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Is there some other way to characterize the functions $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which are expressible as $$f(x,y)=g(x)+g(y)-g(x+y)$$ for some $g:\mathbb Z\to\mathbb Z$?

Easy facts: (1) $f$ must satisfy $f(x,y)=f(y,x)$ and $f(x,0)=g(0)$ for all $x$. (2) Not all functions $f$ are expressible, since for $x,y\in\{1,\ldots,n\}$ the number of choices of $f(x,y)$ the left grows quadratically in $n$ and the number of possible values of $g$ on which they depend, grows just linearly.

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$f(x,0)=g(0)$, not $0$. –  Chris Eagle Apr 5 '11 at 10:34
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Here is a necessary condition: $$f(x,y+z)+f(y,z)=f(y,x+z)+f(x,z).$$ Could it be sufficient ? –  Denis Serre Apr 5 '11 at 11:24
    
@ chris: yes, that's right. I will edit the post. –  Mircea Apr 5 '11 at 12:06
    
@ steve and denis: Thanks Steve, that's an interesting link! In fact my original motivation was apparently different: One has a family of operators $I_x$ with $I_x\circ I_y=F(x,y) I_{x+y}, F>0$, I wanted to know when can I normalize the whole family cancelling away the $F$-factor. The answer of Denis means that associativity of the composition is the requirement! I find that very nice. –  Mircea Apr 5 '11 at 12:41
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3 Answers 3

up vote 3 down vote accepted

The necessary and sufficient condition is that $$f(x,y+z)+f(y,z)=f(y,x+z)+f(x,z),\qquad\forall x,y,z\in\mathbb Z.\qquad (1)$$ On the one hand, this is obviously necessary. On the other hand, the function $g$, if it exists, is given by $$g(k)=kg(1)-\sum_{j=1}^{k-1}f(j,1).$$ Then $g$ is suitable if and only if $$f(k,\ell)=\sum_{j=1}^{k-1}f(j,1)-\sum_{j=\ell}^{k+\ell-1}f(j,1),\qquad k,\ell\in\mathbb Z.$$ There remains to check that these relations are consequences of (1).

Note that (1) contains the fact that $x\mapsto f(x,0)$ is constant.

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Thank you for the fast answer! I think that this settles the question. –  Mircea Apr 5 '11 at 12:25
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This is secretly a question about group cohomology. Suppose we want to compute $\operatorname{Ext}^*_{\mathbb{Z}G}(\mathbb{Z}, \mathbb{Z})$ when $G=\langle g \rangle$ is infinite cyclic and the action on the integers is trivial. We can easily write down a free resolution

$$ 0 \to \mathbb{Z}G(g-1) \to \mathbb{Z}G \to \mathbb{Z} \to 0 $$

and work it out from there: in particular it's clear that $\operatorname{Ext}^n$ vanishes for $n>1$. But we could also try using the bar resolution. On writing this down and applying $\hom_{\mathbb{Z}G} ( -, \mathbb{Z})$ you find that the ext groups are given by the cohomology of the following cocomplex:

$$ \hom_{\mathbb{Z}G} (\mathbb{Z}G, \mathbb{Z}) \to \hom_{\mathbb{Z}G} (\bigoplus_{g \in G} \mathbb{Z}G[g], \mathbb{Z}) \to \hom_{\mathbb{Z}G} (\bigoplus_{g,h \in G} \mathbb{Z}G[g|h],\mathbb{Z}) \to \ldots $$

Of course, $\hom_{\mathbb{Z}G} (\bigoplus_{g \in G} \mathbb{Z}G[g], \mathbb{Z})$ can be identified with the set of maps of sets $G \to \mathbb{Z}$ and so on. On doing this we get that $\operatorname{Ext}^2$ is the kernel of the map

$$\partial_2 : \operatorname{Map}(G\times G, \mathbb{Z}) \to \operatorname{Map}(G\times G\times G, \mathbb{Z}) $$

given by $\partial_2 (f)(g,h,k) = f(h,k) - f(gh,k) + f(g,hk) -f(g,h)$, modulo the image of the map $\partial_1: \operatorname{Map}(G , \mathbb{Z}) \to \operatorname{Map}(G\times G, \mathbb{Z})$ defined by $\partial_1(f)(g,h) = f(h) - f(gh) + f(h)$.

We already know that this cohomology group is zero. But this says exactly that the set of functions $f:G \times G \to \mathbb{Z}$ expressible as $g(x)-g(xy)+g(y)$ is equal to those functions in the kernel of $\partial_2$.

This is equivalent to Denis Serre's answer above - it may look as if there's a discrepancy, but this is fixable by noting that any such f is symmetric.

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Thanks mt! This is certainly a very suggestive interpretation, and it also has a nice side effect: it indicates clearly how one could generalize the problem! –  Mircea Apr 6 '11 at 20:37
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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.

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