# Generalization of the Banach fixed point theorem

## Answer

$$T^m(T(x))=T(x),$$

and hence $T(x)$ is also a fixed point of $T^m$. Since the fixed point of $T^m$ is unique, we must have

\[T(x)=x,\]

so $x$ is also a fixed point of $T$. It remains to show that $T$ does not have any other fixed point. Assume $T(y)=y$, for some $y \in M$. Then

\[T^{m}(y)=T^{m-1}(y)= \dots =T(x)=y,\]

so $y$ is a fixed point of $T^m$, and therefore $y=x$, since $T^m$ has only one fixed point.

$T$ also has the unique fixed point $x$.

The answer is accepted.

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