THE BEST CONSTANT OF THE MOSER-TRUDINGER INEQUALITY ON S 2

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1 THE BEST CONSTANT OF THE MOSER-TRUDINGER INEQUALITY ON S 2 YUJI SANO Abstract. We consider the best constant of the Moser-Trudinger inequality on S 2 under a certain orthogonality condition. Applying Moser s calculation, we construct a counterexample to the sharper inequality with the condition.. Introduction Let (M, g) be a compact smooth Riemannian manifold of dimension n. H n (M) denotes the Banach space obtained from the vector space of C functions ϕ such that ϕ L n is bounded, by completion with the norm ϕ H n := l ϕ L n. l=0 It is well-known as the exceptional case of the Sobolev imbedding theorem that there exists constants C, µ and ν such that all ϕ H n (M) satisfy (.) e ϕ dv C exp(µ ϕ n L +ν ϕ n n L n) M (Trudinger, Aubin, cf. [2], p. 63). For applications, the best constant of µ in (.) is essential. For example, in the case of the sphere, the following result was proved by Moser [5] when n = 2, and Aubin [] for general n. All ϕ H n (S n ) with integral equal zero satisfy (.2) e ϕ dv C exp(µ n ϕ n L n), S n where C depends only on n, and µ n = (n ) n n 2n ω n. Here ω n denotes the volume of S n () := {(x, y, z) R 3 x 2 + y 2 + z 2 = }. Note that ν in (.) equals to zero in this case. In particular, when n = 2, (.2) becomes (.3) log e ϕ dµ grad ϕ 2 dµ + ϕdµ + C S 2 6π S 2 4π S 2 where dµ denotes the volume element of the canonical metric on S 2. It is the best constant of the coefficient of grad ϕ 2 dµ which is useful for Nirenberg s S 2 problem: Characterize all Gauss curvature functions K of Riemannian metrics g 2000 Mathematics Subject Classification. Primary 34A26; Secondary 53C55. Key words and phrases. The Moser-Trudinger inequality. This research was partially supported by the Ministry of Education, Science, Sports and Culture, Grant-in-Aid for JSPS Fellows, 03340, 2003.

2 2 YUJI SANO conformal to the canonical metric g 0, so that g = e ϕ g 0. This requires a solution of the following nonlinear elliptic partial differential equation for a given K. (.4) g0 ϕ = Ke ϕ, where g0 := i i denotes the real Laplacian with respect to g 0. For the variational method we define a functional S(ϕ) by S(ϕ) := log Ke ϕ dµ + grad ϕ 2 dµ + ϕdµ. S 2 6π S 2 4π S 2 The existence of a solution of (.4) is reduced to finding a minimum of S(ϕ). Clearly S(ϕ) is finite, but it is impossible to find a convergent subsequence from a given minimizing sequence when the best constant of µ equals to 6π. So we consider a sharper version of the Moser-Trudinger inequality (.3) under addtional constraints on ϕ. For example, Aubin [] proved that if ϕ H n (S n ) satisfies (.5) ξe ϕ dv = 0, ϕdv = 0 S n S n for all ξ Λ which denotes the first eigenspace of the Laplacian, then e ϕ dv C(ɛ) exp(( µ n S n 2 + ɛ) ϕ n L n) for any ɛ > 0 and some constant C(ɛ). Also when n = 2, Moser [6] proved that we can lower the best constant of µ to 32π for all ϕ C (RP 2 ), i.e., ϕ(ξ) = ϕ( ξ) for ξ =. Note that under these circumstances the minimum of S(ϕ) exists, but these are just only partial answers to Nirenberg s problem. In the context of Kähler geometry, the equation (.4) can be regarded as a - dimensional complex Monge-Ampère equation, since S 2 is a -dimensional Kähler manifold. On this problem, similar results hold as follows. Let be a compact m-dimensional Kähler manifold with the positive first Chern class and its Kähler form ω representing the first Chern class. We define P (, ω) := {ϕ C ω + ϕ > 0}, and a functional F ω on P (, ω) [3] by F ω (ϕ) := J ω (ϕ) ϕω m log( V V J ω (ϕ) := V m ϕ ϕ k=0 where V = ωm, and h ω is uniquely determined by Ric(ω) ω = h ω, (e hω )ω m = 0. e hω ϕ ω m ), m k m + ωm k (ω + ϕ) k, (See [4] for the way how F ω and Mabuchi s K-energy are related to the Futaki character.) Note that the functional S(ϕ) is equivalent to F ωke ( ϕ) by regarding S 2 as CP. It may be commonly known among experts in this problem that the inequality (.3) can be shown from the viewpoint of existence of Kähler-Einstein

