Functional equation with $f(2x)$

Question

Any other solutions(advice) are welcome.

For any $x>0, \;\;\; 2f(\frac{1}{x}+1)+f(2x)=1$
Find all possible f(x).

I wish you luck on a good thing in $2019$.

Answer 1

Yes, but not quite like that.

Say you'd define the function arbitrarily for $x\in(0,2]$. Also, you know how to get $f\left({2\over x}+1\right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $f\left(5\over3\right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.

To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $x\in(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,\infty)$. This in turn gives a piece of $\left(1,{5\over3}\right]$ which sticks nicely to $(0,1]$. Then we get $\left[{11\over5},3\right)$, and so on.

By applying the same equation in reverse, we extend our domain in the different direction, until all $\mathbb R$ is covered.

Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.