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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?

Many thanks in advance for any comments or advice.

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  • $\begingroup$ Is that equivalent that the function is covex on any geodesic? Did you try $f(x)=1/x_{n+1}$ $\endgroup$ – user35593 Feb 28 '19 at 1:09
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    $\begingroup$ 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. $\endgroup$ – Anton Petrunin Feb 28 '19 at 6:04
  • $\begingroup$ 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. $\endgroup$ – John Sung Feb 28 '19 at 8:14
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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.

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  • $\begingroup$ 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? $\endgroup$ – John Sung Mar 1 '19 at 13:38
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    $\begingroup$ yes the function is not defined on the equator but your domain does not contain the equator. It also does not make sense to talk about convexity at the equator. It is probably not possible to find a function which can be extended to the equator because of Anton Petrunin's argument. $\endgroup$ – user35593 Mar 1 '19 at 13:52
  • $\begingroup$ 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. $\endgroup$ – John Sung Mar 1 '19 at 14:12

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