Let
![\mathbb{R}](/media/m/1/4/0/140a3cd0f5aa77f0f229f3ae2e64c0a6.png)
be the set of real numbers. Does there exist a function
![f: \mathbb{R} \mapsto \mathbb{R}](/media/m/8/7/7/877e4fa97f1e620cb7d1089ce437d7fb.png)
which simultaneously satisfies the following three conditions?
(a) There is a positive number
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
such that
![- M \leq f(x) \leq M.](/media/m/c/c/5/cc54354752868fc7882046139d1b0048.png)
(b) The value of
![f(1)](/media/m/b/8/0/b803b92638807d760794bcc6aac87236.png)
is
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
.
(c) If
![x \neq 0,](/media/m/7/7/4/774e8050b60ded39eb925bc0e7cb8e79.png)
then
%V0
Let $\mathbb{R}$ be the set of real numbers. Does there exist a function $f: \mathbb{R} \mapsto \mathbb{R}$ which simultaneously satisfies the following three conditions?
(a) There is a positive number $M$ such that $\forall x:$ $- M \leq f(x) \leq M.$
(b) The value of $f(1)$ is $1$.
(c) If $x \neq 0,$ then
$$f \left(x + \frac {1}{x^2} \right) = f(x) + \left[ f \left(\frac {1}{x} \right) \right]^2$$