This post is devoted to a simple proof of the Bernstein theorem on completely monotone functions. We have already mentioned this theorem in a previous post on the Schoenberg theorem on positive definite functions. Let ${g:[0,\infty)\rightarrow[0,\infty)}$ be a continuous function which is additionally in ${\mathcal{C}^\infty((0,\infty))}$. We say that ${g}$ is completely monotone when

$\forall n\geq0, \forall t>0, \quad (-1)^ng^{(n)}(t)\geq0.$

This means that ${g^{(0)}:=g\geq0}$ on ${[0,\infty)}$, ${g^{(1)}:=g’\leq0}$ on ${(0,\infty)}$, ${g^{(2)}:=g”\geq0}$ on ${(0,\infty)}$, etc. In particular, ${g(+\infty):=\lim_{t\rightarrow+\infty}g(t)=0}$ exists and ${g}$ is bounded. The famous Bernstein theorem on completely monotone functions states that the following are equivalent:

• (k) ${g}$ is completely monotone:

$\forall n\geq0, \forall t>0, \quad (-1)^ng^{(n)}(t)\geq0;$

• (kk) ${g}$ is the Laplace transform of a finite Borel measure ${\mu}$ on ${\mathbb{R}_+}$:

$\forall x\in\mathbb{R}_+, \quad g(x)=\int_0^\infty\!e^{-xt}\,d\mu(t).$

Short proof of the Bernstein theorem. The fact that (kk) implies (k) follows from the fact that ${g^{(n)}(t)=(-1)^n\int_0^\infty\!x^ne^{-tx}\,d\mu(x)}$. Let us show that (k) implies (kk). Indeed, since ${(-1)^ng^{(n)}}$ is non-negative and non-increasing, we have for any ${t>0}$ and ${n\geq1}$,

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

By induction, it follows then that

$\forall n\geq1,\quad g^{(n)}(t)=o_{t\rightarrow+\infty}(t^{-n}).$

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

$\begin{array}{rcl} g(x)-g(+\infty) &=&-\int_x^\infty\!g'(t)\,dt\\ &\vdots&\\ &=&\frac{(-1)^{n+1}}{n!}\int_x^\infty\!(t-x)^ng^{(n+1)}(t)\,dt\\ &=&\frac{(-1)^{n+1}n}{n!}\int_{x/n}^{\infty}\!(1-x/(tn))^n(nt)^ng^{(n+1)}(nt)\,dt\\ &=&\int_0^\infty\!\varphi_n(x/t)d\left(\frac{(-1)^{n+1}n}{n!}\int_0^t\!(nt)^ng^{(n+1)}(nt)\,dt\right)\\ &=&\int_0^\infty\!\varphi_n(xt)\,d\sigma_n(t). \end{array}$

where ${\varphi_n(x):=(1-x/n)^n\mathbf{1}_{[0,n]}(x)}$ and where

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

By integration by parts again,

$\begin{array}{rcl} \frac{1}{n!}\int_0^\infty\!x^n|g^{(n+1)}|(x)\,dx &=&\frac{(-1)^{n}}{(n-1)!}\int_0^\infty\!x^{n-1}g^{(n)}(x)\,dx\\ &\vdots&\\ &=&-\int_0^\infty\!g'(x)\,dx=g(0)-g(+\infty). \end{array}$

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

$\frac{1}{n!}\int_0^\infty\!n(nt)^n|g^{(n+1)}(nt)|\,dt=g(0)-g(+\infty).$

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 ${\varphi_n(x)\rightarrow e^{-x}}$ uniformly on ${[0,\infty)}$ as ${n\rightarrow\infty}$, it follows that

$g(x)-g(+\infty)=\int_0^\infty\!e^{-tx}\,d\sigma(t).$

Finally, if we set ${\mu:=\sigma+g(+\infty)\delta_0}$ we get, then for all ${x\in[0,+\infty)}$,

$g(x)=\int_0^\infty\!e^{-tx}\,d\mu(t)$

This ends the proof of the Bernstein theorem.

Additional notes and further reading. The Bernstein theorem is also known as the (Hausdorff-)Bernstein(-Widder) theorem and can be seen as a characterization of the Laplace transforms of finite Borel measures on ${\mathbb{R}_+}$. Sergei Natarovich Bernstein (1880 – 1968) proved his theorem in a paper published in 1928 in Acta Mathematica, see also theorem 12a in the book The Laplace Transform by Widder. The short proof that we gave is taken from the second section of a paper by Korenblum et al. The Bernstein theorem is at the heart of the famous Bessis-Moussa-Villani conjecture (BMV), which states that for every ${n\times n}$ Hermitian matrices ${A}$ and ${B}$ with ${B}$ semidefinite positive, the function

$t\geq0\mapsto\mathrm{Tr}(\exp(A-tB))$

is the Laplace transform of a positive measure on ${[0,\infty[}$. It seems that Matteo Villani has nothing to do with Cédric Villani. The BMV conjecture comes from mathematical physics and was solved a couple of years ago by Herbert Stahl. For more informations on this conjecture, one may read Lieb and Seiringer.

1. Gérard Letac 2013-03-28

Lieb et Seiringer 2012 continuent à parler de la conjecture, tout en disant que Stahl l’a démontrée, ce qui est bizarre. La démonstration de Stahl sur Arkiv me paraissant difficile à suivre, je me demande si cette démonstration est correcte et authentifiée par publication dans un journal sérieux?