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Just had a midterm with the following problem:

Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.

I was just curious how this would end up.

During the exam I tried setting $f(1) = c$, and end up getting $f(3) = 2c, f(9) = 4c, \ldots$ and from that $c = 3/2$ with the $f(0) = 1$ condition, but I was unable to make more progress on it.

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We have: $f(x) - f\left(\frac{x}{3}\right) = \dfrac{x}{3}$

$f\left(\frac{x}{3}\right) - f\left(\frac{x}{9}\right) = \dfrac{x}{9}$

.....

$f\left(\frac{x}{3^{n-1}}\right) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3^n}$.

Add the above equations:

$f(x) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3}\cdot \left(1 + \frac{1}{3} + ... + \frac{1}{3^{n-1}}\right)$.

Letting $n \to \infty$ in the above equation and using continuity of $f$ at $x = 0$ we have:

$f(x) - f(0) = \dfrac{x}{3}\cdot \dfrac{1}{1-\frac{1}{3}} = \dfrac{x}{2}$. Thus:

$f(x) - 1 = \dfrac{x}{2}$, and $f(x) = 1 + \dfrac{x}{2}$

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Notice that: \begin{align*} f(3x) - f(x) &= x \\ f(9x) - f(3x) &= 3x \\ f(27x) - f(9x) &= 9x \\ &~~\vdots \\ f(3^kx) - f(3^{k-1}x) &= 3^{k-1}x \end{align*} Summing them together, the LHS telescopes, giving us: $$ f(3^kx) - f(x) = x\sum_{i=0}^{k-1}3^i = x \cdot \frac{3^k - 1}{2} $$ Now consider what happens when we take the limit as $k \to -\infty$. By the continuity of $f$, we can move the limit inside the function, giving us: \begin{align*} f\left( \left[\lim_{k\to-\infty}3^k\right]x \right) - f(x) &= x \cdot \frac{\left[\lim\limits_{k\to-\infty}3^k\right] - 1}{2} \\ f(0) - f(x) &= x \cdot \frac{0 - 1}{2} \\ 1 - f(x) &= \frac{-x}{2} \\ f(x) &= 1 + \frac{x}{2} \end{align*}

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An 'obvious' solution is $$ f(x) =\frac x 2 +1 $$

Notice that iterating the relation you get $$ f(3x)=x+f(x)=x+\frac x 3 + f(\frac x 3) = \sum\limits_{k=0}^{+\infty}\frac{ x}{3^k} +f(0)=x\frac{1}{1-\frac 1 3} + f(0)=\frac {3x}2 + f(0) $$ hence the solution is unique.

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