If $x_n \to x$ and $y_n \to y$ weakly. Then $\langle x, y\rangle = \lim_n \langle x_n, y_n\rangle$

Question

I'm trying to show below property, i.e.,

Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space. Let $x,y, x_n, y_n\in H$ such that $x_n \xrightarrow{n \to \infty} x$ and $y_n \xrightarrow{n \to \infty} y$ in the weak topology $\sigma(H, H^*)$. Then $\langle x, y\rangle = \lim_n \langle x_n, y_n\rangle$.

1. Could you have a check on my below attempt?
2. Is there another way that does not use below lemma?

Thank you so much for your help!

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Let $\sigma (H, H^*) \otimes \sigma (H, H^*)$ be the product topology of $\sigma (H, H^*)$ and itself.

Lemma $\sigma (H, H^*) \otimes \sigma (H, H^*)= \sigma (H^2, (H^2)^*)$.

By above Lemma, $(x_n, y_n) \xrightarrow{n \to \infty} (x, y)$ in $\sigma (H^2, (H^2)^*)$. Clearly, the map $(u, v) \mapsto \langle u, v \rangle$ belongs to $(H^2)^*$. The claim then follows.

Answer 1

The statement is not true in general. Consider $L^{2}[0,2\pi]$ . Take $x_{n}=\sin(nx)$ and $y_{n}=(-1)^{n}\sin(nx)$

Then for any $v\in L^{2}[0,1]$ , $\langle v,x_{n}\rangle$ and $\langle v,y_{n}\rangle $ tend to $0$ by Riemann-Lebesgue Lemma. $\langle f,g\rangle =\int_{0}^{2\pi}f(x)g(x)\,\,d\lambda(x)$

But $\langle x_{n},y_{n}\rangle $ is positive for $n$ odd and negative for $n$ even . Hence is not convergent

What is true is that if one of the sequences say $x_{n}\to x$ , i.e. $||x_{n}-x||\to 0$ . Then you have $\langle x_{n},y_{n}\rangle -\langle x,y\rangle = \langle x_{n}-x,y_{n}\rangle +\langle x,y_{n}\rangle-\langle x,y\rangle$ .

Now as $y_{n}$ converges weakly, it is norm bounded by Uniform Boundedness Principle. Thus $||y_{n}||\leq C$

So by Cauchy-Schwartz $|\langle x_{n}-x,y_{n}\rangle|\leq ||x_{n}-x||\cdot||y_{n}||\leq C||x_{n}-x||\to 0$ and also $\langle x,y_{n}\rangle-\langle x,y\rangle \to 0$ .

Thus you have $\langle x_{n},y_{n}\rangle \to \langle x,y\rangle$

Answer 2

Any orthonormal sequence $e_n$ tends to $0$ weakly, due to the Bessel inequality. But $\langle e_n,e_n\rangle =1.$