# Solving $f(x) = f(x/2) + f(x/3) + x$

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.

• What is given about $f$? Is it continuous? Differentiable? Twice-differentiable? Smooth? Is $f:\mathbb R\mapsto\mathbb R$? – Franklin Pezzuti Dyer Oct 18 '18 at 22:59
• Plugging in $x=0$ shows that $f(0)=2f(0)$ and so $f(0)=0$. The only analytic solution is $f(x)=6x$. – Servaes Oct 18 '18 at 23:07
• It would be nice to specify the domain and codomain of $f$, and the constraints on $f$, if any. – Servaes Oct 18 '18 at 23:09

## 1 Answer

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 $$f(x)=|x|^a+6x=\frac{|x|^a}{2^a}+\frac{6x}{2}+\frac{|x|^a}{3^a}+\frac{6x}{3}+x=f(x/2)+f(x/3)+x$$

• 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 $f(x) = 6x$. – Rebellos Oct 19 '18 at 6:08
• @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. – achille hui Oct 19 '18 at 7:32
• 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. – Rebellos Oct 19 '18 at 9:49