Fixed point theorem 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.
 A: 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.
A: 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.
A: 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. 
