3
$\begingroup$

Let $X$ be non empty and $T: X \to X$ a mapping such that $T^n$ has a unique fixed point $x$,

then $x$ is also a unique fixed point of T.

If I assume $T$ has two fixed points $x$ and $v$ $\implies$ $T^nv=v$ so $v$ is also a fixed point of $T^n$ which is a contradiction.

Is this the proper way to do the proof?

Edit: All of the details of the question are below.

Let $X$ be a nonempty set and $T : X \to X$ be a mapping. If, for an $n \in N$, $n \geq 2$, there exists a unique fixed point $x \in X$ for $T^n$ , then $x$ is a also a unique fixed point for $T$.

  • 0
    Yes it seems fine to me if the map is not too strange for example... Any info about T? – 2017-02-06
  • 2
    You have only shown that if $T$ has a fixed point, then it is unique. You didn't show that $T$ actually has a fixed point. – 2017-02-06
  • 0
    Of course Janik is right, but it seems to me enough. Maybe existence actually is not stated above, but some outhors may refer to it as obvious. – 2017-02-06
  • 0
    Uniqueness of the fixed point of $T^n$ is needed to prove the existence of a ficed point of $T$. Indeed, consider the rotation by $\frac{2\pi}{n}$ of a circle around the center. Then $T^n=\text{id}$, while $T$ has no fixed point. – 2017-02-06

2 Answers 2

3

If $T^n(x)=x$, then $T(x)=T\bigl(T^n(x)\bigr)=T^n\bigl(T(x)\bigr)$, so $T(x)$ is a fixed point of $T^n$. By uniqueness, $T(x)=x$.

2

You have already shown the uniqueness of a fixed point of $T$. Assume $x\in X$ is the unique fixed point of $T^n$. Then $$ Tx=T(T^nx)=T^n(Tx), $$ so $Tx$ is a fixed point of $T^n$. But the fixed point of $T^n$ is unique, so you have $Tx=x$ as desired.