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Suppose that

  • $E$ and $F$ are complex Banach spaces and $U\subset E$ and $V\subset F$ are open subses.
  • $f\colon U\to V$ is analytic
  • $f\colon U\to V$ is bijective

Is $f$ bi-analytic? (i.e. is its inverse $f^{-1}\colon V\to U$ analytic?)

I heard this result is known but I could not find a reference

EDIT: As pointed out by @abx my original question in the real analytic setting has trivial counter examples.

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    $\begingroup$ What about $E=F=\mathbb{R}$, $f(x)=x^3$? $\endgroup$
    – abx
    Commented Jul 11, 2016 at 16:52
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    $\begingroup$ For a proof of the inverse function theorem in this setting see Adrien Douady's thesis in Annales Institut Fourier vol 16 1966 page 13 .As abx has pointed you need the differential of the map to be an isomorphism at every point . $\endgroup$
    – Mohan Ramachandran
    Commented Jul 11, 2016 at 17:01
  • $\begingroup$ @abx of course. One should in addition ask $f$ to be locally bijective on $U$. $\endgroup$
    – jan bernlöhr
    Commented Jul 11, 2016 at 20:26
  • $\begingroup$ thank you. I am well aware of the inverse function theorem. In fact as @abx pointed out I should rephrase the question as follows: if $f\colon U\to V$ is analytic an bijective, is it also bi-analytic. $\endgroup$
    – jan bernlöhr
    Commented Jul 11, 2016 at 20:28
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    $\begingroup$ Theorem 15.1.8 on page 302 in Rudin's Function Theory in the Unit Ball of $\mathbb{C}^n$ (1980 edition) provides an affirmative answer to your question when $E=F=\mathbb{C}^n$. $\endgroup$
    – T. Le
    Commented Jul 12, 2016 at 13:51

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