# Show that $(x_n, y_n) → (x, y)$

real-analysis

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

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.