Functional equation $g(2x )= 1/2 g(x)$

Question

I am trying to solve functional equation for $g: \ (0, \infty) \mapsto ( 0, \infty)$ $$ g( 2x ) =\frac 12 g(x)$$ Wolfram claims, and it is intuitive, that the function is $g(x) = C \frac{1}{x}$. But how to prove it?

Answer 1

Even assuming continuity, there are many more solutions: Let $h\colon[1,2]\to(0,\infty)$ be any continuous function with $h(2)=\frac12h(1)$. Then $$ g(x)=2^{\lfloor\log_2 x\rfloor}h(2^{1+\lfloor \log_2 x\rfloor}/x)$$ is continuous and obeys your functional equation. Similarly, we can find many arbitrarily smooth $g$.

Answer 2

Assuming that $g(x)$ has a Laurent series as $\;g(x)=C/x + \sum_{n=0}^\infty c_n x^n\;$ then by comparing coefficients show that $\;c_n=0\;$ for all $\;n\ge0\;$ because $\;g(2x)=C/(2x)+\sum_{n=0}^\infty c_n 2^n x^n\;$ and $\;\frac12g(x)=C/(2x)+\sum_{n=0}^\infty \frac12c_n x^n\;$ implies $\;c_n2^n=\frac12c_n\;$ and $2^n\ne\frac12\;$ for all $\;n\ge0$.