Does a function whose derivatives never attains values $\pm 1$ have a fixed point?

Question

I started wondering if the following is true.

Consider a differentiable function $f:\mathbb R \to \mathbb R$. If $f'(\mathbb R) \cap \{-1,1\} = \emptyset$, then $f$ has a fixed point.

From Darboux theorem it is apparent that one of the following cases holds:

1. $f'(\mathbb R) \subset (-1,1)$;
2. $f'(\mathbb R) \subset (1,\infty)$;
3. $(-f)'(\mathbb R) \subset (1,\infty)$.

However I don't see how to continue from there.

Answer 1

No.

$e^x+x$ has the derivative $e^x+1$ which is a function in case 2. But $e^x+x=x$ implies $e^x=0$.

Answer 2

$$ f(x) = \sqrt{x^2+1} $$ is a counterexample for the first case. The derivative satisfies $$ -1 < f'(x) = \frac{x}{\sqrt{x^2+1}} < 1 \, , $$ but $f$ has no fixed point since $f(x) > |x| \ge x$ for all $x \in \Bbb R$.

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Generally, for case 1 one can start with any (continuous or integrable) function $h:\Bbb R \to \Bbb R$ which satisfies

• $0 < h(x) < 2$ for all $x \in \Bbb R$, and
• $\int_0^\infty h(t)\, dt < 1$.

Then $f(x) = 1+x-\int_0^x h(t)\, dt$ has the derivative $f'(x) = 1-h(x) \in (-1, 1)$, but $$ f(x) - x = 1 - \int_0^x h(t)\, dt $$ is strictly positive for all $x$.

Similarly, for case 2 one starts with a function $h$ satisfiying

• $h(x) > 1$ for all $x \in \Bbb R$, and
• $\int_{-\infty}^0 h(t)\, dt < 1$.

and sets $f(x) = 1+x+\int_0^x h(t)\, dt$.

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Remark: If $|f'(x)| \le k < 1$ or $|f'(x)| \ge k > 1$ for some constant $k$ then $f$ has a fixed point. So any counterexample must have a derivative which takes values arbitrarily close to $1$ or $-1$.