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Given $g(x)=-x\sin^2(\frac{1}{x})$ for $0<x\leq1$

My attempt: let fixed point given by $g(x)-x=-x\sin^2(\frac{1}{x})-x=0$

$$0=-x\left(\sin^2\left(\frac{1}{x}\right)+1\right) $$ Therefore only for $x=0$ will make the entire term $0$. Because the latter term is bounded below by one and is monotonically increasing from $\sin^2(0)$ to $\sin^2(1)$.

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    $\begingroup$ What makes you doubt your argument? It is fine. $\endgroup$ – Kavi Rama Murthy Feb 14 at 5:42
  • $\begingroup$ @KaviRamaMurthy i wasn't sure about my reasoning, thank you $\endgroup$ – Dillain Smith Feb 14 at 5:47
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    $\begingroup$ Your second term is not monotonic (it actually oscillates wildly) but this does not play any role. It is bounded below by 1 and that is what matters. $\endgroup$ – GReyes Feb 14 at 6:14
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The function $g(x)=-x\sin^2(\frac{1}{x})$ has NO fixed point in the interval $(0,1]$. The continuous extension of $g$ to $[0,1]$ has just one fixed point in $[0,1]$, i.e. $x=0=\lim_{x\to 0}g(x)=0$ (here we need the fact that $\sin^2(\frac{1}{x})$ is bounded).

In order to show the first part we don't need to the monotone property (which is false in this case) or the bounded property. The equation $$0=-x\left(h^2(x)+1\right)$$ has NO solutions in $(0,1]$ because because $x>0$ and $h^2(x)\geq 0$ implies $h^2(x)+1\geq 1>0$ for ANY function $h$ defined in $(0,1]$

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