Finding Continuous Functions Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that for all $x \in \mathbb{R}$, $f(x) + f(2x) = 0$ 
I'm thinking; 
Let $f(x)=-f(2x)$ 
Use a substitution $x=y/2$ for $y \in \mathbb{R}$. 
That way $f(y)=-f(y/2)=-f(y/4)=-f(y/8)=....$ 
Im just not sure if this is a good approach. Opinions please..
 A: Hints:


*

*What is $f(0)$?

*What does continuity when $x=0$ mean?

*What bounds does this put on $f(x)$ for general $x$?

A: By induction we prove that
$$f(x)=(-1)^nf\left(\frac x{2^n}\right)$$
but by the sequential characterization of the continuity we have
$$f\left(\frac x{2^n}\right)\xrightarrow{n\to\infty}f(0)$$
so since for $x=0$ we get  $f(0)=0$ so $f(x)$ is constant 
$$f(x)=f(0)=0\quad\forall x\in\Bbb R$$
A: Note that $$f(0)+f(2\cdot 0)=f(0)+f(0)=0\implies f(0)=0.$$
Consider $x_0\ne 0.$ It is
$$f(x)+f(2x)=0\underbrace{\implies}_{x=x_0/2} f(x_0/2)=-f(x_0),$$
$$f(x)+f(2x)=0\underbrace{\implies}_{x=x_0/4} f(x_0/4)=-f(x_0/2)=f(x_0).$$
By induction, $f(x_0/2^n)=-f(x_0)$ if $n$ is odd and $f(x_0/2^n)=f(x_0)$ if $n$ is even. Since $f$ is continuous at $0$ it is:
$$0=f(0)=\lim_{n\to\infty} f(x_0/2^{2n})=f(x_0).$$
Thus, $f\equiv 0$ is the only solution.
Thus 
A: $f(x)=0$ is the only solution.
$$f(0)=-f(2\cdot 0) \iff f(0)=0$$
Suppose that for any $x$,  $f(x) = \epsilon$. But this means that for any $\delta$ and for large $n$ we have:
$$\frac{x}{2^n} < \delta$$
$$\large f\bigg(\frac{x}{2^n}\bigg)= (-1)^n\cdot\epsilon$$
But supposing $f$ was continuous, we could choose a $\delta$ small enough that $|f(x)|<\epsilon$ for $x< \delta$. Thus if this function is continuous, it can have no values other than $0$.
