Suppose we have a measure $\mu(dx)= e^{-V(x)}d\text{vol}$ on a Riemannian manifold M with $V\in C^2(M)$, $\int e^{-V}d\text{vol} = 1$ and $\text{Ric}+\text{Hess(V)}>0$. Is it then true that $\int V^2e^{-V}d\text{vol}<\infty$? This holds on $\mathbb{R}^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.

  • $\begingroup$ May I know how this is true for $\mathbb R^n$? $\endgroup$
    – user99914
    Jul 24, 2018 at 4:09
  • $\begingroup$ My attempt on $\mathbb{R}^n$: $V$ is coercive and thus bounded from below. Let $k_0:= \lfloor \inf V\rfloor>-\infty$ and, for any $k\in\mathbb{Z}$, let $A_k := \left\{k\leq V<k+1\right\}$. \\ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $\text{vol}(A_k)\leq Mk^{n-1}$ for all $k$. Thus, \begin{align*} \int V^2e^{-V} d x \leq \sum_{k=k_0}^\infty (k+1)^2e^{-k}\text{vol}(A_k) \leq M \sum_{k=k_0}^\infty (k+1)^2e^{-k}k^{n-1} < \infty ,\end{align*} where the last sum is finite by the integral test for convergence. $\endgroup$
    – Corram
    Jul 24, 2018 at 8:41


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