1 Introduction

W. Feller and H.P. McKean showed in 1956, see [3], that there exists a Markov chain with all states instantaneous. They considered a diffusion \(x_{t}\) with the state space \(K = [0,1]\) and with some specified transition function, and proved that this diffusion is in fact equivalent to a Markov chain \(\hat{x}_{t}\) with the state space \(\mathbb {N}\), where \(\mathbb {N}\) is the set of natural numbers. In addition, the Q matrix of the chain has \(-\infty \) down the diagonal and zeroes elsewhere. Thus, the Feller–McKean chain is a continuous process which spends almost all of its time on \(\mathbb {N}\) even though it leaves each state instantly.

In 1958, D. Blackwell, see [1], gave a more tractable example of a Markov chain with all states instantaneous. In this situation, the connection between intensity matrices of Markov chains and semigroups of Markov operators in \(l^{1}\) corresponding to them, is of special interest. To be precise, let \(X(t), t \ge 0\) be a Markov chain with values in \(\mathbb {N}\) and \(\{P(t), t\ge 0\}\) be its corresponding Markov semigroup in \(l^{1}\) with generator A. Lemma 2 in [5] says that if a given state \(i\in \mathbb {N}\) is instantaneous then \(e_{i}\notin \mathcal {D}(A)\), where \(e_{i}\) is an element of the standard Schauder basis in \(l^{1}\) and \(\mathcal {D}(A)\) is the domain of A, see Sect. 2 for definitions. Since all states of Blackwell’s chain are instantaneous, \(e_{i}\notin \mathcal {D}(A_{B})\) for every \(i\in \mathbb {N}\), where \(A_{B}\) is the generator of the corresponding Blackwell semigroup. Because \(A_{B}\) is not explicitly given in the literature (as far as I am aware), I became interested in finding a formula for this generator. However, due to difficulties with a precise description of the domains of such generators, I slightly modified my problem. It can be stated as follows: find a formula for a linear operator \(A: \mathcal {D}(A)\rightarrow l^{1}\) that has the following properties

$$\begin{aligned} \text{(i) }\; \mathcal {D}(A)\; \text{ is } \text{ dense } \text{ in }\; l^{1},\quad \text{(ii) }\; e_{i}\notin \mathcal {D}(A)\; \text{ for } \text{ every }\; i\in \mathbb {N}, \end{aligned}$$
(1)

and such that explicit calculations are possible. So now we do not require A to be a generator.

Theorem 1 answers our question and an example of an operator with such properties is given by (12)–(13). Moreover, Theorem 1 implies that the closure of A equals \(A_{B}\), i.e. \(\overline{A} = A_{B}\), see Sect. 4 for details.

In the next section, we introduce basic notions and present the construction of Blackwell’s semigroup. This construction is needed for us to define the operator A in Sect. 3.

2 Preliminaries

2.1 The Blackwell Semigroup

We begin with some definitions. Let \(l^{1}\) be the Banach space of all absolutely summable sequences \(x = (\xi _{i})_{i\in \mathbb {N}}\), i.e. sequences such that \(\sum _{i\in \mathbb {N}}|\xi _{i}|<\infty \), with the norm \( ||x|| = \sum _{i\in \mathbb {N}}\xi _{i}\). In this article, \(\mathbb {N} = \{1,2,3,\ldots \}\). By \(\{e_{i}\}_{i\in \mathbb {N}}\), we denote the standard Schauder basis in \(l^{1}\), i.e. \(e_{i} = (0, \ldots , 0, 1, 0, \ldots )\) with 1 in the ith coordinate. Any \(x\in l^{1}\) can be then written as \(x = \sum _{i\in \mathbb {N}}\xi _{i}e_{i}\).

Throughout we assume that \(\alpha _{n}, \beta _{n}\), \(n\ge 1\) are positive numbers satisfying

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{\beta _{n}}{\alpha _{n}+\beta _{n}} < \infty ,\qquad \sum _{n=1}^{\infty }\beta _{n} = \infty . \end{aligned}$$
(2)

Incidentally, this assumption implies that \(\lim _{n\rightarrow \infty }\frac{\beta _{n}}{\alpha _{n}} = 0\) and \(\sum _{n=1}^{\infty } \frac{\beta _{n}}{\alpha _{n}} < \infty \).

