A Payne-Rayner type inequality for the Robinproblem on arbitrary minimal surfaces in RN
Futoshi Takahashi and Akinobu Uegaki
Abstract. We prove a Payne-Rayner type inequality for the first eigenfunctionof the Laplacian with Robin boundary condition on any compact minimal
surface with boundary in RN . We emphasize that no topological condition isnecessary on the boundary.
Mathematics Subject Classification (2000). Primary 35P15; Secondary 35J25.
Keywords. Robin problem, Payne-Rayner type inequality.

1. Introduction
Let Ω ⊂ R2 be a bounded domain with smooth boundary ∂Ω, and let λ1 (Ω) and ψ
denote the first eigenvalue and the corresponding first eigenfunction, respectively,
to the eigenvalue problem(−∆u = λu in Ω,u=0on ∂Ω.
In [7], Payne and Rayner proved the following inequality
µZ
¶
µZ
¶2
λ1 (Ω)
ψ 2 dx ≤ψ dx .
4πΩΩA remarkable point of this inequality is that it gives an exact lower-bound of thefirst eigenvalue by means of some integral-norms of the first eigenfunction, on onehand, and on the other hand, it also says that the first eigenfunction satisfies areverse Hölder type inequality. Actually, the L2 norm of ψ is bounded by the L1norm of ψ.
In this paper, we extend the above result, known to hold on a flat domain
with the Dirichlet boundary condition, to a more general setting. Namely, let Σ be
The first author acknowledges the support by JSPS Grant-in-Aid for Scientific Research (C), No.
20540216.

2F. Takahashi and A. Uegaki

a compact minimal surface in RN (N ≥ 3) with smooth boundary ∂Σ. We consider
the following eigenvalue problem with the Robin boundary condition:(−∆Σ u = λu in Σ,(1.1)∂uon ∂Σ,∂ν + βu = 0
where ∆Σ is the Laplace-Beltrami operator on Σ, β is a positive constant and ν isthe outer unit normal to ∂Σ. Let λβ1 (Σ) denote the first eigenvalue of (1.1), givenby the variational formulaRR|∇Σ u|2 dH2 + β ∂Σ u2 dH1βΣR,
λ1 (Σ) = minu∈H 1 (Σ)
u2 dH2Σ
where ∇Σ is the gradient operator on Σ and Hk denotes the k-dimensional Hausdorff measure in RN . It is well known that λβ1 (Σ) is simple and isolated, and thecorresponding eigenfunction ψβ is smooth, positive, and unique up to multiplication by constants. (see, for example, [3]).
Now, let us consider the auxiliary problem(∆Σ f = 2 in Σ,(1.2)
f =0on ∂Σ.
Our main result is the following Payne-Rayner type inequality.
Theorem 1.1. Let λβ1 (Σ) be the first eigenvalue of (1.1) and ψβ be the eigenfunction
corresponding to λβ1 (Σ). Thenµ¶µZ¶2ZZ
λβ1 (Σ)∂fΣ11
ψβ2 dH2 ≤ √ψβ2dH1 + √ H1 (∂Σ)2 (M 2 − m2∗ )ψβ dH2 +2∂ν2π2πΣ∂ΣΣholds, where M = max∂Σ ψβ , m∗ = minΣ∪∂Σ ψβ , and fΣ is the unique solution tothe problem (1.2)
As for the Dirichlet eigenvalue problem(−∆Σ u = λu in Σ,
u=0on ∂Σ,

(1.3)

the same proof of Theorem 1.1 works well and we obtain
Theorem 1.2. Let λD
1 (Σ) be the first eigenvalue of (1.3) and ψD be the eigenfunction corresponding to λD
1 (Σ). Then we haveµZ¶2ZλD (Σ)
2
ψDdH2 ≤ 1√ψD dH2 .
2 2πΣΣ
Under the assumption that the boundary ∂Σ is weakly connected (see LiSchoen-Yau [6]), Wang and Xia [8] recently proved the sharp inequalityµZ¶2Z
λD1 (Σ)222
ψD dH
ψD dH ≤4πΣΣ

