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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.

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    One family of solution is $\small f(x)= x - tanh(x) - C $; it suffices $\small C \ge 1 $ . This is constructed as the integral of some function g(x) which is between 0 and 1 (so $\small |f(x+h)-f(x)| \lt |(x+h)-x | \to {|f(x+h)-f(x)| \over | (x+h)-x | } =g(x) \lt 1 $ so the absolute value of the derivative of f(x) must be lower than 1 everywhere. Then the integral allows to vertically shift the function by a parameter C such that the curves of y=f(x) and y=x do not meet.2011-12-06

3 Answers 3

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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.

11

$f(x)=2$ 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 for all $x$, it follows from the mean value theorem that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.

4

(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 ${\mathbb R}\to{\mathbb R}$.

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    @t.b.: You are of course right. I first had ${1\over4}$ and then mistakenly thought that by making it flatter I could bound the derivative below $1$.2011-12-06