Let $f:R_+^n\rightarrow R$
Suppose that $f$ satisfies following three properties:
(a) $f(v+r)=f(v)+r$ for all $v\in R_+^n$ and $r\geq0$
(b) $f(\alpha v)=\alpha f(v)$ for all $v\in R_+^n$ and $\alpha\geq0$
(c) If $u+v=(k,\cdots,k)$ for some $k\geq0$, then $f(u+v)=k=f(u)+f(v)$
Question:
Is it true that $f(u+v)=f(u)+f(v)$ for all $u,v\in R_+^n$? If not, what are additional conditions I need?
Note: By $v+r$ I mean vector $(v_1+r,\cdots,v_n+r)$. By $\alpha v$ I mean vector $(\alpha v_1,\cdots,\alpha v_n)$
I can prove for $n=2$:
Given any $a\geq b$, by (c) we have $f(a,a)=f(a,0)+f(0,a)=f(a,b)+f(0,a-b)$.
This implies $f(a,b)=f(a,0)+f(0,a)-f(0,a-b)=f(a,0)+bf(0,1)=f(a,0)+f(0,b)$. Here I use (b).
Therefore $f(u+v)=f(u_1+v_1,0)+f(0,u_2+v_2)=f(u_1,0)+f(0,u_2)+f(v_1,0)+f(0,v_2)=f(u)+f(v)$
For $n>2$, I do not know how to prove. Does anyone have thoughts? Thank you.
I think for $n>2$ there are counterexamples. I guess the following is a example for dimension $3$. We can partition $R_+^3$ into three parts. The first is the vectors with all the coordinates the same, which has the form $(x,x,x)$. So by the definition we have $f(x,x,x)=x$. The second part consists of the vectors having exact two coordinates the same, and we takes the value under the function $f$ the value of the same coordinates, which means that $f(x,x,y)=f(x,y,x)=f(y,x,x)=x$ when $x\neq y$. And the last part is the vectors with all coordinates with different value. We define the value of $f$ by $f(x,y,z)=x$. It can be verified that the function $f$ satisfies the above properties but is not linear since $f(0,1,1)=1\neq 0= f(0,1,0)+f(0,0,1)$.