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!
 A: I will show that the only such function is $$f(x)=0$$

Consider the case $x=0$, so 
$$f(0) + f(2 \cdot 0 ) = 0$$
This means $f(0) =0$.

Suppose, for contradiction, that there exists a real number $a$ such that $f(a) \neq 0$
Since $a/2 \in \mathbb{R}$, we have
$$f(a/2)+f(a) = 0$$
This means
$$f\left(\frac{a}{2}\right) =-f(a)$$
Likewise we can show
$$f\left(\frac{a}{2^n}\right) = (-1)^n \cdot f(a)$$
Now consider the sequence whose terms are given by$$x_n = f\left(\frac{a}{2^n}\right)$$
Note that $$\lim_{n\to\infty} \frac{a}{2^n} = 0$$
This means that if $f$ is continuous, $$\lim_{n\to \infty} x_n$$
should equal $f(0)=0$. But the terms $x_n$ alternate between $f(a)$ and $-f(a)$ and never decrease in absolute value. Contradiction.

Therefore, $f(x)=0$ is the only continuous function that satisfies $f(x) = f(2x)$ for all $x \in \mathbb{R}$. 
A: That's correct, but you're only writing that f(0)=0 :) , that's not enough.
f is continuous, and using induction you get: $f(x) = (-1)^nf(\frac{x}{2^n}) $ for any positive interger n.
You can set x=0 to get : f(0) = 0 
Using continuity of f, you have : $f(\frac{x}{2^n}) \rightarrow f(0) = 0 $ when $n \rightarrow +\infty$ for any x real number.
Hence for every real x , you have : |f(x)| = $|f(\frac{x}{2^n})| =|\lim_{n \rightarrow +\infty}  f(\frac{x}{2^n})| = f(0) = 0$
So you have found your function
A: 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 $x\in\mathbb{R}$.
