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

Question

As a followup comment to my answer to the question

$\;\;\;\;$Can I split $\frac{1}{a-b}$ into the form $f(a)+f(b)$?

"Lord Shark the Unknown" made the following observation:

$h:\mathbb{R}^2\to\mathbb{R}$

$f,g: \mathbb{R}\to\mathbb{R}$

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

$x,y\in\mathbb{R}$

$$h(a,b)+h(c,d)=h(a,d)+h(c,b)$$

$a,b,c,d\in\mathbb{R}$

My question is about the converse.

To keep it simple, I'll assume the domain of $h$ is $\mathbb{Z}^2$.

Explicitly, the question is this:

$h:\mathbb{Z}^2\to\mathbb{Z}$

$$h(a,b)+h(c,d)=h(a,d)+h(c,b)$$

$a,b,c,d\in\mathbb{Z}$

$f,g:\mathbb{Z}\to\mathbb{Z}$

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

$x,y\in\mathbb{Z}$

score 1 · Answer 1 · answered Jul 30 '19 at 10:52

1

You say that for all $(a,b) \in \mathbb{Z}^2$ $h(a,b) \in \mathbb{Z}^2$. But, the functions $f$ and $g$ have their values in $\mathbb{Z}$. Do you see the non-sense ?

answered Jul 30 '19 at 10:52

kakarotto

11

-1 This is a good comment but a poor answer. – Servaes Jul 30 '19 at 10:52
@kakarotto: It was a typo (now fixed): Thanks for alerting me. – quasi Jul 30 '19 at 10:53
4

@Servaes I cannot comment... so please be kind and remove your -1 – kakarotto Jul 30 '19 at 10:54
@kakarotto: No worries. – quasi Jul 30 '19 at 10:55
@kakarotto Done. – Servaes Jul 30 '19 at 10:59

score 1 · Accepted Answer · answered Jul 30 '19 at 10:59

Some rearranging shows that such a function $h$ also satisfies $$h(a,b)-h(a,d)=h(c,b)-h(c,d),$$ $$h(a,b)-h(c,b)=h(a,d)-h(c,d),$$ for all $a,b,c,d\in\Bbb{Z}$, which means that for all $m,n\in\Bbb{Z}$ the functions $$h_{m,n}(k):=h(k,m)-h(k,n),$$ $$h^{m,n}(k):=h(m,k)-h(n,k),$$ are constant. This means $$h(x,y)=h(x,0)+h_{y,0}(x)=h(x,0)+h_{y,0},$$ so setting $f(x):=h(x,0)$ and $g(y):=h_{y,0}$ yields the desired identity $$h(x,y)=f(x)+g(y).$$

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

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