One-Sided Analyticity Condition Guarantees Analytic Function? Let $f \ \colon \ [0,\infty) \to \mathbb{R}$ be a function satisfying:


*

*$f$ is differentiable infinitely many times in $(0,\infty)$, and has a right-derivative of any order at $0$.

*$f$ satifsfies the condition (condition 3 here) for analyticity: for every compact $K \subset [0,\infty)$ there exists a constant $C_K$ such that 
$$\forall x \in K:\forall n \geq 0:|f^{(n)}(x)|\leq C_K^{n+1}n!$$
where in the last formula, if $K$ contains $0$ and $x=0$, then the $n$-th derivative in the formula is the $n$-th right derivative in $0$.


Is it true in this case that $f$ is analytic in $[0,\infty)$ and that for some $\epsilon > 0$,
$$\forall x \in [0,\epsilon) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
?
If in addition we have a constant $C$ such that
$$\forall x \in [0,\infty):\forall n \geq 0:|f^{(n)}(x)|\leq C^{n+1}n!$$
does the following hold:
$$\forall x \in [0,\infty) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
?
 A: The answer to the first question is yes, as explained by Liviu Nicolaescu.
The answer to the second question is no. Consider a bounded analytic function in the strip
$\{z :|\Im z|<2/C\}$ for which the boundary of the strip is the natural boundary.
Then the radius of convergence of the Taylor series at $0$ is $2/C$, and it cannot
represent the function on the positive ray.
On the other hand,
$$|f^{n}(x)|=\left|\frac{n!}{2\pi}\int_{|z-x|=1/C}\frac{f(z)dz}{(z-x)^{n+1}}\right|\leq C^nn!.$$
A: The Taylor formula  states that,  under your assumptions, for any  $x>0$  and any positive integer $n$ we have
$$
f(x)= f(0)+f'(0)x+\frac{1}{2!}x^2+\cdots +\frac{1}{n!} f^{(n)}(0)x^n+R_n(x),
$$
where according to Lagrange
$$
R_n(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)x^{n+1},
$$
for some $\xi\in (0,x)$. A reference of this fact under your specific assumptions is Theorem 2, Section 5.3.3 of 

V.A. Zorich: Mathematical Analysis I, Springer 2004.

If in your estimate for the derivatives you take $K=[0,1]$ and set $C=C_K$ then   one deduces that the series 
$$
\sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}x^n
$$
converges uniformly to $f(x)$ for $0\leq x\leq \frac{1}{2C}$.
