If $(X^{n},\omega)$ is a complete Kähler manifold with a global potential, i.e. $\omega=i\partial\bar{\partial}f$. There are many articles study the $L^{2}$-cohomology of $X$ under some conditions on $f$, for example, $df$ is bounded or sublinear. The authors stated that the $L^{2}$-cohomology $H^{p,q}_{(2)}(X)=0$ unless $p+q=n$. I want to know if $f$ is bounded, is the vanishing theorem also correct? Thanks a lot.
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1$\begingroup$ Let me make sure I understand your question. Is your $f$ such that $f$ is bounded AND $\omega=i\partial\bar\partial f$ is induced by a complete metric, or $f$ is just bounded, and $\omega$ is not necessarily complete? $\endgroup$– user48958Commented Jul 2, 2018 at 22:18
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1$\begingroup$ If $\omega$ is not complete, then it is trivial to show that the vanishing you mentioned does not hold. Take $X$ to be the unit ball in ${\mathbb C}^n$ and $f=\vert z\vert^2$. Then $H^{0,0}_{(2)}(X)$ is infinite dimensional. $\endgroup$– user48958Commented Jul 2, 2018 at 22:28
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1$\begingroup$ In my question, I suppose that $f$ is bounded and $\omega$ is induced by a complete metric. $\endgroup$– user94640Commented Jul 3, 2018 at 8:20
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$\begingroup$ I want to prove some vanising results by the method of Donnelly-Fefferman Ann.Math.118(1983). But $\bar{\partial}f$ and $\partial{f}$ in the process of calculation are always appearing. In my question, I only suppose $f$ has a bounded. I has no way to bounded $\bar{\partial}f$ and $\partial{f}$ by $f$. $\endgroup$– user94640Commented Jul 3, 2018 at 8:29
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