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Let $B_b(E)$ be the space of bounded measurable functions on some Polish space $E$ endowed with the supremum norm. It seems quite classical that Markov semigroups $P_t:B_b(E)\to B_b(E)$ are in one to one correspondence with Markov processes on $E$.

By Markov semigroup I mean a strongly continuous semigroup $P_t:B_b(E)\to B_b(E)$ satisfying

  • $P_t$ is conservative for all $t$, i.e. $P_t1=1$
  • $\|P_t\|=1$, for all $t$
  • $P_t$ is a positive operator for all $t$

From such an object, the natural way to build the transition functions $p_t(x,dy)$ of an eventually associated Markov process is to set $p_t(x,A):=P_t\mathbb{1}_A(x)$, for all measurable set $A$ and all $x$ in $E$.

However, it does not seem direct that $p_t(x,\cdot)$ is $\sigma$-additive and I was not able to find a proper reference showing this (in the particular context of semigroups on $B_b(E)$).

In fact, the only references I found assume that $P_t$ is bounded pointwise continuous, i.e. for any sequence $(f_n)_n$ of $B_b(E)$ converging pointwisely to a function $f$ and being uniformly bounded, then $$P_t f_n(x)\to P_tf(x)$$ for all $x$ as $n$ goes to infinity. Property which is in fact equivalent to the $\sigma$-additivity of $p_t$.

So, are Markov semigroup bp-continuous ? If yes, how can it be derived from the assumptions? if no, is there a counterexample ?

thanks for your time

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    $\begingroup$ Just two small remarks: (i) Your assumptions which ensure that the semigroup $(P_t)_{t \ge 0}$ is Markov are partially redundant: if a linear operator $P$ on $B_b(E)$ satisfies $P1 = 1$, then it follows that $P$ has operator norm $1$ if and only if $P$ is positive (see for instance [R. Nagel (ed): One-parameter Semigroups of Positive Operators (Springer, 1986), Lemma B-III-2.1]). Hence, you can drop either your second or your third assumption on $P_t$. $\endgroup$ – Jochen Glueck Jan 4 at 22:07
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    $\begingroup$ (ii) The assumption that the semigroup $(P_t)_{t \ge 0}$ be strongly continuous is much stronger than what is usually satisfied in examples. In fact, every $C_0$-semigroup on the space $B_b(E)$ is automatically uniformly continuous (see for instance [op. cit., Theorem A-II-3.6]), so there are only very few $C_0$-semigroups on this space. $\endgroup$ – Jochen Glueck Jan 4 at 22:26
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Some continuity assumption is needed, as shown by the following "Markov" semigroup on the set of natural numbers $\mathbb{N}$.

Let $\omega$ be an ultrafilter in $\mathbb{N}$. Every element $f \in B_b(\mathbb{N})$ (that is, every bounded sequence $f(n)$) has a finite "generalised limit" along $\omega$, that we denote by $\omega(f)$. Furthermore, $\omega(c f) = c \omega(f)$, $\omega(f + g) = \omega(f) + \omega(g)$, if $f$ is non-negative, then $\omega(f) \geqslant 0$, and if $f$ is constant $1$, then $\omega(f) = 1$.

We define $$P_t f(n) = e^{-t} f(n) + (1 - e^{-t}) \omega(f).$$ It is straightforward to see that $P_t$ is a strongly continuous semigroup of operators on $B_b(\mathbb{N})$ with all the properties listed in the question. However, $P_t f(n)$ is not given as an integral of $f$ with respect to a $\sigma$-additive measure (or, in other words, $p_t(n, A)$ is not $\sigma$-additive with respect to $A$).

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  • $\begingroup$ Shouldn't it be $e^{-t}f(n) + (1-e^{-t})\omega(f)$, so that $P_0$ is the identity map? $\endgroup$ – Robert Furber Jan 4 at 14:49
  • $\begingroup$ @RobertFurber: Yes, of course! Thanks. $\endgroup$ – Mateusz Kwaśnicki Jan 4 at 15:15

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