Now we follow the construction of the Blackwell semigroup as described in [2]. Firstly, let \(\mathbb {I}\) be the set of functions \(\widetilde{i}: \mathbb {N}\rightarrow \{0,1\}\) admitting value 1 finitely many times. Elements of \(\mathbb {I}\) may be considered as sequences with a finite number of nonzero terms. Since \(\mathbb {I}\) is countable, the definition of \(l^{1}(\mathbb {I})\) is analogous to that of \(l^{1}(\mathbb {N})\). By \(\{e_{\widetilde{i}}\}_{\widetilde{i}\in \mathbb {I}}\) we denote the natural basis in \(l^{1}(\mathbb {I})\). For \(n \ge 1\), let \(F_{n}\) be the map \(F_{n}: \mathbb {I}\rightarrow \mathbb {I}\) changing the nth coordinate of an \(\widetilde{i}\) from 0 to 1 and vice versa. The map \(G_{n}: \mathbb {I}\rightarrow \{0, 1\}\) assigns to an \(\widetilde{i}\) its nth coordinate.

For \(n\ge 1\), let \(B_{n}\) be the bounded linear operator on \(l^{1}(\mathbb {I})\) determined by its values on \(e_{\widetilde{i}}\), \(\widetilde{i}\in \mathbb {I}\), by

$$\begin{aligned} B_{n}e_{\widetilde{i}} = {\left\{ \begin{array}{ll} -\beta _{n}e_{\widetilde{i}} + \beta _{n}e_{F_{n}(\widetilde{i})}, \quad \text{ if }\; G_{n}(\widetilde{i})=0,\\ - \alpha _{n}e_{\widetilde{i}} + \alpha _{n}e_{F_{n}(\widetilde{i})},\quad \text{ if }\; G_{n}(\widetilde{i})=1.\\ \end{array}\right. } \end{aligned}$$
(3)

Then for any \(x = \sum _{\widetilde{i}\in \mathbb {I}}\delta _{\widetilde{i}}e_{\widetilde{i}}\) from \(l^{1}(\mathbb {I})\) we have \(B_{n}x=\sum _{\widetilde{i}\in \mathbb {I}}\delta _{\widetilde{i}}B_{n}e_{\widetilde{i}}\).

As explained in [2], \(B_{n}\) is the generator of a Markov chain on \(\mathbb {I}\) in which the nth coordinate of an \(\widetilde{i}\in \mathbb {I}\) jumps between 0 and 1. In terms of semigroups, \(B_{n}\) generates a Markov semigroup \(\{e^{tB_{n}}, t\ge 0\}\) which values on \(\{e_{\widetilde{i}}\}_{\widetilde{i}\in \mathbb {I}}\) are given by

$$\begin{aligned} e^{tB_{n}}e_{\widetilde{i}} = {\left\{ \begin{array}{ll} p_{n}(t)e_{\widetilde{i}} + \left( 1-p_{n}(t)\right) e_{F_{n}(\widetilde{i})}, \quad \text{ if }\; G_{n}(\widetilde{i})=0,\\ q_{n}(t)e_{\widetilde{i}} + \left( 1-q_{n}(t)\right) e_{F_{n}(\widetilde{i})},\quad \text{ if }\; G_{n}(\widetilde{i})=1,\\ \end{array}\right. } \end{aligned}$$

where

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

Since \(B_{n}\)’s commute and are bounded, \(T_{n}(t) = \prod _{k=1}^{n}e^{tB_{k}}\) defines a strongly continuous semigroup \(\{T_{n}(t)\), \(t\ge 0\}\), with generator \(A_{n} = \sum _{k=1}^{n}B_{k}\). The semigroup \(\{T_{n}(t), t\ge 0\}\) describes n combined independent Markov chains, each changing one of the first n coordinates of \(\widetilde{i}\in \mathbb {I}\).

