<s>[INST] Application of the Banach fixed point theorem Let $a > 0$. We consider the function: $f: (0, \infty) \to (0, \infty)$, defined by $f(x) = \frac{1}{2}(x + \frac{a}{x})$.
Let $(x_n)_{n \in \mathbb{N}_0}$ be defined by:
$x_0 \in (0, \infty)$, $x_{n+1} := f(x_n)$
What is the smallest $b > 0$ so that f is contracting on $[b, \infty)$, and what is $f's$ Lipschitz-constant $L > 0$?
Also, I want to prove using the Banach fixed point theorem, that $(x_n)$ as defined above converges against $\sqrt{a}$.
Finally, why is it that $|f(x_n) - \sqrt{a}| ≤ \frac{1}{2^n}|\frac{a}{x_0} - x_0|$?
Thanks in advance. I'm not very familiar with the Banach fixed point theorem, so I've been struggling with these questions so far.
 [/INST] *

*To answer that question, it is most convenient to solve $|f'(x)| < 1$. Since $f'(x) = \frac{1}{2} \left[ 1 - \frac{a}{x^2} \right],$ the solution for positive $x$ is $x \in \left( \sqrt{ \frac{a}{3} }, \infty \right)$.
So:

*

*for every $b > \sqrt{\frac{a}{3}}$ there is such $L \in (0, 1)$ that $|f'(x)| \leqslant L$ for $x \geqslant b$ and so by the mean value theorem $|f(x) - f(y)| \leqslant L|x-y|$ for $x, y \in [b, \infty)$, thus $f$ is Lipschitz on $[b, \infty)$

*$f$ is not a contraction on $\left[ \sqrt{\frac{a}{3}}, \infty \right)$, since $f' \left( \sqrt{\frac{a}{3}} \right) = -1$, so $\displaystyle \frac{\left| f \left( \sqrt{\frac{a}{3}} \right) - f(x) \right|}{\left| \sqrt{\frac{a}{3}} - x \right|}$ is arbitrarily close to $1$ for $x$ sufficiently close to $\sqrt{\frac{a}{3}}$.



Therefore there is no smallest $b$, but "a limit value" is $\sqrt{\frac{a}{3}}$.


*Fine. For any $x > 0$ we have $$f(x) = \frac{x+\frac{a}{x}}{2} \geqslant \sqrt{ x \cdot \frac{a}{x} } = \sqrt{a},$$ so we have $x_1 \in [ \sqrt{a}, \infty)$ and $x_{n+1} = f(x_n)$ and $f : [\sqrt{a}, \infty) \to [\sqrt{a} \to \infty)$ is a contraction (as shown above). Just apply Banach theorem and show that $x = \sqrt{a}$ is the only solution of $f(x) = x$. 

*Since $$\left| f(x) - \sqrt{a} \right| = \left| \frac{1}{2} \left( x + \frac{a}{x} \right) - \sqrt{a} \right| = \frac{1}{2} \left| x - \sqrt{a} \right| \cdot \left| 1 - \frac{\sqrt{a}}{x} \right| \leqslant \frac{1}{2} \left| x - \sqrt{a} \right|$$ for $x \geqslant \sqrt{a}$, verify for $n = 0$ and use induction. 
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