1 Introduction

In 1958, D. Blackwell gave an example of a continuous-time Markov chain with all states instantaneous. The nature of this chain allows to perform some explicit or semi-explicit calculations related to it, see e.g. [4, 5, 8, 10].

In [9], I found an explicit formula for an (unbounded) operator A such that its closure is the generator of a strongly continuous semigroup of Markov operators associated with Blackwell’s chain. Based on this article, I give here two examples of semigroups that fit in the framework of [1, Proposition 2.7]. In order to cite this proposition, which I call Theorem 1.1, a short introduction is needed. I use the convention that \({\mathbb {N}} = \{1,2,3,\ldots \}\).

Suppose that \((B_{n})_{n\in {\mathbb {N}}}\) is a sequence of generators of commuting contraction semigroups defined on a Banach space \((X,||\cdot ||)\). This means that for any \(m, n \in {\mathbb {N}}\) and \(x\in X\), the following identity holds

$$\begin{aligned} e^{tB_{n}}e^{s B_{m}}x = e^{s B_{m}}e^{t B_{n}}x,\quad t,s\ge 0. \end{aligned}$$

Furthermore, \(\lim _{t\rightarrow 0+}e^{tB_{n}}x=x\), \(||e^{tB_{n}}x|| \le ||x||\) for every \(x\in X\), \(n\in {\mathbb {N}}\), and \(t\ge 0\). And finally, if n is fixed, then \(e^{tB_{n}}\) is a semigroup, i.e., \(e^{(t+s)B_{n}} = e^{tB_{n}}e^{sB_{n}}\) for all \(t,s\ge 0\). In general, the generators may not be bounded so \(D(B_{n})\) denotes the domain of \(B_{n}\).

If we have a sequence of semigroups described above, then for any \(n\ge 2\), the product

$$\begin{aligned} T_{n}(t)x = \prod _{k=1}^{n}e^{t B_{k}}x,\quad t \ge 0, \end{aligned}$$

is also a strongly continuous contraction semigroup on X and its generator is the closure of \(A_{n}\) given by

$$\begin{aligned} A_{n}x = \sum _{k=1}^{n}B_{k}x,\quad x\in \bigcap _{k=1}^{n}D(B_{k}), \end{aligned}$$
(1.1)

see [2, p. 24]. As in [1], we say that the product \(\prod _{k=1}^{\infty }e^{tB_{k}}\) exists if \(T(t)x:=\lim _{n\rightarrow \infty }T_{n}(t)x\) converges uniformly on compact subsets of \([0,\infty )\) for every \(x\in X\). Then again T(t), \(t\ge 0\), is a \(C_{0}\) semigroup of contractions on X. The following theorem was proved in [1].

Theorem 1.1

Let \((e^{t B_{n}})_{n\in {\mathbb {N}}}\) be a commuting family of contraction semigroups and suppose that

$$\begin{aligned} D_{1} = \left\{ x\in \bigcap _{k=1}^{\infty }D(B_{k}): \sum _{k=1}^{+\infty }||B_{k}x|| <\infty \right\} \end{aligned}$$
(1.2)

is dense in X. Then the product \(\prod _{k=1}^{\infty }e^{tB_{k}}\) exists. Moreover, define A by

$$\begin{aligned} Ax = \lim _{n\rightarrow +\infty }\sum _{k=1}^{n}B_{k}x, \quad x\in D_{1}. \end{aligned}$$

Then A is closable and its closure is the generator of \(\prod _{k=1}^{\infty }e^{tB_{k}}\).

2 The semigroups

Recall that \(l^{1}({\mathbb {N}})\) is the Banach space of all absolutely summable sequences \(x = (\xi _{i})_{i\in {\mathbb {N}}}\). This means that \(\sum _{i\in {\mathbb {N}}}|\xi _{i}|<\infty \) and \(||x||_{l^{1}({\mathbb {N}})} = \sum _{i\in {\mathbb {N}}}|\xi _{i}|\).

