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Fix any $1<p\leq 2$. Let us recall that $E:=(\oplus\ell_2^n)_{\ell_1}$ is just the space of sequences $(x_n)_{n=1}^\infty$, $x_n\in\ell_2^n$, such that $(\|x_n\|_{\ell_2^n})_{n=1}^\infty\in \ell_1$, endowed with the norm $\|(x_n)_{n=1}^\infty\|_E=\sum_{n=1}^\infty\|x_n\|_{\ell_2^n}$. Also, $\ell_1\oplus_\infty\ell_p$ is just the space of ordered pairs $(x,y)$, $x\in\ell_1$, $y\in\ell_p$, endowed with the norm $\|x\oplus y\|_\infty=\max\{\|x\|_{\ell_1},\|y\|_{\ell_p}\}$.

Question. Is the space $E=(\oplus\ell_2^n)_{\ell_1}$ complemented in $\ell_1\oplus_\infty\ell_p$? In other words, does there exists a continuous linear projection $P\in\mathcal{L}(\ell_1\oplus_\infty\ell_p)$ onto a subspace isomorphic to $E$?

Recall that $\ell_2^n$'s are uniformly complemented in $\ell_p^{k_n}$ for some increasing sequence of positive integers $k_1<k_2<k_3<\cdots$, and uniformly embedded (but not complementably) into $\ell_1^{k_n}$. (See Theorem 4.10 here for the first fact. The second fact is just an application of Dvoretsky's Theorem.) Perhaps these two facts might help to construct a projection.

Thanks!

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  • $\begingroup$ I think this should be on Math Overflow. $\endgroup$ Nov 9, 2015 at 14:17
  • $\begingroup$ Is it clear that $E$ is isomorphic to a subspace of $\ell_1\oplus_{\infty}\ell_p$? $\endgroup$ Nov 9, 2015 at 14:27
  • $\begingroup$ If the second fact I mentioned is true, then yes. In other words, if $\ell_2^n$ embeds uniformly into $\ell_p^{k_n}$ then $E$ is indeed a subspace of $\ell_1\oplus\ell_p$. But as I said, this fact should be checked. (I will try to do so now...) EDIT: Yes, it's true. $\endgroup$
    – Ben W
    Nov 9, 2015 at 14:32
  • $\begingroup$ $l_2^n$ is just $\mathbb{C}^n$ space with standart norm? $\endgroup$ Nov 9, 2015 at 15:14
  • $\begingroup$ Yes. More generally, $\ell_p^n=L_p(n)=(\mathbb{K}^n,\|\cdot\|_p)$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – Ben W
    Nov 9, 2015 at 15:28

1 Answer 1

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The answer is no. Since $\ell_1$ and $\ell_p$ ($p\in (1,\infty)$) are totally incomparable (every operator from one space to the other is strictly singular), by a result of Edelstein and Wojtaszczyk (Theorem 3.5 here) infinite-dimensional, complemented suspaces of $\ell_1\oplus \ell_p$ are the obvious ones, that is, they are isomorphic either to $\ell_1, \ell_p$ or $\ell_1\oplus \ell_p$. Neither of those spaces is isomorphic to $(\oplus_n \ell_2^n )_{\ell_1}$.

The moral reason of why $(\oplus_n \ell_2^n )_{\ell_1}$ is not isomorphic to a complemented suspace of $\ell_1\oplus \ell_p$ is that $\ell_2^n$ are not uniformly complemented in $\ell_1$ but they are uniformly complemented in $\ell_p$, so essentially this subspace would have to live in $\ell_p$ which is impossible. This heuristics can be made precise.

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  • $\begingroup$ Thanks! Now that I think about it, it should just be an elementary basic sequence argument. $\endgroup$
    – Ben W
    Nov 9, 2015 at 22:25

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