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Can somenone remind me of the meaning of the following statement:

the family of operator valued functions $A(\omega)$ converges to $A(\omega ')$ in the weak topology of bounded operators from $H_1$ to $H_2$

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up vote 1 down vote accepted

First of all, implicit in this statement is some notion of convergence for $\omega$; although you have not given the context, the statement $\omega \to \omega'$ must make sense.

Here I will assume that for each $\omega \in \mathbb R$, there is a bounded operator $A(\omega):H_1 \to H_2$ and that $\omega \to \omega'$ in $\mathbb R$.

Secondly, what is the weak topology? Sadly, this phrase is a little ambiguous. As noted at, the phrase weak topology is sometimes used to refer to three different things: the weak operator topology, the weak (Banach space) topology, and the ultraweak topology.

However, from the context it seems likely that what is intended here is the weak operator topology, which is defined by the family of seminorms $\{ \langle \cdot \xi, \eta \rangle : \xi \in H_1, \eta \in H_2\}$. (Here $\langle \cdot, \cdot \rangle$ is the inner product on $H_2$.)

So convergence in the weak (operator) topology means that for each $\xi \in H_1$ and $\eta \in H_2$, $$ | \langle A(\omega) \xi, \eta \rangle - \langle A(\omega') \xi, \eta \rangle| =|\langle (A(\omega) -A(\omega'))\xi, \eta \rangle|\to 0, $$ as $\omega \to \omega'$.

In more detail, for each $\xi \in H_1$ and $\eta \in H_2$, given $\epsilon >0$, there exists a $\delta >0$ (depending on $\xi$, $\eta$, and $\epsilon$) such that $$ | \langle ( A(\omega) -A(\omega') ) \xi, \eta \rangle| < \epsilon, $$ whenever $| \omega - \omega' |<\delta$.

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