Find $f(x)$ satisfy $f(2x)=2f(x)+x$

Question

I would appreciate if somebody could help me with the following problem:

Find $f(x)$, given that:

$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous at $x=0$, and $f(2x)=2f(x)+x$

I tried but couldn’t get it that way.

Answer 1

Hint: put $g(x) = \frac{f(x)}{x}$ ($x \ne 0$), and define $f(0)$ suitably.

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To be more explicit, consider the function $f$ defined by: $$ f(x) = \begin{cases} 0 & (x = 0) \\ \frac{x}{2}\log_2\left\lvert{x}\right\rvert & (x \ne 0) \end{cases} $$ Every function $f_a:x\mapsto f(x)+ax$ solves the problem and is continuous everywhere.

Answer 2

I have an ansatz, but I can neither show how to derive it from your functional equation nor prove that it is the only solution. Anyway, here it is:

$$f(x)=ax+\dfrac{x\ln(x)}{\ln{4}}$$

Which gives: \begin{align}f(2x)&=2ax+\dfrac{2x\ln{(2x)}}{\ln{4}}\\&=2ax+\dfrac{2x\ln{x}}{\ln{4}}+\dfrac{2x\ln{2}}{\ln{4}}\\&=2f(x)+2x\cdot\dfrac{1}{2}\\&=2f(x)+x\end{align}

as requested.

(Note that to move from line 1 to 2, we used $\ln{(ab)}=\ln{a}+\ln{b}$, and from line 2 to 3, $\ln{4}=\ln{2^2}=2\ln{2}$).