satisfying , , and , the perturbation will satisfy the two inequalities $$p_-\paren{\norm{u}_\h{1}} \leq \abs{\Delta\calE}\leq p_+\paren{\norm{u}_\h{1}}$$ $p_\pm$ defined in and . Since $\Delta \calE$ is time independent, if the perturbation at time $t=0$ is sufficiently small, $\abs{\Delta\calE}$ may be made arbitrarily small. Provided conditions , , and continue to hold, this will constrain $\norm{u}_\h{1}$ through $p_-$.

Let $\ep_\star$ be defined as $$\ep_\star = \min \left\{\frac{1}{4}, \frac{1}{2}\norm{\f_c-1}_\h{1},\frac{C_2}{3\paren{C_3+2 D_3}} \right\}$$ which depends only on $\f_c$. The significance of the three quantities is:

-   $1/4$ will ensure $\f$ is bounded away from zero, as needed by our existence theory.

-   $1/2 \norm{\f_c-1}_\h{1}$ will ensure the value at which the sliding metric is minimized is finite; see Lemma \[lem:finitemin\].

-   $C_2/(3C_3 + 6 D_3)$ will ensure that the perturbation reamins to the left of the peak of the polynomial $p_-$.

Let $\ep\leq \ep_\star$ and let $\delta>0$ be sufficiently small such that $$\begin{aligned}
\delta < \ep\\
p_+(\delta)< p_-(\ep)\end{aligned}$$

Letting $\f(t=0)\equiv\f_0$, assume there is $x_0\in \R$ such that $$\norm{\f_0-\f_c(\cdot+x_0)}_\h{1} =\norm{u_0}_\h{1}<\delta$$ $\f-1\in C^1\paren{[0,T_{\max});\h{1}}$. With these choices of $\delta$, $\ep$, and $\ep_\star$, we will show that for $t\in [0,T_{\max})$, $\dist(\f(t),\f_c)<\ep$.

Let $$\mathcal{I}=\left\{t: \dist(\f(t'),\f_c)<\eps\quad\mbox{for $t'< t$.}\right\}$$ We will use $\mathcal{I}$ to prove the theorem by contradiction as follows:

-   