Suppose a sequence $\{x_n\}$ converges weakly to $x$ in a Hilbert space $H$ and a sequence

$\{y_n\}$ converges strongly to $y$ in $H$. Show that $(x_n, y_n) → (x, y)$ as $n → ∞$.

Proof idea:

${x_n}$ converges weakly, so $(x_n,a) \rightarrow (x,a)$ for any $a\in H$.

Let $a = y_k\in H$. Then $(x_n,y_k) \rightarrow (x,y_k)$ for any $y_k\in H$.

Does it follow directly that $(x_n,y_n) \rightarrow (x,y)$ ?

$\lim\|x_n-y_k\|^2=\|x\|^2-2\lim \langle x_n,y_k \rangle+\|y_k\|^2\leq\|x\|^2-2\liminf \langle x_n,y_k \rangle+\|y_k\|$

$\leq \|x\|^2-2\|x\|\|y_k\|+\|y_k\|^2=\|x-y\|^2$.

Since $\|x\|\leq \liminf\|x_n\|$.

Thanks

## Best Answer

You need some estimates for the difference $$(x_n, y_n) - (x,y)$$

Some calculations: $$(x_n, y_n) - (x,y) = (x_n, y_n) - (x_n, y) + (x_n, y) - (x,y)$$

Now $$|(x_n, y_n) - (x_n, y)| = |(x_n, y_n-y) | \le \|x_n\| \cdot \|y_n-y\|$$

The second term $(x_n-x, y)$ clearly approaches $0$. It is the first term that is a bit more delicate. Here you have to use the fact that a weakly convergent sequence $(x_n)$ is weakly bounded, so bounded ( Banach-Steinhaus theorem). So some heavy machinery is needed, and with that you are done.