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A question (Are derivatives actually bounded?) has been asked on Stackexchange as to whether there exists $f\in C^{\infty}$ such that $f^{(n)}(0)=(n!)^2$. Obviously $f$ is not analytic but the respondent cites Borel's theorem as implying an affirmative answer.

For fun, I decided to construct one such $f$, and I hit on the following:

$g_n(x)=\exp\frac{\exp\frac{1}{n|x|-1}}{-|x|}$ for $0<|x|<\frac{1}{n}$

$f_n(x)=x^nn![1-g_n(x)]$ for $0<|x|<\frac{1}{n}$; anywhere else, $f_n(x)=0$

$f(x)=\sum_{n=1}^{\infty} f_n(x)$

Are there functions with that property which are easier to verify? Because it took me ages to produce what I believe to be a proof that the k-th derivative of $\sum_{n=1}^{\infty}x^nn!g_n(x)$ is a uniformly convergent series.

And is $f$ the required function?

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    Such $f$ is not unique, since you can add any function $g$ with $g^{(n)}(0)=0$, for example $g(x)=e^{-1/x^2}$ – 2017-01-28
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    Have you seen http://math.stackexchange.com/a/387897/408656 ? – 2017-01-28

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