Operator monotone function

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A function $f : [0, \infty) \to [0, \infty)$ is said to be an operator monotone function (complete Bernstein function, Nevanlinna-Pick function for the half-line) if $A \ge B \ge 0$ implies $f(A) \ge f(B) \ge 0$ for any self-adjoint matrices $A$, $B$. Many equivalent definitions can be given.[1]



A function $f$ is operator monotone if and only if \[ f(z) = a z + b + \int_{(0, \infty)} \frac{z}{z + r} \, \frac{\rho(\mathrm d r)}{r} \] for some $a, b \ge 0$ and a Radon measure $\rho$ such that $\int_{(0, \infty)} \min(r^{-1}, r^{-2}) \rho(\mathrm d r) < \infty$.


The following functions are operator monotone:

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


If $f, f_1, f_2$ are operator monotone, then the following functions are also operator monotone:

  • $c_1 f_1(z) + c_2 f_2(z)$ for $c_1, c_2 > 0$,
  • $((f_1(z))^s + (f_2(z))^s)^{1/s}$ and $(f_1(z^s) + f_2(z^s))^{1/s}$ for $s \in (0, 1]$,
  • $(f_1(z))^s (f_2(z))^{1-s}$ and $f_1(z^s) f_2(z^{1-s})$ for $s \in [0, 1]$,
  • $f_1(f_2(z))$,
  • $z^{1-s} f(z^s)$ and $(f(z^s))^{1/s}$ for $s \in (0, 1]$,
  • $z / f(z)$, $z f(1/z)$ and $1 / f(1/z)$.

Relation to Bernstein functions

Operator monotone functions form a subclass of Bernstein functions. Namely, a Bernstein function $f$ is an operator monotone function if and only if the measure $\mu$ in the Bernstein representation of $f$: \[ f(z) = a z + b + \int_{(0, \infty)} (1 - e^{-t z}) \mu(\mathrm d t) \] has a completely monotone density function. In this case \[ \mu(\mathrm d t) = \left( \int_{(0, \infty)} e^{-t r} \rho(\mathrm d r) \right) \mathrm d t \] This explains the name complete Bernstein functions.

Holomorphic extension

Every operator monotone function $f$ extends to a holomorphic function on $\C \setminus (-\infty, 0]$ such that \begin{align*} \Im f(z) & \ge 0 \qquad && \text{if } \Im z \ge 0 , \\ f(z) & \ge 0 \qquad && \text{if } \Im z = 0 , \\ \Im f(z) & \le 0 \qquad && \text{if } \Im z \le 0 . \end{align*} Conversely, any function $f$ with above properties is an operator monotone function.

Functions with nonnegative imaginary part in the upper half-plane are often called Nevanlinna-Pick functions, or Pick functions.

Operator monotone functions of the Laplacian

Operator monotone functions of the Laplacian are particularly regular examples of translation invariant non-local operators in $\R^n$. More precisely, $A = f(-\Delta)$ for an operator monotone $f$ 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 \int_0^\infty (4 \pi t)^{-n/2} e^{-|z|^2 / (4 t)} e^{-t r} \mathrm d t \rho(\mathrm d r) \\ &= \frac{1}{(2 \pi)^{n/2}} \int_0^\infty \left(\frac{\sqrt{r}}{|z|}\right)^{n/2 - 1} K_{n/2 - 1}(\sqrt{r} |z|) \rho(\mathrm d r) . \end{align*} Here $K_\nu$ is the modified Bessel function of the second kind.

For $n = 1$, the above expression simplifies to \[ k(z) = \int_0^\infty \frac{e^{-\sqrt{r} |z|}}{2 \sqrt{r}} \, \rho(\mathrm d r) ; \] that is, $k(z)$ can be an arbitrary completely monotone function of $|z|$, which satisfies the ususal integrability condition $\int_{-\infty}^\infty \min(1, z^2) k(z) dz < \infty$. In a similar way, for $n = 3$, \[ k(z) = \frac{1}{4 \pi |z|} \int_0^\infty e^{-\sqrt{r} |z|} \rho(\mathrm d r) ; \] hence, $|z| k(z)$ can be an arbitrary completely monotone function of $|z|$, provided that $\int_{\R^3} \min(1, |z|^2) k(z) dz < \infty$.


  1. Schilling, R.; Song, R.; Vondraček, Z. (2010), Bernstein functions. Theory and Applications, Studies in Mathematics, 37, de Gruyter, Berlin, doi:10.1515/9783110215311, http://dx.doi.org/10.1515/9783110215311 

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