Example of $f:\mathbb{R}\to\mathbb{R}$ injective and bounded, but with inverse not bounded or injective.

Question

I am trying to come up with an example of a bounded and injective function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}$ is not injective or bounded.

What are examples that could apply in this situation?

Thanks for helping! :D

Answer 1

As a first thought, I would say $f^{-1}$ is always injective, since a function is invertible if and only if it is injective and surjective. $f^{-1}$ is certainly invertible since its inverse is $f$, thus it is also injective.

For the bounded question, consider the function $$ f(x) = \left\{ \begin{aligned} &\dfrac{1}{1+(\frac{1}{x})^2} &&: x > 0\\ &0 &&: x=0 \\ &-f(-x) &&: x<0 \end{aligned} \right.$$ This function is bounded between -1 and 1 and one-to-one on $\Bbb R$. It has an inverse, at least for its image (which is not all of $\Bbb R$, since it's bounded), $f^{-1}: (-1,1) \longrightarrow \Bbb R$. If you look at the graph of $f$, you can see that this inverse will not be bounded.

Answer 2

Consdier the function $f: \{0,1,2,3,\dots\} \to \Bbb R$ defined by

$\tag 1 n \mapsto {\displaystyle \sum _{i\mathop {=} 1}^{n} 2^{-i}}$

By convention/definition, $f(0) = 0$.

The function $f$ can be extended to all of $\Bbb Z$ by requiring that $f(-x) = -f(x)$.

So we have an injective and increasing function $f: \Bbb Z \to (-1,1)$.

Using linear interpolation (see also the wiki piecewise linear function article) we have an injective extension (might as well keep abusing the $f$ setting),

$\tag 2 f: \Bbb R \to (-1,1)$

By design, $f$ is a bijective function.