When I was reading a paper, I saw something like: If $F$ and $E$ are Banach spaces with symmetric bases (precisely, they are symmetric sequence spaces), and $F$ is isomorphic to a complemented subspace of $l_2\oplus E$, then $F=l_2$ or $F$ is isomorphic to a complemented subspace of $E$.

The author claimed that the result follows by standard elementary arguments. We omit the details. I don't know what the argument is. Any clue?

  • 3
    $\begingroup$ I don't understand the claim. What about $F=l_2 \oplus E$? $\endgroup$ – Jochen Wengenroth May 8 at 14:21
  • $\begingroup$ @JochenWengenroth I guess if $l_2\oplus E$ has a symmetric basis, then $l_2\oplus E \hookrightarrow l_2$ or $E$. I am not sure. My question came from the second sentence in the proof of Corollary 3.3. sciencedirect.com/science/article/pii/0022123681900525 $\endgroup$ – user92646 May 9 at 0:06

This is well known but not trivial. It follows primarily from Theorem 2.c.13 of [LT]. The Theorem says (applied to this situation) if every operator $T:E\to \ell_2$ is strictly singular, then every complemented subspace of $\ell_2\oplus E$ is of the form (up to isomorphism) $X'\oplus E'$ for some complemented subspace $E'$ of $E$, and $X'$ is either finite dimensional or isomorphic to $\ell_2$. So if $E'$ is finite dimensional then of course $F$ is isomorphic to $\ell_2$. In other cases, since $F$ has symmetric basis, $F$ isomorphic to $\ell_2\oplus E'$ implies, $F$ is either isomorphic to complemented subspace of $E'$ (and hence of $E$) or $\ell_2$. For the last statement see Prop 3.b.8 of [LT].

If there is an operator $T:E\to \ell_2$ which is not strictly singular, then, using the definition of not strictly singular and the fact that every subspace of $\ell_2$ is complemented, you get a complemented subspace $E'$ of $E$ isomorphic to $\ell_2$. But then $E\oplus \ell_2$ is isomorphic to $E$ by decomposition method (using $E$ has symmetric basis).

[LT] Lindenstaruss-Tzafriri, Classical Banach spaces vol 1.

| cite | improve this answer | |
  • $\begingroup$ Well, it is a nice proof and I would say that it is truly not trivial. Very appreciate! $\endgroup$ – user92646 May 9 at 2:08

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.