$f(s,u) + f(u,t) = f(s,t)$ implies $f(s,t) = g(t)- g(s)$

Question

Let $f: [0,T]^2 \to \mathbb R$ such that

$$f(s,u) + f(u,t) = f(s,t)$$

for all $u,s,t \in [0,T]$.

Does this implies that all $f$ as above can be written as

$$f(s,t) = g_f(t)- g_f(s)$$

for a certain $g_f:[0,T] \to \mathbb R$

Answer 1

Firstly, take $u=t$ in your constraint equation, and $$ f(s,t)+f(t,t)=f(s,t), $$ which implies that $$ f(t,t)=0 $$ for all $t\in\left[0,T\right]$.

Secondly, take $t=s$ in your constraint equation, and $$ f(s,u)+f(u,s)=f(s,s)=0, $$ which implies that $$ f(s,u)=-f(u,s) $$ for all $u,s\in\left[0,T\right]$.

Finally, take $u=0$ in your constraint equation, and $$ f(s,t)=f(s,0)+f(0,t)=-f(0,s)+f(0,t). $$ Thus if you take $g_f(x)=f(0,x)$, the above equation then reads $$ f(s,t)=g_f(t)-g_f(s). $$

Answer 2

Simply take $g(t)=f(0,t){}{}$.