Suppose that \(\alpha _{n}, \beta _{n}\), \(n\ge 1\), are positive numbers and denote

$$\begin{aligned} S_{n}^{1} = \{1,2,3,\ldots ,2^{n-1}\}, \quad S_{n}^{2} = \{0,2^{n-1}+1,\ldots ,2^{n}-1\}. \end{aligned}$$

For example, \(S_{1}^{1}=\{1\}\), \(S_{1}^{2}=\{0\}\), \(S_{2}^{1}=\{1,2\}\), \(S_{2}^{2}=\{0,3\}\), etc.

For \(n\ge 1\), define \(B_{n}\) on \(l^{1}({\mathbb {N}})\) as follows

$$\begin{aligned} B_{n}x = (\eta _{i})_{i\in {\mathbb {N}}} = {\left\{ \begin{array}{ll} -\beta _{n}\xi _{i} + \alpha _{n}\xi _{i+2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{1},\\ - \alpha _{n}\xi _{i} + \beta _{n}\xi _{i-2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{2}.\\ \end{array}\right. } \end{aligned}$$
(2.1)

It is clear that the \(B_{n}\)’s are linear bounded operators and

$$\begin{aligned} ||B_{n}x||_{l^{1}({\mathbb {N}})}\le 2(\alpha _{n}+\beta _{n})||x||_{l^{1}({\mathbb {N}})}. \end{aligned}$$

In Lemma 2.1, I prove that the \(B_{n}\)’s commute. It may be verified directly that

$$\begin{aligned} \sum _{i=1}^{\infty }\eta _{i}=0 \end{aligned}$$
(2.2)

and this property is a consequence of the fact that the \(B_{n}\)’s are isomorphic images of generators of Markov semigroups, see [9, Corollary 2]. The property (2.2) is also true for A in Theorem 3.1 since the norm convergence in \(l^{1}({\mathbb {N}})\) implies the coordinate-wise convergence.

It is well known (see [3, p. 207]) that the dual space of \(l^{1}({\mathbb {N}})\) can be identified with \(l^{\infty }({\mathbb {N}})\), that is, with the space of all bounded sequences. If \(x = (\xi _{i})_{i\in {\mathbb {N}}}\) is an element of \(l^{\infty }({\mathbb {N}})\), then \(||x||_{l^{\infty }({\mathbb {N}})} = \sup _{i\in {\mathbb {N}}}|\xi _{i}|\).

Thus \(B_{n}\) induces a linear map \(B_{n}^{*}: l^{\infty }({\mathbb {N}})\rightarrow l^{\infty }({\mathbb {N}})\) called the adjoint of \(B_{n}\), see [6, p. 15.] In our case it is given by

$$\begin{aligned} B_{n}^{*}x = (\eta _{i})_{i\in {\mathbb {N}}} = {\left\{ \begin{array}{ll} \beta _{n}(-\xi _{i} + \xi _{i+2^{n-1}}) \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{1},\\ \alpha _{n}(-\xi _{i} + \xi _{i-2^{n-1}}) \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{2}.\\ \end{array}\right. } \end{aligned}$$
(2.3)

Moreover

$$\begin{aligned} ||B_{n}^{*}x||_{l^{\infty }({\mathbb {N}})}\le 2\max \{\alpha _{n},\beta _{n}\}||x||_{l^{\infty }({\mathbb {N}})}. \end{aligned}$$

Since the \(B_{n}\)’s and the \(B_{n}^{*}\)’s are continuous, they are the generators of the following semigroups

$$\begin{aligned} e^{tB_{n}} = \sum _{k=0}^{\infty }\frac{(tB_{n})^{k}}{k!}, \quad e^{tB_{n}^{*}} = \sum _{k=0}^{\infty }\frac{(tB_{n}^{*})^{k}}{k!}, \quad t\ge 0. \end{aligned}$$

