The question in this post is the question below from an article by Rodriguez & Rueda Zoca [1].

Below is a complimentary salad/side dish that accompanies the main course.

Let $B^2(X,Y)$ denote the space of all $\mathbb{C}$-valued bounded bilinear forms. We identify $B^2(X,Y)$ with $(X\mathbin{\hat{\otimes}_\pi}Y)^*$ as it's customary, that is we write $$\beta(x\otimes y) = \beta(x,y)\hspace{6mm} x\in X,\ y\in Y$$ for $\beta\in (X\mathbin{\hat{\otimes}_\pi}Y)^*$. Let $B^2_f(X,Y)$ denote the set of the bilinear maps of finite rank, and $B^2_{wsc}(X,Y)$ denote those that are weakly sequentially continuous.

Let $A\subset X\mathbin{\hat{\otimes}_\pi}Y$ be the set of weak limit points of the sequences of the form $(x_n\otimes y_n)$, where both $(x_n)$ and $(y_n)$ are weakly null. Let $A^\perp$ be the set of all $\beta\in (X\mathbin{\hat{\otimes}_\pi}Y)^*$ such that $\beta(A)=\{0\}$. Clearly $A^\perp$ is weak$^*$-closed and $$B^2_f(X,Y) \subset B^2_{wsc}(X,Y) \subseteq A^\perp$$ Thus, an equivalent formulation of Question 3.9 is

$A^\perp = B^2(X,Y)$?

As exemplary partial answers with an extra condition: if $X$ or $Y$ has the Dunford-Pettis property (DPP), then $B^2_{wsc}(X,Y) = B^2(X,Y)$ (the latter condition is weaker than DPP), so yes. Clearly it is enough to merely assume that $B^2_{wsc}(X,Y)$ is weak$^*$ dense in $B^2(X,Y)$.

If $X$ or $Y$ has the approximation property (AP), then $B^2_f(X,Y)$ is weak$^*$ dense in $B^2(X,Y)$, so the answer is yes.