Is this $C^\infty-$function analytic? Consider a $C^\infty-$function $\,f$ on $[a, b]$. All of its derivatives are non-negative.
I am trying to prove it is an analytic function. 
I have tried to calculate its Taylor remainder 
$$R_n = \int_a^x \frac{f^{(n+1)}(t)}{(n+1)!} (x-t)^{n+1} d t.
$$ 
But I have difficulty estimating the bound. Thanks for any help in advance.
 A: $%Filler text to meet the 6 character minimum edit requirement$
A discussion about this problem appears Here.
The following is based on S.N. Bernstein's proof.
If $t\in (a,b)$, then the Taylor expansion remainder theorem provides that
$$
f(t)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(t-a)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(t-a)^{n+1},
$$
for some $\xi\in (a,t)$. If $\displaystyle P_n(t)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(t-a)^k$, then $P_n$ is an increasing sequence of polynomials, and bounded. In fact,
$f(a)\le P_n(t)\le f(t)\le f(b)$. Therefore $\{P_n\}$ converges
$$
P_n(t)\le P_{n+1}(t)\to P(t)\le f(t).
$$
Since the power series $P(t)=\displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}(t-a)^n$ converges for every $x\in [a,b]$, it consequently has radius of convergence at least $R=b-a$.
It suffices to show that $\,f\equiv P$. 
We shall show that $\,f(t)=P(t)$ holds in the interval $[a,c]$, where $c$ is the midpoint of $[a,b]$. Then, repeating the same argument, by extending this equality in half of the every time remaining subinterval, we cover the interval $[a,b)$, and by continuity the whole interval.   
Expanding $\,f-P_n$ around $a$ for an arbitrary $t\in[a,c]$, we obtain that there is a $\xi\in[a,t]$, such that
$$
0\le f(t)-P_n(t)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(t-a)^{n+1}\le 
\frac{f^{(n+1)}(c)}{(n+1)!}(c-a)^{n+1}=\frac{f^{(n+1)}(c)}{(n+1)!}(b-c)^{n+1}\to 0,
$$
as $n\to\infty$, since the power series $\displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(c)}{n!}(t-c)^n$ converges for all $t\in[c,b]$, if we use the same argument, used to show that the series $\displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}(t-a)^n$ converges, for $t\in[a,b]$.
