Integer translations of $L^2(\mathbb{R})$ functions can never be complete in $L^2(\mathbb{R})$ This is Exercise 10.18 in A Basis Theory Primer, Christopher Heil. I am just reading this book and not a homework.

Given an arbitrary $g\in L^2(\mathbb{R})$, let $T(g):=\{g(x-k)\}_{k\in \mathbb{Z}}$. Then, show that $\overline{\mathrm{span}\{g(x-k)\}_{k\in \mathbb{Z}}}\not= L^2(\mathbb{R})$, that is, the set of all finite linear combinations of elements of $T(g)$ is not dense in $L^2(\mathbb{R})$.

I tried playing around with the ideas in the book, e.g., suppose that $T(g)$ is dense in $L^2(\mathbb{R})$. Then, since the Fourier transform is unitary from $L^2(\mathbb{R})$ to itself, in particular, continuous, the sequence of their Fourier transform 
$M(\hat{g}):=\{\mathrm{e}^{-2\pi\mathrm{i}k\xi}\, \hat{g}(\xi)\}_{k\in\mathbb{Z}}$
is dense in $L^2(\mathbb{R})$, and tried to derive a contradiction but didn't go anywhere.
 A: After the Fourier transform, the problem reduces to show that, for any function $f\in L^2(\mathbb{R})$, the set of products $pf$, where $p$ is a trigonometric polynomial $p(t)=\sum c_k \exp(2\pi i k t)$, is not dense in $L^2(\mathbb{R})$. The key point here is that $p$ is $1$-periodic. 
If $f=0$, the conclusion is obvious, so exclude this case. Then, for a sufficiently small $\epsilon>0$, there exists a set $A$ of positive measure, with $\operatorname{diam} A<1$, and such that  $|f|>\epsilon$ on $A$. 
Since $\sum_{n\in\mathbb{Z}}\int_{A+n}|f|^2 <\infty$, it follows that there exists $n\in\mathbb{Z}$ such that $|f|<\epsilon/2$ on most of the set $A+n$ (that is, on a subset that comprises, say, $99\%$ of the measure of $A+n$). 
It's impossible to approximate the function $\phi = \chi_A + \chi_{A+n}$ by any function of the form $pf$. Indeed, to have small integral $\int_A |pf-1|^2$ we must have $pf$ close to $1$ on most of the set $A$. But then $|pf|$ is less than $2/3$ on most of the set $A+n$, so $\int_{A+n} |pf-1|^2$ cannot be small.
