Is $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$ if $g$ is a Sobolev function on the real line? If we are given a function $g\in W_2^k(\mathbb{R})$ (even consider $k=1$ for simplicity), then is it true or not that $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$?  That is, do we have $$\underset{n\in\mathbb{Z}}\sum|g(n)|^2<\infty$$
?
At first I thought that this must be false, but after trying for a while to construct a counterexample, I found it a little more difficult than I thought.  The conclusion is certainly not true for functions only in $L_2(\mathbb{R})=W_2^0(\mathbb{R})$, because consider the function $g$ so that $g^2$ forms triangles or tent functions at each integer of height 1 and width $2^{-n}$.  The area is square summable, but the above sum is infinite.  However when we have to have control over even the first derivative it becomes a little harder, because such a tent function no longer works.
My idea was to try to set $g(n)=n^{-1/2}$ for an infinite subset of the natural numbers (consider $g(x)=0$ for $x\leq 0$), and try to ensure that the functions derivative was small enough to be square summable, but this didn't seem to work.
Any suggestions or comments would be helpful.  I am sure somebody knows this even though I have never seen it before.
 A: Aha! This is actually a bit of a tricksy problem in Fourier analysis. What you want to do is, consider the Fourier transform of $g$, call it $\hat{g}$. Since $g$ is in $L^2(\mathbb{R})$, so is $\hat{g}$, by the properties of the Fourier transform. 
Now let's periodize $\hat{g}$ into a  function we will suggestively call $\hat{G}$ with
$$ \hat{G} (k) = \sum_{m \in \mathbb{Z}} \hat{g}(m+k)$$
Then $\hat{G}$ is a periodic function on the unit interval, so it has a Fourier series representation. What are the Fourier coefficients of $\hat{G}$? 
The Poisson summation formula tells us that they are precisely the values of our original function! To be precise, 
$$\hat{G}(k) = \sum_{n \in \mathbb{Z}} g(n) e^{-2\pi i n k}$$
From here, it should be clear that 
$$ \sum_n |g(n)|^2 = \|\hat{G}\|_{L^2([0,1])}^2 = \|g\|^2_{L^2(\mathbb{R})}$$
(Response to comment): gerw, you've put your finger on precisely the interesting point of this problem that I neglected in my original answer. My apologies for neglecting it. 
To use the Fourier transform as I have, you need the Poisson summation formulation to be valid, in the sense that $\hat{g}$ needs to be integrable (see, e.g., Stein and Weiss, Chapter VII, Theorem 2.3). This is, to say the least, not true of general $L^2$ functions $g$, but for Sobolev functions it follows from 
$$ \int |\hat{g}(k)| dk \leq \int |\hat{g}(k) (1+|k|^2)^\frac{1}{2}| \frac{1}{(1+|k|^2)^\frac{1}{2}} dk \leq \left(\int |\hat{g}(k)|^2 (1 + |k|^2) dk \right)^\frac{1}{2} \left (\int \frac{1}{1+|k|^2} dk \right)^\frac{1}{2} $$
The first term in the final expression is well known to be the $W^1_2$ norm of $g$, and the second is integrable and a constant. 
A: Theorem
Let $(x_n)_{n\in\mathbb{Z}}$ be a set of separated points in $\mathbb{R}$ such that $\underset{n\in\mathbb{Z}}\inf x_{n+1}-x_n>c$.
If $g\in W_p^k(\mathbb{R})$, $1\leq p\leq\infty$, $k\in\mathbb{N}$, then $(g(x_n))\in\ell_p$.
Proof: 
First, in the $p=\infty$ case, if $g\in W_\infty^k(\mathbb{R})$, then $g$ is Lipschitz, and so the result is obvious. (That $W_\infty^1(\mathbb{R})=Lip(\mathbb{R})$ is a nontrivial result).
So we tackle the case $1\leq p<\infty$.  Let $$I_n:=\left[x_n-\frac{c}{3},x_n+\frac{c}{3}\right],\quad n\in\mathbb{N}$$
We define them this way so that they are disjoint, and also $|I_n|=\frac{2}{3}c$ for all $n$. By disjointness, we have that
$$\underset{n\in\mathbb{Z}}\sum\int_{I_n}|g|^p\leq\int_\mathbb{R}|g|^p<\infty$$
On the other hand, if $y_n:=\underset{x\in I_n}{argmin}|g(x)|^p$, then we have that
$$|g(y_n)|^p\frac{2}{3}c\leq\int_{I_n}|g|^p$$
and consequently, $(g(y_n))\in\ell_p$.
The following Lemma will be very useful presently.
Lemma: [A First Course in Sobolev Spaces, Leoni, Thm 7.13, p.222]
Let $\Omega\subset\mathbb{R}$ be open and $u:\Omega\to\mathbb{R}$.  Let $1\leq p<\infty$.  Then $u\in W_p^1(\mathbb{R})$ if and only if it admits an absolutely continuous representative, $\tilde{u}:\Omega\to\mathbb{R}$ such that $\tilde{u}$ and its classical derivative $\tilde{u}'$ belong to $L_p(\mathbb{R})$.  Moreover, if $p>1$, then $\tilde{u}$ is Holder continuous of exponent $1/p'$.
So since $W_p^k(\mathbb{R})\subset W_p^1(\mathbb{R})$, taking $\Omega=\mathbb{R}$ in the Lemma allows us without loss of generality to assume that $g$ is absolutely continuous.  Therefore by the Fundamental Theorem of Calculus, 
$$g(x_n)=g(y_n)+\int_{y_n}^{x_n}g(t)dt$$
Using the inequality $|a+b|^p\leq 2^p(|a|^p+|b|^p)$, we have
$$|g(x_n)|^p\leq 2^p\left[|g(y_n)|^p+\left|\int_{y_n}^{x_n}g(t)dt\right|^p\right]\leq 2^p\left[|g(y_n)|^p+\left(\frac{2c}{3}\right)^\frac{p}{p'}\int_{I_n}|g(t)|^pdt\right]$$
The second inequality follows by Holder's Inequality:
$$\left|\int_{y_n}^{x_n}g(t)dt\right|^p\leq\left(\int_{y_n}^{x_n}|g(t)|dt\right)^p\leq |I_n|^\frac{p}{p'}\int_{I_n}|g(t)|^pdt$$
We conclude that
$$\underset{n\in\mathbb{Z}}\sum|g(x_n)|^p\leq C\left[\underset{n\in\mathbb{Z}}\sum|g(y_n)|^p+\underset{n\in\mathbb{Z}}\sum\int_{I_n}|g|^p\right]<\infty$$
and the conclusion follows.
