# Smooth functions, zero on a given curve

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 $k_i>0$.

• Please show some efforts first. – Saad Feb 13 '18 at 3:33

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 $$\frac{\partial^k f}{\partial x_1^k}(0)=0\text{ for all }k\geq 1.$$