= Solution
For a <uniform contraction from a comparison function on a bounded set>, the <closed set> $M$ is complete and has diameter at most $2R$. Suppose first $R>0$. The <limit> condition supplies $r_0>0$ and $q_0<1$ such that $\psi(r)/r\le q_0$ for $0<r\le r_0$. On the remaining compact <interval> $[r_0,2R]$, if nonempty, the <continuous> ratio $\psi(r)/r$ attains a maximum $q_1<1$, since it is strictly below one at every point. Thus there is one $q<1$ with $\psi(r)\le qr$ for every distance $0\le r\le2R$ arising in $M$.
Consequently $d(Tx,Ty)\le qd(x,y)$ throughout $M$. <Banach fixed-point theorem>[Banach's theorem] applies and gives \b[a unique <fixed point> in $M$]. No linear structure or <compactness> of the ball is used; the compact object in the argument is only the <scalar> distance <interval>. If the <closed ball> is the singleton at radius zero, its self-map fixes that point. The estimate on the ratio near zero is what prevents the compact-interval supremum argument from approaching one at its missing endpoint.
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