Continuous version of the union-closed sets conjecture

Question

Let $F = \{f_1, \ldots, f_n\}$ be a set of continuous functions $f_i: [0,1] \rightarrow [0,1]$, $i = 1, \ldots, n$, such that $f_i \in F \land f_j \in F \implies \max(f_i,f_j) \in F$.

I would like to just formulate, of course not prove it, a "continuous" version of the Union-closed sets conjecture.

A starting point could be: for any $F$ as defined above there exists $x_0 \in [0,1]$ such that:

$$\sum_{i=1}^n f_i(x_0) \ge \frac{n(\max_{1 \le i \le n} f_i(x_0)-\min_{1 \le i \le n} f_i(x_0))}{2}$$

Could we conjecture a stronger inequality involving all values $f_1(x_0),\ldots,f_n(x_0)$ in the RHS (so that to have a higher RHS) and not just the minimum and maximum?

Answer 1

A conjecture that implies the conclusion of the union-closed sets conjecture and is stronger than your (corrected) "starting point" is as follows:

For any $F$ as defined above there exists some $x_0 \in [0,1]$ such that $$\sum_{i=1}^n f_i(x_0) \ge \frac n2\,\max_{1 \le i \le n} f_i(x_0)>0.$$

(Here, the correction is adding the inequality $\max_{1 \le i \le n} f_i(x_0)>0$, without which we will be unable to deduce the conclusion of the union-closed sets conjecture.)

On the other hand, it is unclear how using intermediate values of the $f_i(x_0)$'s can help strengthen the conjecture.

Answer 2

Not sure whether this might fail, but I have found a possible formulation:

For any $F$ as defined above there exists $x_0 \in[0,1]$ such that: $$\sum_{i=1}^n f_i(x_0) \ge {\frac{n}{2\binom{m}{2}} \sum_{k=2}^m (k-1)g_k(x_0)}$$ where $n \ge m \ge 2$, $g_1(x_0) \lt g_2(x_0) \lt \ldots \lt g_m(x_0)$, and $\{g_1(x_0),\ldots,g_m(x_0)\}$ is the set corresponding to the multiset $\{f_1(x_0),\ldots,f_n(x_0)\}$.