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Given a function $f(x_1,x_2,\ldots,x_n):\Omega\to\mathbb{R}$ which is in $BV(\Omega)$ and has, in $\textbf{0}$, all partial derivatives up to order $n-1$, all equal to 0: $$ \frac{\partial^{|\alpha|} f}{\partial \textbf{x}^\alpha}(\textbf{0})=0,\quad |\alpha|<n, $$

can anything be said about the existence of $f_{x_1x_2\cdots x_n}(\textbf{0})$?

My dream would be to get a Taylor-like expression of the form $$ f(\textbf{x})=0 + 0 + \ldots + 0 + \bar f\Delta x_1\cdots \Delta x_n+o(|\textbf{x}|^{n}). $$ since all terms of lower order cancel, and that $BV$ must be worth something :)

If not, any minimal additional hypothesis that might be required?


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up vote 1 down vote accepted

General remark: $BV$ is worth something when you deal with first-order derivatives. It says that first-order derivatives are measures. This does not help much with higher-order pointwise expansions.

You neglected the possibility that variables can be repeated in monomials of $n$th degree. For example, even the polynomial $f(\mathbf x) = x_1^n$ does not admit an expansion of your kind, because the $n$th degree term here is $x_1^n$, not $x_1\dots x_n$.

If you include other $n$th degree monomials, the answer is still no. The function $f(\mathbf x)=x_1|x_2|$ is in $BV$ (furthermore, is Lipschitz), its first-order derivatives exist and vanish at $0$. But it cannot be approximated by a polynomial up to $o(|\mathbf x|^2)$.

Apart from the familiar theorems about finite Taylor expansions, I don't think you'll find a hypothesis that is easier to check in practice than the asymptotic expansion itself.

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Ok, I definitely should look for my own (very simple) counterexamples before bothering other people :) Thanks! – kenshin Jan 16 '13 at 16:35

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