# Sufficient condition on $f:\mathbb R \to \mathbb R$ to ensure the function $F:x \mapsto f(x^\top a)$ is geodesically convex on $n$-sphere

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 $$H_{-a}$$.

• Good question. Forgive my ignorance, is this notion of "geodesically convex" equivalent to the Hessian matrix being positive definite, in some appropriate sense? Nov 15 '20 at 1:50
• That's definitely a valid question to ask. Yes, indeed it's a theorem: If $M$ is a an open geodesically convex subset of $\mathbb R^n$ and $f$ is $\mathcal C^2$ on $M$, then $F$ is geodesically convex on $M$ iff $\nabla_M^2 f(x)$ is psd for every $x \in M$. See Theorem 11.19 of sma.epfl.ch/~nboumal/book/IntroOptimManifolds_Boumal_2020.pdf Nov 15 '20 at 2:35
• Thank you for recommending that book. It is a really interesting read. Nov 15 '20 at 23:51

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.
• Yes, indeed geo convexity of functions is usually motivated only on open subsets. My question as phrased doesn't quite make much sense. I really on interested in subsets of the sphere on which $F$ would be geo convex. Nov 15 '20 at 7:02
• @dohmatob More specifically, it's common to restrict attention to geodesically convex domains, where convexity of functions behaves much the same as on convex subsets of $\mathbb{R}^n$. Without this restriction, things are not so well behaved. Nov 16 '20 at 0:52