Can $f^{(\infty)}(a)=0$ for almost all $a$?

Question

My question is:

Does there exist an infinitely differentiable function $f$ such that $$\lim_{n\to\infty} f^{(n)}(a)=0\qquad{\text{for almost all } a\in[0,\infty)}$$ ?

($f$ cannot be a constant function or a polynomial.)

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If we restrict us to $C^{\omega}$ (i.e. assuming $f$ is holomorphic), it is likely that the answer is no, since $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz$$ the integrand only decays exponentially in $n$ while there is a $n!$ factor there.

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However, I am not quite sure if the above argument is correct.

For smooth functions, I have no ideas.

Any help will be appreciated.

Thanks in advance.

Answer 1

Try $$f(x) = \sum_{k=0}^\infty \frac{x^k}{(k!)^2} = I_0(2 \sqrt{x})$$ This is an entire function, and $$ f^{(n)}(x) = \sum_{k=n}^\infty \frac{x^{k-n}}{k! (k-n)!} = \sum_{j=0}^\infty \frac{x^j}{(j+n)! j!} $$ so that for all $x$ $$ |f^{(n)}(x)| \le \sum_{j=0}^\infty \frac{|x|^j}{(j+n)! j!} \le \frac{1}{n!} \sum_{j=0}^\infty \frac{x^j}{j!} = \frac{\exp(|x|)}{n!}$$ and thus $f^{(n)}(x) \to 0$ for all $x$.