By a direct computation, we find (see also [8, p.60])

$$\begin{aligned} e^{tB_{n}}x = {\left\{ \begin{array}{ll} p_{n}(t)\xi _{i} + (1-q_{n}(t))\xi _{i+2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{1},\\ q_{n}(t)\xi _{i} + (1-p_{n}(t))\xi _{i-2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{2},\\ \end{array}\right. } \end{aligned}$$
(2.4)

and

$$\begin{aligned} e^{tB_{n}^{*}}x = {\left\{ \begin{array}{ll} p_{n}(t)\xi _{i} + (1-p_{n}(t))\xi _{i+2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{1},\\ q_{n}(t)\xi _{i} + (1-q_{n}(t))\xi _{i-2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{2},\\ \end{array}\right. } \end{aligned}$$
(2.5)

where \(p_{n}(t)\), \(q_{n}(t)\), \(t\ge 0\), are given by

$$\begin{aligned}&p_{n}(t) = \frac{\alpha _{n}}{\alpha _{n}+\beta _{n}} + \frac{\beta _{n}}{\alpha _{n}+\beta _{n}} e^{-(\alpha _{n}+\beta _{n})t},\\&q_{n}(t) = \frac{\beta _{n}}{\alpha _{n}+\beta _{n}} + \frac{\alpha _{n}}{\alpha _{n}+\beta _{n}} e^{-(\alpha _{n}+\beta _{n})t}. \end{aligned}$$

Notice that \(0 < p_{n}(t)\le 1\), \(0 < q_{n}(t)\le 1\) and as a result

$$\begin{aligned} ||e^{tB_{n}}x||_{l^{1}({\mathbb {N}})}\le ||x||_{l^{1}({\mathbb {N}})},\quad ||e^{tB_{n}^{*}}x||_{l^{\infty }({\mathbb {N}})}\le ||x||_{l^{\infty }({\mathbb {N}})} \end{aligned}$$

meaning that \(e^{tB_{n}}\), \(e^{tB_{n}^{*}}\) are contractions.

Lemma 2.1

For the \(B_{n}\)’s defined by (2.1) and \(m,n\in {\mathbb {N}}\), we have

$$\begin{aligned} B_{n} B_{m} x = B_{m} B_{n}x,\quad x\in l^{1}({\mathbb {N}}). \end{aligned}$$
(2.6)

This implies that the \(B_{n}^{*}\)’s also commute and in consequence

$$\begin{aligned} e^{tB_{m}}e^{s B_{n}}x = e^{s B_{n}}e^{t B_{m}}x,\quad x\in l^{1}({\mathbb {N}}), \end{aligned}$$
(2.7)

and

$$\begin{aligned} e^{tB_{m}^{*}}e^{s B_{n}^{*}}x = e^{s B_{n}^{*}}e^{t B_{m}^{*}}x,\quad x\in l^{\infty }({\mathbb {N}}), \end{aligned}$$

for all \(t,s\ge 0\) and \(m,n\in {\mathbb {N}}\).

Proof

It is enough to prove (2.6) since the conditions (2.6) and (2.7) are equivalent if the operators \(B_{n}\), \(B_{m}\) are bounded, see [7, p. 19.] In addition, the equality \((B_{n}B_{m})^{*} = B_{m}^{*} B_{n}^{*}\) implies that the \(B_{n}^{*}\)’s also commute.

Suppose now that \(1 \le m < n\) and denote \(x = (\xi _{i})_{i\in {\mathbb {N}}}\), \((\eta _{i})_{i\in {\mathbb {N}}} = B_{m}x\), \((\eta _{i}')_{i\in {\mathbb {N}}} = B_{n}(\eta _{i})_{i\in {\mathbb {N}}}\). Then by definition

$$\begin{aligned} \eta _{i} = {\left\{ \begin{array}{ll} -\beta _{m}\xi _{i} + \alpha _{m}\xi _{i+2^{m-1}} \quad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{1},\\ - \alpha _{m}\xi _{i} + \beta _{m}\xi _{i-2^{m-1}} \quad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{2},\\ \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \eta _{i}' = {\left\{ \begin{array}{ll} -\beta _{n}\eta _{i} + \alpha _{n}\eta _{i+2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{1},\\ - \alpha _{n}\eta _{i} + \beta _{n}\eta _{i-2^{n-1}} \quad \text{ if } \;\; i\bmod 2^{n} \in S_{n}^{2}.\\ \end{array}\right. } \end{aligned}$$

