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A lower bound for the first eigenvalue of a minimal hypersurface in the sphere

Asun Jiménez, Carlos Tapia Chinchay, Detang Zhou Asun Jiménez
Instituto de Matemática e Estatística, Universidade Federal Fluminense, Campus Gragoatá, São Domingos
24210-200 Niterói, Rio de Janeiro
Brazil.
[email protected] Carlos Tapia Chinchay
Instituto de Matemática e Estatística, Universidade Federal Fluminense, Campus Gragoatá, São Domingos
24210-200 Niterói, Rio de Janeiro
Brazil.
[email protected] Detang Zhou
Instituto de Matemática e Estatística, Universidade Federal Fluminense, Campus Gragoatá, São Domingos
24210-200 Niterói, Rio de Janeiro
Brazil.
[email protected]
Abstract.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1} and let Λ=maxΣ|A|\Lambda=\max\limits_{\Sigma}|A| be the norm of its second fundamental form. In this work we prove that the first eigenvalue of the Laplacian of Σ\Sigma satisfies

λ1(Σ)>m2+m(m+1)32(12Λ+m+11)2+8,\lambda_{1}(\Sigma)>\dfrac{m}{2}+\frac{m(m+1)}{32(12\Lambda+m+11)^{2}+8},

and λ1(Σ)=m\lambda_{1}(\Sigma)=m, when Λm\Lambda\leq\sqrt{m}. In particular, this estimate improves the one obtained recently in [5]. The proof of our main result is based on a Rayleigh quotient estimate for a harmonic extension of an eigenfunction of the Laplacian of Σ\Sigma in the spirit of [3].

Key words and phrases:
minimal hypersurfaces in 𝕊n\mathbb{S}^{n}, first eigenvalue of the laplacian, Yau’s conjecture
2020 Mathematics Subject Classification:
MSC 53C42 and MSC 53C44
The first author is partially suported by Conselho Nacional de Desenvolvimento Científico e Tecnológico - CNPq, code 434253/2018-9. The second author is partially supported by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior - Brasil (CAPES) Finance Code 001. The third author was partially supported by FAPERJ/Brazil [E-26/200.386/2023]) and CNPq/Brazil [403344/2021-2 & 308067/2023-1].

1. Introduction.

Let 𝕊m+1\mathbb{S}^{m+1} denote the unit sphere in m+2\mathbb{R}^{m+2}. and Σ\Sigma a closed embedded hypersurface within 𝕊m+1\mathbb{S}^{m+1}. The eigenvalues of the Laplacian operator \triangle on Σ\Sigma form a discrete set of non-negative real numbers. We denote by λ1(Σ)\lambda_{1}(\Sigma) the first nonzero eigenvalue. It is well known that Σ\Sigma is minimal if and only if all coordinate functions in m+2\mathbb{R}^{m+2} restricted on Σ\Sigma are eigenfunctions corresponding to eigenvalues mm. This implies that λ1(Σ)m\lambda_{1}(\Sigma)\leq m.

On the other hand, it is interesting and important to find the sharp lower bound of λ1\lambda_{1} for minimal hypersurfaces in 𝕊m+1\mathbb{S}^{m+1}. Yau [11] conjectured that λ1(Σ)=m\lambda_{1}(\Sigma)=m. Choi and Wang [3] proved that λ1(Σ)m2\lambda_{1}(\Sigma)\geq\frac{m}{2}. This estimate was later refined in [1] by Barros-Bessa who gave the lower bound

(1.1) λ1(Σ)>m2.\displaystyle\lambda_{1}(\Sigma)>\frac{m}{2}.

Many progress has been made towards proving Yau’s conjecture after Choi-Wang’s paper (see for instance [1], [5], [9], [4], and [12]). Despite an extensive literature relating to the study of λ1(Σ)\lambda_{1}(\Sigma) under additional assumptions on Σ\Sigma, (1.1) has remained the strongest explicit lower bound that is known to hold for a general embedded minimal hypersurface in 𝕊n+1\mathbb{S}^{n+1}. The new estimate that we obtain in this work depends on the geometry of Σ\Sigma as we explain in detail next.

Given xΣx\in\Sigma, let |A|(x):=(i=1mki2(x))1/2,|A|(x):=\left(\sum\limits_{i=1}^{m}k_{i}^{2}(x)\right)^{1/2}, where k1(x),,km(x)k_{1}(x),\ldots,k_{m}(x) are the principal curvatures of Σ\Sigma in xx with respect to ν(x)\nu(x). We call |A||A| the norm of the second fundamental form AνA_{\nu} and we define Λ:=maxΣ|A|\Lambda:=\max\limits_{\Sigma}|A|. It is known (see [8]) that in the case Λm\Lambda\leq\sqrt{m}, then λ1(Σ)=m\lambda_{1}(\Sigma)=m. Therefore our contribution concerns only the case Λ>m\Lambda>\sqrt{m}.

Precisely, we prove

Theorem 1.1.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1} and let Λ=maxΣ|A|\Lambda=\max\limits_{\Sigma}|A| be the norm of its second fundamental form. Then, the first eigenvalue of the Laplacian of Σ\Sigma satisfies

(1.2) λ1(Σ)>m2+m(m+1)32(12Λ+m+11)2+8,\lambda_{1}(\Sigma)>\dfrac{m}{2}+\frac{m(m+1)}{32(12\Lambda+m+11)^{2}+8},

and λ1(Σ)=m\lambda_{1}(\Sigma)=m, when Λm\Lambda\leq\sqrt{m}.

Remark 1.2.

A recent improvement to (1.1) is given by Duncan-Sire-Spruck in [5], where they proved that

λ1m2+a(m)Λ6+b(m),\lambda_{1}\geq\frac{m}{2}+\frac{a(m)}{\Lambda^{6}+b(m)},

for specific functions a(m,Λ)a(m,\Lambda) and b(m)b(m) (see (3.28)). By simple comparison of the order of growth it is easy to see that our estimate is bigger when Λ\Lambda is big enough. Indeed a computation at the end of this paper shows that it is bigger for every mm and Λ\Lambda.

Remark 1.3.

Our estimate depends on the norm of second fundamental form and is not sharp. It would be interesting to find a lower bound which is bigger than m2\frac{m}{2} and depends only on mm.

We can combine (1.2) with the Yang-Yau inequality [10] is an area bound for embedded minimal surfaces in 𝕊3\mathbb{S}^{3} in terms of their genus. This plays a crucial role in the compactness theory of Choi-Schoen in [2]. They find a constant C(χ)C(\chi) which is an upper bound for the norm of the second fundamental form of any compact minimal embedded minimal surface in 𝕊3\mathbb{S}^{3} with Euler number χ\chi.

On the other hand, Yau’s conjecture is true for embedded minimal surfaces in S3S^{3} which are invariant under a finite group of reflections (see [4]). And Zhao [12] proved that there is a lower bound depending on the genus. We also have the following corollary from Theorem 1.1.

Corollary 1.4.

Let C(χ)C(\chi) be the constant in Choi-Schoen’s theorem. Then the first nonzero eigenvalue of the Laplacian of a compact embedded minimal surface Σ\Sigma in 𝕊3\mathbb{S}^{3} with Euler number χ\chi satisfies

λ1(Σ)>1+316(12C(χ)+13)2+4.\lambda_{1}(\Sigma)>1+\frac{3}{16(12C(\chi)+13)^{2}+4}.

The rest of the paper is divided into two sections.

In Section 2, we recall the Reilly’s formula and reformulate a result in [1]. We also give a lower bound for λ1(Σ)\lambda_{1}(\Sigma) in terms of the Rayleigh quotient of its harmonic extension. Namely

λ1(Σ)m1+1m+1Q,\lambda_{1}(\Sigma)\geq\dfrac{m}{1+\sqrt{1-\frac{m+1}{Q}}},

where

Q:=(Ω|¯u|2𝑑V)(Ωu2𝑑V)1Q:=\left(\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV\right)\left(\displaystyle\int_{\Omega}u^{2}dV\right)^{-1}

and uu is a solution to

{¯u=0inΩu=φ,inΣ,\left\{\begin{array}[]{ll}\overline{\triangle}u=0&\textrm{in}\,\Omega\\ u=\varphi,&\textrm{in}\,\Sigma,\end{array}\right.

where the bars in the expressions above refer to operations in the ambient sphere. Here, φ\varphi is an eigenfunction of the Laplacian \triangle associated to the eigenvalue λ1(Σ)\lambda_{1}(\Sigma) and Ω\Omega is a component of 𝕊m+1Σ\mathbb{S}^{m+1}-\Sigma which is chosen appropriately. We call uu the harmonic extension of φ\varphi to Ω\Omega. Then we prove Theorem 1.1 assuming the validity of an appropriate estimate for QQ.

In Section 3 we make a quick review of the normal exponential map and we prove the estimate

(1.3) Ωu2𝑑VC2(m,Λ)Ω|¯u|2𝑑V,\displaystyle\int_{\Omega}u^{2}dV\geq C_{2}(m,\Lambda)\int_{\Omega}|\overline{\nabla}u|^{2}dV,

which may be considered as an inverse Poincaré type inequality, i.e. C2C_{2} is an upper estimate for QQ. We finish the paper by comparing our estimate with the one in [5].

It is important to note that, as in [5], we also use Reilly’s formula and an upper bound of the mean curvature of the parallel surfaces to Σ\Sigma. However, we provide an upper bound of |¯u|L2(Ω)2|\overline{\nabla}u|_{L^{2}(\Omega)}^{2} that depends only on the geometry of Σ\Sigma, via an elementary result on harmonic extensions.