Theorem 4.2 of [2] says that semigroups \(\{T_{n}(t), t \ge 0\}\) converge, when \(n\rightarrow \infty \), to a strongly continuous semigroup \(\{T(t), t \ge 0\}\) composed of Markov operators. In such a case, we say that the infinite product \(\prod _{k=1}^{\infty }e^{tB_{k}}\) exists and denote

$$\begin{aligned} \prod _{k=1}^{\infty }e^{tB_{k}}:= T(t). \end{aligned}$$
(4)

This limit semigroup is just the Blackwell semigroup defined on \(l^{1}(\mathbb {I})\). It is also denoted by \(e^{tA_{B}}\), where \(A_{B}\) is the generator of \(\{T(t), t \ge 0\}\). Although a formula for T(t) is given in Lemma 4.1 of [2] it is not useful here, since it is expressed as an infinite sum of terms which are difficult to handle.

The fact that semigroups \(T_{n}(t)\) converge strongly is also a conclusion from Theorem 1, see Remark 1 in Sect. 4.

2.2 The Binary Bijection

Here we specify a bijection between \(\mathbb {N}\) and \(\mathbb {I}\). By doing so, we are able to write explicit formulae for isomorphic images of \(B_{n}\)’s and \(A_{n}\)’s.

Let \(f(1) = \widetilde{1}:= (0,0,0,\ldots )\in \mathbb {I}\) and for \(i\ge 2\) define

$$\begin{aligned} f(i) = \widetilde{i} := \left( j_{1},\ldots _{}, j_{m},0,0,\ldots \right) , \end{aligned}$$
(5)

where \(j_{k}\in \{0,1\}\), \(k=1,2,\ldots , m\), are such that \(\sum _{k=1}^{m}j_{k} 2^{k-1} = i-1\). This is simply to say that we assign the binary representation of \(i-1\) to a number \(i\ge 2\). For instance, we have

$$\begin{aligned} f(2) = \widetilde{2} = (1,0,0,0,\ldots ),\quad f(8) = \widetilde{8} = (1,1,1,0,0, \ldots ), \quad \text{ etc }. \end{aligned}$$

It is obvious that \(\widetilde{i}\in \mathbb {I}\) and that each element of \(\mathbb {I}\) is a binary representation of a natural number. In consequence, we have a one-to-one correspondence between \(l^{1}(\mathbb {N})\) and \(l^{1}(\mathbb {I})\). This correspondence, written between bases \(\{e_{i}\}_{i\in \mathbb {N}}\) and \(\{e_{\widetilde{i}}\}_{\widetilde{i}\in \mathbb {I}}\), can be defined as follows

$$\begin{aligned} f\left( e_{i}\right) : = e_{f(i)},\quad i \in \mathbb {N}, \end{aligned}$$

where f(i) is given by (5). Then we extend f from \(\{e_{i}\}_{i\in \mathbb {N}}\) to all elements of \(l^{1}(\mathbb {N})\) by linearity, i.e. \(f(x) = \sum _{i\in \mathbb {N}}\xi _{i}f(e_{i})\) if \(x = \sum _{i\in \mathbb {N}}\xi _{i}e_{i}\).

Using (3), we define operators \(\mathcal {B}_{n}\), \(n \ge 1\), on \(l^{1}(\mathbb {N})\) as follows. Firstly, define

$$\begin{aligned} \mathcal {B}_{n}e_{i} := f^{-1}\left( B_{n}f\left( e_{i}\right) \right) ,\quad \text{ for }\; i\in \mathbb {N}, \end{aligned}$$
(6)

and \(\mathcal {B}_{n}x = \sum _{i\in \mathbb {N}}\xi _{i}\mathcal {B}_{n}e_{i}\) for any \(x\in l^{1}(\mathbb {N})\).

Lemma 1

If \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {B}_{n}(\xi _{i})_{i\in \mathbb {N}}\), then

$$\begin{aligned} \eta _{i} = {\left\{ \begin{array}{ll} -\beta _{n}\xi _{i} + \alpha _{n}\xi _{i+2^{n-1}}, \\ \qquad \text{ if } \;\; i\bmod 2^{n} \in \left\{ 1,2,\ldots , 2^{n-1}\right\} ,\\ - \alpha _{n}\xi _{i} + \beta _{n}\xi _{i-2^{n-1}},\\ \qquad \text{ if } \;\; i\bmod 2^{n} \in \left\{ 0,2^{n-1} +1,\ldots , 2^{n}-1\right\} . \end{array}\right. } \end{aligned}$$
(7)

