$f(x) + f(1-x) = f(1)$ 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.
 A: 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.
A: 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 $\displaystyle [0,1]$.
A: In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf.
The general solution is $f(x)=\dfrac{f(1)}{2}+C(x,1-x)$ , where $C(u,v)$ is any antisymmetric function.
A: Whilst this was a post from a while ago; what you are looking at is a symmetric probability function in some cases (under the definition of Segal 1993)
And I also noticed that your edited solution would work, a while ago. ie 
Is the Symmetry a bi-conditional injective claim as well, ie as an iff, claim. ie inverse symmetry as well F-1(y)+F-1(1-y)=f-1(1)=1 ? and is it midpoint convex? That might be weaker, as it may give you the continuity via the connection between midpoint convexity and convexity, with F strictly monotonically increasing and bounded.
Where if F is also midpoint convex,  inverse symmetric and and Strictly monotonic Increasing;F:[0,1] to [0,1], 
Then given your  symmetry, condition,  along with F(1)=1; ,  F(0)=0, F(1/2)=1/2; which just fall out. 
it should effectively be you  jensen's equality Often often a midpoint convex symmetric function, that is strictly monotonic increasing, F(1)=1, F:[0,1] to [0,1] can be only be so, iff it is midpoint concave as well; ie jensen's equality. At least if you have inverse symmetry as well F-1(p)+F-1(1-p)=1 
one can get F(2x)=2F(x), and i think F(nx)=n F(x) which generalizes to all integers if not all rationals, numerically,  sub-additivity and effectively super-addivitiy and cauchy equation, and jensens equality/or rather conditional cauchy F(x+y)=F(x)+F(y) for all pairs, 
Something which will have the same effect under continuity to cauchy equation or to f(x)=x, already I think may be induced by convexity.
Where midpoint convexity given strictly monotone increasing and the other conditions might already give you convexity, and thus continuity etc, F(x)=x already, although I am not sure.
Maybe a mild differentiation or measurability constraint is needed  as well. 
Although strict monotone increasing generally implies strict quasi convexity and quasi convexity (quasi concave and strict as well), where strict quasi convexity and midpoint convexity very often entails convexity, and thus continuity on [0,1] *lipschitz continuity I believe).
And it would presumably be concave as well in virtue of symmetry.
. 
ie, if you have 
(1).F:[0,1] to [0,1]
(1.1) and F(1)=1, {F(0)=0, F(1/2)=1/2);
(1.2) symmetry F(1-x)+F(x)=F(1)=1.
and in addition you have (2.a) (2.b) and (3)
(2a.)  and F is strictly monotonic increasing function ( and thus, injective function) 
(2.b) (inverse symmetry) for all y in range of f,  f-1(y)+F-1(1-y)=F-1(1)=1. 
**(3). F is  midpoint Convex F(x/2+y/2)<=F(x)/2+F(y)/2
A: In fact if F strictly monotonic; with $F(1)=1$,
 and thus $F(0.5)=0.5$ as a result then midpoint convexity just at  at 0 and 1 (mid star convexity at zero or one)
Would be sufficient, if $F:[0,1]\to [0,1]$; or just star convexity, given F(1)=1 and $F(1-x)+F(x)=F(1)=1$ just at $0$ for all real t, at zero alone.
Not sure if strictly monotonic increasing or inverse symmetry or $F^{-1}(1-x)+F^{-1}(x)=F^{-1}(1)=1$ is necessary. The inverse version of your condition essentially, making it into a bi-conditional claim.
with $F$ being$(A)$ 'Locally midpoint Jensen convex', as well as $(B)$ F strict monotone increasing. 
$(A1)$ Almost approximately convex might work with $(1)\,(2)\,(3)\,(4.1), (4.2)$ it may work.F
$$(1)F:[0,1]\to [0,1]\,\text{where}\,F(1)=1,\text{where F(0)=0, F(0.5) (if not already implicit)}$$
$$(2)\forall (p_1,p_2)\,\in \text{Im};\,(F)F^{-1}(1-p)+F^{-1}(p)=F^{-1}(1)=1\,\text{where p_2=1-p}$$
$$(3)\forall (x_1,x_2)\,\in \text{dom}(F);\,F(1-x)+F(x)=F(1)=1\,\text{where x_2=1-x}$$
$$(4.1)\forall (x_1,x_2)\,\in \text{dom}(F);\,[F(x_1)+F(x_2)>1 \leftrightarrow [x_1 +x_2>1]\,\land\quad\,[F(x_1)+F(x_2)<1] \leftrightarrow \,[x_1 +x_2]<1]$$
$$(4.2)\forall (p_1,p_2) \in \text{Im}(F);\,[F^{-1}(x_1)+F^{-1}(x_2)>1] \leftrightarrow\, p_1 +p_2>1\,\land\,[F^{-1}(p_1)+F^{-1}(p_2)<1] \leftrightarrow [p_1 +p_2]<1]$$
$(5)$ where $F$ strictly pseudo-linear or is twice/thrice  differentiable with a non vanishing first derivative. Not sure if log concavity or anything weaker would help replace the convexity condition; mid concavity or local mid concavity (would do the same job) as would star concavity; or super/sub-additivity by itself would work. It may
