Eigenfunctions of quantum systems being uniformly continuous I'm trying to show that eigenfunctions of quantum systems tend to $0$ as $x\to \pm\infty$. 
Working in the Hilbert space $L^2(\mathbb{R})$ even though all functions are normalisable, they may not go to $0$ as $x\to \pm\infty$ - certain nasty examples of functions can be formulated.  
I believe it is the case that all uniformly continuous functions in $L^2$ do tend to $0$ as $x\to \infty$, so I need a proof that all eigenfunctions are uniformly continuous, or failing that, a proof that the eigenfunctions of a quantum system do indeed go to $0$ as $|x|\to \infty$.
 A: In order for a $1d$ Schrodinger problem to be well-posed, it is necessary that the Hamiltonian
$$
               Hf = -\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}f+Vf
$$
be essentially selfadjoint on the domain $\mathcal{C}_{c}^{\infty}(\mathbb{R})$ consisting of compactly-supported infinitely-differentiable functions. (If this were not true, then there would be a non-trivial boundary condition at $\infty$, and conditions at $\infty$ make no sense because they cannot be imposed, and that would result in an ill-posed physical problem.) For real functions $f\in\mathcal{C}_{c}^{\infty}(\mathbb{R})$,
$$
        \int_{-\infty}^{\infty}(Hf)fdx=\frac{\hbar^2}{2\mu}\int_{-\infty}^{\infty}(f')^2dx+\int_{-\infty}^{\infty}Vf^2dx.
$$
If $V$ is uniformly bounded below by some constant $C$, then $2|ab| \le a^2+b^2$ holds for all real numbers $a$ and $b$ and gives
$$
      \frac{\hbar^2}{2\mu}\int_{-\infty}^{\infty}(f')^2dx \le -C\int_{-\infty}^{\infty}f^2dx+2\int_{-\infty}^{\infty}\left(\frac{1}{2\sqrt{|C|+1}}|Hf|\right)\left(\sqrt{|C|+1}|f|\right)dx \\
  \le -C\int_{-\infty}^{\infty}f^2dx+\frac{1}{4(|C|+1)}\int_{-\infty}^{\infty}(Hf)^2dx+(|C|+1)\int_{\infty}^{\infty}f^2dx \\
  \le (2|C|+1)\int_{-\infty}^{\infty}f^2dx+\frac{1}{4(|C|+1)}\int_{-\infty}^{\infty}(Hf)^2dx.
$$
Because $H$ is essentially selfadjoint, the above remains true for all $f \in \mathcal{D}(H)$, which then forces $f'\in L^2(\mathbb{R})$ for all $f\in \mathcal{D}(H)$. Hence, $f'f \in L^1(\mathbb{R})$ for $f\in\mathcal{D}(H)$, which gives the existence of the following limits:
$$
       \lim_{r\rightarrow\pm\infty}2\int_{0}^{r}f(r)f'(r)dr = \lim_{r\rightarrow\pm\infty}f(r)^2dr.
$$
If either limit were non-zero, that would contradict the absolute integrability of $ff'$. Therefore,
$$
          \lim_{r\rightarrow\pm\infty}f(r)=0,\;\;\; f\in\mathcal{D}(H).
$$
In particular this must hold for any $f\in\mathcal{D}(H)$ for which $Hf=\lambda f$. So normalizable eigenfunctions of $H$ must vanish at $\pm\infty$ if, for example, the potential is semibounded below. I suspect this can relaxed somewhat, but it's a beginning.