Proof

Fix \(n\ge 1\). To find an explicit formula for \(\mathcal {B}_{n}\) notice that for \(\widetilde{i} = f(i)\), where \(i \in \mathbb {N}\), we have

$$\begin{aligned} F_{n}(\widetilde{i}) = {\left\{ \begin{array}{ll} \left( \ldots , \underbrace{1}_{n-th}, \ldots \right) , \quad \text{ if }\quad \widetilde{i} = \left( \ldots , \underbrace{0}_{n-th}, \ldots \right) ,\\ \left( \ldots , \underbrace{0}_{n-th}, \ldots \right) ,\quad \text{ if }\quad \widetilde{i} = \left( \ldots , \underbrace{1}_{n-th}, \ldots \right) .\\ \end{array}\right. } \end{aligned}$$

From this, we immediately obtain

$$\begin{aligned} f^{-1}(F_{n}(f(i)) = {\left\{ \begin{array}{ll} i + 2^{n-1}, \quad \text{ if }\quad f(i) = \left( \ldots , \underbrace{0}_{n-th}, \ldots \right) ,\\ i - 2^{n-1},\quad \text{ if }\quad f(i) = \left( \ldots , \underbrace{1}_{n-th}, \ldots \right) .\\ \end{array}\right. } \end{aligned}$$
(8)

Next, we find \(G_{n}(f(i))\). The binary arithmetic with some algebra shows that \(G_{n}(f(i))= 0\) if \(i\bmod 2^{n} \in \{1,2,\ldots , 2^{n-1}\}\) and \(G_{n}(f(i))= 1\) otherwise. Therefore, we conclude from (3) and (8) that the operator \(\mathcal {B}_{n}\) defined by (6) has the form

$$\begin{aligned} \mathcal {B}_{n}e_{i} = {\left\{ \begin{array}{ll} -\beta _{n}e_{i} + \beta _{n}e_{i+2^{n-1}}, \\ \qquad \text{ if } \; i\bmod 2^{n} \in \{1,2,\ldots , 2^{n-1}\},\\ - \alpha _{n}e_{i} + \alpha _{n}e_{i-2^{n-1}},\\ \qquad \text{ if } \; i\bmod 2^{n} \in \{0,2^{n-1} +1,\ldots , 2^{n}-1\}, \end{array}\right. } \end{aligned}$$
(9)

where \(i \ge 1\). Since \(\mathcal {B}_{n}(\xi _{i})_{i\in \mathbb {N}} = \sum _{i\in \mathbb {N}}\xi _{i}\mathcal {B}_{n}e_{i}\), (9) implies directly (7). \(\square \)

For instance, \(\mathcal {B}_{1}\) has the form

$$\begin{aligned} \mathcal {B}_{1}\left( \xi _{i}\right) _{i\in \mathbb {N}} = \left( -\beta _{1}\xi _{1} + \alpha _{1}\xi _{2}, \beta _{1}\xi _{1}-\alpha _{1}\xi _{2}, -\beta _{1}\xi _{3}+\alpha _{1}\xi _{4}, \beta _{1}\xi _{3}-\alpha _{1}\xi _{4}, \ldots \right) . \end{aligned}$$

Denote

$$\begin{aligned} \mathcal {A}_{n} = \sum _{k=1}^{n}\mathcal {B}_{k}, \quad n \ge 1. \end{aligned}$$
(10)

The following corollary is a direct conclusion from Lemma 1.

Corollary 1

If \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {A}_{n}(\xi _{i})_{i\in \mathbb {N}}\), then \(\eta _{i} = \sum _{k=1}^{n}\zeta _{k}\), where \(\zeta _{k}\), \(k=1,\ldots , n\), are given by

$$\begin{aligned} \zeta _{k} = {\left\{ \begin{array}{ll} -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}},\\ \qquad \text{ if } \;\; i\bmod 2^{k} \in \{1,2,\ldots , 2^{k-1}\},\\ - \alpha _{k}\xi _{i} + \beta _{k}\xi _{i-2^{k-1}},\\ \qquad \text{ if } \;\; i\bmod 2^{k} \in \{0,2^{k-1} +1,\ldots , 2^{k}-1\}. \end{array}\right. } \end{aligned}$$

It turns out that \(\mathcal {B}_{n}\)’s and \(\mathcal {A}_{n}\)’s have the following interesting property.