So \((\eta _{i}')_{i\in {\mathbb {N}}} = B_{n}B_{m}(\xi _{i})_{i\in {\mathbb {N}}}\) is given by

$$\begin{aligned} \eta _{i}' = {\left\{ \begin{array}{ll} \beta _{n}\beta _{m}\xi _{i} -\beta _{n}\alpha _{m} \xi _{i+2^{m-1}} - \alpha _{n}\beta _{m}\xi _{i+2^{n-1}} + \alpha _{n}\alpha _{m}\xi _{i+2^{m-1}+2^{n-1}} \\ \qquad \qquad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{1} \;\; \text{ and }\;\; i\bmod 2^{n} \in S_{n}^{1},\\ \beta _{n}\alpha _{m}\xi _{i} -\beta _{n}\beta _{m} \xi _{i-2^{m-1}} - \alpha _{n}\alpha _{m}\xi _{i+2^{n-1}} + \alpha _{n}\beta _{m}\xi _{i-2^{m-1}+2^{n-1}} \\ \qquad \qquad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{2} \;\; \text{ and }\;\; i\bmod 2^{n} \in S_{n}^{1},\\ \alpha _{n}\beta _{m}\xi _{i} -\alpha _{n}\alpha _{m} \xi _{i+2^{m-1}} - \beta _{n}\beta _{m}\xi _{i-2^{n-1}} + \beta _{n}\alpha _{m}\xi _{i+2^{m-1}-2^{n-1}} \\ \qquad \qquad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{1} \;\; \text{ and }\;\; i\bmod 2^{n} \in S_{n}^{2},\\ \alpha _{n}\alpha _{m}\xi _{i} -\alpha _{n}\beta _{m} \xi _{i-2^{m-1}} - \beta _{n}\alpha _{m}\xi _{i-2^{n-1}} + \beta _{n}\beta _{m}\xi _{i-2^{m-1}-2^{n-1}} \\ \qquad \qquad \text{ if } \;\; i\bmod 2^{m} \in S_{m}^{2} \;\; \text{ and }\;\; i\bmod 2^{n} \in S_{n}^{2}.\\ \end{array}\right. } \end{aligned}$$

From \(m<n\), we have \(2^{m}\le 2^{n-1}\). It can be seen now that the condition \(i\bmod 2^{n} \in S_{n}^{1}\) does not depend on whether or not \(i\bmod 2^{m} \in S_{m}^{1}\). Similarly the condition \(i\bmod 2^{n} \in S_{n}^{2}\) is independent of \(i\bmod 2^{m} \in S_{m}^{1}\). Thus if we calculate \(B_{m}B_{n}(\xi _{i})_{i\in {\mathbb {N}}}\), we get the same result as that of \(B_{n}B_{m}(\xi _{i})_{i\in {\mathbb {N}}}\). \(\square \)

3 Main theorems

Similar calculations to those in the proof of Lemma 2.1 can be carried out to find formulae for \(T_{n}(t)=\prod _{k=1}^{n}e^{t B_{k}}\), \(n\ge 2\). These formulae become more and more complicated as n increases. However the generator \(A_{n}\) of \(T_{n}(t)\) has a rather simple form. Denote \((\eta _{i})_{i\in {\mathbb {N}}} = A_{n}(\xi _{i})_{i\in {\mathbb {N}}}\). Then from (1.1), we have that \(\eta _{i} = \sum _{k=1}^{n}\zeta _{k}\), where \(\zeta _{k}\), \(k=1,\ldots , n\), are given by

$$\begin{aligned} \zeta _{k} = {\left\{ \begin{array}{ll} -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}} \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{1},\\ - \alpha _{k}\xi _{i} + \beta _{k}\xi _{i-2^{k-1}} \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{2}.\\ \end{array}\right. } \end{aligned}$$

I show in Theorem 3.1 that \(A_{n}\) still has a manageable form even if \(n=+\infty \). For this, we only need the following condition to be satisfied

$$\begin{aligned} \sum _{n=1}^{\infty } \max \{\alpha _{n},\beta _{n}\} < \infty , \end{aligned}$$
(3.1)

which is equivalent to \(\sum _{n=1}^{\infty } (\alpha _{n}+\beta _{n})<\infty \) for positive \(\alpha _{n}\)’s and \(\beta _{n}\)’s.

