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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 ?

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    $\begingroup$ 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. $\endgroup$ – Dejan Govc Mar 18 '14 at 18:59
  • $\begingroup$ $f(x)+f(2x)+f(4x)$ is continuous. $\lfloor 7x\rfloor$ is not continuous. Contradiction. $\endgroup$ – Guy Mar 18 '14 at 19:04
  • $\begingroup$ Thanks, you're right ! I was thinking too complicated... $\endgroup$ – Odile Mar 19 '14 at 11:17
  • $\begingroup$ @DejanGovc post as answer? $\endgroup$ – user574848 May 24 at 11:21
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    $\begingroup$ @user574848: Done. $\endgroup$ – Dejan Govc May 24 at 12:17
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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.

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