Does Bernstein's theorem on monotone functions hold in an interval?

Question

Suppose that $f:[0,1]\to \mathbb{R}$ is totally monotone, i.e. $(-1)^n \frac{d^n f(t)}{dt^n} \geq 0$ for all $n$ and all $t \in (0,1)$. Does it imply that there exist a PDF $\mu$ such that $ f(t) = \int_0^1 e^{-tx} d \mu(x)? $ In other words does the Bernstein's theorem on monotone functions hold when the domain of the function is an interval instead of $[0,\infty)$?

Answer 1

There doesn't seem to be an extension of Bernstein's theorem quite like this. For example the function $f$ could be given by any power series of the form $$ f(1-t)= \sum_{n=0}^\infty a_n t^n , \quad\mbox{where}\quad a_n\ge0 \quad\mbox{and}\quad \sum_{n=0}^\infty a_n<\infty.$$ However, if $f(t) = \int_0^1 e^{-tx} d\mu(t)$ then $f$ can be extended by this integral formula to be totally monotone on $(0,\infty)$ (a.k.a. completely monotone). Such a function is analytic on $(0,\infty)$ and in general cannot be written as a power series as above. (Note $\mu$ has to be a finite measure on $[0,1]$, so it can be extended to $[0,\infty)$ by making it zero on $(1,\infty)$.)

What is true is that given $f$ totally monotone on $(0,1)$ according to your usage, then $f$ must be analytic on $(0,1)$: The function $g(t)=f(1-t)$ is absolutely monotone, with $g^{(n)}(t)\ge0$ for all $n$ and all $t\in(0,1)$. Then $g$ must be analytic---see Widder's book on Laplace transforms.