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Set $\nabla_{0}$ is the tangential gradient on sphere $S^{N-1}$, and $\Delta_{0}$ is the Laplace-Beltrami operator on $S^{N-1}$, $v,\phi \in C^{2}(S^{N-1})$, $\phi$ is also the first eigenfunction of the Laplace-Beltrami operator $-\Delta_{0}$ in $W_{0}^{1,2}(S_{+}^{N-1})$. where $D^{2}v$ is the Hessian opertor,$g=(g_{i,j})$is the metric tensor on $S^{N-1}$.I want to know why the follownig two identities is satisfied: 1) $D^2 v(\nabla_{0}v,\nabla_{0}\phi)=1/2< \nabla_{0}|\nabla_{0}v|^{2}, \nabla_0\phi > $ 2) $D^{2}\phi + \phi g = 0$

You can read the paper " Ioslated Boundary Singularity of Semilinear Elliptic Equations" appeared in Calculus of Variations and Partial Differential Equations 2011, My question is come from the Theorem 2.1 of this paper, please visit the link http://arxiv.org/pdf/0902.0449.pdf

bigheadliao
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  • It might help you if you used consistent notation. What you write as $D$ and as $\nabla_0$ is the same thing, the covariant derivative. Then it's quite obvious that $\Delta_0$ is actually a trace of the Hessian. – Marek Dec 19 '13 at 09:36
  • http://math.stackexchange.com/questions/81203/definitions-of-hessian-in-riemannian-geometry here is proved your first identity. – Marek Dec 19 '13 at 09:54
  • The second is discussed in the post: http://cuhkmath.wordpress.com/2011/09/29/a-condition-for-a-riemannian-manifold-to-be-isometric-to-a-sphere/, indeed having such a nontrivial $\phi$ implies that the space must be a round sphere. –  Dec 20 '13 at 17:54

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