Suppose $f: X \to B$ is a proper and smooth morphism of complex manifolds, and suppose the fiber $X_0 \subset X$ over $0 \in B$ is a Kähler manifold. How can I show that there is a neighbourhood $U \subset B$ of $0$, such that all fibers $X_t, t \in U$ are also Kähler manifolds?
Here is my attempt:
Let $h = g - i \omega$ be the Hermitian metric on $X$ with Riemannian metric $g = \Re(h)$ and $- \omega = \Im(h)$, such that $\omega_0 = \omega|_{X_0}$ is a Kähler form. I think such a metric can be choosen using a partition of unity on an open cover, which trivializes both the tangent bundle of $X$ and the tangent bundle of $X_0$.
But now I don't know why $\omega_t = \omega|_{X_t}$ would still be closed, for $t \in B$ nearby $0$. If I choose a differential geometric trivialization $f^{-1}(U) \cong X_0 \times U$ (by Ehresmann's theorem), then I can consider $d\omega_t$ as a (continuous?) function $U \to H^3_{\text{dR}}(X_0, \mathbb R)$. However, the zero-set of a continuous function is closed, not open...
I stumbled upon this when reading a proposition by Beauville[1], that claims that if $X_0$ is Kähler and holomorphically symplectic, then the neighbouring fibers are as well Kähler and holomorphically symplectic. The Kähler part is not elaborated on though.
[1] Proposition 9 in Varietés Kähleriennes dont la première Classe de Chern est nulle.
Using Ehresman’s theorem you can assume $X=B \times X_0$ where $X_0$ is compact Kähler – the complex structure on $X_0$ varies with the basis point though (ie the isomorphism $X_t \subset X \rightarrow X_0$ ($t \in B$) is real-smooth, not holomorphic).
Let $\omega_0$ be a Kähler form on $X_0$, $\omega$ its pullback to $X$. Then $\omega$ is a closed $2$-form on $X$, thus so are its restrictions to each $X_t$. And now consider the projection of $\omega_t$ on the subspace of $(1,1)$-forms for $X_t$: it’s still closed and it is a Kähler form.