Let $X$ be a compact Kähler manifold. Show that if two Kähler forms $\omega, \omega'$ satisfy $[\omega]= [\omega'] \in H^2(X,\mathbb R)$, then there exists a real function $f$ such that $\omega = \omega' + i \partial \bar \partial f$.
My idea was to use that locally we can find such a function, i.e. $\omega-\omega'= i \partial \bar \partial f$.
But why does the right side extend globally?
On
You can use the $\partial \bar \partial$ lemma:
Note that $\omega - \omega' = d\alpha$ for some $1$-form $\alpha$.
Further, $d\alpha$ is closed and exact, thus it is $\partial \bar \partial $-exact.
So we get that $\omega-\omega'= \partial \bar \partial \beta$. But $\beta$ has to be a $0$-form, thus a function. Set $\beta= i f$.
$f$ is not necesarrily real; but we can simply take its real part and the demanded condition still holds.
For details, see the last sentence in the answer of Michael Albanese.
As $[\omega] = [\omega']$, $\omega = \omega' + d\alpha$ for some real one-form $\alpha$. As $\omega$ and $\omega'$ are real $(1, 1)$-forms, so is $d\alpha$. Moreover, $d\alpha$ is $d$-closed, but it is also $d$-exact, so by the $\partial\bar{\partial}$-lemma, $d\alpha$ is $\partial\bar{\partial}$-exact. That is, there is a complex-valued function $h$ such that $d\alpha = \partial\bar{\partial}h$. Now let $h = ig$, then $\partial\bar{\partial}h = i\partial\bar{\partial}g$. As $i\partial\bar{\partial}g$ is real, $i\partial\bar{\partial}g = \overline{i\partial\bar{\partial}g} = -i\bar{\partial}\partial\bar{g} = i\partial\bar{\partial}\bar{g}$; note, this does not necessarily mean that the function $g$ is real. Instead, let $f = \frac{1}{2}(g + \bar{g})$, then we have
$$i\partial\bar{\partial}f = \frac{1}{2}(i\partial\bar{\partial}g + i\partial\bar{\partial}\bar{g}) = \frac{1}{2}(i\partial\bar{\partial}g + i\partial\bar{\partial}g) = i\partial\bar{\partial}g = d\alpha = \omega - \omega'.$$