A vanishing density prevents a uniform linear spatial inverse bound

For $0<\epsilon \le 1$, [F1] gives $\widehat{x}=\sqrt{\epsilon }\ge 0$ and ${\widehat{x}}^2=\epsilon >0$, so $\widehat{x}>0$. If $\widehat{x}>1$, then ${\widehat{x}}^2>1$, contradicting $\epsilon \le 1$; hence $\widehat{x}\in [0,1]$ and $F(\widehat{x})={\widehat{x}}^2=\epsilon$. Under the stated endpoint convention, the target is $0$, so the residual is $\epsilon$ while the spatial error is $\sqrt{\epsilon }$.

The ratio $\sqrt{\epsilon }\mathrm{/}\epsilon =1\mathrm{/}\sqrt{\epsilon }$ is unbounded as $\epsilon \downarrow 0$: for any $M>0$, take $0<\epsilon <\min (1,1\mathrm{/}(M+1{)}^2)$, so [F1] gives $\sqrt{\epsilon }<1\mathrm{/}(M+1)$ by comparing nonnegative squares, and the ratio exceeds $M$. Thus no constant $C$ independent of $\epsilon$ gives $\mathrm{\mid }\widehat{x}-Q(0)\mathrm{\mid }\le C\epsilon$ for these points; their nonlinear error is exactly $\sqrt{\epsilon }$. The density $f(x)=2x$ has essential infimum $0$ on every $[0,h]$ with $0<h\le 1$, so the sufficient positive-density condition in [F2] fails near zero. The endpoint computation uses the declared convention $Q(0)=0$, independently of [F2]'s interior-target theorem.