Function which has no fixed points

Question

Problem:

Can anyone come up with an explicit function $f \colon \mathbb R \to \mathbb R$ such that $| f(x) - f(y)| < |x-y|$ for all $x,y\in \mathbb R$ and $f$ has no fixed point?

I could prove that such a function exists like a hyperpolic function which is below the $y=x$ axis and doesn't intersect it. But, I am looking for an explicit function that satisfies that.

Answer 1

(Edited after t.b.'s comment) The function

$$f(x)\ :=\ {1\over4}\bigl(3x +\sqrt{1+x^2}\bigr)$$

is part of a hyperbola having the lines $y={1\over2}x$ and $y=x$ as asymptotes. It is a homeomorphism ${\mathbb R}\to{\mathbb R}$.

Answer 2

$f(x)=2$ when $x\leq 1$, $f(x)=x+\frac{1}{x}$ when $x\geq 1$.

Another example:

Let $f(x)=\log(1+e^x)$. Then $f(x)>x$ for all $x$, and since $0<f'(x)<1$ for all $x$, it follows from the mean value theorem that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.

Answer 3

For an example that doesn't involve defining the function piecewise, how about

$$f(x) = x \Phi(x) + \Phi'(x)$$

where $\Phi$ is the normal cdf and $\Phi'$ is its derivative, the normal pdf. You can view some of its properties on Wolfram Alpha.