Wigner distribution function

Remarks

The integral converges absolutely for each $x$, by the Cauchy–Schwarz inequality. $Wf$ is real-valued, since replacing $t$ by $-t$ turns the integral into its complex conjugate. For $f$ in the Schwartz space, $\int_\R Wf(x, \omega) \, d\omega = \abs{f(x)}^2$ and $\int_\R Wf(x, \omega) \, dx = \abs{\hat f(\omega)}^2$, where $\hat f$ is the Fourier transform. In quantum mechanics the Wigner function of a wave function is usually written with Planck's constant and a normalizing factor.

Example: let $f(t) = 2^{1/4} e^{-\pi t^2}$, so that $\norm{f}_2 = 1$. Then $f(x + t/2) \overline{f(x - t/2)} = \sqrt2 \, e^{-2\pi x^2} e^{-\pi t^2/2}$, and the Fourier transform of $e^{-\pi t^2/2}$ is $\sqrt2 \, e^{-2\pi \omega^2}$. So $Wf(x, \omega) = 2 e^{-2\pi (x^2 + \omega^2)}$, and integrating in $\omega$ gives $\sqrt2 e^{-2\pi x^2} = \abs{f(x)}^2$.

Non-example: the Wigner distribution is quadratic in $f$, not linear. For instance $W(2f) = 4 Wf \ne 2 Wf$ when $f \ne 0$.