2
$\begingroup$

I am following Albiac and Kalton's book Topics in Banach Space Theory. Theorem 2.2.3 is Pelczynski Decomposition: let $X$ and $Y$ be Banach spaces. Suppose that $X$ is isomorphic to a complemented subspace of $Y$, and that $Y$ is isomorphic to a complemented subspace of $X$. Then, under certain conditions, $X$ and $Y$ are isomorphic.

The condition I am having trouble with is asking that $X$ is isomorphic to $\ell_p(X)$. If that is the case, $X$ is isomorphic to $X^2$, and after some manipulation you get that $Y$ is isomorphic to $X \oplus Y$, so far so good.

The problem is the penultimate line of the proof. Suppose $X \cong Y \oplus E$. The penultimate line reads: $$X \cong Y \oplus \ell_p(Y) \oplus \ell_p(E) \cong Y \oplus \ell_p(X) \cong Y \oplus X$$

Where has the first isomorphism come from?

Thank you very much!

$\endgroup$

1 Answer 1

6
$\begingroup$

$$X \cong \ell_p(X) \cong \ell_p(Y \oplus E) \cong \ell_p(Y) \oplus \ell_p(E) \cong Y \oplus \ell_p(Y) \oplus \ell_p(E) \cong Y \oplus \ell_p(X) \cong Y \oplus X$$.

Only the third isomorphism requires thought. That is where the infinite direct sum is in the sense of $\ell_p$ (rather than in some other sense) is used.

$\endgroup$
1
  • $\begingroup$ Thank you very much! Now I see clearly what is happening. $\endgroup$
    – Seven9
    Apr 2 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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