Let $[a,b]\subset[-1,1]$ and $f\in C^{\infty}([a,b],\mathbb{R})$ a function which satisfies for all $n\in\mathbb{N}$ that $\sup\_{x\in[a,b]}|f^{(n)}(x)|\leq n!$.


Does there exist a function $g\in C^{\infty}([-1,1],\mathbb{R})$ which satisfies $g\big|\_{[a,b]}=f$ and for all $n\in\mathbb{N}$ that $\sup\_{x\in[-1,1]}|g^{(n)}(x)|\leq n!$ ?


I feel like this should hold. However, I am unable to find a reference where this is shown (partially because I don't know whether there is a particular name for these functions, i.e. what to look for). 


If someone knows where I could find such a result, or knows a short proof, it would be much appreciated. 



Extension of a certain type of (very) smooth functions to a larger interval
No, this won't work in general. These assumptions make $f,g$ real analytic (by Taylor's theorem), so the only $g$ that could possibly work would be the holomorphic continuation
$$
g(x) = \sum\_{n=0} a\_n x^n, \quad a\_n = \frac{f^{(n)}(0)}{n!} .
$$
However, this need not satisfy your bounds on all of $[-1,1]$ if all we know is that it satisfies them on a smaller interval; for example, you can take something like $a\_n=c(2/3)^n$.


