Kahn-Kalai-Linial for intersecting upsets

Question

Is there any known improvement on the Kahn-Kalai-Linial inequality (on the influences of boolean functions) in the special case in which $f$ is the indicator function of an intersecting monotonic set system? More concretely, is there an absolute constant $C>0$ such that the following statement holds:

If $f:\mathcal{P}([n]) \to \lbrace 0,1\rbrace$ is the indicator of an intersecting upset, then there exists $x\in [n]$ such that $I_x(f)\geq C/\sqrt{n}$.

(Note that the "tribes" example of a half-sized system in which all influences are $\ll \log n /n$ is certainly not intersecting.)

Background: The $x$th influence of a boolean function $f$ is defined as

$$I_x(f) = \mathbf{E}(f(X)\neq f(X \Delta \lbrace x\rbrace),$$

($\Delta$ is symmetric difference) where $X$ is drawn randomly and uniformly from $\mathcal{P}([n])$. In particular, if $f$ is the indicator of a monotonic set system $\mathcal{U}\subset\mathcal{P}([n])$ (monotonic meaning $X\subset Y$ and $X\in\mathcal{U}$ implies $Y\in\mathcal{U}$), $I_x(f)$ is the number of sets $X\in \mathcal{U}$ containing $x$, minus the number of such sets not containing $x$, divided by $2^{n-1}$.

Answer 1

This is a natural question and indeed the property of being intersecting is quite interesting also in various aspect of influences. However, you cannot improve KKL's theorem if f is intersecting: An example is this: consider your variables on a circle and let f=1 if the longest run of 1's is larger than the longest 1's of 0's and in case of equality consider the second longest run (and continue lexicographically). In this case f is intersecting and the influence of every variable is logn/n.

This example is symmetric under rotations and therefore all influences are the same. The sum of influences can be described as the integral over all configurations x of h(x) the number of pivotal variables. Here a variable is pivotal if changing its value changes the value of f. Given x with f(x)=1 typically a variable is pivotal only if it belongs to the largest run which is of expected size logn. There are cases of equality (between 1 runs and 0 euns or between two runs) that the number of pivotal variables will be larger than log n but those are rare. This explains why every influence is proportional to logn/n but for a complete proof some more work is needed.