I was discussing with a friend of mine about her research and I came across this problem.
The problem essentially boils down to this.
$f(x)$ is a function defined in $[0,1]$ such that $f(x) + f(1-x) = f(1)$. I want to find a condition on $f(x)$ so that I can conclude $f(x) = f(1)x$.
Clearly, $f \in C^{0}[0,1]$ alone is insufficient to conclude $f(x) = f(1)x$.
My hunch is if $f(x) \in C^{\infty}[0,1]$, then $f(x) = f(1)x$. However, I am unable to prove it. Further, is there a weaker condition with which I can conclude $f(x) = f(1)x$?
This problem closely resembles another problem:
If $f(x+y) = f(x) + f(y)$, $\forall x,y \in \mathbb{R}$ and if $f(x)$ is continuous at atleast one point, then $f(x) = f(1)x$.
I know how to prove this statement, but I am unable to see whether this will help me with the original problem.
Though these details might not be of much importance to her, I am curious to know.
EDIT:
As Qiaochu pointed out, I need stronger conditions on $f$ to come up with some reasonable answer.
Here is something which I know that $f$ has to satisfy the following:
$\forall n \in \mathbb{Z}^{+}\backslash \{1\}$, $f(x_1) + f(x_2) + \cdots + f(x_n) = f(1)$, where $\displaystyle \sum_{k=1}^{n} x_k = 1$, $x_i \geq 0$.
Note that $n=2$ boils down to what I had written earlier.