Suppose that $X_1$, $X_2$ are Hilbert spaces and $Y=X_1\times X_2$. Let $y\in Y$ be $y=(x_1, x_2)$ and define $F\in Y^*$ by $$\langle F,y\rangle_{Y^*\times Y}=\langle f,x_1\rangle_{X^*_1\times X_1}+\langle g,x_2\rangle_{X^*_2\times X_2},$$ where $f\in X_1^*$ and $g\in X_2^*$are given. Assume we have a sequence $(F_n)\subset Y^*$ which is equivalent that we have sequences $(f_n), (g_n)$ in $X^*_1$ and $X^*_2$, respectively. Is the following statement true: $$F_n\to F\,\,\text{weakly in}\,\,Y^* \Leftrightarrow f_n\to f\,\,\text{weakly in}\,\,X^*_1\,\,\text{and}\,\, g_n\to g\,\,\text{weakly in}\,\,X^*_2\,\,\,?$$
"$\Leftarrow$" $$|\langle F_n-F,y\rangle|\le|\langle f_n-f,x_1\rangle|+|\langle g_n-f,x_2\rangle|\to 0$$ since weak and weak* topologies coincide. What about "$\Rightarrow$"?
Assume that $F_n\to F$ weakly in $Y^*$. Then, for all $y$ we have that $$|\langle F_n-F,y\rangle|=|\langle f_n-f,x_1\rangle+\langle g_n-g,x_2\rangle|\to0.$$ If this is true for all $y$, then in particular for $y=(x_1,0)$ and $y=(0,x_2)$. Such choice gives however the desired convergances.