Let $ f: (a, b) \to\mathbb R $ a function of class $ C^{\infty} $ that satisfies the following condition: given $ K \subset (a, b) $ compact, there exists $ C \gt 0 $ such that $ f^{(n)} (x) \le C^{n + 1} n! $, $ \forall n \in \mathbb Z_+ $, $ x \in K $. Prove that $ f $ is an analytical function.
In my notes we have the following.
Def: Let $I \subseteq \mathbb R$ open range of $f \in C^{\infty}(I)$. We say $f$ is analytic on $I$,if given $\exists r \gt 0$
such that $(a-r, a+r) \subseteq I$ $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}a}{n!}(x-a)^n \forall x\in (a-r, a+r )$
Prop:Let $f \in C^{\infty}(I) (I$ open) suppose that $\exists M \gt 0$ such that $|f^{(n)}(x)| \le M$, $\forall n \in \mathbb Z_+, x\in I$. So $f$ is analytic on $I$.
From the prop and knowing that $C \gt 0$ and $n$ is positive don't we already have that $f$ is analytic? Are the Def and Prop correct? Or did I get something wrong?
Thanks in advance for any help.