Establish the sufficient condition $|g'(x)| < 1$ for convergence of an iteration using the Banach fixed point theorem?

Question

If $x_n = g(x_{n-1})$ is an iteration, it converges if $g$ is continuously differentiable and $|g'(x)| < 1$.

The Banach FPT says that if $T$ is a contraction on a complete metric space $X$ then it has a unique fixed points.

I'm having trouble with relating the two. I suppose it would be easier if we knew $g$ is a contraction because then $\frac{|g(x) - g(y)|}{|x-y|} \le \alpha < 1$ so that by taking $\lim y \to x$ we would have $|g'(x)|<1$.

Answer 1

By the MVT, you have $|g(x)-g(y)| = |g'(c)| |x-y| \leq \alpha |x-y|$, for every $x$ and $y$, and $\alpha <1$, so that $g$ is a contraction. Assuming the domain of $g$ is a complete metric space, then you can apply Banach fixed point theorem.

Answer 2

Let $X=[1,\infty)\subset \Bbb R$, which is a complete metric space. The function $g\colon X\to X$, $x\mapsto x+\frac1x$ has $g'(x)=1-\frac1{x^2}$ so $|g'(x)|<1$ for all $x\in X$. But $g$ has no fixed point.