PomeLemma 11.15.4.319641577F72 Let $T$ be smooth and irreducible, let $U\subset T$ be a nonempty Zariski-open subset, and let $\mathcal E$ be a vector bundle on $X\times T$. For every $t_0\in T$, there are points $t_ j\in U$ tending to $t_0$ in the analytic topology, together with smooth complex-linear identifications of the bundles underlying $\mathcal E_{t_ j}$ and $\mathcal E_{t_0}$, under which the Dolbeault operators of the fibers converge in the $C^ m$ norm for every integer $m\geq 0$. The dual operators converge in the same sense.
Lemma 11.15.4319641577F72
Proof. The proper algebraic closed subset $T\setminus U$ has empty interior in the analytic topology. In an analytic coordinate ball $B$ around $t_0$, choose $t_ j\in U\cap B$ with $t_ j\to t_0$. Choose a smooth complex connection on $\mathcal E|_{X\times B}$. Parallel transport along radial paths in $B$, keeping $x\in X$ fixed, gives identifications
\[ \Phi _ t:\mathcal E_{t_0}^{\mathrm{sm}}\longrightarrow \mathcal E_ t^{\mathrm{sm}},\qquad \Phi _{t_0}=\operatorname {id}, \]
depending smoothly on $(x,t)$, where the superscript $\mathrm{sm}$ denotes the underlying smooth complex bundle. These identifications need not be holomorphic. Transport the Dolbeault operators by these maps and write
\[ \bar\partial _ t=\Phi _ t^{-1}\bar\partial _{\mathcal E_ t}\Phi _ t, \qquad A_ t=\bar\partial _ t-\bar\partial _{\mathcal E_{t_0}}. \]
Since both operators satisfy the same Leibniz rule, $A_ t$ is an endomorphism-valued $(0,1)$-form. Its coefficients are smooth in $(x,t)$, and $A_{t_0}=0$. Fix background metrics and a smooth connection on the central bundle. Compactness of $X$ and continuity of every finite collection of derivatives give
\[ \| A_{t_ j}\| _{C^ m(X)}\longrightarrow 0 \qquad \text{for every integer }m\geq 0. \]
The dual identifications are $(\Phi _ t^{-1})^*$. In transported dual frames the perturbation is $-A_ t^{\mathsf T}$, so the same convergence holds for the dual Dolbeault operators. $\square$
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