Solving Functional Equations by Limits ...

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Let $f(x)$ be a continuous function and satisfying the equation :

$f(2x) - f(x) = x$. Given $f(0)=1$ ; Find $f(3)=?$

My teacher solves this as :

$$f(x) - f(x/2) = x/2$$ $$f(x/2) - f(x/4) = x/4$$ .

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$$f(x/2^{n-1}) - f(x/2^{n}) = x/ 2^{n-1}$$

...........…...... Add them up : And let $n\rightarrow \infty$

$$f(x) - f(x/2^{n}) =(x/2)/(1-(1/2))$$

Thus $f(x) - f(0) = x$.

Thus $f(x) = x+1$. Thus $f(3)= 4$.

However my problem is that I didn't find it intuitive ; I didn't understand how to get such an idea.

So is there an alternate way to go about this problem ?

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I would say,

search function $f(x) = x+\mu(x), \quad \mu(x)$ continuous for $x\in R,\,\mu(0)=1$:

$f(2x)-f(x)=x\Rightarrow 2x + \mu(2x) - x - \mu(x)=x\Rightarrow \mu(2x)=\mu(x)$

$\Rightarrow \mu(x) \equiv$ const $= \mu(0) = 1$

$\Rightarrow f(x)=x+1\Rightarrow f(3)=4$