Can we expect that $\sum_{n\in \mathbb Z^d} |\hat{f}(x-An)|^2 \leq C \|f\|_{L^2}$? Let $f\in L^2(\mathbb R^d)$ and $A$ be a real  $d\times d$ matrix,  and $\det  A \neq 0.$

Can we expect that $\sum_{n\in \mathbb Z^d} |\hat{f}(x-An)|^2 \leq C \|f\|_{L^2}$ 

for some constant $C$?
Side note: By Plancheel theorem we have $\|f\|_{L^2}=\|\hat{f}\|_{L^2}$ and $n\in \mathbb Z^d$ we may think of as a $d\times 1$ matrix, and so $An$ is $d\times 1$ matrix. And thus we regard $An\in \mathbb R^d.$
 A: No, not at all. First of all, the Fourier transform is irrelevant here: by writing $g=\hat f$, we can restate the conjectural inequality as 
$$
\sum_{n\in \mathbb Z^d} |g(x-An)|^2 \leq C \|g\|_{L^2},\quad \forall \, g\in L^2(\mathbb{R}^d) \tag1
$$
But there is no reason for (1) to be true. The $L^2$ norm does not control the pointwise values; in fact they are not even well-defined for a function in $L^2$ (which is an equivalence class of functions that agree a.e.). One can let $g = 0$ except at the points $-An$, for example. The same example can be made continuous by replacing each discontinuity at $-An$ by a sharp peak of height $1$: such a peak can contribute an arbitrarily small amount to the $L^2$ norm.
A: As already pointed out, the answer to the question as stated is of course not.
But it's true if you assume in addition that $f$ has compact support (which, to start, implies that $f\in L^1$, so that $\hat f$ is at least defined pointwise...)
Briefly: Fix $x$. Say $$\chi_{n}(t)=e^{i(x+An)\cdot t}.$$Say $f$ is supported in the compact set $K$. Then there exists a compact set $K'$ with $K\subset K'$ such that the functions $\chi_n$ are orthogonal on $K'$. Since $\hat f(x+An)=\langle f,\chi_n\rangle$ the inequality is just Bessel's inequality.
