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I wonder if it is possible to find an interesting example of a function that is K-Lipschitz ( $ K \in [0;1] $ ) and that has no fixed point.

So I gleaned what I know :

If a function is K-Lipschitz ( $ K \in [0;1[ $ ) it has a unique fixed point.

The example of affine functions such as $ d(x) = x + 1 $ is easy and effective. But I'm searching for something funnier and richer.

Moreover, as this question shows : Every 1-Lipschitz function in the closed unit ball has a fixed point, we need to look at infinity to find something (perhaps? I'm not sure to understand this example ; I have not yet study $\mathbb{R}^n$, however I'm open to any example, I'll take a look at it, even if first I can't understand it).

We can look at function $ f\in C^1$ that are more powerful than just Lipschitz continuous and I've found this example : $f(x) = x + \frac{1}{1+x^2} $ or in a more general way : $g(x) = x + \frac{1}{1+e^x} $. Maybe it is possible to find another example which is not $C^1$?

So to sum-up my question would be to find another interesant function that is K-Lipschitz ( $ K \in [0;1] $ ) and that has no fixed point, and if it is possible to find a function like this that is not differantiable ?

Finaly, is it possible to find sufficient conditions to create a function like this?

Thank you very much :)

Marine Galantin
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  • What does "interesting" mean? Any of your examples can be made non-differenentiable at some point. For instance, consider $f(x) = \min{x+1/2,x+1/(1+x^2)}.$ – Jeff Feb 27 '18 at 21:35
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    If a Lipschitz-continuous function with $K\in[0,1]$ is defined over a compact domain it has a fixed point by the Banach-Caccioppoli theorem. Hence all the counterexamples you are looking for need to be defined over non-compact domains. – Jack D'Aurizio Feb 27 '18 at 21:36
  • interesting means something less intuitive than just an affine function but not something that would be complicated to describe – Marine Galantin Feb 27 '18 at 22:03

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A more general form of the examples you mentioned: let $g:[0, \infty)\to (0, \infty)$ be a nonincreasing $2$-Lipschitz function. Then the function $$f(x) = x + g(x)$$ maps $[0, \infty)$ into $[0, \infty)$, is $1$-Lipschitz and has no fixed point. Nothing here forces $g$ to be differentiable, you can start with something like $g(x)=e^{-x}$ and add corners to its graph.

Proof: if $a<b$, then $$ f(b)-f(a) = (b-a) + (g(b)-g(a)) \le b-a $$ and $$ f(a)-f(b) = (a-b) + (g(a)-g(b)) \le a-b + 2|a-b| = b-a $$ So $f$ is $1$-Lipschitz. Also, $f(x)>x$ by construction.