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]](http://link.springer.com/article/10.1007%2FBF03007667) 
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 $1\leq p<2$)?. 



Uncomplemented copies of $\ell_2$ in $L_p(0,1)$ for $2-\varepsilon<p<2$
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 $\{1,2,\dots, n\}$.


