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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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    All great ideas, thank you.2012-01-18

1 Answers 1

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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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    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.2012-01-18