An Interesting High-School Maths Problem

Question

Given $$f\left ( x \right )-\frac{1}{3}f\left ( \frac{x}{3} \right )=x,$$
find $f\left ( x \right )$.

Alternative solution:

We derive the following equations,
$$f\left ( x \right )-\frac{1}{3}f\left ( \frac{x}{3} \right )=x$$
$$\frac{1}{3^{1}}f\left ( \frac{x}{3} \right )-\frac{1}{3^{2}}f\left (\frac{x}{3^{2}}\right )=\frac{x}{3^{2}}$$
$$\frac{1}{3^{2}}f\left ( \frac{x}{3^{2}} \right )-\frac{1}{3^{3}}f\left (\frac{x}{3^{3}}\right )=\frac{x}{3^{4}}$$
$$\vdots$$
$$\frac{1}{3^{n}}f\left ( \frac{x}{3^{n}} \right )-\frac{1}{3^{n+1}}f\left (\frac{x}{3^{n+1}}\right )=\frac{x}{3^{2n}}$$
where $n\in \mathbb{N}$.

By adding up both sides of the equation, we obtain
$$f\left ( x \right )-\frac{1}{3^{n+1}}f\left (\frac{x}{3^{n+1}}\right )=x+\frac{x}{3^{2}}+\frac{x}{3^{4}}+...+\frac{x}{3^{2n}}$$
Taking $n\to +\infty$,
$$f\left ( x \right )=\frac{x}{1-\frac{1}{3^{2}}}=\frac{9}{8}x$$

Solution:

From the original equation, let $x=0$, we obtain $$f\left ( 0 \right )=0$$
Construct $\alpha$ such that $$f\left ( x \right )+\alpha\cdot x=\frac{1}{3}\left [ f\left ( \frac{x}{3} \right )+\alpha\cdot \frac{x}{3} \right ]$$

We attain $$\alpha\ = -\frac{9}{8}$$

Let $g\left ( x \right ) = f\left ( x \right )-\frac{9}{8}x$, therefore

$$g\left ( x \right ) = \frac{1}{3}g\left ( \frac{x}{3} \right )=\cdots=\frac{1}{3^{n}}g\left ( \frac{x}{3^{n}} \right ), n\in \mathbb{Z^{+}}$$

If $n\to +\infty$,

$$\lim_{n \to +\infty}g\left ( x \right )=g\left ( 0 \right )=f\left ( 0 \right )-0=0$$

Thus, $$f\left ( x \right )=\frac{9}{8}x$$