6.1 Ultraweakly Continuous Implies Normal
In what follows, let \(T\) denote the set of all \(\sigma \)–continuous positive linear functionals on \(M\), and \(E\) the linear space of all finite linear combinations of elements of \(T\). Let \(P\) denote the set of positive elements of \(M\).
Every continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
Omitted. In Mathlib.
(Banach-Alaoglu) The closed unit ball of the dual space of a normed vector space is compact in the weak-\(*\) topology.
Omitted. In Mathlib.
(Sak 1.7.4) Every uniformly bounded, increasing net in \(M^{s}\) converges to its least upper bound in the \(\sigma \)–topology. Further, if \(x=\sup _{\lambda }x_{\lambda }\) then \(a^{*}xa=\sup _{\lambda }a^{*}x_{\lambda }a\).
Letting \(T\) denote the set of \(\sigma \)–continuous positive functionals on \(M\) and \(E\) its linear span, by Lemma 38, the subspace \(E\) separates the points of \(M\) and so the \(\sigma (M,E)\) is Hausdorff. By the Banach-Alaoglu Theorem (44) the closed unit ball \(B\) of \(M\) is compact in the \(\sigma (M,M_{*})\)–topology. Since \(\sigma (M,E)\) is weaker than \(\sigma (M,M_{*})\), the identity map on \(B\) is \(\sigma (M,M_{*})-\sigma (M,E)\)–continuous, and therefore a homeomorphism, by Lemma 43. So, to prove that a uniformly bounded net \((x_{t})\) is \(\sigma \)– Cauchy, it’s enough to check that for all \(\varepsilon {\gt}0\) and \(\varphi \in T\) that there is a \(\Lambda \) such that \(\varphi (x_{\lambda }-x_{\mu })\le \varepsilon \) for \(\lambda \ge \mu \ge \Lambda \).
If \((x_{\lambda })\) is a uniformly bounded increasing net in \(M^{s}\), and \(\varphi \in T\), then \((\varphi (x_{\lambda }))\) is a uniformly bounded increasing net of real numbers and therefore is Cauchy. By the above paragraph, \((x_{\lambda })\) is \(\sigma \)–Cauchy and by the \(\sigma \)–compactness of \(B\) and the fact that \((x_{\lambda })\) is increasing there exists \(x\in B\) so that \(x_{\lambda } \to x\) in the \(\sigma \)–topology. Since for any \(\varphi \in T\) and any \(\lambda \) we have \(0 \le \varphi (x-x_{\lambda })\), by Lemma 37 we have \(0\le x - x_{\lambda }\). Therefore \(\sup (x_{\lambda })\le x\). For a given \(\varphi \in T\) and \(\varepsilon = \varphi (x-\sup (x_{\lambda }))\) one find \(x_{\lambda }\) so that \(\varphi (x-x_{\lambda }){\lt}\varphi (x - \sup (x_{\lambda }))\), which is a contradiction, therefore \(x=\sup (x_{\lambda })\).
By Lemma 28 and the fact that \((u^{-1})^{*}=(u^{*})^{-1}\), then \(u^{*}x_{\lambda }u\le u^{*}xu\) is equivalent to \(x_{\lambda }\le x\) and so \(\sup (u^{*}x_{\lambda }u)=u^{*}\sup (x_{\lambda })u=u^{*}xu\) if \(u\in M\) is invertible.
If \(a\in M\) is not invertible, one can shift the spectrum horizontally to avoid zero, i.e. there is a \(c{\gt}0\) so that \(c1+a\) is invertible. Indeed observe that \(a-t1=(a+c1)-(t+c)1\) and therefore \(t\in \sigma (a)\iff t+c \in \sigma (a+c1)\). In particular if we let \(t=-c\) here we find that \(0\in \sigma (a)+c \iff 0 \in \sigma (a+c1)\). By the invertible case above,
for all \(\varphi \in T\). It suffices to show that \(\varphi (a^{*}x_{\lambda }+x_{\lambda }a)\to \varphi (a^{*}x+xa)\) for all \(\varphi \in T\), because then \(\varphi (a^{*}x_{\lambda }a)\to \varphi (a^{*}xa)\) by the above. We can use Cauchy-Schwartz for this as follows. If \(\alpha \le \beta \), we have \(x_{\beta }-x_{\alpha }\ge 0\), and since the net \((x_{\lambda })\) is uniformly bounded there exists \(M{\gt}0\) such that (using 29)
and by the positivity of \(\varphi \) we have \(|\varphi (a^{*}(x_{\beta }-x_{\alpha }))|=|\varphi ((x_{\beta }-x_{\alpha })a)|\). It follows that \(\varphi (a^{*}x_{\lambda }+x_{\lambda }a)\to \varphi (a^{*}x+xa)\), and the proof is finished.