Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g=cf$ ?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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