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.

Let $f\,$$\in$$\,C^\infty[\mathbb{R},\mathbb{R}]$ . Apparently the only functions $f$ for which there exists $n\in\mathbb{N}$ such that $f^{(n)}=0$ are polynomials in $\mathbb{R}[x]$.

Is it possible to characterize the functions $f\,$$\in$$\,C^\infty[\mathbb{R},\mathbb{R}]$ for which $\lim_{n\to \infty} f^{(n)}=0$, but $f^{(n)}\neq 0$ for all $n\in\mathbb{N}$. For example are they dense in ($C^\infty[\mathbb{R},\mathbb{R}]$,$||.||_{\infty}$)?

EDIT: It maybe easier if we resctrict to $C^\infty[(0;1),\mathbb{R}]$. Any discussion is welcome for this one too.

share|improve this question
2  
Which topology do you put on $C^\infty[\Bbb R,\Bbb R]$? –  Davide Giraudo Jan 7 '13 at 10:31
1  
Yes take $f(x)=\sin(\frac{x}{n})$ with $n>1$. –  Lucien Jan 7 '13 at 11:09
1  
Since the derivatives $f^{(n)}$ are uniformly bounded independent of $n$, we can conclude that $f$ is real analytic. –  Willie Wong Jan 7 '13 at 11:41
5  
Note that $f(x):=\cos(\lambda x)$ has this property when $|\lambda|<1$, and doesn't when $|\lambda|\geq1$. This shows that the property is not scaling invariant. –  Christian Blatter Jan 7 '13 at 12:08
2  
For the interval case, if a function is continuously differentiable with bounded derivative, it is continuous on the closure of the interval. Then you can apply Stone-Weierstrass to get density of polynomials. Density of your set would follow, as "polynomial + $\epsilon \sin(\epsilon x)$" approximates polynomials. –  Willie Wong Jan 7 '13 at 16:52
show 4 more comments

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.