Use continuous dependence upon the data of solutions of to prove that $\mathcal{I}$ is not empty.

-   Seek the maximal time $T_0$ in $\mathcal{I}$. If it is not $T_{\max}$, we will show that there is some time interval beyond $T_0$ for which $\dist(\f(t),\f_c)<2\ep$.

-   Prove that for any $t$ such that $\dist(\f(t),\f_c)<2\ep$, in fact $\dist(\f(t),\f_c)<\ep$, producing the contradiction.

First we prove that $\mathcal{I}$ is not empty. This is an application of Theorem \[thm:local-existence\]. $$\norm{\f_0-1}_\h{1} \leq \frac{3}{2}\norm{\f_c-1}_\h{1} \quad\mbox{and}\quad \norm{\frac{1}{\f_0}}_{L^\infty} \leq \frac{4}{3}$$ Taking $R=3/2\norm{\f_c-1}_\h{1} $, $\alpha = 3/8$, we know from part (a) of Theorem \[thm:local-existence\], that $\norm{\f(t)-1}\leq 2 R$ and $\norm{1/\f(t)}_{L^\infty} \leq 1/\alpha$ up till at least $T_\exist (R, \alpha)>0$ and $T_{\max}\geq T_\exist$.

By part (c) of the same Theorem, we have a constant $K_\cont=K_\cont (R, \alpha)$, such that $$\norm{\f(\cdot,t)- \f_c(\cdot -ct +x_0)}_\h{1} \leq \norm{\f_0-\f_c(\cdot+x_0)}_\h{1} e^{K_\cont t}< \delta  e^{K_\cont t}$$ Taking $t$ sufficiently small, there is some time interval over which $\f$ is within $\ep$ of $\f_c$.

Let $$T_0 = \sup \mathcal{I}$$ Suppose $T_0 < T_{\max}$. For any $t< T_0$, $$\dist(\f(t),\f_c)<\eps\leq  \frac{1}{2}\norm{\f_c-1}_\h{1}$$ Lemma \[lem:finitemin\] asserts there exists $x_0(t)\in \R$ such that $$\dist(\f(t), \f_c)= \norm{\f(\cdot,t)-\f_c(\cdot +x_0(t))}_\h{1}< \ep$$ Furthermore, for all $t< T_0$, we have again $$\norm{\f(t)-1}_\h{1} \leq \frac{3}{2}\norm{\f_c-1}_\h{1}\quad \mbox{and}\quad \norm{\frac{1}{\f(t)}}_\h{1} \leq \frac{4}{3}$$ 