# n'th derivative does not vanish, but $\lim_{n\to \infty} f^{(n)}=0$.

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.

-
Which topology do you put on $C^\infty[\Bbb R,\Bbb R]$? – Davide Giraudo Jan 7 '13 at 10:31
Yes take $f(x)=\sin(\frac{x}{n})$ with $n>1$. – Lucien Jan 7 '13 at 11:09
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
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
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

First, a few remarks:

1. $(C^\infty(\mathbb R,\mathbb R),\|\cdot\|_\infty)$ is problematic: there are many smooth functions from $\mathbb R$ to itself that are unbounded, such as $f(x)=x$. Thus, $\|\cdot\|_\infty$ does not define a norm on $C^\infty(\mathbb R,\mathbb R)$.
2. The same problem arises on $(C^\infty((0,1),\mathbb R),\|\cdot\|_\infty)$, where $(0,1)$ denotes the open interval. Consider $f(x)=1/x$.
3. $\|\cdot\|_\infty$ does indeed define a norm on $C^\infty([0,1],\mathbb R)$, where $[0,1]$ denotes the closed (compact) interval. However, it makes no sense to ask if a subset $A$ is dense in $(C^\infty([0,1],\mathbb R),\|\cdot\|_\infty)$: Since we're considering the norm topology induced by $\|\cdot\|_\infty$ (which is in fact a metric topology with $d(x,y)=\|x-y\|_\infty$), the set $A$ is dense in $C^\infty([0,1],\mathbb R)$ if and only if $\overline A=C^\infty([0,1],\mathbb R)$ (where $\overline A$ denotes the closure). However, this can never be satisfied since $C^\infty([0,1],\mathbb R)$ is not itself closed under $\|\cdot\|_{\infty}$. Indeed, according to the Weierstrass Approximation theorem, one can find a sequence of polynomial functions $\{p_n\}\subset C^\infty([0,1],\mathbb R)$ such that $\|p_n-f(x)\|_{\infty}\to0$, where $f(x)=|x-1/2|$, yet $f$ is not differentiable.

An alternative question in the same spirit as what you're asking would be the following:

Question. What is the closure of the set $A$ of functions $f\in C^{\infty}([0,1],\mathbb R)$ such that \begin{align}f^{(n)}\neq0 \text{ for every $n$ and }\lim_{n\to\infty}\|f\|_\infty=0.\tag{1}\end{align}

The answer to this question was provided in a comment by Willie Wong: The set of functions $f:[0,1]\to\mathbb R$ defined as $$f(x)=p(x)+\epsilon\sin(\epsilon x)$$ (where $p$ is a polynomial function and $0<\epsilon<1$) is a subset of the functions that satisfy conditions $(1)$, and it approximates the polynomial functions, which are dense in $C[0,1]$. Thus, the closure of your set is the continuous functions on the compact [0,1].

Note. If you ask the same question as above but with the condition \begin{align}f^{(n)}\neq0 \text{ for every $n$ and }\lim_{n\to\infty}f(x)=0 \text{ pointwise},\tag{2}\end{align} then the result still holds, since uniform convergence implies pointwise convergence, hence the set of every function satisfying $(2)$ contains the set of every function satisfying $(1)$.

-