This is a follow-up question to this question. I read a more elegant looking proof in [1] Proposition 8.44, p208. So it goes like this.
Proposition If $x_n \rightharpoonup x$ then
$$\left\Vert x \right\Vert \leq \liminf_{n \to \infty}\left\Vert x_n \right\Vert $$.
Proof. Using the weak convergence of $(x_n)$ and th Cauchy-Schwarz inequality, we find that $$\left\Vert x \right\Vert^2 = \langle x, x \rangle =\lim_{n \to \infty}\langle x, x_n \rangle \leq \left\Vert x \right\Vert\liminf_{n\to \infty}\left\Vert x_n \right\Vert.$$
What I don't understand is that how after the Cauchy-Schwarz the lim becomes liminf and still the inequality still holds. I would think that replacing the limit with liminf should have a less than or equal. But ofcourse I know the result is correct, infact if we use the weak lower semicontinuity of the norm then we are good. As weak semicontinuity is defined a bit after the proof in the book.
So is there a way to do it without invoking sequential weak lower semicontinuity?
[1] J. K. Hunter and B. Nachtergaele, Applied Analysis. World Scientific Publishing Company, 2001.
As mentioned in the comment, since the strong limit of $x_n$ might not exists, one cannot write $\lim_{n\to \infty} \|x_n\|$. Instead, the Cauchy Schwarz gives
$$ \langle x, x_n\rangle \le \|x\| \cdot \|x_n\|,$$ now take liminf on both sides:
$$\liminf_{n\to\infty} \langle x, x_n\rangle \le \|x\| \cdot \liminf_{n\to \infty} \|x_n\|,$$ since $\langle x, x_n\rangle$ does has a limit, so
$$\liminf _{n\to\infty}\langle x, x_n\rangle = \lim_{n\to\infty} \langle x, x_n\rangle,$$ which is what you want.