Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Notice that neither I nor Dixmier have said anything about symmetry of $B$. – Jeff Jul 12 '12 at 18:03
up vote 4 down vote accepted

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.

share|cite|improve this answer
Ah okay I see what you're saying, and thanks for the response. But Dixmier is talking about Normed vector spaces, so hopefully someone can clear that up. – Jeff Jul 12 '12 at 19:18
Actually, Hahn Banach works for NVS I thought. It actually works on general vector spaces, where the statement is if you have a partially defined linear functional and a dominating convex symmetric map defined on the whole space, then the linear functional extends to the whole space. So that ought to do it for here. To get continuity into this whole mess, the dominating function is $||L||*||.||$ if L is the partial linear functional. – Jeff Jul 12 '12 at 19:23
@Jeff: I am sorry, I wasn't thinking: Hahn-banach does not need completeness, so the same argument applies. I have edited my answer accordingly, and hope you are convinced. – wildildildlife Jul 12 '12 at 19:24
@Jeff: we realized this at the same time! Problem solved :) – wildildildlife Jul 12 '12 at 19:24

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.

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.

share|cite|improve this answer
@ Giuseppe Tortorella Thank you for your answer as well. I have a followup question. From Dixmier, the page after this one, same setup, he argues for the proof of Lemma 1. The canonical $\theta$ he defines that is to be the isometric isomorphism induced by the canonical bilinear map is easily (for me) seen to be norm to norm continuous, linear, and injective. He then tries to show it's a surjective isometry, and uses polars. I don't get that part. Why does $\theta(L_1)=E_1$? Thanks! Using the definition of polars on wikipedia, I can see the polars are equal, just not why that matters. – Jeff Jul 12 '12 at 21:05
Actually the wikipedia definition of polars is the "dual" flavor but I think the definition is intended to allow for the "predual" flavor as well, where what is meant by polar say of $E_1$ is the set of elements $x$ in $L_{\tilde}$ for which $|E_1(x)|\subset B_1$ – Jeff Jul 12 '12 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.