# 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.

- answered
- 121 views
- $25.00

### Related Questions

- Find $\lim\limits _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}$
- Calculus / imaginary numbers and S^2
- real analysis
- [Real Analysis] Show that $B$ is countable.
- A problem on almost singular measures in real analysis
- Prove that $A - B=A\cap B^c$
- Integrate $\int x^2(1-x^2)^{-\frac{3}{2}}dx$
- True-False real analysis questions