*Note: This is a generalisation of [an earlier problem](https://mathoverflow.net/questions/397726/on-equibounded-sequences-in-l-infty) as suggested by user Jochen Glueck in the comments.*


Let $1 \leq p < q \leq \infty$, and $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions in the closed unit ball of $L^q$.


**Question:** Is it true that there exists a constant $C < 2$, depending only on $p$ and $q$ such that


$$\inf\_{n\_k} \sup\_{i,j \in\mathbb N} \|f\_{n\_i} - f\_{n\_j}\|\_{L^p} \leq C?$$


Where the infimum is taken over all increasing sequences of natural numbers $n\_k$.


If so, what is the sharpest such constant for each $p, q$?


*Remark: In the linked problem, the sharp constant $C = 1$ for $p = 1, q = \infty$ is obtained in the answer by Yuval Peres.*



$L^p$ bounds on tails of bounded $L^q$ sequences
Let me prove that such constant $C$ always exists.


It is not hard to find such $\alpha$, $\beta$ that the inequality $$x^p\leqslant \alpha x^q+\beta$$
holds for all positive $x$ and turns into equality if and only if $x=2$. Then $$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\beta.$$
Since all three differences $|f-g|$, $|g-h|$, $|f-h|$ can not be equal to $\pm 2$, we actually have
$$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\tilde{\beta}, \quad \text{with some}\,\,\tilde{\beta}<\beta$$
Then integrating against $[0,1]$ we get $$\|f-g\|\_p^p+\|g-h\|\_p^p+\|f-h\|\_p^p\leqslant \alpha (\|f-g\|\_q^q+\|g-h\|\_q^q+\|f-h\|\_q^q)+3\tilde{\beta}.$$
If all $f,g,h$ are in the closed unit ball in $L^q$, this yields
$$\|f-g\|\_p^p+\|g-h\|\_p^p+\|f-h\|\_p^p\leqslant 3(2^q\alpha+\tilde{\beta}).$$
Since $2^q\alpha+\beta=2^p$, we see that at least one of expressions $\|f-g\|\_p, \|g-h\|\_p, \|f-h\|\_p$ is at most $(2^q\alpha+\tilde{\beta})^{1/p}=:C<2$.


Now join $n$ and $m$ by a red edge if $\|f\_n-f\_m\|\leqslant C$ and by a blue edge otherwise. By infinite Ramsey theorem, there exists either a blue triangle or an infinite red clique. The second case is impossible, the first case is what we need.


