Given a function $f:\mathbb{Z}^{d} \to \mathbb{C}$, we define its Fourier transform as: \begin{eqnarray} \hat{f}(k) := \sum_{x\in \mathbb{Z}^{d}}f(x)e^{-ik\cdot x} \tag{1}\label{1} \end{eqnarray} where $k \in B:= [-\pi, \pi]^{d}$ and $k\cdot x := k_{1}x_{1}+\cdots k_{d}x_{d}$. Now, I know that the following inversion formula holds: \begin{eqnarray} f(x) = \frac{1}{(2\pi)^{d}}\int_{B}e^{ik\cdot x}\hat{f}(k)d^{d}k \tag{2}\label{2} \end{eqnarray}
Now, if $f \in \mathcal{l}^{2}(\mathbb{Z}^{d}) := \{\varphi:\mathbb{Z}^{d}\to \mathbb{C}: \sum_{x\in \mathbb{Z}^{d}}|\varphi(x)|^{2}<+\infty\}$, we can define the Discrete Laplacian $\Delta$ by defining: \begin{eqnarray} (\Delta f)(x) := \sum_{e\in \mathcal{E}}f(x+e)-f(x) \tag{3}\label{3} \end{eqnarray} where $\mathcal{E}$ is the standard basis for $\mathbb{Z}^{d}$. I also know that the Green function for $m^{2}-\Delta$ (defined to be the function $G(xy)$ satisfying $(m^{2}-\Delta)G(x,y) = \delta_{xy}$, with $\delta_{xy}$ being the Kronecker delta) is given by: \begin{eqnarray} G(x,y) = \frac{1}{(2\pi)^{d}}\int_{B}\frac{1}{m^{2}+\mu(k)}e^{ik\cdot (x-y)}d^{d}k \tag{4}\label{4} \end{eqnarray}
I would like to prove the inversion formula (\ref{2}) and the Green function formula (\ref{4}) but I'm a little confused. Can someone help me please?
The Fourier inversion formula is just the usual formula for the sum of a Fourier series. Concerning the Green function, I assume that $\mu(k)$ stands for $$ \mu(k_1, \ldots, k_d)=\sum_{j=1}^d e^{ik_j}-d.$$ Then I would prove (4) by computing that $$ -\Delta_x(e^{ik\cdot x})=(e^{ik_1}+\ldots+e^{ik_d}-d)e^{ik\cdot x}, $$ which implies $$ \begin{split} (m^2-\Delta_x)G(x, y)&= \int_B\frac{m^2+e^{ik_1}+\ldots+e^{ik_d}-d}{m^2+\mu(k)} e^{ik\cdot(x-y)}\, \frac{dk}{(2\pi)^d}\\&=\int_B e^{ik\cdot (x-y)}\,\frac{dk}{(2\pi)^d} \\&= \delta_{x, y}. \end{split}$$