0
$\begingroup$

I see examples on Stack Exchange and elsewhere of Banach spaces with non-complemented subspaces (examples: 1, 2 [Remark 8], 3 [a remarkable example of a Banach space with no complemented closed subspaces]). My question is thus:

We know that every vector space (and therefore every Banach space) has a vector space basis. Therefore, for a Banach space $X$ and a subspace $E$, we can write a basis $\mathcal{B}_E$ for $E$ and extend it to a basis $\mathcal{B}$ for $X$. Consider $F$ to be the subspace spanned by $\mathcal{B}\setminus \mathcal{B}_E$. Why is $F$ not necessarily a complementary subspace to $E$? It seems that for $v\in X$, we should have that $v$ is uniquely represented by an element of $E$ plus an element of $F$, since it is uniquely represented by a finite linear combination of elements in $\mathcal{B}$ (and we can therefore segregate those elements into elements of $\mathcal{B}_E$ and $\mathcal{B}\setminus \mathcal{B}_E$). Also, it seems that $E\cap F = \{0\}$, since the basis vectors for $E$ and $F$ are linearly independent from each other. Could somebody please help by pointing out where I might have made a faulty assumption?

$\endgroup$
  • 1
    $\begingroup$ There is no reason $F$ should be closed. $\endgroup$ – Qiaochu Yuan Mar 7 '17 at 20:16
  • $\begingroup$ Does $F$ need to be closed to be complementary to $E$? $\endgroup$ – Michael Lee Mar 7 '17 at 20:21
  • 1
    $\begingroup$ Yes, in the context of Banach spaces we only consider closed subspaces (eg because they are the subspaces which are themselves Banach spaces), or else as you say you could just consider a vector space complement. The first example you link to also explicitly states this as part of the definition. $\endgroup$ – Qiaochu Yuan Mar 7 '17 at 20:23
  • $\begingroup$ I understand. Although, I disagree that we only ever consider closed subspaces of Banach spaces. Rudin and the textbook I'm working through right now (Folland) have plenty of examples and problems that refer to nonclosed subspaces of Banach spaces. $\endgroup$ – Michael Lee Mar 7 '17 at 20:25
  • $\begingroup$ There are times when it is appropriate to consider non-closed subspaces (for example, when we want to look at dense subspaces), however in this context the restriction to closed subspaces is what makes the theory interesting. For example, when one wants to extend a bounded linear operator defined on a subspace of a Banach space, extending to its closure is trivial, but then extending to the whole space depends on whether or not its closure is complemented (as defined in your first link). $\endgroup$ – Aweygan Mar 7 '17 at 20:39
4
$\begingroup$

Let me summarise some known facts related to your question.

  1. If $X$ is isomorphic to a Hilbert space then every closed subspace of $X$ is complemented.

  2. By a very deep result of Lindenstrauss and Tzafriri, a Banach space whose every closed subspace is complemented is necessarily isomorphic to a Hilbert space. Thus, spaces not isomorphic to Hilbert spaces always contain uncomplemented subspaces.

  3. Complementability is related to the possibility of extension of operators. If $Y$ is a closed subspace of $X$ then $Y$ is complemented if and only if the inclusion map extends to a bounded operator $X\to Y$.

  4. If $X$ is $\ell_\infty(\Gamma)$, $L_\infty(\mu)$, or more generally an injective space then whenever we have a bounded operator from a subspace $Y$ of some Banach space $Z$ to $X$, this operator can be extended to a bounded operator $Z\to X$. The same is true for $X=c_0$ as long as the space $Z$ is separable.

$\endgroup$
0
$\begingroup$

The point is that once in infinite-dimensions, a subspace can be somewhat wrapped up by its surrounding space to prevent unique representation of vectors by components in and out of the subspace. I think the best example to understand this phenomenon is in the non-locally-convex Ribe space which is a quasi-Banach topology on $R+L$ where $L$ is the space little-$L_1$ of absolutely summable sequences. It is constructed so that $R+\{0\}$ is equal to dual annihilated subspace and is contained in every closed infinite-dimensional closed subspace. Thus, it is a one-dimensional subspace that is not complemented. In Banach spaces, one-dimensional spaces are always complemented.

This is presented in a cleaned up manner in https://www.cambridge.org/core/books/an-fspace-sampler/917AAA51A5370CD786D32448FE828810 but A later article by N.J. Kalton shows this space has no strictly weaker Hausdorff topology, i.e. is an F-space with no basic sequences. This can be found as article 150. at the cite: https://kaltonmemorial.missouri.edu/publications.html

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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