1. Introduction
Let
and
be groupoids (i.e.,
X is a nonempty set and
are binary operations in
X). If there exists a bijection
such that
then we say that the pair
induces ∘. It is easily seen that (
1) is equivalent to the following property
which means that
h is an isomorphism of
into
.
Recall yet that a groupoid
is cancellative if operation ∘ is cancellative, i.e.,
for all
with
. A groupoid
is a semigroup if operation ∘ is associative, i.e.,
for every
.
Let S denote the unit circle of the complex plane and · be the usual multiplication of complex numbers. The main result of this paper is an elementary construction of homeomorphisms such that the pair induces a given binary, continuous, associative, and cancellative operation on S, where the topology in S is induced by the usual topology of the plane. As far as we know it is the only such proof.
This paper is partly of expository type because we also show several interesting applications of the main outcome and underline some symmetry issues concerning functional equations, which arise between those applications and some consequences of the following well known result of J. Aczél [
1,
2] (analogous result was obtained in Reference [
3]; also see Reference [
4] for a simpler proof).
Theorem 1. Let I be a real nontrivial interval and be a binary operation that is continuous, associative, and cancellative. Then, there exist an infinite real interval J and a homeomorphism with Note that property (
4) means, in particular, that
for every
, i.e.,
for every
. Therefore,
J must be either the set of reals
or of one of the following forms:
,
,
,
with some real
.
Next, it is clear that every operation of form (
4) must be cancellative. So, without the cancellativity assumption, the result is not true, and, for instance, the natural operation
which is associative and continuous, but not cancellative, cannot be represented in form (
4). The same is true for
for
.
In the terms of functional equations, Theorem 1 is about continuous solutions
(if we write
) of the following associativity equation:
In this case, the cancellativity of ∘ is equivalent to the injectivity of
A with respect to either variable, which means the strict monotonicity with respect to either variable (because of the assumed continuity of
A). Similar problem without the assumption of strict monotonicity (i.e., the cancellativity of the corresponding binary operation), was considered in Reference [
5] (also see Reference [
6]), and, under some additional assumptions, the following representation was obtained
with some continuous injection
, where
is a pseudoinverse of
H (see Reference [
5] for more details). Clearly, representation (
4) is actually (
7) with
. For further, more general investigations of that subject, we refer to References [
7,
8,
9].
Let us add that solutions
to the associativity Equation (
6) are important in statistical metric spaces and are called triangular norms or shortly
t-norms (cf., e.g., References [
6,
10,
11]); we also refer to Reference [
12] for the notion of copulas.
A result analogous to Theorem 1, for the unit circle, has the subsequent form.
Theorem 2. Let a binary operation be continuous, associative, and cancellative. Then, there exist exactly two homeomorphisms such thatIn particular, for , where is the complex conjugate of x. Theorem 2 can be derived (with some additional reasoning) from Reference [
13] (Theorems 1.10 and 1.13) (continuous, compact and cancellative semigroup is a topological group) and Reference [
14] (Theorem 2) (connected topological group, which contains a neighbourhood of the neutral element that is homeomorphic to an open real interval, is isomorphic either to the additive group of reals
or to the factor group
, where
denotes the set of integers).
However, as we will see in the section with applications, it is useful to know how to construct the function
(and thus also
). In the case of Theorem 1, the form of
H can be deduced from the proofs in References [
2,
4]. We show in
Section 5 that a symmetric reasoning, with a reasonably elementary and simple construction of the functions
and
, works for Theorem 2.
Finally, let us mention that the form of
A in (
7) reminds about a well-known problem of representing functions with several variables by functions in one variable, whether under some regularity assumptions or not (see References [
5,
15,
16,
17] for more details).
For further information on topological semigroups (also on historical background) we refer to References [
13,
18,
19].
The paper is divided into 6 sections. The next section contains some observations concerning symmetry issue arising between Theorems 1 and 2. In
Section 3, we present several applications of both theorems.
Section 4, titled
Auxiliary Results, contains a series of definitions, lemmas, remarks, and corollaries, leading to the final reasoning contained in
Section 5, titled
The Proof of Theorems 2, and being the proper proof of the theorem with a description of function
. The sixth section (the last one) presents some concluding remarks.
