# Subordination

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

If $T_t$ is a strongly continuous semigroup of contraction operators on a Banach space and $\mu_t$ is a semigroup of sub-probabilistic measures on $[0, \infty)$, then $\tilde{T}_t u = \int_{[0, \infty)} T_r u \mu_t(\mathrm d r)$ defines another semigroup of operators $\tilde{T}_t$, which is said to be subordinate (in the sense of Bochner) to $\tilde{T}_t$.[1]

## Relation to Bernstein functions

For some Bernstein function $f$, $\int_{[0, \infty)} e^{-r z} \mu_t(\mathrm d r) = e^{-t f(z)} .$ Conversely, for any Bernstein function $f$ there exists a semigroup of sub-probabilistic measures $\mu_t$ satisfying the above equality.

## Functional calculus

Let $-L$ and $-\tilde{L}$ be the generators of $T_t$ and $\tilde{T}_t$, respectively. If $f$ is the Bernstein function described above, then it is customary to write $\tilde{L} = f(L)$.

This notation agrees with spectral-theoretic definition of $f(L)$ if the underlying Banach space is a Hilbert space. Furthermore, if $f_1$, $f_2$ are Bernstein functions and $c > 0$, then \begin{align*} c f_1(L) & = (c f_1)(L) , \\ f_1(L) + f_2(L) & = (f_1 + f_2)(L) , \\ f_1(f_2(L)) & = (f_1 \circ f_2)(L) , \end{align*} and if $f_1 f_2$ is also a Bernstein function, then $f_1(L) f_2(L) = (f_1 f_2)(L) .$

Since $z^s$ is a Bernstein function for $s \in (0, 1)$, subordination gives a way to define the fractional power $L^s$ of an operator $L$ on a Banach space, provided that $-L$ generates a strongly continuous semigroup of contractions.

## References

1. Schilling, R. (1998), "Subordination in the sense of Bochner and a related functional calculus", J. Aust. Math. Soc. Ser. A 64: 368–396