Let $X,Y$ be Banach spaces. In V. Paulsen's book on completely positive maps, he shows that $B(X,Y^\ast)$ is a dual space as follows: For $x \in X, y \in Y$ let $x \otimes y \in B(X, Y^\ast)^\ast$ be given by $(x \otimes y)(L) = L(x)(y)$. Let $$Z = \overline{\operatorname{span}}\{x\otimes y: x \in X, y \in Y\}$$ He shows that $B(X, Y^\ast) \cong Z^\ast$ with the pairing $$(L, x\otimes y) = (x \otimes y)(L)$$ The topology on $B(X,Y^\ast)$ inherited from $Z^\ast$ is called the BW topology. This all makes sense to me so far. He then claims the following:
Let $(L_\lambda)$ be a bounded net in $B(X,Y^\ast)$. Then $(L_\lambda)$ converges in the BW topology to $L$ iff $(L_\lambda(x))$ converges weakly to $L(x)$ for all $x \in X$.
One direction of his proof goes as follows:
Suppose $L_\lambda \to L$ in BW and let $x \in X$. Then $$L_\lambda(x)(y) = (L_\lambda, x\otimes y) \to (L, x\otimes y) = L(x)(y) \quad\text{for all $y \in Y$}$$ Thus $L_\lambda (x) \to L(x)$ weakly.
The very last sentence of the proof is where I am confused. Hasn't he just shown that $L_\lambda(x) \to L(x)$ in weak* (i.e. pointwise) rather than weakly? Why don't we need to take an arbitrary $f \in Y^{\ast\ast}$ and show that $f(L_\lambda(x)) \to f(L(x))$? Of course I know that $Y$ canonically embeds into $Y^{\ast\ast}$, but if $Y$ is not reflexive then there might be more elements in $Y^{\ast\ast}$ than just in $Y$, and so showing pointwise convergence isn't enough to show weak convergence right? Perhaps this has something to do with the net being bounded?
Any help is appreciated.