Normalizable wavefunction that does not vanish at infinity

Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.

Let $$ \psi(x) = \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases} = \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) , $$ where $\mathbf 1_A$ is the characteristic function of the set $A$. Then $$ \int_{-\infty}^\infty |\psi(x)|^2 dx = \sum_{n=1}^\infty \frac 1 {n^2} < \infty, $$ but $\psi(x)$ doesn't converge to zero as $x \to +\infty$.

Note that $\psi \in L^2(\mathbb R)$, but it is not twice (weakly) differentiable and can therefore not be the solution to Schrödinger's equation with $H = -\Delta + V$. However, that problem can easily be solved by replacing the rectangle function with a smooth pulse with compact support. Alternatively, use $$ \psi(x) = x^2 \mathrm e^{-x^8 \sin^2 x} ,$$ as discussed in Example 2 of §2.1 in arXiv:quant-ph/9907069 -- this is even analytical.

Emilio Pisanty and Eckhard Giere have already given discontinuous, piecewise constant counterexamples in their answers. Here we provide for-the-fun-of-it a smooth infinitely-many-times-differentiable counterexample $f\in C^{\infty}(\mathbb{R})$ of a square integrable function $f:\mathbb{R} \to [0,1]$ that does not satisfy $\lim_{|x|\to \infty}f(x)=0$. Our counterexample is

$$\tag{1} f(x)~:=~ e^{- g(x)} ~\in ~]0,1], \qquad g(x)~:=~x^4 \sin^2 x~\in ~[0,\infty[. $$

Intuitive idea: If we imagine $x$ as a time variable, then the function $f$ returns periodically to its maximum value

$$\tag{2} f(x) =1 \quad\Leftrightarrow\quad g(x) =0 \quad\Leftrightarrow\quad \frac{x}{\pi}\in \mathbb{Z} ,$$

but spends most if its time close to the $x$-axis in order to be square integrable.

Proof: We leave a detailed rigorous epsilon-delta mathematical proof to the reader, but a sketched heuristic proof goes like this. For each very large integer $|n|\gg 1$, define a shifted variable

$$\tag{3} y~:=~x-\pi n.$$

For the fixed integer $n\in\mathbb{Z}$, always assume from now on that the $y$-variable belongs to the interval

$$\tag{4} |y|~\leq~ \frac{\pi}{2}.$$

For $|y|\ll\frac{\pi}{2}$ very small, we may approximate $g(x) \approx (\pi n)^4y^2$, so that in the interval (4), we have

$$\tag{5} g(x)~\lesssim~ \pi^4 |n| \quad \Leftrightarrow\quad |y| ~\lesssim~ |n|^{-\frac{3}{2}}.$$

Thus we may form a square integrable majorant function $h\geq f$ (outside a compact region on the $x$-axis) by defining

$$\tag{6} h(x)~:=~\left\{\begin{array}{lcl} 1 &{\rm for}& |y| ~\lesssim~ |n|^{-\frac{3}{2}}, \cr e^{-\pi^4 |n|}&{\rm for}& |n|^{-\frac{3}{2}}~\lesssim~ |y| ~\leq~ \frac{\pi}{2}, \end{array} \right. \qquad |n|\gg 1. $$

The function $h\in {\cal L}^2(\mathbb{R})$ is square integrable on the whole $x$-axis, since

$$\tag{7} \sum_{n\neq 0} |n|^{-\frac{3}{2}} ~<~ \infty$$

and

$$\tag{8} \pi \sum_{n\in\mathbb{Z}}e^{-2\pi^4 |n|}~<~\infty$$

are convergent series.