If $(M,d)$ is a metric space, we say that $T:M\to M$ is a contraction if $d(T(x),T(y))\le d(x,y)$.
The question seems simple however I’ve not been able to furnish examples for the following contraction mappings on the Euclidean spaces which
1) which has no fixed points
2) which has infinitely many fixed points
3) has exactly 2 fixed points.
For the second case, the identity mapping seems to work, but I’m not sure. Any hints or suggestions will be appreciated.
For 1) a tranlsation works, for 2) the identity works. 3) is not possible:
Claim: The fixed locus of a contraction of an Euclidean space is convex.
Proof. Suppose $x,z$ are two fixed points for a contraction $T$ of an Euclidean space $E$. Let $y$ be a point in the segment $[x,z]$. By triangular inequality and by contrcactivity we have $$ d(y,x)+d(y,z)\geq d(T(y),x)+d(T(y),z)\geq d(x,z)=d(x,y)+d(y,z)$$
hence $d(T(y),x)+d(T(y),z)=d(y,x)+d(y,z)$. If $d(T(y),x)<d(y,x)$ then $d(T(y),z)>d(y,z)$, contradicting the contracion hypothesis. Thus $d(T(y),x)=d(y,x)$ and $d(T(y),z)=d(y,z)$ and this, in a Euclidean space, forces $T(y)=y$. So the whole segment $[x,z]$ is point-wise fixed by $T$.