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: $$\forall x \in [0,\infty) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

?