MathDB
Totally convex => real analytic - OIMU 2005 Problem 6

Source:

September 3, 2010
functioncalculusderivativereal analysisreal analysis unsolved

Problem Statement

A smooth function f:IRf:I\to \mathbb{R} is said to be totally convex if (1)kf(k)(t)>0(-1)^k f^{(k)}(t) > 0 for all tIt\in I and every integer k>0k>0 (here II is an open interval).
Prove that every totally convex function f:(0,+)Rf:(0,+\infty)\to \mathbb{R} is real analytic.
Note: A function f:IRf:I\to \mathbb{R} is said to be smooth if for every positive integer kk the derivative of order kk of ff is well defined and continuous over R\mathbb{R}. A smooth function f:IRf:I\to \mathbb{R} is said to be real analytic if for every tIt\in I there exists ϵ>0\epsilon> 0 such that for all real numbers hh with h<ϵ|h|<\epsilon the Taylor series k0f(k)(t)k!hk\sum_{k\geq 0}\frac{f^{(k)}(t)}{k!}h^k converges and is equal to f(t+h)f(t+h).