On ultraweak continuity

Question

Let $A$ be a C$^*$-algebra, $f$ a representation of $A$, $F$ the universal representation of $A$, and $g=f\circ F^{-1}$. For an ultraweakly continuous linear functional $w$ on $f(A)$, $w\circ g$ is bounded and hence according to a well-known theorem, is ultraweakly continuous linear functional on $F(A)$. My question: it results that g is ultraweakly continuous. How is its proof? Thank you for consideration.

Answer 1

What do you think about the following argument?

Given $x, y$ in $H$, let $W_{x, y}(S)=(Sx, y)$ defined on $f(A)^-$ (closure respect to WOT) and $G$ from $F(A)^-$ to $f(A)^-$, i.e. $W_{x, y}$ and $G$ are extensions of $w_{x, y}$ and $g$, respectively. Since $W$ is ultraweakly continuous on $f(A)^-$ then is weak-operator continuous on the unit ball of $f(A)^-$. Similarly, $W\,o\,G$ is weak-operator continuous on the unit ball of $F(A)^-$. It results that $G$ is ultraweakly continuous.

Answer 2

You give a reason why something is true, and then you ask for a proof. Perhaps you are asking for a simpler or more direct proof, or an intuitive explanation?

The universal representation $F: A \to \bigoplus B(H_i)$ of $A$ is obtained by taking the direct sum of all representations $f_i: A \to B(H_i)$. Thus your map $g$ taking $F(A) \cong A$ into $B(H_{i_0})$, where $f_{i_0} = f$, is just the restriction to $F(A)$ of the projection map taking $\bigoplus B(H_i) \subset B(\bigoplus H_i)$ onto a single factor $B(H_{i_0})$. Ultraweak continuity of this projection really should be an easy exercise.