We have $f(x)+g(x)=g(2x)$. We assume that $f$ hasn't root in (0,2016) so f keep constant sign.WLOG $f>0$. We have $f(x)+g(x)=g(2x)$ $f(2x)+g(2x)=g(4x)$ . . $f(2^{n-1}x)+g(2^{n-1}x)=g(2^nx)$ so we have $f(x)+f(2x)+...+f(2^{n-1}x)=g(2^nx)-g(x)$.We put $2^nx_{n}=2016$ so we have $f(x_{n})+f(2x_{n})+...+f(2^{n-1}x_{n})=g(2^nx_{n})-g(x_{n})=-g(x_{n})$and when $\lim_{n\to \infty}g(x_{n})=0$ but $f(x_{n})+f(2x_{n})+...+f(2^{n-1}x_{n})>f(1008)>0$ so we arrive at a contradiction.

Nice solution.