A Payne-Rayner type inequality for the Robin problem

3for the first eigenfunction to (1.3), with the equality holds if and only if Σ is a flatdisc on an affine 2-plane in RN .
Our method of proof is strongly related to that of [8], which in turn goes backto the work [7]. However, in our case, we cannot apply the sharp isoperimetricinequality by Li-Schoen-Yau [6] directly to level sets of the first eigenfunction,
since we put no topological assumptions on the boundary. Instead, we use a weakerversion of the isoperimetric inequality due to A. Stone ([1]: Lemma 4.3):
Let Σ be a compact minimal surface in RN with boundary ∂Σ. Let A denotethe area of Σ and L the length of ∂Σ. Then the inequality√2 2πA ≤ L2(1.4)
holds.
√Though the constant 2 2π in front of A is not the best possible value 4π, thisweaker inequality is valid for any compact minimal surface in RN with boundary.
Thanks to this, we do not need any topological assumption such as weak connectedness on the boundary in Theorem 1.1 and Theorem 1.2.
In case Σ = Ω ⊂ R2 is a bounded smooth domain in (1.1), we can appeal tothe classical sharp isoperimetric inequality 4πA ≤ L2 on the plane, then we obtain
Theorem 1.3. Let Σ = Ω is a smooth bounded domain in R2 . Then we haveµZ¶2µ¶ZZ
λβ1 (Ω)
1∂fΩ
1 122
ψβ dx ≤ψβ dx +ψβ
dH1 +H (∂Σ)2 (M 2 − m2∗ )2π2 ∂Ω∂ν2πΩΩ
We do not repeat the proof of Theorem 1.2 and Theorem 1.3 here, since itneeds only a trivial change in the proof of Theorem 1.1.

2. Proof of Theorem 1.1
First, we set
U (t) = {x ∈ Σ : ψβ (x) > t},S(t) = Σ ∩ ∂U (t),Γ(t) = ∂Σ ∩ ∂U (t)for t > 0. Then ∂U (t) = S(t) ∪ Γ(t) is a disjoint union. Since ψβ is smooth up
to the boundary ([5]), Sard’s lemma implies that |∇Σ ψβ | 6= 0 on S(t), S(t) is asmooth hypersuraface and can be written as S(t) = {x ∈ Σ : ψβ (x) = t} for a.e.
t > 0. Recall M = max∂Σ ψβ and m∗ = minΣ∪∂Σ ψβ . We claim that min∂Σ ψβ > 0.
Indeed, if ψβ (x0 ) = 0 for some x0 ∈ ∂Σ, then the boundary condition implies that∂ψβ∂ν (x0 ) = 0 also holds. On the other hand, by the positivity of ψβ and Hopf’s∂ψlemma, we have ∂νβ (x0 ) < 0, which is a contradiction. Since ψβ is positive on Σ,the above claim yields m∗ > 0, and then U (t) = Σ for any 0 < t < m∗ . Also wenote that Γ(t) = φ if t > M .

4F. Takahashi and A. Uegaki

As in the proof of [2], [3], [8], our main tool is the following co-area formula,asserting that for every w ∈ L1 (Σ), it holds
ZZ ∞Zw
wdH2 =dH1 dτ,|∇ψ|ΣβU (t)tS(τ )
ZZdw
2
wdH = −dH1 .
dt U (t)|∇ψ|Σ βS(t)See, for instance, [4]. Note that in the right hand side, the integral over Γ(t) does
not appear.
We define the following two functions g and h asZZ ∞Z
ψβψβ dH2 =g(t) =dH1 dτ,U (t)tS(τ ) |∇Σ ψβ |¿µ¶
ÀZ
1 2h(t) = −∇Σψβ , ∇Σ f dH22U (t)Z ∞Zψβ h∇Σ ψβ , ∇Σ f i 1=−dH ds,|∇Σ ψβ |tS(τ )where f is the unique solution of the problem (1.2).
Differentiating g and h, we haveZ1g 0 (t) = −t
dH1 ,(2.1)
S(t) |∇Σ ψβ |ZZh∇Σ ψβ , ∇Σ f ih0 (t) = t
dH1 = −th∇Σ f, νi dH1|∇ψ|Σ βS(t)S(t)Z∂f= −tdH1(2.2)
S(t) ∂ν¯∇ ψ ¯for a.e. t > 0, since − |∇ΣΣ ψββ | ¯is outward unit normal vector field ν of S(t).
S(t)

On the other hand, integrating both sides of −∆Σ ψβ = λβ1 (Σ)ψβ over U (t),we haveZZ
λβ1 (Σ)g(t) = λβ1 (Σ)ψβ dH2 = −∆Σ ψβ dH2U (t)U (t)ZZ∂ψβ=|∇Σ ψβ | dH1 −dH1S(t)Γ(t) ∂νZZ1=|∇Σ ψβ | dH + β
ψβ dH1S(t)Γ(t)
Z≥|∇Σ ψβ | dH1 ,(2.3)S(t)∂ψsince − ∂νβ = βψβ > 0 on Γ(t) ⊂ ∂Σ.

