wave $\f_c$, assume that a solution to $\eqref{eq:magma}$ $\f$, $\f-1\in C^{1}([0,T):H^{1}(\mathbb{R}))$, satisfies $$\dist(\f(t),\f_c) < \norm{\f_c-1}_\h{1}$$ for $t<T$. Then the infimum of the function $\rho$, $$\rho(y) =  \norm{\f(\cdot)-\f_c(\cdot+y)}_\h{1}^2$$ is achieved at a finite value of of $y\in \R$. \[lem:finitemin\]

$\rho(y)$ is obviously continuous, and because $\phi(\cdot+y)-1 \rightharpoonup 0$ in $H^1$ as $y\to \pm \infty$, $$\lim_{y\to \pm \infty} \rho(y) = \norm{\f-1}_\h{1}^2 + \norm{\f_c-1}_\h{1}^2 > \norm{\f_c-1}_\h{1}^2$$ But by assumption, $$\inf_{y\in \R}\rho(y) = \inf_{y\in \R} \norm{\f(\cdot)-\f_c(\cdot+y)}_\h{1}^2 = d(\f,\f_c)^2 <  \norm{\f_c-1}_\h{1}^2$$ So there must be some finite $y_0$ such that $\rho(y_0)<\rho(\pm\infty)$. By the continuity of $\rho$, there then exists an $x_0(t)$ at each $t<T$ such that $\rho(-x_0(t))\leq \rho(y)$ for all $y\in\R$.

There therefore exists a function $x_0(t)$ such that $$\dist(\f,\f_c) = \norm{\f(\cdot,t)-\f_c(\cdot - x_0(t))}_\h{1}$$ Since $\f_c$ is in fact smooth, so is $\rho(y)$, and hence $\rho'(-x_0(t))=0$, $$\label{eq:constraint2}
\int_{-\infty}^{\infty} \bracket{I - \dx^2}\dx\f_c u dx=\inner{ \bracket{I - \dx^2}\dx\f_c }{u}=0$$ where we used the decomposition . is a second constraint on the perturbation to $\f_c$, which together with the near-orthogonality condition , will be shown to yield local convexity of $\calE$ near $\f_c$.

Properties of the Linear Operator, $L_c$
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Here we summarize properties of $L_c$, and exhibit the non-positivity of $L_c$.

(Properties of the Linear Operator $L_c$) 