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Is there any example of a function $f:M\rightarrow\mathbb{R}$ such that the following condition holds $$\nabla^2f_p(X,Y)\geq Ric_p(X,Y)f(p)\ \forall p\in M \text{ and } X,Y\in T_pM$$ where $\nabla^2$ is the Hessian operator and $Ric_p$ is the Ricci tensor at $p$. I know that every convex function in Euclidean space satisfies the condition but I can not find any other example in other Riemannian manifold. Please help.

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  • $\begingroup$ Well $f(m)=0$ for all $m \in M$ always works but it's not very interesting. $\endgroup$ – Maik Pickl Sep 11 '17 at 14:49
  • $\begingroup$ Existence of some non trivial function will be more interesting $\endgroup$ – chandan mondal Sep 11 '17 at 14:55
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    $\begingroup$ I think your condition is a little strong. Suppose that $f(p)$ is not zero and that either $Ric_p$ or $\Delta^2f_p$ is not constant zero. Then your condition can never be fulfilled since you can always replace $X$ with $-X$ and hence your inequality can not be true for all $X,Y \in T_pM$. Well maybe it can but only if you get equality. $\endgroup$ – Maik Pickl Sep 11 '17 at 15:57
  • $\begingroup$ Yes, surely you mean $\nabla^2 f \ge f \cdot \mathrm{Ric}$ in the sense of quadratic forms? Otherwise even your convex/Euclidean example fails as described by Maik. $\endgroup$ – Anthony Carapetis Sep 12 '17 at 0:27
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Let $M$ be a complete manifold with sectional curvature $L\leq 0$. The function $f$ defined on the warped product manifold $R\times_{e^t} M$ by $$f(t,x)=e^t$$ is positive and has a positive definite Hessian. I think this function satisfies the condition. For more details, you may see the interesting article Here

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