Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks!

share|improve this question

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.