Let $\mathbb S_{n-1}$ be the $(n-1)$-dimensional sphere in $\mathbb R^n$. Fix $a \in \mathbb R^n$ and consider a function of the form $F(x) = f(x^\top a)$, for some one-dimensional function $f:\mathbb R \to \mathbb R$.
Question. What is a sufficient (and hopefully, necesssary) condition on $f$ for which $F$ is geodesically convex on $\mathbb S_{n-1}$ ?
You may assume $f \in \mathcal C^2(\mathbb R)$.
Definition. Recall that if $\mathcal M$ is a manifold, then a function $G:\mathcal M \to \mathbb R$ is said to be geodescially convex iff $G\circ \gamma:[0,1] \to \mathbb R$ is convex for every geodesic curve $\gamma:[0,1] \to \mathcal M$. Geodesic strong-convexity, geodesic concavity, and geodesic strong-concavity are defined similarly.
Examples of functions that I have in mind
- Linear function. $f(z) \equiv z$
- Nonlinear functions. $f(z) \equiv e^{z}$ or $f(z) \equiv \log(1 + e^{z})$.
Note that all of the above functions are convex in the usual sense in $\mathbb R^n$.
Update: solution to case of linear function
Consider the case where $f(z) \equiv z$. One computes
- Euclidean gradient: $\nabla_{\mathbb R^n} F(x) = a$.
- Euclidean hessian: $\nabla_{\mathbb R^n}^2 F(x) = 0$.
Therefore, using standard formula, the riemannian hessian is $$ \nabla_{\mathbb S_{n-1}}^2 f(x)[u] = P_{x}(\nabla_{\mathbb R^n}^2 f(x)[u]) - \mbox{trace}(x^\top \nabla_{\mathbb R^n}f(x))u = -F(x)u. $$
Therefore if we define the spherical cap $H_a := \{x \in \mathbb S_{n-1} \mid x^Ta < 0\}$, then $$ \nabla_{\mathbb S_{n-1}}^2 f(x)[u,u] = -F(x)\|u\|^2 = \begin{cases}\le 0,&\mbox{ if }x \in H_{a},\\\ge 0,&\mbox{ if }x \in H_{-a}\end{cases} $$
and so $F$ is geodesically convex on $H_a$ and geodescially concave on $H_{-a}$.
The resulting function must be constant, since all geodesically convex functions on $S^n$ are constant.
for any function $f:S^1\to\mathbb{R}$, there is a corresponding periodic function $\tilde{f}:\mathbb{R}\to\mathbb{R}$. $f$ is geodesically convex iff $\tilde{f}$ is convex on any closed interval, which, by periodicity, is only the case if $\widetilde{f}$ is constant.
A geodesically convex function $f:S^n\to\mathbb{R}$ must be constant along all great circles by the previous argument, and thus $f$ must be constant on $S^n$, since any two points are joined by a great circle.
This result isn't specific to $S^n$: If I'm not mistaken, all geodeiscially convex functions are constant on any compact, connected Riemannian manifold without boundary.