Schroeder's equation: simple observation

[phoenixthoth]

i was wondering if you could check this simple observation; there must be something wrong with it.

schroeder's equation is a functional equation. one version is given f, say a self-mapping of R, find bijection g (defined on R) and nontrivial constant r such that g\circ f=rg.

one utra nice result of solving this equation is that the n-th iterate of f can be calculated:
f^{n}\left( x\right) =g^{-1}\left( r^{n}g\left( x\right) \right).

for example, if one is looking to solve the difference equation
a_{n+1}=f\left( a_{n}\right)
then the solution is
a_{n}=f^{n}\left( a_{0}\right), where the nth iterate of f is given above.

clearly, if schroeder's equation is solvable, then the difference equation is too. it isn't as clear to me that if the difference equation is solvable then schroeder's equation is solvable.

turning to conjugacy, let's say for the sake of argument that two self-mappings of R, f and h, are conjugate if there is a bijection g such that
f\circ g=g\circ h.

let Fix(f) denote the set of fixed points for the function f with respect to a given domain (let's stick to R). one can note that if f and h are conjugate, then Fix(f) has the same cardinality as Fix(h).

going back to the main equation, g\circ f=rg, let's assume that there are function g and constant r that satisfies this equation. then compose both sides with g^{-1} to get
g\left( f\left( g^{-1}\left( x\right) \right) \right) =rx. this equation holds for all x in R.

this means that f is conjugate to the linear function h(x)=rx which has one fixed point (we rule out the case when r=1).

since Fix(f) has the same cardinality as Fix(h), they being conjugates, f also has one fixed point.

in short, if schroeder has a solution for the given f then f has one fixed point.

the contrapositive of that is that if f does not have one fixed point then schroeder does not have a solution for the given f.

example: f(x)=2^x.

i'm mainly trying to show that 2^x doesn't admit a solution on all of R. this seems too trivial to be right because only a few functions would admit solutions to schroeder's equation...

WHAT GIVES!?

 

[fresh_42]

$f(x)=2x$ admits any linear $g$ as solution with $r=2$: $(g\circ f)(x)=g(f(x))=g(2x)=2g(x)$.