Generalization of the Banach fixed point theorem
Answer
Suppose $T^m(x)=x$, and assume $x$ is the only fixed point of $T^m$. Then
$$T^m(T(x))=T(x),$$
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 $x$.

4.4K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to 50% commission on every question your affiliated users ask or answer.
- answered
- 488 views
- $25.00
Related Questions
- Is it possible to transform $f(x)=x^2+4x+3$ into $g(x)=x^2+10x+9$ by the given sequence of transformations?
- Pathwise connected
- Limits : $x^{-1} \sin(x) $ as x -> 0 and $\tfrac{\ln(x)}{1-x}$ as x-> 0
- Early uni/college Calculus (one question)
- Use Stokes's Theorem to evaluate $\iint_S ( ∇ × F ) ⋅ d S$ on the given surface
- Does $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$ exist?
- Is it true almost all Lebesgue measurable functions are non-integrable?
- Assume there is no $x ∈ R$ such that $f(x) = f'(x) = 0$. Show that $$S =\{x: 0≤x≤1,f(x)=0\}$$ is finite.