On the output function in a Ginsburg’s machine

We give the form of the output function in Ginsburg’s machine in which the input and output dictionaries are abelian groups and the transition function is of a special form.


Introduction
By Ginsburg [4] the 5-tuple (A,B,S,δ, λ) is called a machine if -A (the input dictionary) is a semigroup (with additive notation), -B (the output dictionary) is a semigroup with additive notation and with the left cancellation property, -S is an arbitrary set (the set of states), and δ:S × A → S is the transition function such that δ(δ(s, x), y) = δ(s, x + y) for s ∈ S, x, y ∈ A, λ:S × A → B is the output function such that λ(s, x + y) = λ(s, x) + λ(δ(s, x), y) for s ∈ S, x, y ∈ A.
In the paper [4] the notations in the semigroups A and B are multiplicative. We use the additive notation for convenience.
Eq. (1) is called the translation equation and it has many applications [8]. The deterministic automaton [12] is an example of this machine. Here the input and output dictionaries are monoids in which the operation is the concatenation of words. for s ∈ S k , x ∈ A, a(s, x) for s ∈ f(S)\ S k , x ∈ A, g(s) + Λ(δ(f(s), x)) − Λ(f(s)) for s ∈ S\f(S), f(s) ∈ S k , x ∈ A, g(s) + a(f(s), x) where Λ is an arbitrary function from S k to B, a(s, .) for s ∈ f (S)\ S k is an arbitrary additive function from A to B and g is an arbitrary function from S\f(S) to B.
Proof. of the "only if " part.
Assume that the function λ:S × A → B is a solution of Eq. (2). Case s ∈ S k . Thus s ∈ S l for a l ∈ K. Let s l be the point of the set S l such that δ l (s l ) = 0. We have λ(s l , x + y) = λ(s l , y) + λ(δ(s l , y), x).

Zenon Moszner AEM
Example 1. Let S be a group and let ϕ:A → S be an additive function. Then the function δ(s, x) = s + ϕ(x) is the solution of (1). If ϕ is an injection, then A(s) = {x ∈ X : δ(s, x) = s} = {0}. The solution λ of (2) is of the form λ(s,x) = Λ(δ(s,x))-Λ(s), where Λ is an arbitrary function from S to B. If A is a subgroup of B and ϕ is not an injection, then λ(s,x)=x is the solution of (2) and there does not exists a function Λ such that λ(s, Remark. The condition (3)  Remark. In applications it is assumed that δ(s,0) = s for s ∈ S. In this case the functions δ and λ in (4) and (5) have the following more simple form From here λ(s,0) = 0 for every s in S.
Remark. Let δ ∈ (s, .) : R → S ⊂ R for s ∈ S be continuous functions and let δ(s, 0) = s for s ∈ S. The solution δ : S × R → S of (1), where S is a nondegenerated interval in R, in this case is of the form (7), where S k are open disjoint subintervals in S and δ k are homeomorphisms from S k onto R [11]. The function δ is then continuous and it is also called a dynamical system. Eq. (2) for A = (R, +) and B = (0, +∞) with the usual multiplication has the form thus the function lnλ(s,x) is of the form (8). From here we have the form of the function λ(s,x) = exp[lnλ(s,x)].
The function λ(s, .) : R → R is continuous for every s ∈ I if and only if every function Λ k for k ∈ K is continuous and a(s,x)=b(s)x for a function b : S → R.

