<s>[INST] Functional Equation satisfying f(2x)=f(x) Determine all contimouse functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f\left(2x\right)=f\left(x\right)$
Attempt
Let $a$ and $b$ be two distinct points on $\mathbb{R}$.
Then consider the sequences of real numbers
$$\{x_n\}=\frac{a}{2},\frac{a}{2^2},\frac{a}{2^3},\frac{a}{2^4} \dots\frac{a}{2^n}$$
$$\{y_n\}=\frac{b}{2},\frac{b}{2^2},\frac{b}{2^3},\frac{b}{2^4} \dots\frac{b}{2^n}$$
We now have 
$$f\left(a\right)=f\left(\frac{a}{2}\right)=f\left(\frac{a}{2^2}\right)=f\left(\frac{a}{2^3}\right) \dots f\left(\frac{a}{2^n}\right) \dots= f\left(0\right)$$
Simmilarly
$$f\left(b\right)=f\left(\frac{b}{2}\right)=f\left(\frac{b}{2^2}\right)=f\left(\frac{b}{2^3}\right) \dots f\left(\frac{b}{2^n}\right) \dots= f\left(0\right)$$
Hence $f\left(a\right)=f\left(b\right)=f\left(0\right)$
Hence $f$ is constant function.
Note we use the continuity to establish that $$\lim_{n \to \infty} f\left(x_n\right)=f\left(\lim_{n \to \infty}x\right)=f\left(0\right)$$
 [/INST] Let $x\ne 0$. for $n\in\mathbb N,$
$$f (x)=f (\frac {x}{2^n} )$$
$f $ is continuous at $0$ and $$\lim_{n\to+\infty}\frac {x}{2^n}=0$$
$$\implies \lim_{n\to+\infty}f (\frac {x}{2^n})=f (0) $$
thus
$$(\forall x\in\mathbb R) \;\;f (x)=f (0) $$
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