weak convergence in the banach algebra of continuous operators on a hilbert space I'm attending a course named "Operator theory on Hilbert spaces" that is close to the book "Spectral Theory of Self-Adjoint Operators in Hilbert Space", M.S Birman and M.Z Solomjak.
On Chapter 2, Section 5 they define a weak-convergence on the algebra of continuous linear operators $\bf{B}$ on a Hilbert space $H$ as follows:
Let $(T_n)_{n\in\mathbb{N}}$ be a sequence of continuous linear operators. We say
$$T_n \stackrel{w}{\rightarrow} T$$ iff
$$ \langle T_nx,y \rangle \rightarrow \langle Tx,y\rangle\ \forall x,y \in H.$$
Now to my question:
$\bf{B}$ is endowed with the weak topology where $(T_n)_{n\in\mathbb{N}}$ converges if and only if
$$ l(T_n) \rightarrow l(T)\ \forall l \in \bf{B'}$$ where $\bf{B'}$ is the dual space of $\bf{B}$.
It is not clear to me why this two definitions of weak convergece are equivalent. I see this would be the case if $\bf{B'}$ was a Hilbert space, which to my knowledge is not the case.
Thank you!

The first topology is the weak operator topology. It's different from the weak topology.
Because the dual of $B(H)$ is not really tractable , the weak topology is never used in Operator Theory/Operator Algebras. The weak operator topology (wot), on the other hand, is  used frequently. Possibly because of both these facts, many authors talk about "weakly convergence" when talking about the wot; there is no ambiguity because the weak topology is never used.  One of the biggest advantages is that the wot has a Banach-Alaoglu theorem: the closed unit ball is compact.
