Bernstein function - Revision history

imported>Mateusz: added "Examples" and "Properties"

2012-07-20T11:56:33Z

added "Examples" and "Properties"

@@ Line 8: / Line 8: @@
 \[ f(z) = a z + b + \int_{(0, \infty)} (1 - e^{-t z}) \mu(\mathrm d t) \]
 for some $a, b \ge 0$ and a Radon measure $\mu$ such that $\int_{(0, \infty)} \min(1, t) \mu(\mathrm d t) < \infty$.
 ==Subordination==

imported>Mateusz: typos

2012-07-19T10:41:35Z

typos

← Older revision		Revision as of 05:41, 19 July 2012
Line 1:		Line 1:
	A 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 $k = 1, 2, ...$.<ref name="SSV"/>		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 $k = 1, 2, ...$<ref name="SSV"/>

	==Relation to complete monotonicity==		==Relation to complete monotonicity==

imported>Mateusz: Created page with "A 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 comp..."

2012-07-19T10:30:33Z

Created page with "A 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 $k = 1, 2, ...$.<ref name="SSV"/> ==Relation to comp..."

New page

A 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 $k = 1, 2, ...$.<ref name="SSV"/>

==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 $\int_{(0, \infty)} \min(1, t) \mu(\mathrm d t) < \infty$.

==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==
{{reflist|refs=
<ref name="SSV">{{Citation | last1=Schilling | first1=R. | last2=Song | first2=R. | last3=Vondraček | first3=Z. | title=Bernstein functions. Theory and Applications | year=2010 | publisher=de Gruyter, Berlin | series=Studies in Mathematics | volume=37 | url=http://dx.doi.org/10.1515/9783110215311 | doi=10.1515/9783110215311}}</ref>
}}

{{stub}}