# $\lambda$- geodesic convexity

Let $$(X,d)$$ be a geodesic space, suppose $$g:X \to \mathbb{R}$$ is geodesically $$\lambda$$ convex with $$\lambda >0$$, that is for all $$x,y \in X$$ there exists a constant speed geodesic $$\gamma$$ from $$x$$ to $$y$$ such that for all $$s\in [0,1]$$ $$g(\gamma_s) \leq s g(x) + (1-s) g(y) -\frac{\lambda}{2}s(1-s)d(x,y)^2$$ Let $$x_0$$ be the unique minimizer of $$g$$. Show that $$\forall x$$ $$g(x) \geq g(x_0) + \frac{\lambda}{2}d(x,x_0)^2$$

EDIT: thanks for the comment. Honestly I did not do much because I was stuck from the beginning, since I had no idea how to use the geodesic property. I just picked a geodesic between $$x$$ and $$x_0$$ and tried to plug in the property but did not see how that could help! Any hint or suggestions?

Choose your geodesic with $$\gamma_{0}=x_{0}$$ and $$\gamma_{1}=x$$. Since $$x_{0}$$ is minimal we have $$$$0=\frac{d}{ds}|_{s=0}g(\gamma_{s}).$$$$ Then, using your first inequality it follows $$$$0=\lim_{h\to 0}\frac{1}{h}(g(\gamma_{h})-g(x_{0}))\le\lim_{h\to 0}\frac{1}{h}\left(hg(x)+(1-h)g(x_{0})-\frac{\lambda}{2}h(1-h)d^{2}(x,x_{0})-g(x_{0})\right)$$$$ which yields the claim.
• That's essentially correct, but in general you only have $0\leq \liminf_{h\to 0}(g(\gamma_h)-g(x_0))$ - think of $g(x)=|x|$ on $\mathbb{R}$ for example. Commented Nov 30, 2020 at 17:13