Does T have a unique fixed point in X?

[Samuel Williams]

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?

 

[mathwonk]

If you meant: T^2(T(x)) = T^3(x) = T(T^2(x)) = T(x), then I agree.

 

[WWGD]

I would interpret the fixed point property for T^2 to mean that $T^2(x)=(T \circ T)(x)=T(T(x))=x$ . The relation $T(T(x))=T(x)$ is often called an involution, but not every contraction is an involution, e.g., $f(x)=x/2$ is a contraction but not an involution.

 

[HallsofIvy]

At some point you should say "because T^2 is a contraction, T^2 has a unique fixed point". Also "T^2= T(T(x))= T(x)" does not make sense for two reasons. The first part, T^2 is an operator while the other two T(T(x)) and T(x) are points. I presume you meant to say T^2(x)= T(T(x))= T(x). But even then the second equation, T(T(x))= T(x) assumes that x is a fixed point of T, not T^2 and that assumes what you are trying to prove. What you have proved is that "if x is a fixed point of T then it is a fixed point of T^2". That is not what you want to prove.

 

[mathwonk]

let me elaborate my reply; As you said, every fixed point of T is a fixed point of T^2, hence T has at most one fixed point since T^2 has only one.

So it remains to show that T has also a fixed point at x, i.e. to show that T(x) = x. For this, since T^2 fixes x and has only one fixed point, it suffices to show that T^2 also fixes T(x).

Then since T^2(x) = x, we get T^2(T(x)) = T^3(x) = T(T^2(x)) = T(x), and that does it.

I assume you had some such idea but mistyped it somehow.