Proving that a sequence in $L^2(\mathbb R)$ is relatively compact I have a bounded sequence $\{f_n\}_n$ in $L^2(\mathbb R)$ such that $\mbox{supp } f_n$ is uniformly bounded and
$$ \int_{\mathbb R} x^2 |\Theta_n(x) (F f_n)(x)|^2 dx \leq C^2 $$
for all $n$, where $\Theta_n(x) := \frac{\sin(x/n)}{x/n}$ and $Ff_n$ is the Fourier transform of $f_n$.
I'd like to prove that $\{f_n\}_n$ is relatively compact in $L^2(\mathbb R)$. From the condition on the supports of $f_n$, i know that $Ff_n$ is smooth and has uniformly bounded derivatives.

The following characterization of relatively compact subsets of $L^2(\mathbb R)$ holds: a set $A \subset L^2(\mathbb R)$ is relatively compact if and only if:


*

*$A$ is bounded;

*$\int_{\mathbb R-(-R,R)} |f|^2 \to 0$ as $R \to \infty$, uniformly with respect to $f \in A$;

*$\int_{\mathbb R} |f(x-y)-f(x)|^2 dx \to 0$ as $y \to 0$, uniformly with respect to $f \in A$.



The first condition holds by hypothesis, and the second condition is easily verified. How can I check the third one?
 A: Contrary to what I tried to do first, I think I now have a (complete) argument that your claim is false.
Let me rewrite your last assumption (the displayed equation).
Since $\int x^2|\widehat{g}(x)|^2 = \int |g'|^2$, this really says that certain derivatives are uniformly bounded in $L^2$ (originally, I may have to define these as distributional derivatives). Now since $\theta_n(x)$ is the Dirichlet kernel $D(x)=\sin x/x$, evaluated at $x/n$, and $\widehat{D}=\chi_{(-1,1)}$, and since multiplication becomes convolution, the role of $g$ above is actually taken by
$$
F_n(x)= n\int_{-1/n}^{1/n} f_n(x+t)\, dt .
$$
So, to sum this up, we have that
$$
\|F'_n\|_2\lesssim 1 ; \quad\quad\quad\quad (1)
$$
this is equivalent to your assumption.
If the $f_n$ are also smooth, then $F'_n=n(f_n(x+1/n)-f_n(x-1/n))$, so (1) becomes
$$
\|\tau_{2/n}f_n -f_n\|_2 \lesssim 1/n \quad\quad\quad\quad (2)
$$
($\tau_y$ denotes the shift by $y$). Let's now consider functions $f_n$ supported by $[-3,3]$, that have bumps of width $1/n$ alternating with intervals of the same length on which I put $f_n=0$. The bump at $x=0$ has height $1$, and as we move away from $x=0$ towards the endpoints $\pm 3$, the heights decrease (linearly) by $1/n$ from one bump to the next (OK, this is not smooth, but of course I can approximate, so this is not an issue).
Notice, first of all, that we do have (2): when I shift by $2/n$, each bump will be moved exactly to the location of the next one, so after taking the difference, $\tau_{2/n}f_n-f_n$ will have bumps of height $1/n$ only. There are $\sim n$ of these, so $\|\tau_{2/n}f_n-f_n\|^2\lesssim n\cdot (1/n)\cdot (1/n)^2 = 1/n^2$, as required by (2).
Moreover, by similar reasoning, we also have that $\|f_n\|^2 \lesssim n\cdot (1/n)\lesssim 1$, again as required.
However, if I now shift only by $1/n$, then all bumps involved will have pairwise disjoint supports after shifting, so $\|\tau_{1/n}f_n-f_n\|^2 \sim n\cdot 1/n =1$, and this doesn't get small. The third condition from your set of necessary and sufficient conditions for relative compactness fails.
