A mapping may fail to be a contraction while one of its iterates is a contraction. The focus keyword iterated contractions is especially natural for integral operators because repeated integration introduces factorial decay. This lesson examines operators whose powers have exact contraction estimates and explains the fixed point consequence. :::example[Operator Whose Square is a Contraction] Let $C[0,\pi/2]$ have the uniform metric and define $(Tx)(t)=\int_0^t x(u)\sin u\,du$. Then $T^2$ is a contraction. ::: :::proof Given the operator $T$. To prove that $T^2$ is a contraction. For $x,y\in C[0,\pi/2]$, $|(Tx)(t)-(Ty)(t)|\leq d(x,y)\int_0^t\sin u\,du=d(x,y)(1-\cos t)$. Applying $T$ again, $|(T^2x)(t)-(T^2y)(t)|\leq d(x,y)\int_0^t(1-\cos u)\sin u\,du\leq d(x,y)/2$. Hence $d(T^2x,T^2y)\leq d(x,y)/2$, so $T^2$ is a contraction. $\square$ ::: :::solved-problem Let $A>0$ and define $T:C[0,A]\to C[0,A]$ by $(Tx)(t)=\int_0^t x(s)\,ds$. Show that $T^n$ is a contraction if and only if $A^n/n!<1$. ::: :::solution With the uniform metric, induction gives $|(T^nx)(t)-(T^ny)(t)|\leq t^n d(x,y)/n!$. Taking the supremum over $0\leq t\leq A$ gives $d(T^nx,T^ny)\leq A^n d(x,y)/n!$. Hence $T^n$ is a contraction if $A^n/n!<1$. Conversely, take $x(t)=1$ and $y(t)=0$. Then $(T^nx)(t)=t^n/n!$ and $(T^ny)(t)=0$, so the best Lipschitz constant is $A^n/n!$. ::: :::remark The factorial in the denominator is the reason Volterra-type operators eventually become contractions on bounded intervals. ::: :::solved-problem Choose $A$ such that $T^n$ is not a contraction but $T^{n+1}$ is a contraction for the Volterra operator. ::: :::solution Choose $A$ satisfying $(n!)^{1/n}<A<((n+1)!)^{1/(n+1)}$. Then $A^n/n!>1$ but $A^{n+1}/(n+1)!<1$. Hence $T^n$ is not a contraction, while $T^{n+1}$ is a contraction. ::: :::exercise 1. Explain why an iterate can be a contraction when the original operator is not. 2. Compute the Lipschitz constant for $T^3$ when $A=1$. 3. State the fixed point consequence when $T^k$ is a contraction and $T$ is continuous. ::: :::answer 1. Repeated application can introduce additional decay. 2. The constant is $1/3!=1/6$. 3. $T$ has a unique fixed point and its iterates converge to it. ::: :::call-to-action[Revise Fixed Point Methods] subtitle: Review contractions, iterated contractions, and the applications to linear, integral, and differential equations before moving to the next metric space topic. button: Practise Contractions | /functional-analysis/metric-spaces/continuous-mappings-and-fixed-point-principles/contractions-and-fixed-points button-ghost: Review Uniform Limits | /functional-analysis/metric-spaces/continuous-mappings-and-fixed-point-principles/uniform-limits-of-function-sequences :::