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Show that the function $f:\mathbb{R}^{3} \rightarrow \mathbb{R}^3$ given by $(x,y,z) \mapsto \bigg(\frac{\sin y}{4},\frac{\sin z}{3}+1,\frac{\sin y}{4}+2 \bigg)$ has a unique fixed point.

Attempt

By the contraction mapping theorem if $\|{f(x)-f(y)}\| \leq K\|{x-y}\|$ for $x,y \in \mathbb{R}^{3}$ and $0\leq K < 1$ then there exists a unique fixed point. I've tried showing this using the norms such as the euclidean and max norms but couldn't get it to come out. Can someone suggest a better norm to try.

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Let's try with the maximum norm. \begin{align} \|f(x,y,z)-f(x',y',z')\|&=\max\Bigl(\frac{|\sin y-\sin y'|}{4},\frac{|\sin z-\sin z'|}{3},\frac{|\sin y-\sin y'|}{4}\Bigr)\\ &\le \frac13\max(|\sin y-\sin y'|,|\sin z-\sin z'|)\\ &\le \frac13\max(|y-y'|,|z-z'|)\\ &\le\frac13\,\|(x,y,z)-(x',y',z')\|. \end{align} Similar calculations prove it for any other norm.

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  • $\begingroup$ Thank you , looks like I just did something stupid when trying the max norm before. $\endgroup$ – Roger Feb 13 at 18:50

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