FBI transform

Remarks

The factor $e^{-\frac12 \abs{\xi} \abs{x - y}^2}$ is a Gaussian window centred at $x$ of width of order $\abs{\xi}^{-1/2}$, and $e^{i\xi \cdot (x - y)}$ selects the frequency $\xi$. Authors use different normalizations of the window and of the constant in front; the qualitative properties do not depend on these choices. The main property concerns the analytic wave front set. Cut $f$ off near a point $x_0$, and let $\xi_0 \ne 0$. Then $(x_0, \xi_0)$ lies outside the analytic wave front set of $f$ exactly when $Tf(x, \lambda \xi)$ decays like $e^{-c\lambda}$ as $\lambda \to \infty$. Here $c > 0$ is a constant and the bound is uniform for $(x, \xi)$ near $(x_0, \xi_0)$.

Example: $n = 1$ and $f(y) = e^{-y^2/2}$. For $\xi > 0$ the exponent is a quadratic polynomial in $y$. The Gaussian integral gives $$Tf(x, \xi) = \sqrt{\frac{2\pi}{1 + \xi}}\, \exp\Big( i\xi x - \frac{\xi x^2}{2} + \frac{(\xi x - i\xi)^2}{2(1 + \xi)} \Big),$$ whose modulus is $\sqrt{2\pi/(1+\xi)}\, \exp\big(-\frac{\xi (x^2 + \xi)}{2(1 + \xi)}\big)$; it decays like $e^{-\xi/2}$ as $\xi \to \infty$, as it must for a real analytic function.

Non-example: $f(y) = e^{\abs{y}^3}$ is not integrable, and the integral defining $Tf(x, \xi)$ diverges for every $(x, \xi)$, because $e^{\abs{y}^3}$ grows faster than any Gaussian decays.