Mathematics Asked by Dave Lunal on October 25, 2020
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 $fin L^p$ for some $pgeq 1$. Then $|f(x)|to 0$ as $|x| to infty$.
Another interesting answer is this one.
My questions are:
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 $$f(x)=f(0)+int_0^xf'(t),dttext,$$ and $f'in L_1(mathbb R)$ implies that $$lim_{xtopminfty},int_0^xf'(t),dt$$ exist which can be seen from a Cauchy-like criterion, observing that $$lim_{Ntoinfty},int_{mathbb Rsetminus[-N,N]}lvert f'(t)rvert,dt=0.$$ (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 $fin L_p(mathbb R)$ would be used here only to verify that the limits are not different from zero...
Answered by Martin Väth on October 25, 2020
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