The Feichtinger's algebra
$S\_0(\mathbb{R^d})=M^{1,1}(\mathbb{R^d}):=\{f\in L^2(\mathbb{R^d}):V\_g(f)\in L^1(\mathbb{R^{2d}})\}$, where
$V\_g(f)(x,\omega)$ is the short-time Fourier transform of $f$ with respect to $g \in S(\mathbb{R^d})$, which is the Schwartz space, defined by
$$
V\_g(f)(x,\omega)=\int\_\mathbb{R^d} f(t)\, \overline{g(t-x)}\, e^{-2\pi i\omega \cdot t} \, dt.
$$ 
Then we define the norm on $S\_0(\mathbb{R^d})$:
$$
||f||\_{S\_0}=\int\_{\mathbb{R^d}} \left(\int\_{\mathbb{R^d}}|V\_gf(x,\omega)| \, dx\right)\, d\omega
$$
It's known that $(S\_0(\mathbb{R^d}),||\cdot||\_{S\_0})$ is a Banach space, and $S(\mathbb{R^d})$ is a dense subspace of $S\_0(\mathbb{R^d}).$ My question: is $(S\_0(\mathbb{R^d}),||\cdot||\_{S\_0})$ reflexive?



Is the Feichtinger's algebra $(S_0(\mathbb{R^d}),||\cdot||_{S_0})$ reflexive?
No. Wilson bases (or more generally local Fourier bases) allow to identify $S\_0$ with the sequence space $\ell^1$ (i.e. the elements of $S\_0$ are just the functions in $L^2$ having an absolutely convergent Wilson series expansion). HENCE the space is a dual space (the pre-dual is so to say $c\_0$ at the sequence space level), resp. the closure of the test functions in the dual space $S\_0'$, resp. the space of all tempered distributions which have STFTs (short time Fourier transforms) which vanish at infinity (in phase space). So, NO, it is not a reflexive space. HGFei 


