Find all continuous function over real numbers such that $f(f(x+y))=f(x)+f(y)$
We have $$f(f(x+y))=f(x)+f(y)\tag{1}$$Also $$f(f(x+y))=f(x+y)+c \tag{2}$$ where $c$ is $f(0)$.From $(1)$ and $(2)$ we have, $$f(x)+f(y)=f(x+y)+c$$ the solution to this function equation can be found as $f(x)=kx-(x-1)c$. Here $k$ is $f(1)$ Now putting this the original functional equation, $$f(f(x+y))=(k^2-2kc+c^2)(x+y)-c(c-k-1) \tag{3}$$ $$f(x)+f(y)=(k-c)(x+y)+2c \tag{4}$$ For problem condition $(3)$ and $(4)$ must be equal ie $$(k-c)(k-c-1)(x+y)-c(c+1-k)=0$$ which implies$(k-c)(k-c-1)=0$ and $c(c-k+1)=0$. The solution to these equations are (0,0) ,and all number of the form (u+1,u) for all real u. Thus we conclude that the functions satisfying this functional equation are $f(x)=x,f(x)=c,f(x)=x+c$ where c is a real constant.
My question: Is my proof correct? If yes, am I just overdoing it?
Edit 1
2 comments and 1 answer later I have not received answer to my original question. Is my proof correct?
Umm... In fact, you can set $g(x)=f(x)+c$ which leads the F.E. to $g(x)+g(y)=g(x+y)$. So using the continuous hint, you can use the Cauchy function to $g$, so we get $g(x)=kx$. So you can say that $f(x)=kx-c$. Putting this to the original F.E., we get $k(k(x+y)-c)-c=k(x+y)-2c$, so $(k^2-k)(x+y)=(k-1)c$. $k^2-k$ and $(k-1)c$ should be $0$, so it leads to $(k, c)=(1, c), (0, 0)$.