let $f : \mathbb R \longrightarrow \mathbb R$ be a function such that $f (f(x)) = -x$ , $x \in \mathbb R$ then is $f$ continuous over $\mathbb R$?
I have observed that $f$ is a bijection and so $f(x) = f^{-1} (-x)$.So, $f$ and $f^{-1}$ have the same range.Is this fact helpful?I hard to find any such.Now how can I proceed?I have also failed to find out a function having the above property.So please help me.
Thank you in advance.
Notice that a continuous bijection is either strictly increasing or decreasing.
However $f\circ f$ is strictly increasing in both of these cases, which contradicts the fact that $-x$ is decreasing. Therefore $f$ cannot be continuous.
A slightly different proof just for fun:
Note that $f(-x) = f(f(f(x))) = -f(x)$, so $f$ is odd.
Suppose that $f$ is continuous.
Let $p(x)=(x,f(x))$ and consider the points $p(f^k(1))$, which are $P=\{ (1,f(1)), (f(1), -1), (-1,-f(1)), (-f(1),1) \}$ (they repeat, are orthogonal or opposite to each other and lie on a circle of radius $\sqrt{1+f(1)^2}$).
Note that $f(1)$ is neither of $\pm 1$, otherwise we would contradict $f(f(x)) = -x$, and $f(1) \neq 0$ for the same reason.
Hence two of the points of $P$ have positive and distinct first coordinates, and the second coordinates have opposite sign. By continuity, there is some positive $x_0$ such that $f(x_0) = 0$.
By applying $f$ to both sides we get $f(f(x_0)) = -x_0 = f(0) = 0$, a contradiction.
