Notation: Let $H$ be a Hilbert space and let $\mathcal{L}$ denote the set of bounded linear operators from $H$ to itself. On $\mathcal{L}$ we have have the ultra-weak topology which is induced by the local sub-basis elements: \begin{equation} \left\lbrace T \in \mathcal{L} : \left|\sum_{i \in \mathbb{N}} (Tx_i | y_i)\right| < 1\right\rbrace \end{equation} (where $(x_i), (y_i)$ are sequences in $H$ with $\sum \|x_i\|^2, \sum_i \|y_i\|^2 < \infty$).
Say we have an ultra-weak closed subspace $M\subset \mathcal{L}$ and a convex set $K \subset M$. We denote by $M_r$ the elements of $M$ with norm $\leq r$.
Question: How do I show the equivalence \begin{equation} \left(K \cap M_r \text{ is ultra-weak closed for every } r>0\right) \iff \left( K \text{ is ultra-weak closed}\right)? \end{equation}
Attempt: $M_r$ is the intersection of $M$ with a norm ball in $\mathcal{L}$ and so the right-to-left implication is ok. I don't know how to show the left-to-right implication. This is part of the proof of Theorem 1 on page 42 of Dixmier's book 'Von Neumann algebras.'
There is a reference to theorem 5, section 2, chapter IV in Bourbaki's Espaces vectoriels Topologiques which I do not believe exists (Perhaps I cannot understand the indexing of Bourbaki books at all). I'm further supposed to use the fact that $M$ is the dual of its own dual (in the ultra-weak topology). This is part (iii) of the Theorem mentioned above.
Can someone please point me to right page in Bourbaki's book or provide some helpful comments on how to prove the equivalence?
Ok, I guess to close the question I should just copy down the guided exercise in Rudin's 'Functional Analysis' book, chapter 4, exercise 21. Here $X$ is a Banach space with its dual of continuous linear functionals $X^*$. I use the notation $B_X(x,r), B_{X^*}(f,r)$ for open norm-balls in $X$ and $X^*$. Further, $B_X[x,r], B_{X^*}[f,r]$ denotes closed norm-balls (radius up to and including $r$ so that $B_X[x,0] = \{x\}$).
The hypothesis is that we have a convex $K\subset X^*$ such that, for every $r>0$, $K \cap {B_{X^*}[0,r]}$ is weak*-closed (and hence weak*-compact).
(a) First note that $K$ is norm closed: For, if $g \in X^* \smallsetminus K$, we can find some $r>0$ with $g \in B_{X^*}(0,r)\smallsetminus K$. Since $ g\notin {B_{X^*}[0,r]} \cap K$ with this set being weak*-closed and hence norm-closed, we can find an $\varepsilon >0$ with $$B_{X^*}(g,\varepsilon) \subset B_{X^*}(0,r) \smallsetminus \left(K \cap {B_{X^*}[0,r]}\right).$$ Thus, $$ B_{X^*}(g,\varepsilon) \cap K = \varnothing. $$
(b) If we have, in addition to the hypothesis on $K$, that $K \cap {B_{X^*}[0,1]} = \varnothing$, then there exists an $x \in X$ with $\Re(f(x)) > 1$ for all $f \in K$: For $S \subset X$, we define the polar $$ P(S) := \bigcap_{s \in S} \left\lbrace f\in X^*: |f(s)| \leq 1 \right\rbrace. $$ It is easy to check that, for any $r>0$, \begin{equation}\label{1}\tag{1} \bigcap_{S \subset {B_{X}[0,r]}\text{ is finite.}} P(S) = {B_{X^*}\left[0,r^{-1}\right]}. \end{equation} Claim: We can choose a family of finite subsets $(S_i)_{i \in \mathbb{Z}_{\geq 0}}$ of $X$ such that, for each $k \in \mathbb{Z}_{\geq 0}$, $$ P(S_0) \cap \dots \cap P(S_{k-1}) \cap K \cap {B_{X^*}[0,k]} = \varnothing $$ and such that, for each $k \in \mathbb{Z}_{\geq 0}$, $kS_k \subset B_{X}[0,1]$.
