Norm convergence to $0$ of Markov semigroup

Question

Let $P : (0, \infty) \to B(\ell^1)$ be a norm-continuous Markov operator semigroup on $\ell^1$. Is $P(t)$ (norm-)convergent as $t \to 0$? I have only found in the literature the statement that if $P(t)$ is strongly continuous (i.e. when $B(\ell^1)$ is equipped with its strong operator topology) then $P(t)$ strongly converges as $t \to 0$. The limit is then a projection operator (it need not be the identity as is often demanded for a Markov chain). Formulated the question differently: does the strictly stronger uniform continuity of $P$ imply that the strong convergence of $P(t)$ as $t \to 0$ can be strengthened to norm convergence?

Answer 1

Here is a counterexample:

Let $B := \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ and set $P(t)$ as the block diagonal matrix with $2\times2$-blocks $e^{n Bt}$ for $n \in \mathbb{N}$ and $t \geq 0$. Each component of the matrix $P(t)$ is continuous for $t \geq 0$, meaning that $P(t)$ as an operator semigroup is weakly continuous and hence (since it is a semigroup) also strongly continuous for $t \geq 0$, i.e. $P : [0, \infty) \to B(\ell^1)$ is continuous where $B(\ell^1)$ is equipped with the strong operator topology. Moreover, it is easy to see that $P(t)$ is an analytic semigroup and hence an immediately norm-continuous semigroup [Engel, Nagel "A short course on semigroups" (2006)], meaning that $P : (0, \infty) \to B(\ell^1)$ is continuous where $B(\ell^1)$ is equipped with the operator norm topology. But the generator of $P(t)$ is not bounded and therefore $P(t)$ is not a norm-continuous semigroup for $t \geq 0$ meaning that the strong limit $I = P(0)$ of $P(t)$ as $t \to 0$ cannot be strengthened to be a norm-limit.