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.


  • $\begingroup$ $g([a,b]) \subset [a,b]$ is a further condition that may require choosing a smaller $\lambda$. Try to show that you have $g([a,b]) \subset [a,b]$ for all small enough $\lambda > 0$ by the compactness of $[a,b]$ and the continuity of $H(x,\lambda) = x - \lambda f(x)$. $\endgroup$ – Daniel Fischer Nov 8 '13 at 20:42
  • $\begingroup$ Should the inequality with $k_1$ and $k_2$ have $f'(x)$ instead of $f(x)$? $\endgroup$ – user103402 Nov 9 '13 at 6:07
  • $\begingroup$ @user103402 Yes. $\endgroup$ – Pedro Nov 9 '13 at 21:44
  • $\begingroup$ @DanielFischer By the compactness of $[a,b]$ and the continuity of $H$ we can conclude that $H$ attains maximum and minimum in $[a,b]$. Is this property of $H$ that should we use? How? $\endgroup$ – Pedro Nov 10 '13 at 0:39

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.

  • $\begingroup$ The other inequality is equivalent to $f(x)\geq\lambda^{-1}(x-b)$. So, how can we evaluate $\lambda$ if we there is $\geq$ instead of $\leq$? $\endgroup$ – Pedro Nov 10 '13 at 0:04
  • $\begingroup$ @Pedro You don't evaluate it; you choose it so that it satisfies the inequalities. Follow the same reasoning as above, but notice that $x-b$ is negative. The inequality will be of the form $\lambda^{-1}\epsilon \ge \max(-f)$. $\endgroup$ – user103402 Nov 10 '13 at 1:07

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