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I am searching for all functional solutions of equation $$(f(x))^2-(f(y))^2 = (f(x)-f(y))(f(x)+f(y)) =f(x-y)f(x+y),\quad x,y \in \mathbb{R}.$$ A few properties can be derived from the equation:

  1. If $f$ is a solution then $\lambda f$ is also a solution for $\lambda \in \mathbb{R}$.
  2. Plugging $y=0$, $(f(x))^2-(f(0))^2=(f(x))^2$, implying $f(0)=0$.
  3. Plugging $y=-x$ implies $f(x)=-f(-x)$ (odd) or $f(x)=f(-x)$ (even).
  4. For $x=0$, $-(f(y))^2 = f(-y)f(y)$ forces that $f$ is an odd function from 3.
  5. $f(x)=x$ and $f(x) = \sin(x)$ are two solutions.

Are there other solutions to this equation ?

Edit 1+2: I am looking for all measurable solutions of this functional equation.

Sam
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  • @KBDave This is different, I am not seeking continuous and bounded solutions. I require all solutions of this functional equation (even discontinuous). – Sam Apr 29 '24 at 20:48
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    One example of a discontinuous function that solves the equation is a discontinuous additive function ($f(x+y)=f(x)+f(y)$), the existence of such a function however requires the Axiom of Choice: https://math.stackexchange.com/questions/269434/non-continuous-function-satisfies-fxy-fxfy – Vit Apr 29 '24 at 21:33
  • Thanks @Vit. This is interesting, so a discontinuous function $f$ satisfying $f(x+y) = f(x)+f(y)$ is non-mesaurable. Does a similar result also hold true for this functional equation? – Sam Apr 29 '24 at 22:01
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    @Sam Any additive function will satisfy your constraint as $ f(x+y) \times f(x-y) = [f(x) + f(y) ] \times [f(x) - f(y) ] = f^2(x) - f^2 (y) $. $\quad$ Are you asking for non-additive functions? – Calvin Lin Apr 29 '24 at 22:38
  • Yes, @CalvinLin I am looking for measurable non-additive solutions. – Sam Apr 29 '24 at 22:45
  • I think all additive measurable functions are of the form $f(x)=\lambda x$ – Sam Apr 29 '24 at 22:50
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    @Sam One of the sufficient conditions for Cauchy's functional equation is that $f$ is Lebesgue measurable, then it is linear. – Sil Apr 29 '24 at 23:09
  • yeah, thanks @Sil. This is why I am interested in non-additive solutions. – Sam Apr 29 '24 at 23:14

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