Exercise 24, § 2

Let $A\subset \mathscr{L}(E)$ be a unital self-adjoint subalgebra of $\mathscr{L}(E)$. Denote by $A''$ the bicommutant of A. Denote by $\mathscr{L}(E)_s$ the space $\mathscr{L}(E)$ endowed with the topology of simple convergence.

a) The space $A''$ is a unital self-adjoint subalgebra of $\mathscr{L}(E)$ which is closed in $\mathscr{L}(E)_s$.

b) Let $v\in A''$ and $x\in E$. The vector $v(x)$ belongs to the closure in E of the set of the $u(x)$ for $u\in A$. (Let F be this closure; observe that the orthogonal projector of E with image F belongs to $A.'$)

c) The algebra $A''$ coincides with the closure of A in $\mathscr{L}(E)_s$. (For every finite family $(x_i)_{i\in I}$ of elements of E, apply the preceding question to the space $E^I$, to the image $\widetilde{A}$ of A under the unital involutive morphism $f:\mathscr{L}(E)\rightarrow$ $\mathscr{L}(E^I)$ such that $f(u)(y_i) = (u(y_i))_{i\in I}$, and to the element $(x_i)$ of $E^I.$)

The equality of $A''$ and of the closure of A in $\mathscr{L}(E)_s$ is von Neumann's bicommutant theorem.