Consider the function $f:(a,b) \rightarrow \Bbb R $, which is $C^\infty$ class. It is required to show that if the function $f$ satisfies: $\forall_{n\in\Bbb N_+} \forall_{x\in(a,b)} f^{(n)}(x) \ge 0 $, then it is analytic on $(a,b)$.
I add that the function is called analytic if and only if it can be expressed as a sum of some power series in some neighborhood of any point in the domain. I was considering the rest in Taylor series and thought about showing that it converges to $0$ but without succes.
Best Answer
Since $f^{(n)}\ge 0$ on $(a,b)$ for every $n\ge 1$, $f^{(n)}$ is increasing on $(a,b)$ for every $n\ge 0$.
Now given $c\in (a,b)$, let us estimate $f^{(n)}(c)$ for every $n\ge 1$. Fix some $0<h<b-c$. By Taylor's expansion with Lagrange form of the remainder, for every $n\ge 1$, there exists $\xi_n \in(c,c+h)$, such that $$f(c+h)=f(c)+\sum_{k=1}^{n-1}\frac{f^{(k)}(c)}{k!}h^k+\frac{f^{(n)}(\xi_n)}{n!}h^n.\tag{1}$$ Since $f^{(k)}(c)\ge 0$, $1\le k\le n-1$ and $f^{(n)}(\xi_n)\ge f^{(n)}(c)$, it follows that $$\frac{f^{(n)}(c)}{n!}\le\frac{f^{(n)}(\xi_n)}{n!}\le{h^{-n}}[f(c+h)-f(c)]. \tag{2}$$
Claim: If $x_0,x\in (a,c)$ and $|x-x_0|<h$, then $$\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\tag{3}$$ converges to $f(x)$.
Proof: Since $x_0,x\in(a,c)$, similar to $(1)$, for every $n\ge 1$, there exists $\xi_n\in(a,c)$, such that $$f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+\frac{f^{(n)}(\xi_n)}{n!}(x-x_0)^n.$$ By $(2)$ and noting that $0\le f^{(n)}(\xi_n)\le f^{(n)}(c)$ and $|x-x_0|<h$, we have $$0\le \frac{f^{(n)}(\xi_n)}{n!}|x-x_0|^n\le \frac{f^{(n)}(c)}{n!}|x-x_0|^n\le[f(c+h)-f(c)]\cdot\left|\frac{x-x_0}{h}\right|^n\to 0$$ as $n\to\infty$, so the series in $(3)$ converges to $f(x)$. $\quad\square $
The claim implies that $f$ is analytic on $(a,c)$. Since $c\in(a,b)$ is arbitrary, the proof is completed.