We need to point out that our techniques can be improved and generalized in order to obtain an estimate for the first eigenvalue of a minimal surface embedded in an ambient of bounded sectional curvature. However, the sharp bound can only be achieved by proving that Q=m+1Q=m+1. Further work will be part of the PhD thesis of the second author. In fact, this work was in preparation before we had access to [5].

2. A first eigenvalue estimate via Rayleigh quotient

In this section we will review Reilly’s formula and give a lower bound for the nonzero first eigenvalue λ1(Σ)\lambda_{1}(\Sigma) in terms of the Rayleigh quotient of the harmonic extension of the corresponding eigenfunction. As a consequence, we will prove our main result, Theorem 1.1, assuming the inequality (1.3) which will be explicitly stated in Theorem 3.10.

From now on, Σ\Sigma will denote a closed embedded hypersurface of 𝕊m+1\mathbb{S}^{m+1}. It follows that Σ\Sigma divides the sphere into two components Ω1\Omega_{1} and Ω2\Omega_{2}, where Ω1=Ω2=Σ\partial\Omega_{1}=\partial\Omega_{2}=\Sigma (see [3]). Set ν\nu as the unit normal of Σ\Sigma pointing outward to Ω1\Omega_{1} (ν-\nu as the unit normal of Σ\Sigma and pointing outward to Ω2\Omega_{2}) and AνA_{\nu} the second fundamental form of Σ\Sigma with respect to ν\nu.

Let φC(Σ)\varphi\in C^{\infty}(\Sigma). We can assume, without loss of generality, it satisfies the property

ΣAνφ,φ𝑑S0\int_{\Sigma}\langle A_{\nu}\nabla\varphi,\nabla\varphi\rangle dS\geq 0

and denote Ω:=Ω1\Omega:=\Omega_{1}. Otherwise we can choose Ω=Ω2.\Omega=\Omega_{2}. Let us denote all the functions of class C2C^{2} that extend the function φ\varphi over Ω¯\overline{\Omega} as Cφ2(Ω¯)C^{2}_{\varphi}(\overline{\Omega}). The following equation is known as Reilly’s formula (see [7]).

Lemma 2.1.

For all uCφ2(Ω¯)u\in C^{2}_{\varphi}(\overline{\Omega}) we have

(2.1) Ω[(¯u)2|¯2u|2Ric𝕊m+1(1)(¯u,¯u)]𝑑V=Σ[Aνφ,φ+2uνφ+mHΣ(uν)2]𝑑S.\displaystyle\int_{\Omega}\left[(\overline{\triangle}u)^{2}-|\overline{\nabla}^{2}u|^{2}-\textrm{Ric}_{\mathbb{S}^{m+1}(1)}(\overline{\nabla}u,\overline{\nabla}u)\right]dV=\displaystyle\int_{\Sigma}\left[\langle A_{\nu}\nabla\varphi,\nabla\varphi\rangle+2\frac{\partial u}{\partial\nu}\triangle\varphi+mH_{\Sigma}(u_{\nu})^{2}\right]dS.

where ¯u,¯u\overline{\triangle}u,\overline{\nabla}u and ¯2u\overline{\nabla}^{2}u denote the Laplacian, gradient and Hessian of uu in Ω\Omega, while φ\triangle\varphi and φ\nabla\varphi denote the Laplacian and the gradient of φ\varphi in Σ\Sigma with respect to the induced metric of Ω\Omega. On the other hand, uν:=u,ν\frac{\partial u}{\partial\nu}:=\langle\nabla u,\nu\rangle denotes the outward normal derivative of uu in Σ\Sigma, Ric𝕊m+1\textrm{Ric}_{\mathbb{S}^{m+1}} is the Ricci tensor of 𝕊m+1\mathbb{S}^{m+1} and HΣ:=trace(Aν)H_{\Sigma}:=\textrm{trace}(A_{\nu}) is the mean curvature of Σ\Sigma.

The following corollary follows from (2.1) by using that |¯2u|21m+1(¯u)2|\overline{\nabla}^{2}u|^{2}\geq\frac{1}{m+1}(\overline{\triangle}u)^{2}.

Corollary 2.2.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1}. For any vC2(Ω¯)v\in C^{2}(\overline{\Omega}) we assume Ω\Omega is chosen so that ΩAν((v|Ω)),(v|Ω)𝑑S0\int_{\partial\Omega}\langle A_{\nu}(\nabla(v|_{\partial\Omega})),\nabla(v|_{\partial\Omega})\rangle dS\geq 0. Then

(2.2) Ω[mm+1(¯v)2m|¯v|2]𝑑V2Ωvν(v|Ω)𝑑S0.\displaystyle\int_{\Omega}\left[\frac{m}{m+1}(\overline{\triangle}v)^{2}-m|\overline{\nabla}v|^{2}\right]dV\displaystyle-2\int_{\partial\Omega}\frac{\partial v}{\partial\nu}\triangle(v|_{\partial\Omega})dS\geq 0.

The following result is proven by Barros and Bessa in [1]. We include here a simpler proof.

Lemma 2.3.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1}. Assume that φ\varphi is a eigenfunction of \triangle on Σ\Sigma corresponding to the eigenvalue λ1(Σ)\lambda_{1}(\Sigma) and that uu is its harmonic extension to Ω\Omega, i.e.

(2.3) {¯u=0inΩ,u=φinΣ.\left\{\begin{array}[]{ll}\overline{\triangle}u=0&\textrm{in}\,\Omega,\\ u=\varphi&\textrm{in}\,\Sigma.\end{array}\right.

Then for all tt\in\mathbb{R},

(2.4) (2λ1(Σ)m)Qt2+2λ1(Σ)t+mm+10,(2\lambda_{1}(\Sigma)-m)Qt^{2}+2\lambda_{1}(\Sigma)t+\dfrac{m}{m+1}\geq 0,

where QQ is defined by

(2.5) Q:=Ω|¯u|2𝑑VΩu2𝑑V.Q:=\frac{\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV}{\displaystyle\int_{\Omega}u^{2}dV}.
Proof.

For t0t\neq 0, let gg be the solution of the problem

{g=u,inΩ,g=tφ,inΣ,\left\{\begin{array}[]{ll}\triangle g=u,&\textrm{in}\,\Omega,\\ g=t\varphi,&\textrm{in}\,\Sigma,\end{array}\right.

where uu is a solution to (2.3). Then, from (2.2) applied to gg, we have

(2.6) Ω(mm+1u2m|¯g|2)𝑑V+2tλ1(Σ)Ωφgν𝑑S0.\displaystyle\int_{\Omega}\left(\frac{m}{m+1}u^{2}-m|\overline{\nabla}g|^{2}\right)dV\displaystyle+2t\lambda_{1}(\Sigma)\int_{\partial\Omega}\varphi\frac{\partial g}{\partial\nu}dS\geq 0.

On the other hand, by (2.3) and Stokes’ formula

(2.7) Ω¯g,¯u𝑑V=ΩgΔ¯u𝑑V+Σguν𝑑S=tΣφuν𝑑S=tΩ|¯u|2𝑑V.\displaystyle\int_{\Omega}\langle\overline{\nabla}g,\overline{\nabla}u\rangle dV=-\displaystyle\int_{\Omega}g\overline{\Delta}u\,dV+\displaystyle\int_{\Sigma}g\frac{\partial u}{\partial\nu}dS=t\displaystyle\int_{\Sigma}\varphi\frac{\partial u}{\partial\nu}dS=t\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}\,dV.

Hence, by (2.7),

0Ω|¯gt¯u|2𝑑V=Ω(|¯g|22t¯g,¯u+t2|¯u|2)𝑑V=Ω|¯g|2𝑑Vt2Ω|¯u|2𝑑V.\begin{array}[]{rcl}0\leq\displaystyle\int_{\Omega}|\overline{\nabla}g-t\overline{\nabla}u|^{2}\,dV&=&\displaystyle\int_{\Omega}\left(|\overline{\nabla}g|^{2}-2t\langle\overline{\nabla}g,\overline{\nabla}u\rangle+t^{2}|\overline{\nabla}u|^{2}\right)dV\\ &=&\displaystyle\int_{\Omega}|\overline{\nabla}g|^{2}dV-t^{2}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV.\end{array}

Therefore

(2.8) Ω|¯g|2𝑑Vt2Ω|¯u|2𝑑V.\displaystyle\int_{\Omega}|\overline{\nabla}g|^{2}dV\geq t^{2}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV.

Similarly

(2.9) Σφgν𝑑S=Ω¯u,¯g𝑑V+ΩuΔ¯g𝑑V=tΩ|¯u|2𝑑V+Ωu2𝑑V.\displaystyle\int_{\Sigma}\varphi\frac{\partial g}{\partial\nu}dS=\displaystyle\int_{\Omega}\langle\overline{\nabla}u,\overline{\nabla}g\rangle dV+\displaystyle\int_{\Omega}u\overline{\Delta}g\,dV=t\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}\,dV+\displaystyle\int_{\Omega}u^{2}\,dV.

