Let $L^2(X)$ be the set of Lebesgue square-integrable functions on a locally compact Hausdorff space $X$. Define $\mathfrak{A}:=\{M_f:f\in L^{\infty}(X), f=\overline{f}\}$, where $M_f$ is the the multiplication operator on $L^2(X)$ defined by $M_f(g)=fg$. In particular all elements of $\mathfrak{A}$ are self-adjoint since $M_f^*=M_{\overline{f}}$. I know, if one omit the property $f=\overline{f}$, that $\mathfrak{A}=\mathfrak{A}'$ (hence $\mathfrak{A}$ is its own commutant and hence a von Neumann algebra). But what can be said about our $\mathfrak{A}$? Is it also a von Neumann algebra? Whats about $\mathfrak{A}'$? Someone an idea?
Thanks a lot.