# Is there exists (strictly) convex function on hemisphere?

Given $$\mathbb{S}^n_+:=\{x\in \mathbb{R}^{n+1}: |x|=1,x_{n+1}>0\}$$ be the open domain in $$\mathbb{S}^n$$, or be viewed as the geodesic ball centered at the pole with radius $$\frac{\pi}{2}$$ in $$\mathbb{S}^n$$, where $$\mathbb{S}^n$$ equiped with the standard spherical metric. My question is: Whether there exists a smooth positive strictly convex function $$f$$ defining in $$\mathbb{S}^n_+$$? Here strictly convex means the Hessian of function$$f:\mathbb{S}^{n}_+ \to \mathbb{R}^+$$ is positive definite with respect to the spherical metric $$g_{\mathbb{S}^n}$$, i.e. there exists $$c_0>0$$ such that $$\nabla^2f\geq c_0 g_{\mathbb{S}^n}$$. Another further questions： Is there exists a strictly convex function defining on the geodesic sphere (within injectivity radius) located in ambient Riemannian manifold space with nonnegative curvature?

• Is that equivalent that the function is covex on any geodesic? Did you try $f(x)=1/x_{n+1}$ – user35593 Feb 28 '19 at 1:09
• Note that, the circle $\sigma_\varepsilon$ of radius $\tfrac\pi2-\varepsilon$ in the hemisphere is closed curve and its curvature is small if $\varepsilon$ is. If there is a strongly convex function $f$ then the restriction of $f|_{\sigma_\varepsilon}$ would be strictly convex --- a contradiction. – Anton Petrunin Feb 28 '19 at 6:04
• Thanks, Yes, it is equivalent to the function is strictly convex when restricted it to any geodesic $\gamma(t)$ be strictly convex, $\frac{d}{dt^2} f(\gamma(t))\geq c_0>0$, for any geodesic $\gamma(t)\subset \mathbb{S}^n_+$. But the circle $\sigma_{\epsilon}$ of radius $\frac{\pi}{2}$ in the hemisphere is not a geodesic? how does the contradiction coming out? Thanks for more explanation. – John Sung Feb 28 '19 at 8:14
The function $$f(x)=1/x_{n+1}$$ is strictly convex. A geodesic curve on the sphere can be written as $$\gamma(t)=\sin(t)u+\cos(t)v$$ with $$|u|=|v|=1$$ and $$\langle u,v\rangle=1$$. Hence $$f(\gamma(t))=1/(\sin(t)u_{n+1}+\cos(t)v_{n+1})=c/ \sin(t+a)$$, for some $$a$$ and $$c\geq 1$$. Hence it remains to prove that $$1/\sin$$ is strictly convex on $$[0,\pi]$$. The second derivative of $$1/\sin$$ is $$(2-\sin^2)/\sin^3\geq 1$$ hence we have strict convexity.
• Thanks for the detail. Since the function $f=\frac{1}{x_{n+1}}$ has no definition on the equator (the boundary of hemisphere)? When restricting it to the boundary of $\mathbb{S}^n_+$, how to see it is strict convexity on the equator? – John Sung Mar 1 '19 at 13:38
• Initially, I want to find a function which is well-defined in the whole domain $\overline{\mathbb{S}}^n_+$ and strictly convex in the interior. So there is impossible. But it is true on the spherical cap due to your example. Many thanks. – John Sung Mar 1 '19 at 14:12