Completions of $C(X)$ with respect to the topologies generated by states I have no intuition in this field so excuse me if this is trivial.
Let $X$ be a compact Hausdorff space, and $C(X)$ the algebra of continuous functions on $X$ with the usual $\sup$-norm. This is a $C^*$-algebra, and each state $\omega:C(X)\to{\mathbb C}$ generates a GNS-representation
$$
\pi_\omega: C(X)\to B(H_\omega).
$$
Let us consider the $\sigma$-strong* topology on $B(H_\omega)$, i.e. the topology defined in the Bratteli–Robinson book Operator Algebras and Quantum Statistical Mechanics 1 by the seminorms of the form
$$
A\mapsto \sqrt{\sum_{n=1}^\infty \lVert A\xi_n\rVert^2+\sum_{n=1}^\infty \lVert A^*\xi_n\rVert^2}
$$
where $\xi_n$ are various sequences in $H_\omega$ with the property
$$
\sum_{n=1}^\infty \lVert\xi_n\rVert^2<\infty.
$$
Let $\tau$ denote this $\sigma$-strong* topology on $B(H_\omega)$, and $B_\tau(H_\omega)$ denote this space with this topology. So we have a (continuous) mapping
$$
\pi_\omega: C(X)\to B_\tau(H_\omega).
$$
Each state $\omega:C(X)\to{\mathbb C}$ generates such a mapping, and together these mappings generate a topology on the linear space $C(X)$, i.e. the initial topology on $C(X)$ generated by the family of mappings $\pi_\omega$. And since $\tau$ are locally convex topologies on $B(H_\omega)$, and $\pi_\omega$ are linear mappings, the corresponding initial topology on $C(X)$ is locally convex. My question is the following:

What is the completion of $C(X)$ with respect to the initial topology generated by the mappings $\pi_\omega$?

I have a suspicion that this must be the space $U(X)$ of universally integrable functions (see Leader - On Universally Integrable Functions), but up to now I did not manage to prove it. Is this true?
Actually, if we replace this $\sigma$-strong* topology with the other standard operator topologies on the spaces $B(H_\omega)$, the corresponding completions of $C(K)$ are also interesting for me. I would be grateful if somebody could cast a light on this.
 A: Every state on $C(X)$ is of the form $$\mu(g) = \int_Xg\ d\mu$$ for some positive measure $\mu$, $\mu(X)=1$. Thus, $H_{\mu}=L^2(X,\mu)$ and $\pi_{\mu}:C(X)\to B(H_{\mu})$ is the multiplication operator $\pi_{\mu}(g)f = gf$. If $\sum_n\|\xi_n\|_{L^2_\mu}<\infty$, then define $$f_n = \sum_{k=0}^n |\xi_k|^2$$ so $(f_n)$ is a Cauchy sequence in $L^1(X,\mu)$, and so converges to some $f\in L^1(X,\mu)$ with $f\geq 0$.
Conversely, clearly every $f\in L^1(X,\mu)$, $f\geq 0$ is a limit of a series this form (take $\xi_0=\sqrt{|f|}$ and $\xi_n = 0$ otherwise). Thus,
$$
\sum_k\|\pi_\mu(g)\xi_k\|_{L^2_\mu}^2 = \lim_{n\to\infty}\int_X |g|^2f_n\ d\mu = \int_X |g|^2f\ d\mu
$$
Consequently, every seminorm above on $C(X)$ is of the form $$|g|_{\nu} = \sqrt{\int_X |g|^2\ d\nu} = \|g\|_{L^2_\nu}$$ for some positive bounded measure $\nu$ on $X$. Hence, a net $(g_i)_{i\in I}$ is Cauchy in this topology if and only if converges to some $G_{\nu}\in L^2(X,\nu)$ for each $\nu$. This induces an element $G$ in the space $$\prod_{\nu\in M_b(X)} L^2(X,\nu) $$ with the product topology. The closure of $C(X)$ in this space is the completion that you're looking for.
