Modulation space

Remarks

The space does not depend on the choice of the window $g$, and different windows give equivalent norms. With this norm $M^{p,q}(\R^d)$ is a Banach space. One writes $M^p = M^{p,p}$. Special cases are $M^{2} = L^2(\R^d)$ with an equivalent norm, and $M^{1}$, the Feichtinger algebra. Weighted versions $M^{p,q}_m$ insert a weight $m(x, \omega)$ into the integrand.

Example: let $d = 1$ and $\varphi(t) = e^{-\pi t^2}$, and take $g = \varphi$. Completing the square in $t$ gives $$\abs{V_\varphi \varphi(x, \omega)} = 2^{-1/2} e^{-\pi (x^2 + \omega^2)/2}.$$ This is integrable to every power in $x$ and in $\omega$, so $\varphi \in M^{p,q}(\R)$ for all $p, q$.

Non-example: the constant function $f = 1$ is not in $M^{p,q}(\R)$ for $p < \infty$. Here $\abs{V_g 1(x, \omega)} = \abs{\hat{\overline{g}}(\omega)}$ does not depend on $x$, so the inner integral is infinite whenever $\hat{\overline{g}}(\omega) \ne 0$. It does lie in $M^{\infty, 1}(\R)$.