From (2.6) and (2.9) we have

Ω(mm+1u2m|¯g|2)𝑑V+2t2λ1(Σ)Ω|¯u|2𝑑V+2tλ1(Σ)Ωu2𝑑V0,\displaystyle\int_{\Omega}\left(\frac{m}{m+1}u^{2}-m|\overline{\nabla}g|^{2}\right)dV\displaystyle+2t^{2}\lambda_{1}(\Sigma)\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}\,dV+2t\lambda_{1}(\Sigma)\displaystyle\int_{\Omega}u^{2}\,dV\geq 0,

and so from (2.8),

(2.10) (2λ1(Σ)m)t2Ω|¯u|2𝑑V+2λ1(Σ)tΩu2𝑑V+mm+1Ωu2𝑑V0.(2\lambda_{1}(\Sigma)-m)t^{2}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV+2\lambda_{1}(\Sigma)t\displaystyle\int_{\Omega}u^{2}dV+\dfrac{m}{m+1}\displaystyle\int_{\Omega}u^{2}dV\geq 0.

Then (2.4) follows by dividing this last inequality by Ωu2𝑑V\displaystyle\int_{\Omega}u^{2}dV and using the definition of QQ in (2.5). ∎

Next we obtain a first estimate for λ1(Σ)\lambda_{1}(\Sigma) which is a corollary of Barros-Bessa’s theorem.

Theorem 2.4.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1}. Then

(2.11) λ1(Σ)m1+1m+1Q.\lambda_{1}(\Sigma)\geq\dfrac{m}{1+\sqrt{1-\frac{m+1}{Q}}}.
Proof.

First note that, choosing any tt in (2.4) such that 2λ1(Σ)t+mm+1<02\lambda_{1}(\Sigma)t+\dfrac{m}{m+1}<0, then we trivially have λ1(Σ)>m2\lambda_{1}(\Sigma)>\frac{m}{2}. On the other hand, by choosing t=λ1(Σ)(m+1)(2λ1(Σ)m)t=\frac{-\lambda_{1}(\Sigma)}{(m+1)(2\lambda_{1}(\Sigma)-m)} in (2.4) we have that

Qλ12(Σ)(m+1)2(2λ1(Σ)m)2λ12(Σ)(m+1)(2λ1(Σ)m)mm+1=1(m+1)(2λ1(Σ)m)(2λ12(Σ)m(2λ1(Σ)m))=1(m+1)(2λ1(Σ)m)(λ12(Σ)+(λ1(Σ)m)2)λ12(Σ)(m+1)(2λ1(Σ)m).\begin{split}\frac{Q\lambda^{2}_{1}(\Sigma)}{(m+1)^{2}(2\lambda_{1}(\Sigma)-m)}&\geq\frac{2\lambda^{2}_{1}(\Sigma)}{(m+1)(2\lambda_{1}(\Sigma)-m)}-\frac{m}{m+1}\\ &=\frac{1}{(m+1)(2\lambda_{1}(\Sigma)-m)}\left(2\lambda^{2}_{1}(\Sigma)-m(2\lambda_{1}(\Sigma)-m)\right)\\ &=\frac{1}{(m+1)(2\lambda_{1}(\Sigma)-m)}\left(\lambda^{2}_{1}(\Sigma)+(\lambda_{1}(\Sigma)-m)^{2}\right)\\ &\geq\frac{\lambda^{2}_{1}(\Sigma)}{(m+1)(2\lambda_{1}(\Sigma)-m)}.\end{split}

Hence Qm+1Q\geq m+1 (see also [1]) and so (2.11) is well defined.

Again from (2.4) we have

2λ1(Σ)(Qt2+t)mQt2mm+1.2\lambda_{1}(\Sigma)(Qt^{2}+t)\geq mQt^{2}-\dfrac{m}{m+1}.

Then

(2.12) λ1(Σ)m2maxt(Qt+1)>0β(t)\lambda_{1}(\Sigma)\geq\frac{m}{2}\max\limits_{t(Qt+1)>0}\beta(t)

where β(t):=11Qt+11(m+1)(Qt2+t)\beta(t):=1-\frac{1}{Qt+1}-\frac{1}{(m+1)(Qt^{2}+t)}. Note that β(t)=Q(m+1)t2+2Qt+1(m+1)t2(Qt+1)2\beta^{\prime}(t)=\frac{Q(m+1)t^{2}+2Qt+1}{(m+1)t^{2}(Qt+1)^{2}}. We have that at the points where t(Qt+1)>0t(Qt+1)>0, β(t)=0\beta^{\prime}(t)=0 if and only if

t=2Q4Q24Q(m+1)2Q(m+1)=11m+1Qm+1=:t0.t=\frac{-2Q-\sqrt{4Q^{2}-4Q(m+1)}}{2Q(m+1)}=\dfrac{-1-\sqrt{1-\frac{m+1}{Q}}}{m+1}=:t_{0}.

Note that, in particular, β\beta has no critical points in the interval (0,+)(0,+\infty). It follows that

(2.13) maxt(Qt+1)>0β(t)=β(t0)=Q(m+1)t021(m+1)t0(Qt0+1)=2t0(m+1)=21+1m+1Q.\max\limits_{t(Qt+1)>0}\beta(t)=\beta(t_{0})=\dfrac{Q(m+1)t_{0}^{2}-1}{(m+1)t_{0}(Qt_{0}+1)}=\frac{-2}{t_{0}(m+1)}=\frac{2}{1+\sqrt{1-\frac{m+1}{Q}}}.

The above equality is valid from the fact that β(t0)=0\beta^{\prime}(t_{0})=0, i.e. Q(m+1)t02+2Qt0+1=0Q(m+1)t_{0}^{2}+2Qt_{0}+1=0.

Therefore, from (2.12) and (2.13) we have (2.11).

Next we prove Theorem 1.1 assuming Theorem 3.10.

Proof of Theorem 1.1.

It is well known that when Λm\Lambda\leq\sqrt{m}, Σ\Sigma is either a great sphere 𝕊n\mathbb{S}^{n} or a Clifford torus and so λ1(Σ)=m\lambda_{1}(\Sigma)=m. Therefore, we can now assume that Λm\Lambda\geq\sqrt{m}. Let φC(Σ)\varphi\in C^{\infty}(\Sigma) be the eigenfunction corresponding to the first nonzero eigenvalue λ1(Σ)\lambda_{1}(\Sigma) and uCφ2(Ω¯)u\in C^{2}_{\varphi}(\overline{\Omega}) its harmonic extension to Ω\Omega as before. It follows from Theorem 3.10 that Q<4(12Λ+m+11)2+1.Q<4(12\Lambda+m+11)^{2}+1. Therefore we have from (2.11) that

(2.14) λ1(Σ)m1+1(m+1)Q1=m2+m(11+1(m+1)Q112).\lambda_{1}(\Sigma)\geq\dfrac{m}{1+\sqrt{1-(m+1)Q^{-1}}}=\dfrac{m}{2}+m\left(\dfrac{1}{1+\sqrt{1-(m+1)Q^{-1}}}-\dfrac{1}{2}\right).

On the other hand, since for all 0x<10\leq x<1,

x8<11+1x12,\frac{x}{8}<\frac{1}{1+\sqrt{1-x}}-\frac{1}{2},

we can consider x=(m+1)Q1x=(m+1)Q^{-1} in (2.14) and deduce from Theorem 3.10 that

(2.15) λ1(Σ)>m2+m(m+1)8Q1>m2+m(m+1)32(12Λ+m+11)2+8.\lambda_{1}(\Sigma)>\frac{m}{2}+\frac{m(m+1)}{8}Q^{-1}>\frac{m}{2}+\frac{m(m+1)}{32(12\Lambda+m+11)^{2}+8}.

The proof is then complete.

\square

3. Gradient estimate via an inverse Poincaré - type inequality.

The aim of this section is to prove an inverse Poincaré-type inequality in Theorem 3.10. To do that we recall first some preliminary results concerning the normal exponential map.

In what follows NΣ:={(x,v):xΣ,vTxΣ}N\Sigma:=\{(x,v):\,x\in\Sigma,v\in T_{x}^{\bot}\Sigma\} will denote the normal bundle of Σ\Sigma, UΣ:={(x,v)NΣ:|v|=1}U\Sigma:=\{(x,v)\in N\Sigma:\,|v|=1\} will denote the normal unit bundle of Σ\Sigma and exp:NΣ𝕊m+1\textrm{exp}^{\bot}:N\Sigma\to\mathbb{S}^{m+1} defined by exp(x,v):=expx(v)\textrm{exp}^{\bot}(x,v):=\exp_{x}(v) will denote the normal exponential map on Σ\Sigma. Such map is well defined in Σ\Sigma since Σ\Sigma is embedded with compact closure on the sphere 𝕊m+1\mathbb{S}^{m+1} (see for instance [6] ). Let θΣ:NΣ\theta_{\Sigma}:N\Sigma\to\mathbb{R} denote the Jacobian determinant of the normal exponential map exp\textrm{exp}^{\bot}. On the other hand, let Φt:Σ𝕊m+1\Phi_{t}:\Sigma\to\mathbb{S}^{m+1} be defined by Φt(x):=exp(x,tν(x))\Phi_{t}(x):=\textrm{exp}^{\bot}(x,t\nu(x)) and

Σt:=Φt(Σ)={exp(x,tν(x)):xΣ}.\Sigma_{t}:=\Phi_{t}(\Sigma)=\{\textrm{exp}^{\bot}(x,t\nu(x)):x\in\Sigma\}.

From Proposition 2.1 in [5], if we define kmax:=maxΣ,i|ki|k_{\textrm{max}}:=\max\limits_{\Sigma,i}|k_{i}| and

(3.1) TΣ:=arctan(kmax1),T_{\Sigma}:=\arctan(k_{\textrm{max}}^{-1}),

we have

TΣsup{t>0|Φt:ΣΣtis a diffeomorphism}=:t.T_{\Sigma}\leq\sup\{t>0|\,\Phi_{t}:\Sigma\to\Sigma_{t}\,\textrm{is a diffeomorphism}\}=:t_{\ast}.

