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 $x$.
Erdos
4.7K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1144 views
- $25.00
Related Questions
- Calculus 2 / Calculate the surface of F
- What is the Lebesgue density of $A$ and $B$ which answers a previous question?
- Use Stokes's Theorem to evaluate $\iint_S ( ∇ × F ) ⋅ d S$ on the given surface
- (a) Find the coordinates (x,y) which will make the rectangular area A = xy a maximum. (b) What is the value of the maximum area?
- Mechanical principle help (maths)
- Question for KAV1
- Solutions to Stewart Calculus 8th edition
- Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$