This is an old revision of the document!

Tema 7: Hadamard manifolds - part 2

Convexity properties

We continue exploring the convexity properties on Hadamard manifolds.

Convex function: A function $f:M\to\mathbb R$ is convex if for every geodesic $\gamma$ the composition $f\circ\gamma:\mathbb R\to\mathbb R$ is convex.

Convex set: A set $C\subset M$ is convex if for every $x,y\in C$ the segment $xy\subset C$.

Here are examples of convex sets: if $f:M\to\mathbb R$ is a convex function then $f^{-1}(-\infty,\alpha]$ is a convex set for all $\alpha\in\mathbb R$. Indeed, let $f(x)\leq \alpha$ and $f(y)\leq\alpha$, and let $\gamma:[0,t]\to M$ be the segment $xy$. Since $f\circ\gamma:[0,t]\to \mathbb R$ is convex, it is bounded by $\alpha$ and so $xy\subset f^{-1}(-\infty,\alpha]$. The main result of this subsection is the following.

Proposition 1. If $\gamma_1,\gamma_2$ are geodesics on $M$, then $t\mapsto d(\gamma_1(t),\gamma_2(t))$ is convex.

Proof. Fix $t_1,t_2$, let $t=\tfrac{t_1+t_2}{2}$, and let $x$ be the midpoint of the segment $\gamma_1(t_1)\gamma_2(t_2)$. By the triangle inequality, $$ d(\gamma_1(t),\gamma_2(t))\leq d(\gamma_1(t),x)+d(x,\gamma_2(t)). $$ By Corollary 2 of the previous post, $d(\gamma_1(t),x)\leq \tfrac{d(\gamma_1(t_2),\gamma_2(t_2))}{2}$ and $d(x,\gamma_2(t))\leq \tfrac{d(\gamma_1(t_1),\gamma_2(t_1))}{2}$, thus $$ d(\gamma_1(t),\gamma_2(t))\leq \frac{d(\gamma_1(t_1),\gamma_2(t_1))+d(\gamma_1(t_2),\gamma_2(t_2))}{2}\cdot $$

Corollary 1. If $C$ is a convex set, then $x\mapsto d(x,C)$ is a convex function.

Proof. For simplicity, assume that $C$ is closed (the general case follows by approximating $d(x,C)$ by $d(x,y)$ for $y\in C$). Fix a geodesic $\gamma$, and $x=\gamma(t_1)$, $y=\gamma(t_2)$. Let $\overline{x},\overline{y}$ s.t. $d(x,C)=d(x,\overline{x})$ and $d(y,C)=d(y,\overline{y})$. If $z,\overline{z}$ are the midpoints of $xy,\overline{x}\overline{y}$ then by the previous proposition: $$ d(z,C)\leq d(z,\overline{z})\leq \frac{d(x,\overline x)+d(y,\overline y)}{2}=\frac{d(x,C)+d(y,C)}{2}\cdot $$