Equivalence between two properties related to a pair of operators Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F, \langle \cdot,\cdot\rangle)$. 

If $A,B\in \mathcal{B}(F)$. Why the following two assertions are equivalent?
  
  
*
  
*(1) There exists $(x_n)_n\subset F$ such that $\|x_n\|=1$ and
  \begin{equation*}
\lim_{n\to \infty}|\langle A x_n, Bx_n\rangle|=\|A\|\|B\|.
\end{equation*}
  
*(2) There exist $\alpha \in \mathbb{C}$ with $|\alpha|=1$, a sequence $(x_n)_n\subset F$ such that $\|x_n\|=1$ and 
  \begin{equation*}
\lim_{n\to \infty}\langle A x_n, Bx_n\rangle=\alpha\|A\|\|B\|.
\end{equation*}

I ask this question since I read this paragraph

However, Theorem 3.3 states

 A: These properties are indeed equivalent. Suppose that property (1) holds and let $(x_{n})$ be a sequence in $F$ such that
$$\lim_{n\rightarrow\infty}|\langle Ax_{n},Bx_{n}\rangle|=\|A\|\|B\|.$$
Note that the unit sphere of $\mathbb{C}$ is compact, so by taking a further subsequence we may assume that there is some $\alpha\in\mathbb{C}$ with $|\alpha|=1$ such that
$$\lim_{n\rightarrow\infty}\frac{\langle Ax_{n},Bx_{n}\rangle}{|\langle Ax_{n},Bx_{n}\rangle|}=\alpha.$$
It follows that 
$$\lim_{n\rightarrow\infty}\langle Ax_{n},Bx_{n}\rangle=\lim_{n\rightarrow\infty}\frac{\langle Ax_{n},Bx_{n}\rangle}{|\langle Ax_{n},Bx_{n}\rangle|}|\langle Ax_{n},Bx_{n}\rangle|=\alpha\|A\|\|B\|,$$
so property (2) holds.
To show that (2)$\Rightarrow$(1) simply requires taking the absolute values on LHS and the RHS.
A: EDIT: The essence of the issue is cleared up in Floris' answer. So I recommend checking his. However, the following might be interesting to get a feeling about what the conditions actually mean geometrically.
WLOG I assume that $\|A\|=\|B\|=1$. I'll show that both conditions are equivalent to the following one:

(3) There exist a sequence $(x_n)$ with $\|x_n\|=1$ and $z\in\mathbb T$ such that $\|Ax_n\|\to 1$, $\|Bx_n\|\to 1$, and $z\in\mathbb T$ such that $\|Bx_n-zAx_n\|\to 0$.

First of all (3) implies (2):
$$
(Ax_n,Bx_n) = (Ax_n,Bx_n-zAx_n) + \overline z\|Ax_n\|^2\,\to\,\overline z.
$$
Clearly, (2) implies (1). So, let's show that (1) implies (3). If (1) holds, then
$$
1\leftarrow |(Ax_n,Bx_n)|\le\|Ax_n\|\|Bx_n\|\le 1,
$$
so $\|Ax_n\|\to 1$ and $\|Bx_n\|\to 1$. Now, let $z_n\in\mathbb T$ be such that $z_n(Ax_n,Bx_n) = |(Ax_n,Bx_n)|$. Then
$$
\|Bx_n-z_nAx_n\|^2 = \|Bx_n\|^2 - 2|(Ax_n,Bx_n)| + \|Ax_n\|^2\to 0.
$$
Now, choose a convergent subsequence $(z_{n_k})$ of $(z_n)$ such that $z_{n_k}\to z\in\mathbb T$. WLOG, $z_{n_k} = z_n$. Then
$$
\|Bx_n - zAx_n\|\le\|Bx_n-z_nAx_n\| + |z-z_n|\|Ax_n\|\to 0.
$$
