Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Hahn–Banach theorem states that: Given a sublinear functional $S: V \rightarrow \mathbb R$, if $T: U \rightarrow \mathbb R$ is a linear functional on a linear subspace $U \subseteq V$ that is dominated by $S$ on $U$, then there exists a linear extension of $T$ to $V$ that is dominated by $S$ on $V$.

Now, let us consider a symmetric multisublinear (positively homogeneous and subadditive in every component) continous functional $S: V\times\cdots\times V \rightarrow\mathbb R$ satisfying some good additionnal assumptions and a symmetric multilinear continous functional $T: U\times\cdots\times U \rightarrow \mathbb R$ that is dominated by $S$ on $U\times\cdots\times U$. Does there exist an extension of $T$ to $V\times\cdots\times V$ that is dominated by $S$ on $V\times\cdots\times V$?

share|improve this question
2  
Probably I am being dense... but I'm finding it very hard to understand what you mean. Could you please give some more explanation? –  Matthew Daws Feb 22 '12 at 14:07
2  
Hi Yves, you can edit your question instead of posting an answer. Just delete the answer and include the new info into the question! –  Jon Bannon Feb 22 '12 at 19:02
    
I took the liberty of doing this for Yves, and also added LaTeX. I have no good thoughts on the problem though... –  Matthew Daws Feb 22 '12 at 20:22
add comment

3 Answers

This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$


I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.

Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.

(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)

Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.

(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)

Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.


Edit: Since there is a conflict between this and Kofi's answer, let me make the following observation. The argument given by Kofi would, if valid, apply equally well to the problem of extending bilinear functionals defined on $E\times Y$ to bilinear functionals on $X\times Y$, while still preserving domination by the given bi-sublinear functional. (Here $E$ is a closed subspace of $X$, as above, and $Y$ is another Banach space.)

Take the sub-bilinear functional on $X\times Y$ to be $(x,y)\mapsto \norm{x}\norm{y}$. Then, translated into the language of tensor products as above, this would say that the natural restriction map $(X\ptp Y)^\ast \to (E \ptp Y)^\ast$ is surjective. But by the Hom-tensor duality for the proj tp, this is equivalent to saying that every bounded linear operator from $E$ to $Y^\ast$ extends to a bounded linear operator from $X\to Y^\ast$. In particular, taking $E=Y^*$ and $X=\ell^\infty(\hbox{suitable set})$, this would say that $Y^\ast$ is complemented in any copy of $\ell^\infty(\hbox{suitable set})$ that contains it as a closed subspace. This is not the case.

share|improve this answer
    
Your second claim is not correct. If you have a linear map $f: X \longrightarrow Y$ between Banach normed vector spaces that is bounded, injective and isometric (for example if $X$ is a subspace of $Y$ equipped with the same norm), then the dual map $f': Y' \longrightarrow X'$ is always surjective. Also, I gave a proof of the positive answer below, I believe. –  Kofi Feb 24 '12 at 12:43
    
@kofi, the point is that the norms do not match - if you take an element of the completed projective tensor product $E \hat{\otimes} E$, and view it as an element of $X\hat{\otimes} X$, then the norm can strictly decrease. This is well known in Banach-space theory, see examples in Defant and Floret's book for example. –  Yemon Choi Feb 24 '12 at 17:53
    
@Yemon: An off-topic question: How do you create bold text in comments ? Thanks. –  Ralph Feb 25 '12 at 3:09
    
@Ralph: I think that enclosing the words in double asterisks works (markdown syntax) –  Yemon Choi Feb 25 '12 at 3:22
    
Works Thank you very much –  Ralph Feb 25 '12 at 3:46
add comment

In this answer I made more or less the same mistake over and over again. It turns out that sublinear functions are not as nice as I believed. I thought it would be best to delete it.

share|improve this answer
2  
Could you give details of how the inductive step (adding an additional vector) works in the multisublinear case? –  Yemon Choi Feb 22 '12 at 21:03
1  
Kofi, if E is closed in F then $E\hat\otimes E$ need not be closed in $F\hat\otimes F$ ... –  Yemon Choi Feb 22 '12 at 23:09
    
Also, it's not obvious to me how to extend a multisublinear (whatever precisely this actually means) functional to a tensor product. I mean, this might work, but I suspect the devil is in the detail... –  Matthew Daws Feb 23 '12 at 9:22
    
@Yemon Choi: I don't understand what you mean by "closed". The Hahn-Banach Theorem is purely algebraic and has nothing to do with topology. @ Matthew: Actually, you are right. It might work, it might not, but it seems to be complicated if possible at all. Therefore I chose another route. –  Kofi Feb 23 '12 at 22:29
    
@Kofi: I have tried to explain my reasoning in a separate "answer" to this question –  Yemon Choi Feb 24 '12 at 7:26
show 2 more comments

First of all, I would like to thank you for your detailed answers.

On the following web page, I found a thesis entitled "Sur les opérateurs multisouslinéaires " by TALLAB Abdelhamid :

http://www.univ-msila.dz/theses/index.php?option=com_docman&task=cat_view&gid=46&limit=5&order=date&dir=ASC&Itemid=1

The author presents an "Extension du théoreme de Hahn-Banach aux opérateurs multisouslinéaires"

which is extracted from the paper

L.Mezrag and K.Saadi, On the multilisublinear operators. Begehr, H.G.W.(ed.) et al. Further progress in analysis. Proceedings of the 6th internationl ISAAC congress, Ankara, Turkey, August 13-18 (2007), 806-813.

The conditions are rather restrictives and I am still wondering whether my original formulation correspond to a true statement or not.

Do you know some counter-example (satisfying the symmetry assumptions) ?

share|improve this answer
1  
Yves: it is better practice to add updates to your original question, suing the "edit" functionality, rather than posting them in the box reserved for "answers". MO does not work like a blog comment thread, and if you write comments in the answer boxes it is not clear for new readers of the question what is going on. –  Yemon Choi Feb 27 '12 at 22:56
add comment

Your Answer

 
discard

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.