |<\infty \}. \end{aligned}$$

  (42)

It is straightforward to verify that the quantities defined in (39) and (40) are invariant under homotheties of $\varOmega$. Below we characterize those pairs $\{(p,q):p\ge 1\}$ for which the sharp constants defined in (41) and (42) are finite.

Theorem 4

• (i) For $1\le p<\infty$, $\mathfrak {F}_{p,q}<\infty$ if and only if $q\le 1$.

• (ii) For $p=\infty$, $\mathfrak {F}_{\infty ,q}<\infty$ if and only if $q\le 1$.

Proof

(i) We first suppose $q>1,\, 1\le p<\infty$. Let $\varOmega _n$ be the disjoint union of n balls with equal radii $r_n$, where $|\varOmega _n|=\omega _mnr_n^m=1$. Then $\lambda _1(\varOmega _n)=r_n^{-2}\lambda _1(B_1)$. By scaling we have that

$$\begin{aligned} T_p^p(\varOmega _n)=r_n^{2p}|B_1|^{-1}T_p^p(B_1). \end{aligned}$$

(43)

Hence by (43),

$$\begin{aligned} \mathfrak {F}^p_{p,q}\ge F_{p,q}^p(\varOmega _n)=|B_1|^{-1}r_n^{2p-2pq}T_p^p(B_1)\lambda _1^{pq}(B_1). \end{aligned}$$

(44)

Since $q>1$ and $r_n\downarrow 0$ as $n\rightarrow \infty$, we have that the right-hand side of (44) tends to infinity as $n\rightarrow \infty$.

Next suppose $q\le 1,\, 1\le p<\infty$. By (39), Faber–Krahn, and Theorem 1

$$\begin{aligned} F_{p,q}(\varOmega )&=\frac{T_p(\varOmega )\lambda _1(\varOmega )}{|\varOmega |^{\frac{1}{p}}}\lambda ^{q-1}_1(\varOmega ) |\varOmega |^{\frac{2}{m}(q-1)}\nonumber \\&\le \mathfrak {F}_p\lambda ^{q-1}_1(B_1) |B_1|^{\frac{2}{m}(q-1)}\nonumber \\&\le \mathfrak {F}_{\infty }\lambda ^{q-1}_1(B_1) |B_1|^{\frac{2}{m}(q-1)}. \end{aligned}$$

This proves part (i).

(ii) We first suppose $q>1$, and let $\varOmega _n$ be the set as in the proof of part (i) above. Then $\Vert v_{\varOmega _n}\Vert _{L^{\infty }(\varOmega _n)}=\frac{r_n^2}{2m}$. Hence

$$\begin{aligned} \mathfrak {F}_{\infty ,q}&\ge \frac{r_n^{2-2q}\lambda _1^q(B_1)}{2m}, \end{aligned}$$

which tends to infinity as $r_n$ tends to 0.

Next suppose $q\le 1.$ By (40), (4) and Faber–Krahn,

$$\begin{aligned} F_{\infty ,q}(\varOmega )&=\frac{\Vert v_{\varOmega }\Vert _{L^{\infty }(\varOmega )}\lambda ^q_1(\varOmega )}{|\varOmega |^{\frac{2}{m}(1-q)}}\nonumber \\&\le \mathfrak {F}_{\infty }\lambda ^{q-1}_1(\varOmega )|\varOmega |^{\frac{2}{m}(q-1)}\nonumber \\&\le \mathfrak {F}_{\infty }\lambda ^{q-1}_1(B_1)|B_1|^{\frac{2}{m}(q-1)}. \end{aligned}$$

This proves part (ii). $\square$

In general it looks very difficult to compute $\mathfrak {F}_{p,q}$ or even $\mathfrak {F}_p=\mathfrak {F}_{p,1}, p>2$, with the exception of $\mathfrak {F}_{p,0}$. G. Talenti in [15] obtained a pointwise estimate between the rearrangement of the torsion function of a generic set with finite measure and the torsion function of the ball with the same measure. In particular this estimate implies that the $L^p$ norm of the torsion function is maximised by the $L^p$ norm of the torsion function for the ball with the same measure. Hence, by (39) and (40) we have

$$\begin{aligned} \mathfrak {F}_{p,0}=\frac{T_p(B_1)}{|B_1|^{\frac{1}{p}+\frac{2}{m}}}. \end{aligned}$$

However, in the one-dimensional case we have the following result.

