Fixed point theorem knowing that ##T^n## is a contraction

[mahler1]

Homework Statement .
Let $X$ be a complete metric space and let $T:X \to X$ such that there exists $n \in \mathbb N$: $T^n$ is a contraction. Prove that there is a unique $x \in X$ such that $T(x)=x$.

The attempt at a solution.
Sorry but I am completely lost with this exercise and I am a little bit confused about the following: if $T^n$ is a contraction, would this mean that for any $x,y \in X$ $d(T^n(x), T^n(y))<αd(x,y)$ or $d(T^n(x), T^n(y))<αd(T^{n-1}(x),T^{n-1}(y))$, for $0<α<1$?. How could I begin the exercise to find the $x$ such that $T(x)=x$?.

 

[Dick]

Homework Statement .
Let $X$ be a complete metric space and let $T:X \to X$ such that there exists $n \in \mathbb N$: $T^n$ is a contraction. Prove that there is a unique $x \in X$ such that $T(x)=x$.

The attempt at a solution.
Sorry but I am completely lost with this exercise and I am a little bit confused about the following: if $T^n$ is a contraction, would this mean that for any $x,y \in X$ $d(T^n(x), T^n(y))<αd(x,y)$ or $d(T^n(x), T^n(y))<αd(T^{n-1}(x),T^{n-1}(y))$, for $0<α<1$?. How could I begin the exercise to find the $x$ such that $T(x)=x$?.

Let's just write $C$ instead of $T^n$ so we can skip some exponents. Here's a hint. Pick $x_0$ to be any point. Now define $x_1=C(x_0), x_2=C(x_1)$, etc. Now you've got things like $d(x_1,x_2)=d(C(x_0),C(x_1)) \le αd(x_0,x_1)$. The first step is going to be trying to prove that sequence is Cauchy.

 

[HallsofIvy]

Homework Statement .
Let $X$ be a complete metric space and let $T:X \to X$ such that there exists $n \in \mathbb N$: $T^n$ is a contraction. Prove that there is a unique $x \in X$ such that $T(x)=x$.

The attempt at a solution.
Sorry but I am completely lost with this exercise and I am a little bit confused about the following: if $T^n$ is a contraction, would this mean that for any $x,y \in X$ $d(T^n(x), T^n(y))<αd(x,y)$ or $d(T^n(x), T^n(y))<αd(T^{n-1}(x),T^{n-1}(y))$, for $0<α<1$?.

BOTH. The definition of "contraction map" is that for any x, y, $d(T(x), T(y))\le \alpha d(x, y)$ (with $\alpha< 0$). For any positive integer, $d(T^n(x), T^n(y))<\alpha d(T^{n-1}(x), T^{n-1}(y))$ is just that same statement with $T^{n-1}(x)$ in place of x and $T^{n-1}(y)$ in place of y.

How could I begin the exercise to find the $x$ such that $T(x)=x$?.

Take $x_0$ to be any point in X. Look at the sequence $\{T^n(x_0)\}$. Show that it is a Cauchy sequence so converges. See what happens when you apply T to the limit of that sequence.

 

[Dick]

BOTH. The definition of "contraction map" is that for any x, y, $d(T(x), T(y))\le \alpha d(x, y)$ (with $\alpha< 0$). For any positive integer, $d(T^n(x), T^n(y))<\alpha d(T^{n-1}(x), T^{n-1}(y))$ is just that same statement with $T^{n-1}(x)$ in place of x and $T^{n-1}(y)$ in place of y.

Not quite, I don't think. The problem (for some reason) didn't say T is a contraction, it says $T^n$ is a contraction.

 

[mahler1]

Let's just write $C$ instead of $T^n$ so we can skip some exponents. Here's a hint. Pick $x_0$ to be any point. Now define $x_1=C(x_0), x_2=C(x_1)$, etc. Now you've got things like $d(x_1,x_2)=d(C(x_0),C(x_1)) \le αd(x_0,x_1)$. The first step is going to be trying to prove that sequence is Cauchy.

