Confused about fixed point method condition

Question

In my numerical analysis class, and specifically in the section regarding fixed point iteration, the hypothesis of the fixed point theorem states:

Let $g \in C[a,b]$ be such that $g(x)\in[a,b]\ \forall\ x\in(a,b)$. Suppose, in addition, that $g'$ exists on $(a,b)$ and that a constant $0 < k < 1 $ exists with $|g'(x)| \le k, \forall\ x \in(a,b)$.

I'm confused about the last part. What exactly does "a constant $0 < k < 1 $ exists with $|g'(x)| \le k, \forall\ x \in(a,b)$" mean? Why does the absolute of the derivative of $g$ have to be less than 1 for the fixed point to exist?

Edit:

I realized that including the conclusion is probably a good idea:

Then, for any number $p_0$ in $[a,b]$, the sequence defined by $p_n = g(p_{n - 1}), n \ge 1$, converges to the unique fixed point $p$ in $[a,b]$.

Thank you

Answer 1

It is related to Banach's fixed point theorem which imposes $g:[a,b]\rightarrow [a,b]$ and $\exists k: 0\leq k< 1$ such that $\forall x,y \in [a,b]$: $$|g(x)-g(y)| \leq k |x-y|$$ also known as a contraction mapping.

On the other hand mean value theorem states that, if the function is differentiable on $(a,b)$, then $\exists c\in (x,y)$ (let's assume $x < y$) such that $$g(x)-g(y)=g'(c)(x-y)$$ Combining this with the fact that $|g'(x)| \leq k < 1, \forall x \in (a,b)$ we have $$|g(x)-g(y)|=|g'(c)||x-y| \leq k|x-y|$$ which satisfies Banach's fixed point theorem, thus there $\exists!x^* \in [a,b]: g(x^*)=x^*$.

Answer 2

Actually, it suffices that $g$ is continuous (not differentiable) to show that $g$ has a fixed point in $[a,b]$: If $g(a)=a$ or $g(b)=b$, we are done. So assume $g(a)-a>0$ and $g(b)-b<0$. By the IVT, $x\mapsto g(x)-x$ has a zero within $(a,b)$.

Answer 3

The condition on the derivative assures that the Banach-Picard fixed point scheme works. The fixed point is unique and you may obtain it numerically very efficiently simply using $g$ to iterate, i.e.

Let $x_0\in [a,b]$ and define $x_{n+1}=g(x_n)\in [a,b]$.

The MVT implies that the sequence $(x_n)$ is Cauchy: First, $$|x_{n+1}-x_n|\leq k|x_n-x_{n-1}| \leq ... \leq k^n |x_1-x_0|\leq k^n (b-a).$$ This implies for $m>n>1$ that $|x_m-x_n|\leq k^n(b-a)/(1-k)$ whence convergence to some (unique) $x^*\in [a,b]$ as well as an error estimate: $$ |x^*-x_n | \leq \frac{k^n (b-a)}{1-k} .$$ I can add details if needed.