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Let $X$ be a metric space. Suppse $T:X \to X $ is a contraction. I have shown that $T^n$ where $n $ is positive is a contraction:

Question: If $T^n$ is contraction for $n > 1$, do we have that $T$ is a contraction? I cannot find a counterexample to this statement...

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up vote 4 down vote accepted

Consider $X=C([0,2],\Vert\cdot\Vert_\infty)$. Let $T:X\to X$ be defined by $$T(f)(x)=\int_0^xf(t)dt.$$ Now, $T$ is a linear operator which is not a contraction since $\Vert T\Vert=2$. Moreover, we can show by an easy induction that, for $n\geq1$ we have $$ T^n(f)(x)=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1}f(t)dt.$$ From this we see easily that $\Vert T^n\Vert=\frac{2^n}{n!}$. Therefore, $T^n$ becomes a contraction starting from $n=4$.

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It's a bit unfortunate that $\lVert T\rVert = 1$ here, which some may believe has significance. If you take $X = [0,2]$, you have $\lVert T\rVert = 2$, and you need to go to $T^4$ to reach a contraction, perhaps that's more illustrative. – Daniel Fischer Apr 14 '14 at 10:45
You are right, but i think the idea is there. – Omran Kouba Apr 14 '14 at 11:57
Oh, yes. I just think changing it would make the matter stand out even clearer. Not that it's terribly important. – Daniel Fischer Apr 14 '14 at 12:03
I reformulated the answer, as you suggested, I hope this is OK. – Omran Kouba Apr 14 '14 at 12:35
It was good enough for me to upvote it in the original version. I think it's even better now, and that you gave an explicit formula for $T^n$ is great. I can't upvote more than once, though. – Daniel Fischer Apr 14 '14 at 13:27

Consider the matrix $$ T=\begin{bmatrix}0&2\\0&0\end{bmatrix}, $$ seeing as an operator in $\mathbb C^2$ with the operator norm. Then $\|T\|=2$, while $T^n=0$ for all $n>1$.

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