We use these bounds to control how far $\f$ can deviate from $\f_c$ beyond $T_0$. Taking $R=3/2 \norm{\f_c-1}_\h{1}$ and $\alpha= 3/8$, there exists $T_\exist (R, \alpha)>0$ such that $\f(t)$, for $t< T_0$, may be continued in time by the amount $T_\exist$ and will satisfy $\norm{\f(t)-1}\leq 2R $, and $\norm{1/\f(t)}_{L^\infty} \leq 1/\alpha$. Because the solution is unique, these bounds apply to $\f$ for $t< T_0+ T_\exist$.

With this control on norms, we let $t_1< T_0$ be a new starting point, and apply part (c) of Theorem \[thm:local-existence\] $$\norm{\f(\cdot, t)-\f_c(\cdot -c t+ x_0(t_1))}_\h{1}\leq \norm{\f(\cdot,t_1) - \f_c(\cdot+x_0(t_1)}_\h{1}e^{K_\cont (t-t_1)}< \eps e^{K_\cont (t-t_1)}$$ Making $t_1$ sufficiently close to $T_0$, we can find $T_2 > T_0$ for which $$\norm{\f(\cdot, t)-\f_c(\cdot -c t+ x_0(t_1))}_\h{1}=\norm{u(t)}_\h{1}< 2 \ep$$ for $t< T_2$. We claim this implies the stricter estimate $\norm{\f(\cdot, t)-\f_c(\cdot -c t+ x_0(t_1))}_\h{1}<\ep$ for $T_0<t<T_2$.

$2\ep\leq 2\ep_\star \leq \norm{\f_c-1}_\h{1}$; applying Lemma \[lem:finitemin\] to find $x_0(t)$ again, and then decompose our solution via . As noted in the remark following Lemma \[lem:finitemin\], this perturbation satisfies $\inner{\xi_2}{u}=0$. $\calN[\f] = \calN[\f_c]$ and $$\norm{u}_\h{1}<2\ep\leq 2\ep_\star \frac{1}{2}$$ Therefore $u$ satisfies the neccessary conditions to apply Proposition \[prop:lowerbound\], $$p_-\paren{\norm{u}_\h{1}}\leq \abs{\Delta \calE}\leq p_+(\delta)< p_-(\ep)$$ Because $\norm{u}_\h{1}< 2\eps\leq 2\ep_\star \leq 2C_2/(3 C_3 + 6 D_3)$, it sits 