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I have seen at several places, incl. some notes and books, the following inference of the Holomorphic Implicit Function Theorem from the Smooth Real Function Theorem, but I believe this proof to be incorrect or rather incomplete in that it seems to be missing a non-obvious key step. I would like to know how to complete the proof if possible at all. Please note that there are of course other proofs of the Holomorphic Implicit Function Theorem that work, but my question is not about them, but about fixing this one. So, let me illustrate what I have in mind in the case of 2 complex variables:

Hol. Impl. Funct. Thm. in 2 var.-s: Let $U,V \subseteq \mathbb{C}$ be open subsets and let $f:U \times V \to \mathbb{C}$ be a holomorphic function. Let $(z_0,w_0) \in U \times V$ be a point such that $f(z_0,w_0) = 0$ and $\frac{\partial f}{\partial w}(z_0,w_0) \neq 0$. Then $z_0$ has an open neighbourhood $\widetilde{U} \subseteq U$ such that there exists a holomorphic function $g: \widetilde{U} \to \mathbb{C}$ with the property $g(z_0)=w_0$ and $\forall z\in \widetilde{U}: f(z,g(z)) = 0$.

Proof: By the Real Smooth Implicit Function Theorem there exist an open neighbourhood $\widetilde{U}\ni z_0$, $\widetilde{U}\subseteq U$, and a smooth $g:\widetilde{U} \to \mathbb{C}$ such that $\forall z \in \widetilde{U}: f(z,g(z))=0$ as smooth functions. Thus we only need to show that $g$ is holomorphic in $\widetilde{U}$. Since $f$ is holomorphic in both variables, one computes $$ 0 = \frac{\partial}{\partial\bar{z}} f(z,g(z)) = \frac{\partial f}{\partial w}(z,g(z)) \frac{\partial g}{\partial\bar{z}}, $$ hence at $(z_0,w_0)$ $$ 0 = \frac{\partial f}{\partial w}(z_0,w_0) \frac{\partial g}{\partial\bar{z}}(z_0), $$ from where it follows that $\frac{\partial g}{\partial\bar{z}}(z_0)=0$ since $\frac{\partial f}{\partial w}(z_0,w_0) \neq 0$ by hypothesis. $\Box$

The problem: this only shows that $g$ is complex-differentiable at the point $z_0\in\widetilde{U}$ rather than in all of $\widetilde{U}$, and none of the proofs I have seen actually justifies why the reasoning should extend to the whole neighbourhood.

Attempt to rectify the problem: by continuity of $\frac{\partial f}{\partial w}$ there are neighbourhoods $U'\ni z_0$ and $V'\ni w_0$ such that $\forall (z,w)\in U'\times V': \frac{\partial f}{\partial w}(z,w)\neq 0$. So we can take $\widetilde{U}\cap U'$ instead, but this does not suffice because we only know that $g(\widetilde{U}\cap U')\cap V' \ni w_0$, so we don't actually have that $$ \forall z\in \widetilde{U}\cap U': \frac{\partial f}{\partial w}(z,g(z))\neq 0. $$

Is there a way to salvage this proof without resorting to a completely different proof strategy?

(For example, a completely different strategy would be to invoke the Holomorphic Inverse Function Theorem.)

Feel free to add or remove tags as you see fit.

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    $\begingroup$ You get existence of a smooth $g$ mapping an appropriate neighborhood to an appropriate neighborhood. Now shrink $U$ by intersecting with $g^{-1}(V')$? (This is all in the standard implicit function statement/proof, actually.) $\endgroup$ – Ted Shifrin Sep 5 '18 at 21:53
  • $\begingroup$ @TedShifrin: but of course! $w_0\in V'$ and $g$ is continuous with $g(z_0)=w_0$. You are right on both points. I feel so silly now. Thanks for leaving this comment! $\endgroup$ – M.G. Sep 6 '18 at 9:11
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The smooth Implicit Function Theorem tells you that $g$ is smooth and $$ \frac{\partial g}{\partial z} = -\frac{\partial f}{\partial w}(z,g(z))^{-1} \frac{\partial f}{\partial z}(z,g(z)) $$ in a neighborhood of $z_0$, and the right hand side satisfies the Cauchy-Riemann equations, since $f$ is holomorphic. Hence $g$ is holomorphic.

To be rigorousm the equation above should be properly stated with the jacobian matrices and then see that the RHS satisfies Cauchy-Riemann.

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  • $\begingroup$ Let me make sure I understand the difference between both approaches correctly. The trick is to borrow both the existence of the neighbourhood and the validity of the expression for the Jacobians over said neighbourhood from the smooth Implicit Function Theorem. That works. $\endgroup$ – M.G. Sep 1 '18 at 13:12
  • $\begingroup$ BTW, shouldn't $\mathbb{C}$-linearity of the Jacobians be immediate from this expression (rather than having to verify the CR-equations)? $\endgroup$ – M.G. Sep 1 '18 at 13:14
  • $\begingroup$ Yes, it is immediate. $\endgroup$ – Alan Muniz Sep 1 '18 at 13:18
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    $\begingroup$ Actually I was thinking on preserving the complex structure when I wrote CR-equations. $\endgroup$ – Alan Muniz Sep 1 '18 at 13:19
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    $\begingroup$ Using the complex structure it is raher easy to extend this type theorem since it is just a linear condition on the Jacobians. $\endgroup$ – Alan Muniz Sep 1 '18 at 13:30

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