# Another functional equation: $f(x) + f(2x) + f(4x) = \lfloor 7x \rfloor$

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 ?

• Unless I'm missing something, there is no such function. From continuity of $f$ and the functional equation, it would follow that $x\mapsto\lfloor7x\rfloor$ is also continuous, which is not the case. – Dejan Govc Mar 18 '14 at 18:59
• $f(x)+f(2x)+f(4x)$ is continuous. $\lfloor 7x\rfloor$ is not continuous. Contradiction. – Guy Mar 18 '14 at 19:04
• Thanks, you're right ! I was thinking too complicated... – Odile Mar 19 '14 at 11:17
• @DejanGovc post as answer? – user574848 May 24 at 11:21
• @user574848: Done. – Dejan Govc May 24 at 12:17

There is no such function. From continuity of $$f$$ and the functional equation, it would follow that $$x\mapsto\lfloor7x\rfloor$$ is also continuous, which is not the case.