3. Applications
Now, we show some simple applications of Theorems 1 and 2 in functional equations. They show some symmetries and some lack of symmetry between those two cases of the continuous cancellative semigroups on a real interval and on the unit circle. As they are clearly visible, we do not discuss them in detail.
Let us begin with the following auxiliary useful result in Reference [
21] (p. 155).
Lemma 1. Let . Then, a continuous function satisfies the functional equationif and only if - (I)
in the case , there is with for ;
- (II)
in the case , there are with for .
The next lemma seems to be well known, but, for the convenience of readers, we present a short proof of it.
Lemma 2. Let be a continuous solution of the equationThen, there is a real constant d such that for . Proof. Define a function
by
for
. Then,
B is continuous and, by (
14),
for
. Consequently, by Lemma 1, there is
such that
for
. Since
, the real part of
a must be equal 0. So
and consequently
for
. Hence,
for
. □
Now, we are in a position to present the mentioned before applications. As far as we know, they are new results, never published so far.
Proposition 1. Let a binary operation be associative, continuous and cancellative. Let and be a homeomorphism withThen, a continuous function fulfills the equationif and only if there is such that, - (a)
in the case , for ;
- (b)
in the case , for .
Proof. Let
be a continuous function that fulfills Equation (
16). Then, by (
15),
whence the function
,
for
, fulfills Equation (
13). Hence, by Lemma 1, conditions (I) and (II) are valid. Clearly,
for
if and only if
. Moreover, it is easy to check that
for
if and only if
. This implies statements (a) and (b).
Now, suppose that
has the form depicted by (a). Then, in view of (
15),
The case of (b) is analogous. □
Proposition 2. Let I be a nontrivial real interval and a binary operation be associative, continuous, and cancellative. Let and be a continuous injection satisfying (4). Then, the functional equationhas a non-constant continuous solution if and only if H is bijective. Moreover, if H is bijective, then a continuous function fulfills Equation (19) if and only if there is such that, - (i)
in the case , for ;
- (ii)
in the case , for .
Proof. Let
be a continuous solution of Equation (
19). Then, the function
,
for
, fulfills the equation
Moreover, h is continuous.
In the case there is such that for , which implies that either or H is surjective and for .
In the case there are such that for , where and denote the real and imaginary parts of a complex number s. Write . Then, it is easily seen that for , which implies that either or H is surjective and for .
Now, we see that, in the case where
H is bijective, we obtain statements (i) and (ii). Moreover, it is easy to check functions depicted in statements (i) and (ii) are solutions to (
19).
Next, suppose that Equation (
19) has a non-constant continuous solution
. Then, as we have already observed, in the case
there is
such that
for
. Since
f is not constant and
H is injective, this means that
and consequently
H is surjective. If
, then we show the surjectivity of
H in a similar way.
Finally, note that, if
H is bijective, then functions depicted by statements (i) and (ii), with
, are non-constant continuous solutions to (
19). □
Proposition 3. Let I be a nontrivial real interval and binary operations and be associative, continuous and cancellative. Let and be such that (4) and (15) hold. Then, the following two statements are valid. - (A)
A continuous function fulfills the functional equationif and only if there is such that for . - (B)
Every continuous function fulfilling the functional equation is constant.
Proof. First we prove (A). So, let
be a continuous function that fulfills functional Equation (
21). Then, in view of (
4) and (
15),
which means that the function
,
for
is continuous and satisfies the equation
Let
be the solution of
such that
for
(we define
by:
for
). According to Proposition 1 there is
such that
for
, whence
for
. Consequently,
The converse is easy to check.
Now, we prove (B). Let
be a continuous solution to (
22). Then, in view of (
4) and (
15),
whence the function
,
for
, is continuous and satisfies the equation
Thus, the set is compact (because S is compact and h is continuous) and it is a subgroup of . The only such subgroup is the trivial one , which means that . Since is bijective and H is injective, f must be constant. □
Using Theorem 2 and some results from Reference [
22] we also get the following proposition on the minimal homeomorphisms on
S. Let us recall (see Reference [
22] (Ch.1 §1)) that, if
, then a set
is called
f-invariant provided
. A homeomorphism
is minimal if
S does not contain any non-empty, proper, closed
f-invariant subset (see References [
23,
24] for some related results).
