Throughout this note, assume that $\mathbb{C}$ is the field of
all complex numbers, $G$ is a subsemigroup of $\mathbb{C}$ and $M(X)$
is all complex valued functions with domain $X$. Also $\Delta:
M(G)\rightarrow M(G^{2})$ is an operator where it define as
follows:

$$(\Delta f)(x,y)=f(x+y)-f(x)-f(y)$$

for all $x,y \in G$ and $f\in M(G)$.

**Definition.**
The function $g\in M(G^{2})$ is a Cuachy error function, if there
exists $f\in M(G)$ such that
$$\Delta f=g.$$
We denote the set of all Cuachy error functions with $\Delta(G)$.

**Proposition.**
The set $\Delta(G)$ is $\mathbb{C}$-linear space under pointwise
addition and scalar multiplication.

We define an equivalence relation on $M(G)$ as follows:

$$f\sim g\ \Leftrightarrow\ \Delta f=\Delta g\ for\ all\ f,g\in
M(G).$$

Let $\overline{f}$ be equivalence class under relation $\sim$ for
any $f\in Map(G)$ and $M^{*}(G)$ be set of all equivalence
classes. Now we define addition and scalar multiplication with
$\overline{f}+\overline{g}=\overline{f+g}$ and $\alpha
\overline{f}= \overline{\alpha f}$ for all $f, g\in M(G)$ and
$\alpha \in \mathbb{C}$. These operations are well-defined and so
that $M^{*}(G)$ is $\mathbb{C}$-linear space.

**Proposition.**
There exist an isomorphism $\bar{I}$ between $\mathbb{C}$-linear
spaces $M^{*}(G)$ and $\Delta(G)$ such that
$\bar{I}(\bar{f})=\Delta f$ for all $f\in M(G)$.

Our aim is to study an equation of the following type

$$f(x+y)=f(x)+f(y)+g(x,y)$$

for given function $g\in M(G^{2})$ and unknown function $f\in
M(G)$. By the operator $\Delta$, the above equation can be
rewritten as

$$\Delta f=g$$

Therefore, the equation $f(x+y)=f(x)+f(y)+g(x,y)$ have a solution if and only if
$g\in \Delta(G)$. Thus, the set $\Delta^{-1}g$ is all solution of
the equation $f(x+y)=f(x)+f(y)+g(x,y)$. If $f\in \Delta^{-1}g$, then for any
$h\in \Delta^{-1}g$, $\Delta (f-h)=0$, so we have

$$\Delta^{-1}g=\overline{f}=\overline{0}+f$$

which says that the solutions of $f(x+y)=f(x)+f(y)+g(x,y)$ with approximation of
linear maps are unique. There exists a question: Whether for any
$g\in M(G^{2})$, $g$ is a Cuachy error function (or equivalently
$g\in \Delta(G)$)? We show that the answer is negative. Moreover,
in the following we find that some necessary conditions for any
Cuachy error function.

**Proposition.**
Let $g\in \Delta(G)$, then

$$g(x,0)=g(0,y)=g(0,0)\ and\ g(x,y)=g(y,x)$$

for any $x, y\in G$.

**Proof.** Since $g\in \Delta(G)$, so there exists
$f\in M(G)$ such that
$$f(x+y)-f(x)-f(y)=g(x,y)$$
for all $x, y\in G$. Setting $x=y=0$, we obtain $-f(0)=g(0,0)$
and again setting $y=0$ in $f(x+y)-f(x)-f(y)=g(x,y)$, we get $-f(0)=g(x,0)$ for
all $x\in G$. Similarly, we can to obtain $-f(0)=g(y,0)$ for all
$y\in G$. The proof in complete.

Note that $G\subset\Delta(G)$, because for any $\lambda\in G$,
$T-\lambda\in \Delta^{-1}\lambda$, where $T\in \overline{0}$.
According to above proposition, for any $f, h\in M(G)$, if the
function $h(x)+f(y)$ is a Cuachy error function, then the
functions $f$ and $h$ must be constant. In the first, our aim is
to find that some sufficient conditions and necessary conditions
for any Cuachy error function. In the following, we obtain some necessary conditions for
any Cuachy error function.

**Proposition.**
Let $\Delta f=g$ such that

$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=0}^{n-1}g(a+bi,b)=0$$ and
$$\lim_{n\rightarrow \infty} \frac{1}{n}g(x+an,y+bn)=0$$
for any fixed $a, b, x, y\in \mathbb{R}$. Then $g=0$.

**Proof.** Let $b$ be any fixed element of
$\mathbb{R}$. Since $\Delta f=g$, with induction, we can to show
that the following equality
$$f(x+nb)-nf(b)-f(x)=\sum_{i=0}^{n-1}g(x+ib,b)$$
for each fixed $x\in \mathbb{R}$ and $n\in \mathbb{N}$. Now bye
assumption $\frac{1}{n}\sum_{i=0}^{n-1}g(a+bi,b)=0$, so
$$f(b)=\lim_{n\rightarrow \infty}\frac{f(x+nb)}{n}$$
for any fixed $x\in \mathbb{R}$. Let $a,b$ be any two fixed
element of $\mathbb{R}$, then from $\Delta f=g$, we obtain
$$f(x+y+n(a+b))-f(x+na)-f(y+nb)=g(x+na,y+nb)$$
for any fixed $x, y\in \mathbb{R}$. Now since $\lim_{n\rightarrow
\infty} \frac{1}{n}g(x+bn,y+nd)=0$, thus
$$f(a+b)=f(a)+f(b),$$
which says that $f$ is an additive mapping and so $g=\Delta f=0$.

Let $g\in M(\mathbb{R}^{2})$, where
$g(x,y)=\frac{1}{x^{2p}}+y^{2q}$ for all $x,y\in \mathbb{R}$ and
for some reals $p>0$ and $0<q<\frac{1}{2}$, then by the above
Proposition, $g$ isn't a Cauchy error function. In the
following, we present a sufficient condition for any Cuachy error
function under a suitable condition.

**Proposition.**
Let $g\in M(G^{2})$ such that

$$\phi(x,y):=\sum_{k=0}^{\infty}\frac{g(2^{k}x,2^{k}y)}{2^{k+1}}$$

be converge for any $x, y\in G$. Let $h(x):=-\phi(x,x)$ for any
$x\in G$, then $\Delta h=g$.

In the above, if $g$ is a bounded function, then the solution
$h\in \Delta^{-1} g$ is a unique bounded function.

associativityof the composition is the requirement! I find that very nice. – Mircea Apr 5 '11 at 12:41