Functional equation (with composition) Suppose that $f(1-f(x))=x$  for all $x \in \mathbb{R}$.   There exists a function $f$ with such a characteristic? Is it unique?
My progress: I got the following results but I'm not able to conclude or continue.
-> $f(f(x))=1-x$ and $f(f(1-x))=x$ for all $x$
-> $f(x)+f(1-x)=1$ for all $x$
-> $f^n(1/2)=1/2$
-> $f^n(x)+f^n(1-x)=1$
(Power means composition)
Any suggestions?
 A: There is no continuous solution.
Let $g(x) = 1 - f(x)$.  Then the equation can be written as
$$ g(g(x)) = 1 - x $$
Since $x \to 1-x$ is one-to-one, so is $g$.
Now if $f$ is continuous, so is $g$.  A continuous one-to-one function on $\mathbb R$  is monotonic, either increasing or decreasing.  But then $g \circ g$ is increasing, while $x \to 1-x$ is decreasing.  
EDIT:
There are discontinuous solutions.  Let $\alpha$ be a one-to-one function
from $(-\infty,0]$ onto $(0,1/2)$, and take $g(x) = \alpha(x)$ for
$x \in (-\infty,0]$.  
For $x \in (0,1/2)$ take $g(x) = 1 - \alpha^{-1}(x)$.
This maps $(0,1/2)$ one-to-one onto $[1,\infty)$.
Thus for $x \in (-\infty, 0]$ we have $g(g(x)) = 1 - \alpha^{-1}(\alpha(x)) = 1-x$.  
For $x \in [1,\infty)$ we take $g(x) = 1 - \alpha(1-x)$.  This maps $[1,\infty)$ one-to-one onto $(1/2,1)$.
Thus for $x \in (0,1/2)$ we have $g(g(x)) = 1 - \alpha(1-(1-\alpha^{-1}(x))) = 1 - \alpha(\alpha^{-1}(x)) = 1 - x$.
For $x \in (1/2, 1)$ we take $g(x) = \alpha^{-1}(1-x)$.  This maps $(1/2,1)$ one-to-one onto $(-\infty,0]$.  Thus for $x \in [1,\infty)$ we have
$g(g(x)) = \alpha^{-1}(1-(1-\alpha(1-x))) = \alpha^{-1}(\alpha(1-x))= 1-x$.
For $x \in (1/2, 1)$, we have $g(g(x)) = \alpha(\alpha^{-1}(1-x))=1-x$.
Finally, take $g(1/2)=1/2$.