Corollary 2

Suppose that \(x\in l^{1}(\mathbb {N})\) and denote \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {B}_{n}x\), \((\eta _{i}')_{i\in \mathbb {N}} = \mathcal {A}_{n}x\). Then

$$\begin{aligned} \sum _{i=1}^{\infty }\eta _{i} = \sum _{i=1}^{\infty }\eta _{i}' = 0. \end{aligned}$$
(11)

Proof

Due to (10), it is enough to prove \(\sum _{i=1}^{\infty }\eta _{i} = 0\). Since \(\mathcal {B}_{n}\) is bounded, i.e. \(||\mathcal {B}_{n}x||\le \max \{\alpha _{n},\beta _{n}\}||x||\), we have

$$\begin{aligned} \sum _{i=1}^{\infty }\eta _{i}= \lim _{m\rightarrow \infty }\sum _{i=1}^{m}\eta _{i} = \lim _{m\rightarrow \infty }\left( \sum _{i=1}^{k2^{n}}\eta _{i}+ \sum _{i=k2^{n}+1}^{m}\eta _{i}\right) \end{aligned}$$

for any \(k\in \mathbb {N}\). The first sum in the above limit is 0 by Lemma 1. Furthermore, for any \(\varepsilon >0\) there exists \(k_{0}\) such that \(\sum _{i=k_{0}}^{\infty }|\eta _{i}|<\varepsilon \). In consequence, \(\left| \sum _{i=k_{0}}^{\infty }\eta _{i}\right| < M\varepsilon \) for a certain constant \(M>0\) and \(\varepsilon >0\). Since \(\varepsilon \) is arbitrary (11) is proved. \(\square \)

3 The Operator

We define \(\mathcal {A}\) as

$$\begin{aligned} \mathcal {A}x:= \sum _{k=1}^{+\infty }\mathcal {B}_{k}x =\lim _{n\rightarrow +\infty }\mathcal {A}_{n}x \end{aligned}$$
(12)

with domain

$$\begin{aligned} \mathcal {D}(\mathcal {A})= \left\{ x\in l^{1}(\mathbb {N}): \sum _{k=1}^{+\infty }||\mathcal {B}_{k}x|| <\infty \right\} . \end{aligned}$$
(13)

Let \(x=(\xi _{i})_{i\in \mathbb {N}}\) be an element of \(\mathcal {D}(\mathcal {A})\) and denote \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {A}x\). Since the norm convergence in \(l^{1}(\mathbb {N})\) implies the coordinate-wise convergence, we conclude from Corollary 1 that

$$\begin{aligned} \eta _{1} = \lim _{n\rightarrow +\infty }\sum _{k=1}^{n}\left( -\beta _{k}\xi _{1} + \alpha _{k}\xi _{1+2^{k-1}}\right) , \end{aligned}$$
(14)

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

$$\begin{aligned} \eta _{i} = \sum _{k=1}^{l+1}\zeta _{k} + \lim _{n\rightarrow +\infty }\sum _{k= l+2}^{n}\left( -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}}\right) , \end{aligned}$$
(15)

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

$$\begin{aligned} \zeta _{k} = {\left\{ \begin{array}{ll} -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}},\\ \qquad \text{ if } \;\; i\bmod 2^{k} \in \{1,2,\ldots , 2^{k-1}\},\\ - \alpha _{k}\xi _{i} + \beta _{k}\xi _{i-2^{k-1}},\\ \qquad \text{ if } \;\; i\bmod 2^{k} \in \{0,2^{k-1} +1,\ldots , 2^{k}-1\}. \end{array}\right. } \end{aligned}$$

Notice that a necessary condition for an x to be in \(\mathcal {D}(\mathcal {A})\) is

$$\begin{aligned} \lim _{n\rightarrow +\infty }\left( -\beta _{n}\xi _{i}+\alpha _{n}\xi _{i+2^{n-1}}\right) = 0, \end{aligned}$$

for every \(i\ge 1\).

