Fourier transform localisation (still unanswered, but apparently off-topic?) In the context of Pólya's theorem I was reading these notes here on p. 19. In the last paragraph the authors claim (it is the sentence starting like "standard Fourier theory shows...")  that the Fourier transform of the function
$$f(z)= \frac{1}{1- \frac{\sum_{i=1}^{d}\cos(z_i)}{d} },f \in L^1{[-\pi,\pi]^{d}}$$ for $d \ge 3,$
is supported close to the zero for large vectors $||k||$, i.e.
$$ \int_{[-\pi,\pi]^d} f(x)e^{-i \langle k,x\rangle} dx \approx \int_{B_{\varepsilon}(0)} f(x)e^{-i \langle k,x\rangle} dx$$
for $||k||$ large.
Now, when I read this phrase (standard Fourier theory shows...), the Riemann-Lebesgue Lemma came to my mind first, as $k$ is large and so the Fourier transform will be close to zero. 
This would certainly be an easy fact, but arguing that the support of the integrand is localized around zero, seems to be more difficult and I cannot really tell where this could follow from. 
Moreover, all the simple decay conditions we know from Sobolev space theory do not seem to apply, as $f$ is not in any $H^s$ or even $C_c^{\infty}.$
Edit: I received two answers that do not provide an answer to my question, as their answers do not work for the function I am talking about. 
Remark devoted to the closers: I would like to improve my question if there is anything wrong about it, i.e. I appreciate comments critizising my question. But putting a question on-hold without any advice on what you don't like about my question is pointless in my eyes.
 A: (This is posted as CW as it is mostly an explanation of alpoge's comment.)
Observe that your $f$ is smooth away from the origin. 
Write $f = f_1 + f_2$ where $f_1 = f \chi$ where $\chi$ is a smooth bump function localized near the origin with support roughly in $[-\epsilon,\epsilon]^d$. Then $f_2$ is $C^\infty$. And standard Fourier theory (integrating by parts) tells you that the Fourier transform of $f_2$ decays faster than any polynomial rate. 
However, we know that the Fourier transform of $f$ cannot decay too fast: if it does (such that the Fourier transform is absolutely summable), then we can apply the Fourier inversion formula and get that $f$ is continuous and bounded, which is false due to the singularity at zero. This means that asymptotically (as $|x| \to \infty$) $f_2$ must decay faster than $f_1$. (If both decay fast then $f$ must decay fast.) 
Hence we have that for large $x$ it must be that $f_1 \gg f_2$, which is what is asked. 
A: The Riemann-Lebesgue lemma says that for $u\in L^1(\mathbb R^n)$, 
$
\lim_{\vert \xi\vert\rightarrow\infty}\hat u(\xi)=0.
$
The proof is simple: take an arbitrary $\phi\in \mathscr S(\mathbb R^n)$, then
$$
\vert\hat u(\xi)\vert\le \vert\widehat{u-\phi}(\xi)\vert+\vert\hat{\phi}(\xi)\vert
\le \Vert u-\phi\Vert_{L^1(\mathbb R^n)}+\vert\hat{\phi}(\xi)\vert,
$$
so that, since $\hat \phi\in\mathscr S(\mathbb R^n)$,
$$
\limsup_{\vert \xi\vert\rightarrow\infty}\vert\hat u(\xi)\vert\le\Vert u-\phi\Vert_{L^1(\mathbb R^n)},
\text{ and thus   } 
\limsup_{\vert \xi\vert\rightarrow\infty}\vert\hat u(\xi)\vert\le \inf_{\phi\in\mathscr S(\mathbb R^n)}\Vert u-\phi\Vert_{L^1(\mathbb R^n)}=0,
$$
by density of $\mathscr S(\mathbb R^n) $ in $L^1(\mathbb R^n).$
Now, assuming $\psi\in C^1_c(\mathbb R^n)$, you get
$$
\hat \psi(\xi)=\int e^{-2iπ x\cdot \xi} \psi(x) dx=
\int (1-2iπ\vert \xi\vert^2)^{-1}(1+\sum_j\xi_j\partial_{x_j})\bigl\{e^{-2iπ x\cdot \xi} \bigr\}\psi(x) dx,
$$
so that
$$
\hat \psi(\xi)=(1-2iπ\vert \xi\vert^2)^{-1}
\int e^{-2iπ x\cdot \xi} (1-\sum_j\xi_j\partial_{x_j})\psi(x) dx,
$$
which implies
$
\vert\hat \psi(\xi)\vert\le (1+4π^2\vert \xi\vert^4)^{-1/2}(1+\vert\xi\vert)\Vert\nabla \psi\Vert_{L^1}=O((1+\vert \xi\vert)^{-1}).
$
Assuming $\psi\in C^k_c$ will provide
$
\vert\hat \psi(\xi)\vert =O((1+\vert \xi\vert)^{-k}).
$
Similarly, for $s\in \mathbb R$, the Sobolev space $H^s(\mathbb R^n)$ is defined as the tempered distributions $u$ such that
$$
\hat u(\xi) (1+\vert\xi\vert^2)^{s/2}\in L^2(\mathbb R^n).
$$
Roughly speaking $k$ derivatives for a function $u$ will provide decay like $\vert \xi\vert^{-k}$ for the Fourier transform. What is nice about Hilbertian Sobolev spaces is that you get a true characterization with the Fourier transformation. 
