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Is the square root a contraction? If yes, there should be only one fixed point; yet, there are two of them, 0 and 1! Thanks for help!

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    $\begingroup$ You have essentially answered your own question. What happens between $0$ and $1$? $\endgroup$ – Jonas Meyer Oct 23 '12 at 8:00
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    $\begingroup$ But, for $x\ge 1$, it is a contraction. $\endgroup$ – Berci Oct 23 '12 at 8:52
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    $\begingroup$ @Berci: For $x\geq \frac{1}{4}$ it is a contraction, too. $\endgroup$ – Jonas Meyer Oct 23 '12 at 15:33
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You’ve actually answered your own question: it’s not a contraction. I suspect, however, that what you’d really like is an example showing that it’s not a contraction. If $f(x)=\sqrt x$, how does $$\left|f\left(\frac14\right)-f(0)\right|$$ compare with $$\left|\frac14-0\right|\,?$$

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