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Is $|g'(x)|<1\ \forall x\in(a,b)$ is one of the hypothesis of the Fixed-Point Theorem?

The answer is NO. Can someone please enlightened me about this? My teacher reason is this...

Note that the condition must be $|g'(x)| \leq k < 1\ \forall x\in(a,b)$. This condition is equivalent to $g'(x)\in[-k,k]\ \forall x\in(a,b)$. The condition $|g'(x)|<1\ \forall x\in(a,b)$ is equivalent to $g'(x)\in(-1,1)\ \forall x\in(a,b)$. Observe that the two conditions are not the same.

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    Which fixed-point theorem? http://en.wikipedia.org/wiki/Fixed_point_theorem mentions about half a dozen.2011-07-18
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    Let $g\in C[a,b]$ be such that $g(x)\in [a,b]$, for all $x\in [a,b]$. Suppose, in addition, that $g'$ exists on $(a,b)$ and that a constant $0 exists with \[ |g'(x)| \leq k \] for all $x\in (a,b)$. Then, for any number $p_0$ in $[a,b]$, the sequence defined by \[ p_n = g(p_{n-1}), n\geq 1, \] converge to a unique fixed point $p$ of $g$ in $[a,b].$\\2011-07-18
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    I just want to know why this is the answer by my teacher? Note that the condition must be $|g'(x)| \leq k < 1\ \forall x\in(a,b)$. This condition is equivalent to $g'(x)\in[-k,k]\ \forall x\in(a,b)$. The condition $|g'(x)|<1\ \forall x\in(a,b)$ is equivalent to $g'(x)\in(-1,1)\ \forall x\in(a,b)$. Observe that the two conditions are not the same.2011-07-18

3 Answers 3