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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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Hmm, $\small f(x)= \exp(x)+x $ has no fixpoint (which satisfies your requirement). Unfortuately it is not $\small |f(x)-f(y)| \lt |x-y| $ but $\small |f(x)-f(y)| \ge |x-y| $ instead. But perhaps one can proceed from here? –  Gottfried Helms Dec 6 '11 at 10:03
Jonas's example below is correct. You can salvage this statement by requiring $|f(x)-f(y)| < k|x-y|$ for some constant $k<1$, in which case $f$ is guaranteed a fixed point by the Banach fixed-point theorem. –  user7530 Dec 6 '11 at 11:01
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. –  Gottfried Helms Dec 6 '11 at 13:34
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3 Answers

up vote 3 down vote accepted

(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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There is something wrong here. The condition on the derivative shows there's a unique fixed point (plotting shows that it is approximately at $0.204$). However, $\dfrac{1}{4}\left(3x+\sqrt{1+x^2}\right)$ seems to work. –  t.b. Dec 6 '11 at 12:07
@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 $1$. –  Christian Blatter Dec 6 '11 at 12:43
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$f(x)=2$ 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 $y$.

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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.

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