I have a question about weak convergence in Hilbert spaces. Suppose we have a Hilbert space $H$ equipped with a scalar product $(\cdot / \cdot)$, and that in this space we have $x_n \rightharpoonup x$ for some $x \in H$.
Then I know that $\forall \lambda \in H^{*}$ we have $\lambda(x_n) \rightarrow \lambda(x)$, which by applying the Riesz representation theorem yields:
$$ (x_n / y) \rightarrow (x / y) \forall y \in H \quad \quad (1) $$
Question: is it true that $(x_k / x_l) \rightarrow (x/x) = ||x||^2$ as $k,l \rightarrow \infty$?
My proof idea: I think this result is true. Consider the quantity $R = (x/x) - (x_k / x_l)$. Add and take away $(x / x_l)$:
$$ R = [(x/x) - (x / x_l)] + [(x / x_l) - (x_k / x_l)] $$
And both brackets go to 0 by letting $k,l$ go to $\infty$, by (1). However I know we don't necessarily have that $(x_n / x_n) \rightarrow (x / x)$... I am not sure if this means the proof I gave is wrong or not...
Thank you for your help!
In general your claim is wrong. See that if you assume $(x_k,x_l) \rightarrow (x,x)$ you get with $k=l$ that $$ ||x_k||^2 = (x_k,x_k) \rightarrow (x,x) = ||x||^2. $$ Now you may know that weak convergence plus convergence of the norm gives strong convergence, this is true in an arbitrary Hilbert space and it even holds in uniformly convex spaces. So for a sequence that converges weakly and not strongly your claim cannot hold.