# $f(x,y) - f(x,z) = g(y,z)$ implies $f(x,y) = a(x) + b(y)$

A result I need is:

If $$f(x,y) - f(x,z) = g(y,z)$$ for all $$(x,y,z)$$, then $$f(x,y) = a(x) + b(y)$$ for some functions $$(a,b)$$.

This seems almost obvious, and I've constructed a proof, but that proof seems unnecessarily complicated and is remarkably tedious (and so isn't included here). I'd like pointers or ideas leading to something more simple and elegant. Surely this is a known result, or a special case of a well-known result?

• What do you want to know by your statement ? – nmasanta May 31 at 17:38

Plugging $$z=0$$ into equation gives that $$f(x,y)-f(x,0)=g(y,0),$$ so $$f(x,y)=f(x,0)+g(y,0).$$ Now, just denote $$a(x):=f(x,0)$$ and $$b(y):=g(y,0)$$ and we get the desired result.
take $$a(x)=f(x,z)$$, $$b(x)=g(x,z)$$ for some $$z$$. then $$f(x,y)=f(x,z)+g(y,z)=a(x)+b(y)$$.
• I think it's easier to understand if we choose some fixed value of $z$. – richrow May 31 at 17:41