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 🙂
Best Answer
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.