Can deformation equivalent Kähler manifolds always be obtained by a deformation where all the fibers are Kähler?

Question

Given compact Kähler manifolds $X$ and $X'$ deformation equivalent over the unit disk $\Delta \subset \mathbb{C}$. More precisely, there is a proper holomorphic surjective map \begin{align*} \pi\colon \mathcal{X}\to \Delta \end{align*} and $t,t' \in \Delta$ such that $X$ and $X'$ are biholomorphic to the fibers $\pi^{-1}(t), \pi^{-1}(t')$ respectively. Is there a deformation of $X$ and $X'$ over $\Delta$ such that every fiber is Kähler?

I am specially interested in the case where $X$ and $X'$ are of hyperkähler type, i.e. irreducible holomorphic symplectic. In other words, simply connected and admitting a unique holomorphic symplectic form. I know that there are (large) deformations where the deformed space is not Kähler.

Answer 1

I don't think this is known. For hyperkahler manifolds, conjecturally, all smooth complex deformations are class C and birational to hyperkahler. If this is true, your conjecture would follow automatically. The only relevant publication that I am aware of is

https://arxiv.org/abs/1703.02001

Arvid Perego

Kählerness of moduli spaces of stable sheaves over non-projective K3 surfaces

We show that a moduli space of slope-stable sheaves over a K3 surface is an irreducible hyperkähler manifold if and only if its second Betti number is the sum of its Hodge numbers h2,0, h1,1 and h0,2.

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Perego proves that a (smooth) limit of hyperkahler manifolds is Fujiki class C if $b_2=h^{2,0}+ h^{1,1} + h^{0,2}$. This is a bit weaker than what you need, of course.

All the best
Misha