# Which real functions have their higher derivatives tending pointwise to zero?

Let $\mathrm C^\infty\!(\Bbb R)$ be the space of infinitely differentiable functions $f:\Bbb R\rightarrow\Bbb R$, and define the subspace$$A:=\{f\in\mathrm C^\infty\!(\Bbb R):(\forall x\in \Bbb R)\lim_{n\rightarrow\infty} f^{(n)}(x)=0\},$$where $f^{(n)}$ is the $n$th derivative of $f\;(n=0,1,\dots).$ Clearly all polynomial functions are in $A$. Are any others?

Edit: Alfonso has answered this question well, but is there any characterization of $A$ in terms of familiar types of function?

• Some uniformly convergent example: $f(x)=\sin(a x)$, where $0<a<1$. – 23rd May 19 '13 at 17:44
• Hagen von Eitzen has just classified $A$ in the best way possible. If you want to find a different way to express the elements of $A$, start with Hagen von Eitzen's result. – Patrick Da Silva May 19 '13 at 18:19
• @PatrickDaSilva No, mine are just the analytic examples. I'm only almost sure that this is complete – Hagen von Eitzen May 19 '13 at 18:25
• @HagenvonEitzen : Hm, nice remark, I didn't pay attention. So you found the intersection of $A$ with the space of analytic functions, we need to find what's outside of it now. – Patrick Da Silva May 19 '13 at 18:36
• @jkn: I don't see that $A$ is closed under multiplication. For example, take Alfonso's function: $f(x)f(x)=\mathrm e^x,$ so $ff\notin A.$ – John Bentin May 19 '13 at 18:41

Consider an analytic function $$\tag0f(x)=\sum_{n=0}^\infty \frac{a_n}{n!}x^n.$$ As $f^{(n)}(0)=a_n$, a necessary condition for such $f$ to be an example is that $a_n\to 0$.

But $a_n\to 0$ is also sufficient. Indeed, with $b_n:=\sup_{k\ge n} |a_k|$ we have $b_n\to 0$ and hence \begin{align}|f^{(n)}(x)|&= \left|\sum_{k=0}^\infty \frac{a_{n+k}}{k!}x^k\right|\\ &\le\sum_{k=0}^\infty\frac{|a_{n+k}|}{k!}|x|^k\\ &\le b_n\sum_{k=0}^\infty\frac{1}{k!}|x|^k\\&=b_ne^{|x|}\to 0. \end{align} (Note that the calculation for $f^{(0)}(x)$ shows that $f$ is entire to begin with).

• This looks pretty close to being a complete answer. There is just a slight niggle about whether the elements of $A$ are analytic. It seems obvious, but the proof eludes me. Anyway, it is time for me to accept your answer! – John Bentin May 23 '13 at 16:32

$$f(x)=e^{x/2}$$ $$f^{(n)}(x)=2^{-n}e^{x/2}$$

• Of course, this is true for any $e^{\alpha x}$, $-1<\alpha<1$, and also for a more uniform result $$\left| \frac{d^n}{dx^n} \sin (\alpha x) \right| \le \alpha^n$$ – Alfonso Fernandez May 19 '13 at 17:44

Another example would be the reciprocal function:

$$f(x)=\frac{1}{x}$$

$$\frac{d^n}{dx^n}\frac{1}{x}=(-1)^n\frac{n!}{x^n}$$

By extension, $\ln(x)$ is another example, because $\frac{d}{dx}\ln(x)=\frac{1}{x}$

• What is $f(0)$? – Hagen von Eitzen May 19 '13 at 17:55
• What if $x=0.00000000000000000001$? – Andrés E. Caicedo May 19 '13 at 17:55
• @HagenvonEitzen I see. It's not continuous or differentiable at 0. – Ataraxia May 19 '13 at 17:58
• Not only that. The limit never goes to zero. – Martin Argerami May 19 '13 at 18:09
• The function is not defined at zero, therefore does not belong to the space. This should be the first problem you notice. – Patrick Da Silva May 19 '13 at 18:17