Use the Contraction Mapping Principle to show that $x=\frac19\sin\left(3x\right) + \sqrt{x}$ has exactly one solution $x\geqslant\frac{8}{9}$.

I have literally no idea if this is right, please could someone check my answer?

Let $T$ be defined by $Tx=\frac19\sin\left(3x\right) + \sqrt{x}$. To prove that $T$ is a contraction: $|Tx-Ty| = \left|\frac{\sin(3x)}{9} + \sqrt{x} - \frac{\sin(3y)}{9} - \sqrt{y}\right| \le \max\left|\frac {\ 3cos(3c)}9 + \frac 1{2\sqrt{c}}\right| \times |x-y|$

by the Mean value theorem

$= \left(\frac{1}{2}+\frac{1}{3}\right) |x-y| = \frac{5}{6} |x-y| $

$\lambda = 5/6 $

Which proves the Contraction (although I haven't shown that the metric space $X$ is complete! How would I do this?)

Using iterative methods, with $x_0 = 1$, the fixed point converges quite quickly to $1.01877$.

Thank you!

  • $\begingroup$ Where is the $\frac12$ part of $\lambda$ coming from? The inequality $\frac1{2\sqrt{z}}\leqslant\frac12$ does not hold for every $z$ in $[\frac89,\infty)$. $\endgroup$ – Did Jan 19 '15 at 7:53
  • $\begingroup$ As Did commented, you also need to show that $T$ maps $[8/9,\infty)$ into itself. $\endgroup$ – Braindead Jan 19 '15 at 15:31
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    $\begingroup$ @Surb Watch out your edits: $\sin\left(\frac{3x}9\right)\ne\frac19\sin(3x)$. $\endgroup$ – Did Jan 19 '15 at 20:38
  • $\begingroup$ @Did Sorry, I'll be more careful in the future. $\endgroup$ – Surb Jan 19 '15 at 20:41

$[\frac 8 9,\infty)$ is a closed subset of a complete metric space ($\mathbb R)$, hence it is complete. Simple proof that closed subsets of complete spaces are complete: The space itself is complete, so any cauchy sequence in the subspace has a limit in the space itself. But then it's a convergent sequence in the big space, so since the subset is closed, it contains all of its limit points.

  • $\begingroup$ Thank you for clearing that up, I'd seen that theorem but it still isn't obvious to me why an infinite subset is closed. I think I'm just being stupid! $\endgroup$ – Kate Sinclair Jan 18 '15 at 12:48
  • $\begingroup$ It's closed because it's complement is open. $\endgroup$ – Alan Jan 18 '15 at 12:51
  • $\begingroup$ This assumes crucially, but does not prove, that the interval $[\frac89,\infty)$ is invariant by $T$, that is, that $Tx\geqslant\frac89$ for every $x\geqslant\frac89$. $\endgroup$ – Did Jan 19 '15 at 7:48

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