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Let us consider a compact Riemann manifold $(M,g)$ and a function $u$ on $(M,g)$. $u$ also satisfies that $|\triangle_g u|\leq \varepsilon$ where $\varepsilon$ is a small constant. Can we obtain that $\operatorname{\mathop{osc}}(u)\leq C_\varepsilon$ for some small constant $C_\varepsilon$ dependent only on $\varepsilon$ and the geometry of $(M,g)$?

If this is not true, how to construct a counterexample?

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  • $\begingroup$ No, even if your manifold has dimension 1 (that is a circle). $\endgroup$ Dec 12, 2019 at 5:02
  • $\begingroup$ @AlexandreEremenko Thank you first! But in dimension 1, I think this is right. Since in dimension 1, the smallness of $\triangle_g u$ is equivalent to the smallness of the hessian $D^2 u$. Then by differential mean value theorem, we know $|Du(x)-Du(y)|\leq C_\varepsilon$ and if we take $y$ the maximal value point of $u$, we get the smallness of $Du$. And hence, we get the smallness of $osc (u)$ also by differential mean value theorem. Indeed, the question holds if we replace $\triangle u$ by $D^2 u$ for all dimension. $\endgroup$
    – Xiao Cao
    Dec 12, 2019 at 6:09
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    $\begingroup$ Up to addition by a constant, $u = -G \Delta u$, where $G$ is the Green operator for $-\Delta$. Known bounds for the kernel of $G$ imply that indeed the oscillation of $u$ is small. $\endgroup$ Dec 12, 2019 at 7:56
  • $\begingroup$ Your constant $C_\epsilon$ depends not only on $\epsilon$ as you required but also on the manifold (on the length of the circle in case $n=1$. The mean value theorem involves the length of the interval! $\endgroup$ Dec 12, 2019 at 12:30
  • $\begingroup$ @AlexandreEremenko I have edited the question such that $C_\varepsilon$ can dependent on the bounded geometry of $(M,g)$. $\endgroup$
    – Xiao Cao
    Dec 17, 2019 at 8:36

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