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If $f$ and $g$ are two smooth functions in $\mathbb R^n$ such that if ${\partial ^\alpha }f(x)=0$ for arbitrary index $\alpha$ and $x \in \mathbb R^n$, then ${\partial ^\alpha }g(x) = 0$. Then is there a smooth function $c(x)$ such that $g=cf$ ?

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No. Consider $n=1$ and

$$f(x)=\begin{cases}f_-(x)\mathrm e^{-1/x^2}&x\lt0\;,\\0&x=0\;,\\f_+(x)\mathrm e^{-1/x^2}&x\gt0\end{cases}$$

with $f_\pm$ smooth and non-zero, and $g$ defined likewise. Then if $c$ existed, it would have to take the form


with $c_0$ to be determined. But we can give the left-hand and right-hand quotients whatever limits we want, and if they differ we can't make $c$ continuous by any choice of $c_0$.

I suspect this might be true if you require that only a finite number of derivatives vanish at any given point; in that case it might be possible to prove this by induction using l'Hôpital's rule.

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What are ${g_ + }$ and ${g_ - }$? Do you mean ${g_ + }(x) = g(x){e^{1/{x^2}}}{\kern 1pt} {\kern 1pt} {\kern 1pt} (x > 0)$? If so, I cannot figure out how we can give whatever limits we want. – Hezudao Nov 2 '11 at 8:24
@Adterram: You can choose them to be almost anything. For instance constants, where the ratios of the constants on the left and right differ. – joriki Nov 2 '11 at 11:53

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