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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.

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  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.

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  • $\begingroup$ 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. $\endgroup$ – moran May 14 '15 at 14:16
  • $\begingroup$ 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. $\endgroup$ – Adayah May 14 '15 at 19:50

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