# Give $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $|f'(x)|<1$ $f(x) \neq x$ for all $x \in \mathbb{R}$

Problem

Give a function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ , $$C^\infty$$ such that

$$1) |f'(x)|<1$$

$$2) f(x) \neq x$$ for all $$x \in \mathbb{R}$$

My ideia

The idea is to get a function that tends asymptotically to $$y=x$$. If you reduce the condition to $$|f'(x)|\le 1$$, it's easy but I don't know how to proceed with the condition $$|f'(x)|<1$$.

• Take the graph of $y=e^{-x}$ and rotate it clockwise about $(0,0)$ thru an angle $3\pi/4.$ Mar 21 '20 at 6:29
• $f(x)=\frac{1}{2}x$ also works. Mar 21 '20 at 7:04
• @C.Brendel No, this doesn't work since $f(0)=0$ Mar 21 '20 at 7:11
• ahh, you're right Mar 21 '20 at 10:35

The following function fulfills the requirements (and asymptotically tends to $$y=x$$):
$$f(x)=x+\frac{1}{1+e^{x}}$$
$$\frac{1}{1+e^{x}}$$ resembles a sigmoid function, that takes values in range $$(0, 1)$$ and is monotonically decreasing, i.e. $$\frac{d}{dx}\frac{1}{1+e^{x}}=\frac{-e^x}{(e^x+1)^2}<0$$, so $$\frac{d}{dx}f(x)<1$$.