Proposition 4. A function is a minimal homeomorphism if and only if there exist and a continuous, associative, and cancellative binary operation such thatwhere and for ( denotes the set of positive integers). Proof. Assume that
is a minimal homeomorphism. Then, there exist an irrational real number
c and a homeomorphism
with
for
(see Reference [
22] (Ch.3 §3 Th.1,3)). Define a binary operation
by
and put
. Then,
for
. Moreover, since
is injective and
c is irrational, we have
which means that (
30) holds.
Now, suppose that
is of form (
29), where
is a continuous, associative, and cancellative binary operation. In view of Theorem 2, there exists a homeomorphism
such that (
15) is valid. Consequently,
T is a homeomorphism and
for
. Let
be such that
. Then,
Now, we show that
c is irrational. So, for the proof by contradiction suppose that there are
with
. Then,
. This is a contradiction to (
30).
Thus, we have proved that
c is irrational. Hence, it follows that the set
is dense in
S; thus, by (
33), the set
is dense in
S for every
, which means that
T is a minimal homeomorphism. This ends the proof. □
Before our last proposition in this section, let us remind the notion of continuous flow on a topological space
X. Namely (cf. Reference [
13,
25]), if
for
is a family of maps such that
for all
,
is the identity map on
X and the mapping
, given by
, is continuous, then we say that the family
for
is a continuous flow on
X.
Observe that, if
for
is a continuous flow on a topological space
X, then
,
, is a continuous solution of the translation functional equation
The last proposition in this section shows that the continuous, associative, and cancellative binary operations on
S can be used in a description of the continuous flows on
S (cf. References [
26,
27]) and therefore also in a description of the continuous solutions
to the translation functional equation.
Proposition 5. A family of continuous functions is a continuous flow such that either (the identity function on S) or for if and only if there exist a continuous, associative, and cancellative operation and a continuous solution of the functional equationsuch that Proof. Let
be a continuous flow such that either
or
for
. Then, by [
13] (Theorem 2), there exist
and an orientation preserving homeomorphism
with
for
,
.
Let
be the operation given by (
31). Then, ★ is continuous and
is a group. Next, the function
,
for
, fulfills Equation (
35) (because
for
) and
Conversely, suppose that (
36) holds. Then, it is easily seen that
is a continuous flow. Further, by Theorem 2,
is a group. So, if there is
such that
, then, by (
36),
, whence
is the neutral element of
. Thus, (
36) implies that
. This completes the proof. □
Finally, let us mention that somewhat related issues in particle physics can be found in Reference [
28]. Moreover, it seems that Theorem 2 can be applied in finding continuous solutions
of some suitable functional equations analogously as Theorem 1 has been used in References [
29,
30,
31] for real functions.
4. Auxiliary Results
In this section, we provide information and several observations necessary in the proof of Theorem 2 for the construction of function . They are presented in a series of lemmas, remarks, and corollaries. We start with some notations and definitions.
We define an order in
S as follows:
for
, where
stands for the argument of the complex number
u. For every
,
, we write
It is known (cf., e.g., Reference [
22] (Chap. 2)) that, for every homeomorphism
, there exists a unique homeomorphism
satisfying
for
. If
f is increasing (decreasing), then we say that homeomorphism
T preserves (reverses) orientation. Note that every homeomorphism
preserves or reverses orientation.
Now, we will prove several auxiliary lemmas, corollaries and remarks.
So, let
be as in Theorem 2. Then,
is a topological group (see Reference [
13] (Theorems 1.10 and 1.13)). Denote by
e the neutral element of
and define
by
Remark 4. It is easily seen that the mapping , for , is an isomorphism from onto . So, is a group with the neutral element equal 1. Next, γ is a homeomorphism andwhence is a topological group. For every , by , we always denote the inverse element to u in the group and is defined by for . In what follows,
and
for
and
. Clearly, by the associativity of the operation, we have
Next, given
, we define functions
by the formulas:
Remark 5. Let . Since is a topological group, and are homeomorphisms without fixed points and therefore preserve orientation (see Reference [13] (Remark 3)). Lemma 3. There exists such that .
