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On page 39 of Dixmier's text on Von Neumann Algebras, he argues for Lemma 1, in which he tries to see that $\theta(L_1)=E_1$ using an argument about polars from functional analysis. I was hoping someone could explain why the asserted fact about polars there, which I agree is true, proves the desired claim that the continuous linear map taking $x \in B(H)$ to "evaluation at x" is surjective and isometric.

Edit: here is a paraphrase of the relevant excerpts from the book:

Notation: Let $E, F$ be Normed Vector Spaces, and $B$ a bilinear form on $E \times F$. Then if $y \mapsto B(. , y)$ is an isometric isomorphism then we say that $F$ is the dual of $E$ via $B$. Also, let $L_r$ denote the set of continuous bilinear functionals on $B(H)$ in the span of the ones taking the form $<\xi, (.)\eta>$ where $\xi, \eta \in H$.

Lemma 1 from Dixmier, which is what I want to prove. Please help me to fill in the details surrounding surjectivity and the polars: The canonical form on $L^* \times L$ induces on $L_r \times L$ a bilinear form for which $L$ is the dual of $L_r$. (All instances of dual mean norm-continuous, and all instances of "continuous" means from norm to norm unless I say otherwise.)

In other words, we want to see that the map $f$ taking elements of $L$ to the corresponding evaluational functionals defined on $L_r$ is an isometric isomorphism. Injectivity, continuity, and linearity are within my grasp. I need surjectivity and isometric of $f$. He notes $f$ is continuous from the weak-operator topology on $L$ to the weak-* topology on $L_r^*$ Then he notes that the unit ball $B_1$ in the weak-operator topology is compact and so its image under $f$ is compact. He then states that $f(B_1)$ and the unit ball of $L_r^*$ have the same polars. From here, and this is where I get lost, he says $f(B_1)=$ the unit ball of $L_r^*$. After that, I think I would understand why our map is surjective and isometric.

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I may have introduced the question quickly, but here is more background information if you need it. I thought it prudent to start a new thread for this related, but different question: math.stackexchange.com/questions/170025/… –  Jeff Jul 13 '12 at 0:43
Can you please make your question self-contained and explain the notations? I'm sure many people here would be able to answer the question without even having to consult Dixmier if only you phrased the problem in such a way that it is understandable without additional (superfluous) effort on the answerer's part. –  t.b. Jul 13 '12 at 2:25
@t.b. Okay done, I hope someone can help. –  Jeff Jul 13 '12 at 3:41
By the double polar theorem and $f(B_1)$ and $B_{L_r^*}$ having the same polars, it follows $\overline{\mathrm{conv}} f(B_1) = \overline{\mathrm{conv}} B_{L_r^*}$ (closure with respect to the $\sigma(L_r,L)$-topology). As both sets are convex and closed, you have equality. –  martini Jul 13 '12 at 6:47
So why is the compactness argument needed? The polar stuff here is being done on the norm topology. Am I missing something? –  Jeff Jul 13 '12 at 7:13

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