Functional Equation satisfying f(2x)=f(x)

Question

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 $$\lim_{n \to \infty} f\left(x_n\right)=f\left(\lim_{n \to \infty}x\right)=f\left(0\right)$$

Answer 1

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

$$(\forall x\in\mathbb R) \;\;f (x)=f (0) $$

Answer 2

Not an answer, but too long for a comment.

Notice that the answer would be very different with $f$ continuous everywhere but at $x=0$.

We have (with $\lg x:=\log_2 x$)

$$f(2^{\lg x+1})=f(2^{\lg x}).$$

Then let

$$g:[1,2]\to\mathbb R:t\to g(t)$$ be an arbitrary continuous function such that $g(2)=g(1)$.

Now,

$$f(x):=g(2^{\{\lg x\}})$$ is a continuous solution for the positive $x$ (the curly braces denote the fractional part). A similar solution can be written for the negative $x$.

A simple example below: