I'll denote by $\mathbb{C}^*$ the punctured complex plane $\mathbb{C} \setminus \{0\}$. Say that I've got some open cover $\{V_j\}_{j \in J}$ of the closed unit interval $[0,1]$, and continuous functions $w_j: V_j \times \mathbb{C}^* \to \mathbb{C}^*$ such that, for any $t \in V_j$,

(a) $w_j(t, \cdot)$ is a holomorphic function $\mathbb{C}^* \to \mathbb{C}^*$ and

(b) $w_j(t, \cdot)$ has a holomorphic antiderivative, say $F: \mathbb{C}^* \to \mathbb{C}$.

I want to find a function $u: [0,1] \times \mathbb{C}^* \to \mathbb{C}^*$ that such that, for fixed $t \in [0,1]$, $u(t, \cdot)$ is still holomorphic and still has a holomorphic antiderivative.

My first thought was to use partitions of unity with respect to this cover, say $\{p_j\}_{j \in J}$ and to set $u(t, \cdot) = \sum_{j} p_j(t)w_j(t, \cdot)$. Of course, such a $u$ does still have an antiderivative for each $t$ but now we can't be sure whether the image of $u$ still lies in $\mathbb{C}^*$.

I get that this is a bit vague, but I'd appreciate any good advice and/or ideas...can someone think of another way of constructing $u$. Or perhaps I can pursue the partitions of unity idea, provided $w_j$ and/or the open cover have some additional features, and if so, what kinds of features would enable this? I'd appreciate any ideas.

  • $\begingroup$ You don't seem to require any compatibility relations between this $u$ and the previous data, so it's unclear to me what kind of $u$ do you want to construct. Could you give some additional precisions ? $\endgroup$ – Nicolas Hemelsoet Dec 20 '18 at 11:57
  • $\begingroup$ P.S I think your previous idea should work with some additional modifications, for example you can assume that $J$ is countable and fix an arbitrary ordering on it. Now it should be possible to inductively build the $p_j$ so that the function is still $\Bbb C^*$ valued. $\endgroup$ – Nicolas Hemelsoet Dec 20 '18 at 11:59
  • $\begingroup$ Thanks for your remarks! Is it actually possible to build the partition of unity inductively so that the function is still nonzero? I'm not quite sure if I understand what you mean! $\endgroup$ – Acton Dec 21 '18 at 2:00

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