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A solution to $f(2x)=\alpha f(x)$ with a boundary condition $f(\beta) = \beta$ is $$ f(x) = \left( \frac{\beta}{\alpha^{\log_2 \beta}} \right) \alpha^{\log_2 x}. $$

Do we know whether or not the solution is unique?

Edit: the function should be real-valued of a real variable, continuous, and monotonic.

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    $\begingroup$ Not unique without additional constraints. The function could be zero at those x not beta times an integer power of two. $\endgroup$ Dec 2, 2014 at 5:02
  • $\begingroup$ Good point. It needs to be an $\mathbb R \to \mathbb R$ function that is continuous and monotonic. $\endgroup$ Dec 2, 2014 at 7:41

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No, it is not unique. To obtain the general solution, set $g(t)=\log f(e^t)$. Then your equation is equivalent to $g(t+\log 2)=g(t)+\log\alpha$. One solution is an affine function, but any periodic function can be added to it. So there are infinitely many solutions and continuity and monotonicity do not help.

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