# If $f$ vanishes sufficiently fast at $x_0$ does this imply that all derivatives vanish as well?

Let $f: U \rightarrow \mathbb{R}$, with $U \subset \mathbb{R}^n$ open, be infinitely differentiable.

Suppose there is a $x_0 \in U$ such that

$$\lim_{x \rightarrow x_0} \frac{f(x)}{|x-x_0|^k}=0$$

for all $k \in \mathbb{N}$.

How can we prove that every derivative $D^{\alpha} u(x_0)$, $\alpha \in \mathbb{N}^n$, is zero?

For $|\alpha|=1$ we can just use the definition of the partial derivative. But in the case $n=2$ this already gets messy.

This is a consequence of Taylor's theorem. Suppose we're on the line and $x_0=0.$ Suppose we have $f(0) = f'(0) = 0$ but $f''(0) \ne 0.$ By Taylor we have

$$f(x) = f(0) +f'(0)x + f''(0)x^2/2 +o(x^2)= f''(0)x^2/2 +o(x^2).$$

That implies $|f(x)| \ge (|f''(0))|/3 )x^2$ for small nonzero $x,$ contradicting the hypothesis. Hence $f''(0)=0.$ This reasoning can be continued to show all derivatives of $f$ at $0$ are $0.$

For $n>1$ the same ideas apply, although Taylor polynomials get more complicated. Suppose we know $D^\alpha f(0,0) =0$ for $|\alpha| \le 1.$ Then

$$f(x) = \sum_{|\alpha| = 2} [D^\alpha f(0,0)]/\alpha !]x^\alpha + o(|x|^2).$$

This follows from the one variable result. Let $T_2(x)$ denote the above sum. Note that $T_2$ is homogeneous of degree $2,$ so for $r> 0$ and $|\zeta| =1,$ $T_2(r\zeta) = r^2T_2(\zeta).$ Suppose $T(\zeta_0) \ne 0.$ Then $|T_2(r\zeta_0)|$ is on the order of $r^2$ as $r\to 0.$ This implies the same is true of $f.$ That contradicts the given hypothesis. Thus we have $T_2(\zeta)=0$ for all $\zeta, |\zeta| =1.$ That implies $T_2\equiv 0,$ which in turn implies $D^{\alpha}f(0) = 0$ for $|\alpha|=2.$

For more on this topic, search on Taylor polynomials in several variables, either on MSE or google.

• I've changed my answer. – zhw. Nov 10 '16 at 18:58
• How does this generalize to the $n$-dimensional case? There we have several derivatives of order 2 in the Taylor expansion and $x$ doesn't factor out nicely. – NiU Nov 11 '16 at 9:29
• @NiU I added to my answer. – zhw. Nov 12 '16 at 16:11

I take it from your comment about $|\alpha| = 1$, that you already know the first derivatives vanish quickly at $x_0$. You do realize that all the $|\alpha| = 2$ derivatives are first derivatives of things you have already shown vanish quickly...

• I know that the first derivatives vanish. But the limit of $\dfrac{\partial_{x_i} f(x)}{|x-x_0|^k}$ as $x \rightarrow x_0$ is actually a double limit. Maybe it's just a notational problem but I don't see how this is zero. – NiU Nov 10 '16 at 18:44