I have a bounded bilinear operator $B:X\times Y \to Z$, where $X,Y$ and $Z$ are Banach spaces. That is linear in both variables and satisfies: $$\|B(x,y)\|_Z \le \|B\|\ \|x\|_X\ \|y\|_Y, \quad \forall (x,y) \in X\times Y.$$
For some sequences $(x_n) \subset X$ such that $x_n \rightharpoonup \bar x$ in $X$, and $(y_n) \subset Y$ such that $y_n \rightharpoonup \bar y$ in $Y$.
My objective is to establish that $B(x_n,y_n) \rightharpoonup B(\bar x,\bar y)$ in $Z$. I would greatly appreciate any references or suggestions to help me prove this.
I tried with \begin{align*} B(x_n,y_n)-B(\bar x,\bar y)= B(x_n,y_n-\bar y)-B(x_n-\bar x,\bar y). \end{align*} The second term of the r.h.s converges weakly to zero in $Z$, as the operator $B(.,\bar y): X \to Z$ is linear and bounded. But, I can not see why $B(x_n,y_n-\bar y)$ converges weakly to zero in $Z$.
Thank you in advance for your assistance!
The claim is false generally as indicated in the comments. However, we can show it holds under additional hypotheses on $X$ and/or $Y$.
A Banach space $B$ is said to have the Dunford-Pettis property (DPP for short) if $x_n\to 0$ weakly in $B$ and $f_n\to 0$ weakly in $B^*$, then $f_n(x_n)\to 0$. We claim that if either $X$ or $Y$ (or both) has DPP, then $B(x_n,y_n)\to 0$ whenever $B$ is bounded bilinear, $x_n\to 0$ weakly in $X$, $y_n\to 0$ weakly in $Y$.
First, some notation: Let $f\in Z^*$, $B:X\times Y\to Z$ be a bounded bilinear operator.
For each $x\in X$, we can define $\gamma_x\in Y^*$ by $\gamma_x(y) = f(B(x,y))$,
For each $y\in Y$, we can define $\xi_y\in X^*$ by $\xi_y(x) = f(B(x,y))$.
It is up to the reader to show that if $x_n\to 0$ weakly in $X$ (resp. $y_n\to 0$ weakly in $Y$), then $\gamma_{x_n}\to 0$ weakly in $Y^*$ (resp. $\xi_{y_n}\to 0$ weakly in $X^*$). Hint: a linear operator is norm-norm continuous iff it is weak-weak continuous. Apply this to the map $X\to Y^*$ (resp. $Y\to X^*$) defined by $x\to \gamma_x$ (resp. $y\to\xi_y$)
Now, suppose $X$ has DPP. Let $f\in Z^*$ be arbitrary. Let $x_n\to 0$ weakly in $X$ and $y_n\to 0$ in $Y$. Then, $$f(B(x_n,y_n)) = \xi_{y_n}(x_n)\to 0$$ since $\xi_{y_n}\to 0$ weakly in $X^*$.
If $Y$ has DPP, the result follows verbatim similar as above.
Examples of Banach spaces with DPP are $L^1(X,\mu)$, $L^{\infty}(X,\mu)$ for $\sigma$-finite measure spaces $(X,\mu)$, and $C(K)$ for compact Hausdorff topological spaces $K$. Reflexive Banach spaces have DPP if and only if they're finite dimensional.
As a final note, if $x_n\to x$ weakly, $y_n\to y$ weakly, and $B(x_n-x,y_n-y)\to 0$ weakly in $Z$, then
\begin{eqnarray} f(\hspace{2mm} B(x_n,y_n) - B(x,y)\hspace{2mm}) &=& f(\hspace{2mm} B(x_n-x,y) + B(x,y_n-y) + B(x_n-x,y_n-y) \hspace{2mm}) \\ &=& \xi_y(x_n-x) + \gamma_x(y_n-y) + f(B(x_n-x,y_n-y))\to 0 \end{eqnarray} for each $f\in Z^*$; that is $B(x_n,y_n)\to B(x,y)$ weakly in $Z$.