# Fixed Point For A Contraction

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1. Sep 18, 2015

### 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?

2. Sep 18, 2015

### mathwonk

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

3. Sep 19, 2015

### 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.

4. Sep 19, 2015

### 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.

5. Sep 19, 2015

### 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.