Abstract
In the present paper we are concerned with the problem of characterization of maps which can be expressed as an affine difference i.e. a map of the form
where \(t\in (0,1)\) is a given number. We give a general solution of the functional equation associated with this problem.
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1 Introduction
The well-known cocycle functional equation
and its applications have a long history in connection with many areas of mathematics, as discussed for example in [1, 5,6,7]. It has occurred in different fields, including homological algebra, the Dehan theory of polyhedra, statistics and information theory. The general solution of the cocycle functional equation on abelian groups has been known for about half a century (see [1, 6, 7]). It turns out that a function \(F:G\rightarrow H\) (where G and H stand for abelian and a divisible abelian group, respectively) is a solution to the system of functional equations
if and only if the function F is representable as a Cauchy difference i.e. it has the form
for some one-place function \(f:G\rightarrow H\). The above characterization have been proved first by Erdös in [6] for real functions, then by Jessen, Karpf and Thourup in [7] on abelian groups. An interesting method for finding the symmetric solutions of the cocycle functional equation on commutative semigroups was given by B. Ebanks in [4].
A similar characterization for differences of the form
where f is arbitrary function and \(\lambda \) and \(\mu \) are given parameters has been given by Ebanks in [3]. In particular, he has characterized a Jensen differences
as a special case \(\lambda =2, \mu =\frac{1}{2}\). The above form is called Jensen difference because it vanishes exactly when f is a solution of the Jensen functional equation. In our main result we generalize the following statement which is a particular case of result proved by Bruce Ebanks [3, Corollary 7]
Theorem 1
Let G be a uniquely 2-divisible abelian group, and let X be a rational vector space. Then a map \(\Delta :G\times G\rightarrow X\) satisfies conditions:
-
(a)
\(\Delta (x,x)=0,\quad x\in G\),
-
(b)
\(\Delta (x,y)=\Delta (y,x),\quad x,y\in G\),
-
(c)
\(\Delta (x,y)+\Delta (z,w)+2\Delta \Big (\frac{x+y}{2},\frac{z+w}{2}\Big )= \Delta (x,z)+\Delta (y,w)+2\Delta \Big (\frac{x+z}{2},\frac{y+w}{2}\Big ), x, y, w, z\in G,\)
if and only if there exists a function \(f:G\rightarrow X\) such that
A key tool we are going to use in the proof of our main result is the following theorem from [5, Theorem 3.3]. This theorem gives a general solution of the most general form of the cocycle functional equation.
Theorem 2
Let G be an abelian group and X a rational vector space. The general solution \(F_i:G\times G\rightarrow X\) \((i=1,\ldots ,6)\) of
is given by
where \(A_1, B_1, C_1: G\times G\rightarrow X\) are additive in the first variable, \(B_2:G\times G\rightarrow X\) is additive in its second variable, and \(f_i:G\rightarrow X\) \((i=1,\ldots ,11)\) is arbitrary.
2 Results
Let \(t\in (0,1)\) be a fixed number. Throughout this paper X and Y stand for linear spaces over the field \(\mathbb {K}\) such that \(\mathbb {Q}(t)\subseteq \mathbb {K} \subseteq \mathbb R\), where \(\mathbb {Q}(t)\) is the smallest field containing a singleton \(\{t\}\). Clearly, \(\mathbb {Q}\subseteq \mathbb {Q}(t)\), where \(\mathbb {Q}\) denotes the field of rational numbers. The assumptions about X and Y will not be repeated in the sequel. The purpose of the present paper is to characterize a difference of the form
where \(t\in (0,1)\) is a given number. Let us recall that a function \(f:X\rightarrow \mathbb R\) is said to be t-convex, t-concave, t-affine if
respectively. The above difference we will call a t-affine difference because it vanishes exactly when f is a t-affine function (see [8] for more information about t-affine and t-convex functions). It can be easily seen that the t-affine difference has the following two properties:
-
(i)
\(a_f(x,x,t)=0,\quad x\in X\),
-
(ii)
\(a_f(x,y,t)=a_f(y,x,1-t),\quad x,y\in X\).
Furthermore, let us observe that \(a_f\) satisfies the following functional equation:
Indeed, for arbitrary \(u, x, y, v \in X\) we get
In our main result we show that conditions (i)–(iii) characterize exactly those maps \(\omega : X\times X\times \{t,1-t\}\rightarrow [0,\infty )\) which can be expressed as a t-affine difference \(a_f\) for some t-convex function \(f:X\rightarrow \mathbb {R}\). In [9] we have proved the following result in this spirit.
Theorem 3
Let D be a t-convex subset of a real linear space i.e. \(tD+(1-t)D\subseteq D\) and let the maps \(f, g:D\rightarrow \mathbb R\) and \(\omega :D\times D\times [0,1]\rightarrow \mathbb {R}\) satisfy the inequalities
where
Then there exists a function \(h:D\rightarrow \mathbb R\) such that
if and only if for all \(x, y, u, v \in D\) the map \(\omega \) satisfies the functional equation:
The following theorem gives a general solution of the functional equation corresponding to equation (iii).