For now, one can merely infer that \(0\in \mathcal {D}(\mathcal {A})\). However, in the proof of Theorem 1 we will produce some non-trivial elements of \(\mathcal {D}(\mathcal {A})\). The following theorem is the main result of this article.

Theorem 1

For the operator \(\mathcal {A}\) defined by (12)–(13), the conditions in (1) are satisfied.

Proof

First we show (ii). Fix \(i\in \mathbb {N}\) and notice that by (2) we have

$$\begin{aligned} \lim _{n\rightarrow +\infty }\sum _{k= l+2}^{n}\left( -\beta _{k}\xi _{i} + \alpha _{k}\xi _{i+2^{k-1}}\right) = -\lim _{n\rightarrow +\infty }\sum _{k= l+2}^{n}\beta _{k}=-\infty , \end{aligned}$$

proving that \(e_{i}\notin \mathcal {D}(\mathcal {A})\).

Next we prove (i). Let \(x = (\xi _{1},\xi _{2},\ldots )\) be defined as follows. First we put

$$\begin{aligned} \xi _{1} = 1,\quad \xi _{1+2^{k-1}} = \frac{\beta _{k}}{\alpha _{k}},\quad k\ge 1, \end{aligned}$$
(16)

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

$$\begin{aligned} \xi _{i+2^{k-1}} = \frac{\beta _{k}}{\alpha _{k}}\xi _{i},\quad k\ge l+2. \end{aligned}$$
(17)

From this, we infer that for \(i = 1+2^{l_{1}-1} + \ldots + 2^{l_{m}-1}\) with \(l_{1}<l_{2}<\ldots < l_{m}\) one can write

$$\begin{aligned} \xi _{i+2^{k-1}} = \frac{\beta _{k}}{\alpha _{k}}\cdot \frac{\beta _{l_{m}}\beta _{l_{m}-1} \ldots \beta _{l_{1}}}{\alpha _{l_{m}} \alpha _{l_{m}-1}\ldots \alpha _{l_{1}}},\quad k\ge l_{m}+1. \end{aligned}$$

Then, the norm of x can be expressed as

$$\begin{aligned} ||x|| = 1 + \sum _{l=1}^{\infty } \frac{\beta _{l}}{\alpha _{l}} + \sum _{l_{1}< l_{2}}^{\infty } \frac{\beta _{l_{1}}\beta _{l_{2}}}{\alpha _{l_{1}}\alpha _{l_{2}}} + \sum _{l_{1}< l_{2} < l_{3}}^{\infty } \frac{\beta _{l_{1}}\beta _{l_{2}}\beta _{l_{3}}}{\alpha _{l_{1}}\alpha _{l_{2}}\alpha _{l_{3}}} + \ldots . \end{aligned}$$
(18)

From the Problem 3.8.26 in [4] and the fact that \(\sum _{l=1}^{\infty } \frac{\beta _{l}}{\alpha _{l}} < \infty \) (which follows from (2)), we have

$$\begin{aligned} ||x|| = \prod _{l=1}^{\infty } \left( 1+\frac{\beta _{l}}{\alpha _{l}}\right) < \infty , \end{aligned}$$

meaning that \(x\in l^{1}(\mathbb {N})\). To conclude that \(x\in \mathcal {D}(\mathcal {A})\) we need to show that condition (13) is satisfied. To this end, denote \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {B}_{k}x\) and notice that (7) implies \(\eta _{i} = -\beta _{k}\xi _{i} + \alpha _{k} \frac{\beta _{k}}{\alpha _{k}}\xi _{i} = 0\) or \(\eta _{i} = -\alpha _{k}\xi _{i} + \beta _{k}\frac{\alpha _{k}}{\beta _{k}}\xi _{i} = 0\). In other words,

$$\begin{aligned} \sum _{k=1}^{+\infty }||\mathcal {B}_{k}x|| = 0. \end{aligned}$$

The above means that \(x\in \mathcal {D}(\mathcal {A})\) and \(\mathcal {A}x=0\).

