Most algebras of observables from quantum mechanics are closed. For example, fix a separable Hilbert space $H$, and consider the algebra of bounded operators on it. This is a Banach space.
In classical mechanics, the algebra of observables is typically the Poisson algebra of smooth functions $C^\infty(X)$ where $X$ is the classical phase space.
I am wondering how this fact that spaces of "quantum observables" are larger than their classical counterparts is reconciled in quantization. For example, consider the machinery of deformation quantization, where there is supposed to be a star product $$f\star_\hbar g\in C^\infty(X)$$ which makes the classical observable algebra into a quantum one. This leads to formulations of quantum mechanics in phase space (e.g. see https://en.wikipedia.org/wiki/Phase-space_formulation).
If I am to be told that the phase-space formulation of quantum mechanics is to be entirely on equal footing with the Hilbert space formulation, then there must be a way to assign to arbitrary quantum observables (say $A,B$) completely equivalent classical counterparts $f_A,g_A$. However, if this is true, what is the classical analogue of the theorem
The space of trace class operators forms an even subalgebra of the space of Hilbert-Schmidt operators
According to the Weyl-Wigner transform, the space of Hilbert-Schmidt operators corresponds to the closed space $L^2(X)$. However, then the trace-class operators would be
$$\text{trace-class operators}=\{f\star_\hbar g~~|~~f,g\in L^2(X)\}$$ So, just by two seconds of naivety, we already are in the uncomfortable situation where we have to define the Moyal product of non-differentiable functions.
Is the Moyal product uniquely extendable to $L^2$ spaces, or even spaces of distributions? If not, then I would argue that the phase-space picture of quantum mechanics is incomplete, as there is no way of performing functional analysis.