Similarly, we have

TΣinf{t<0|Φt:ΣΣtis a diffeomorphism}:=t.T_{\Sigma}\leq-\inf\{t<0|\,\Phi_{t}:\Sigma\to\Sigma_{t}\,\textrm{is a diffeomorphism}\}:=t^{\ast}.

Defining minfoc(Σ):=min{t,t}\textrm{minfoc}(\Sigma):=\min\{t_{\ast},t^{\ast}\}, we have that

TΣminfoc(Σ).T_{\Sigma}\leq\textrm{minfoc}(\Sigma).

The following lemma correspond to Lemma 10.9 in [6].

Lemma 3.1.

For all 0t<minfoc(Σ)0\leq t<\textrm{minfoc}(\Sigma)

ddtlnθΣ(x,tν(x))i=1mki(x)=0,\dfrac{d}{dt}\ln\theta_{\Sigma}(x,t\nu(x))\leq\sum\limits_{i=1}^{m}k_{i}(x)=0,

where k1(x),,km(x)k_{1}(x),\ldots,k_{m}(x) are the principal curvatures of Σ\Sigma in xx with respect to ν(x)\nu(x). In addition

(3.2) θ(t,x):=θΣ(x,tν(x))θΣ(x,0)=1.\theta(t,x):=\theta_{\Sigma}(x,t\nu(x))\leq\theta_{\Sigma}(x,0)=1.

The proof of the following Lemma is straightforward from Lemma 8.1 in [6] and the definition of θΣ\theta_{\Sigma}.

Lemma 3.2.

For all 0<b<minfoc(Σ)0<b<\textrm{minfoc}(\Sigma), the application Φ:[0,b]×ΣΦ([0,b]×Σ)\Phi:[0,b]\times\Sigma\to\Phi([0,b]\times\Sigma) defined by Φ(t,x):=Φt(x)\Phi(t,x):=\Phi_{t}(x) is a diffeomorphism. In particular for any continuous function FF over Φ([0,b]×Σ)\Phi([0,b]\times\Sigma) we have

(3.3) Φ([0,b]×Σ)F(y)𝑑V=0bΣF(Φ(t,x))θ(t,x)𝑑S𝑑t,\int_{\Phi([0,b]\times\Sigma)}F(y)dV=\int\limits_{0}^{b}\int\limits_{\Sigma}F(\Phi(t,x))\theta(t,x)dSdt,

where θ(t,x)=θΣ(x,tν(x))\theta(t,x)=\theta_{\Sigma}(x,t\nu(x)).

Lemma 3.3.

For each 0t<TΣ0\leq t<T_{\Sigma} and xΣx\in\Sigma we have that

costki(x)sint=sin(θi(x)t)sinθi(x)sin(TΣt)sinTΣ>0,\cos t-k_{i}(x)\,\sin\,t=\dfrac{\sin\,(\theta_{i}(x)-t)}{\sin\,\theta_{i}(x)}\geq\dfrac{\sin\,(T_{\Sigma}-t)}{\sin\,T_{\Sigma}}>0,

where cotθi(x)=ki(x)\cot\theta_{i}(x)=k_{i}(x).

Proof.

By the definition of TΣT_{\Sigma} in (3.1) we have that ki(x)|ki(x)|kmax=cot(TΣ).k_{i}(x)\leq|k_{i}(x)|\leq k_{\textrm{max}}=\cot(T_{\Sigma}). Therefore

costki(x)sintcostcotTΣsint=sin(TΣt)sinTΣ>0,\cos t-k_{i}(x)\,\textrm{sin}\,t\geq\cos t-\cot T_{\Sigma}\,\sin\,t=\dfrac{\sin\,(T_{\Sigma}-t)}{\sin\,T_{\Sigma}}>0,

where we have used that 0<TΣtTΣ<π/20<T_{\Sigma}-t\leq T_{\Sigma}<\pi/2. ∎

Given φC(Σ)\varphi\in C^{\infty}(\Sigma), we define φ~(Φt(x)):=φ(x)\widetilde{\varphi}(\Phi_{t}(x)):=\varphi(x). The function φ~\widetilde{\varphi} is called the normal extension of φ\varphi and it is well defined in the set

{y𝕊m+1:y=Φt(x),xΣ,|t|<TΣ}.\{y\in\mathbb{S}^{m+1}:y=\Phi_{t}(x),\,x\in\Sigma,\,|t|<T_{\Sigma}\}.
Lemma 3.4.

For each 0t<TΣ0\leq t<T_{\Sigma} and xΣx\in\Sigma we have

|Tφ~|2(Φt(x))sin2TΣsin2(TΣt)|φ|2(x),|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))\leq\dfrac{\sin^{2}\,T_{\Sigma}}{\sin^{2}\,(T_{\Sigma}-t)}|\nabla\varphi|^{2}(x),

where φ~\widetilde{\varphi} is the normal extension of φ\varphi into Ω\Omega, Tφ~(Φt(x))\nabla^{T}\widetilde{\varphi}(\Phi_{t}(x)) denotes the gradient of φ~|Σt\widetilde{\varphi}|_{\Sigma_{t}} in y=Φt(x)y=\Phi_{t}(x) and Φt(x)=Φ(t,x)\Phi_{t}(x)=\Phi(t,x).

Proof.

For xΣx\in\Sigma, let {Ei(t):=Ptei}\{E_{i}(t):=P_{t}e_{i}\} be an orthonormal basis of TΦt(x)ΣtT_{\Phi_{t}(x)}\Sigma_{t} where {ei}\{e_{i}\} is an orthonormal basis of TxΣT_{x}\Sigma such that Aν(x)ei=ki(x)eiA_{\nu(x)}e_{i}=k_{i}(x)e_{i} and PteiP_{t}e_{i} the parallel transport of eie_{i} along the geodesic γx(t):=Φt(x)\gamma_{x}(t):=\Phi_{t}(x) from γx(0)=x\gamma_{x}(0)=x to γx(t)=Φt(x)\gamma_{x}(t)=\Phi_{t}(x). It follows that

(3.4) d(Φt)xei=Pt((cost)ei+(sint)ν(x)ei)=Pt((cost)ei(sint)Aν(x)ei)=(costki(x)sint)Ei(t).\begin{array}[]{rcl}d(\Phi_{t})_{x}e_{i}&=&P_{t}((\cos t)\,e_{i}+(\sin\,t)\,\nu^{\prime}(x)e_{i})\\ &=&P_{t}((\cos t)\,e_{i}-(\sin\,t)\,A_{\nu(x)}e_{i})\\ &=&(\cos t-k_{i}(x)\sin\,t)E_{i}(t).\end{array}

We have from (3.4) that

|Tφ~|2(Φt(x))=iTφ~(Φt(x)),Ei(t)2=iTφ~(Φt(x)),d(Φt)xeicostki(x)sint2=iTφ~(Φ(t,x)),dΦ(t,x)(0,ei)2(costki(x)sint)2=i(d(φ~|ΣtΦ)(t,x)(0,ei))2(costki(x)sint)2=i(0,φ(x)),(0,ei)2(costki(x)sint)2.\begin{array}[]{rcl}|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))=\sum\limits_{i}\langle\nabla^{T}\widetilde{\varphi}(\Phi_{t}(x)),E_{i}(t)\rangle^{2}&=&\sum\limits_{i}\langle\nabla^{T}\widetilde{\varphi}(\Phi_{t}(x)),\dfrac{d(\Phi_{t})_{x}e_{i}}{\cos t-k_{i}(x)\sin\,t}\rangle^{2}\\ &&\\ &=&\sum\limits_{i}\dfrac{\langle\nabla^{T}\widetilde{\varphi}(\Phi(t,x)),d\Phi_{(t,x)}(0,e_{i})\rangle^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}\\ &&\\ &=&\sum\limits_{i}\dfrac{(d\left(\widetilde{\varphi}|_{\Sigma_{t}}\circ\Phi\right)_{(t,x)}(0,e_{i}))^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}\\ &&\\ &=&\sum\limits_{i}\dfrac{\langle(0,\nabla\varphi(x)),(0,e_{i})\rangle^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}.\end{array}

And so

(3.5) |Tφ~|2(Φt(x))=iφ(x),ei2(costki(x)sint)2.|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))=\sum\limits_{i}\dfrac{\langle\nabla\varphi(x),e_{i}\rangle^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}.

Using Lemma 3.3 in formula (3.5), the proof of the lemma is concluded. ∎

Next we are going to construct a transition function wich will be a key technical tool in the proof of our main result. For any a<ba<b we define

(3.6) ψa,b(t):=1g(t2a2b2a2),\psi_{a,b}(t):=1-g\left(\dfrac{t^{2}-a^{2}}{b^{2}-a^{2}}\right),

where the function g:[0,1]g:\mathbb{R}\to[0,1] is defined by

g(t):=f(t)f(t)+f(1t),f(t):={e1/t,t>00,t0.g(t):=\dfrac{f(t)}{f(t)+f(1-t)},\qquad f(t):=\left\{\begin{array}[]{ll}e^{-1/t},&t>0\\ 0,&t\leq 0\end{array}\right..

It follows that

  1. (1)

    g(t)0g(t)\geq 0 t.\forall t\in\mathbb{R}.

  2. (2)

    g(t)=0g(t)=0 t(,0].\forall t\in(-\infty,0].

