Let $M$ be a riemannian manifold. A function $f: M \mapsto \mathbb{R} $ is called spherical convex, if \begin{equation} \sin(\lvert xz \rvert) f(y) \leq \sin(\lvert xy\rvert) f(z) + \sin( \lvert yz\rvert ) f(x) \end{equation} For every 2 points $x,z$ and every third point $y$ lying on a shortest path between $x$ and $z$.

For a regular convex function one can say: A function $f$ is convex iff its graph is below the straight line connecting 2 points on the graph of $f$.

Is there a similar way to describe spherical convex functions?

  • $\begingroup$ What does the notation $|xy|$ stand for? $\endgroup$ – Harald Hanche-Olsen Oct 16 '12 at 9:34
  • $\begingroup$ The distance between the points x and y. Where the distance is given by $\vert xy \vert = d(x,y):= \inf \lbrace L(c) \vert c(0)= x , c(1)=y , c \quad smooth \rbrace$. L(c) denotes the length of the curve c $\endgroup$ – asterisk Oct 16 '12 at 9:39
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    $\begingroup$ Hmm. To me, this seems like a truly bizarre notion. In what connection does it arise? $\endgroup$ – Harald Hanche-Olsen Oct 16 '12 at 11:17
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    $\begingroup$ It arises in Alexandrov Geometry and spaces with curvature bounded below (There you work with spherical concave functions but the difference is just the sign). If u Start with a Alexandrov space A and a concave function $f:A \mapsto \mathbb{R}$ then u can define a directional derivative in every point $p \in A$ on $\Sigma A_p$. The space $\Sigma A_p$ is an Alexandrov space with dimension 1 less then A and curvature $\geq 1$ and the directional derivative is spherically concave. $\endgroup$ – asterisk Oct 16 '12 at 11:59

For your characterisation of convex functions, you are using the fact that the straight line is an optimiser of the inequality. More precisely, we try to solve

$$ |xz| f(y) = |xy| f(z) + |yz|f(x) $$

where $x,y,z$ are collinear. Without loss of generality, we can assume that $x,y,z\in \mathbb{R}$. Write $z = y + \delta y$ and $x = y - \delta y$ and doing a second order Taylor expansion we get that

$$ f'' = 0 $$

and so a necessary condition is that $f$ is linear. (We are making assumption of differentiability etc.) We then check that all linear functions $f$ satisfy this condition. And we can use it as an upper envelope of "convexity" between two points.

Now let us try to do the same with spherical convexity. We have $$ \sin( 2 \delta y) f(y) = \sin (\delta y) f(y + \delta y) + \sin (\delta y) f(y - \delta y) $$ Taking the Taylor expansion to $O(\delta y^3)$ on both sides we get $$ \left[ 2 \delta y - \frac{8}{6} (\delta y)^3\right] f(y) =_{O(\delta y^3)} \left[ \delta y - \frac{1}{6} (\delta y)^3\right] \left[ 2 f(y) + f''(y) (\delta y)^2\right] $$ which simplifies to $$ -f(y) = f''(y) $$ or that $f(y) = A \sin (y + B)$. We check that these functions indeed verify the hypothesis: assuming $z \geq y \geq x$, and $A = 1$ since the expression is scale invariant,

$$ \sin( z - x) \sin (y + B) \overset{?}{=} \sin( y - x) \sin (z + B) + \sin(z - y) \sin(x + B) $$ or $$ \left( \sin z \cos x - \cos z \sin x \right) \left(\sin y \cos B + \cos y \sin B\right) \overset{?}{=} \left( \sin y \cos x - \cos y \sin x\right) \left( \sin z \cos B + \cos z \sin B\right) + \left( \sin z \cos y - \cos z \sin y\right) \left(\sin x \cos B + \cos x \sin B\right) $$ which one can simply check to hold.

Therefore, the interpretation of "spherical convex" that is analogous to the characterisation of convex function is exactly like the standard convex case, except where the comparison is made with the straightline through $(x,f(x))$ and $(z,f(z))$, we compare against the unique function $g(s) = A \sin(s + B)$ defined over the geodesic connection $x$ and $z$, with $s$ an arclength parametrisation, such that $g(s(x)) = f(x)$ and $g(s(z)) = f(z)$.


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