I would like to find all continuous functions $f \, : \, \mathbb{R} \, \longrightarrow \, \mathbb{R}$ such that :
$$ \forall x \in \mathbb{R}, \; f(x) + f(2x) + f(4x) = \lfloor 7x \rfloor \tag{1}$$
Follow-up :
Now that I know there are no continuous functions satisfying $(1)$, I would like to find all functions $f \, : \, \mathbb{R} \, \longrightarrow \, \mathbb{R}$, continuous at $0$, such that :
$$ \forall x \in \mathbb{R}, \; f(x) + f(2x) + f(4x) = x \varphi(x) \tag{2} $$
where $\displaystyle \forall x \in \mathbb{R}, \; \varphi(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$. I feel like there exist no such functions (but I might be mistaken). Here, both RHS and LHS are continuous at $x=0$. My try :
It is clear that $f(0)=0$. Since $\varphi(qx)=\varphi(x)$ for all $x \in \mathbb{R}$ and all $q \in \mathbb{Q}$, I think the following is true :
$$ f(8x) - f(x) = x \varphi(x) $$
which leads to :
$$ f(x) - f \Big( \frac{x}{8} \Big) = \frac{x}{8} \varphi(x) $$
which would lead to
$$ f(x) - f \Big( \frac{x}{2^{3k}} \Big) = \frac{x}{2^{3k}} \varphi(x) $$
Am I on the right track or is there an easier way ?