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In Hormander's Introduction to Complex Analysis in Several Variables, I'm confused by the usage of subordinate partitions of unity in the proof of a strengthening of the Mittag-Leffler theorem. In Theorem 1.4.3', we have $\Omega=\bigcup _j\Omega _j$ where the $\Omega _j$'s are open subsets of $\mathbb C$. Then, in the Proof of Theorem 1.4.5 we have

"We can choose a partition of unity subordinate to the covering $\left\{ \Omega _j \right\}$..."

What does this mean? The $\Omega_j$ are open subsets of $\mathbb C$ after all, and $\mathbb C$ does not have partitions of unity..

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  • $\begingroup$ I don't understand the question. What's your objection here? $\endgroup$ – user98602 Aug 18 '15 at 22:36
  • $\begingroup$ @MikeMiller the author only mentions smooth partitions of unity for real functions, so I'm confused as to choosing one subordinate to subsets of the complex numbers. $\endgroup$ – user153312 Aug 18 '15 at 22:54
  • $\begingroup$ Given a topological space $X$ and an open cover $U_j$, one defines partitions of unity subordinate to the cover $U_j$. $\Omega$ is certainly a topological space and $\Omega_j$ an open cover. If this doesn't clear the question up for you I can look at the theorem. $\endgroup$ – user98602 Aug 18 '15 at 23:17
  • $\begingroup$ @MikeMiller what confuses me is the fact the author freely chooses a (smooth) partition of unity subordinate to an open cover of a subset of $\mathbb C$, not $\mathbb R^n$. I found this confusing because smooth in the complex case means analytic, and there are no analytic bump functions. I think the author means to look at the sets $\Omega _i$ as subsets of $\mathbb R^2$. Is this right? $\endgroup$ – user153312 Aug 18 '15 at 23:23
  • $\begingroup$ No, smooth does not mean 'analytic'. Smooth means smooth as a real function from a subset of the reals. (That is to say, the word 'smooth' should always be read as 'smooth as a real function'.) You're free to describe it that way, but I would rather say to look at the sets $\Omega_i$ as open subsets of a real smooth manifold, which $\Bbb C$ certainly is. $\endgroup$ – user98602 Aug 18 '15 at 23:24
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When Hormander says 'smooth partition of unity', he doesn't mean 'holomorphic partition of unity'. $\Bbb C$ (and $\Omega$) is a real smooth manifold, hence has smooth partitions of unity corresponding to any open cover.

This is just a terminology thing: even though one often hears 'a smooth function is one that is infinitely differentiable', when one says smooth, "infinitely real differentiable" is always meant, even for complex-valued functions on a complex manifold. "Complex differentiable" is just 'analytic' or 'holomorphic'. And never the twain shall meet.

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