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I want a $C^\infty$ function $f:U\rightarrow\mathbb{C}$, where $U\subset\mathbb{C}$ is a neighborhood of $0$, such that there exists a sequence of points $z_n\in U-\{0\}$ and $\lim_{n\rightarrow\infty} z_n=0$, satisfy $(f(z_n))^2=z_n^3$.

I faced this problem in my research. I want to find a smooth substite for $z^{3/2}$, since $z^{3/2}$ cannot be defined as a single-value function and has bad behavior at $0$ (only $C^{1,1/2}$, not smooth).

By inverse function theorem, if $f$ satisfies the condition, then the Jacobian of $f$ at $0$ must be degenerate. It's quite difficult for me to really find a $f$, since it seems very hard for a real-analytic $f$ to satisfy those conditions, besides I have limited knowledge about non-real-analytic $C^\infty$ functions.

Any comments and answers are welcome. You can also replace $3/2$ by other half-integers, like $5/2$, $7/2$, etc. I'll really appreciate your help.

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How about $f(x)= \mathop{\rm sgn} x\, |x|^{3/2}$, then any sequence $x\in \mathbb{R}$ fulfills the property... – Fabian Aug 26 '12 at 12:43
@Fabian Note that $f$ needs to be defined on $U$, how do you define the sign of a complex number? – N. S. Aug 26 '12 at 14:27

Partial Answer

It is easy to see that a real Analytic function cannot exists. Indeed if $f$ is analytic, then $f^2$ is also analytic and the set $\{ f^2(z)=z^3 \}$ has an accumulation point. Then it would follow that $f^2(z)=z^3$.

Same argument can be made about any other half integer.

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Thanks! Then we only need to check non-real-analytic functions. Do you have any idea about these functions? – Yuchen Liu Aug 26 '12 at 14:54
For non-analytic functions, one idea which might or might not work is the Mean value theorem. Try applying the MVT to $f$ on the line segment $[z_n, z_{n+1}]$ to get some point $y_n$ for which the MVT works, and then maybe to the partial derivatives on the segment $[y_n,y_{n+1}]$. It looks to me that this way you might get some points where the second derivatives of $f$ are about the same as the second derivatives of $z^{\frac{3}{2}}$, which contradicts that $f$ is twice differentiable at $z=0$. – N. S. Aug 26 '12 at 15:10

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