To begin with, the following is something I need to prove in order to solve a bigger problem of mine rather than something my professor assigned to me and this is why I put "Question" instead of "Exercise" in the title of my post. I'm not even sure if the answer is positive.
Given a function $\;f :\mathbb R \to \mathbb R^n\;$ which is a classical solution of the PDE:
$\;f_{xx}=G_f(f)\;$ where $\;f_{xx}=\frac{{\partial}^2 f}{\partial x^2}\;$ and $\;G_f=(\frac{\partial G}{\partial f_1}, \dots, \frac{\partial G}{\partial f_n})^{T}\;$ for some $\;G \in \mathcal C^3(\mathbb R^n;{\mathbb R}_{+})\;$
denote the set $\;\mathcal A=\{f(x-τ)|τ\in \mathbb R\}\subset L^2_{loc}\;$.
If $\;\vert f(x)-l_{\pm} \vert \le Ce^{-\vert x \vert}\;$ as $\;\vert x \vert \to \infty\;$ where $\;\vert \cdot \vert \;$ stands for the Euclidean norm, then is it true that $\;\mathcal A\;$ is a compact set?
The reason I concluded to this thought, is that since $\;f\;$ approaches $l_{\pm}\;$ as $\;\vert x \vert \to \infty\;$ exponentially, then this should imply some sort of compactness for $\;\mathcal A\;$ but I have no idea how to prove it. I tried to take a sequence in $\;\mathcal A\;$ and find a convergent subsequence by Ascoli-Arzela Theorem but then I need to show $\;\mathcal A\;$ is also closed in order for the limit to belong in the set...
Last but not least, I 'd like to ask one more thing. Since $\;f\;$ is a classical solution, doesn't that mean $\;f\;$ is also continuous? Why $\;\mathcal A\;$ is a subset only of $\;L^2_{loc}\;$?
I don't really ask for a detailed solution (if does exist one) rather than some hints on how to proceed. Any help would be valuable.
Thanks in advance!
$\mathcal{A}$ is not closed, unless $f$ is a constant function. Take any sequence $\tau_n\rightarrow \infty$ and a compact set $K \subset \mathbb{R}$. Then $$ \int_K \vert f(x-\tau_n)-l_+\vert^2 dx \le C \int_K e^{-\vert x - \tau_n\vert^2} dx \rightarrow 0, $$ hence the function $L_+\in L^2_\text{loc}\mathbb{R}$ with $L_+(x) \equiv l_+$ is in the closure of $\mathcal{A}$. But it does not lie in $\mathcal{A}$, because (unless $f\equiv l_+$), there is no $\tau$ such that $f(\cdot + \tau) = L_+$.
$\mathcal{A}$ has a compact closure: Take a homeomorphism $\sigma\colon(-1,1)\rightarrow \mathbb{R}$ (say $\sigma(t)=\tan(\pi t/2))$ and define $\Psi\colon[-1,1]\rightarrow L^2_\text{loc}\mathbb{R}$ by $\Psi(t) = f(\cdot - \sigma(t))$ for $-1< t < 1$ and $\Psi(\pm1)=L_\pm.$ Then the argument from above shows that $\Psi$ is continuous, hence $\Psi([-1,1])$ is compact. We have $\mathcal{A}\subset \Psi[-1,1]$ and hence it has compact closure.