I need help finding a map which fails to be a contraction but satisfies the following properties. The map $T: X \to X $ has to be such that given any pair of real numbers $a, b$ such that $0<a<b$, there exists some function $\alpha(a,b)$ depending on $a, b$ with $0 \le \alpha(a, b) < 1$ such that $d(Tx, Ty) \le \alpha(a, b) d(x, y)$ holds true whenever $a<d(x,y)<b$. I've been embarrassingly stuck on this for days now and can't find a single example. Here $X$ is complete.
By a contraction, I mean a map $T:X\to X$ such that there is some fixed $\alpha \in [0, 1)$ such that for all $x,y \in X$, $d(Tx, Ty)\le \alpha d(x,y)$.
 A: Let $X=[0,\infty)$ and let $g:X\to X$ be the function such that $g(t)=1-1/n$ for $t\in(n,n+1/n)$ for each positive integer $n$ and $g(t)=0$ otherwise.  Define $T(x)=\int_0^x g(t)dt$; then $T:X\to X$ is continuous.  Note that $T$ is not a contraction, since $d(T(n),T(n+1/n))/d(n,n+1/n)=1-1/n$ for each $n$.
But $T$ satisfies your conditions: given $0<a<b$, let $n>2$ be such that $a>2/n$.  Then it is easy to see that whenever $d(x,y)>a$, $d(T(x),T(y))<(1-1/2n)d(x,y)$.  Indeed, if $x<y$ and $y-x>a$, then for at least half of the measure of the interval $(x,y)$, either $t<n$ or $g(t)=0$ (since the intervals beyond $n$ on which $g(t)\neq 0$ are less than half as long as $a$, and they are separated by much longer intervals on which $g(t)=0$).  Furthermore, when $t<n$, $g(t)<1-1/n$, and $g(t)<1$ for all $t$.  So on at least half of $(x,y)$, $g(t)<1-1/n$, and on the other half we still have $g(t)<1$, giving the bound $$T(y)-T(x)=\int_x^y g(t)dt\leq\frac{y-x}{2}\left(1-\frac{1}{n}\right)+\frac{y-x}{2}\cdot1=(y-x)\left(1-\frac{1}{2n}\right).$$
