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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?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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