4
$\begingroup$

Consider the following functional equation:

$$f(f(x)-x)=f(f(x))$$

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?

$\endgroup$
  • $\begingroup$ 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
9
$\begingroup$

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 $f(f(x)-x)=f(1-x)=-1=f(1)=f(f(x))$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.