Find all functions by condition

Question

How to find all functions continuous on $R$ by condition: $$f(x)+f(2x)=6x+1$$

If I assume that the function is linear. So $f(3x)=6x+1$. But then how to find the coefficients? Through derivative?
And if the function is nonlinear? How then to act?

Answer 1

You should not start by assuming that $f$ is linear.

hint

First observe that $f(0)=1/2$. Let $g(x)=f(x)-2x-1/2$. Then the given functional equation can be written as $$g(x)+g(2x)=0$$ Note $g(x)$ is also continuous. So try to show that $g(c)=\lim_{n \to \infty}(-1)^ng\left(\frac{c}{2^n}\right)=g(0)=0$ for any $c \in \Bbb{R}$.

Answer 2

$$\begin{align} f(x)&=3x+1-f(x/2)= \\ &=3x+1-(3x/2+1-f(x/4))=3(x-x/2)+f(x/4) \\ &=3(x-x/2)+(3x/4+1-f(x/8))=3(x-x/2+x/4)+1-f(x/8) \\ &=3(x-x/2+x/4)+1-(3x/8+1-f(x/16))=3(x-x/2+x/4-x/8)+f(x/16)... \,. \end{align} $$ In general for odd $N$ $$f(x)=3x\sum_{n=0}^N \frac{(-1)^n}{2^n}+f(x/2^{N+1}), $$ and letting $N \to \infty$ we get $$f(x)=2x+f(0). $$