Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?


If $\dim\_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim\_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.


If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?



Can a metric conformal to a Kahler metric be Kahler?
You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$. 


This is a linear algebra fact, valid pointwise : if $\alpha \in T\_x^\*X$ satisfies $\omega\_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim\_R X \geq 4$. 


The short argument is that, $\omega\_x^{n-1}\wedge : T^\*\_x X\to \bigwedge^{2n-1} T^\*\_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).


This said, as in Francesco's answer, you can have non proportional conformal *riemannian* metrics that are Kähler with respect to *different* complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.


