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Let $X$ be a complete metric space . Suppose $T:X\to X$ is a function and $T^n$ is a contraction for some positive integer n. Here $T^n$ is the composition of $T$ with itself $n$ times. Must $T$ have a fixed point? Must $T$ be a contraction ?

Cornman
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Gül
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1 Answers1

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$T$ doesn't have to have fixed point. Let's consider space $X=\{1\} \times I \cup \{2\} \times I$, where $I$ is a unit compact interval, and $T \colon X \rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,\frac{1}{2}]$ and moves to the other interval, such that $T(\{1\} \times I)=\{2\} \times[0,\frac{1}{2}]$ and $T(\{2\} \times I)=\{1\} \times[0,\frac{1}{2}]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

pem
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  • Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :) – Cornman Aug 04 '18 at 10:24
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    We can say, that $T((1,x))=(2,\frac{x}{2})$ and $T((2,x))=(1,\frac{x}{2})$, for $x \in I$, then $T^2((i,x))=(i,\frac{x}{4})$ for $i=1,2$. – pem Aug 04 '18 at 10:29
  • Well, there is mistake somewhere, look here: https://math.stackexchange.com/questions/1447255/when-k-times-composition-of-a-function-is-a-contraction?rq=1 – pem Aug 04 '18 at 10:36
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    Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there. – pem Aug 04 '18 at 10:40