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.