BernsteinsTheorem

Let $g:[0,\infty) \to [0,\infty)$ be a continuous function which is additionally in $C^\infty((0,\infty))$. We say that $g$ is completely monotone when $(-1)^n g^{(n)}(t) \geq 0 \forall n \geq 0, t > 0$.

This means that

$$g^{(0)}:=g \geq 0 \in [0,\infty)$$

$$g^{(1)}:=g' \leq 0 \in (0,\infty)$$

$$g^{(2)}:=g'' \geq 0 \in (0,\infty)$$ etc.

In particular,

$$g(+\infty):=\lim_{t \to +\infty} g(t) = 0$$

exists and $g$ is bounded. The famous Bernstein theorem on completely monotone functions states that the following are equivalent:

1. $g$ is completely monotone:

$$(-1)^n g^{(n)}(t) \geq 0 \forall n \geq 0, t > 0$$

2. $g$ is the Laplace transform of a finite Borel measure $\mu$ on $\mathbb{R}^+$:

$$g(x) = \int_{0}^{\infty} e^{-xt} d\mu(t) \forall x \in \mathbb{R}^+$$

Short proof of the Bernstein theorem. The fact that (1) implies (2) follows from the fact that

$$g^{(n)}(t)=(-1)^n \int_{0}^{\infty} x^n e^{-tx} d\mu(x)$$

Let us show that (1) implies (2). Indeed, since $(-1)^n g^{(n)}$ is non-negative and non-increasing, we have for any $t>0$ and $n \geq 1$,

$$|g^{(n)}(t)|=(-1)^n g^{(n)}(t) \leq 2t \int_{t/2}^{t} (-1)^n g^{(n)}(u) du = 2t |g^{(n-1)}(t) - g^{(n-1)}(t/2)|$$

By induction, it follows then that

$$g^{(n)}(t)=o_{t \to +\infty}(t^{-n}) \forall n \geq 1$$

By integration by parts (the boundary terms vanish thanks to the former result),

$$g(x) - g(+\infty) = -\int_{x}^{\infty} g'(t) dt \dots = (-1)^{n+1} n! \int_{x}^{\infty} (t - x)^n g^{(n+1)}(t) dt = \int_0^\infty \phi_n(tx) \left(\frac{(-1)^{n+1}}{n!} \int_0^t n(nt)^n g^{(n+1)}(nt) dt \right) dt = \int_{0}^{\infty} \phi_n(x t) d \sigma_n(t)$$

where

$$\phi_n(x):=(1-x/n)^n \mathbb{1}_{[0,n]}(x)$$

and where

$$\sigma_n(t):=\frac{(-1)^{n+1}}{n!} \int_{ 1/t}^{\infty} n (n t)^n g^{(n+1)}(n t) dt$$

By integration by parts again,

$$1/n! \int_{0}^{\infty} x^n |g^{(n+1)}|(x) dx = (-1)^n (n-1)! \int_{0}^{\infty} x^{n-1} g^{(n)}(x) dx \dots = -\int_{0}^{\infty} g'(x) dx = g(0) - g(+\infty)$$

Therefore, the total variation of $\sigma_n$ on $[0,+ \infty)$ is

$$1/n! \int_{0}^{\infty} n (n t)^n |g^{(n+1)}(n t)| dt = g(0) - g(+\infty)$$

By the Helly selection theorem, there exists a sub-sequence $\sigma_{n_k}(t)$ that converges almost everywhere to a bounded non-negative non-decreasing function $\sigma(t)$ on $[0,+ \infty)$. Since $\phi_n(x) \to e^{-x}$ uniformly on $[0,\infty)$ as $n \to \infty$, it follows that

$$g(x) - g(+\infty) = \int_{0}^{\infty} e^{-t x} d\sigma(t)$$

Finally, if we set

$$\mu:=\sigma+g(+\infty)\delta_0$$

then

$$g(x)=\int_{0}^{\infty} e^{-t x} d\mu(t) \forall x \in [0,+ \infty)$$

This ends the proof of the Bernstein theorem.

Integration By Parts

The progression from the integral of $x^n |g^{(n+1)}|(x)$ to the result $g(0) - g(+\infty)$ involves multiple applications of integration by parts, reducing the power of $x$ in each step. The "..." here represents these steps being repeated.

For the process to become clear, it's best to step through a couple iterations of the process. Let's start from the second step:

1. Apply integration by parts:

$$\int_{0}^{\infty} x^n |g^{(n+1)}|(x) dx = [x^n g^{(n+1)}(x)]_{0}^{\infty} - \int_{0}^{\infty} n x^{n-1} g^{(n+1)}(x) dx$$

The first term on the right-hand side vanishes because we are dealing with a completely monotone function that goes to 0 at infinity.

2. We then repeat the process for the new integral on the right-hand side:

$$- \int_{0}^{\infty} n x^{n-1} g^{(n+1)}(x) dx = - [n x^{n-1} g^{(n+1)}(x)]_{0}^{\infty} + \int_{0}^{\infty} n (n-1) x^{n-2} g^{(n+1)}(x) dx$$

Again, the first term on the right-hand side vanishes.

This process is repeated until the power of $x$ in the integral is reduced to zero, at which point we obtain:

$$(-1)^n n! \int_{0}^{\infty} g^{(n+1)}(x) dx = (-1)^n n! [g^{(n+1)}(x)]_{0}^{\infty} = (-1)^n n! [g(0) - g(+\infty)]$$

Which simplifies to:

$$g(0) - g(+\infty)$$

This progression assumes that $g(+\infty)$ is finite, and hence it vanishes, which is given in the problem statement.