Non-reflexivity of $B(\mathbb H)$ How does one prove that the space $B(\mathbb H)$ of bounded operators on a infinite dimensional (separable) Hilbert space is not reflexive?
I guess this should go along the lines of the non-reflexivity of $l_\infty$, but I was unable to find this written down somewhere.
 A: It seems to be a good idea to extend my comment to an answer.
First of all, pick any orthonormal basis of $\mathbb{H}$ so that $B(\mathbb{H})$ can be identified with $B(\ell_2)$. The subspace of diagonal operators is clearly isometric to $\ell_{\infty}$. It is a general truth that if $Y$ is a closed subspace of a reflexive Banach space $X$ then $Y$ is reflexive itself; e.g. its unit ball is weakly compact. It means that $B(\mathbb{H})$ cannot be reflexive.
Another route, already suggested by Marc Palm, is to use the fact that the reflexivity of $X$ is equivalent to the reflexivity of $X^{\ast}$. Since $(K(\ell_2))^{\ast\ast}$ (bidual of the space of compact operators) is isometric to $B(\ell_2)$, the reflexivity of $B(\ell_2)$ would imply that $K(\ell_2)$ is isomorphic to $B(\ell_2)$; the former space is separable, so it is not possible.
A: The predual of $B(H)$ are the trace class operators. It is a general result that if $X$ is not isomorphic to $X^{**}$ via the canonical embedding, so is $X^{*}$ not isomorphic to $X^{***}$ via the canonical embedding. The last fact is an exercise in Simon-Reed Mathematical Prinicple of Physics Vol 1.