  3. (3)

    limt0+g(t)=0\lim\limits_{t\to 0^{+}}g(t)=0 and limt1g(t)=1\lim\limits_{t\to 1^{-}}g(t)=1.

  4. (4)

    The function g(t)=e1t(1t)(12t+2t2)(e1tt+e1t)2(t1)2t2,g^{\prime}(t)=\dfrac{e^{\frac{1}{t(1-t)}}(1-2t+2t^{2})}{(e^{\frac{1}{t-t}}+e^{\frac{1}{t}})^{2}(t-1)^{2}t^{2}}, is such that 0<g(t)20<g^{\prime}(t)\leq 2, for all t(0,1)t\in(0,1) and limt0+g(t)=limt1g(t)=0.\lim\limits_{t\to 0^{+}}g^{\prime}(t)=\lim\limits_{t\to 1^{-}}g^{\prime}(t)=0.

Refer to caption
Figure 1. Graph of ψa,b\psi_{a,b}.

For t[a,b],t\in[a,b], it follows from (3.6) that

|ψa,b(t)|=g(t2a2b2a2)2tb2a22(2tb2a2)=4tb2a2.|\psi_{a,b}^{\prime}(t)|=g^{\prime}\left(\dfrac{t^{2}-a^{2}}{b^{2}-a^{2}}\right)\dfrac{2t}{b^{2}-a^{2}}\leq 2\left(\dfrac{2t}{b^{2}-a^{2}}\right)=\dfrac{4t}{b^{2}-a^{2}}.

Therefore

ab(ψa,b(t))2𝑑t16(b2a2)2abt2𝑑t=16(b3a3)3(b2a2)2.\displaystyle\int_{a}^{b}(\psi_{a,b}^{\prime}(t))^{2}dt\leq\frac{16}{(b^{2}-a^{2})^{2}}\displaystyle\int_{a}^{b}t^{2}dt=\dfrac{16(b^{3}-a^{3})}{3(b^{2}-a^{2})^{2}}.

In particular, if we denote ψ¯ρ,c:=ψa,b\overline{\psi}_{\rho,c}:=\psi_{a,b} for the special choices a=bca=\frac{b}{c} for some c>1c>1, and b=ρTΣb=\rho T_{\Sigma} for some 0<ρ<10<\rho<1, we have that

(3.7) ρTΣcρTΣ(ψ¯ρ,c(t))2𝑑t163ρTΣc(c31)(c21)2.\displaystyle\int_{\frac{\rho T_{\Sigma}}{c}}^{\rho T_{\Sigma}}(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}dt\leq\dfrac{16}{3\rho T_{\Sigma}}\dfrac{c(c^{3}-1)}{(c^{2}-1)^{2}}.
Lemma 3.5.

For any 0<ρ<10<\rho<1 and c>1c>1, the function vρ,c:Ω¯v_{\rho,c}:\overline{\Omega}\to\mathbb{R} defined by

(3.8) vρ,c(Φt(x))=ψ¯ρ,c(t)φ(x)v_{\rho,c}(\Phi_{t}(x))=\overline{\psi}_{\rho,c}(t)\varphi(x)

satisfies

(3.9) Ω|¯vρ,c|2𝑑V=0ρTΣΣ(ψ¯ρ,c(t))2φ2(x)θ(t,x)𝑑S𝑑t+0ρTΣΣ(ψ¯ρ,c(t))2|Tφ~|2(Φt(x))θ(t,x)𝑑S𝑑t.\displaystyle\int_{\Omega}|\overline{\nabla}v_{\rho,c}|^{2}dV=\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}\varphi^{2}(x)\theta(t,x)dS\,dt+\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}(\overline{\psi}_{\rho,c}(t))^{2}|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))\theta(t,x)dS\,dt.
Proof.

Let v¯ρ,c:[0,ρTΣ]×Σ\overline{v}_{\rho,c}:[0,\rho T_{\Sigma}]\times\Sigma\to\mathbb{R} be the function defined by v¯ρ,c(t,x):=ψ¯ρ,c(t)φ(x)\overline{v}_{\rho,c}(t,x):=\overline{\psi}_{\rho,c}(t)\varphi(x). It follows that v¯ρ,c=vρ,cΦ\overline{v}_{\rho,c}=v_{\rho,c}\circ\Phi and

(3.10) ¯v¯ρ,c(t,x)=(ψ¯ρ,c(t)φ(x),ψ¯ρ,c(t)φ(x)).\overline{\nabla}\overline{v}_{\rho,c}(t,x)=(\overline{\psi}_{\rho,c}^{\prime}(t)\varphi(x),\overline{\psi}_{\rho,c}(t)\nabla\varphi(x)).

For t[0,ρTΣ]t\in[0,\rho T_{\Sigma}] and xΣx\in\Sigma, consider {Ei(t)}\{E_{i}(t)\} be as in the proof of Lemma 3.4. Then

|¯vρ,c|2(Φt(x))=i¯vρ,c(Φt(x)),Ei(t)2+¯vρ,c(Φt(x)),γx(t)2=i¯vρ,c(Φ(t,x)),dΦ(t,x)(0,ei)costki(x)sint2+¯vρ,c(Φ(t,x)),dΦ(t,x)(1,0)2,\begin{array}[]{rcl}|\overline{\nabla}v_{\rho,c}|^{2}(\Phi_{t}(x))&=&\sum\limits_{i}\langle\overline{\nabla}v_{\rho,c}(\Phi_{t}(x)),E_{i}(t)\rangle^{2}+\langle\overline{\nabla}v_{\rho,c}(\Phi_{t}(x)),\gamma_{x}^{\prime}(t)\rangle^{2}\\ &&\\ &=&\sum\limits_{i}\langle\overline{\nabla}v_{\rho,c}(\Phi(t,x)),\dfrac{d\Phi_{(t,x)}(0,e_{i})}{\cos t-k_{i}(x)\sin\,t}\rangle^{2}+\langle\overline{\nabla}v_{\rho,c}(\Phi(t,x)),d\Phi_{(t,x)}(1,0)\rangle^{2},\end{array}

where the last equality is a consequence of (3.4). It follows that

|¯vρ,c|2(Φt(x))=i¯vρ,c(Φ(t,x)),dΦ(t,x)(0,ei)(costki(x)sint)22+¯vρ,c(Φ(t,x)),dΦ(t,x)(1,0)2=i(d(vρ,cΦ)(t,x)(0,ei))2(costki(x)sint)2+(d(vρ,cΦ)(t,x)(1,0))2=i¯v¯ρ,c(t,x),(0,ei)2(costki(x)sint)2+¯v¯ρ,c(t,x),(1,0)2.\begin{array}[]{rcl}|\overline{\nabla}v_{\rho,c}|^{2}(\Phi_{t}(x))&=&\sum\limits_{i}\dfrac{\langle\overline{\nabla}v_{\rho,c}(\Phi(t,x)),d\Phi_{(t,x)}(0,e_{i})}{(\cos t-k_{i}(x)\sin\,t)^{2}}\rangle^{2}+\langle\overline{\nabla}v_{\rho,c}(\Phi(t,x)),d\Phi_{(t,x)}(1,0)\rangle^{2}\\ &&\\ &=&\sum\limits_{i}\dfrac{\left(d(v_{\rho,c}\circ\Phi)_{(t,x)}(0,e_{i})\right)^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}+\left(d(v_{\rho,c}\circ\Phi)_{(t,x)}(1,0)\right)^{2}\\ &&\\ &=&\sum\limits_{i}\dfrac{\langle\overline{\nabla}\overline{v}_{\rho,c}(t,x),(0,e_{i})\rangle^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}+\langle\overline{\nabla}\overline{v}_{\rho,c}(t,x),(1,0)\rangle^{2}.\end{array}

From (3.10) we have

(3.11) |¯vρ,c|2(Φt(x))=i(ψ¯ρ,c(t)φ(x),ei)2(costki(x)sint)2+(ψ¯ρ,c(t)φ(x))2=(ψ¯ρ,c(t))2iφ(x),ei2(costki(x)sint)2+(ψ¯ρ,c(t))2φ2(x)=(ψ¯ρ,c(t))2|Tφ~|2(Φt(x))+(ψ¯ρ,c(t))2φ2(x)\begin{array}[]{rcl}|\overline{\nabla}v_{\rho,c}|^{2}(\Phi_{t}(x))&=&\sum\limits_{i}\dfrac{\left(\overline{\psi}_{\rho,c}(t)\langle\nabla\varphi(x),e_{i}\rangle\right)^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}+\left(\overline{\psi}_{\rho,c}^{\prime}(t)\varphi(x)\right)^{2}\\ &&\\ &=&(\overline{\psi}_{\rho,c}(t))^{2}\sum\limits_{i}\dfrac{\langle\nabla\varphi(x),e_{i}\rangle^{2}}{(\cos t-k_{i}(x)\sin\,t)^{2}}+(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}\varphi^{2}(x)\\ &&\\ &=&(\overline{\psi}_{\rho,c}(t))^{2}|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))+(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}\varphi^{2}(x)\end{array}

where the last equality is a consequence of (3.5). On the other hand, from (3.3) we have

(3.12) Ω|¯vρ,c|2𝑑V=0ρTΣΣ|¯vρ,c|2(Φt(x))θ(t,x)𝑑S𝑑t.\displaystyle\int_{\Omega}|\overline{\nabla}v_{\rho,c}|^{2}dV=\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}|\overline{\nabla}v_{\rho,c}|^{2}(\Phi_{t}(x))\theta(t,x)\,dS\,dt.

