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How would one proceed to proving that the solution to the functional equation $$f(x) = f(x/2) + f(x/3) + x$$ is $f(x)=6x$ which is also unique?

To clarify, I am not aware of neither the proof of the solution, neither the uniqueness or not.

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    $\begingroup$ What is given about $f$? Is it continuous? Differentiable? Twice-differentiable? Smooth? Is $f:\mathbb R\mapsto\mathbb R$? $\endgroup$ – Franklin Pezzuti Dyer Oct 18 '18 at 22:59
  • $\begingroup$ Plugging in $x=0$ shows that $f(0)=2f(0)$ and so $f(0)=0$. The only analytic solution is $f(x)=6x$. $\endgroup$ – Servaes Oct 18 '18 at 23:07
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    $\begingroup$ It would be nice to specify the domain and codomain of $f$, and the constraints on $f$, if any. $\endgroup$ – Servaes Oct 18 '18 at 23:09
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Unfortunately, $f(x)=6x$ is not the unique solution to your functional equation. Let us define the constant $a\approx 0.787885$ as the unique real solution to the equation $$\frac{1}{2^a}+\frac{1}{3^a}=1$$ Then we may see that the function $$f(x)=|x|^a+6x$$ also satisfies the given functional equation, since $$f(x)=|x|^a+6x=\frac{|x|^a}{2^a}+\frac{6x}{2}+\frac{|x|^a}{3^a}+\frac{6x}{3}+x=f(x/2)+f(x/3)+x$$

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    $\begingroup$ Hello, good point. But when inserting the problem in WolframAlpha or Matlab, the solution that the pre-determined algorithms yield is only one which is $f(x) = 6x$. $\endgroup$ – Rebellos Oct 19 '18 at 6:08
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    $\begingroup$ @Rebellos WA can be wrong, this isn't the first time nor it will be the last. e.g. if you throw solve f(x) = f(3*x/5) + f(4*x/5) + x to WA, it only return the solution f(x) -> -5x/2 without telling you Ax^2 - 5x/2 is another valid solution. $\endgroup$ – achille hui Oct 19 '18 at 7:32
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    $\begingroup$ Okay, so figured out that mathematical packages, given no further constraints, assume that the deriving function is linear, because they seem to always produce a linear result. I also figured out that proving $f(x) = 6x$ comes from a limit manipulation. In the case of linearity, this is a unique solution. Non-linearity though, there's no uniqueness and definitely stems from an initial value. $\endgroup$ – Rebellos Oct 19 '18 at 9:49

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