Denote by $S^n$ the unit sphere in $\mathbb{R}^{n+1}$, and consider the coordinate function $x_{n+1}$ on it, i.e. the function $(x_1, \ldots, x_{n+1}) \mapsto x_{n+1}$. Denoting by $\mathrm{Hess}(x_{n+1})$ the Hessian of $x_{n+1}$ with respect to the round metric $g$ on $S^n$, it is claimed in the paper that I am reading that $\mathrm{Hess}(x_{n+1}) = -x_{n+1} \cdot g$.
I have to admit that I have trouble varifying this identity. How can one prove it?
edit: As Deane suggests, I am posting my computation. I am parametrizing the upper hemisphere as $$ \phi(x_1, \ldots, x_n) := (x_1, \ldots, x_n, \sqrt{1-|x|^2}), $$ where $|x| := |(x_1, \ldots, x_n)|$ is the usual Euclidean norm. The coordinate function $f := x_{n+1}$ is then given by $f(x_1, \ldots, x_n) := \sqrt{1-|x|^2}$ and therefore $$ \partial_i \partial_j f = \frac{-x_i x_j}{(1-|x|^2)^{3/2}} $$ for $i \not= j$ (let me just treat this case here). On the other hand, to compute $g_{ij}$ we first compute $\partial_i \phi = \frac{-x_i}{(1-|x|^2)^{1/2}}$ and therefore $$ g_{ij} = g(\partial_i \phi,\partial_j \phi) = \frac{x_i x_j}{1-|x|^2} $$ for $i \not= j$. For $-f\cdot g$ we finally get, in the case $i \not= j$, $$ (-f \cdot g)_{ij} = \frac{-x_i x_j}{(1-|x|^2)^{1/2}}. $$ This is, as I said in the comments, off by the factor $\frac{1}{1-|x|^2}$ from the claimed answer.
edit 2: I think I found my error. I forgot to evaluate the matrix $(\partial_i\partial_j f)_{ij}$ on the basis vectors. And indeed, I get $$(\partial_i\partial_j f)_{ij}(\partial_i \phi,\partial_j \phi) = \frac{-x_i x_j}{(1-|x|^2)^{1/2}}$$ (for $i \not= j$) as claimed.