Proof. First, suppose that for every , that is for every . Let and . Then, by and Remark 5, we have , where . This means that .
On the other hand, , and, consequently, , a contradiction because . So, there exists such that . Putting , we have either or . □
Lemma 4. If and , then .
Proof. It follows from Lemma 3 that there exists such that . Since and , J reverses orientation. Thus, the assertion follows from the fact that . □
Lemma 5. Let , , and . Then, and .
Proof. Lemma 4 implies that . Thus, and . The lemma results now from Remark 5. □
From Lemma 5 (for ), we obtain the following
Corollary 1. If and , then and .
Lemma 6. Let , . Then:
- (a)
;
- (b)
if , , then .
Proof. Assertion (a) results from Corollary 1 (with
). For the proof of (b), fix
. Then, by Lemma 4, we get
. Hence, Lemma 6 gives
This ends the proof. □
Lemma 7. Assume that and . Then, for every , , there exists exactly one , , such that . Moreover, .
Proof. Define a function , . Then, q is continuous and, on account of Lemma 6b, q is injective. Thus, and for every . Fix . By Lemma 4, . Consequently, in view of Lemma 6a, , which ends the proof. □
Lemma 8. Let , . Then, for every , the set is non–empty and has the smallest element . Moreover, and for .
Proof. The proof is by induction with respect to n.
The case results from Lemma 7. Fix and assume that and there exists the smallest element of with . Then, according to Lemma 7, there is exactly one element such that . Note that . Thus, .
Suppose that there is with . Then, , whence . Moreover, by Lemma 6b, , a contradiction. Consequently, y is the smallest element of , , and . This completes the proof. □
Remark 6. Fix with (see Lemma 3). From Lemma 8, by induction, we get for , where . Hence, for , , we haveLet Note that, for every , there are unique , and such that and . Thus, we can define a function byfor , , , , where , , for and . In what follows, we assume that is fixed and . Moreover, the function d is defined as above. Now, we prove some properties of d. We start with the following
Lemma 9. Let D and be defined as in Remark 6. Then, for every .
Proof. In view of Remark 6, for every , there exist unique , , and with and . If , then we write and .
The proof is by induction with respect to
. The case
is trivial because, then,
So, fix
and assume that the assertion holds for every
with
. Take
with
. Since, by (
47),
, it suffices to consider the following two cases.
Case I:
Then,
with
. Write
Clearly
. So, in view of Remark 6, Lemma 8, and the induction hypothesis, we have
Case II:
. Let
and
. Then, there exist
,
, and
for
,
such that
. Hence, by Remark 6 and the induction hypothesis, we obtain
□
Lemma 10. Let D and be the same way as in Lemma 9. Then, for every , .
Proof. Clearly, . Next, for every , , there are unique , , , with .
The proof is by induction with respect to
. The case
results from Lemma 8. Fix
and assume that the assertion holds for every
,
, with
. Let
,
, and
. From the induction hypothesis and Lemma 4, we obtain
because Lemma 8 yields
. By Lemma 8, we also get
Hence, in view of the induction hypothesis, Lemma 5 and Lemma 9, we have
This completes the proof. □
Lemma 11. Let D and be the same way as in Lemma 9, and . Then, .
Proof. By Lemmas 10 and 4, we have . Thus, on account of Corollary 1 and Lemma 9, . This ends the proof. □
Lemma 12. If M is defined by (57) with D and d the same way as in Lemma 9, then . Proof. From Lemma 11, we get immediately . We prove that M is a finite number.
Suppose that for every . Then, on account of Corollary 1, for and consequently there exists .
If
, then, in view of the continuity of the operation, we have
a contradiction.
If , then, by Corollary 1, , a contradiction, too.
In this way we have proved that there is with . Thus, writing , we get for and . Observe that the set is connected, , and . Hence, and consequently, by Remark 5, , which means that . This ends the proof. □
Lemma 13. Let D and be the same way as in Lemma 9 and M be given by (57). Then, the following two statements are valid. - (a)
For every , there is , , with .
