Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of positive functions in $L^\infty$ (hence a fortiori in $L^1$) that are equibounded in $L^\infty$ norm - that is $\sup\_{n \in \mathbb N} \|f\_n\|\_{L\_\infty} \leq M$ for some $M > 0$.


Is it true that there exists some absolute positive constant $c < 1$ such that


$$\inf\_{n\_k} \sup\_{i, j > N} \|f\_{n\_i} - f\_{n\_j}\|\_{L^1} \leq cM$$


for all such sequences $f\_n$?


Where the first infimum is taken over all increasing sequences $n\_k$ of naturals.



On equibounded sequences in $L^\infty$
**Edit:** I improved the constant to $c = \frac{2}{3}$. (Later edit: But the optimal constant turns out to be $c = \frac{1}{2}$, see [Yuval Peres' answer](https://mathoverflow.net/a/397845/102946).)


**Answer:** Yes, we have
$$
 \inf\_{(n\_k)} \sup\_{i,j \in \mathbb{N}} \|f\_{n\_i} - f\_{n\_j}\|\_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$}
$$
for each sequence $(f\_n)$ in $(L^1)\_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.


To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.


**Lemma.** Three functions $f\_1, f\_2, f\_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.


*Proof.*
Set $g\_1 = |f\_1 - f\_2|$, $g\_2 = |f\_1 - f\_3|$ and $g\_3 = |f\_2 - f\_3|$.
For any three numbers $r\_1,r\_2,r\_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.


Hence, $\int g\_1 + \int g\_2 + \int g\_3 \le 2$, which shows that it can't happen that all three functions $g\_k$ have norm strictly larger than $\frac{2}{3}$. $\square$


*Proof of the claim.*
We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f\_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.


Then there exists $n\_0$ such that $\|f\_{n\_0} - f\_n\|\_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f\_{n\_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.


For any two $j,k \in J$, it follows from the lemma that $\|f\_j - f\_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f\_{n\_k})$. Contradiction, since we assumed no such subsequence exists. $\square$


**Remark.** It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I\_k$ of measure $\frac{1}{3}$ and define $f\_k = \mathbf{1} - \mathbf{1}\_{I\_k}$), but I don't know whether it is optimal for the answer to the question.


