$f(x) + f(1-x) = f(1)$

Question

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.

Answer 1

I presume $\displaystyle f(1) \neq 0$, in which case we can assume $\displaystyle f(1) = 1$.

If $\displaystyle f$ is such a function, then $\displaystyle g(x) = \sin^{2}\left(\frac{\pi f(x)}{2}\right)$ is also such a function.

If $\displaystyle f(1) = 0$, take $\displaystyle g(x) = \sin(f(x))$.

So you will really need much stronger restrictions than infinite differentiability etc.

As to your edit, continuity is enough.

We can first show $\displaystyle f(1/q) = f(1)/q$ for integral $\displaystyle q$.

This can easily be extended to $\displaystyle f(p/q)$ (for instance $\displaystyle f(2/q) + f(1/q) + \dots + f(1/q) = f(1)$) and by continuity, to the whole of $\displaystyle [0,1]$.

Answer 2

Let $g(x) = f(x + 1/2)$, which is defined on $[-1/2, 1/2]$ and which satisfies $g(x) + g(-x) = g(1/2)$. Now, any function $g$ on $[-1/2, 1/2]$ can be written

$$g(x) = \frac{g(x) + g(-x)}{2} + \frac{g(x) - g(-x)}{2} = \frac{g(1/2)}{2} + \frac{g(x) - g(-x)}{2}$$

where the first term is the even part and the second term is the odd part, both of which can be chosen arbitrarily. The problem conditions stipulate the even part and the odd part $h(x)$ is subject to the condition $h(1/2) = \frac{g(1/2)}{2}$. So we can pick an arbitrary odd function for $h$ and then everything else is determined, e.g. if $h(x) = x^3$ then $g(x) = \frac{1}{8} + x^3$.

So the meta-problem is "what conditions on an odd function are strong enough to make it equal to $h(x) = x$?" and I don't think this is a reasonable or interesting question to ask unless you have something very specific in mind.

This problem is not actually much like the Cauchy functional equation because there is only one free parameter instead of two.