Continuous functions satisfying $f(x)+f(2x)=0$? I have to find all the continuous functions from $\mathbb{R}$ to $\mathbb{R}$ such that for all real $x$,
$$f(x)+f(2x)=0$$
I have shown that $f(2x)=-f(x)=f(x/2)=-f(x/4)=\cdots$ etc. and I have also deduced from the definition of continuity that for any $e>0$, there exists a $d>0$ so if we have that:
$|2x-x|=|x|< d$, this implies that
$|f(2x)-f(x)|=|-2f(x)| < e$.
Is this the correct way to begin? And if so, how should I continue?
Thank you!

first of all for $x=0$ we have
$$f(0)+f(2\cdot0)=0\Leftrightarrow f(0)=0$$
On the other hand 
$$f(x)+f(2x)=0$$
$$-f(2x)-f(4x)=0$$
$$f(4x)+f(8x)=0$$
$$.......$$
$$f(2^nx)+(-1)^nf(2^{n+1}x)=0$$
Adding both sides respectively yields
$$f(x)+(-1)^nf(2^{n+1}x)=0\Rightarrow f(x)=(-1)^{n+1}f(2^{n+1}x)$$
The LHS is continuous by the hypothesis of the problem, however the RHS is alternating in sign depending on the value of $n$ and hence not continuous, unless $f(x)\equiv0$ for all $x\in\mathbb{R}$.
