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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?

John Bentin
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    Some uniformly convergent example: $f(x)=\sin(a x)$, where $0<a<1$. – 23rd May 19 '13 at 17:44
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    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
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    @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
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    @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
  • @JohnBentin yes you are right - I dunno what I was thinking. – jkn May 19 '13 at 18:49
  • @jkn : When speaking of such questions when the domain is $\mathbb R$, multiplication is usually not behaving nicely. It is an exercise in a first course in analysis to show that the product of continuous functions remains continuous, but you cannot add "uniformly"s in this sentence. – Patrick Da Silva May 19 '13 at 19:17

3 Answers3

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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).

Hagen von Eitzen
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  • 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
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$$f(x)=e^{x/2}$$ $$f^{(n)}(x)=2^{-n}e^{x/2}$$

Alfonso Fernandez
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  • 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
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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 $\frac{d}{dx}\ln(x)=\frac{1}{x}$

rurouniwallace
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