= Solution
A <contraction mapping> $T$ on a <metric space> satisfies $d(Tx,Ty)\leq qd(x,y)$ for all $x,y$, with one constant $0\leq q<1$. The <contraction mapping theorem> says that on a nonempty <complete metric space> such a map has a unique <fixed point>; successive iterates from any starting point converge to it. More precisely, if $x_n=T^nx_0$, then $d(x_n,x_*)\leq q^nd(x_1,x_0)/(1-q)$.
The <Volterra integration operator> maps <continuous functions> to <continuous functions>. In the <uniform norm>,
$$
\|Af-Ag\|_\infty\leq\|f-g\|_\infty.
$$
This constant cannot be decreased: for $f=1$ and $g=0$, $Af(x)=x$, and both <uniform norms> are one. Hence \b[$A$ is not a contraction].
Changing the order of <integration> over the triangular region gives
$$
A^2f(x)=\int_0^x\int_0^t f(s)\,ds\,dt
=\int_0^x(x-s)f(s)\,ds.
$$
Thus
$$
\|A^2f-A^2g\|_\infty\leq\sup_{0\leq x\leq1}\frac{x^2}{2}\|f-g\|_\infty
=\frac12\|f-g\|_\infty.
$$
\b[$A^2$ is a contraction, with sharp constant $1/2$.] This is an example of an <iterated contraction>.
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