3 THE BEST CONSTANT OF THE MOSER-TRUDINGER INEQUALITY ON S 2 3 metrics. Concretely, if admits a Kähler-Einstein metric ω KE, F ωke ( ϕ) is bounded from below by some absolute constant for all ϕ P (, ω KE ). Namely (.6) e ϕ ωke m C exp(j ωke ( ϕ) + ϕω m V V KE). When the complex dimension of equals to one, so that = CP = S 2, (.6) is equivalent to (.3). So (.6) can be regarded as a fully nonlinear generalization of the Moser-Trudinger inequality. In [7], Tian proved the sharper form of (.6): if (, ω KE ) is a Kähler-Einstein manifold, there is positive constants δ, C such that J ωke ( ϕ) F ωke ( ϕ) δ ( + osc ϕ) C 2m+2/2m+2+β for any ϕ P (, ω KE ) perpendicular to Λ (i.e., ξ ( ϕ)ωm KE = 0 for any ξ Λ ). Here β = 4 e m and Λ denotes the space of the eigenfunctions of the complex Laplacian KE with eigenvalue with respect to ω KE (if this space is not empty, it is equivalent to the first eigenspace). Note that δ and C may depend on the dimension of, the first eigenvalue of KE, which is greater than one, and Sobolev constant with respect to ω KE. Moreover, Tian posed the following conjecture (see [7], p. 20, conjecture 5.5): if is a Kähler-Einstein manifold, there are constants δ > 0, C δ which may depend on δ, such that namely (.7) V F ωke ( ϕ) δ J ωke ( ϕ) C δ, e ϕ ω m KE C δ exp(( δ)j ωke ( ϕ) + V ϕωke) m for any ϕ P (M, ω KE ) perpendicular to Λ. When = S 2, (.7) is equivalent to (.8) log e ϕ dµ ( δ) grad ϕ 2 dµ + ϕdµ + C δ. S 2 6π S 2 4π S 2 In a word, Tian s conjecture sugguests that the perpendicularity to Λ could make the best constant of the Moser-Trudinger inequality on S 2 smaller than 6π. If this conjecture is true, it will give a new partial answer to Nirenberg s problem. Note that this perpendicularity is different from Aubin s one (.5) and is essential. In fact, it is known (Ding [3]) that we cannot make δ in (.7) larger than zero without the perpendicularity if admits a nonzero holomorphic vector field (i.e., Λ is not empty). The purpose of this paper is to construct a counterexample to (.8) for δ > 0 even if ϕ is perpendicular to Λ. Note that Λ is equivalent to Λ. In other words, the perpendicularity to Λ cannnot make the best constant of the Moser-Trudinger inequality smaller than 6π. Our counterexample is based on Moser s proof in [5]. Remark.. Our example is not a counterexample to Tian s conjecture, because it does not satisfy the Kähler condition. The author would like to thank A. Futaki for leading him to this problem and for valuable advices, S. Bando and Y. Nakagawa for checking his paper and G. Tian for valuable indications.

4 4 YUJI SANO 2. Construction of the counterexample As mentioned, our example closely follows the calculations of [5]. It suffices to find a sequence of functions {ϕ} on S 2 such that for any β > (2.) log e ϕ dµ grad ϕ 2 dµ + ϕdµ S 2 6πβ S 2 4π S 2 tends to infinity and that ϕ is perpendicular to Λ. Theorem 2 in [5] implies that if ϕ is a smooth function defined on S 2 satisfying grad ϕ 2 dµ, ϕdµ = 0, S 2 S 2 then there exists an absolute constant C such that 2 dµ C. S 2 e 4πϕ First, we will construct a sequence of functions {ϕ} such that (i) grad ϕ 2 dµ =, S 2 (ii) ϕdµ = 0, S 2 (iii) ϕ is perpendicular to Λ, 2 (iv) dµ tends to infinity for any β >. S 2 e 4πβϕ We introduce longitude φ and latitude θ on S 2 so that the canonical metric is given by ds 2 = dθ 2 + cos 2 θdφ 2, θ < π 2 and θ = ±π/2 corresponds to the two poles. Let ϕ(θ) be a radially symmetric function, that is, independent of φ. Moreover, introducing the variable t and the functions w(t), ρ(t) by e t/2 := tan( θ 2 + π 4 ), w(t) := (4π) /2 ϕ(θ), ρ(t) := e t + e t + 2, it suffices to find a sequence of functions such that (See [5]) (i ) (ii ) + + ẇ 2 dt =, wρdt = 0, (iii) ϕ is perpendicular to Λ, (iv ) + e βw2 ρdt tends to infinity for any β >. Regarding S 2 as {(x, y, z) R 3 x 2 + y 2 + z 2 = }, the basis of the first eigenfunctions are given by the coordinate functions x, y and z. Since ϕ(θ) is radially symmetric, we may assume that ϕ(θ) is perpendicular to both x and y. Then, we

