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Assume that $f:\mathbb{C}^n\rightarrow \mathbb{C}$ is a $C^{\infty}$ function such that $f^2$ is (complex) analytic. Then one can show that $f$ is analytic. (Note: Liviu Nicolaescu and Alexandre Eremenko have given elegant proofs of this below; my own proof involved the Weierstrass preparation theorem.)

Now, when we consider the question for functions in some quasi-analytic Denjoy-Carleman class these proofs don't carry over:

Question: If $f^2$ is a function in some quasi-analytic Denjoy-Carleman class, then $f$ is quasi-analytic belonging to the same class?

Weierstrass preparation theorem doesn't hold, for $n\geq2$, in quasi-analytic Denjoy-Carleman classes, and it is an open problem whether a $C^\infty$ function that belongs to a quasi-analytic Denjoy-Carleman class along every line belongs to that class. This is part of the difficulty of this problem for $n\geq2$.

Another open question in Denjoy-Carleman classes is about whether ideals are closed. For principal ideals this is related to solving for $f$ in $gf=h$, where $g$ and $h$ are known to belong to the Denjoy-Carleman class. The ideal generated by $g$ would not be closed if we can find a smooth $f$ that doesn't belong to the class such that it gets pushed into the class by the multiplication by $g$. In this way, the question above is about understanding whether a smooth function, not in the class, can be pushed into the class by multiplying by itself. If the square appears composed with $f$ on the other side then it is known to occur. This is, $f(x^2)$ may be in a quasi-analytic Denjoy-Carleman class while $f$ is not.

Note: I put in the tags model-theory and o-minimal because people that work in those areas sometimes have also worked with quasi-analytic functions enough to maybe have an idea for proving it. Subject classification in mathematics doesn't play the same role for exposition than for finding proofs.

Note: Joris' theorem states that if $f^2$ and $f^3$ are smooth functions then $f$ is smooth. This holds for several variables functions. Also, for the quasianalytic case, in the case of one variable, it is again easy to show that if $f$ is smooth and $f^2$ and $f^3$ belong to some quasianalytic Denjoy-Carlmenan class, then $f$ also belong to the same class. For several variables ... who know!? The problem with just $f^2$ is also a subproblem steaming form asking if Joris' theorem is true for quasianalytic DC classes.

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    $\begingroup$ And where does the question come from? $\endgroup$
    – Igor Rivin
    Nov 20, 2013 at 20:09
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    $\begingroup$ The phrasing of the first question suggests homework (thus, someone else's head...) -- if you say: "prove that..." that indicates you already know the answer, which indicates that this is not an appropriate question. $\endgroup$
    – Igor Rivin
    Nov 20, 2013 at 20:17
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    $\begingroup$ To prove that $f^2$-holomorphic $\Rightarrow$ $f$ holomorphic is an easy exercise. $\endgroup$ Nov 20, 2013 at 21:05
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    $\begingroup$ If $f^2$ is holo, then $0=\bar{\partial}(f^2)=2f\bar{\partial} f$ so $f$ is holo in the open set $\{f\neq 0\}=\{f^2\neq 0\}$. This is a dense open set because otherwise the holo function $f^2$ will be identically zero. Hence $\bar{\partial} f$ is zero on a dense open set, thus zero everywhere. $\endgroup$ Nov 20, 2013 at 21:31
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    $\begingroup$ I have edited the question to remove impressions it may have given of being offensive or competitive or the like. I'm going to see now about removing or editing comments that are contentious. Let's please stick to mathematics now; I think there's a question here and it would be a shame to close it on account of bickering. (Comments between ABC and BS have now been removed, and comments by Paul and Liviu on the exchange between ABC and BS have been removed. Apologies for the heavy moderator intervention here; I'll assume there were misunderstandings due to language.) $\endgroup$
    – Todd Trimble
    Nov 20, 2013 at 23:40

1 Answer 1

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Here is a simple proof for complex-analytic case. If restrictions of $f$ on all complex lines are analytic, then $f$ is analytic. This reduces the problem to the case $n=1$. Now $f^2$ is analytic so near every point, so it has a representation $f^2(z)=(z-a)^mg(z),$ where $g(a)\neq 0$. If $m$ is even, we obtain analyticity of $f$. If $m$ is odd, $f$ cannot be $C^\infty$; some derivative blows up.

The argument also works for real-analytic, as a real-analytic function extends to complex analytic in some complex neighborhood.

EDIT. This proof extends to quasianalytic functions of $DC$ (Denjoy-Carleman) class. One needs two facts:

  1. If $f\in DC$ and $f(a)=0$ then $f(x)=(x-a)g(x)$ with $g\in DC$. The proof is based on the formula $$g(x)=\int_0^1f'(tx)dt.$$ By induction, this gives $f(x)=(x-a)^mg(x)$, for every $f\in DC$ and every $a$, with some non-negative integer $m$ and $g\in DC$, $g(a)\neq 0$.

  2. If $f\in DC$ and $F$ is analytic on the range of $f$, then $F\circ f\in DC$. This follows from the great theorem of E. M. Dyn'kin (J.d'Analyse 60 (1993)) that characterizes $DC$ in terms of pseudoanalytic extension to a complex semi-neighborhood of a real interval.

In the proof above, $F$ is a branch of the square root.

Validity of facts 1 and 2 was explained to me by MO user fedja.

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  • $\begingroup$ Cool! I didn't see this solution. I used Weierstrass preparation, and that is why I could not apply it to quasi-analytic functions. The interesting thing is that if I adapt this proof by lines I can get that the function is in the quasi-analytic class in each line. But then it is an open problem to show that if a smooth function is in certain quasianalytic Denjoy-Carleman class on each line then it is in the class. Thanks. $\endgroup$
    – O.R.
    Nov 20, 2013 at 22:05
  • $\begingroup$ Oh yes, I should have said that for one variable quasi-analytic functions it is trivial too. The analogous of extracting the factor $(x-a)$ in several variables would be Weierstrass preparation. That is why I had proved the analytic case of several variables using Weierstrass. But Weierstrass doesn't work anymore in quasi-analytic DC classes. The reduction to lines also faces a problem. From knowing a function is in a DC class along lines it is not known if one can conclude the function is in the class. Actually, one would have to know it is in the class for every line, not only $\endgroup$
    – O.R.
    Nov 23, 2013 at 18:23
  • $\begingroup$ passing through the origin to conclude the function is in the DC class. $\endgroup$
    – O.R.
    Nov 23, 2013 at 18:24

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