We conclude the proof of the lemma by replacing (3.11) into (3.12). ∎

In order to get an upper estimate for |¯u|L2(Ω)2|\overline{\nabla}u|_{L^{2}(\Omega)}^{2}, we will use the fact that the harmonic extension uu minimizes the Dirichlet energy in Cφ(Σ)C^{\infty}_{\varphi}(\Sigma).

Lemma 3.6.

Let uu be the harmonic extension of φC(Σ)\varphi\in C^{\infty}(\Sigma). For all vCφ(Ω¯)v\in C^{\infty}_{\varphi}(\overline{\Omega}) we have

Ω|¯u|2𝑑VΩ|¯v|2𝑑V.\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV\leq\displaystyle\int_{\Omega}|\overline{\nabla}v|^{2}dV.
Proof.

Since uu is harmonic, by Stokes’ theorem

Ω¯v,¯u𝑑V=Σvuν𝑑S=Σuuν𝑑S=Ω|¯u|2𝑑V.\displaystyle\int_{\Omega}\langle\overline{\nabla}v,\overline{\nabla}u\rangle dV=\int_{\Sigma}v\frac{\partial u}{\partial\nu}dS=\int_{\Sigma}u\frac{\partial u}{\partial\nu}dS=\int_{\Omega}|\overline{\nabla}u|^{2}dV.

And so

0Ω|¯(uv)|2𝑑V=Ω|¯u|2𝑑V+Ω|¯v|2𝑑V2Ω¯u,¯v𝑑V=Ω|¯u|2𝑑V+Ω|¯v|2𝑑V2Ω|¯u|2𝑑V,\begin{array}[]{rcl}0&\leq&\displaystyle\int_{\Omega}|\overline{\nabla}(u-v)|^{2}dV\\ &&\\ &=&\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV+\int_{\Omega}|\overline{\nabla}v|^{2}dV-2\int_{\Omega}\langle\overline{\nabla}u,\overline{\nabla}v\rangle dV\\ &&\\ &=&\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV+\int_{\Omega}|\overline{\nabla}v|^{2}dV-2\int_{\Omega}|\overline{\nabla}u|^{2}dV,\end{array}

and so we are done.

Proposition 3.7.

Let uu be the harmonic extension of φC(Σ)\varphi\in C^{\infty}(\Sigma). Then, if φ\varphi is a first eigenfunction of the laplacian satisfying Σφ2𝑑S=1\displaystyle\int\limits_{\Sigma}\varphi^{2}\,dS=1, we have

(3.13) Ω|¯u|2C1,\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}\leq C_{1},

where C1=C1(kmax):=323arctan(1/kmax)+λ1(Σ)1+kmax2C_{1}=C_{1}(k_{\textrm{max}}):=\dfrac{32}{3\arctan(1/k_{\textrm{max}})}+\dfrac{\lambda_{1}(\Sigma)}{\sqrt{1+k_{\textrm{max}}^{2}}}.

Proof.

Let 0<ρ<10<\rho<1, c>1c>1 and let vρ,c:Ω¯v_{\rho,c}:\overline{\Omega}\to\mathbb{R} be the function defined by (3.8).

From Lemma 3.5 and Lemma 3.6 we have

(3.14) Ω|¯u|2𝑑VΩ|¯vρ,c|2𝑑V=0ρTΣΣ(ψ¯ρ,c(t))2φ~2(Φt(x))θ(t,x)𝑑S𝑑t+0ρTΣΣ(ψ¯ρ,c(t))2|Tφ~|2(Φt(x))θ(t,x)𝑑S𝑑tρTΣcρTΣ(ψ¯ρ,c(t))2𝑑t+0ρTΣΣ(ψ¯ρ,c(t))2|Tφ~|2(Φt(x))𝑑S𝑑t,\begin{array}[]{rcl}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV&\leq&\displaystyle\int_{\Omega}|\overline{\nabla}v_{\rho,c}|^{2}dV\\ &=&\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}\widetilde{\varphi}^{2}(\Phi_{t}(x))\theta(t,x)dS\,dt+\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}(\overline{\psi}_{\rho,c}(t))^{2}|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))\theta(t,x)dS\,dt\\ &&\\ &\leq&\displaystyle\int_{\frac{\rho T_{\Sigma}}{c}}^{\rho T_{\Sigma}}(\overline{\psi}_{\rho,c}^{\prime}(t))^{2}\,dt+\displaystyle\int_{0}^{\rho T_{\Sigma}}\int_{\Sigma}(\overline{\psi}_{\rho,c}(t))^{2}|\nabla^{T}\widetilde{\varphi}|^{2}(\Phi_{t}(x))\,dS\,dt,\end{array}

where the last inequality is a consequence of the condition Σφ2𝑑Σ=1\displaystyle\int\limits_{\Sigma}\varphi^{2}\,d\Sigma=1 and (3.2).

Then, by (3.7) and Lemma 3.4 we can rewrite (3.14) as

(3.15) Ω|¯u|2𝑑V163ρTΣc(c31)(c21)2+sin2(TΣ)0ρTΣcsc2(TΣt)𝑑tΣ|φ|2𝑑S.\begin{array}[]{rcl}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV&\leq&\dfrac{16}{3\rho T_{\Sigma}}\dfrac{c(c^{3}-1)}{(c^{2}-1)^{2}}+\sin^{2}(T_{\Sigma})\displaystyle\int_{0}^{\rho T_{\Sigma}}\csc^{2}(T_{\Sigma}-t)dt\displaystyle\int_{\Sigma}|\nabla\varphi|^{2}dS.\end{array}

On the other hand, note that λ1(Σ)=Σ|φ|2𝑑S\lambda_{1}(\Sigma)=\displaystyle\int_{\Sigma}|\nabla\varphi|^{2}dS. Then it follows that

sin2(TΣ)0ρTΣcsc2(TΣt)𝑑tΣ|φ|2𝑑Sλ1(Σ)sin2(TΣ)(cot((1ρ)TΣ)cot(TΣ))=λ1(Σ)sin(TΣ)sin(ρTΣ)sin((1ρ)TΣ),\begin{array}[]{rcl}\sin^{2}(T_{\Sigma})\displaystyle\int_{0}^{\rho T_{\Sigma}}\csc^{2}(T_{\Sigma}-t)dt\displaystyle\int_{\Sigma}|\nabla\varphi|^{2}dS&\leq&\lambda_{1}(\Sigma)\,\sin^{2}(T_{\Sigma})\left(\cot((1-\rho)T_{\Sigma})-\cot(T_{\Sigma})\right)\\ &&\\ &=&\lambda_{1}(\Sigma)\,\dfrac{\sin\,(T_{\Sigma})\sin\,(\rho T_{\Sigma})}{\sin\,((1-\rho)T_{\Sigma})},\end{array}

where in the last equality we have used that cot(AB)=cotAcotB+1cotBcotA\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}.

Then (3.15) stands as

Ω|¯u|2𝑑V163ρTΣc(c31)(c21)2+λ1(Σ)sin(TΣ)sin(ρTΣ)sin((1ρ)TΣ).\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV\leq\dfrac{16}{3\rho T_{\Sigma}}\dfrac{c(c^{3}-1)}{(c^{2}-1)^{2}}+\lambda_{1}(\Sigma)\,\dfrac{\sin\,(T_{\Sigma})\sin\,(\rho T_{\Sigma})}{\sin\,((1-\rho)T_{\Sigma})}.

Making c+c\to+\infty, choosing ρ=1/2\rho=1/2 and from the definition of TΣT_{\Sigma} in (3.1), it follows that

(3.16) Ω|¯u|2𝑑V323TΣ+λ1(Σ)sin(TΣ)=323arctan(1/kmax)+λ1(Σ)sin(arctan(1/kmax))=323arctan(1/kmax)+λ1(Σ)1+kmax2.\begin{array}[]{rcl}\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV&\leq&\dfrac{32}{3T_{\Sigma}}+\lambda_{1}(\Sigma)\,\sin\,(T_{\Sigma})=\dfrac{32}{3\arctan(1/k_{\textrm{max}})}+\lambda_{1}(\Sigma)\,\sin\,(\arctan(1/k_{\textrm{max}}))\\ &&\\ &=&\displaystyle{\dfrac{32}{3\arctan(1/k_{\textrm{max}})}+\frac{\lambda_{1}(\Sigma)}{\sqrt{1+k_{\textrm{max}}^{2}}}}.\end{array}

Lemma 3.8.

For all 0t<TΣ0\leq t<T_{\Sigma} and fC1(Ω¯)f\in C^{1}(\overline{\Omega}) we have

ddt(Σtf(y)𝑑St)=Σtf(y),d(y)𝑑StΣtf(y)HΣt𝑑St,\dfrac{d}{dt}\left(\int_{\Sigma_{t}}f(y)dS_{t}\right)=\int_{\Sigma_{t}}\langle\nabla f(y),\nabla d(y)\rangle dS_{t}-\int_{\Sigma_{t}}f(y)H_{\Sigma_{t}}dS_{t},

where dd is the signed distance to Σ\Sigma in 𝕊m+1\mathbb{S}^{m+1}, i.e.

d(y)={dist (y,Σ) ifxΩ¯dist(y,Σ) ifxΩ¯c,d(y)=\left\{\begin{array}[]{ll}\textrm{dist }\,(y,\Sigma)&\textrm{ if}\,x\in\overline{\Omega}\\ -\textrm{dist}\,(y,\Sigma)&\textrm{ if}\,x\in\overline{\Omega}^{c},\end{array}\right.

and HΣtH_{\Sigma_{t}} the mean curvature of the hypersurface Σt\Sigma_{t}.

