Abstract
I provide an example of a family of commuting contraction semigroups \((e^{t B_{n}})_{n\in {\mathbb {N}}}\) defined on \(l^{1}({\mathbb {N}})\) such that the product semigroup \(\prod _{n=1}^{\infty }e^{tB_{n}}\) exists and has bounded generator. The infinite product of the corresponding family of adjoint semigroups \((e^{t B_{n}^{*}})_{n\in {\mathbb {N}}}\) defined on \(l^{\infty }({\mathbb {N}})\) also exists and its generator is bounded. I give explicit formulae for these generators. The results follow from a general convergence theorem for such semigroups proved in Arendt et al. (J Funct Anal 160: 524–542, 1998).
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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
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
is also a strongly continuous contraction semigroup on X and its generator is the closure of \(A_{n}\) given by
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
is dense in X. Then the product \(\prod _{k=1}^{\infty }e^{tB_{k}}\) exists. Moreover, define A by
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
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
It is clear that the \(B_{n}\)’s are linear bounded operators and
In Lemma 2.1, I prove that the \(B_{n}\)’s commute. It may be verified directly that
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
Moreover
Since the \(B_{n}\)’s and the \(B_{n}^{*}\)’s are continuous, they are the generators of the following semigroups
By a direct computation, we find (see also [8, p.60])
and
where \(p_{n}(t)\), \(q_{n}(t)\), \(t\ge 0\), are given by
Notice that \(0 < p_{n}(t)\le 1\), \(0 < q_{n}(t)\le 1\) and as a result
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
This implies that the \(B_{n}^{*}\)’s also commute and in consequence
and
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
and
So \((\eta _{i}')_{i\in {\mathbb {N}}} = B_{n}B_{m}(\xi _{i})_{i\in {\mathbb {N}}}\) is given by
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
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
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
and if \(i = 2^{l} + m\) for \(l\ge 0\) with \(m\in \{1,2,\ldots , 2^{l}\}\), we have
where \(\zeta _{k}\), \(k=1,\ldots , l+1\), are given by
Proof
We use Theorem 1.1. In our case, \(D_{1} = l^{1}({\mathbb {N}})\) because
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
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
and
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
and if \(i = 2^{l} + m\) for \(l\ge 0\) with \(m\in \{1,2,\ldots , 2^{l}\}\), we have
where \(\zeta _{k}\), \(k=1,\ldots , l+1\), are given by
Proof
The proof is analogous to that of Theorem 3.1. For every \(x\in l^{\infty }({\mathbb {N}})\), we have
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
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)
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),
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).
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Nieznaj, E. The infinite product of contraction semigroups on \(l^{1}({\mathbb {N}})\) and \(l^{\infty }({\mathbb {N}})\). Arch. Math. 119, 593–600 (2022). https://doi.org/10.1007/s00013-022-01794-2
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DOI: https://doi.org/10.1007/s00013-022-01794-2