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.



Non-reflexivity of $B(\mathbb H)$
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.


