I want to be able to show that the function $g(x) = 2^{-x}$ has a unique fixed point contained within the interval $[1/3,1]$, and that the fixed-point iteration $$x_{n+1} = g(x_n), \qquad n \in \mathbb{N}$$
converges to this unique fixed point for all starting points $x_0 \in [1/3, 1]$.
How can one go about demonstrating something like this? Any help/ guidance would be appreciated.
Best Answer
Relying on Banach-Caccioppoli fixed-point theorem, the following result can be proven:
Theroem: Let $f$ be a function which is continuous on a closed interval $[a,b]$, and such that $f(x) \in [a,b]$ $\forall x \in [a,b]$. Then $f$ has at least one fixed-point $\alpha \in [a,b]$.
If $f \in C^{1}(a,b)$ and there exist a constant $\beta <1$ such that $|f'(x)| \leq \beta <1 \forall x \in (a,b)$ then we can conclude that there is a single fixed-point in $[a,b]$. In this case, for every $x_{0} \in [a,b]$ the iteration $x_{n+1}=f(x_{n})$ converges to the unique fixed-point $\alpha$.
So, in your case: $g(x)=2^{-x}$ is continuous in the interval $[\frac{1}{3},1]$. We know that: $g(\frac{1}{3})=2^{-\frac{1}{3}}<1$ and $g(1)=\frac{1}{2}>\frac{1}{3}$. The function $g$ is streactly decreasing, so $g(x)\in [\frac{1}{3},1]$ $\forall x \in [\frac{1}{3},1]$. Furthermore: $|g'(x)|=|-\frac{log(2)}{2^x}|\leq \frac{log(2)}{2^{\frac{1}{3}}}<1$ $\forall x \in [\frac{1}{3},1]$. You can conclude applying the theorem we've seen before.