"$(x_n, y_n)$ is weakly convergent to $(x,y)$" implies "$(x_n)$ is weakly converging to $x$"?

Question

Let $H_1$, $H_2$ and $H_1 \times H_2$ be three Hilbert spaces, let the sequence $(x_n, y_n)$ be weakly convergent to $(x,y)$ in $H_1 \times H_2$. Then, when do we have " the sequence $(x_n)$ is weakly converging to $x$ and the sequence $(y_n)$ is weakly converging to $y$"?

Answer 1

Recall that the inner product on $H_1\times H_2$ is given by $\langle(x_1,y_1),(x_2,y_2)\rangle=\langle x_1,x_2\rangle+\langle y_1,y_2\rangle.$

Now to the problem at hand: If $\{(x_n,y_n)\}$ converges weakly to $(x,y)$, then for all $x'\in H_1$, we have $$\langle x_n,x'\rangle=\langle x_n,x'\rangle+\langle y_n,0\rangle\to \langle x,x'\rangle+\langle y,0\rangle=\langle x,x'\rangle$$ as $n\to\infty$. Thus $\{x_n\}$ converges weakly to $x$, and similarly we see that $\{y_n\}$ converges weakly to $y$.