Open-closed bootstrap continuation

Fix a finite $T<T_{\max }$ and let $S_T=\{t\in [0,T]:E(t)\ge 2A{ϵ}^2\}$. If $S_T$ is nonempty, [F3] gives $\tau =\inf S_T$. Since $0$ is a lower bound and every element of $S_T$ is at most $T$, [F2] gives $0\le \tau \le T$; continuity and $E(0)<A{ϵ}^2$ then give $\tau >0$. Since $\tau$ is a lower bound, $E(t)<2A{ϵ}^2$ for every $t<\tau$. If $E(\tau )>2A{ϵ}^2$, continuity would put points just below $\tau$ in $S_T$, contradicting that it is a lower bound; hence $E(\tau )\le 2A{ϵ}^2$. If $\tau =T$, nonemptiness and $\inf S_T=T$ force $T\in S_T$, so equality holds. If $\tau <T$ and $E(\tau )<2A{ϵ}^2$, continuity gives $0<\delta <T-\tau$ such that $E(t)<2A{ϵ}^2$ on $[\tau ,\tau +\delta )$; no point of $S_T$ lies below $\tau$ or in this neighborhood, so $\tau +\delta \mathrm{/}2$ is a larger lower bound, contradicting [F2]. Thus in either case $E(\tau )=2A{ϵ}^2$. The improvement in [F1] on $[0,\tau ]$ gives $E(\tau )\le A{ϵ}^2$, a contradiction. Hence $S_T=\mathrm{\varnothing }$.

Since $S_T=\mathrm{\varnothing }$, $E(t)<2A{ϵ}^2$ on $[0,T]$. Applying [F1] gives $E(t)\le A{ϵ}^2$ on $[0,T]$. The choice of finite $T<T_{\max }$ was arbitrary, so this bound holds throughout the maximal interval.

If $T_{\max }<\mathrm{\infty }$ and the margins persist, step 2.1 gives $E(t)\le A{ϵ}^2<\eta$ for every $t<T_{\max }$. The continuation criterion in [F1] then extends the solution past $T_{\max }$, contradicting maximality. Therefore $T_{\max }=\mathrm{\infty }$.