Cocycles
The solution λ of Eq. (9) if δ is a dynamical system and B = [0, +∞) (with 0) with the usual multiplication, i.e., B is a abelian group (with multiplicative notation) with 0, is said to be the cocycle of δ. The 5-tuple (R, [0, +∞), S, δ, λ) is not a Ginsburg's machine since the cancellation property is not true in the interval [0, +∞) with the usual multiplication. We have the following corollary.
Corollary. Let S, A, B, δ be an arbitrary set, an arbitrary abelian group (with additive notation), an abelian group G (with multiplicative notation) with 0 (0.a = a.0 = 0) and the solution of (1) satisfying (3) and δ(s, 0) = s for s in S, respectively. The function λ : S × A → B is the solution of (9) if and only if it is of the form where S k are as in Theorem 1, Λ : S k \S 0 → G is an arbitrary function, E(s, .) for s ∈ S\[ S k \S 0 ] is an arbitrary exponential function from A to G (E(s,x+y)=E(s,x)E(s,y)) and S 0 = ∅ or it is the union of transitive fibres δ(s,A) of δ (not necessarily all).
We obtain (10) by the form of λ(s, x + y) and λ(s,x) in this case. If S 0 = ∅ and s ∈ S\ S k we obtain (10) since here δ(s,x) = s. Assume that S 0 = ∅. The equality (10) is evident for s ∈ S 0 . If s ∈ S k \S 0 ⊂ S\S 0 , then S\S 0 = ∅, thus it is the union of transitive fibres of δ, too. From here δ(s,x) ∈ S k \S 0 and the proof of (10) is as above.
Proof. of the "only if" part. Assume that the function λ : S × A → B is the solution of (9). Let S 0 be the set {s ∈ S : ∃x ∈ A : λ(s,x) = 0}. This set is equal to the set {s ∈ S : ∀x ∈ A : λ(s,x) = 0} since if λ(s, x 0 ) = 0 for a x 0 in A, then λ(s, x 0 + x) = 0 for every x in A by (9). We prove that if S 0 = ∅, then it is the union of transitive fibres of δ, i.e., we prove that if S 0 ∩ S k = ∅, then S k ⊂ S 0 . Assume for the indirect proof that there exists s 0 ∈ S k which is not in S 0 . From here λ(s 0 , x) =0 for every x in A. There exists s 1 ∈ S 0 ∩ S k , thus λ(s 1 , x) = 0 for every x in A. Since s 0 , s 1 ∈ S k , there exists x 0 such that s 1 = δ(s 0 , x 0 ). By (9) we have By the above λ(s,x) = 0 for s in S 0 and x in A. For s ∈ S\S 0 the function λ(s, .) has its values in the abelian group G (with multiplicative notation!) thus we have the conclusion by Remark 2.

Problem. For which functions Λ and E in
Remark. I announced the above results for A=B=R in [9] (for continuous cocycles see [1,5] too).

Vol. 92 (2018)
On the output function in a Ginsburg's machine 623

Stability
Let the output dictionary B in the machine be a metric space with metric d.
there exists a solution λ of (2) such that Definition 3. (inverse stability) Let δ be the fixed input function in the machine (A,B,S, δ, λ). Eq. (2) for the output function is said to be inversely stable if for every β > 0 there exists ε > 0 such that for every function Λ : S × A → B for which the condition (12) is true for a solution λ of (2) we have (11).
Remark. These definitions are not equivalent. Indeed, e.g., let S = {0} and let A be free group generated by two elements, B = Z with the usual addition and usual metric and δ(0, x) = 0 for x ∈ A. Eq. (2) has in this case the form λ(0, x + y) = λ(0, x) + λ(0, y), thus this equation is the equation of a homomorphism from A to B. Such an equation is not b-stable [2], thus Eq. (2) is not b-stable either. This equation is stable since it is sufficient to put β < 1 for every ε.
Example 5. Eq. (2) is stable with δ = ε 7 if A is the interval (0, +∞) with the usual addition, B is a Banach space, S is a closed interval in R and δ:S×A → A is a continuous solution of (1) [1].
We have here the following.
Remark. This stability depends on the function δ. Indeed, let N be a noncomplete normed space and let S=A=B=N. Assume that there exists s 0 such that δ(s 0 , x) = x for x ∈ A, e.g., δ(s,x) = s+x, and that for the function Λ:N × N → N we have |Λ(s, x + y) − Λ(s,x) − Λ(s + x, y)| ≤ ε for ε > 0. From here for s = s 0 we obtain |Λ(s 0 , x + y) − Λ(s 0 , x) − Λ(x,y)| ≤ ε. Since the function λ(s,y) = Λ(s 0 , s+y) + Λ(s 0 , s) is the solution of (2), Eq. (2) is b-stable and stable with β = ε. Assume at present there exists s 1 such that δ(s 1 , x) = s 1 for x ∈ A and suppose that Eq. (2): λ(s 1 , x + y) = λ(s 1 , x) + λ(s 1 , y) is bstable. This equation is the equation of homomorphism from A to B and it is b-stable too. Since the space B is not complete we have a contradiction with the following theorem [3]: -Let A be an abelian group with an element of infinite order. Assume that the homomorphism equation from A to a normed space B is b-stable. Then B is complete.
The last example is good for stability too since the above theorem is true for stability in place of b-stability on the basis of Theorem 2.
Problem. Characterize the input function for which the output function is stable.
We have for the inverse stability the following. Let ε = β 3 for β > 0. If for a function Λ there exists a solution λ of (2) such that (12) is true, then (11) is true, too.
Remark. It is possible to consider other stabilities (see [10]).