Before stating the first of two main theorems, notice that any natural number \(n \ge 2\) can be expressed in a unique way as \(2^{l}+m\) for some \(l\ge 0\) with \(m\in \{1,2,\ldots ,2^{l}\}\). For example, \(2=2^{0}+2^{0}\), \(7 = 2^{2}+3\), \(8=2^{2}+2^{2}\), etc.

Theorem 3.1

Assume that \(\alpha _{n}, \beta _{n}\), \(n\ge 1\), are positive numbers satisfying (3.1). Then the infinite product T(t), \(t\ge 0\), of semigroups given by (2.4) exists. Let A be the generator of T(t) and denote \((\eta _{i})_{i\in {\mathbb {N}}} = Ax\), where \(x = (\xi _{i})_{i\in {\mathbb {N}}}\in l^{1}({\mathbb {N}})\). Then

$$\begin{aligned} \eta _{1} = \sum _{k=1}^{\infty }(-\beta _{k}\xi _{1} + \alpha _{k}\xi _{1+2^{k-1}}), \end{aligned}$$
(3.2)

and if \(i = 2^{l} + m\) for \(l\ge 0\) with \(m\in \{1,2,\ldots , 2^{l}\}\), we have

$$\begin{aligned} \eta _{i} = \sum _{k=1}^{l+1}\zeta _{k} + \sum _{k= l+2}^{\infty }(-\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}}), \end{aligned}$$
(3.3)

where \(\zeta _{k}\), \(k=1,\ldots , l+1\), are given by

$$\begin{aligned} \zeta _{k} = {\left\{ \begin{array}{ll} -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}} \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{1},\\ - \alpha _{k}\xi _{i} + \beta _{k}\xi _{i-2^{k-1}} \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{2}.\\ \end{array}\right. } \end{aligned}$$

Proof

We use Theorem 1.1. In our case, \(D_{1} = l^{1}({\mathbb {N}})\) because

$$\begin{aligned} \sum _{k=1}^{+\infty }||B_{k}x||_{l^{1}({\mathbb {N}})} \le 2||x||_{l^{1}({\mathbb {N}})} \sum _{k=1}^{\infty } (\alpha _{k}+\beta _{k}) < \infty \end{aligned}$$

for every \(x\in l^{1}({\mathbb {N}})\). The norm convergence in \(l^{1}({\mathbb {N}})\) implies the coordinate-wise convergence, hence components of Ax are limits of components of \(A_{n}x\), where \(A_{n}=\sum _{k=1}^{n}B_{k}x\). Thus (3.2) and (3.3) follow and

$$\begin{aligned} ||A|| \le 2\sum _{k=1}^{\infty } (\alpha _{k}+\beta _{k}). \end{aligned}$$

This completes the proof. \(\square \)

Any \(x\in l^{1}({\mathbb {N}})\) can be written as \(\sum _{i\in {\mathbb {N}}}\xi _{i}e_{i}\), where \(\{e_{i}\}_{i\in {\mathbb {N}}}\) is the standard Schauder basis in \(l^{1}({\mathbb {N}})\), i.e., \(e_{i} = (\ldots , 0, 1, 0, \ldots )\) with 1 in the i-th coordinate. From Theorem 3.1, we have for example

$$\begin{aligned} A e_{1} = (\eta _{i})_{i\in {\mathbb {N}}} = {\left\{ \begin{array}{ll} -\sum _{k=1}^{\infty }\beta _{k},\quad i=1,\\ \beta _{k},\quad \text{ for }\; i=1+2^{k-1}\; \text{ and }\; k\ge 1,\\ 0,\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} A e_{2} = (\eta _{i})_{i\in {\mathbb {N}}} = {\left\{ \begin{array}{ll} \alpha _{1},\quad i=1,\\ -\alpha _{1}-\sum _{k=2}^{\infty }\beta _{k},\quad i=2,\\ \beta _{k},\quad \text{ for }\; i=2+2^{k-1}\; \text{ and }\; k\ge 2,\\ 0,\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Theorem 3.2

