# contraction mapping theorem for 'non-strict' contraction

Given a map $F:X \to X$ on a complete metric space $(X,d)$, and let $K<1$ such that:

$$d(F(x), F(y)) \le K d(x,y), \quad \forall x,y \in X$$

then the contraction mapping theorem tells us that $F$ has a unique fixed point, and we can iteratively solve for this fixed point.

My question is, if we take $K=1$, then it is no longer a contraction, I've seen this being called a 'non-strict' contraction. I'm wondering if there are any results regarding this case and fixed points? Do they exist but aren't unique, or do they not exist at all?

• In general, no (see Severin Schraven's answer). If, however, $X$ is compact and we have $d(F(x),F(y))<d(x,y)$ for any $x,y\in X$ (without Lipschitz property), consider $f \colon X \to [0, \infty)$ defined as $f(x):=d(x,F(x))$. $f$ is continuous (can you prove that?), so it attains its global minimum somewhere in $X$, at $x^*$, say. Suppose $f(x^*)>0$. Is that possible? – user539887 Apr 22 '18 at 9:26
• – Lutz Lehmann Apr 22 '18 at 9:29
• A small remark: I forgot to mention in my comment that $d(F(x),F(y))<d(x,y)$ for any $x, y \in X$ with $x \ne y$. – user539887 Apr 22 '18 at 11:18

In general, they do not need to admit fixed points. Take for example any translation (different from the identity) in $\mathbb{R}^n$. It has Lipschitz constant equal to 1, but has no fixed point.
Let $X$ any set and endow it with the discrete metric, then any injective map $f: X \rightarrow X$ has Lipschitz constant equal 1. Thus, you can have any number of fixed points you like (just fix the points of your choice and make sure that you have an bijection without fixed points on the complement. You can actually do that, see for example here Existence of a bijective function with no fixed points).
• @user539887 : To be more specific, with $x_{n+1}=\sin(x_n)$ set $u_n=x_n^{-2}$ then you get $$u_{n+1}=u_n/(1-x_n^2/6+O(x_n^4))^2=u_n+1/3+O(u_n^{-1})$$ so that $u_n\approx u_0+n/3$ or $$x_n\approx x_0/\sqrt{1+nx_0^2/3}$$ which is far away from being geometrically convergent. – Lutz Lehmann Apr 22 '18 at 11:00
• @user539887 Thanks for pointing out my terrible mistake. The problem was, that of course we could have $x=y$ in my argument and for this case the inequality $d(F(x),F(y))<d(x,y)$ does not apply. – Severin Schraven Apr 22 '18 at 11:10