What function satisfy: $f(x)+f^{-1}(x)=2x$? What function satisfy: $f(x)+f^{-1}(x)=2x$?
I have tried to substitute $x=f(x)$ to get $f^{(2)}(x)+1=2f(x)$ and subsequently plug in values to try to find $f(x)$ but to no avail.
Please help thank you in advance!
 A: I think that I can prove that (under some conditions),
linear functions $f(x) = x + c$ are the only solutions.
The equation (implicitly) assumes that
$f$ has an inverse function $f^{-1}$ with the same domain as $f$.
I am only considering bijective functions $f :  \mathbb R \to \mathbb R$ here. Note that such functions are strictly monotonic if and only if they are continuous.
My claim is: 

Let $f : \mathbb R \to \mathbb R$ be continuous and bijective
  such that
  $$ \tag 1
f(x) + f^{-1}(x) = 2x \text{ for all } x \in \mathbb R \, .
$$
Then $f(x) = x+f(0)$ for all $x \in \mathbb R$.

If $f(x) \equiv x$ then we are done, so let's assume that $f(a) \ne a$
for some $a \in \mathbb R$. Without loss of generality we can
assume that $f(a) > a$, otherwise consider $g := f^{-1}$ instead of $f$.
So
$$
  d := f(a) - a > 0 \, .
$$
It follows from $(1)$ that
$$
f^{(2)}(x) = 2 f(x) - x \tag 2
$$
and then via induction for all $n \in \mathbb N$
$$
f^{(n)}(x) = n f(x) - (n-1)x \, .
\tag 3 
$$
In particular,
$$
   f^{(n)}(a) = n\,f(a) - (n-1) a = n (a+d) - (n-1) a = a + nd \, .
$$
Applying the same
calculation to $f^{-1}$ gives
$$
f^{(-n)}(a) = a - nd \, .
$$
Together it follows that
$$
 f(a + kd) = a + (k+1)d \tag 4
$$
for all $k \in \mathbb Z$.
$f$ is strictly monotonic, so $(4)$ implies that each
interval from $a + kd$ to $a + (k+1)d$ is mapped onto the "next" interval:
$$
 f \bigl([a + kd, a + (k+1)d]\bigr) = [a + (k+1)d, a + (k+2)d] \tag {5}
$$
Now let $x \in \mathbb R$ and choose $k \in \mathbb Z$ such that
$$
 a + kd \le x < a + (k+1)d \, . \tag 6
$$
For all $n \in \mathbb N$ it follows from $(5)$ and $(6)$ that
$$
x + (n-1)d \le a + (k+n)d \le f^{(n)}(x) \le a + (k+n+1)d \le x + (n+1) d \, .
$$
Substituting this in $(3)$ gives
$$
x + (n-1)d \le n f(x) - (n-1)x \le x + (n+1) d
$$
or 
$$
\frac{nx + (n-1)d}{n} \le f(x) \le \frac{n x + (n+1)d}{n} \, .
$$
Finally, $n \to \infty$ gives
$$
 x + d \le f(x) \le x + d \, .
$$