Theorem 4
The general solution \(\omega :X\times X\rightarrow Y\) of the functional equation
is given by
where \(d, r:X\rightarrow Y\) are additive functions satisfying the condition
\(\bar{c}\in Y\) is an arbitrary constant and \(n:X\rightarrow Y\) is an arbitrary function.
Proof
It is easy to verify that \(\omega \) given by (3) with (4) is a solution of (2). Conversely, suppose that (2) holds. Put \(y=0\) in (2) to get
This equation can be rewritten in the form
Putting in (5) \(u=\frac{1-t}{t}y\) we get
We can rewrite the above equation in the form
where,
Choosing the fourth line of the solution in Theorem 2, we see that
where \(B_1, C_1: X\times X\rightarrow Y\) are additive functions in the first variable and \(f_1, f_2, f_3: X\rightarrow Y\) are arbitrary. If we replace \(-\frac{1}{t}B_1\) by \(A_1\), \(-\frac{1}{t}C_1(y,x)\) by \(A_2(x,y)\) and \(\frac{1}{t}f_1, \frac{1}{t}f_2, \frac{1}{t}f_3\) by g, f, h, respectively, we get
and finally by putting \(\alpha :=\frac{t}{1-t}\) we obtain
where \(A_i:X\times X\rightarrow Y\) is an additive map with respect to the i-th variable, for \(i=1, 2\) and \(f, g, h: X\rightarrow Y\) are arbitrary. Now, we substitute (6) into (2) and obtain after rearrangement
Put \(u=0\) in (7) to obtain
Subtracting (8) from (7) we get
We see that the right hand side of (9) is independent of y. Therefore the left-hand side left unchanged upon setting \(y=0\). That is, after some calculation, we get
Setting u, v, y, x in the palace of \(\alpha tu, tv, (1-t)y,\) and tx, respectively, we obtain
From this we see that for arbitrary fixed \(u, v, x\in X\) the map
is additive, consequently
for all \(u, v, x, y, z\in X\). Putting \(z:=y\) and \(v:=x+y\) we get
for any \(x, y, u\in X\), hence the function \(x\longrightarrow A_1(u,x)\) is a polynomial function of second order for arbitrary \(u\in X\). Therefore, by a result of Djoković [2], this map can be written as the sum of a constant, an additive map and the diagonalization of a symmetric bi-additive map i.e.
It is easy to observe that \(a_0, a_1, a_2\) are additive with respect to the first variable. By a similar argument we deduce that
where \(b_0(y)\) is constant (with respect to x), \(x\longrightarrow b_1(x,\cdot )\) is additive, and \(x\longrightarrow b_2(x,x,\cdot )\) is a diagonalization of a symmetric bi-additive map. It is not hard to check that \(b_0, b_1, b_2\) are additive with respect to y.
Inserting (11) and (12) back into (10) and simplifying we get
hence, \(a_2=b_2\), in particular; \(a_2\) is a 3-additive and symmetric function.
Put \(v=0\) in (9) and rewrite this equation using the representations (11) and (12). After rearrangement we get
where here and in the sequel \(c(x,y):=a_1(x,y)+b_1(x,y),\ x, y \in X.\) Replace in the above equation \(\alpha u\) by u and define the functions \(b, p:X\rightarrow Y\) by the formulas
to get the form
Since the right hand side of (16) is symmetric in x and u, the function b is bi-additive and symmetric. Observe that the function \(\delta :X\rightarrow Y\) given by the formula
is additive. Indeed, for arbitrary \(u, x \in X\) by virtue of (16), bi-additivity and symmetry of b, the 3-additivity and symmetry of \(\gamma \) we have
Therefore on account of (13) and (17) f has the form
where \(c_1=-p(0)=f(0)-th(0)\).
Now, we return to equation (7). Put \(y=0\) in (7) to obtain
Subtracting the resulting equation from (7), we get
Put \(u=v=0\) in Eq. (19). Having the forms (11) and (12) in mind we obtain
Replace in the above equation \(\alpha x\) by x and define the functions \(\bar{b}, l, q:X\rightarrow Y\) by the formulas
to get the form
Similarly as before one can check that the function \(k:X\rightarrow Y\) given by the formula
is additive, moreover, we get the representation
where \(c_2=g(0)-(1-t)h(0)\).
Now, put (11), (12), (18) and (20) into Eq. (6) to get
where
and
Finally, set \(u=y=0\) in (2) and substitute the new form of \(\omega \) into Eq. (2) to obtain after rearrangement
Put \(\beta x\) and \(\beta v\) in the place of x and v, respectively, for \(\beta \in \mathbb {Q}\setminus \{0\}.\) Since any additive function is \(\mathbb {Q}\)-homogeneous we get
consequently,
and hence
This completes the proof of the theorem.\(\square \)
Remark 1
It follows from the Proof of Theorem 4 that it is also true in the case where \(t\in \mathbb {R}\setminus \{0,1\}\).
