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Prove: For a function $f$, if $f^2$ and $f^3$ are analytic in G (a set in C), then $f$ is analytic in C.

I don't get the question. Why would I need to know that $f^3$ is analytic if I already know that $f^2$ is. Can't I just derive the Taylor series expansion? Any help with this would be appriciated!

John
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    For example, let $f(x)=1$ when $x\neq 0$ and $f(0)=-1$. Then $f^2$ is analytic, since it's the constant function $1$, but $f$ is not even continuous. – Nishant Feb 03 '15 at 17:16
  • Google search of the title on this site yields http://math.stackexchange.com/q/348627/; – Jonas Meyer Feb 03 '15 at 17:17
  • Mathematicians didn't get the question for quite a while either. It wasn't until 1982 (I was alive!) when Henri Joris proved it. – Pp.. Feb 03 '15 at 17:31
  • @Pp..: What are you referring to? I don't think it is the question here. – Jonas Meyer Feb 03 '15 at 17:37
  • @JonasMeyer The original problem. – Pp.. Feb 03 '15 at 17:39
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    @Pp..: This basic problem would have been solved long before you were born. Fortunately Googling revealed what you must have meant, a theorem about real differentiability. http://projecteuclid.org/euclid.kmj/1138038992 – Jonas Meyer Feb 03 '15 at 17:40
  • @JonasMeyer I know what I meant. The complex and real cases are as simple as each other. That has nothing to do with the theorem not being proven until 1982. The quasianalytic case (in several variables) remains unsolved. – Pp.. Feb 03 '15 at 17:48
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    @Pp..: I don't know why you think the case of real smooth functions would be as simple as the case of complex analytic functions. If it were, Henri Joris wouldn't have had work to do. You can find elementary solutions to this problem at the linked duplicates, not using any results found since the birth of any math.se user. But thanks for sharing the info about Henri Joris, etc. – Jonas Meyer Feb 03 '15 at 20:20
  • @JonasMeyer Because the solution for the real case is the same as the one linked. Joris, however, didn't proved that way. The result seems to have just been late to appear. Maybe not so many people looked at it. – Pp.. Feb 03 '15 at 20:25
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    @Pp..: No, Joris's work applies to real smooth functions, and for these, the arguments in the linked duplicates would not apply. If they did, Joris would not have proved something new. You cannot say that if $g$ is a real $C^\infty$ function with $g(x_0)=0$, then there exists a positive integer $n$ and a real $C^\infty$ function $h$ such that $g(x)=(x-x_0)^nh(x)$ and $h(x_0)\neq 0$. Zeros are much more complicated for real smooth functions, and that is why Joris's result is not trivial. [I will comment no more here.] – Jonas Meyer Feb 03 '15 at 20:34
  • @JonasMeyer That proof only needs to be modified minimally to escape that problem. Google "A elementary proof of Joris theorem". – Pp.. Feb 03 '15 at 20:44

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