Bernstein function

A continuous function $f : [0, \infty) \to [0, \infty)$ is said to be a Bernstein function if $(-1)^k f^{(k)}(x) \le 0$ for $x > 0$ and $k = 1, 2, ...$

Relation to complete monotonicity

Clearly, $f$ is a Bernstein function if and only if it is nonnegative, and $f'$ is a completely monotone function.

Representation

By Bernstein's theorem, $f$ is a Bernstein function if and only if: \[ f(z) = a z + b + \int_{(0, \infty)} (1 - e^{-t z}) \mu(\mathrm d t) \] for some $a, b \ge 0$ and a Radon measure $\mu$ such that $\int_{(0, \infty)} \min(1, t) \mu(\mathrm d t) < \infty$.

Examples

The following functions are Bernstein functions of $z$:

• $z^s$ for $s \in [0, 1]$,
• $\log(1 + z)$,
• $\frac{z}{r + z}$ for $r > 0$,
• $1 - e^{-t z}$ for $t > 0$.

All but the last one are in fact complete Bernstein functions.

Properties

If $f_1, f_2$ are Bernstein functions and $c > 0$, then $c f_1$, $f_1 + f_2$ and $f_1 \circ f_2$ are Bernstein functions.

Subordination

Bernstein functions are closely related to Bochner's subordination of semigroups. Namely, for a nonnegative definite self-adjoint operator $L$ and a Bernstein function $f$, the operator $-f(L)$ (defined by means of spectral theory) is the generator of some semigroup of operators which is subordinate to the semigroup $e^{-t L}$ generated by $-L$. Conversely, every generator of a semigroup subordinate to $e^{-t L}$ is equal to $-f(L)$ for some Bernstein function $f$.

Bernstein functions of the Laplacian

Bernstein functions of the Laplacian are translation invariant non-local operators in $\R^n$. More precisely, $A = f(-\Delta)$ for a Bernstein function $f$ if and only if \[ -A u(x) = a \Delta u(x) + b u(x) + \int_{\R^n} (u(x + z) - u(x) - z \cdot \nabla u(x) \mathbf{1}_{|z| < 1}) k(z) \mathrm d z \] for some $a, b \ge 0$ and $k(z)$ of the form \begin{align*} k(z) &= \int_0^\infty (4 \pi t)^{-n/2} e^{-|z|^2 / (4 t)} \mu(\mathrm d t) . \end{align*}

References

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