Banach Fixed Point Theorem: Consider a metric space $X = (X, d)$, where $X\neq \varnothing$. Suppose that $X$ is complete and let $T: X \to X$ be a contraction on $X$. Then $T$ has precisely one fixed point $\tilde{x}$. Furthermore, for any $x_0\in X$ we have $\tilde{x}=\lim\,x_n$ where $(x_n)$ is given by $x_1=x_0$ and $x_{n+1}=T(x_n)$.
Problem: Using Banach's theorem, set up an iteration process for solving $f(x) = 0$ if $f$ is continuously differentiable on an interval $J = [a,b]$, $f(a)<0<f(b)$ and $0<k_1\leq f'(x)\leq k_2$ ($x\in J$); use $g(x) = x-\lambda f(x)$ with a suitable $\lambda$. (This exercise was taken of Kreyszig's Functional Analysis book).
What I have tried: if we take $$\lambda=\left\{\frac{1}{2k_1},\frac{1}{2k_2}\right\}>0$$ and $\alpha=1-\lambda k_1\in(0,1)$ then (by Mean Value Theorem in $g$) we get $$|g(x)-g(y)|\leq\alpha|x-y|$$ for all $x,y\in[a,b]$. Hence, with this choose for $\lambda$ the function $g:[a,b]\to\mathbb{R}$ becomes a contraction on the metric space $[a,b]$. However, to use Banach's Theorem it's necessary to get $g([a,b])\subset[a,b]$. It seems my choice for $\lambda$ doesn't implies it. So, I need help.
Thanks.
You want to have $ x-\lambda f(x)\ge a$ for all $x\in [a,b]$. This is equivalent to $$ f(x) \le \lambda^{-1} (x-a), \quad x\in [a,b] \tag{1} $$ By assumption, $f<0$ in some neighborhood $[a,a+\epsilon)$; in particular, (1) holds there no matter what (positive) $\lambda$ you take. Outside of this neighborhood, the right hand side of (1) is at least $\lambda^{-1}\epsilon$. Choose $\lambda$ so that $$\lambda^{-1}\epsilon>\max_{[a,b]} f$$ and you are done.
The other inequality, $ x-\lambda f(x)\le b$, is similar.