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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

$c(x)=\begin{cases}g_-(x)/f_-(x)&x\lt0\;,\\c_0&x=0\;,\\g_+(x)/f_+(x)&x\gt0\end{cases}$

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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    @Adterram: You can choose them to be almost anything. For instance constants, where the ratios of the constants on the left and right differ.2011-11-02