Proof.

Making the change of variable y=Φt(x)y=\Phi_{t}(x),

ddt(Σtf(y)𝑑St)=ddt(Σf(Φt(x))θ(t,x)𝑑S)=Σ¯f(Φt(x)),¯d(Φt(x))θ(t,x)𝑑S+Σf(Φt(x))θ(t,x)𝑑S=Σ¯f(Φt(x)),¯d(Φt(x))𝑑SΣf(Φt(x))HΣt(Φt(x))θ(t,x)𝑑S=Σt¯f(y),¯d(y)𝑑StΣtf(y)HΣt(y)𝑑St.\begin{array}[]{rcl}\dfrac{d}{dt}\left(\displaystyle\int_{\Sigma_{t}}f(y)dS_{t}\right)&=&\dfrac{d}{dt}\left(\displaystyle\int_{\Sigma}f(\Phi_{t}(x))\theta(t,x)dS\right)\\ &&\\ &=&\displaystyle\int_{\Sigma}\langle\overline{\nabla}f(\Phi_{t}(x)),\overline{\nabla}d(\Phi_{t}(x))\rangle\theta(t,x)dS+\displaystyle\int_{\Sigma}f(\Phi_{t}(x))\theta^{\prime}(t,x)dS\\ &&\\ &=&\displaystyle\int_{\Sigma}\langle\overline{\nabla}f(\Phi_{t}(x)),\overline{\nabla}d(\Phi_{t}(x))\rangle dS-\displaystyle\int_{\Sigma}f(\Phi_{t}(x))H_{\Sigma_{t}}(\Phi_{t}(x))\theta(t,x)dS\\ &&\\ &=&\displaystyle\int_{\Sigma_{t}}\langle\overline{\nabla}f(y),\overline{\nabla}d(y)\rangle dS_{t}-\displaystyle\int_{\Sigma_{t}}f(y)H_{\Sigma_{t}}(y)dS_{t}.\end{array}

Here we have used that HΣt(Φt(x))=θ(t,x)θ(t,x)H_{\Sigma_{t}}(\Phi_{t}(x))=-\frac{\theta^{\prime}(t,x)}{\theta(t,x)} (see Lemma 10.9 in [6]), where θ(t,x)\theta^{\prime}(t,x) denotes the derivative of θ(t,x)\theta(t,x) with respect to the first variable.

The following result is a consequence of Lemma 3.5 in [5].

Lemma 3.9.

Let 0<εΛ20<\varepsilon\leq\frac{\Lambda}{2}. Then for t[0,arctan(εΛ2)]t\in[0,\arctan(\frac{\varepsilon}{\Lambda^{2}})],

HΣt2Λ.H_{\Sigma_{t}}\leq 2\Lambda.
Proof.

Let 0<εΛ20<\varepsilon\leq\frac{\Lambda}{2} and t[0,arctan(εΛ2)]t\in[0,\arctan(\frac{\varepsilon}{\Lambda^{2}})]. From Lemma 3.5 in [5] it follows that

HΣtΛεΛε(mΛ2+1).H_{\Sigma_{t}}\leq\dfrac{\Lambda\varepsilon}{\Lambda-\varepsilon}\left(\dfrac{m}{\Lambda^{2}}+1\right).

On the other hand, since εΛ/2\varepsilon\leq\Lambda/2 and mΛ2m\leq\Lambda^{2} we have that

ΛεΛε(mΛ2+1)2ΛεΛεΛ2Λε2Λ2Λ=2Λ.\frac{\Lambda\varepsilon}{\Lambda-\varepsilon}\left(\frac{m}{\Lambda^{2}}+1\right)\leq\dfrac{2\Lambda\varepsilon}{\Lambda-\varepsilon}\leq\frac{\Lambda^{2}}{\Lambda-\varepsilon}\leq\frac{2\Lambda^{2}}{\Lambda}=2\Lambda.

We conclude that HΣt2ΛH_{\Sigma_{t}}\leq 2\Lambda. ∎

Theorem 3.10.

Let Σ\Sigma be a closed embedded minimal hypersurface in the unit sphere 𝕊m+1\mathbb{S}^{m+1} and let Λ=maxΣ|A|\Lambda=\max\limits_{\Sigma}|A| be the norm of its second fundamental form. Assume that Λ>m\Lambda>\sqrt{m} and φ\varphi is a first eigenfunction of the laplacian satisfying Σφ2𝑑S=1\displaystyle\int\limits_{\Sigma}\varphi^{2}\,dS=1. Then the harmonic extension uu of φ\varphi satisfies

Ωu2𝑑V>14(12Λ+m+11)2+1Ω|¯u|2𝑑V.\displaystyle\int_{\Omega}u^{2}dV>\dfrac{1}{4(12\Lambda+m+11)^{2}+1}\int_{\Omega}|\overline{\nabla}u|^{2}dV.
Proof.

For t0t\geq 0, let Ω(t):={yΩ:d(y)>t}\Omega(t):=\{y\in\Omega:d(y)>t\} and η(t):=Ω(t)u2\eta(t):=\displaystyle\int_{\Omega(t)}u^{2}. It follows that η(0)=Ωu2𝑑V\eta(0)=\displaystyle\int_{\Omega}u^{2}dV. Moreover, from the coarea formula

η+(t)=limε0+η(t+ε)η(t)ε=limε0+1ε(Ωt+εu2𝑑VΩtu2𝑑V)=limε0+1ε{td(y)t+ε}u2𝑑V=limε0+1εtt+εΣsu2𝑑Ss𝑑s=Σtu2𝑑St,\begin{array}[]{rcl}\eta^{\prime}_{+}(t)&=&\lim\limits_{\varepsilon\to 0^{+}}\dfrac{\eta(t+\varepsilon)-\eta(t)}{\varepsilon}\\ &&\\ &=&\lim\limits_{\varepsilon\to 0^{+}}\dfrac{1}{\varepsilon}\left(\displaystyle\int_{\Omega_{t+\varepsilon}}u^{2}dV-\displaystyle\int_{\Omega_{t}}u^{2}dV\right)\\ &&\\ &=&-\lim\limits_{\varepsilon\to 0^{+}}\dfrac{1}{\varepsilon}\displaystyle\int_{\{t\leq d(y)\leq t+\varepsilon\}}u^{2}dV\\ &&\\ &=&-\lim\limits_{\varepsilon\to 0^{+}}\dfrac{1}{\varepsilon}\displaystyle\int_{t}^{t+\varepsilon}\int_{\Sigma_{s}}u^{2}dS_{s}ds\\ &&\\ &=&-\displaystyle\int_{\Sigma_{t}}u^{2}dS_{t},\end{array}

(analogously η(t)=Σtu2𝑑St\eta^{\prime}_{-}(t)=-\displaystyle\int_{\Sigma_{t}}u^{2}dS_{t}) and so η(0)=Σφ2=1\eta^{\prime}(0)=-\displaystyle\int_{\Sigma}\varphi^{2}=-1.

From Lemma 3.8 we have

η′′(t)=ddt(Σtu2𝑑St)=Σt¯(u2),¯d𝑑St+Σtu2HΣt𝑑StΣt¯(u2),¯d𝑑St+2ΛΣtu2𝑑St=Ω(t)¯(u2)𝑑V2Λη(t),\begin{array}[]{rcl}\eta^{\prime\prime}(t)&=&-\dfrac{d}{dt}\left(\displaystyle\int_{\Sigma_{t}}u^{2}dS_{t}\right)\\ &&\\ &=&-\displaystyle\int_{\Sigma_{t}}\langle\overline{\nabla}(u^{2}),\overline{\nabla}d\rangle dS_{t}+\displaystyle\int_{\Sigma_{t}}u^{2}H_{\Sigma_{t}}dS_{t}\\ &\leq&-\displaystyle\int_{\Sigma_{t}}\langle\overline{\nabla}(u^{2}),\overline{\nabla}d\rangle dS_{t}+2\Lambda\displaystyle\int_{\Sigma_{t}}u^{2}dS_{t}\\ &&\\ &=&\displaystyle\int_{\Omega(t)}\overline{\triangle}(u^{2})dV-2\Lambda\eta^{\prime}(t),\end{array}

where in the last two lines we have used Lemma 3.9 and Stokes’ formula respectively. Therefore, by Proposition 3.7

(3.17) η′′(t)+2Λη(t)2Ω(t)|¯u|22Ω|¯u|22C1.\eta^{\prime\prime}(t)+2\Lambda\eta^{\prime}(t)\leq 2\displaystyle\int_{\Omega(t)}|\overline{\nabla}u|^{2}\leq 2\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}\leq 2C_{1}.

Multiplying by e2Λte^{2\Lambda t} both sides of (3.17) and integrating from 0 to tt we have

(3.18) η(t)2C1(1e2Λt2Λ)e2Λt.\eta^{\prime}(t)\leq 2C_{1}\left(\dfrac{1-e^{-2\Lambda t}}{2\Lambda}\right)-e^{-2\Lambda t}.

Now we can integrate from 0 to Tε:=arctan(εΛ2)T_{\varepsilon}:=\arctan(\frac{\varepsilon}{\Lambda^{2}}) in (3.18) and deduce that

(3.19) η(Tε)η(0)1ΛC1(Tε+e2ΛTε12Λ)+e2ΛTε12Λ.\eta(T_{\varepsilon})-\eta(0)\leq\frac{1}{\Lambda}C_{1}\left(T_{\varepsilon}+\dfrac{e^{-2\Lambda T_{\varepsilon}}-1}{2\Lambda}\right)+\dfrac{e^{-2\Lambda T_{\varepsilon}}-1}{2\Lambda}.

