Iterated contraction

If a self-map $T$ of a nonempty complete metric space has an iterate $T^n$ that is a contraction mapping, the contraction mapping theorem gives a unique fixed point $a$ of $T^n$. The maps commute, so $T^n(Ta)=T(T^{n}a)=Ta$, and uniqueness forces $Ta=a$. Conversely every fixed point of $T$ is fixed by $T^n$. Thus $a$ is the unique fixed point of $T$.

Let $T$ be a self-map of a nonempty closed subset of a Banach space, with the displayed bounds. Then $T$ has a unique fixed point $p$ and every orbit $x_n=T^n{x}_0$ converges to it. Indeed, $\Vert x_{n+1}-x_n\Vert \le {\alpha }_n\Vert x_1-x_0\Vert$ makes the orbit Cauchy by summing successive increments. The first bound makes $T$ continuous, so the limit is fixed. Since ${\alpha }_n\to 0$, the bounds also imply uniqueness. The tail sum gives $\Vert p-x_n\Vert \le {\sum }_{k\ge n}{\alpha }_k\Vert x_1-x_0\Vert$ for $n\ge 1$. A Volterra integral equation supplies factorially decreasing bounds even when the first iterate is not a contraction mapping.