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 $H_{-a}$.

  • $\begingroup$ Good question. Forgive my ignorance, is this notion of "geodesically convex" equivalent to the Hessian matrix being positive definite, in some appropriate sense? $\endgroup$ Nov 15 '20 at 1:50
  • $\begingroup$ 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 $\endgroup$
    – dohmatob
    Nov 15 '20 at 2:35
  • 1
    $\begingroup$ Thank you for recommending that book. It is a really interesting read. $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$
    – dohmatob
    Nov 15 '20 at 7:02
  • $\begingroup$ @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. $\endgroup$
    – Kajelad
    Nov 16 '20 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.