Proof of Claim: We proceed by induction with the base step $P(S_0) \cap K \cap B_{X^*}[0,1]$ being clear for the set $S_0:= \{0\}$. We also have $0S_0 \subset B_X[0,1]$ Assume we have defined $S_0, \dots, S_{m}$ for some $m \geq 0$ with the required properties. That is, they are finite sets with $iS_i \subset {B_{X}[0,1]}$ for each $i=1,\dots,m$, and we have \begin{equation}\label{2}\tag{2} P(S_0) \cap \dots \cap P(S_m) \cap K \cap {B_{X^*}[0, m+1]} = \varnothing. \end{equation} Consider the intersection $$ Q:= P(S_0) \cap \dots \cap P(S_{m}) \cap K \cap {B_{X^*}[0, m+2]} $$ which is weak*-compact by hypothesis on $K$. By \eqref{1}, we see that $$ \bigcup_{S \subset B_X[0,(m+1)^{-1}]\text{ is finite}} P(S)^c = B_{X^*}[0,(m+1)]^c $$ with the sets $P(S)^c$ occurring in the union being weak*-open. In light of \eqref{2}, the above equation shows that the collection $$ \left\lbrace P(S)^c \subset X^*: S \subset B_{X}[0,(m+1)^{-1}]\text{ is finite}\right\rbrace $$ is an open cover of $Q$. Weak*-compactness and the natural order of the open family over shows that there is some finite $S_{m+1} \subset B_{X}[0,(m+1)^{-1}]$ with $$ Q \cap S_{m+1} = \varnothing. $$ Thus the induction step is concluded and the proof of the Claim is complete. $\square$
We return to the proof of (b). Enumerate the elements of $\bigcup_{k \geq 0} S_k$ as a sequence $(x_k)_{k \in \mathbb{N}} \subset X$. By construction $$ \lim_{k \to \infty} \|x_k\| = 0. $$ Let $c_0 \subset \mathbb{C}^{\mathbb{N}}$ denote the Banach space of functions vanishing at infinity (with supremum norm). Define a function $\Lambda: X^* \to c_0$ by $$ \Lambda(f) := (f(x_k))_{k \in \mathbb{N}}. $$ Since, by construction, $K \cap \left(\bigcap_{k \in \mathbb{Z}_{\geq 0}} P(S_k)\right) = \varnothing$, we have that, for every $f \in K$, $$ \|\Lambda(f)\| > 1. $$ Using the convexity of $K$, the Hahn-Banach Theorem (cf. Theorem 3.4 (a) in Chapter 3 of the same book) then guarantees the existence of an element in the dual of $c_0$, that is some $(\alpha_k)_{k \in \mathbb{N}} \subset \ell^1(\mathbb{N})$, such that $$ 1 < \Re \sum_{k \in \mathbb{N}} \alpha_k f(x_k)\ \text{ for every } f \in K. $$ Putting $x:= \sum_{k \in \mathbb{N}} \alpha_k x_k \in X$ establishes part (b).
(c) We have the Krein-Smulian theorem now. Given only the assumption that the convex set $K \subset X^*$ satisfies $K\cap B_{X^*}[0,r]$ is weak*-closed for every $r>0$, we show that $K$ itself must be weak*-closed: Say $g \notin K$. By part (a), there is some $\varepsilon >0$ such that $$ B_{X*}[g, \varepsilon] \cap K = \varnothing. $$ Thus, we also have $$ B_{X*}[0,1] \cap \varepsilon^{-1}\left(K - g\right) = \varnothing. $$ We see that part (b) applies to the set $\varepsilon^{-1} (K - g)$ so that there is some $x \in X$ with $$ \Re f(x) > \Re g(x) \ \text{ for all } f \in K. $$ Thus $g$ is not in the weak*-closure of $K$ and $K$ is thus weak*-closed.
I would still like to see the page in Bourbaki's book where this theorem occurs.