Is it possible to prove that if $f\in L^2(\mathbb {R}) $ then $\exists\lim_{x\to\pm\infty}\lvert f\rvert^2$ and $\lim_{x\to\pm\infty}\lvert f\rvert^2=0$? If not, is it easy to find a counterexample?
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5Let $f(x)=n$ on $[n,n+{1\over n^4}]$, $n$ a positive integer, and $0$ otherwise. – David Mitra Jul 16 '15 at 13:41
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If we add the hypothesis $f\in\mathcal{C}^0$ is it still so easy? I was thinking to something like a wavefunction in QM for example. Thanks! – Red Lex Jul 16 '15 at 13:52
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2Instead of steps over $[n,n+1/n^4]$, take "spikes". (You can smooth things out as nicely as you wish.) – David Mitra Jul 16 '15 at 13:53
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A standard condition to have a zero limit is $f \in W^{1,2}(\mathbb{R})$, i.e. $f \in L^2$ and $f' \in L^2$. – Siminore Jul 16 '15 at 14:00
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3The shortest answer to the title: may be nasty. – A.Γ. Jul 16 '15 at 14:00
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If you assume $f$ to be uniformly continuous, your claims holds. Also, one can show that $f(x+n)\to0$ for almost all $x$ (for arbitrary $f\in L^p$, $p<\infty$). – PhoemueX Jul 16 '15 at 15:07
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possible duplicate of Integrable function $f$ on $\mathbb R$ does not imply that limit $f(x)$ is zero. See also http://math.stackexchange.com/questions/543866/finding-examples-for-a-non-negative-and-continuous-function-for-which-the-infini and http://math.stackexchange.com/q/92105/. – Jonas Meyer Jul 16 '15 at 17:48