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It was proved in [Bennett, G.; Dor, L.E.; Goodman, V.; Johnson, W.B.; Newman, C.M. On uncomplemented subspaces of $L_p$, $1<p<2$. Israel J. Math. 26 (1977), 178–187] that, for $1 <p < 2$, there is an uncomplemented subspace of $L_p(0,1)$ that is isomorphic to $\ell_2$.

Moreover it is well-known that if $1<p<q<2$ and $M$ is an uncomplemented copy of $\ell_2$ in $L_q(0,1)$, then $M$ is also an uncomplemented copy of $\ell_2$ in $L_p(0,1)$.

My question: Is it possible to find a subspace $M$ of $L_2(0,1)$ such that, for some $\varepsilon>0$, $M$ is an uncomplemented copy of $\ell_2$ in $L_p(0,1)$ for $2-\varepsilon <p<2$ (hence for $1\leq p<2$)?.

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Yes. The idea is to find a sequence of finite dimensional subspaces $E_n$ of $L_n$ on which the $L_1$ and $L_2$ norms are uniformly equivalent and such that for all $p<2$ the norm of the best projection from $L_p$ onto $E_n$ tends to infinity as $n\to \infty$, and piece these together to get an infinite dimensional Hilbert space. By Kashin or Figiel, Lindenstraus, and Milman (or a remark in the BDGJN paper you cited) you can take $E_n\subset L_2\{-1,1\}^n$ of proportional dimension $c 2^n$ ($c>0$ independent of $n$) on which $\|\cdot \|_2 \le 2\|\cdot\|_1$, and all the functions in $E_n$ have mean zero. Take an independent sum of these in $L_2\{-1,1\}^{N}$, where $N $ is a disjoint union over $n=1,2,3,\dots$ of $\{1,2,\dots, n\}$ and $L_2\{-1,1\}^n$ is regarded as the functions in $L_2\{-1,1\}^{N}$ that depend only on the $n$th set $\{1,2,\dots, n\}$.

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  • $\begingroup$ I suppose that, in the concrete version $E_n\subset L_2\{-1,1\}^n$, the proportionality of dimension of $E_n$ implies that the $E_n$'s are not uniformly complemented in $L_p$ for $p<2$, and that the independence of the sum $E=\oplus E_n$ implies that $E$ is isomorphic to $\ell_2$ in $L_p$ for $p<2$. Is $E$ also a subspace of $L_q$ isomorphic to $\ell_2$ for $q>2$? $\endgroup$ – M.González Oct 14 '16 at 10:52
  • $\begingroup$ No, because for $q>2$it is proved in the paper that you reference that the spaces $\ell_q^m$ do not contain spaces of uniformly proportional dimension that are uniformly Hilbertian ($m^{2/q}$ is the limit). That is why $E$ is uncomplemented in $L_p$ for $p<2$. $\endgroup$ – Bill Johnson Oct 14 '16 at 17:33

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