- (b)
The set is dense in S.
Proof. (a) Put (with respect to the order ≺) and suppose that . From Lemma 8 we get and, by Lemma 7 with , . Thus, there exists with . Hence, on account of Lemma 6b with , . By the definition of r and Lemma 8, there exists such that . Next, in view of Lemma 7, with , there exists with , whence, by Lemma 8, , which contradicts the definition of r.
(b) For the proof by contradiction, suppose that there exist
such that
and
. First, consider the case where
for every
. Put
(with respect to the order ≺) and
. By a) and Lemma 11, there exists
such that
. Let
and fix
such that
,
. Put
. Then, by Lemma 12, there exist
and
such that
and
,
, which, in virtue of the definition of
M and Lemma 4, means that
,
and
. Thus, on account of Lemmas 5 and 9, we obtain
Since, according to the definition of M, for , , in this way, we have proved that for every with . This contradicts the definition of M.
Now, consider the case where there is such that . Put , (with respect to ≺), and . Let V be a neighbourhood (in S) of such that for every . Choose a neighbourhood of 1 such that . It results from a) and Lemma 11 that there exists with , , and . Thus, . Moreover, Lemma 4 and Corollary 1 give . Let be a neighbourhood of such that for every . By the definition of , there is with and , which means that . Clearly, for some . Since and, by Lemma 9, , we obtain a contradiction with the definition of and . □
Define a function
by the formula
where
D and
are the same way as in Lemma 9 and
M is given by (
57).
As an immediate consequence of the definition and Lemma 13, we have the following.
Corollary 2. Let be defined as above, where M is given by (57). Then, is a strictly increasing function (with respect to the order ≺ in S). Lemma 14. Let M and be as in Corollary 2. Then, is a continuous function.
Proof. Using Lemma 13a and Corollary 2, it is easily seen that is continuous at 0. Next, we show that is continuous at M. So, fix an increasing sequence in D such that . Then, according to the definitions of M and and Lemma 13b, for every , there is with , which means that . Hence, by Corollary 2, is continuous at M.
To complete the proof suppose that
is discontinuous at a point
. Then, by Corollary 2 and Lemma 13b, there exist
with
and
, and a sequence
in
such that
and
Hence,
for
. Since
we obtain a contradiction. This ends the proof. □
Lemma 15. Let M and be as in Corollary 2 and . Then:
- (a)
if , then ;
- (b)
if , then .
Proof. (a) Let
be sequences in
such that
for
and
,
. By the continuity of ∘ and Lemmas 9 and 14, we have
(b) Let
be increasing sequences in
with
,
. Take a sequence
in
with
and write
for
. Then,
for
and, by Lemmas 9 and 14, we may write
This ends the proof. □
6. Conclusions
Given a binary, continuous, associative, and cancellative operation
, we presented an elementary construction of all continuous isomorphisms from the group
onto the semigroup
, where
S is the unit circle on the complex plane and · is the usual multiplication of complex numbers. There are exactly two such isomorphisms
, and they uniquely determine the form of operation ★ in the following way:
Moreover, for all .
Using this result, we have easily determined all continuous solutions
of the functional equation
where
is either the set of reals
or the set of complex numbers
.
We also provided some further applications of that result in functional equations and showed how to use it in the descriptions of the continuous flows and minimal homeomorphisms on S. In particular, we underlined some symmetry issues, which arise between the consequences of the result and of the analogous outcome for the real interval.
It would be interesting to investigate in the future to what extent the statements of Theorems 1 and 2 remain valid if the cancellativity is replaced by the one side cancellativity; for instance, by the left-cancellativity (a groupoid is left-cancellative if for all with ).
The next step might be to investigate how much the associativity assumption can be weakened; for instance, to what extent the assumption can be replaced by the square symmetry defined by the formula:
. A natural example of square symmetric operation in
, which is not associative, is given by:
for
, where
are fixed and
Such new results would have interesting applications in functional equations in a similar way as Theorem 1 in References [
29,
30,
31].