# Subsequences of an orthonormal basis generating a strongly embedded subspace in $L_2(0,1)$

A closed subspace $$M$$ of $$L_2(0,1)$$ is said to be strongly embedded if the norms $$\|\cdot\|_2$$ and $$\|\cdot\|_1$$ are equivalent on $$M$$.

Let $$(f_n)_{n\in \mathbb N}$$ be a orthonormal basis of $$L_2(0,1)$$. Suppose that $$\limsup_{n\to\infty}\|f_n\|_1>0$$. Is it possible to find a subsequence $$(f_{n_k})_{k \in \mathbb N}$$ which generates a strongly embedded subspace?

The answer is positive for $$(e^{i2\pi nt})_{n\in \mathbb Z}$$ and for the Walsh functions (finite products of Rademacher functions). It is negative for the Haar system $$(h_n)$$ because $$\lim_{n\to\infty} \|h_n\|_1=0$$.

• My first thought/question is: what about the Haar system? Commented Oct 18, 2019 at 14:52
• Manuel, to make this a reasonable question you should add the condition that $\inf \|f_n\|_1 >0$, or at least $\lim\sup \|f_n\|_1 >0$. Commented Oct 18, 2019 at 16:00

The answer is yes". Slightly more generally, if $$(f_n)$$ is an orthonormal sequence in $$L_2:= L_2(0,1)$$ whose $$L_1$$ norms are bounded away from zero, then there is a subsequence that spans a strongly embedded subspace. It is equivalent (by extrapolation) to get a subsequence that in the $$L_p$$ norm with $$p:= 3/2$$ is equivalent to an orthonormal basis in a Hilbert space. The upper $$\ell_2$$-estimate in $$L_p$$ for linear combinations of the $$f_n$$ comes for free because $$(f_n)$$ is orthonormal and $$p<2$$. For the lower estimate, use the fact that $$L_p$$ has an unconditional basis to pass to a subsequence of $$(f_n)$$ that is unconditional. By the cotype $$2$$ property of $$L_p$$ when $$p<2$$, linear combinations of that subsequence of $$(f_n)$$ has a lower $$\ell_2$$ estimate.
EDIT Oct. 21, 2019: This question is closely related to the Maurey--Rosenthal [MR] problem whether there is a normalized weakly null sequence in $$L_1$$ that has no unconditionally basic subsequence. An example was given in [JMS]. It is natural to ask whether such a sequence can be bounded in $$L_p$$. The simple answer above says no" if $$p\ge 2$$. However, there are examples that are bounded in $$L_p$$ for all $$p<2$$--see the conditions (7) and (8) in [JMS].