Folland PDE chapter 6.C problem 1 Problem 6.C.1: Suppose $0 \neq \phi \in C^\infty_c(\mathbb{R}^n)$ and $\{ a_j \}$ is a sequence in $\mathbb{R}^n$ with $|a_j| \to \infty$, and let $\phi_j(x) = \phi(x - a_j)$. Show that $\{\phi_j\}$ is bounded in $H_s$ for every $s \in \mathbb{R}$ but has no convergent subsequence in $H_t$ for any $t \in \mathbb{R}$.
Info: Folland says $H_s$ is $L^2$-Sobolev space with norm
$$||f||_s^2 = \int_{\mathbb{R}^n} |\widehat{f}(\xi)|^2(1+|\xi|^2)^s d\xi$$
First part is easy but I am stuck on second part. I want to show sequence $\{\phi_j\}$ cannot have Cauchy subsequence, but trouble finding lower bound for integral
$$||\phi_k - \phi_l||^2_t = \int_{\mathbb{R}^n} |1-e^{-2\pi i (a_k - a_l)\cdot \xi}|^2|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xi$$
Help please, I cannot find good inequality to use anywhere. Reverse triangle inequality is not good here because modulus of complex exponential is 1. 
 A: 1 - First prove that for each subsequence $b_k$ of $a_k$, there is a subsequence $c_k$ of $b_k$ such that $$|c_k-c_l|\geq \delta,\ \forall k,l$$
for some $\delta>0$.
2 - Write $$I(1,k,l)=\int_{|(a_k-a_l)\cdot\xi|<1 } |1-e^{-2\pi i (a_k - a_l)\cdot \xi}|^2|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xi$$ 
$$I(2,k,l)=\int_{|(a_k-a_l)\cdot\xi|\geq1 } |1-e^{-2\pi i (a_k - a_l)\cdot \xi}|^2|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xi$$
Note that $$\|\phi_k-\phi_l\|_t^2=I(1,k,l)+I(2,k,l)\geq I(2,k,l)$$
Define $A(k,l)=\{\xi\in\mathbb{R}^n:\ |(a_k-a_l)\cdot\xi|\ge 1\}$. Because $\phi\in C_0^\infty(\mathbb{R}^n)$ we have that $\hat{\phi}$ is a entire function, and hence, because $\phi$ is not identically zero, it cannot be compactly supported (see here).
Now we use item 1 to conclude that there is $\epsilon>0$ such that if $$A_0(k,l)=\{\xi\in\mathbb{R}^n:\ (a_k-a_l)\cdot\xi=0\}$$
then $$A(k,l)\subset \{\xi\in\mathbb{R^n}:\ \operatorname{dist}(\xi,A_0(k,l))\leq\epsilon\}\tag{1}$$
where $\operatorname{dist}$ denotes distance.
$(1)$ combined with the fact that $\hat{\phi}$ is entire, implies that $$\int_{|(a_k-a_l)\cdot\xi|\ge 1}|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xi$$
is bounded away from zero independently of $k$ and $l$.
On the other hand, the function $g(a_k,a_l,\xi)=|1-e^{-2\pi i (a_k - a_l)\cdot \xi}|$ will be bounded away from zero independently of $k$ and $l$ in the set $A(k,l)$, let's call this bound $c$, hence, $$\|\phi_k-\phi_l\|_t^2\geq c^2\int_{|(a_k-a_l)\cdot\xi|\ge 1}|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xi,\ \forall k,l$$
which completes the proof.
