Does there exist a sequence with $ \int_{[0,1]^2}| f_i - f_j |\, {\rm d}x = {\rm const.} > 0 $ for $ \forall i \neq j $?

Question

Let $A = [0,1]^2$. Is there a such a sequence of functions $ f_i\colon A \rightarrow \mathbb{R} $ with $ \int_{A}| f_i | \ {\rm d}x \leqslant 1$ and $ \int_{A} | f_i - f_j | \ {\rm d}x = {\rm const.} > 0 $ for $ \forall i \neq j $ ? $ x \in \mathbb{R}^2 $

I have found this thread here: Is this a totally bounded set in the space of continuous functions? , which was very interesting, but not yet enough to give an example. At the moment I am still trying to figure out, whether such a sequence even exists. So I would be very happy for any constructive hint, comment, answer or recommendation for further reading. Thanks in advance.

Answer 1

If $\mu$ is an atomless measure then $L_1(\mu)$ contains an isometric copy of $\ell_2$ (actually of any $\ell_p$ for $p\in [1,2]$); for example take a sequence of independent Gaussian random variables, their span is isometric to $\ell_2$. Certainly, it is easy to find such a set in $\ell_2$ (any orthonormal basis would do). In conclusion, you can get $c=2^{1/p}$ for any $p\in [1,2]$.

Answer 2

Let $1=a_1 > a_2 > \cdots \to 0.$ Let $S_n$ be the square $[a_{n+1},a_n]^2.$ Define $f_n = 1/(a_n-a_{n+1})^2$ on the interior of $S_n,$ $f=0$ elsewhere. Then $\int_A |f_n| = 1$ for all $n,$ and because the interiors of the $S_n$ are pairwise disjoint, we have

$$\int_A |f_n-f_m| = \int_A |f_n| +\int_A |f_m| = 2 \,\text { for all } m,n, m\ne n.$$