2
$\begingroup$

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 $g=cf$ ?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hezudao Nov 2 '11 at 8:24
  • $\begingroup$ @Adterram: You can choose them to be almost anything. For instance constants, where the ratios of the constants on the left and right differ. $\endgroup$ – joriki Nov 2 '11 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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