A Payne-Rayner type inequality for the Robin problem
Also, we seeZZ

2H2 (U (t)) =Z

2dH2 =U (t)∆f dH2 =U (t)ZZ∂f∂fdH1 +dH1
∂ν∂ν
S(t)
Γ(t)
Z
∂f−1 01≥dH =h (t)tS(t) ∂ν5

∂fdH1
∂U (t) ∂ν=(2.4)

by (2.2). The last inequality follows by the fact ∂f∂ν > 0 on Γ(t) ⊂ ∂Σ, which inturn is assured by the Hopf lemma.
From the weak isoperimetric inequality (1.4) applied to U (t), we have√
2 2πH2 (U (t)) ≤ H1 (∂U (t))2¡¢2≤ H1 (S(t)) + H1 (Γ(t))≤ 2H1 (S(t))2 + 2H1 (Γ(t))2 .
Now, Schwarz’s inequality, (2.1) and (2.3) imply
ÃZ!2 ÃZ
12H (S(t)) =1 dHS(t)1≤! ÃZ|∇Σ ψβ | dH

1dH1S(t) |∇Σ ψβ |1S(t)≤ λβ1 (Σ)g(t) ·(2.5)µ 0 ¶g (t)−.
t!

Therefore, by (2.4) and (2.5), we obtain√µ 0 ¶√2π 0g (t)−h (t) ≤ 2 2πH2 (U (t)) ≤ 2λβ1 (Σ)g(t) · −+ 2H1 (Γ(t))2 ,ttor equivalently,ddt½λβ1 (Σ)g(t)2 −√¾Z t
2πh(t) −122τ H (Γ(τ )) dτ

≤ 0.

(2.6)

0for a.e t > 0. Note that the function l(t) = 2tH1 (Γ(t))2 is integrable on the interval
Rtdt ∈ (0, kψβ kL∞ (∂Σ) ), and thus l(t) = dtl(τ )dτ .
0Fix ε > 0 so small such that ε < m∗ . Integrating (2.6) from mε = m∗ − ε tot, we haveZ tZ mε√√λβ1 (Σ)g(t)2 − 2πh(t)−2τ H1 (Γ(τ ))2 dτ ≤ λβ1 (Σ)g(mε )2 − 2πh(mε )−2τ H1 (Γ(τ ))2 dτ,00which implies√

2πh(mε ) ≤ λβ1 (Σ)g(mε )2 − λβ1 (Σ)g(t)2 +√Z t2πh(t) +mε2τ H1 (Γ(τ ))2 dτ.

6

F. Takahashi and A. Uegaki

We easily see thatZ tZ M¡¢2τ H1 (Γ(τ ))2 dτ ≤ H1 (∂Σ)22τ dτ = H1 (∂Σ)2 M 2 − m2εmεmεfor any t > mε . Letting t → +∞, and noting that U (t) is empty for sufficiently
large t, we obtain¡¢
λβ1 (Σ) 21h(mε ) ≤ √g (mε ) + √ H1 (∂Σ)2 M 2 − m2ε .
2π2πg(mε ) and h(mε ) are given byZ
ψβ dH2 ,g(mε ) =Σµ¶ÀZ ¿1 2∇Σh(mε ) = −ψβ , ∇Σ f dH22Z ΣZ1 21∂f2=ψβ ∆f dH −
ψβ2dH1 .
2 ∂Σ∂νΣ 2Since ∆Σ f = 2 by (1.2), we haveµZ¶2ZZ¡¢
1λβ1 (Σ)
1222 ∂f12
ψβ dH −
ψβdH ≤ √ψβ dH+ √ H1 (∂Σ)2 M 2 − m2ε .
2
∂ν
2π2πΣ
∂Σ
Σ
Finally letting ε → 0, we obtain the result.
Remark 2.1. In the case that Ω = BR ⊂ R2 is a disc of radius R, then the inequality
in Theorem 1.3 becomes the equalityµZ¶2ZZλβ (Ω)
Rψβ2 dx = 1(2.7)ψβ dx +ψ 2 dH1 .
4π2 ∂Ω βBRBRThis is because, first, ψβ is positive, radial and decreasing in the radial direction on BR ([3]:Proposition 2.6). Therefore ψβ ≡ c > 0 on ∂BR and U (c) = BR ,∂U (t) = S(t) for any t > c. Also |∇ψβ | is constant on S(t). Secondly, we can use
the sharp isoperimetric inequality as the equality 4πH2 (U (t)) = H1 (S(t))2 in (2.5)
in this case. Finally, the unique solution fBR of (1.2) is fBR = 12 |x|2 − 12 R2 . By
these reasons, we see all inequalities in the proof of Theorem 1.1 are equalities andwe obtainod n β
λ1 (BR )g(t)2 − 4πh(t) = 0dt
for a.e. t > c, instead of (2.6). Integrating this from t = c to t, and letting t → ∞,we obtain (2.7).

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A Payne-Rayner type inequality for the Robin problem

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Futoshi Takahashi
Department of Mathematics, Osaka City University,
3-3-138 Sugimoto, Sumiyoshi-ku, Osaka
558-8585, Japane-mail: futoshi@sci.osaka-cu.ac.jp
Akinobu Uegaki
Department of Mathematics, Osaka City University,
3-3-138 Sugimoto, Sumiyoshi-ku, Osaka
558-8585, Japane-mail: u.akinobu@gmail.com