Why is the algebra of bounded (left) $F$-equivariant operators weakly closed in $B(\ell^2(F))$? Let $F$ be a free group with finite rank at least two. The Hilbert space of square-summable functions $f:F\to\mathbb{C}$ is denoted $\ell^2((F))$.
Define the weak (operator) topology on $B(\ell^2(F))$ as the topology induced by the family of complex valued functionals $f:B(\ell^2(F))\to\mathbb{C}$ s.t. 
$$f:T\mapsto \langle Tx,y\rangle\in\mathbb{C}$$ 
is continuous for any $x$ and $y$ in $\ell^2(F)$. Explicitly, the weak topology may be described as the topology generated by sets of the form $f^{-1}(U)$, where $U$ is an open set in $\mathbb{C}$.
Then how to see the algebra of bounded (left) $F$-equivariant operators weakly closed in $B(\ell^2(F))$?
 A: Let $T$ be an $F$-invariant linear map, this is equivalent to
$$\langle gTx,y\rangle=\langle Tgx,y\rangle$$
for all $x,y\in \ell^2(F)$ and all $g\in F$. Note that the adjoint of the multiplication with $g$ is multiplication with $g^{-1}$:
$$\langle g\cdot x,y\rangle = \sum_{a\in F}\overline{x_{ga}} y_{a}=\sum_{a\in F}\overline{x_{gg^{-1}a}}y_{g^{-1}a}=\langle x, g^{-1}\cdot y\rangle$$
It follows
$$\langle Tx,g^{-1}y\rangle=\langle Tgx,y\rangle$$
for all $x,y\in\ell^2(F)$, $g\in F$ is equivalent to $T$ being $F$-invariant.
Now let $T_\alpha$ be a net of $F$-invariant operators converging to $T$ in weak topology. Since $T\mapsto\langle Tx,g^{-1}y\rangle$ etc are all continuous it follows that
$$\langle Tx,g^{-1}y\rangle=\langle \lim_{\alpha}T_\alpha x,g^{-1}y\rangle=\lim_\alpha \langle T_\alpha x,g^{-1}y\rangle =\lim_{\alpha}\langle T_\alpha gx,y\rangle=\langle Tgx,y\rangle$$
And $T$ is also $F$-invariant. However the closure of a set is the same as all limit points of nets of that set and it follows that the closure of $F$-invariant linear maps in weak topology is itself.
