how to prove translation identity: $(\mathcal{F}^{-1}m\mathcal{F}f)(x)= e^{ikx}[\mathcal{F}^{-1}m(\cdot+k)\mathcal{F}(e^{-iky}f(y))](x)$? Let $m, f\in S(\mathbb R^{n})$=The Schwartz space.

My question: How to prove:
  $$(\mathcal{F}^{-1}m\mathcal{F}f)(x)= e^{ikx}[\mathcal{F}^{-1}m(\cdot+k)\mathcal{F}(e^{-iky}f(y))](x);$$

where $\mathcal{F}$ denotes the Fourier transform and 
$\mathcal{F}^{-1}$ the inverse Fourier transform and $k\in \mathbb Z^{n}.$
Thanks,
 A: This is a straightforward computation. Since the Fourier transform and its inverse map $\mathcal S$ to $\mathcal S$ we can use the integral representation of the Fourier transform. Let $g(y) = e^{-i k \cdot y} f(y)$. Then
$$\cal F g(z) = \int_{\mathbb R^n} e^{-i z \cdot y} g(y) \, dy = \int_{\mathbb R^n} e^{-i z \cdot y} e^{-i k \cdot y} f(y) = \int_{\mathbb R^n} e^{-i(k + z)\cdot y} f(y) \, dy,$$
$$m(z + k) \mathcal F g(x) = m(z + k) \int_{\mathbb R^n} e^{-i(k + z)\cdot y} f(y) \, dy,$$
$$\mathcal F^{-1}[m(\cdot + k) \mathcal Fg](x) = \int_{\mathbb R^n} e^{i x \cdot z}m(z + k) \mathcal Fg(z) \, dz = \int_{\mathbb R^n} \int_{\mathbb R^n} e^{ix\cdot z}m(z+k)e^{-i(k+z)\cdot y} f(y) \, dy \, dz.$$
Now use the change of variable $w = z + k$. Then \begin{align*}\mathcal F^{-1}[m(\cdot + k) \mathcal Fg](x) &= \int_{\mathbb R^n} \int_{\mathbb R^n} e^{ix\cdot (w-k)}m(w)e^{-iw\cdot y}f(y) \, dy \, dw \\ &= e^{-i k\cdot x} \int_{\mathbb R^n} e^{i x \cdot w} m(w)\int_{\mathbb R^n} e^{-iw\cdot y}f(y) \, dy \, dw \end{align*} The last expression is just $e^{-i k\cdot x} \mathcal F^{-1} [m \mathcal F f](x)$.
