Showing a function with bounded derivatives converges to it's taylor series

Question

The original question:

Let $f:[0,1]\to\mathbb{R}$ be infinitely differentiable, and suppose that $$\max_{\xi\in[0,1]}|f^{(n)}(\xi)|\leq C^n\cdot n!$$ for some $C\in\mathbb{R}$. Show there is a region $G$ such that $[0,1]\subset G$ and $f$ is holomorphic on $G$.

It seems similar to this: Prove Taylor series converges to $f$. but the conclusion there contradicts what I was asked so I'm not sure I understood everything correctly.

My idea, and the only way I know how approach this, is to show that the bound on the derivatives is enough to guarantee that the Tayloe series of $f$ converges to it, and then use that to define $f$ on it. But I'm encountering a problem with that approach, as developing around $0$, the error term becomes

$$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}\cdot x^{n+1}$$

which by the bound on the derivative decreases to

$$R_n(x) = (Cx)^{n+1}$$

but wouldn't then the series converge only if $C<1$ ($C\leq 1$ would suffice for $x\in(0,1)$, but then can I show holomorphism at $x=1$)? Is there a mistake or is there a way to proceed from here?

Answer 1

Note that $C \ge 0$ and, if $C=0$, then $f(x) = 0$ which is entire. Hence we can assume $C>0$.

First we find a suitable $G$ and an analytic function $g$.

Note that ${1 \over R} = \limsup_n \sqrt[n]{|{f^{(n)}(z_0) \over n! }|} \le C$, hence $R \ge {1 \over C} > 0$. Then for each $z_0 \in [0,1]$, the function $g_{z_0}(z) = \sum_n {f^{(n)}(z_0) \over n! } (z-z_0)^n$ is analytic on $B(z_0, R)$. Since $[0,1]$ is compact, we can cover $[0,1]$ by a finite number of these balls, and the corresponding functions agree on the real part of the overlap of any two overlapping balls , and hence there is some $g$ analytic on some open $U$ that contains $[0,1]$.

To see that $f=g \mid_{[0,1]}$, pick $x \in [0,1]$, then note that $g(x) = g_x(x) = f(x)$.