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

It is an easy undergraduate exercise to show that (finite) direct sums are preserved under dualisation. Thus, it is natural to ask if we the following holds:

is it true that if $X$ is a subspace of $Y$, then $X^* $ is a subspace of $Y^*$?

In many cases this is certainly not true (one can construct relevant subspaces of $Y=\ell^\infty$) but... let me say that $Y$ has property (D) if the above hypothesis holds for $Y$.

Is it true that the only spaces with property (D) are Hilbert spaces?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

$Y=L_1[0,1]$ has the property (D) since it is separable and the dual of any separable space embeds into $Y^\ast = L_\infty[0,1]$.

Of course, any separable space with a complemented subspace whose dual is isomorphic to $L_\infty[0,1]$ will have the property too.

If I think of other examples that are fundamentally different I'll add them later.

share|improve this answer
    
ok, it was easy, thanks! –  Tomek Kania Jul 24 '11 at 13:39
1  
I probably should have just answered in a comment, but in any case my initial attempts to find a counterexample (before I thought of $L_1$) involved trying to construct a space $Y$ with the formally stronger property $(E)$ (say) that if $X$ is a subspace of $Y$, then $X$ is isomorphic to $X^\ast$... this seems to be not so easy if one excludes Hilbert spaces I wonder if having the property (E) is equivalent to being isomorphic to Hilbert space? Maybe the answer to this also follows easily from known results, but off the top of my head I can't think of a good argument or counterexample. –  Philip Brooker Jul 24 '11 at 14:04
    
Nice question, Phil. It is related to an old unsolved problem of Pelczynski's: If $X$ is separable and every subspace of $X$ that has a basis is isomorphic to $\ell_2$, must $X$ be isomorphic to $\ell_2$? –  Bill Johnson Jul 24 '11 at 15:34
    
Phil, what about a local version of your problem? Suppose you have a family of finite dimensional spaces that is closed under taking of subspaces. Assume that every space in the collection is $C$-isomorphic to its dual. Is every space in the collection $D$-isomorphic to a Hilbert space for some $D=D(C)$? This looks to me easier than your question. –  Bill Johnson Jul 24 '11 at 15:39
    
Pelczynski's question is quite intriguing, Bill. I guess one could write down many properties for which the only known spaces with the property are those isomorphic to $\ell_2$; e.g., I wonder if the only minimal Banach space isomorphic to its dual is $\ell_2$? In a sort of opposite direction, I think it is (still) an open question whether an indecomposable space can be isomorphic to its dual. The local question you pose above seems like it is probably true... Is it known to be true for $C=1$? –  Philip Brooker Jul 26 '11 at 14:24

There are separable reflexive examples also, such as the $\ell_2$ sum $X$ of a sequence of finite dimensional spaces that is dense (in the sense of the Banach-Mazur distance) in the collection of all finite dimensional spaces. See my 1974 paper with Zippin in the Israel Journal of Mathematics.

Another example is a separable reflexive space $X$ s.t. every subspace of every quotient of $X$ is isomorphic to a complemented subspace of $X$, yet $X$ is not isomorphic to a Hilbert space. This is in a recent paper with Szankowski and can be downloaded from

http://www.math.tamu.edu/~bill.johnson/selpubs.html

(no. 117).

The example with Zippin lacks this property since it has subspaces that fail the approximation property (we proved in an earlier paper that each of its subspaces that has the approximation property is isomorphic to a complemented subspace).

share|improve this answer
    
Awesome examples, Bill. It's a shame that my answer was already accepted when you posted yours, because the content of mine was really only comment-worthy. –  Philip Brooker Jul 26 '11 at 14:33

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.