Ok, but with that sequence, what I am goint to prove is that $T^n$ has a fixed point. If I call that point $x$, would $x$ help me to find the fixed point of $T$ that I am looking for?

 

[HallsofIvy]

In my first response I said "(with $\alpha< 0$)" when, of course, I meant "(with $\alpha< 1$)".

Now, in your first post you said "there exist $n\in N$ such that $T^n$ is contraction". The first thing I would do is define $S= T^n$ for that n so that S is a contraction. You can then show that the sequence $T^m x_0$ is a Cauchy sequence and so converges to some point $x$.

The point is that applying T to that x, Tx, is the same as applying T to each term in the sequence. But that just shifts the sequence by one place. It is still the same sequence and still converges to $x$. That is, Tx=x.

 

[Dick]

Ok, but with that sequence, what I am goint to prove is that $T^n$ has a fixed point. If I call that point $x$, would $x$ help me to find the fixed point of $T$ that I am looking for?

First you want to show that the fixed point is unique. If T^n(x)=x and T^n(y)=y, then x=y. Now can you show that if x is a fixed point then T(x) is also a fixed point?

 

[mahler1]

First you want to show that the fixed point is unique. If T^n(x)=x and T^n(y)=y, then x=y. Now can you show that if x is a fixed point then T(x) is also a fixed point?

Suppose there are $x,y \in X$ such that $T^n(x)=x$ and $T^n(y)=y$, $d(x,y)=d(T^n(x),T^n(y))\leq \alpha d(x,y)$ and this inequality holds if and only if $x=y$.

Now let $x$ be the fixed point of $T^n$, $T^n(T(x))=T(T^n(x))=T(x)$. It follows that $T(x)$ is a fixed point but by unicity we have that $T(x)=x$, which means $x$ is a fixed point of $T$. One can prove unicity using the fact that $T^n$ is a contraction.

Thank you very much!

 

[Dick]

BTW it might be tempting to think that if T^n is a contraction, then T must be a contraction. That's not true. Take X=R^2 and T to be the linear mapping represented by the matrix [[0,1],[1/2,0]]. T is not a contraction. T^2 is a contraction. Show that if you are interested.

 

[Dick]

Suppose there are $x,y \in X$ such that $T^n(x)=x$ and $T^n(y)=y$, $d(x,y)=d(T^n(x),T^n(y))\leq \alpha d(x,y)$ and this inequality holds if and only if $x=y$.

Now let $x$ be the fixed point of $T^n$, $T^n(T(x))=T(T^n(x))=T(x)$. It follows that $T(x)$ is a fixed point but by unicity we have that $T(x)=x$, which means $x$ is a fixed point of $T$. One can prove unicity using the fact that $T^n$ is a contraction.

Thank you very much!

Very welcome. Nicely done!

 

[mahler1]

BTW it might be tempting to think that if T^n is a contraction, then T must be a contraction. That's not true. Take X=R^2 and T to be the linear mapping represented by the matrix [[0,1],[1/2,0]]. T is not a contraction. T^2 is a contraction. Show that if you are interested.

Thanks for the remark, I appreciate it. Another example could be $f:[\frac{1}{4},\frac{1}{2}] \to [\frac{1}{4},\frac{1}{2}]$ defined as $f(x)=x$. $f$ is clearly not a contraction but $f^2$ is a contraction.

 

[Dick]

Thanks for the remark, I appreciate it. Another example could be $f:[\frac{1}{4},\frac{1}{2}] \to [\frac{1}{4},\frac{1}{2}]$ defined as $f(x)=x$. $f$ is clearly not a contraction but $f^2$ is a contraction.

f^2(x) means f(f(x)). f(f(x))=f(x)=x. I'm not sure that's a good example. You really have to go to more than one dimension to get a linear example.

 

[mahler1]

f^2(x) means f(f(x)). f(f(x))=f(x)=x. I'm not sure that's a good example. You really have to go to more than one dimension to get a linear example.

Oops.