# Solving functional equation $f(4x)-f(3x)=2x$

Given that $f(4x)-f(3x)=2x$ and that $f:\mathbb{R}\rightarrow\mathbb{R}$ is an increasing function, find $f(x)$. My thoughts so far: subtituting $\frac{3}{4}x$, $\left(\frac{3}{4}\right)^2x$, $\left(\frac{3}{4}\right)^3x$, $\ldots$, we get that: $$f(4x)-f(3x)=2x$$ $$f\left(4\cdot\frac{3}{4}x\right)-f\left(3\cdot\frac{3}{4}x\right)=2\cdot\frac{3}{4}x$$ $$f\left(4\cdot\left(\frac{3}{4}\right)^2x\right)-f\left(3\cdot\left(\frac{3}{4}\right)^2x\right)=2\cdot\left(\frac{3}{4}\right)^2$$ $$f\left(4\cdot\left(\frac{3}{4}\right)^3x\right)-f\left(3\cdot\left(\frac{3}{4}\right)^3x\right)=2\cdot\left(\frac{3}{4}\right)^3$$ $$\ldots$$ Then note that after adding all these equations we get: $$f(4x)=\sum_{k=0}^{\infty}\left(\frac{3}{4}\right)^k2x$$ And this series obviously converges to $8x$. Substituting $\frac{1}{4}x$ we get: $$f(x)=2x.$$ Is this correct? ;)

• This is only one solution, not all possible solution. eg. $f(x) = 2x+ 1$ or even strange beast like $f(x) = \begin{cases} 2x + 1,& x > 0\\ 0, & x = 0\\ 2x -1,& x < 0\end{cases}$. – achille hui Dec 30 '15 at 21:43
• So it should be $f(x)=x+c$, where $c$ is some constant. So... what should I modify in my solution? – VanDerWarden Dec 30 '15 at 21:44
• Your solution assumes that $\lim_{x\to 0}f(x)=0$ when "adding up all these equations". – DanielWainfleet Dec 30 '15 at 22:56

It's almost correct. You are correct that we can deduce that $$f(4x)=f(3x)+2x$$ and by repeatedly applying this we get $$f(4x)=f\left(4\left(\frac{3}4\right)^kx\right)+\sum_{n=0}^{k-1}2\left(\frac{3}4\right)^{n}x$$ You make an error on the next step, however. You mean to take a limit as $k$ goes to infinity, but you to this incorrectly. In particular, the correct expression would be: $$f(4x)=\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)+8x$$ where we have a term of $\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)$ that you omitted; in particular, this can be any constant, and the constant can be different for positive and negative numbers. It does, however, exist since $f$ is increasing. Thus, the solutions are of the form: $$f(x)=\begin{cases}2x+c_1&&\text{if }x>0\\c_2&&\text{if }x=0\\2x+c_3&&\text{if }x<0\end{cases}$$ for some constants $c_1\geq c_2 \geq c_3$.