0
$\begingroup$

Let $c:I=[0,a) \rightarrow \mathbb{R}^n$ be a smooth curve with $c(0) = 0$. Is there an open neighborhood $U \subset \mathbb{R}^n$ of $0$, and a smooth function (not necessarily real analytic) $f:U \rightarrow \mathbb{R}$, such that $f \equiv 0$ on $c\cap U$, and is of finite order in every argument at $0$? That is, for every $x_i, i=1,\cdots,n$, $$\dfrac{\partial^{k_i}f}{\partial x_i^{k_i}}(0)\neq 0$$ for some $k_i>0$.

$\endgroup$
  • $\begingroup$ Please show some efforts first. $\endgroup$ – Saad Feb 13 '18 at 3:33
1
$\begingroup$

Not in general. Consider the case $c:[0,\infty)\to\mathbb R^n$ defined by $c(t)=te_1$ where $e_1=(1,0,\ldots,0)\in\mathbb R^n$. Let be $f:\mathbb R^n\to\mathbb R$ be smooth with $f\equiv 0$ on the image of $c$. For all $\xi\geq 0$ we get $$ \frac{\partial f}{\partial x_1}(\xi,0,\ldots,0)=\lim_{t\to 0}\frac{f(te_1)-f(0)}{t} $$ Since $f$ is smooth, we can consider the limit for $t>0$. But $te_1$ is in the image of $c$ such that $f(te_1)=0$ and we get $$ \frac{\partial f}{\partial x_1}(\xi,0,\ldots,0)=0. $$ We can iterate this argument to prove $$ \frac{\partial^k f}{\partial x_1^k}(0)=0\text{ for all }k\geq 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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