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

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