The necessary condition for $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ to be integrable is that $\psi(x)\rightarrow 0$ as $x\rightarrow\pm\infty$. But this is not the sufficient condition. For example, $\delta(x)$ vanishes as $x\rightarrow \pm \infty$, but it is not square-integrable. What is the sufficient criterion for $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ to converge? In the previous question $\psi(x)$ was an arbitrary function. If now $\psi(x,t)$ be an arbitrary solution (not necessarily a stationary state) of the time-dependent Schrodinger equation what is the sufficiency condition for $\int\limits_{-\infty}^{+\infty}|\psi(x,t)|^2dx$ to be finite? Does this conditions remain valid in three-dimensions?   EDIT: Can we have a continuous function $\psi(x)$ which does not go to zero as $x\rightarrow \pm\infty$ and yet the integral $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ converges?
I'm studying Quantum Mechanics and I came across this which I don't quite understand: For a vector space of functions $f(x)$ to be square integrable (i.e $\int{|f(x)|^2dx &lt; \infty)}$, the necessary condition is $\lim_{|x| \to \infty}f(x) \to 0$. Can someone help me understand why that must be the case? I sort of understand that it must not blow up at the infinities but can a (continuous or piece-wise continuous) function have a non-zero finite value at the infinities and still be square integrable?
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