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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.

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    $\begingroup$ Completion of a topological space (as opposed to metric space) doesn't seem to be a well defined notion, can you please elaborate on what you mean here. $\endgroup$ Commented Apr 18, 2022 at 16:43
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    $\begingroup$ @ChristianRemling Christian, all these topologies are locally convex, so $C(X)$ becomes a locally convex space with this topology. I'll correct this. $\endgroup$ Commented Apr 18, 2022 at 18:39
  • $\begingroup$ Do you think the initial topology above is the same as the initial topology given only by the pure states? An affirmative answer would considerably simplify the problem for $C(X)$. $\endgroup$
    – Onur Oktay
    Commented Apr 19, 2022 at 23:34
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    $\begingroup$ @OnurOktay I would think that these are different topologies. $\endgroup$ Commented Apr 20, 2022 at 8:05
  • $\begingroup$ @OnurOktay but I don't know... And what will the answer be if we rerplace all states by pure states? $\endgroup$ Commented Apr 20, 2022 at 8:25

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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.

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    $\begingroup$ Onur, this is vague for me. Is it possible that this closure of $C(X)$ in $\prod L^2(X,\nu)$ is exactly the space $U(X)$ of universally integrable functions on $X$ as I conjectured in the question? Or perhaps there is another explicit description. $\endgroup$ Commented Apr 21, 2022 at 17:49
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    $\begingroup$ Onur, you should remove the limit in $\lim_{n\to\infty}\int_X |g|^2f\ d\mu$. $\endgroup$ Commented Apr 21, 2022 at 18:06
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    $\begingroup$ @SergeiAkbarov I do not know the answer to your question, but I can say this much: If $(g_n)$ is a bounded sequence in $C(X)$, then $(g_n)$ is Cauchy in the topology that you'd given iff $(g_n)$ converges pointwise. Thus, the space of Baire-1 functions $B_1(X)$ is contained in the completion of $C(X)$ in this topology. $\endgroup$
    – Onur Oktay
    Commented Apr 22, 2022 at 19:53

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