I am looking for a proof that the solution of the functional equation:
$$f(x+y)=f_1(x)+f_2(y)$$
is:
$$f(z)=az+(b_1+b_2)$$ $$f_1(z)=az+b_1$$ $$f_2(z)=az+b_2$$
Assume that $x,y,z$ are real numbers, that the domain of $f$ is the real numbers, and that $f$ is continuous. Feel free to substitute some other domain/regularity assumption if they are necessary.
I am interested in this functional equation because it comes up in social choice theory. An accessible proof would be an advantage. Thanks!
Suppose the functions $f,f_1,f_2$ are all $C^0$. Then by setting $y=0, x=x$ and $x=0, y=y$
$$f(x)=f_1(x)+f_2(0)~~,~~f(x)=f_1(0)+f_2(x)$$
We substitute back and after setting $g(x)=f(x)-f_1(0)-f_2(0)$ we see that it satisfies the well known functional relation
$$g(x+y)=g(x)+g(y)$$
Since $g$ is continuous, it can be shown using some standard manipulations (see some related tricks and results for reference in the question here) that $g(x)=ax$. Thus we conclude that with $f_{1,2}(0)=b_{1,2}$
$$f(x)=ax+b_1+b_2~~,~~f_1(x)=ax+b_1~~,~~ f_2(x)=ax+b_2$$
NOTE: If the functions are supposed to be $C^1$ showing this becomes trivial since we can rearrange the last equation, divide by $y$ and take the limit $y\to 0$
$$\lim_{y\to 0}\frac{g(x+y)-g(x)}{y}=\lim_{ y\to 0} \frac{g(y)}{y}\Rightarrow g'(x)=g'(0)=a$$
which in turn implies that $g$ is linear, as above.
