Does $f(f(x))=-x \Rightarrow $ $f$ is bijective?

Question

Does $f(f(x))=-x \Rightarrow $ $f$ is bijective?

Is there any continuous map $f:\mathbb{R}\to \mathbb{R}$ such that $f(f(x))=-x, $ for all $x\in \mathbb{R}$?

---

For injectivity $x_1,x_2 \in \mathbb{R}$ such that $x_1=x_2\Rightarrow f(f(x_1))=f(f(x_2)) $

Is it right process?

Now if $x_1>x_2$ then $f(f(x_1))>f(f(x_2))\Rightarrow x_1<x_2$. From here what can I say?

What about the existence of a continuous function?

Answer 1

There is no such continuous functions. Every continuous injective function from $\mathbb R$ into $\mathbb R$ is either increasing or decreasing. In any case, $f\circ f$ is increasing. But $x\mapsto-x$ is decreasing.

Note tha $f$ is surjective beacause, given $y\in\mathbb R$, $f\bigl(f(-y)\bigr)=y$. And it is injective because\begin{align}f(x)=f(y)\implies&f\bigl(f(x)\bigr)=f\bigl(f(y)\bigr)\\\iff&-x=-y\\\iff&x=y.\end{align}

Answer 2

It is impossible for a continuous function to satisfy $f\circ f=-\text{Id}_\mathbb{R}$.

First, it is easy to see that $f$ would have to be bijective, hence a homeomorphism.

Second, whether $f$ respects or reverse the orientation of $\mathbb{R}$, $f\circ f$ has to respect it. $-\text{Id}_\mathbb{R}$ doesn't.

Answer 3

Let $p(x)=\sum_{i=1}^{n}a_{i}x^{i}$ be a polynomial, and $g(x)$ be a bijection.

Denote $F=\{f|f:\mathbb{R} \to \mathbb{R} \}$

Suppose $p$ acts on $f$, $P: F \rightarrow F$ by $P(f)(x) = \sum_{i=0}^{n}a_{i}f^{i}(x)$ where $f^{i}$ denotes composition.

Theorem: If $P(f)=g$ then $f$ is an injection. If $f$ is further assumed to be continuous then $f$ is surjective as well.

Proof: Suppose $f(x_1)=f(x_2)$ we will show that $x_1=x_2$. Since $f(x_1)=f(x_2)$, $$f^{i}(x_1))=f^{i-1} \circ(f(x_1))=f^{i-1} \circ(f(x_2))=f^{i}(x_2))$$

Therefore, $P(f)(x_1)=P(f)(x_2) \Rightarrow g(x_1)=g(x_2)\Rightarrow x_1=x_2$ Hence, $f$ is an injection.

We will show that $f$ is a surjection when $f$ continuous. We will use that any continuous injection is monotone. Since $g$ is continuous $g$ must be monotone, WLOG assume that $g$ is increasing. Then $\lim_{x\rightarrow \infty} g(x)=\infty$ and $\lim_{x\rightarrow -\infty} g(x)=-\infty$.

Again $f$ is monotone so $\lim_{x\rightarrow \pm\infty}f$ exists. Since $\lim_{x\rightarrow \infty}P(f)(x)=\infty$, the limit $\lim_{x\rightarrow \infty}f$ must be infinite. Same with $\lim_{x\rightarrow -\infty}f$ since $f$ monotone one the limits must be $\infty$ and the other must be $-\infty$, so by the intermediate value property $f$ is a surjective, hence $f$ is a bijection.

Theorem: Suppose $p$ is an even polynomial with positive coefficients, and $g(x)$ is decreasing. Then there are no continuous functions satisfying $P(f)(x)=g(x)$

Proof: Since $g$ is injective $f$ must be monotone therefore $f^{i}(x)$ is increasing when $i$ is even. Since $p$ is an even polynomial with positive coefficients $P(f)(x)$ is increasing, contradicting the fact that $g(x)$ is decreasing.

Corollary: There are no continuous functions satisfying $f(f(x))=-x$.

Notice, that $p(x)=x^2$ is an even polynomial with positive coefficients and $g(x)=-x$ is decreasing, and then apply the previous theorem.