If a real function $f$ on an interval $I$ has derivatives of all orders which are positive at every point of $I$, how can I prove that $f$ is real analytic? (That is, it equals the Taylor expansion of itself)
Edited: The following is a kind of hint provided by the textbook. Think about a neighborhood of $a$ and Taylor expansion about $a$. For $x>a, f(x) = \sum_{k=0}^{n}{f^k(a){\frac{(x-a)^k}{k!}}} + R_{n,a}(x) > \sum_{k=0}^{n}{f^k(a){\frac{(x-a)^k}{k!}}}$ because all derivatives are positive(therefore the remainder term also positive). Then $\sum_{k=0}^{n}{f^k(a){\frac{(x-a)^k}{k!}}}$ is an increasing sequence of $n$ that is bounded above by $f(x)$, therefore converges.
My question about this is:
- I can understand that $\sum_{k=0}^{n}{f^k(a){\frac{(x-a)^k}{k!}}}$ converges, but how do I know that it converges 'to $f(x)$'?
- How about the case when $x < a$? Each term has alternating sign, so I can't use the above method.