Proving the existence of analytic continuing of a function Let $f$ be a real smooth function (infinitely differentiable) on $[0,1]$ and suppose that there exists some $C>0$ such that the maximum absolute value of every $n$th derivative of $f$ is smaller than $C^{n+1}n!$ on $[0,1]$.
I need to show that there exists a domain $G$ which includes $[0,1]$ such that there is an analytic function $g$ on $G$ which equals $f$ on $[0,1]$.
The material we've studied so far was up to Cauchy's formula and Cauchy's Integral theorem, though I read a little bit further on Taylor series but I guess they expect us to use what we've studied.
Any direction?
 A: The condition $|f^{(n)}(x)| < C^{n+1}n!$ for all $x\in [0, 1]$ guarantees that for each $a \in [0, 1]$, $f(x)$ is equal to its Taylor series at $x=a$:
$$
 f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n
$$
for $x \in [0, 1]$, $|x - a| < \frac 1C$. (Use e.g. Taylor's theorem with the Lagrange remainder.)
That Taylor series has a radius of convergence of (at least) $1/C$. If we define
$$
 g_a(z) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z-a)^n
$$
then $g_a$ is analytic in the disc $U_a = B_{1/C}(a)$, and equal to
$f$ in $U_a \cap [0, 1]$.
Now cover the interval $[0, 1]$ by finitely many such disk $U_a$, and define $g$ in the union $G$ of those disks as
$$
 g(z) = g_a(z) \text{ if } z \in U_a \, .
$$
Use the identity theorem to show that $g$ is unambiguously defined, and equal to $f$ on $[0, 1]$.
A: Various answers and comments point out that the Taylor series has a positive radius of convergence, which is clear, and then say we're done. No, that's not enough.
Example Define $f:\Bbb R\to\Bbb R$ by $$f(t)=\begin{cases}e^{-1/t^2},&(t\ne0),
\\0,&(t=0).\end{cases}$$This is a standard example; it's well known that $f$ is in fact infinitely differentiable on the line, and that $f^{(n)}(0)=0$ for every $n$. So the Taylor series centered at the origin has infinite radius of convergence. But $f$ certainly does not extend to a holomorphic function in a neighborhood of the origin. The "extension"  given by those other answers is identically zero, which is not an extension at all.
To give a correct solution: Say $a\in[0,1]$, and consider the Taylor series centered at $a$. You need to show that if $|a-t|<1/C$ then the series converges to $f(t)$.
Not just that it converges, which is all you get from the radius of convergence; it has to converge to $f(t)$. This is easy from Taylor's Theorem, with, say, the Lagrange form of  the remainder. Look up "Taylor's Theorem"  on Wikipedia...
Regarding "I guess they expect us to use what we've studied": I gather  this is a class in complex analysis. Hence "they" expect you to know calculus already, and Taylor's Theorem is part of any decent calculus class.
But in fact calculus students never know Taylor's Theorem! I suppose there may be counterexamples, but I've never seen one - every calculus student I've ever seen who thinks  he or she knows Taylor's Theorem thinks it's this:


NOT Taylor's Theorem. If $f$ is infinitely differentiable near $a$ then $f(t)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(t-a)^n$ near $a$.


That's not Taylor's Theorem, and in fact the example above shows that it's false.
A: The inequality guarantees that the Taylor series of $f$ converges on the unit disk. That's your extension.