Now we construct a sequence of elements of \(\mathcal {D}(\mathcal {A})\) converging to \(e_{1}\). Fix \(p \ge 2\) and define \(y_{p} = (\xi _{1},\xi _{2},\ldots )\) as follows

$$\begin{aligned} \xi _{1}=1,\quad \xi _{1+2^{k-1}} = {\left\{ \begin{array}{ll} 0 &{}, \quad k = 1,2,\ldots , p-1,\\ \frac{\beta _{k}}{\alpha _{k}}&{}, \quad k \ge p , \end{array}\right. } \end{aligned}$$
(19)

and if \(i = 2^{l} + m\) for some \(l\ge 0\) and \(m\in \{1,2,\ldots , 2^{l}\}\), we define \(\xi _{i+2^{k-1}}\) recursively as in (17). Observe that \(||y_{p}|| \le ||x||\) for every \(p\ge 2\). Additionally, from (19) we have \(\xi _{i}=0\) for \(i=2,3,\ldots ,1+2^{p-1}\). It is clear that the coordinate-wise limit of \(y_{p}\) is \(e_{1}\).

Now choose any \(\varepsilon >0\) and fix \(M\in (0,1)\). We conclude from (2) that there exists \(l_{0}\) such that \(\sum _{l=l_{0}}^{\infty } \frac{\beta _{l}}{\alpha _{l}} < \varepsilon \) and \(\frac{\beta _{l}}{\alpha _{l}} < M\), for all \(l\ge l_{0}\). This, together with (18), implies

$$\begin{aligned} ||y_{p} - e_{1}|| \le \varepsilon \sum _{k=0}^{\infty }M^{k} = \frac{\varepsilon }{1-M},\quad \text{ for }\; p \ge l_{0}. \end{aligned}$$

Since \(\varepsilon \) is arbitrary, \(\lim _{p\rightarrow +\infty }||y_{p} - e_{1}|| = 0.\)

What remains to show is \(\mathcal {A}y_{p}\in l^{1}(\mathbb {N})\). From Lemma 1, we have

$$\begin{aligned} \mathcal {B}_{k}y_{p} = \left( -\beta _{k},0,\ldots ,0,\underbrace{\beta _{k}}_{1+2^{k-1}-th},0,0,\ldots \right) ,\quad \quad k=1,2,\ldots ,p-1 \end{aligned}$$

and \(\mathcal {B}_{k}y_{p} = 0\) for \(k \ge p\). This implies

$$\begin{aligned} \sum _{k=1}^{+\infty }||\mathcal {B}_{k}y_{p}|| = 2 \sum _{k=1}^{p-1}\beta _{k} < \infty ,\quad p\ge 2. \end{aligned}$$

So \(y_{p}\in \mathcal {D}(\mathcal {A})\) for every \(p\ge 2\) and

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

Incidentally, notice that \(\mathcal {A}y_{p}\) has only a finite number of nonzero components and \(\sum _{k=1}^{\infty }\eta _{k}=0\). To summarize, we have constructed a sequence \(y_{p}\), \(p \geqslant 2\), of elements of \(\mathcal {D}(\mathcal {A})\) converging to \(e_{1}\).

Now we show that a similar construction can be carried out for any \(e_{i}\), \(i\ge 2\). Fix \(i\ge 2\) and observe that for x defined by (16)–(17) we have

$$\begin{aligned} ||\tfrac{1}{\xi _{i}}x|| = ||\left( \tfrac{1}{\xi _{i}}, \tfrac{\xi _{2}}{\xi _{i}},\ldots ,\tfrac{\xi _{i-1}}{\xi _{i}}, \underbrace{1}_{i-th},\tfrac{\xi _{i+1}}{\xi _{i}},\ldots \right) || = \tfrac{1}{\xi _{i}}||x|| < \infty . \end{aligned}$$
(20)

