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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? en.wikipedia.org/wiki/Fixed_point_theorem mentions about half a dozen. –  joriki Jul 18 '11 at 8:13
    
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 $0<k<1$ 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 $[a,b].$\\ –  ron Jul 18 '11 at 8:15
    
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. –  ron Jul 18 '11 at 8:39
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3 Answers

Let $K$ be an closed set and $g \in C^1(K), g(K) \subseteq K$ with $|g'(x)|<1\ \forall x\in K$. Then $g$ doesn't neccessiarily have a fixed point. Look for example at $g(x) = x+1/x$ on $[1,\infty)$. Then $|g'(x)|=1 - 1/x^2<1$ but you have no fixed point. I hope this is what you are looking for. If you can otherwise find a $\lambda$ such that $|g'(x)|<\lambda<1\ \forall x\in K$ then you can use the convergence of the geometric sum to show there is a fixed point.

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before you can use the conclusion of the theorem i believe you should satisfy all the assumption first not the only one... –  ron Jul 18 '11 at 8:27
    
@rockabhai what do you mean? –  Listing Jul 18 '11 at 8:32
    
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. –  ron Jul 18 '11 at 8:41
    
This is exactly what I explain in my answer. –  Listing Jul 18 '11 at 8:44
    
By the way, you can transform this into an example on a finite interval by transforming from $x$ to $u=1/x$; then $f(u)=u/(1+u^2)$ has no fixed point on $(0,1)$ even though $|f'(u)|=(1-u^2)/(1+u^2)<1$. A simpler example for this interval would be $g(x)=x/(1+x)$. –  joriki Jul 18 '11 at 8:46
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People gave reasons why it is not true if you consider the two different hypothesis... but I will tell why the two hypothesis are not the same!

If you say that $|g'(x)| < 1$, that is saying that $g'(x)$ lies in the interval $(-1,1)$, but might be arbitrarily close to $-1$ or $1$. On the other hand, the second statement says that $|g'(x)| \le k < 1$, which means $g'(x)$ $\textbf{cannot}$ get arbitrarily close to $-1$ or $1$, but is somewhat trapped in a compact interval which is itself in $(-1,1)$. In that sense it is a stronger statement than the one you suggested (i.e. that $|g'(x)| < 1$).

I felt like this was one of the points you were missing so I mentioned it.

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Well, the reason is implicitly covered by the answers above. The underlying theoretical reason (if you know about Cauchy sequences), is that to be sure that the sequence $(p_n)$ is Cauchy (and hence definitely convergent), you need the constant $k$ in the formula $|g(x) - g(y)| \leq k|x-y|$ (for all $x,y \in [a,b]$) to be definitely less than one.

The basic idea is that for large $r$ and $s$ with $s >r$, we have $|p_s - p_r| \leq k^{r}(1 + k + \ldots + k^{s-r-1})|p_1 -p_0|$, and we need to be sure that the partial sums of the geometric series converge, and that $k^r \to 0$, so we need $k$ strictly less than one. The condition on derivatives allows the Mean Value Theorem to be applied to get $|g(x)-g(y)| \leq k|x-y|$ for all $x,y \in [a,b]$, as others have already commented.

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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. –  ron Jul 18 '11 at 8:40
    
Yes, I intended to make it clear that the fixed $k$ which works for all $a,b$ in the interval should be less than $1$. I hope that my rewrite above makes it even clearer. –  Geoff Robinson Jul 18 '11 at 10:09
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