5 THE BEST CONSTANT OF THE MOSER-TRUDINGER INEQUALITY ON S 2 5 have only to require ϕ to be perpendicular to z. Since z is the eigenfunction of eigenvalue (we are considering the complex Laplacian), 0 = ϕ zdµ = S 2 ϕ zdµ = S 2 i ϕ i zdµ. S 2 Note also that the condition (iii) reduces to (iii ) + ẇ(t)ż(t)dt = 0. Here we regard z as a function of t. Now we define w (t) as follow ( > 0). ε : t < ε, t : ε t < 0, w (t) = ε t : 0 t <, ( ε) : t. 0 < ε < is determined by as indicated below. It is obvious that w (t) as above satisfies (i ). To fulfil (iii ), w (t) has to satisfy 0 ε ż(t)dt + ż(t)dt = 0, ε 0 hence ε z( ε) = z(). Note that ε 0 ( + ), since 0 < ε < and z() tends to as +. Let A and w (t) denote + w (t) ρ(t)dt and w (t) A. Therefore w (t) satisfies (i ), (ii ), (iii ). Note that A tends to 0 as +, since 0 ε 0 A ε ρdt + te t dt + + ( ε) e t dt. In what follows, we will prove that as +. For > 0 we have + + e β w2 ρdt 4 0 e β w2 ρdt + exp(β w 2 t)dt = 4 exp(β( ( ε) A ) 2 ) (2.2) = 4 exp((β( ε A ) 2 )). Since β > and A 0 as +, there is a positive constant c such that β( ε A ) 2 > c for large enough. Then + eβ w2 ρdt tends to infinity as +.

6 6 YUJI SANO Next we will construct a counterexample by making use of w (t), or rather ϕ := (4π) /2 w (t). Let us fix β > β >. Now applying a ϕ to (2.) where a is a positive constant depending only on, what we should show is that by taking a appropiately (2.3) exp(a ϕ S 2 6πβ a2 )dµ as +. Let M() be the maximum of ϕ. Now we choose a such that = M() for given. Since a 8πβ 4πβ ϕ 2 a ϕ + 6πβ a2 = 4πβ (ϕ 8πβ a ) 2 + a2 6π ( β β ), we deduce (2.4) a ϕ 6πβ a2 4πβ ϕ 2 on {p S 2 ϕ (p) = M()}. From (2.4) we obtain exp(a ϕ S 2 6πβ a2 )dµ (2.5) = 4 {ϕ =M()} + + exp(4πβ ϕ 2 )dµ exp(β w (t) 2 )ρ(t)dt exp(β w (t) 2 t)dt. Since β >, the right hand side of (2.5) tends to infinity as + by the same argument as in (2.2). So (2.3) is proved. Thus we conclude that {a ϕ } are the counterexample we want. Consequently we find that the best constant of the Moser- Trudinger inequality cannot be lowered only by requiring the perpendicularity to Λ. References. T. Aubin: Meilleures constantes dans le théorème d inclusion de Sobolev et un théorème de Fredholm non linéaire pour la transformation conforme de la courbure scalaire. J. Funct. Anal. 32 (979), no. 2, T. Aubin: Some nonlinear problems in Riemannian geometry. Springer-Verlag, Berlin, W. Ding: Remarks on the existence problem of positive Kähler-Einstein metrics. Math. Ann. 282 (988), no. 3, A. Futaki and Y.Nakagawa: Characters of automorphism groups associated with Kähler classes and functionals with cocycle conditions. Kodai Math. J. 24 (200), no., J. Moser: A sharp form of an inequality by N. Trudinger. Indiana Univ. Math. J. 20 (970/7), J. Moser: On a nonlinear problem in differential geometry. Dynamical systems (Proc. Sympos., Univ. Bahia, Salvador, 97), pp Academic Press, New York, G. Tian: Kähler-Einstein metrics with positive scalar curvature. Invent. Math. 30 (997), no., 37. Department of Mathmatics, Tokyo Institute of Technology Oh-okayama, Meguroku, Tokyo , Japan address: ysano@math.titech.ac.jp