# Showing uniqueness of a fixed point on $[0,1]$

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

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 $$\sin^2(1)$$.

• What makes you doubt your argument? It is fine. – Kavi Rama Murthy Feb 14 at 5:42
• @KaviRamaMurthy i wasn't sure about my reasoning, thank you – Dillain Smith Feb 14 at 5:47
• 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. – GReyes Feb 14 at 6:14

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 $$(0,1]$$