Theorem 5

If $m=1,\, q\le 1,\, 1\le p<\infty$, then

$$\begin{aligned} \mathfrak {F}_{p,q}=\frac{\pi ^{(4pq+1)/(2p)}}{2^{(1+3p)/p}} \bigg (\frac{\varGamma (p+1)}{\varGamma (p+\frac{3}{2})}\bigg )^{1/p}, \end{aligned}$$

(45)

and

$$\begin{aligned} \mathfrak {F}_{\infty ,q}=\frac{\pi ^{2q}}{8}. \end{aligned}$$

(46)

Proof of Theorem 5

Since $\varOmega \subset \mathbb {R}^1$ is open it is a countable union of open intervals. Since $|\varOmega |<\infty$, we let $2a_1\ge 2a_2\ge \cdots$ be the lengths of these intervals. Without loss of generality we may assume that $|\varOmega |=2\sum _{j=1}^{\infty }a_j=1$. By the first equality in (25) we have by scaling for a single interval $B_a$ of length 2a that

$$\begin{aligned} T_p(B_a)&=\frac{a^{(2p+1)/p}}{2}\bigg (2\int _{[0,1]}dr\, (1-r^2)^p\bigg )^{1/p}\nonumber \\&=\frac{a^{(2p+1)/p}\pi ^{1/(2p)}}{2}\bigg (\frac{\varGamma (p+1)}{\varGamma (p+\frac{3}{2})}\bigg )^{1/p}\nonumber \\&=a^{(2p+1)/p}c_p, \end{aligned}$$

(47)

where $c_p$ can be read-off from (47). Since $T_p^p$ is additive on disjoint open sets we have that

$$\begin{aligned} T^p_p(\varOmega )&=c^p_p\sum _{j=1}^{\infty }a_j^{2p+1}\le c^p_pa_1^{2p}\sum _{j=1}^{\infty }a_j=2^{-1}c^p_pa_1^{2p}. \end{aligned}$$

Since

$$\begin{aligned} \lambda _1(\varOmega )=\frac{\pi ^2}{4a_1^2}, \end{aligned}$$

$q\le 1$, and $2a_1\le 1$, we have that $(2a_1)^{2-2q}\le 1$. Hence

$$\begin{aligned} F_{p,q}(\varOmega )\le 2^{-1/p}c_p\bigg (\frac{\pi ^2}{4}\bigg )^qa_1^{2-2q} \le 2^{-(1+2p)/p}\pi ^{2q}c_p. \end{aligned}$$

By taking the supremum over all $\varOmega \subset \mathbb {R}^1$ with measure 1 we obtain that

$$\begin{aligned} \mathfrak {F}_{p,q}\le 2^{-(1+2p)/p}c_p\pi ^{2q}. \end{aligned}$$

(48)

To obtain a lower bound for $\mathfrak {F}_{p,q}$ we make the particular choice of $\varOmega =B_1$. This gives that

$$\begin{aligned} \mathfrak {F}_{p,q}\ge F_{p,q}(B_1)=2^{-(1+2p)/p}\pi ^{2q}c_p. \end{aligned}$$

(49)

By (48) and (49) we conclude that

$$\begin{aligned} \mathfrak {F}_{p,q}= F_{p,q}(B_1)=2^{-(1+2p)/p}\pi ^{2q}c_p. \end{aligned}$$

(50)

and (45) follows from (50) and the definition of $c_p$ in (47).

To prove (46) we just observe that the maximum of the torsion function and the first Dirichlet eigenvalue are determined by the largest interval in $\varOmega$, i.e. $a_1$. Since $q\le 1$ we maximise the resulting expression by taking $a_1=\frac{1}{2}$. $\square$

Note that as $B_1$ is convex we also have that

$$\begin{aligned} \mathfrak {F}_p^{\text {convex}}=\mathfrak {F}_{p,1}=\mathfrak {F}_p, \end{aligned}$$

(51)

and recover the known values $\mathfrak {F}_1=\frac{\pi ^2}{12}, \mathfrak {F}_{\infty }=\frac{\pi ^2}{8}$ [4, 5]. Note that $\mathfrak {F}_1<\mathfrak {F}_2=\frac{\pi ^2}{\sqrt{120}}<1$, which is in contrast with the higher dimensional situation $m\ge 2$, where $\mathfrak {F}_p=1,\,1\le p\le 2$. $\square$