Given $f \in L^2(\mathbb{R})$, show that $\operatorname{span}\{f(x-n)\}_{n \in \mathbb{Z}}$ is not dense in $L^2(\mathbb{R})$? Given any $f \in L^2(\mathbb{R})$, show that the span of $\{f(x-n)\}_{n \in \mathbb{Z}}$ is not dense in $L^2(\mathbb{R})$?
This problem comes from this presentation with slides here. It is claimed under the "Remark" section that this set cannot be dense. They actually deal with a more general case but I choose $p=2$ for simplicity.
Since $L^2(\mathbb{R})$ is a Hilbert space, then the sequence above being dense in the space is equivalent to the following condition: If we have $g \in L^2(\mathbb{R})$ with $(g,f(x-n)) = 0$ (inner-product) for every $n \in \mathbb{Z}$, then $g = 0$ (in the $L^2(\mathbb{R})$ - sense).
The idea is to find a nonzero $g \in L^2(\mathbb{R})$ that satisfies the hypothesis above. However, I am unable to show this. Is there a simple function that gets the job done?
Thanks :)!
 A: I think the easiest proof of this doesn't rely on the result you mentioned. Simply look at the Fourier Transform of $f$. Define $e_n(\omega) := e^{2\pi i n \omega}$.
Suppose the span of $\{f(\cdot - n )\}_{n \in \mathbb{Z}} $ is dense in $L^2(\mathbb{R})$. After taking Fourier transforms, we have span$\{e_n \widehat{f} \}_{n \in \mathbb{Z}}$ is also dense. This means that for any $G \in L^2(\mathbb{R})$ , we have a sequence of trigonometric polynomials $P_k = \sum_{n \in I_k}a_ne_n$ such that $P_k\widehat{f}$ converges to $G$ in $L^2$.
$L^2$ convergence implies point-wise a.e convergence of a sub-sequence. So replacing $P_k $ with a subsequence, we get that for any $G \in L^2(\mathbb{R}) $, there exists a sequence of trigonometric polynomials $P_k $ such that $P_{k}(x)\widehat{f}(x) \to G(x)$ for almost all $x \in \mathbb{R}$. For this to be true, we must have $\widehat{f}$ nonzero almost everywhere since there are $G \in L^2({\mathbb{R}})$ nonzero a.e.
Let $G = \chi_{[0,1]}$ (Letting $G$ be any function which is $0$ on a sufficiently large set would work). Then $P_{k} \to \chi_{[0,1]}/\widehat{f}$ a.e. implies $P_k \to 0 $ a.e since it's periodic, thereby giving us a contradiction.
A: Take $a$ such that $$\int_a^{a+1}|\hat{f}(y)|^2dy\ne 0$$ then look at
$g$ such that $$\hat{g}(y)= 1_{y\in [a-1,a]} \overline{\hat{f}(y+1)} -
1_{y\in [a,a+1]} \overline{\hat{f}(y-1)}$$
Using unitary-ness of the Fourier transform we get that
$$\begin{eqnarray}\overline{\langle g,f(x-n)\rangle} &=&\overline{\langle \hat{g},e^{-2i\pi n y} \hat{f}\rangle} \\&=& \int_{a-1}^a 
\hat{f}(y+1) \hat{f}(y) e^{-2i\pi ny}dy-\int_a^{a+1}\hat{f}(y-1) \hat{f}(y) e^{-2i\pi ny}dy\\&=&0\end{eqnarray}$$
A: Not a full answer, I leave you to fill details. This idea works for $p \neq 2$ as well.
Suppose that $\mathrm{span} \{ f(x-n) : n \in \Bbb Z\}$ is dense in $L^2(\Bbb R)$. Then you could restrict everything to the domain $[0;1)$ and conclude that the span of $\{ f|_{[0;1)}\}$ is dense in $L^2([0;1))$. This is impossible since a one-dimensional vector space is not dense in $L^2([0;1))$.
For sake of simplicity, let's consider a specific example $f(x)$
$$f(x)= \mathbf{1}_{[0;1)}(x)= \cases{ 1 \qquad x \in [0;1) \\ 0 \qquad \mathrm{otherwise}}$$
A function belonging to the span of $\{ f(x-n)\}_{n \in \Bbb Z}$ has the form
$$\sum_{k=a}^b \lambda_k \mathbf{1}_{[a;a+1)}(x)$$
which is represented by some steps of length $1$.
These are not good to approximate functions which are not constant in the intervals $[a;a+1)$, for example this one
$$g(x)= x \cdot \mathbf{1}_{[0;1)}(x)= \cases{ x \qquad x \in [0;1) \\ 0 \qquad \mathrm{otherwise}}$$
