# bp continuity of Markov operators / semigroups

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 ?

• 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 $P_t$. – Jochen Glueck Jan 4 at 22:07
• (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. – Jochen Glueck Jan 4 at 22:26

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 $$A$$).
• Shouldn't it be $e^{-t}f(n) + (1-e^{-t})\omega(f)$, so that $P_0$ is the identity map? – Robert Furber Jan 4 at 14:49