- #1
Samuel Williams
- 20
- 3
Let (X, d) be a complete metric space, and suppose T : X → X is a function such that T^2 is a contraction. [By T^2, we mean the function T^2 : X → X given by T^2(x) = T(T(x))]. Show that T has a unique fixed point in X.
So I have an answer, but I am not sure whether it is correct. It goes as follows :
Any fixed point of T is also a fixed point of T^2, and there is only one of these. Let x∈X be the unique fixed point of T^2, and consider T^2=T(T(x))=T(x). But now T(x) is a fixed point of T^2, so T(x)=x and x is also a fixed point of T. Since x∈X, T has a unique fixed point in X
Would that be sufficient or am I missing something?
So I have an answer, but I am not sure whether it is correct. It goes as follows :
Any fixed point of T is also a fixed point of T^2, and there is only one of these. Let x∈X be the unique fixed point of T^2, and consider T^2=T(T(x))=T(x). But now T(x) is a fixed point of T^2, so T(x)=x and x is also a fixed point of T. Since x∈X, T has a unique fixed point in X
Would that be sufficient or am I missing something?