Bernstein's theorem on monotone functions

Remarks

So the completely monotone functions are exactly the Laplace transforms of positive measures. The "if" direction is direct: differentiating under the integral sign gives $(-1)^k f^{(k)}(x) = \int t^k e^{-tx} \, d\mu(t) \ge 0$. The "only if" direction is the hard part; one proof approximates $f$ by sums built from its derivatives and uses a compactness argument for measures. Example: $f(x) = 1/x$ is completely monotone on $(0, \infty)$, and $1/x = \int_0^\infty e^{-tx} \, dt$, so $\mu$ is Lebesgue measure on $[0, \infty)$. By $x \mapsto -x$, the theorem also describes absolutely monotonic functions on $(-\infty, 0)$ as integrals $\int e^{tx} \, d\mu(t)$.

Going deeper

For a student

The theorem links a condition on derivatives with a positivity condition on a measure. It is used in probability, where completely monotone functions are Laplace transforms of distributions on $[0, \infty)$, and in the theory of positive definite functions. If $f$ is also continuous on $[0, \infty)$, then $\mu$ is finite and $\mu([0, \infty)) = f(0)$.