Let $f$ be a smooth function on $[a,b]$ such that $f^{(k)}(x) \ge 0$ for all $x \in [a,b]$ and $k \ge 0$. Show that $$ f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k , \forall x \in [a,b] $$
That is $f$ is analytic on $[a,b]$. The hint in to use Taylor expansion of integral form. But since $f^{(k)}$ can be very large, I couldn't find a suitable estimation.
Hint: show that for all $a < x < y < b$, $$f(y) \geq \sum_{k=0}^{\infty}{\frac{(y-x)^k}{k!}f^{(k)}(x)}.$$
As a consequence, $f^{(n+1)}(x) \leq \frac{(n+1)!}{(y-x)^{n+1}}f(y)$.
Therefore, if $z < x < y$, then
$$\sum_{k=0}^n{\frac{f^{(k)}(z)}{k!} (x-z)^k} \leq f(x) \leq \sum_{k=0}^n{\frac{f^{(k)}(z)}{k!} (x-z)^k} + \left(\frac{x-z}{y-x}\right)^{n+1}f(y).$$
Thus, let $$C_x = \{x \leq y \leq b,\, f(y)=\sum_{k=0}^{\infty}{\frac{f^{(k)}(x)}{k!}(y-x)^k}\}.$$
Standard manipulations show that if $y \in C_x$ and $z \in C_y$, then $z \in C_x$.
Moreover, from what is before, $C_x \supset [x, (x+b)/2[$, and it follows easily that $C_a=[a,b]$, hence the conclusion.
