# 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.

1. 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 $\sqrt{\frac{a}{3}}$.
1. 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 $f(x) = x$.
2. 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.
• Thanks, that helped quite a lot. The only thing I'm still struggling with: why is it that in 3., the inequality holds true for n = 0? When writing down the inequality $|\sqrt{a} - \frac{x_0 + \frac{a}{x_0}}{2}| ≤ |\frac{a}{x_0} - x_0|$, I couldn't rearrange it to see why it's true for all $x \in (0, \infty)$. How can I show this? Once it's shown for $n = 0$, the rest works fine per induction. It would be appreciated if you could take a look at this. – moran May 14 '15 at 14:16
• OK. Note that $\displaystyle \left| \frac{x_0 + \frac{a}{x_0}}{2} - \sqrt{a} \right| = \frac{1}{2x_0} \cdot \left| x_0 - \sqrt{a} \right|^2$ and $\displaystyle \left| \frac{a}{x_0} - x_0 \right| = \frac{1}{x_0} \cdot \left| x_0 - \sqrt{a} \right| \left| x_0 + \sqrt{a} \right|$. There's a common factor that can be pulled out. In fact, the $\frac{1}{2}$ is even unnecessary, so a stronger inequality with $\frac{1}{2^{n+1}}$ also holds. – Adayah May 14 '15 at 19:50