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 \(n\to +\infty \),
\[f\left ( x \right )=\frac{x}{1-\frac{1}{3^{2}}}=\frac{9}{8}x\]


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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