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.
$\endgroup$ 33 Answers
$\begingroup$$f(x)=2$ when $x\leq 1$, $f(x)=x+\frac{1}{x}$ when $x\geq 1$.
Another example:
Let $f(x)=\log(1+e^x)$. Then $f(x)>x$ for all $x$, and since $0<f'(x)<1$ for all $x$, it follows from the mean value theorem that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.
$\endgroup$ $\begingroup$(Edited after t.b.'s comment) The function
$$f(x)\ :=\ {1\over4}\bigl(3x +\sqrt{1+x^2}\bigr)$$
is part of a hyperbola having the lines $y={1\over2}x$ and $y=x$ as asymptotes. It is a homeomorphism ${\mathbb R}\to{\mathbb R}$.
$\endgroup$ 2 $\begingroup$For an example that doesn't involve defining the function piecewise, how about
$$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.
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