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!


1 Answer 1


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

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.

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

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