Assume that \(\alpha _{n}, \beta _{n}\), \(n\ge 1\), are positive numbers satisfying (3.1). Then the infinite product \(T^{*}(t)\), \(t\ge 0\), of semigroups given by (2.5) exists. Let \(A^{*}\) be the generator of \(T^{*}(t)\) and denote \((\eta _{i})_{i\in {\mathbb {N}}} = A^{*}x\), where \(x = (\xi _{i})_{i\in {\mathbb {N}}}\in l^{\infty }({\mathbb {N}})\). Then

$$\begin{aligned} \eta _{1} = \sum _{k=1}^{\infty }\beta _{k}(-\xi _{1} + \xi _{1+2^{k-1}}), \end{aligned}$$
(3.4)

and if \(i = 2^{l} + m\) for \(l\ge 0\) with \(m\in \{1,2,\ldots , 2^{l}\}\), we have

$$\begin{aligned} \eta _{i} = \sum _{k=1}^{l+1}\zeta _{k} + \sum _{k= l+2}^{\infty } \beta _{k}(-\xi _{i}+\xi _{i+2^{k-1}}) \end{aligned}$$
(3.5)

where \(\zeta _{k}\), \(k=1,\ldots , l+1\), are given by

$$\begin{aligned} \zeta _{k} = {\left\{ \begin{array}{ll} \beta _{k}(-\xi _{i} + \xi _{i+2^{k-1}}) \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{1},\\ \alpha _{k}(-\xi _{i} + \xi _{i-2^{k-1}}) \quad \text{ if } \;\; i\bmod 2^{k} \in S_{k}^{2}.\\ \end{array}\right. } \end{aligned}$$

Proof

The proof is analogous to that of Theorem 3.1. For every \(x\in l^{\infty }({\mathbb {N}})\), we have

$$\begin{aligned} \sum _{k=1}^{+\infty }||B_{k}^{*}x||_{l^{\infty }({\mathbb {N}})} \le 2||x||_{l^{\infty }({\mathbb {N}})} \sum _{k=1}^{\infty } \max \{\alpha _{k},\beta _{k}\} < \infty . \end{aligned}$$

So \(D_{1} = l^{\infty }({\mathbb {N}})\) and the operator \(A^{*}\) is bounded. The norm convergence in \(l^{\infty }({\mathbb {N}})\) implies the coordinate-wise convergence, thus components of \(A^{*}x\) are limits of \(A_{n}^{*}x\), where \(A_{n}^{*}=\sum _{k=1}^{n}B_{k}^{*}x\) and

$$\begin{aligned} ||A^{*}|| \le 2\sum _{k=1}^{\infty } \max \{\alpha _{k},\beta _{k}\}. \end{aligned}$$

This completes the proof. \(\square \)

4 Remarks

I proved in [9] that if \(\alpha _{n}, \beta _{n}\), \(n\ge 1\), are positive numbers satisfying the following conditions (introduced by D. Blackwell in [4] to secure the existence of a Markov process with countably many states all of which are instantaneous)

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{\beta _{n}}{\alpha _{n}+\beta _{n}} < \infty ,\qquad \sum _{n=1}^{\infty }\beta _{n} = \infty , \end{aligned}$$
(4.1)

then the infinite product T(t), \(t\ge 0\), of semigroups given by (2.4) exists and is composed of Markov operators associated with Blackwell’s chain. However then the generator of T(t) is unbounded and is the closure of A given by (3.2)–(3.3). In this case, A is defined on a dense subset D(A) of \(l^{1}({\mathbb {N}})\) and interestingly \( e_{i}\notin D(A)\) for every \(i\in {\mathbb {N}}\). To see it, suppose that \(i = 2^{l} + m\) for some \(l\ge 0\) with \(m\in \{1,2,\ldots , 2^{l}\}\). Then by (2.1) and (4.1),

$$\begin{aligned} \sum _{k=1}^{+\infty }||B_{k}e_{i}||_{l^{1}({\mathbb {N}})} \ge \lim _{n\rightarrow +\infty }\sum _{k= l+2}^{n} \beta _{k} = +\infty . \end{aligned}$$

As a result, any \(x= (\xi _{i})_{i\in {\mathbb {N}}}\) with a finite number of non-zero components does not belong to D(A).