How Does Sincov's Functional Equation Validate the Condition F(x,y) = F(0, y-x)?

[filip97]

I read Aczel book "Lectures of functional equations an their applications".

On page 223. (Sincov's equation) is equation :

$F(x,y)+F(y,z)=F(x,z)$

and general solution of this

$F(x,y)=g(x)−g(y)$

, but how I prove that this function satisfies conditions

$F(x,y)=F(0,y−x)$

??

 

[mitochan]

I am skeptical whether your last equation holds.
For example
$$F(a,b)=\int_a^b f(u)du$$
satisfies the first and the second equations but
$$F(a,b)=\int_a^b f(u)du \neq \int_0^{b-a} f(u)du=F(0,b-a)$$
unless f(u) is periodic, i.e.
$$f(u)=f(u+a)$$
for any a, so constant.
$$F(a,b)=c(b-a)$$

 

[Stephen Tashi]

and general solution of this

$F(x,y)=g(x)−g(y)$

My copy of Aczel says the general solution is $F(x,y) = g(y) - g(x)$.

 

[Stephen Tashi]

, but how I prove that this function satisfies conditions

$F(x,y)=F(0,y−x)$

??

Aczel's book doesn't say explicitly that the equation

$F(x,y) + F(y,z) = F(x,z)$ implies $F(x,y) = F(0,y-x)$

The book states

Equation (1) can also be considered as a generalization or as an inhomogeneous form of Cauchy's basic equation 2.2.1 (1) into which it is transformed with:
$F(x,y) = F(0,y-x) = f(y-x)$
In fact, this involves $f(y-x) + f(z-y) = f(z-x)$, that is 2.1.1(1).

Cauchy's equation 2.2.1 (1) is $f(x+y) = f(x) + f(y)$

 

[Stephen Tashi]

I think Aczel asserts implications in the following direction:

Assume $F(A,B) = f(B-A)$

Then $F(0, y-x) = f(y-x-0) = f(y-x) = F(x,y)$
and
$F(A,B) + F(B,C) = f(B-A) + f(C-B)$.

If we also assume $F(A,B) + F(B,C) = F(A,C)$ then we must have
$f(B-A) + f(C-B) = f(C-A)$.
Letting $x = B-A, y = C-B$ this implies
$f(x) + f(y) = F( A,C) = f(C-A) = f( (y+B) - (B-x)) = f(y+x) = f(x+y)$
So $f$ satisifes $f(x+y) = f(x) + f(y)$