# complemented subspace of the direct sum of two Banach spaces

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?

• I don't understand the claim. What about $F=l_2 \oplus E$? – Jochen Wengenroth May 8 at 14:21
• @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 – 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).