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One of the basic nonlinear functional equations is the following one:
$$(f(x))^{2}=xf(2x),\ \ \ x>0.$$ I found out that functions $f(x)=2^{1-x}x\exp(cx)$ form the family of solutions of this equation. But do this family cover all possible solutions to this equation? Truly speaking, I have no idea how to answer to this question. Thank you.

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    $\begingroup$ Usually functional equations tend to have non continuous solutions as well, which might turn out weird. $\endgroup$ – mvw Mar 29 '16 at 19:45
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    $\begingroup$ Small detail: The $2^{-x} = e^{-\log(2) x}$ can be combined with the $e^{cx}$ (since $c$ is arbitrary) so your solution can simply be written $2xe^{cx}$. $\endgroup$ – Winther Mar 29 '16 at 19:48
  • $\begingroup$ I'm aware of the fact that non continuous solutions may exist but do they exist in this case? $\endgroup$ – Peter95 Mar 29 '16 at 21:07
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Indeed, this family does not cover all solutions, nor even all continuous ones. The general solution is: take any function defined for $x\in[1,2)$ (which means an awful lot of possibilities, mind you!) and continue it both ways, up and down, using the expressions for $f(2x)$ via $f(x)$ and vice versa.

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  • $\begingroup$ Oh, I see now! Its' just so simple. Thanks! $\endgroup$ – Peter95 Apr 1 '16 at 14:52

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