Decay properties of Fourier transform Suppose $\psi : \Bbb R^2 \to \Bbb C$ is the characteristic function of the diamond-shaped set $D\subset\Bbb R^2$ delimited by the four lines $y=\pm 1 \pm x$. I have calculated its Fourier transform to be
$$\widehat\psi(\xi_1,\xi_2) = \int_D e^{i(\xi_1 x_1 + \xi_2x_2)} d\xi_1 d\xi_2 = \frac{4}{\xi_1^2-\xi_2^2}\left(\cos\xi_2-\cos\xi_1\right).$$
This function is ill-defined at $\xi_2=\pm\xi_1$, but can be continuously extended to a smooth function on $\Bbb R^2$. If the frequency set of $\psi$ is defined as
$$\Sigma(\psi) = \complement\{\xi \in \Bbb R^2\setminus\{0\}\ |\  \text{in conic nbhd of }\xi,\ \forall m\in\Bbb N\ \exists c_m\geq 0\ :\ |\widehat\psi(\eta)|\leq c_m (1+|\eta|^2)^{-m/2}\}, $$
is it correct to say that $\Sigma(\psi)=\Bbb R^2\setminus\{0\}$? I would think so, because the numerator of $\widehat \psi$ is bounded, whereas the denominator is a polynomial of degree 2, and so the decay condition would fail as soon as $m=3$. Any flaws in my reasoning?
Follow-up question: if $g\in C^\infty_c(\Bbb R^2)$ has a "small" support around $x\in\partial D$, is there a way to see intuitively what happens to the Fourier transform, and thus frequency set, of $g\psi$? Alternatively, consider a sequence of compactly supported functions with support decreasing around $x$. Does $\Sigma(g\psi)$ change if $x$ is a vertex of $D$ or lies on the edge, and if so, how?

$\psi(x,y) = 1_{(x,y)\in D}= f((x,y)P^{-1})$ where $P=\pmatrix{1&1\\1&-1}$ and $f(x,y)=1_{x\in [-1,1],y\in [-1,1]}$ gives that $\hat{\psi}(u,v) = |\det(P)|\hat{f}((u,v)P^{\top})$ where $\hat{f}(u,v) = \frac{4 \sin(u)\sin(v)}{uv}$ is smooth and has decay $O((1+u^2+v^2)^{-1})$  on all (closed) cones except the ones containing an horizontal or vertical semi-axis where the decay is $O((1+u^2+v^2)^{-1/2})$,
whence the same holds for $\psi$ with decay $O((1+u^2+v^2)^{-1})$ on all cones except the ones containing one of the semi-diagonals.
For your smoothed version, the Fourier transform of $\psi \ast \phi$ is $\hat{\psi}\hat{\phi}$, if $\phi\in C^\infty_c$ then $\hat{\psi}\hat{\phi}$ is  Schwartz.