Suppose that \(i = 2^{l} + m\) for some \(l\ge 0\) and \(m\in \{1,2,\ldots , 2^{l}\}\). We define \(z_{p} = (\xi _{1}, \xi _{2},\ldots )\) as follows. At first, let

$$\begin{aligned} \xi _{i} = 1, \quad \xi _{i+2^{k-1}} = {\left\{ \begin{array}{ll} 0 &{}, \quad k = l+2,l+3,\ldots , p-1,\\ \frac{\beta _{k}}{\alpha _{k}}&{}, \quad k \ge p, \end{array}\right. } \end{aligned}$$
(21)

where \(p > i+2^{l+1}\). For \(k=1,2,\ldots ,l+1\), define

$$\begin{aligned} {\left\{ \begin{array}{ll} \xi _{i+2^{k-1}} = 0, &{} \quad \text{ if }\; i\bmod 2^{k} \in \{1,2,\ldots , 2^{k-1}\},\\ \xi _{i-2^{k-1}} = 0, &{} \quad \text{ if }\; i\bmod 2^{k} \in \{0,2^{k-1} +1,\ldots , 2^{k}-1\}. \end{array}\right. } \end{aligned}$$

In all other cases, elements of \(z_{p}\) are defined recursively by (17).

Notice that \(\xi _{j}=0\) for \(j\in \{1,2,\ldots ,i+2^{p-1}\}\ \setminus \{i\}\) and that the coordinate-wise limit of \(z_{p}\) is \(e_{i}\). Furthermore, it is clear from (20) that \(||z_{p}|| \le \frac{1}{\xi _{i}}||x||\) and that \(\lim _{p\rightarrow \infty }||z_{p} - e_{i}|| = 0\). Computing \(\mathcal {A}z_{p}\) in a similar way as we did \(\mathcal {A}y_{p}\), we see that it has only a finite number of nonzero terms, meaning that \(z_{p}\in \mathcal {D}(\mathcal {A})\). This completes the proof. \(\square \)

4 Conclusions

Recall that \((\mathcal {B}_{n})_{n\in \mathbb {N}}\) is a sequence of bounded operators defined on \(l^{1}(\mathbb {N})\) and \((e^{t \mathcal {B}_{n}})_{n\in \mathbb {N}}\) is the corresponding sequence of commuting Markov semigroups, i.e. for any \(m, n \in \mathbb {N}\) and \(x\in l^{1}(\mathbb {N})\), we have

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

The above follows from the fact that \(\mathcal {B}_{n}\)’s are isomorphic images of \(B_{n}\)’s, see (6). Hence, for every \(n\in \mathbb {N}\), \(\mathcal {T}_{n}(t) = \prod _{k=1}^{n}e^{t\mathcal {B}_{k}}\) is a strongly continuous semigroup with the generator \(\mathcal {A}_{n}\) given by (10).

The following is a consequence of Theorem 1.

Corollary 3

The semigroups \(\{\mathcal {T}_{n}(t), t\ge 0\}\) converge, when \(n\rightarrow \infty \), to a strongly continuous Markov semigroup \(\{\mathcal {T}(t), t\ge 0\}\). This limit semigroup is the Blackwell semigroup on \(l^{1}(\mathbb {N})\). Furthermore, the generator \(\mathcal {A}_{B}\) of \(\{\mathcal {T}(t), t\ge 0\}\) is the closure of \(\mathcal {A}\) given by (12)–(13). In other words, \(\overline{\mathcal {A}} = \mathcal {A}_{B}\).

Proof

Theorem 2.1 of [2] specifies conditions under which an infinite product of commuting contraction semigroups exists. All those conditions are satisfied in our case (thanks to Theorem 1), especially the key one that \(\mathcal {D}(\mathcal {A})\) is dense in \(l^{1}(\mathbb {N})\). \(\square \)

Remark 1

Theorem 1 may be also considered as an alternative proof of existence of (4) to the proof presented in [2]. Since semigroups \(\{\mathcal {T}_{n}(t), t\ge 0\}\) and \(\{T_{n}(t), t\ge 0\}\) are isomorphic, the strong convergence of one of them implies the strong convergence of the other and vice versa. The proof of convergence of \(T_{n}(t)\) presented in [2] can be termed as “a direct proof”.

Corollary 4

Suppose that \(x\in \mathcal {D}(\mathcal {A})\) and denote \((\eta _{i})_{i\in \mathbb {N}} = \mathcal {A}x\), Then

$$\begin{aligned} \sum _{i=1}^{\infty }\eta _{i} = 0. \end{aligned}$$
(22)

Proof

Notice that \(\eta _{i}\)’s are the coordinate-wise limits of \(\mathcal {A}_{n}x\), see (14) and (15). Hence, (22) is a consequence of (11). \(\square \)