Let $(x\_{n})\_{n}$ be a sequence in $B\_{L\_{1}}$ that has no weakly Cauchy subsequences. Here $B\_{L\_{1}}$ is the closed unit ball of $L\_{1}$. By Rosenthal's $l\_{1}$-theorem, the sequence $(x\_{n})\_{n}$ has a subsequence that is equivalent to the unit vector basis of $l\_{1}$. But I want a stronger result.


Question. Does the sequence $(x\_{n})\_{n}$ admit a subsequence that is $C$-equivalent to the unit vector basis of $l\_{1}$? where $C$ is a universal constant.


Thank you!



Bounded sequences in $L_{1}$ that has no weakly Cauchy subsequences
The answer is "No", and can be shown as follows. Let $\{e\_i\}$ be a sequence which is $1$-equivalent to the unit vector basis of $\ell\_1$ and is contained in $L\_1[0,1/2]$ (that is, functions are equal to $0$ on $[1/2,1]$). Let $\{f\_i\}$ be a sequence which is $1$-equivalent to the unit vector basis of $\ell\_2$ and is contained in $L\_1[1/2,1]$. For each $n\in \mathbb{N}$ consider the sequence $\{g\_i^n\}$ given by $g\_i^n=\frac1n\, e\_i+\frac{n-1}n\,f\_i$. It is clear that $\{g\_i^n\}$ is $n$-equivalent to the unit vector basis of $\ell\_1$, but passing to a subsequence does not improve the constant of equivalence. 


