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I cannot seem to find a contraction factor such that $$Tx = \frac{x}{2}+\frac{1}{x}$$ is a contraction on the whole set $[1,\infty)$ in the complete normed space $(\mathbb{R}, |\cdot|)$.

My argument for $x,y\in [1,\infty)$:

\begin{align} d(Tx, Ty) &= \left|\frac{x}{2}+\frac{1}{x} - (\frac{y}{2}+\frac{1}{y})\right| \newline \newline &= \left|\frac{x-y}{2}+\frac{1}{x} - \frac{1}{y} \right| \newline \newline &=\left|\frac{x-y}{2}+\frac{y-x}{xy} \right| \newline \newline &\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{xy} \right| \newline \newline &\leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{2} \right|,\quad \forall x,y \geq \sqrt{2} \newline \newline &=\frac{1}{2}|x-y| +\frac{1}{2}|x-y| \newline \newline &= |x-y|. \end{align}

Hence, restricting $T$ to $[\sqrt{2},\infty)$ yields a non-expansive mapping. $T$ also has a fixed point at $x=\sqrt{2}$. However, the book in which the exercise was found states that $T$ is a contraction with a minimal contraction factor $\lambda \in [0,1)$. Can I take better estimates to show that $T$ is a contraction on $[1,\infty)$?

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    $\begingroup$ Just before you apply your inequality, have you tried putting over a common denominator? $\endgroup$ – Patrick Da Silva Jan 18 '12 at 4:57
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    $\begingroup$ The mean value theorem is nice here too. $\endgroup$ – Ragib Zaman Jan 18 '12 at 5:01
  • $\begingroup$ All great ideas, thank you. $\endgroup$ – Henry Shearman Jan 18 '12 at 5:04
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When you wrote

$$\left|\frac{x-y}{2}+\frac{y-x}{xy} \right| \leq\left|\frac{x-y}{2}\right| +\left|\frac{x-y}{xy} \right|,$$

you didn't take advantage of the fact that $x-y$ and $y-x$ have opposite signs, so there is cancellation that will give you a better estimate. Instead you could note that

$$\left|\frac{x-y}{2}+\frac{y-x}{xy} \right|=|x-y|\left|\frac{1}{2}-\frac{1}{xy}\right|,$$

and that $$-\frac{1}{2}\leq \frac{1}{2}-\frac{1}{xy}<\frac{1}{2}$$

when $x,y\geq 1$.

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  • $\begingroup$ Heh, nice follow-up of my comment =) But I wouldn't be surprised if you just thought about it yourself. +1 $\endgroup$ – Patrick Da Silva Jan 18 '12 at 5:03
  • $\begingroup$ Thanks Patrick. Indeed I saw your comment after I posted my answer (I can't type that fast!), but I'm not sure it would be a follow up, as I did not follow your approach of finding a common denominator. $\endgroup$ – Jonas Meyer Jan 18 '12 at 5:05

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