What does "weak continuity of the norm" means?

Question

I am trying to read an article about $C^*$-algebras but I am stuck in one passage in wich the autor is estimating the value of sequences of linear functionals $\omega_{1,n}$ and $\omega_{2,n}^A$ that converge weakly to $\omega_{1}$ and $\omega_{2}^A$ respectively, being that

$$\omega_{1,n}(b) = \langle \psi_{1,n}\,\, ,\pi_n(b)\psi_{1,n}\rangle$$

$$\omega_{2,n}^A(b) = \langle \psi_{2,n}\,\, ,\frac{\pi(A^*bA)}{\omega_{2,n}(A^*A)}\psi_{2,n}\rangle$$

$$\omega_{2,n}(b)= \langle \psi_{2,n}\,\, ,\pi_n(b)\psi_{2,n}\rangle$$

so that

$$\omega_{2,n}(A^*A) = \langle \psi_{2,n}\,\, ,\pi_n(A^*A)\psi_{2,n}\rangle=\langle \pi_n(A)\psi_{2,n}\,\, ,\pi_n(A)\psi_{2,n}\rangle$$

and knowing that $||\omega_{1} - \omega_{2}^A|| = 2$

he says that "by the weak continuity of the norm":

$$||\omega_{1,n} - \omega_{2,n}^A||^2\leq 4 - 4\frac{|\langle\psi_{1,n}\,\,,\pi_n(A)\psi_{2,n}\rangle|^2}{||\pi_n(A)\psi_{2,n}||^2}$$

I imagine that $||\pi_n(A)\psi_{2,n}||^2= \left(\sqrt{\omega_{2,n}(A^*A)}\right)^2 = \langle \pi_n(A)\psi_{2,n}\,\, ,\pi_n(A)\psi_{2,n}\rangle$

But how can I get the cross-term ?

I have tried to do this in the following way:

$$||\omega_{1,n} - \omega_{2,n}^A||^2 \leq |\omega_{1,n}(b)-\omega_{2,n}^A(b) |^2 = \Bigg\vert\langle \psi_{1,n}\,\, , \pi_n(b)\psi_{1,n}\rangle - \frac{\langle\psi_{2,n}, \pi_n(A^*bA)\psi_{2,n} \rangle}{\langle\psi_{2,n}, \pi_n(A^*A)\psi_{2,n} \rangle}\Bigg\vert^2$$

I then thoght of using something like an polarization identity like $||u-v||^2 = ||u||^2 + ||v||^2 - 2\langle u,v\rangle$ and ignoring one of the squared norms in the right hand side to get an inequality, but I don't think that is the way.

Does the term "weak continuity of the norm" mean anything? I googled it and didn't find anything except the very article I am reading.

(FYI the article is "Quantum theory of measurement and macroscopic observables" by Klaus Hepp)

Answer 1

When he says "weak continuity of the norm", it looks like he doesn't refer to the inequality, but rather to the other side, $$ \|\omega_{1,n}-\omega_{2,n}^A\|\to2. $$ Still, the terminology is at least unusual.

It's a bit annoying to write all the details. But basically what happens is that because $\omega_1$ and $\omega_2$ are disjoint, there "almost" exists a projection $p$ with $\omega_1(p)=1$, $\omega_2(1-p)=1$. Properly, this projection is guaranteed to exist if $\mathcal A$ is a von Neumann algebra; but in general it exists in $\mathcal A''$ and one can work with approximations in $\mathcal A$. Then \begin{align} \|\omega_{1,n}-\omega_{2,n}\|&\geq|\omega_{1,n}(2p-1)-\omega_{2,n}(2p-1)|\to|\omega_1(2p-1)-\omega_2(2p-1)|\\[0.3cm]&=|\omega_1(p)+\omega_2(1-p)|=2. \end{align}

The inequality, on the other hand, is standard but not trivial. Basically, what happens is that if $\alpha(X)=\langle \psi, X\psi\rangle$, $\beta(X)=\langle \phi, X\phi\rangle$ are states (that is, $\|\psi\|=\|\phi\|=1$), then $$ \|\alpha-\beta\|=2\sqrt{1-|\langle \psi,\phi\rangle|^2}. $$

Edit: I have included a proof of the norm equality here.