As an immediate consequence of the previous theorem we obtain the following corollary.
Corollary 1
A map \(\omega :X\times X\rightarrow Y\) satisfies the functional Eq. (2) and vanishes on the diagonal i.e.
if and only if it has the form
where \(d:X\rightarrow Y\) is an additive function and \(f:X\rightarrow Y\) is an arbitrary map.
Proof
Assume that \(\omega \) satisfies Eq. (2) and vanishes on the diagonal. On account of Theorem 4\(\omega \) has the form
where \(\bar{c}\in Y\) is a constant, \(d,r:X\rightarrow Y\) are additive and \(f:X\rightarrow Y\) is arbitrary. Since the t-affine difference vanishes on the diagonal,
Because \(d+r\) is an additive map, by putting \(x=0\) we get \(\bar{c}=0\) and consequently
\(\square \)
In order to present our next result we need some kind of symmetry of \(\omega \) and this leads us to the consideration of \(\omega \) as a map of three variables. The following theorem generalizes Corollary 8 from [3] in the case where G is a linear space.
Theorem 5
A map \(\omega :X\times X\times \{t,1-t\}\rightarrow Y\) satisfies the functional equation
vanishes on the diagonal and is symmetric i.e.
if and only if there exist an additive function \(d:X\rightarrow Y\) and a map \(g:X\rightarrow Y\) such that
Proof
Obviously, the function \(\omega \) of the form (24) is symmetric, vanishes on the diagonal and satisfies Eq. (22).
Conversely, assume that \(\omega \) vanishes on the diagonal and satisfies (22) and (23). According to Corollary 1 there exist functions \(g,h:X\rightarrow Y\) and additive maps \(a, b:X\rightarrow Y\) such that
for \(x,y \in X\). Since
then
By putting
we can rewrite the above equation in the form
It follows from the above identity and the additivity of m that
Putting \(z=0\) and replacing tx by x and \((1-t)y\) by y we have
so subtracting 2p(0) from the both sides of this equation we get
Therefore, a function \(r:X\rightarrow Y\) given by the formula
is additive and consequently,
moreover,
where \(\bar{c}=p(0)\). Hence we have
This implies that
Finally, defining a new additive function \(d:X\rightarrow Y\) by the formula
we get a desired form of \(\omega \)
The proof of the theorem is completed. \(\square \)
Now, we present a particular case of Theorem 9 from [10] (for \(D=X\)) which we are going to use in our last result.
Theorem 6
Assume that for some point \(y\in X\) a map \(\omega :X\times X\times \{t,1-t\}\rightarrow [0,\infty )\) satisfies the following three conditions:
-
(a)
\(\omega (y,y,t)=0,\)
-
(b)
\(\omega (x,z,t)=\omega (z,x,1-t),\)
-
(c)
\(s\omega (u,z,s)+(1-s)\omega (v,z,s)-\omega (su+(1-s)v,z,s) \le s\omega (u,v,s)-\omega (su+(1-s)z,sv+(1-s)z,s)\),
for all \(x, z\in X\) and \(s\in \{t,1-t\}\). Then for arbitrary \(c\in \mathbb {R}\) there exists a t-concave function \(g_y:X\rightarrow \mathbb {R}\) such that \(g_y(y)=c,\ g_y(x)\le c,\ x\in X,\) and
Our main result reads as follows
Theorem 7
A map \(\omega :X\times X\times \{t,1-t\}\rightarrow [0,\infty )\) is symmetric (i.e. satisfies (23)), vanishes on the diagonal and satisfies the functional Eq. (22) if and only if there exists a t-convex function \(f:X\rightarrow \mathbb R\) such that
Proof
It is easy to see that an affine difference \(a_f\) satisfies conditions (a)–(c) from Theorem 6. Conversely, assume that \(\omega \) is symmetric, vanishes on the diagonal and satisfies the functional Eq. (22). On account of Theorem 5 there exist a function \(h:X\rightarrow \mathbb R\) and an additive function \(a:X\rightarrow \mathbb R\) such that
As it can be easily checked the function \(\omega \) of the above form satisfies conditions (a)–(c) from Theorem 6 then there exists a concave function \(g:X\rightarrow \mathbb R\) such that
Finally, by putting \(f:=-g\) we see that f is t-convex, moreover,
\(\square \)
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Olbryś, A. Characterization of t-affine differences and related forms. Aequat. Math. 95, 985–1000 (2021). https://doi.org/10.1007/s00010-021-00800-2
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DOI: https://doi.org/10.1007/s00010-021-00800-2