Functional Equation Mood II

It is easy to see that any function of the form $f(x)=\frac{2}{3}x+g(x)$ is a solution of the given functional equation, where $g(x)+g(\frac{1}{2}x)=0.\quad\quad(*)$ It can be proved that there are infinitely many possible functions $g(x).$ For example, for any real number $c\neq 0,$ we can define a function $g_c$ as a piecewise function in the following way: $g_c(2^n)=(-1)^n c,$ for any integer $n,$ and $g_c(x)=0,$ if $x$ is any real number that cannot be represented as $2^n,$ where $n$ is any integer number. It is obvious that any $g_c$ satisfies the equation $(*)$ and the number of possible function of this type is infinite, because if $c_1\neq c_2,$ then $g_{c_1}\neq g_{c_2}.$

For any problem there’s always infinite amount of solutions. Some of us just can’t find it.

Infinite solution, because the variable x can be any numbers, any random numbers.