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I am asked to prove that for $p\in (1, \infty)$, $L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$ on a homework assignment, and I think I can show using results from class that $\ell_2\oplus \ell_2\cong \ell_2$.

From this I could say that $L_{p}[0,1]\oplus \ell_{2}\cong L_{p}[0,1]\oplus \ell_{2}\oplus\ell_{2}$

From here I feel like I should be able to conclude that $L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$

But I know of no such result that allows me to do this. Can anyone tell me if it's true or false?

EDIT: Definitely false. (See counter example below from Arturo Magidin).

That is, if $B\oplus A\cong C\oplus A$ can I conclude that $B\cong C$?

Proper Solution (based on hints below from t.b.):

1) Prove that $\ell_2\oplus \ell_2\cong \ell_2$

2) Use the fact that $\ell_2$ is complemented in $L_p[0,1]$ to write $L_p[0,1] = \ell_2\oplus (\ell_2)^{c}$.

3) Then I combine these to obtain:

$L_p[0,1]\cong \ell_2\oplus (\ell_2)^{c}\cong (\ell_2\oplus \ell_2) \oplus (\ell_2)^{c} \cong \ell_2 \oplus (\ell_2\oplus (\ell_2)^{c})\cong \ell_2\oplus L_p[0,1]$.

I skipped some pieces of your more general argument. I was just wondering if I did anything illegal, so to speak.

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    Your added argument looks fine (though you have an extra "$\oplus$" in item 1.).2012-03-19

1 Answers 1

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Since Arturo explained why your idea doesn't work, here's a

Hint:

Let $X = L^p$ and let $Y = L^p \oplus \ell^2$.

Prove:

  1. $X^2 \cong X$ and $Y^2 \cong Y$.

  2. $X$ is isomorphic to a complemented subspace of $Y$ and vice versa (you probably know that $L^p$ has a complemented subspace isomorphic to $\ell^2$, see e.g. Proposition 6.4.2 in Albiac-Kalton or follow David Mitra's suggestion, Corollary 9.2 on page 90 in Carother's A Short Course on Banach Space Theory).

  3. Now you are in position to apply Pełczyński's argument to conclude that $X \cong Y$ (see point 3. of my question here for the details).

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    That's why I wrote: yes, that's it! (that was directed at you) :)2012-03-19