Considering that η(Tε)>0\eta(T_{\varepsilon})>0 in (3.19) it follows that

(3.20) η(0)>1e2ΛTε2Λ1ΛC1(Tε+e2ΛTε12Λ)=1eΛTε2ΛTεTεC12(2ΛTε+e2ΛTε1Λ2Tε2)Tε2\begin{array}[]{rcl}\eta(0)&>&\dfrac{1-e^{-2\Lambda T_{\varepsilon}}}{2\Lambda}-\dfrac{1}{\Lambda}C_{1}\left(T_{\varepsilon}+\dfrac{e^{-2\Lambda T_{\varepsilon}}-1}{2\Lambda}\right)\\ &&\\ &=&\dfrac{1-e^{-\Lambda T_{\varepsilon}}}{2\Lambda T_{\varepsilon}}T_{\varepsilon}-\dfrac{C_{1}}{2}\left(\dfrac{2\Lambda T_{\varepsilon}+e^{-2\Lambda T_{\varepsilon}}-1}{\Lambda^{2}T_{\varepsilon}^{2}}\right)T_{\varepsilon}^{2}\end{array}

On the other hand, for all x0x\geq 0

(3.21) x+ex1x212.\dfrac{x+e^{-x}-1}{x^{2}}\leq\frac{1}{2}.

Using the inequality (3.21) and the fact that Tε<εΛ2T_{\varepsilon}<\frac{\varepsilon}{\Lambda^{2}} in (3.20) we have

(3.22) Ωu2𝑑V=η(0)>Tε(1ΛTε)C1Tε2=Tε(1(Λ+C1)Tε)>Tε(1εΛ2(Λ+C1))=Tε(1ε(1Λ+C1Λ2)).\begin{array}[]{rcl}\displaystyle\int_{\Omega}u^{2}dV&=&\eta(0)\\ &&\\ &>&T_{\varepsilon}(1-\Lambda T_{\varepsilon})-C_{1}T_{\varepsilon}^{2}\\ &&\\ &=&T_{\varepsilon}\left(1-(\Lambda+C_{1})T_{\varepsilon}\right)\\ &&\\ &>&T_{\varepsilon}\left(1-\frac{\varepsilon}{\Lambda^{2}}(\Lambda+C_{1})\right)\\ &&\\ &=&T_{\varepsilon}\left(1-\varepsilon\left(\frac{1}{\Lambda}+\frac{C_{1}}{\Lambda^{2}}\right)\right).\end{array}

From (3.22), it follows that for any 0<ε12(1Λ+C1Λ2)10<\varepsilon\leq\dfrac{1}{2}\left(\dfrac{1}{\Lambda}+\dfrac{C_{1}}{\Lambda^{2}}\right)^{-1},

Ωu2𝑑V>Tε2=12arctan(εΛ2).\displaystyle\int_{\Omega}u^{2}dV>\dfrac{T_{\varepsilon}}{2}=\dfrac{1}{2}\arctan(\dfrac{\varepsilon}{\Lambda^{2}}).

In particular, for ε=12(1Λ+C1Λ2)1\varepsilon=\dfrac{1}{2}\left(\dfrac{1}{\Lambda}+\dfrac{C_{1}}{\Lambda^{2}}\right)^{-1} we have

Ωu2𝑑V>12arctan(12(Λ+C1)).\displaystyle\int_{\Omega}u^{2}dV>\dfrac{1}{2}\arctan\left(\dfrac{1}{2(\Lambda+C_{1})}\right).

Finally, this last inequality joint with Proposition 3.7 lead us to

(3.23) Ωu2𝑑VΩ|¯u|2𝑑V>arctan(12(Λ+C1))2C1.\dfrac{\displaystyle\int_{\Omega}u^{2}dV}{\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV}>\dfrac{\arctan\left(\dfrac{1}{2(\Lambda+C_{1})}\right)}{2C_{1}}.

On the other hand, note that 1kmax2m1mΛ21\leq k_{\textrm{max}}^{2}\leq\frac{m-1}{m}\Lambda^{2}. Then, using the fact that λ1(Σ)m\lambda_{1}(\Sigma)\leq m and since arctanxx1+x2\arctan x\geq\dfrac{x}{1+x^{2}} for x[0,1]x\in[0,1], we have

(3.24) C1=323arctan(1/kmax)+λ1(Σ)1+kmax232(1+kmax2)3kmax+λ1(Σ)1+kmax2<32kmax3+m+11kmax<32Λ3+m+11kmax<11Λ+m+11.\begin{split}C_{1}&=\dfrac{32}{3\arctan(1/k_{\textrm{max}})}+\dfrac{\lambda_{1}(\Sigma)}{\sqrt{1+k_{\textrm{max}}^{2}}}\\ &\leq\dfrac{32(1+k_{\textrm{max}}^{2})}{3k_{\textrm{max}}}+\dfrac{\lambda_{1}(\Sigma)}{\sqrt{1+k_{\textrm{max}}^{2}}}\\ &<\dfrac{32k_{\textrm{max}}}{3}+\dfrac{m+11}{k_{\textrm{max}}}\\ &<\dfrac{32\Lambda}{3}+\dfrac{m+11}{k_{\textrm{max}}}\\ &<11\Lambda+m+11.\end{split}

We now define

(3.25) C2=C2(m,Λ):=arctan(12(Λ+C1))2C1.\begin{split}C_{2}=C_{2}(m,\Lambda):&=\dfrac{\arctan\left(\dfrac{1}{2(\Lambda+C_{1})}\right)}{2C_{1}}.\end{split}

From (3.24) and (3.25) we have

(3.26) C2=arctan(12Λ+2C1)2C12Λ+2C12C1[(2Λ+2C1)2+1]1(2Λ+2C1)2+1>14(12Λ+m+11)2+1.\begin{split}C_{2}&=\dfrac{\arctan\left(\dfrac{1}{2\Lambda+2C_{1}}\right)}{2C_{1}}\\ &\geq\dfrac{2\Lambda+2C_{1}}{2C_{1}[(2\Lambda+2C_{1})^{2}+1]}\\ &\geq\dfrac{1}{(2\Lambda+2C_{1})^{2}+1}\\ &>\dfrac{1}{4(12\Lambda+m+11)^{2}+1}.\end{split}

Combining (3.23) and (3.26) we have

Ωu2𝑑VΩ|¯u|2𝑑V>14(12Λ+m+11)2+1.\dfrac{\displaystyle\int_{\Omega}u^{2}dV}{\displaystyle\int_{\Omega}|\overline{\nabla}u|^{2}dV}>\dfrac{1}{4(12\Lambda+m+11)^{2}+1}.

This completes the proof the theorem.

We conclude the paper by comparing our estimate and the lower bound for λ1(Σ)\lambda_{1}(\Sigma) obtained by Duncan, Sire and Spruck in [5]. In their work, it is established that given a closed and embedded minimal hypersurface Σ\Sigma in 𝕊m+1\mathbb{S}^{m+1} with Λ=maxΣ|A|m\Lambda=\max\limits_{\Sigma}|A|\geq\sqrt{m}, then

(3.27) λ1(Σ)m2+a(m)Λ6+b(m),\lambda_{1}(\Sigma)\geq\dfrac{m}{2}+\dfrac{a(m)}{\Lambda^{6}+b(m)},

where

(3.28) a(m):=3m(m1)3200(marctan(13m))3andb(m):=5(m1)8m(marctan(13m))3.\begin{array}[]{rcl}a(m)&:=&\displaystyle{\frac{3\sqrt{m}(m-1)}{3200}}\left(m\arctan\left(\frac{1}{3\sqrt{m}}\right)\right)^{3}\,\quad\textrm{and}\\ &&\\ b(m)&:=&\displaystyle{\frac{5(m-1)}{8\sqrt{m}}}\left(m\arctan\left(\frac{1}{3\sqrt{m}}\right)\right)^{3}.\end{array}

Since x2arctanxx\frac{x}{2}\leq\arctan x\leq x when x[0,1]x\in[0,1], we have m6marctan(13m)m3\frac{\sqrt{m}}{6}\leq m\arctan(\frac{1}{3\sqrt{m}})\leq\frac{\sqrt{m}}{3}, then from (3.28) we deduce

(3.29) a(m)(m1)m228800and b(m)5(m1)m1728.\begin{array}[]{rcl}a(m)&\leq&\frac{(m-1)m^{2}}{28800}\,\quad\textrm{and }\\ &&\\ b(m)&\geq&\frac{5(m-1)m}{1728}.\end{array}

Then, since m2m\geq 2 and Λm\Lambda\geq\sqrt{m} we trivially obtain from (3.29) that

a(m)Λ6+b(m)<(m1)m228800Λ6+28800175<(m+1)mΛ228800Λ6+164<(m+1)m32(12Λ+Λ2+11)2+8<(m+1)m32(12Λ+m+11)2+8.\dfrac{a(m)}{\Lambda^{6}+b(m)}<\dfrac{(m-1)m^{2}}{28800\Lambda^{6}+\frac{28800}{175}}<\dfrac{(m+1)m\Lambda^{2}}{28800\Lambda^{6}+164}<\frac{(m+1)m}{32(12\Lambda+\Lambda^{2}+11)^{2}+8}<\frac{(m+1)m}{32(12\Lambda+m+11)^{2}+8}.

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