Consider the following functional equation:


where $f: \mathbb{R} \rightarrow \mathbb{R}$.

Obviously $f(x)$ cannot be inverted.

One solution is $f(x) = k$ where $k \in \mathbb{R}$ is a constant.

Are there any other solution?

  • $\begingroup$ Are there any assumptions on $f$? Is it continuous, differentiable, anything? $\endgroup$ – Servaes Aug 22 '18 at 10:01
  • $\begingroup$ No there is no assumption, but if with an assumption we can get a partial result it would be nice. $\endgroup$ – mbjoe Aug 22 '18 at 10:06
  • $\begingroup$ Note that if $f$ has a fixed point, then it is $f(0)$. $\endgroup$ – Servaes Aug 22 '18 at 10:15

Define $f(x)$ by

$$ f(x)= \begin{cases} \ \ \,1&\text{if $x<0$} \\ \ \ \,0&\text{if $x=0$} \\ -1&\text{if $x>0$} \\ \end{cases} $$ The equation holds for $x=0$ as $f(f(0)-0)=f(f(0))$.

The equation holds for $x>0$ as $f(f(x)-x)=f(-1-x)=1=f(-1)=f(f(x))$.

The equation hold for $x<0$ as $f(f(x)-x)=f(1-x)=-1=f(1)=f(f(x))$.

  • $\begingroup$ Thank you. This can be generalized with two arbitrary constant values, one positive and one negative, instead of 1 and -1, respectively. $\endgroup$ – mbjoe Aug 22 '18 at 11:41
  • $\begingroup$ Note also that the value at 0 is completely arbitrary. $\endgroup$ – patrik Aug 23 '18 at 7:22

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