If $H$ is Hilbert space, prove that $$ \|A\| \leq \liminf_{n \to \infty} \|A_n\|,$$ when $A$ is weak limit of $A_n$, for some $A_n \in B(H)$.
I know that $A_n$ converges weakly to $A$ iff $\lim\langle A_nf,g\rangle = \langle Af,g\rangle, \forall f,g \in H$. Also, $\|A\| = \underset{\|f\|=1}{\sup}|\langle Af,f\rangle|$. Further,
$$|\langle Af,f\rangle |=|\lim \langle A_nf,f\rangle| = \lim |\langle A_nf,f\rangle|.$$
Let $\varepsilon > 0$, arbitary. $\langle A_nf,f\rangle$ converges, so I can take subsequence $\langle A_{nk}f,f\rangle$, such that $$|\langle A_{nk}f,f\rangle| \leq \varepsilon + \liminf_{n \to \infty} \|A_n\|.$$ Since $\lim |\langle A_nf,f\rangle| = \lim |\langle A_{nk}f,f\rangle|$, I get the result.
I am not sure about part: $|\langle Af,f\rangle |=|\lim \langle A_nf,f\rangle| = \lim |\langle A_nf,f\rangle|$, because it's weak limit. Can someone tell me if this is wrong and help me if it is?
Thanks!