Duals via a Bilinear map

Question

Let $E$ and $F$ be normed vector spaces. Then if $B$ is a bounded bilinear form on $E \times F$ then every $y \in F$ defines a bounded linear functional $f_y$ where $f_y(x)=B(x, y) \forall x \in E$. When the bounded linear map $y \mapsto f_y$ of $F$ into $E^*$ the continuous dual of $E$ happens to be an isometric isomorphism, we say that $F$ is the dual of $E$ via $B$. (This can all be found in Dixmier Von Neumann Algebras) Why is it then true that

$||x||=sup_{||y|| \leq 1} |B(x, y)| \forall x \in E$? He states both this and its analog with x, y interchanged, and E, F interchanged. That version is pretty much the definition of what it means to be an isometry, but this one here is not.

Answer 1

In general, for $E$ a normed space with continuous dual $E^*$, one has for all $x\in E$ that

$$\|x\|=\sup_{\alpha\in E^*,\|\alpha\|=1}|\alpha(x)|$$

by a standard Hahn-Banach argument. Since by assumption $F\to E^*$ is an isometric isomorphism, this answers your question.

Answer 2

The constuction of $f$ which you report implies, for any $y\in F$, that $$f_y\in E^{\ast},\text{ with }||f_y||_{E^{\ast}}=\sup\{B(x,y):x\in E,\ \||x||_E=1\}$$ This constitues one of the equalities stated in Dixmier.

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The further hypothesis that $y\in F\to f_y\in E^{\ast}$ is an isometric isomorphism, means that $$f(F)=E^{\ast}\quad\text{and}\quad||f_y||_{E^{\ast}}=||y||_F,\ \forall y\in F\tag{1}.$$ A corollary of Hahn-Banach theorem states that $$||x||_E=\sup\{\langle\phi,x\rangle:\phi\in E^{\ast}\, ||\phi||_{E^{\ast}}=1\}, \forall x\in E.\tag{2}$$ From $(1)$ and $(2)$ we conclude that, as searched, $$||x||_E=\sup\{\langle f_y,x\rangle=B(x,y):y\in F,\ ||y||_{F}=||f_y||_{E^{\ast}}=1\}.$$ This is the second equality stated in Dixmier.