In quantum mechanics we use quadratically integrable functions ($\psi \in L^2$). This means $$ \int_{-\infty}^\infty |\psi(x)|^2 \mathrm{d}x < \infty. $$
I'm interested in the question when those function vanish at infinity, i.e. $$ \lim_{x \rightarrow \pm \infty} \psi(x) = 0. $$
I know that this is not the case for every function in $L^2$, see for example this answer or this answer.
I found in a similar question something interesting:
Suppose $f : \mathbf R \to \mathbf R$ is uniformly continuous, and $f\in L^p$ for some $p\geq 1$. Then $|f(x)|\to 0$ as $|x| \to \infty$.
Another interesting answer is this one.
My questions are:
- How can one prove the given statement?
- What are other cases where quadratically integrable functions vanish at infinity?
- Which cases are relevant in physics (for quantum mechanics)?
Edit:
My first question was answered in the comments by @reuns.
My remaining question is:
What criteria (beside uniform continuity) do exist, so that quadratically integrable functions vanish (or not) at infinity?
A trivial sufficient condition is that $f$ is absolutely continuous with $f'\in L_1(\mathbb R)$. Indeed, the absolute continuity means that $$f(x)=f(0)+\int_0^xf'(t)\,dt\text,$$ and $f'\in L_1(\mathbb R)$ implies that $$\lim_{x\to\pm\infty}\,\int_0^xf'(t)\,dt$$ exist which can be seen from a Cauchy-like criterion, observing that $$\lim_{N\to\infty}\,\int_{\mathbb R\setminus[-N,N]}\lvert f'(t)\rvert\,dt=0.$$ (The latter follows for instance from Lebesgue's dominated convergence theorem.)
Note that $f'\in L_1(\mathbb R)$ does even for continuous $f'$ not imply the boundedness of $f'$ and thus also does not imply that $f$ is uniformly continuous.
On the other hand, the assumption $f\in L_p(\mathbb R)$ would be used here only to verify that the limits are not different from zero...