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For an antisymmetric function $f:\mathbb R\rightarrow \mathbb R$ (i.e. f(x)=-f(-x)) we have: necessary condition for the differential of f of order $r$ to not vanish at $0$ is that $r$ is odd.

My question is: what if I consider an antisymmetric $f:\mathbb R^n\rightarrow \mathbb R$ with $n\geq 2$ ?

Here antisymmetric means: $f(x)= sign(\pi) f(\pi x)$ for every signed permutation $\pi$ of the coordinates ( a signed pemutation is a permutation which is also allowed to change the sign of the coordinate, I hope it's clear what I mean, i don't know if it's standard terminology).

I think that now we have: necessary condition for the differential of f order $r$ to not vanish at $0$ is that $r=k n$ with $k$ odd.

Is this true? In case, how to prove it? Any help would be appreciated.

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That's interesting. Your functions are smooth, are they? And could you please define what a signed permuation and its signum mean? – shuhalo Apr 7 '12 at 23:09
Okay, maybe a nice definition of signed permuation of coordinates is a permuation matrix times a ${-1,1}$-diagonal matrix, and the signum is the determinant of that transform. – shuhalo Apr 7 '12 at 23:19
up vote 1 down vote accepted

This is not true. For example, $f(x_1,x_2)=x_1^3x_2-x_1x_2^3$ is antisymmetric in your sense [if I understood it correctly] but has nonzero derivatives of order $4$ at the origin.

A more straightforward generalization would be to consider the condition $f(-x)=-f(x)$, which easily implies that all even-order derivatives vanish at the origin (by the uniqueness of the Taylor approximation).

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