# Functional Equation simple problem

How do I show that if there are functions $$f,g$$ such that$$f(g(x)+g(y))=bx+cy$$holds for all $$b,c\in\mathbb{R}$$, then we necessarily have $$b=c$$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?

Extra: In addition, I'd like to prove that there are no pair of functions $$f,g$$ such that$$y^2+z^2=f(x,g(y-x)+g(z-x)).$$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?

## 1 Answer

For every pair $$(x,y)$$, $$bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$$.

Thus $$b(y-x)=c(y-x)$$. Take $$x=0,y=1$$.

Edit: for the second question, the answer is, in general, no. Take $$g$$ to be a bijection from $$\mathbb{R}$$ into a $$\mathbb{Q}$$-basis of $$\mathbb{R}$$, which we denote as $$\mathcal{B}=(g(x))_x$$.

Note that $$g(a)+g(b)$$ determines the unordered pair $$(a,b)$$ (because the $$g(x)$$ are all linearly independent over the rationals).

Take now $$f(x,y)=0$$ if $$y \notin \mathcal{B} + \mathcal{B}$$, and $$f(x,y)=(a+x)^2+(b+x)^2$$ otherwise where $$y=g(a)+g(b)$$.

• Thank you! I have added an extra question, which I can't find a way to prove, if you have any insights about it, let me know! – sam wolfe Jan 10 at 15:25
• So $f$ is a two-variable function? – Mindlack Jan 10 at 16:35
• In the extra case, yes. – sam wolfe Jan 10 at 16:47
• With a proper change of variable, we want some universal function $g$ and a family $f_x$ of functions such that $f_x(g(y)+g(z))=(y+x)^2+(z+x)^2$, thus $g(y)+g(z)$ determines the unordered pair $(y,z)$. This seems a very strong property, so it could imply a contradiction. – Mindlack Jan 10 at 16:47
• What I did prove is that there is no contradiction (at least if the axiom of choice is granted). – Mindlack Jan 17 at 7:35